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Philippine Copyright 2015 Philippine Copyright by Rex Book Store, 2015 Inc. byPhilippine Rex BookCopyright Store, Inc. 2015



RBS Mathematics Series by Rex Book Store, Inc. RBS New Mathematics Philippine Copyright 2017 The Syllabus Series Primary Mathematics by Rex Book Store,2016 Inc. The NewEdition Syllabus Primary Mathematics Revised 2015 Philippine Copyright RBS Mathematics Series Revised Edition 2015 ISBN 978-971-23-6915-5 by Rex Book Store, Inc. The New Syllabus Primary Mathematics RBS Mathemati cs Series (02-MB-00250-0) ISBN 978-971-23-6915-5 Classification: Worktext Revised Edition 2015 The New Syllabus Mathemati cs Classification: Worktext (02-MB-00250-0) RBSISBN Mathemati cs Series 978-971-23-6915-5 Revised Editi on 2017 The New Syllabus Primary Mathemati cs 1 Published, copyrighted 2015, and distributed by Rex Store, Inc. (RBSI) withoffice mainatoffice at 856 Reyes Nicanor Reyes Classification: Worktext (02-MB-00250-0) Published, copyrighted 2015, and distributed by Rex BookBook Store, Inc. (RBSI) with main 856 Nicanor Sr. St., Sampaloc, ISBN 978-971-23-8834-7 Revised Editicopyrighted on 2016 Published, 2015, and distributed Rex Book Store, (RBSI) 856 Nicanor Published, copyrighted 2015, and distributed byby Rex Book Store, Inc.Inc. (RBSI) withwith mainmain officeoffice at 856atNicanor Reyes Sr.Reyes St., Sampaloc, Sr. St., Sampaloc, Manila / Tel. Nos. 735-1364, 736-0567 Manila/Tel. Nos.: 735-1364, 736-0567 (83-MB-00149-0A) Classifi cati on: Worktext ISBN 978-971-23-7887-4 Sr. St., Sampaloc, Manila / Tel. Nos. 735-1364, 736-0567



Manila/Tel. Nos.: 735-1364, 736-0567 Classifi caticopyrighted on: Worktext (02-MB-00265-0) PRBSI ublished, 2017, and distributed by Rex Book Store, Inc. (RBSI) with main office at 856 Nicanor Reyes Sr. St., Sampaloc, Manila/Tel. Nos.: 735RBSI Branches: Branches: RBSI Branches: 1364, 736-0567 RBSI Branches: LUZON Published, copyrighted 2016, and distributed by Rex Book Store, Inc. (RBSI) with main office at 856 Nicanor Reyes Sr. St., Sampaloc, Manila/Tel. LUZON LUZON •MORAYTA: 736-0169, 733-6746; Telefax: 736-4191 •RECTO: 2161-65 Freedom RBSI Branches: 856 N. Reyes Sr. St., Sampaloc, Manila / Tel. Nos.: Nos.: 735-1364, 736-0567 LUZON •MORAYTA: 856 Reyes Sr. Sampaloc, / 522-4521, Tel. Nos.: 736-0169, 733-6746; Telefax: 736-4191 2161-65 Freedom •MORAYTA: N.N. Reyes Sr.Sampaloc, St.,St., Sampaloc, / Tel. Nos.: 736-0567, 735-1364; Telefax: 736-4191 •RECTO: 1977 C.M. Recto Building, C.M.856 Recto Avenue, ManilaManila / Manila Tel. Nos.: 522-4305, 522-4107, 733-8637 •RECTO (La•RECTO: Consolacion): Mendiola, LUZON •MORAYTA: 856Recto N. Reyes Sr. St., Sampaloc, Manila Tel. Telefax: •RECTO: 1977 C.M. Recto Building, C.M. Avenue, Sampaloc, Manila /Bldg., Tel./Nos.: 522-4521, 522-4305, 522-4107, 733-8637 •RECTO (La Consolacion): Mendiola, RBSI Branches: Ave., Sampaloc, Manila /UG-2, Tel. Nos.: 735-5527, 736-3063; Telefax: 735-5534 •MAKATI: UG-2, Star Centrum Bldg., Sen. Gil Manila • MAKATI: Unit Star Centrum Sen.Nos.: Gil 736-0567, Puyat Ave.,735-1364; Makati City /Unit Tel.736-4191 No.: 818-5363; Telefax: 893-3744 •MORAYTA: 856 N. Reyes Sr. St., Sampaloc, Manila / Tel. Nos.: 736-0169, 733-6746; Telefax: 736-4191 •RECTO: 2161-65 Freedom Building, C.M. LUZON Manila • MAKATI: Unit UG-2, Star Centrum Bldg., Sen. Gil Puyat Ave., Makati City / Tel. No.: 818-5363; Telefax: 893-3744 Ave., Sampaloc, Manila / Tel. Nos.: 735-5527, 736-3063; Telefax: 735-5534 •MAKATI: Unit UG-2, Star Centrum Bldg., Sen. Gil Recto Puyat Ave., Makati CityAteneoNo.: 818-5363; Telefax: 893-3744 •ROCKWELL: 1st City Floor, Ateneo Professional School, Rockwell •ROCKWELL: 1st Floor, Professional School, Rockwell Center, Bel-Air, Makati / Tel. No.: 729-2015 •CUBAO: Unit 10 UGF, •2161-65 MAKATI: Unit UG-2, Star Centrum Avenue, Sampaloc, Manila / Tel. Nos.: 522-4521, 522-4305, 522-4107, 733-8637 •RECTO (La Consolacion): Mendiola, Manila Freedom Building, •MORAYTA: 856 N. Reyes Sr. St., Sampaloc, Manila /Center, Tel. Nos.: 736-0169, 733-6746; Telefax: 736-4191 •RECTO: 10 UGF, •ROCKWELL: 1st Floor, Professional School, Rockwell Center, Bel-Air, Makati City /Ateneo Tel. No.: 729-2015 •CUBAO: Puyat Ave., Makati City /Ateneo Tel. No.: 818-5363; Telefax: 893-3744 •ROCKWELL: 1st Floor, Professional School, Rockwell Center, Bel-Air, Makati City /Santos Tel. No.: 729-2015 •CUBAO: 36 Quezon Shopwise Araneta Center, Cubao, City /Unit Telefax: •ORTIGAS: G/FQuezon East Tower, Philippine Doña Consolacion Bldg., Gen. Ave., Araneta Cubao, CityArcade, /Telefax: 911-1070 Bldg., Sen. Gil PuyatSampaloc, Ave., Makati City /Tel. Tel. No.:Araneta 818-5363; Telefax: 893-3744 •ROCKWELL: 1st 911-1070 Floor, Ateneo Professional School, Rockwell Center, BelC.M. Recto Avenue, Manila / Nos.: 522-4521, 522-4305, 522-4107, 733-8637 •RECTO (La Consolacion): Mendiola, Manila • MAKATI: •ORTIGAS: G/F East Tower, Philippine Doña Consolacion Bldg., Gen. Santos Ave., Center, Cubao, Quezon City /Telefax: Center, Bel-Air, Makati City / Tel.Center No.: •CUBAO: 36 Shopwise Arcade, Araneta Center, Cubao, Quezon Telefax: Stock Exchange Center, Exchange Road,729-2015 Ortigas Center, Pasig City / Tel. No.: (02) 650-4347 •CAVITE: Block 4, Quezon Lot 20City Don//Telefax: Gregorio 911-1070 •SHAW: 548729-2015 Facilities Bldg., Blvd., Mandaluyong City / Tel. No.:Araneta 531-1306; Telefax: 531-1339 •CAVITE: Air, Makati City Tel. No.: •CUBAO: UnitAve., 10 Shaw UGF, Doña Consolacion Gen. Santos Ave., Center, Cubao, City 911-1070 Unit UG-2, Star/ Centrum Bldg., Sen. Gil Puyat Makati City / Tel. 818-5363; Telefax: 893-3744 •ROCKWELL: 1st20 Floor, Ateneo Stock Exchange Center, Exchange Road, Ortigas Center, Pasig CityNo.: /Bldg., Tel. No.: (02) 650-4347 •CAVITE: Block 4,531-1339 Lot Don Gregorio 911-1070 •SHAW: 548 Facilities Center Bldg., Shaw Blvd., Mandaluyong City / Cavite Tel. No.: 531-1306; Telefax: •CAVITE: Heights 2,G/F Zone 1-A Aguinaldo Highway, Dasmariñas, Cavite / Telefax: (046) 416-1824 •CAVITE (Tanza): (Display Area) Block 5,Block Lot 4, Lot Block 4, Lot 20 Don Gregorio Heights 2, Zone 1-A Aguinaldo Hi-way, Dasmariñas, / Telefax: (046) 416-1824 •NAGA: Rodson •ORTIGAS: East Tower, Philippine Stock Exchange Center, Exchange Road, Orti gas Center, Pasig City / Tel. No.: (02) 650-4347 •CAVITE: Professional School, Rockwell Center, Bel-Air, Makati City / Tel. No.: 729-2015 •CUBAO: Unit 10 UGF, Doña Consolacion Bldg., Gen. Santos Ave., 1-A Aguinaldo Highway, Dasmariñas, / Geronimo Telefax: (046) 416-1824 (Tanza): (Display Area) Block 5,LotLot Block 4,Gregorio Lot 20 Don Gregorio Heights 2, Zone 1-A Aguinaldo Hi-way, Dasmariñas, Cavite Telefax: 416-1824 •NAGA: Rodson 1-1A Bldg., Barlin St.,/•CAVITE Sta. Cruz, Naga City, Camarines Sur/Telefax: 6,Heights CityI-II, View 4Zone and 5, Brgy. Tanauan, Tanza, Cavite •NAGA: Bldg. J.2,Hernandez Naga City, Camarines Sur Cavite / Telefax: (054) 811-6878 •LEGAZPI: 3rd(046) Floor Bichara Mall, Magallanes 20 Don Heights 2,Ave., Zone 1-A Aguinaldo Highway, Dasmariñas, Cavite /Tower, Telefax: (046) 416-1824 •CAVITE (Tanza): (Display Area) Block 5,gas 6, City Araneta Center, Cubao, Quezon City /Telefax: 911-1070 •ORTIGAS: G/F East Philippine Stock Exchange Center, Exchange Road, Orti 6, 4City View 4•LEGAZPI: and 5, Brgy. Tanauan, Tanza, Cavite •NAGA: 1-1A Geronimo Bldg., Barlin St.,Camarines Sta.3rd Cruz, Naga City, Camarines Sur/Telefax: (054) 811-6878 Unit 6, 3rd Floor, A. Bichara Silverscreen, Legazpi City, Albay / Telefax: (052) 480-2244 •CALAPAN: Brgy. Salong, Bldg. I-II, J. Hernandez Ave., Naga City, Camarines Sur / Telefax: (054) 811-6878 •LEGAZPI: Floor Bichara Mall, Magallanes 1-1A Geronimo Bldg., Barlin St., Sta. Cruz, Naga City, Sur/Telefax: (054) 811-6878 •LEGAZPI: View and 5, Brgy. Tanauan, Tanza, Cavite •NAGA: cor. Alonzo St., Legazpi City, Albay / Telefax: (052)Block 480-2244 •CALAPAN: Brgy. Salong, National Hi-way, Calapan City, Oriental Unit Center, Pasig City / Tel. No.: (02) 650-4347 •CAVITE: 4, Lot 20 Don Gregorio Heights 2, Zone 1-A Aguinaldo Highway, Dasmariñas, •LEGAZPI: Unit 6,Oriental 3rd Floor, A. (052) Bichara Silverscreen, Legazpi City, Albay / Salong, Telefax: (052) 480-2244 •CALAPAN: Brgy. Salong, (054) 811-6878 6, 3rd Floor, A. Bichara Silverscreen, Legazpi City, Albay /L.Telefax: (052) 480-2244 •CALAPAN: Brgy. Nati onal Highway, Calapan City, Oriental National Calapan City, Mindoro /480-2244 Telefax: (043) 288-1650 •BATANES: L.Brgy. Lopez St., Kayvalugan, Basco, Batanes cor. Alonzo St., Legazpi City, Albay / Telefax: •CALAPAN: Brgy. Salong, Hi-way, Calapan City, Oriental Mindoro /Highway, Telefax: (043) 288-1650 •BATANES: Lopez St., Block Kaywalungan, Basco, Batanes •TUGUEGARAO: 10 Arellano St., Brgy. Cavite / Telefax: (046) 416-1824 •CAVITE (Tanza): (Display Area) 5, Lot 6, City View 4 and 5,National Tanauan, Tanza, Cavite •NAGA: 1-1AMindoro National Highway, Calapan City, Oriental Mindoro / Telefax: (043) 288-1650 •BATANES: L. Lopez St., Kayvalugan, Basco, Batanes /Mindoro Telefax: (043) 288-1650 •BATANES: L. Lopez St., Kayvaluganan, Basco, Batanes •TUGUEGARAO: 10 Arellano Ext., Brgy. Ugac Sur, Tuguegarao, Cagayan •TUGUEGARAO: 10 Arellano Ext., Brgy. Ugac Sur, Tuguegarao, Cagayan / Telefax: (078) 844-8072 •CABANATUAN: Fontelera Building, /Tuguegarao, Telefax: 288-1650 •BATANES: L.844-8072 Lopez St., Kaywalungan, Batanes •TUGUEGARAO: 10Ext., Arellano St., Brgy. Geronimo Bldg., Barlin(043) St., Cagayan Sta. Cruz, Naga City, Camarines Sur/Telefax: (054) 811-6878Basco, •LEGAZPI: Unit 6, 3rd1271 Floor,Del A. Bichara Silverscreen, Legazpi Ugac Sur, / Telefax: (078) •CABANATUAN: Fontelera Building, Pilar Sangitan East, •TUGUEGARAO: 10 Arellano Ext.,Cabanatuan Brgy. Ugac Sur, Tuguegarao, / Sangitan Telefax: (078) 844-8072 •CABANATUAN: Building, /1271 Telefax: (078) 844-8072 •CABANATUAN: Fontelera Building, 1271 Ecija DelCagayan Pilar Ext., East, Cabanatuan City,(044) Nueva600-5684 Ecija / Fontelera Tel. •URDANETA: No.: (044) 464-2151; Del Pilar Ext., Sangitan East, City, Nueva / Tel. No.: (044) 464-2151; Telefax: UgacAlbay Sur, /Tuguegarao, / •CALAPAN: Telefax: (078) 844-8072 •CABANATUAN: Fontelera Building, Del Pilar Ext.,(043) Sangitan East, Brgy. Salong, Nati onal (044) Highway, Calapan City, Oriental1271 Mindoro / Telefax: 288-1650 City, Telefax: (052)Cagayan 480-2244 Cabanatuan City,Ext., Nueva Ecija / East, Tel. No.: (044) 464-2151; Telefax: •URDANETA: Zone 6, Pinmaludpod, Urdaneta Telefax: 600-5684 •URDANETA: Zone 6, Pinmaludpod, Urdaneta City,568-3975 / Telefax: (075) 568-3975 •ANGELES: Unit H,Highway, JMS Bldg.,Brgy. MacArthur 12716,(044) Del Pilar Sangitan Cabanatuan City, Nueva Ecija /Pangasinan Tel. 600-5684 No.:•ANGELES: (044) 464-2151; (044) 600-5684 •URDANETA: UnitSur, H,Telefax: JMS Bldg., MacArthur Zone Pinmaludpod, Urdaneta City, Pangasinan / Telefax: (075) •BATANES: L. Lopez St., Kayvalugan, Basco, Batanes •TUGUEGARAO: 10 Arellano Ext., Brgy. Ugac Tuguegarao, Cagayan / Telefax: (078) Cabanatuan City,/Nueva Ecija / Tel. No.: (044) 464-2151; Telefax: (044) 600-5684 •URDANETA: Zone 6,City, Pinmaludpod, Urdaneta City, Pangasinan Telefax: (075) 568-3975 •ANGELES: 259 (Stall B) Sto. Rosario St., San Jose, Angeles Pampanga / Telefax: Highway, Brgy. Salapungan, Angeles City, Pampanga/Telefax: (045) 887-5371 • BAGUIO: Rex Hall Student Residences, Upper Gen. Luna cor. A. Bonifacio St., Baguio Zone 6, Pinmaludpod, Urdaneta City, Pangasinan / Telefax: (075) 568-3975 •ANGELES: Unit H, JMS Bldg., MacArthur Highway, Brgy. Salapungan, Angeles City, Pampanga/Telefax: (045) 887-5371 • BAGUIO: Rex Hall Student Residences, Upper Gen. Luna cor. A. Bonifacio 844-8072 •CABANATUAN: Fontelera Building, •ANGELES: 1271 Del Pilar Ext., Sangitan East, Cabanatuan Nueva Ecija City, / Tel.Pampanga No.: (044) 464-2151; City, Pangasinan / Telefax: (075) 568-3975 259 (Stall B) Sto. Rosario St., SanCity, Jose, Angeles /A.Telefax: City, Benguet / Tel. No.: (074) 422-0574 (045) 887-5371 Salapungan, Angeles City, Pampanga/Telefax: (045) 887-5371 • BAGUIO: Rex Hall Student Residences, Upper Gen. Luna cor. Bonifacio St., Baguio City, Benguet•URDANETA: / Tel. No.: (074) VISAYAS Telefax: (044) 600-5684 Zone422-0574 6, Pinmaludpod, Urdaneta City, Pangasinan / Telefax: (075) 568-3975 •ANGELES: Unit H, JMS (045) 887-5371 St.,MacArthur Baguio City, Benguet / Tel. No.: (074)Angeles 422-0574 Bldg., Brgy. Salapungan, (045) 887-5371 • BAGUIO: Rex Hall Student Residences, Upper Gen. •TACLOBAN: Brgy.Highway, 74 Marasbaras, Tacloban City, LeyteCity, / Tel.Pampanga/Telefax: No.: (053)VISAYAS 323-8976; VISAYAS Telefax: (053) 523-1784 •ILOILO: 75 Lopez Jaena St., Brgy. San Isidro, Jaro, Luna cor. Bonifacio St., Baguio City, Benguet / Tel. No.: (074) 422-0574 VISAYAS •TACLOBAN: 74 (033) Marasbaras, Tacloban City, Leyte /•BACOLOD: Tel. No.: (053) Telefax: (053) 523-1784 •ILOILO: 75 Lopez Iloilo City,A. Iloilo /Brgy. Tel. 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No.: (033) 329-0332; Telefax: (033) 329-0336 •BACOLOD: 28(032) Brgy. 36, Immaculada, Quezon Ave., Bacolod City, Negros Occidental •CEBU: Sanciangko Cebu City /28Tel. Nos.: 416-9684, MINDANAO Brgy.St., San Isidro, Iloilo City, Iloilo / Tel. No.: (033) 329-0332; Telefax: (033) 329-0336 •BACOLOD: Brgy. 36, Purok Immaculada, •TACLOBAN: Brgy. 74Jaro, Marasbaras, Tacloban City, Leyte / Tel. No.: (053) 323-8976; Telefax: 523-1784 •ILOILO: 75 Lopez Jaena St.,Purok Brgy. Quezon Ave., Bacolod City, Negros Occidental •CEBU: 11 Sanciangko St.,11Cebu City / (053) Tel.St., Nos.: (032) 416-9684, 254-6773, 505-4313; Immaculada, Quezon Ave., Bacolod City, Negros Occidental •CEBU: 11Misamis Sanciangko St., Cebu City /416-9684, Tel. Nos.: (032) 416-9684, 254-6773; Telefax: 254-6466 San Isidro,(032) Jaro, Iloilo City, Iloilo /Negros Tel. No.: (033) 329-0332; Telefax: (033) •BACOLOD: 28 Brgy. 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i i



PREFACE New Syllabus Mathematics (NSM) is a series of worktexts specially designed to provide valuable learning experiences to engage the hearts and minds of students in the learning process. Included in the worktexts are Investigation, Class Discussion, Thinking Time, Journal Writing, Performance Task and Problems in Real-World Contexts to support the teaching and learning of Mathematics.



Every chapter begins with a chapter opener which motivates students in learning the topic. Interesting stories about mathematicians, real-life examples and applications are used to arouse students’ interest and curiosity so that they can appreciate the beauty of Mathematics in their surroundings.



The use of ICT helps students to visualize and manipulate mathematical objects more easily, thus making the learning of Mathematics more interactive. Ready-to-use interactive ICT templates are available at http://www.shinglee.com.sg/StudentResources/.



Preface



iii



KEY FEATURES CHAPTER OPENER Each chapter begins with a chapter opener to arouse students’ interest and curiosity in learning the topic.



LEARNING OBJECTIVES Learning objectives help students to be more aware of what they are about to study so that they can monitor their own progress.



RECAP Relevant prerequisites will be revisited at the beginning of the chapter or at appropriate junctures so that students can build upon their prior knowledge, thus creating meaningful links to their existing schema.



WORKED EXAMPLE This shows students how to apply what they have learned to solve related problems and how to present their working clearly. A suitable heading is included in brackets to distinguish between the different Worked Examples.



PRACTICE NOW At the end of each Worked Example, a similar question will be provided for immediate practice. Where appropriate, this includes further questions of progressive difficulty.



SIMILAR QUESTIONS A list of similar questions in the Exercise is given here to help teachers choose questions that their students can do on their own.



EXERCISE The questions are classified into three levels of difficulty – Basic, Intermediate, and Advanced.



SUMMARY At the end of each chapter, a succinct summary of the key concepts is provided to help students consolidate what they have learned.



REVIEW EXERCISE This is included at the end of each chapter for the consolidation of learning of concepts.



CHALLENGE YOURSELF Optional problems are included at the end of each chapter to challenge and stretch high-ability students to their fullest potential.



REVISION EXERCISE This is included after every few chapters to help students assess their learning.



iv



Preface



Learning experiences have been infused into Investigation, Class Discussion, Thinking Time, Journal Writing, and Performance Task.



Class Discussion



Investigation



Questions are provided for students to discuss in class, with the teacher acting as the facilitator. The questions will assist students to learn new knowledge, think mathematically, and enhance their reasoning and oral communication skills.



Activities are included to guide students to investigate and discover important mathematical concepts so that they can construct their own knowledge meaningfully.



Journal Writing



Thinking Time



Key questions are also included at appropriate junctures to check if students have grasped various concepts and to create opportunities for them to further develop their thinking.



Opportunities are provided for students to reflect on their learning and to communicate mathematically. It can also be used as a formative assessment to provide feedback to students to improve on their learning.



Performance Task Mini projects are designed to develop research and presentation skills in the students.



MARGINAL NOTES AT



TE



NTI



ON



This contains important information that students should know.



RE



P So roblem lvin g T ip



INF



This guides students on how to approach a problem.



r Fun



CAL



L



This contains certain mathematical concepts or rules that students have learned previously.



Just Fo



This contains puzzles, fascinating facts and interesting stories about Mathematics as enrichment for students.



OR



MA



TIO N



This includes information that may be of interest to students.



Internet Resources



This guides students to search on the Internet for valuable information or interesting online games for their independent and self-directed learning.



Preface



v



Contents CHAPTER 1 Sequences and Series



001 003 005



1.1 Number Sequences 1.2 General Term of a Number Sequence 1.3 Arithmetic and Geometric 009 Progressions 1.4 Number Patterns in Real-World 015 Contexts Summary 020 Review Exercise 1 021



vi



Contents



New Syllabus Mathematics (NSM)



CHAPTER 2 Polynomials



023 2.1 Polynomials 025 2.2 Remainder Theorem 036 2.3 Factor Theorem 039 2.4 Cubic Expressions and Equations 042 2.5 Partial Fractions 051 Summary 063 Review Exercise 2 064



CHAPTER 3 Graphs of Functions and Graphical Solution



067



3.1 Graphs of Cubic Functions 069 3.2 Graphs of Reciprocal Functions 072 3.3 Graphs of Exponential Functions 079 3.4 Slope of a Curve 083 3.5 Applications of Graphs 088 in Real-World Contexts Summary 106 Review Exercise 3 108



Revision Exercise A



111



CHAPTER 4 Arc Length, Area of Sector and Radian Measure



113



4.1 Length of Arc 115 4.2 Area of Sector 124 4.3 Radian Measure 131 4.4 Arc Length and Area of Sector 140 using Radian Measure Summary 150 Review Exercise 4 151



New Syllabus Mathematics (NSM)



Contents



vii



CHAPTER 5 Geometrical Properties of Circles 155 5.1 Symmetric Properties of Circles 157 5.2 Angle Properties of Circles 176 Summary 197 Review Exercise 5 199



CHAPTER 6 Coordinate Geometry 6.1



6.2 Midpoint of a Line Segment 210 6.3 Equation of a Circle 215 Summary 226 Review Exercise 6 226



CHAPTER 7 Permutations and Combinations



Revision Exercise B 231 233



7.1 Addition and Multiplication Principles 7.2 Permutations 237 7.3 Combinations 243 Summary 248 Review Exercise 7 248



viii



Contents



New Syllabus Mathematics (NSM)



Length of a Line Segment



203 205



229



CHAPTER 8 Probability of Combined Events



251 253



8.1 Addition Law of Probability and Mutually Exclusive Events 8.2 Multiplication Law of 257 Probability and Independent Events Summary 270 Review Exercise 8 271



CHAPTER 9 Statistical Data Analysis



273 275 276 278 285



9.1 Statistical Diagrams 9.2 Dot Diagrams 9.3 Stem-and-Leaf Diagrams 9.4 Cumulative Frequency Table and Curve 9.5 Median, Quartiles, Percentiles, 296 Deciles, Range and Interquartile Range 9.6 Box-and-Whisker Plots 310 9.7 Standard Deviation 321 Summary 339 Review Exercise 9 340



Revision Exercise C Problems in Real-World Contexts Practice Now Answers Answers



New Syllabus Mathematics (NSM)



345 349 356 360



Contents



ix



Sequences and Series The Fibonacci sequence is interesting as it is not merely a theoretical sequence but is also prevalent in nature such as in the number of petals of a flower and in the breeding patterns of rabbits. Can you think of other examples that exhibit the Fibonacci sequence?



One



LEARNING OBJECTIVES At the end of this chapter, you should be able to: • recognize simple patterns from various number sequences, • determine the next few terms and find a formula for the general term of a number sequence, • recognize and solve problems involving arithmetic and geometric progressions, • solve problems involving number sequences and number patterns.



1.1



Number Sequences



Consider the following whole numbers: 1st term 2,



2nd term 5,



3rd term 8,



+3



+3



4th term 11, +3



5th term 14,… +3



2, 5, 8, 11, 14, … forms a number sequence. The numbers in the number sequence are known as the terms of the sequence. The numbers are governed by a specific rule, i.e. start with 2, then add 3 to each term to get the next term.



Class Discussion Number Sequences



Work in pairs. 1. Table 1.1 shows a few examples of number sequences. Complete Table 1.1. Sequence Positive even numbers Positive odd numbers Multiples of 3



Powers of 2



Powers of 3



2,



4,



+2 1,



+2 3,



+2 3,



1, ×3



5,



6,



7,



9,



2, 3,



8,



9,



15, ____, ____, …



+ 3 ____ ____



×2



×3



9, ____, ____, …



12,



4,



____ ____



+ 2 ____ ____



+3



×2



10, ____, ____, …



+2



+2



+3



×2



8, +2



+2



+3 1,



6,



Rule



16, ____, ____, … × 2 ____ ____



27, ×3



81, ____, ____, …



× 3 ____ ____



Start with _____, then add _____ to each term to get the next term. Start with _____, then add _____ to each term to get the next term. Start with _____, then add _____ to each term to get the next term. Start with _____, then multiply each term by _____ to get the next term. Start with _____, then multiply each term by _____ to get the next term.



Table 1.1 2. The sequence of positive even numbers can also be obtained by multiplying each term of the sequence 1, 2, 3, 4, 5, … by 2. Can you think of a rule, different from that in Table 1.1, to obtain the sequence of positive odd numbers? 3. For each of the following sequences, state a rule and write down the next two terms.



003



(a) Perfect squares: 1, 4, 9, 16, 25, …



Chapter 1



Sequences and Series



(b) Perfect cubes: 1, 8, 27, 64, 125, …



Worked Example



1



(Rules of Sequences) For each of the following sequences, state a rule and write down the next two terms. (a) 42, 39, 36, 33, 30, …



(b) –22, –18, –14, –10, –6, … (c) 256, 128, 64, 32, 16, … (d) –1, 1, –1, 1, –1, …



Solution: (a) 42,



39, 36,



– 3



– 3



– 3



–3



27 and 24. + 4



–10,



+ 4 + 4



–6, … +4



Check if • the sequence is a common one that you recognize, • two consecutive terms in the sequence are related by a constant value, • you can add/subtract/ multiply/divide consecutive terms to get the next term.



Rule: Add 4 to each term to get the next term. The next two terms are –2 and 2.



(c) 256, 128, 64, ÷ 2



30, …



Rule: Subtract 3 from each term to get the next term. The next two terms are



(b) –22, –18, –14,







33,



P So roblem lvin g T ip



32,



÷ 2 ÷ 2



16, … ÷2



Rule: Divide each term by 2 to get the next term. The next two terms are 8 and 4.



(d) –1,



1,



–1,



1,



–1, …



× (–1) × (–1) × (–1) × (–1)



Rule: Multiply each term by (–1) to get the next term. The next two terms are



1 and –1.



SIMILAR QUESTIONS



PRACTICE NOW 1



1. For each of the following sequences, state a rule and write down the next two terms. (a) 3, 8, 13, 18, 23, … (b) –20, –26, –32, –38, –44, … (c) 5, 15, 45, 135, 405, … (d) 4374, –1458, 486, –162, 54, …



Exercise 1A Questions 1(a)-(h), 2(a)-(d), 3(a)-(d)



2. Write down the next two terms of each of the following sequences. (a) 1, 2, 4, 7, 11, 16, … (b) 26, 25, 21, 20, 16, …



Sequences and Series



Chapter 1



004



Exercise



1A



BASIC LEVEL



1.



INTERMEDIATE LEVEL



2.



For each of the following sequences, state a rule and write down the next two terms. (a) 14, 19, 24, 29, 34, … (b) 80, 72, 64, 56, 48, … (c) 6, 12, 24, 48, 96, 192, … (d) 1600, 800, 400, 200, 100, … (e) –16 384, 4096, –1024, 256, –64, … (f) 9, –18, 36, –72, 144, … (g) –52, –59, –66, –73, –80, … (h) –100, –90, –80, –70, –60, …



1.2



Write down the next two terms of each of the following sequences. (a) –6, –5, –3, 0, 4, … (b) 47, 38, 30, 23, 17, … (c) –50, –45, –44, –39, –38, … (d) 100, 98, 95, 93, 90, …



ADVANCED LEVEL



3.



Write down the next two terms of each of the following sequences. (a) –5, –7, –11, –19, –35, … (b) 1, 1, 2, 3, 5, … (c) 4, 16, 36, 64, 100, … (d) 1, –8, 27, –64, 125, …



General Term of a Number Sequence



Simple Sequences Consider the sequence of positive even numbers: 2, 4, 6, 8, 10, … The terms of the sequence are denoted by T1, T2, T3, T4, T5, …, Tn, where T1 = 1st term = 2, T2 = 2nd term = 4, T3 = 3rd term = 6, T4 = 4th term = 8, T5 = 5th term = 10,  Tn = nth term (general term). From Table 1.2, observe that each term in the sequence can be obtained by multiplying its position n by 2. Position n



1



2



3



Term Tn



2×1=2



2×2=4



2×3=6



4



5



…



n



2 × 4 = 8 2 × 5 = 10 … 2n



Table 1.2 Hence, the general term of the sequence is Tn = 2n.



Note: n is a variable, i.e. by substituting different values of n, we are able to



005



generate the terms of the sequence. For example, the 68th term of the sequence is given by T68 = 2(68) = 136.



Chapter 1



Sequences and Series



Class Discussion



Generalizing Simple Sequences



Work in pairs. For each of the following sequences, use the table provided to find a formula for the general term and hence, state the 100th term, T100. (a) Multiples of 3: 3, 6, 9, 12, 15, … From Table 1.3, observe that each term in the sequence can be obtained by multiplying its position n by 3. Position n



1



2



3



4



5



…



Term Tn



3×1=3



3×2=6



3×3=9



3 × 4 = 12



3 × 5 = 15



…



n



Table 1.3



Hence, Tn = _____. 100th term, T100 = _____



(b) Perfect squares: 1, 4, 9, 16, 25, … From Table 1.4, observe that each term in the sequence can be obtained by squaring its position n . Position n



1



2



3



4



5



…



Term Tn



12 = 1



22 = 4



32 = 9



42 = 16



52 = 25



…



n



Table 1.4



Hence, Tn = _____. 100th term, T100 = _____



(c) Perfect cubes: 1, 8, 27, 64, 125, … From Table 1.5, observe that each term in the sequence can be obtained by cubing its position n. Position n



1



2



3



4



5



…



Term Tn



1 =1



2 =8



3 = 27



4 = 64



5 = 125



…



3



3



Hence, Tn = _____. 100th term, T100 = _____



Worked Example



2



Solution: (i) T3 = 5(3) – 3 = 15 – 3 = 12



3



3



3



n



Table 1.5



(Finding a Specific Term) Given that the nth term, Tn, of a sequence is Tn = 5n – 3, find (i) the 3rd term, (ii) the difference between the 3rd term and the 5th term, of the sequence.



(ii) T5 = 5(5) – 3 = 25 – 3 = 22 Difference between the 3rd term and the 5th term of the sequence = T5 – T3 = 22 – 12 = 10 Sequences and Series



Chapter 1



006



PRACTICE NOW 2



SIMILAR QUESTIONS



Given that the nth term, Tn, of a sequence is Tn = 4n + 7, find (i) the 4th term, (ii) the sum of the 4th term and the 7th term, of the sequence.



More Complicated Sequences Consider the sequence 2, 5, 8, 11, 14, … How do we find a formula for the general term?



Looking for a Pattern



Notice that the differences between consecutive terms are all equal to a constant. Thus we say that the common difference of this sequence is equal to 3. 2



Position n 1 Term Tn



2,



5,



3 8,



+ 3 + 3



4



11,



+ 3 + 3



5



14,



… +3



Therefore, we can express each term as follows: T1 = 2 = 2 =2+ 0 ×3 T2 = 5 = 2 + 3 =2+ 1 ×3 T3 = 8 = 2 + 3 + 3 =2+ 2 ×3 T4 = 11 = 2 + 3 + 3 + 3 =2+ 3 ×3 T5 = 14 = 2 + 3 + 3 + 3 + 3 =2+ 4 ×3  Tn = 2 + 3 + 3 + 3 + ... + 3 = 2 + (n − 1) × 3 (n − 1) terms By looking at the above pattern, we can infer that Tn = 2 + (n − 1) × 3 = 2 + 3n – 3 = 3n − 1.



Transformation to Another Sequence



Let us try another method to find a formula for the general term of the same sequence 2, 5, 8, 11, 14, … Since the common difference is equal to 3, we can transform this sequence to another sequence which consists of terms that are multiples of 3 (the first term must be 3 so that we know that the general term is given by 3n immediately): Position n



1



2



3



4



Term Tn



2,



5,



8,



11,



14, ...



3n − 1



12,



15, ...



3n



↓+1



↓+1



↓+1



3,



6,



9,



↓+1



5 ↓+1



n ↑ −1



Since we add 1 to each term of the first sequence to obtain the corresponding term in the second sequence, we will have to subtract 1 from 3n (as we are moving backwards) to get 3n − 1. Therefore, the general term of the original sequence is Tn = 3n − 1.



007



Chapter 1



Sequences and Series



Exercise 1B Question 1



Connecting the General Term to a Graph



Tn



The explanation for this method is long, but after understanding how it works, we will be able to write down the formula for the general term of a sequence



14 11



straightaway without drawing the graph. This is by far the fastest method. Let us consider the same sequence 2, 5, 8, 11, 14, …



8



If we plot Tn against n, we will get a straight line as shown in Fig. 1.1(a).



5



The equation of the straight line is Tn = mn + c (in the form of y = mx + c). rise 3 = = 3 = d , where d is the common difference From Fig. 1.1(b), slope m = run 1



2 n



0



1



2



3



and Tn – intercept c = 2 − 3 = −1.



When n = 0, T0 = c, where T0 is the term just before the first term T1.



14



Thus we can find c = T0 = 2 − 3 = −1 mentally as follows:



T0



T1



T2



T3



T4



T5







?,



2,



5,



8,



11,



14,



− 3



+ 3



+ 3



+ 3 + 3



5



(a)



Tn



There is another way to find c without looking at the graph.



4



11



1 3



8



…



5



+3



Hence, the general term is Tn = 3n − 1. Worked Example 3 shows how to use the above concept to find a formula for



2 –3 0 c?



3



–1 1



1 3



1 3



1



2



the general term of a sequence without drawing the graph.



3



(b)



4



5



n



Fig. 1.1



Worked Example



3



(Finding the General Term by Observation) Find a formula for the general term of the sequence 8, 12, 16, 20, 24, …



Solution:



T0



?,



T1



T2



8,



− 4



T3



12, 16,



+ 4



+ 4



T4



20,



+ 4



Since the common difference is 4, Tn = 4n + ?.



+4



T5



24, +4



…



P So roblem lvin g T ip



You can work out the solution mentally and write down the formula straightaway.



The term before T1 is c = T0



= 8 − 4



= 4.



∴ General term of the sequence, Tn = 4n + 4



Sequences and Series



Chapter 1



008



SIMILAR QUESTIONS



PRACTICE NOW 3



1. Find a formula for the general term of each of the following sequences. (a) 5, 9, 13, 17, 21, … (b) 7, 12, 17, 22, 27, … (c) 2, 8, 14, 20, 26, … (d) 1, 4, 7, 10, 13, … 2.



Consider the sequence 3, 7, 11, 15, 19, … (i) Write down the next two terms of the sequence. (ii) Find, in terms of n, a formula for the nth term of the sequence. (iii) Hence, find the 50th term.



1.3



Arithmetic and Geometric Progressions



Arithmetic Progression



An arithmetic progression (AP) progresses from one term to the next by adding or subtracting the same value, which is known as the common difference. An arithmetic progression can be represented by a general term as follows: Tn = T1 + (n – 1)d Where Tn = nth term in a sequence n = number of term d = common difference The sum of a certain number of terms in an arithmetic progression can also be determined. We can apply the following equation to find the first n term of the sequence, Sn. Sn =



Worked Example



009



Chapter 1



4



n(T1 + Tn ) 2



(Solving Problem involving Arithmetic Progression) Consider the sequence 5, 9, 13, 17, ... (i) Find the common difference, d, of the sequence. (ii) Hence, find the 25th term. (iii) Find the sum of the first 10 terms of the sequence.



Sequences and Series



Exercise 1B Questions 2(a)-(d), 3, 10



Solution:



(i) Tn = T1 + (n – 1)d Since T3 = 13, T3 = T1 + (3 – 1)d T −T d= 3 1 2 13− 5 = 2 =4



∴ Common difference of the sequence, d = 4



(ii) T25 = 5 + (25 – 1)(4) = 101 (iii) T10 = 5 + (10 – 1)(4) = 41



S10 =



10(5 + 41) 2



= 230



SIMILAR QUESTIONS



PRACTICE NOW 4



Consider the sequence −7, − 4, −1, 2, ... (i) Find the common difference, d, of the sequence. (ii) Hence, find the 16th term. (iii) Find the sum of the first 16 terms of the sequence.



Exercise 1B Questions 4-5, 11



Geometric Progression Let us consider the sequence 1, 4, 16, 64, 256, ... Notice that the division between consecutive terms gives the common value, which are equal to 4. Thus we say that the common ratio of this sequence is equal to 4. Therefore, we can express each term as follows: T1 = 1 = 1 = 1(4)(0) T2 = 4 = 1 × 4 = 1(4)(1) T3 = 16 = 1 × 4 × 4 = 1(4)(2) T4 = 64 = 1 × 4 × 4 × 4 = 1(4)(3) T5 = 256 = 1 × 4 × 4 × 4 = 1(4)(4) Tn



= 1 × 4 × 4 × 4 × … × 4 = 1(4)(n−1) (n – 1) terms



Sequences and Series



Chapter 1



010



The sequence above is known as a geometric progression. A geometric progression (GP) progresses from one term to the next by multiplying or dividing by the same value. It can be represented by a general term as follows: Tn = T1r(n – 1) Where Tn = nth term in a sequence n = number of term r = common ratio Similarly, the sum of the first n terms of a geometric progression can also be determined using the following equation. Sn =



5



Worked Example



(Solving Problem using Geometric Progression) 1 1 1 Consider the sequence 1, − , ,– , … 2 4 8 (i) Find the common ratio, r, of the sequence. (ii) Hence, find the 11th term. (iii) Find the sum of the first 8 terms of the sequence.



Solution: (i) Tn = T1 r(n – 1)



1 Since T4 = − , 8 T4 = T1 r(4 − 1) 1 r= − 8 T = 3 4 T1







=



3



−



= −



1 2



1 8



(11−1) 1 ⎛ 1⎞ (ii) T11 = − ⎜− ⎟ 2 ⎝ 2⎠ 1 = − 2048



8 1 ⎡ ⎛ 1⎞ ⎤ − ⎢1 − ⎜− ⎟ ⎥ 2 ⎢⎣ ⎝ 2 ⎠ ⎥⎦ (iii) S8 = ⎛ 1⎞ 1 − ⎜− ⎟ ⎝ 2⎠







011



=−



85 256



Chapter 1



T1 (1− r n ) 1− r



Sequences and Series



PRACTICE NOW 5



SIMILAR QUESTIONS



Consider the sequence 0.5, 3.5, 24.5, 171.5, ... (i) Find the common ratio, r, of the sequence. (ii) Hence, find the 9th term. (iii) Find the sum of the first 9 terms of the sequence.



Exercise 1B Questions 6, 12-13



INF



A geometric progression is finite when its first term, last term and the number of terms are known. Hence, we can calculate the sum of a finite geometric progression using the formula T (1− r n ) given earlier, i.e. Sn = 1 . 1− r However, when the last term and the number of terms are unknown, a geometric progression becomes infinite and the sum of an infinite geometric progression will not be determined using the same formula. For an infinite geometric progression that has −1 , r , 1, the sum of the sequence can be calculated using the following equation. Sn =



Worked Example



6



OR



MA



TIO N



We say that an infinite sequence is convergent when its partial sums get closer to a single value as the number of terms increases. For example, the sequence with the 1 1 1 1 1 terms 1, , , , , ,… 2 4 8 16 32 1 has the partial sum of 1048576 at 20th term. The sum does not increase much and gives a limit as more terms are involved. The sequence diverges when the limit does not exist in the sum.



T1 (1− r)



(Determining the Sum of an Infinite Geometric Progression) Consider the sequence 27, 18, 12, 8, ... (i) Find the common ratio, r, of the sequence. (ii) Hence, find the sum of the sequence.



Solution: T2 T1 18 = 27 2 = 3



(i) r =



(ii) Since −1 , r , 1, T1 S∞ = (1− r) S∞ =







27 ⎛ 2⎞ ⎜1 − ⎟ ⎝ 3⎠



= 81



PRACTICE NOW 6



Consider the sequence 250, 100, 40, 16, ... (i) Find the common ratio, r, of the sequence. (ii) Hence, find the sum of the sequence.



SIMILAR QUESTIONS



Exercise 1B Questions 7, 14



Sequences and Series



Chapter 1



012



Geometric Mean In Grade 7, we have learned about the measures of central tendency. In this section, we will look at another measure of central tendency. Geometric mean is defined as the nth root of the product of the n items. Geometric Mean = n Product of n items For example, the geometric mean for 4 and 9 is and 9 is 3 4 × 6 × 9 = 6.



2



INF



OR



MA



TIO N



The ‘average‘ we learned in Grade 7 is also known as the arithmetic mean.



4 × 9 = 6 and that of 4, 6



The geometric mean only works with positive numbers and is frequently used in fields of science, finance, and statistics. It is sometimes preferred for averaging ratios of two variables such as rates of population growth, rates of interest and rates of depreciation.



Journal Writing Find out what are the other applications for the geometric mean.



Class Discussion Geometric Mean



Given a geometric progression 2, 4, 8, 16, 32 … 1. Find the geometric mean for the following pairs.



(i) 2 and 8



(ii) 4 and 16







(iii) 8 and 32







What do you observe from the above calculations?



2. Based on your observation in Question 1, can you determine the 6th term of the progression? Hence, find the 7th term.



From the class discussion, we observe that the geometric mean of T1 and T3 is T2. Similarly, the geometric mean of T2 and T4 is T3. Therefore, we deduce that the geometric mean of Tn and Tn + 2 is Tn + 1. We can thus build the geometric progression using geometric mean.



013



Chapter 1



Sequences and Series



INF



OR



MA



TIO N



The geometric mean has the following properties: • The geometric mean is less t h a n a r i t h m e t i c m e a n , i. e. G.M < A.M. • The product of the items remains unchanged if each item is replaced by the geometric mean. • The geometric mean of the ratio of corresponding observations in two series is equal to the ratios their geometric means. • The geometric mean of the products of corresponding items in two series is equal to the product of their geometric mean.



Worked Example



7



(Determining Geometric Mean) Find the geometric mean for each of the following. (i) 13 and 39 (ii) 1, 3, 5, 7 and 9



Solution:



(i) Geometric mean = (ii) Geometric mean =



13× 39 = 22.5 (to 3 s.f.) 5



1× 3× 5 × 7 × 9 = 3.94 (to 3 s.f.)



PRACTICE NOW 7



SIMILAR QUESTIONS



Find the geometric mean for each of the following. (i) 35 and 51



Worked Example



8



Exercise 1B Question 8



(ii) 2, 4, 6, 8 and 10



(Determining Term from Geometric Mean) The geometric mean is 16. The larger number is twice the smaller. What are the two numbers in simplified radical form?



Solution:



Let x be the smaller number. Since geometric mean = 16, x × 2x = 16



2x2 = 256 x2 = 128



x = ± 128 x = 8 2 or x = −8 2 (rejected since geometric mean involves positive numbers)



∴ The smaller number is 8 2 and the larger number is 16 2 .



SIMILAR QUESTIONS



PRACTICE NOW 8



The geometric mean is 48. The smaller number is one fourth that of the larger. What are the two numbers?



Exercise 1B Question 9



Sequences and Series



Chapter 1



014



1.4



Number Patterns in Real-World Contexts



In this section, we shall take a look at number patterns that can be found in the real world. v



Investigation Fibonacci Sequence



As mentioned at the start of this chapter, the number of petals of a flower follows a special type of sequence known as the Fibonacci sequence. In this investigation, we shall explore this by taking a look at some flowers. 1. Write down the number of petals for each of the following flowers on the line beside its name. Some of them have been done for you.



Picture B: Euphorbia 2 (Note: There are 8 flowers in the picture.)



Picture A: White Calla Lily _____



Picture C: Mariposa Lily



3



Picture E: Moonbeam Coreopsis



Picture G: Shasta Daisy _____



015



Chapter 1



Sequences and Series



Picture D: Madagascar Periwinkle _____



8



Picture F: Black-eyed Susan _____



Picture H: Sunflower 34



2.



The number of petals for each of the flowers forms a sequence. Fill in the next 6 terms of the sequence. 1, 1, 2, _____, _____, _____, _____, _____, _____. This is known as the Fibonacci sequence.



3. A sunflower has 34 petals. Using your answer in Question 2, predict the number of petals for the flower next in the sequence. The name of this flower is Michaelmas Daisy (see photo below).



4. Note that there are four common exceptions to the Fibonacci sequence. Write down the number of petals for each of the following flowers on the line beside its name.



Picture I: Ixora _____



Picture J: Daylily _____



Picture K: Anemone Nemorosa _____



Picture L: Passion Flower _____



Journal Writing Find out more about Pascal’s Triangle. Illustrate clearly how the Fibonacci sequence is found in Pascal’s Triangle.



Sequences and Series



Chapter 1



016



Worked Example



9



(Number Patterns in Chemistry) The members in a family of chemical compounds are made up of carbon atoms and hydrogen atoms. (i) The number of carbon atom(s) and hydrogen atoms of the first four members in the family are given in the table. Complete the table. Member Number 1 2 3 4 5 6  n



Number of carbon atom(s) 1 2 3 4



Number of hydrogen atoms 4 6 8 10











(ii) If a member of the family has 30 carbon atoms, how many hydrogen atoms does it have? (iii) If a member of the family has 52 hydrogen atoms, how many carbon atoms does it have?



Solution: (i)



Member Number 1 2 3 4 5 6  n



Number of carbon atom(s) 1 2 3 4 5 6  n



To find a formula for the general term of the number of hydrogen atoms, consider the sequence







4, 6, 8, 10, 12, 14, …















T0



?,



− 2



T1



4,



+ 2



T2



T3



T4



T5



T6



6, 8, 10, 12, 14, … + 2



+ 2



+ 2



+2



Since the common difference is 2, Tn = 2n + ?. The term before T1 is c = T0 = 4 − 2 = 2. ∴ General term of the sequence, Tn = 2n + 2 (ii) When n = 30, T30 = 2(30) + 2 = 62 The member of the family has 62 hydrogen atoms. (iii) Let 2n + 2 = 52. 2n = 52 – 2 = 50 ∴ n = 25 The member of the family has 25 carbon atoms. 017



Number of hydrogen atoms 4 6 8 10 12 14  2n + 2



Chapter 1



Sequences and Series



SIMILAR QUESTIONS



PRACTICE NOW 9



The members in a family of chemical compounds are made up of carbon atoms and hydrogen atoms.



Exercise 1B Questions 15-16



(i) The number of carbon atom(s) and hydrogen atoms of the first four members in the family are given in the table. Complete the table. Member Number 1 2 3 4 5 6  n



Number of carbon atoms 2 3 4 5



Number of hydrogen atoms 4 6 8 10











(ii) If the hth member of the family has 55 carbon atoms, find the value of h. Hence, find the number of hydrogen atoms the member has. (iii) If the kth member of the family has 120 hydrogen atoms, find the value of k. Hence, find the number of carbon atoms the member has.



Exercise



1B



BASIC LEVEL



1. Given that the nth term, Tn, of a sequence is



Tn = 2n + 5, find







(i) the 5th term,







(ii) the 8th term,







(iii) the lowest common multiple of the 5th term







and the 8th term,



of the sequence.



2. Find a formula for the general term of each of



3. Consider the sequence 3, 6, 9, 12, 15, …



(i) Write down the next two terms of







the sequence.



(ii) Find, in terms of n, a formula for the



nth term of the sequence.



(iii) Hence, find the 105th term.



4. Consider the sequence 10, 14, 18, 22, 26, …



(i) Find the common difference, d, of the sequence.







(ii) Hence, find the 200th term.







the following sequences.







(a) 7, 13, 19, 25, 31, … (b) –4, –1, 2, 5, 8, …







(i) the common difference,







(c) 60, 67, 74, 81, 88, … (d) 14, 11, 8, 5, 2, …







(ii) the 10th term,



5. In each of the following AP, find







(iii) the sum of the first 10 terms.



(a) 10, 9, 8, 7, … 1 1 (b) 1, 2 , 4, 5 , … 2 2 (c) 20, 18, 16, 14, …



(d) −25, −20, −15, −10, … 1 1 3 1 (e) − , − , − , − , … 8 4 8 2 Sequences and Series



Chapter 1



018



6. In each of the following GP, find



12. The fifth term of an AP is x and the ninth term is y. Find the common difference and the third term



(i) the common ratio,



(ii) the 10 term, th



in terms of x and y.



(iii) the sum of the first 15 terms. 1 (a) 4, 2, 1, , … 2 (b) −2, 4, −8, 16, … (c) 5, 20, 80, 320, … 1 1 1 1 (d) , − , ,− ,… 3 9 27 81 8 4 (e) , , 2, 3, … 9 3



2 7. The third and sixth terms of a GP are 9 and 2 3 respectively. Find



13. The first three terms of a GP of positive terms are x − 3, x + 2 and 2x + 4. Find







(a) the value of x,







(b) the value of the fifth term.



14. Given that x + 18, x + 4 and x − 8 are the first three terms of a GP, find the value of x. Hence, find



(a) the common ratio,







(a) the common ratio,







(b) the fifth term,







(b) the first term,







(c) the sum to infinity of the progression.







(c) the sum to infinity of the progression.



8. Find the geometric mean of each of the following.



(a) 6, 24 1 9 (b) , 4 64 (c) 0.2, 3.2



15. The members in a family of chemical compounds



are made up of carbon atoms and hydrogen atoms.







(i) The number of carbon atom(s) and hydrogen















9. The geometric mean is 6. The smaller number is one third that of the larger. What are the two numbers?



Number of carbon atoms



Number of hydrogen atoms



2



4



6



4



5



2n2 + 1. Write down the first four terms of



3 5



4



8



6



10











6



the sequence.







(b) The first four terms of another sequence



n



are 1, 7, 17, 31.



(i) By comparing this sequence with the







(ii) If the hth member of the family has 25 carbon



sequence in (a), write down, in terms of











atoms, find the value of h. Hence, find the



n, a formula for the n term of the sequence.







number of hydrogen atoms the member has.



(ii) Hence, find the 388th term.







(iii) If the kth member of the family has 64 hydrogen











th



11. The sixth term of an AP is 234 and the sixty-eighth term is −156. Calculate



019



Member Number



3



10. (a) The n th term of a sequence is given by







are given in the table. Complete the table.



1



INTERMEDIATE LEVEL







atoms of the first four members in the family







(a) the first term and the common difference,







(b) the value of the first negative term of the AP.



Chapter 1



Sequences and Series







atoms, find the value of k. Hence, find the number of carbon atoms the member has.



16. It is a curious biological fact that a male bee (M) has



(i) Complete the model and show all the 4th







only one parent (the mother) while a female bee (F)



generation ancestors. How many 4 th







has two parents (both mother and father).











The figure shows part of the bee ancestry model



(ii) The number of nth generation ancestors forms







for a male bee up to his 4th generation ancestors. M F



generation ancestors does a male bee have?







a sequence with an interesting pattern. How do







you obtain the next term in the sequence?



(iii) Predict the number of 5th generation ancestors 1st Generation Ancestor



M



F



2nd Generation Ancestors



M



3 Generation Ancestors rd



F







that a male bee has. Verify your answer by







drawing the 5th generation ancestors in the







above model.



(iv) Predict the number of 10th generation ancestors



that a male bee has.



4th Generation Ancestors



1. A number sequence is formed by a set of numbers. These numbers, known as the terms of the sequence, are governed by a specific rule. 2. The general term Tn of a number sequence can be represented by an algebraic expression. 3. Arithmetic progression (AP)



• AP progresses from one term to the next by adding or subtracting the common difference, d and represented



by a general term Tn = T1 + (n – 1)d .



• The sum of first n term in an AP can be determined by







Sn =



n(T1 + Tn ) 2



.



4. Geometric progression (GP)



• GP progresses from one term to the next by multiplying or dividing by the common ratio, r and represented



by a general term Tn = T1r(n – 1) .



T1 (1− r n ) . 1− r







• The sum of the first n terms of a finite geometric progression can be determined by Sn =







• The sum of the first n terms of an infinite geometric progression can be determined by Sn =



−1 , r , 1.



Sequences and Series



T1 , where 1− r



Chapter 1



020



1 1. Write down the next two terms of each of the



following sequences.







(a) 98, 89, 80, 71, 62, ...







(b) −2, 0, 4, 10, 18, ... 1 1 (c) 9, 3, 1, 3 , 9 , ... (d) 1, 9, 25, 49, 81, ...







Figure 1



2. Consider the sequence 9, 16, 25, 36, 49, …



5. The first four figures of a sequence are as shown.



Figure 2



(i) Draw the next figure of the sequence.







(ii) Complete the table. Figure Number



(i) Write down the next two terms of the (ii) Find, in terms of n, a formula for the nth term







of the sequence.



(iii) Hence, find the 25th term.



3. The arithmetic mean of a and b is 6 and the 5 1 1 geometric mean of and is . Find the value a b 2 of a and of b. 4. Given that 2x − 14, x − 4 and terms of a sequence,







(a) find the value of x when the sequence is



(i) an AP, (ii) a GP.



(b) if 2x − 14 is the third term of a GP with infinite











terms, find



Number of Triangles



2







3







3× 4 6 = 2



4







10 =



2×3 3 = 2



4×5 2











n







(iii) Find the number of triangles in the 77th















(iv) Find the value of n for which the figure has



figure.



66 triangles. 6. Consider the following number pattern:



(i) the common ratio,



13 = 1 = 12



(ii) the sum to infinity, S∞ ,



13 + 23 = 9 = (1 + 2)2



(iii) the sum of the first 15 terms, S15.



Hence, express S15 as a percentage of S∞ , giving







13 + 23 + 33 = 36 = (1 + 2 + 3)2







13 + 23 + 33 + 43 = 100 = (1 + 2 + 3 + 4)2 



your answer correct to 2 decimal places.



021



1× 2 1 = 2



1



5



1 x are successive 2







Figure 4







sequence.



Figure 3



Chapter 1



Sequences and Series







(i) Write down the 7 line in the pattern.







(ii) Find the value of 13 + 23 + 33 + 43 + … + 153.







(iii) Given that 13 + 23 + 33 + 43 + … + k3 = 1296,











th



find the value of k.



7. Kate uses buttons to make a sequence of figures.



8. Three different sequences are shown in the table.



The first four figures are as shown.



Sequence A B C



Figure 1 Figure 2



Figure 3







(i) Draw the 5th figure.







(ii) Complete the table. Figure Number 1 2 3 4 5  n



Figure 4



2nd term 6 8 10



3rd term 8 15 17



4th term 10 24 26



5th term 12 35 a



(i) Find the value of a. Hint: The first term of sequence C is obtained



Number of Buttons 5×1+1=6 5 × 2 + 1 = 11 5 × 3 + 1 = 16 5 × 4 + 1 = 21







from the first terms of sequences A and B as



follows:



4 2 + 32 = 5 .



(ii) Find, in terms of n, a formula for the nth







term of sequence A.







(iii) If the nth term of sequence B is given by n2 + 2n,











find the 18th term of sequence C.











(iii) Find the number of buttons in the 56th figure.







(iv) Is it possible for a figure in the sequence to be







1st term 4 3 5



made up of 583 buttons? Explain your answer.



Challenge 1. Determine the last digit of 32015. 2. There are n people at a party. If each person shakes hands



with each of the other people only once, find an expression,







in terms of n, for the number of handshakes that will take place.



3. Consider the sequence 1, 4, 9, 7, 7, 9, 4, 1, 9, 1, …



(i) Write down the next two terms of the sequence.







(ii) State the rule of the sequence.



4. Consider the sequence 2, 1, 3, 4, 7, …



(i) Write down the next two terms of the sequence.







(ii) State the rule of the sequence.







(iii) There is a name for this sequence. Search on the Internet to find out its



name. 5. Consider the sequence 3, 0, 2, 3, 2, 5, 5, 7, …



(i) Write down the next two terms of the sequence.







(ii) State the rule of the sequence.







(iii) There is a name for this sequence. Search on the Internet to find out its



name. Sequences and Series



Chapter 1



022



Polynomials



The Van der Waals' equation is an equation of state for a fluid, in which forces of attraction, such as Van der Waals' forces, exist between the particles. When the density of the gas is equal to that of the liquid, both the gas and the liquid are said to have reached critical temperature and pressure. At this point, the Van der Waals' equation becomes a cubic equation in v, where v is the volume of the fluid. In this chapter, we will learn about polynomials and how to solve cubic equations.



Two



LEARNING OBJECTIVES At the end of this chapter, you should be able to: At the end of this chapter, you should be able to: • use the Remainder Theorem to find the remainder when an expression is divided by a linear factor, • factorize cubic expressions or polynomials of higher degree using the Factor Theorem, • solve cubic equations and problems involving cubic equations, • express improper fractions as the sum of a polynomial and a proper fraction, • perform partial fraction decomposition of an algebraic fraction.



2.1



Polynomials



What is a polynomial? Recall that we have learned about algebraic expressions such as 2a + b (linear 1 in two variables), x2 − 4x + 3 (quadratic in one variable) and (algebraic 5−x fraction). In this section, we will study a certain type of algebraic expression called polynomials.



Investigation Polynomials and Non-polynomials



The table below shows some examples and non-examples of polynomials in one variable. Polynomials



Non-polynomials



x − 4x + 3



2x2 − 7x + x−1



2x5 + 7x4



− x 2 + x −3



2



5 x3 + 6 −



8



5 x 7



1



x2 − x 1 2 3 − − x x x4



1. Compare the powers of x in polynomials with those in non-polynomials. What do you notice? 2. Polynomials cannot contain terms such as other terms that polynomials cannot include?



1



x = x 2 . Can you think of



3. Explain to your classmates some features of a polynomial.



From the above investigation, a polynomial in x is an algebraic expression consisting of terms with non-negative powers of x only. Its general form is: anxn + an − 1xn − 1 + an − 2xn − 2 + ... + a2x2 + a1x + a0, where n is a non-negative integer, the coefficients an, an − 1, an − 2, ... a2, a1, a0 are constants and x is a variable. 025



Chapter 2



Polynomials



Linear and quadratic expressions are examples of polynomials. Polynomials cannot contain terms involving fractional powers or negative powers of the variable. For example, polynomials in x cannot contain 3



1 x3



( ) , 1x (= x



x =



−1



) and



non-negative integers.



1 x2



( ),



x =



1 (= x −2 ) , because the powers of x must be x2



We usually arrange the terms of a polynomial in decreasing powers of the variable, e.g. 2x3 + 7x2 − 4x + 3. Sometimes, the terms can be arranged in ascending powers of the variable, especially when the leading coefficient is negative, e.g. 2 + 3x − x4.



INF



OR



MA



TIO N



A zero polynomial is denoted by 0 (i.e. all the coefficients are 0) and has no degree.



AT



TE



NTI



ON



The term with the highest power is called the leading term and its coefficient is called the leading coefficient.



Degree of a Polynomial If an ≠ 0, then the highest power of x in the polynomial is n and it is called the degree of the polynomial. For example, x3 + 5 is a polynomial of degree 3. What is the degree of each polynomial in the investigation on the previous page? In particular, state the degree of a (i) linear expression, (ii) quadratic expression, (iii) non-zero constant polynomial.



Thinking Time Can any or all of the coefficients an, an − 1, an − 2, ... a2, a1, a0 of a polynomial be equal to 0? Explain your answer.



Evaluating Polynomials We can also denote a polynomial in x by P(x), Q(x), g(x) and so on. To find the value of P(x) = 2x5 + 7x4 when x = 3, we substitute x = 3 into P(x): P(3) = 2(3)5 + 7(3)4 = 1053. What is the value of the polynomial function P(x) = 2x5 + 7x4 when x = −2? For simplicity, we will just call P(x) = 2x5 + 7x4 a ‘polynomial’.



Polynomials



Chapter 2



026



Addition and Subtraction of Polynomials To find the sum and the difference of polynomials, we add and subtract like terms.



Worked Example



1



(Addition and Subtraction of Polynomials) If P(x) = 2x3 + x2 − 4x + 5 and Q(x) = x2 − 6x + 5, find an expression for (ii) P(x) − Q(x). (i) P(x) + Q(x), State the degree of each expression.



Solution:



(i) P(x) + Q(x) = (2x3 + x2 − 4x + 5) + (x2 − 6x + 5) = 2x3 + 2x2 − 10x + 10



Degree of P(x) + Q(x) = 3



(ii) P(x) − Q(x) = (2x3 + x2 − 4x + 5) − (x2 − 6x + 5) = 2x3 + x2 − 4x + 5 − x2 + 6x − 5 = 2x3 + 2x



TE



NTI



ON



The sum or the difference of polynomials is a polynomial.



RE



CAL



L



−(a + b − c) = −a − b + c



Degree of P(x) − Q(x) = 3



PRACTICE NOW 1



If P(x) = 4x3 + 3x2 − 5x + 5 and Q(x) = 4x3 − 2x + 6, find an expression for (ii) P(x) − Q(x). (i) P(x) + Q(x), State the degree of each expression.



Thinking Time P(x) and Q(x) are polynomials. 1. How is the degree of P(x) + Q(x) related to the degree of P(x) and of Q(x)? 2. Can the degree of P(x) − Q(x) be less than the degrees of both P(x) and Q(x)? Explain your answer.



027



AT



Chapter 2



Polynomials



SIMILAR QUESTIONS



Exercise 2A Questions 1(a)-(d), 2(a)-(d)



Multiplication of Polynomials We have learned how to find the product of algebraic expressions using the distributive law.



Worked Example



2



(Multiplication of Polynomials) If P(x) = 2x4 − 4x and Q(x) = x3 + 3x2 − 5, find (ii) P(2) × Q(2), (i) P(x) × Q(x), (iii) the relationship between the degrees of P(x), Q(x) and P(x) × Q(x).



Solution: AT



(i) P(x) × Q(x) = (2x − 4x) (x + 3x − 5) = 2x7 + 6x6 − 10x4 − 4x4 − 12x3 + 20x = 2x7 + 6x6 − 14x4 − 12x3 + 20x (ii) P(2) × Q(2) = 2(2)7 + 6(2)6 − 14(2)4 − 12(2)3 + 20(2) = 360 (iii) Degree of P(x) = 4 Degree of Q(x) = 3 Degree of P(x) × Q(x) = 7 ∴ Degree of P(x) × Q(x) = Degree of P(x) + Degree of Q(x) 4



3



2



PRACTICE NOW 2



If P(x) = 5x3 + 2x − 3 and Q(x) = x2 − 3x + 6, find (ii) P(−1) × Q(−1), (i) P(x) × Q(x), (iii) the relationship between the degrees of P(x), Q(x) and P(x) × Q(x).



TE



NTI



ON



The product of polynomials is a polynomial.



SIMILAR QUESTIONS



Exercise 2A Questions 3(a)-(e), 4(a)-(e)



In general, if P(x) and Q(x) are non-zero polynomials, then Degree of P(x) × Q(x) = Degree of P(x) + Degree of Q(x)



Equality of Polynomials Two polynomials P(x) and Q(x) are equal if and only if the coefficients of the terms with same powers of x are equal. Given that Ax3 + Bx2 + Cx + D = 2x3 + 3x2 − x + 4 for all values of x, where A = 2 and B = 3, what is the value of C and of D?



Polynomials



Chapter 2



028



3



Worked Example



(Finding Unknown Coefficients) Given that 5x2 – 7x + 3 = A(x – 1)(x – 2) + B(x – 1) + C for all values of x, find the values of A, B and C.



Solution:



Method 1: Substitute suitable values of x



Since 5x2 – 7x + 3 = A(x – 1)(x – 2) + B(x – 1) + C for all values of x, we can select any value of x to find the unknowns A, B and C.



Let x = 1. 5(1)2 – 7(1) + 3 = A(1 – 1)(1 – 2) + B(1 – 1) + C C = 1 Let x = 2. 5(2)2 – 7(2) + 3 = A(2 – 1)(2 – 2) + B(2 – 1) + 1 B = 8 Let x = 0. 3 = A(0 – 1)(0 – 2) + 8(0 – 1) + 1 2A – 8 + 1 = 3 A = 5 ∴ A = 5, B = 8, C = 1 Method 2: Equate coefficients 5x2 – 7x + 3 = A(x2 – 3x + 2) + B(x – 1) + C = Ax2 + (−3A + B)x + 2A – B + C Equating coefficients of x2: 5 = A Equating coefficients of x: −7 = −3A + B −7 = −3(5) + B −7 = −15 + B B=8 Equating coefficients of x0: 3 = 2A – B + C 3 = 2(5) – 8 + C 3 = 10 – 8 + C C=1



∴ A = 5, B = 8, C = 1 PRACTICE NOW 3



SIMILAR QUESTIONS



If 3x2 + 2x − 9 = A(x – 2)(x + 1) + B(x – 2) + C for all values of x, find the values of A, B and C using both of the above methods.



Exercise 2A Questions 5(a)-(c), 6



Class Discussion Finding Unknown Coefficients



For Worked Example 3 and Practice Now 3, discuss with your classmates which method you prefer, stating your reason(s) clearly.



029



Chapter 2



Polynomials



Worked Example



4



(Equating Coefficients) If 2x3 − 11x − 6 = (x + 2)(ax2 + bx + c), find the values of a, b and c.



Solution:



2x3 − 11x − 6 = (x + 2)(ax2 + bx + c) By observation, a = 2 (because only x(ax)2 will give the term in x3) c = −3 (because only 2(c) will give the constant term)



P So roblem lvin g T ip



If we substitute x = −2, we will get 0 on both sides. If we substitute x = 0, we can obtain c, but c and a can be found more easily by observation. To find b, we can equate the coefficients of x2 or of x.



∴ 2x3 − 11x − 6 = (x + 2)(2x2 + bx − 3) Equating coefficients of x2: 0 = b + 4 (0x2 on LHS) b = −4



∴ a = 2, b = − 4, c = −3



SIMILAR QUESTIONS



PRACTICE NOW 4



1. For all values of x, 3x3 + 4x2 – 17x – 11 = (Ax + 1)(x + B)(x – 2) + C. Find the values of A, B and C.



Exercise 2A Questions 7(a)-(d), 11



2. For all values of x, 6x3 + 7x2 – x + 5 = (Ax + B)(2x + 1)(x – 1) + C. Find the values of A, B and C.



Division of Polynomials Before we learn how to divide a polynomial by another polynomial, let us recall the long division of numbers, e.g. divide 7 by 3: 2 3) 7 −6 1



divisor



quotient dividend remainder



We can express 7 in terms of the divisor, the quotient and the remainder as follows:



7



=



3



×



2



+



1



dividend = divisor × quotient + remainder This is called the Division Algorithm for Positive Integers.



Polynomials



Chapter 2



030



Class Discussion Division of Polynomials



Discuss with your classmates: When you divide one positive integer by another, (a) is the dividend always larger than or equal to the divisor? Explain your answer. (b) is the remainder always smaller than the divisor? Explain your answer.



Similarly, we can divide a polynomial by another polynomial using long division then express the relationship between the two polynomials using the corresponding division algorithm for polynomials.



Worked Example



5



(Long Division of Polynomials) Divide x3 + 2x − 7 by x − 2 and state the remainder.



Solution: divisor



x−2



x2 + 2x + 6



x 3 + 0x 2 + 2 x − 7



2 −(x3 − 2 x )



quotient dividend



2 x2 + 2 x 2 − (2 x − 4 x ) 6x − 7 − (6 x − 12) 5



remainder



∴ The remainder is 5.



PRACTICE NOW 5



Divide 4x3 + 3x2 − 16x − 12 by x + 2. Hence, express 4x3 + 3x2 − 16x − 12 in terms of x + 2.



031



Chapter 2



Polynomials



SIMILAR QUESTIONS



Exercise 2A Questions 8(a)-(e), 9



In general, if a polynomial P(x) is divided by another polynomial D(x), the resulting quotient is Q(x) and the remainder is R(x), then the Division Algorithm for Polynomials states that P(x) = D(x) × Q(x) + R(x) dividend = divisor × quotient + remainder and the degree of R(x) < the degree of D(x).



We can also perform synthetic division on polynomials using a linear or quadratic expression. The leading coefficient (first number) of a divisor must be a 1. For example, we can use synthetic division to divide by x + 3 or x – 6. If the leading coefficient of a divisor is not a 1, then we must divide by the leading coefficient to turn the leading 1 coefficient into a 1. For example, 3x – 1 would become x − and 2x + 7 would 3 7 become x + . 2 The following example shows the steps required for synthetic division of a polynomial using a linear expression.



Worked Example



6



(Synthetic Division of Polynomials Using Linear Expression) Divide 4x3 – 8x2 – x + 5 by 2x – 1 using synthetic division and state the remainder.



Solution:



Step 1: Rewrite 2x – 1 as x –



1 . 2



Then set the divisor equal to zero to find the number to put in the division box, 1 i.e. x – =0 2 1 x= 2



Next, make sure the dividend is written in descending order and if any terms are missing you must use a zero to fill in the missing term, finally list only the coefficient in the division problem.



So, 4x3 – 8x2 – 1x + 5 divide by x – 1 2



4



–8



1 can be written as 2



–1



5



Polynomials



Chapter 2



032



Step 2: Bring the leading coefficient straight down. 1 2







4



–8



–1



5



4 Step 3: Multiply the number in the division box with the number you brought



down and put the result in the next column. Add the two numbers together







and write the result in the bottom of the row.



1 4 2 ⎛1 ⎞ 4⎜ ⎟ ⎝2 ⎠ 4



–8



–1



5



2



–6



Step 4: Repeat step 3 until you reach the end of the problem. 1 2







4



–8 2



4



⎛1 ⎞ –6 ⎜ ⎟ ⎝2 ⎠



–6



–1 –3



⎛1 ⎞ 4⎜ ⎟ ⎝2 ⎠



–4



5 –2



3



remainder P So roblem lvin g T ip



Step 5: Write the final answer. 4x2 – 6x – 4 +







3



The final answer is written as the coefficients of the variables with one power less than the original dividend.



1 x– 2



Quotient Remainder Divisor



∴ 4x2 – 6x – 4 +



3 x–



1 2



and the remainder is 3. SIMILAR QUESTIONS



PRACTICE NOW 6



Divide 4x3 + 7x − 26 by 2x – 3 using synthetic division and state the remainder. Synthetic division of polynomials by a quadratic expression can be done in a similar way. The quadratic expression is factorized before the division is performed.



Worked Example



7



(Synthetic Division of Polynomials Using Quadratic Expression) Divide 3x3 – 2x2 – 7x + 6 by x2 + 3x + 2 using synthetic division and state the remainder.



Solution:



Step 1: Factorize the quadratic expression. x2 + 3x + 2 = (x + 1)(x + 2)



033



Chapter 2



Polynomials



Exercise 2A Questions 10(a)-(c)



Step 2: Perform the synthetic division of the polynomial using one of the factors. Divide 3x3 – 2x2 – 7x + 6 by x + 1 first to obtain the first remainder.







–1



3



–2



–7



6



–1



3



–2



–7



6



3(–1)



3



∴ 3x2 – 5x – 2 +



–3



–5(–1)



–5











–2(–1)



–2



2



8



remainder



8 x +1



Step 3: Divide the expression 3x2 – 5x – 2 + x + 2, i.e.



5



8 using the second factor, x +1



3x 2 – 5x – 2 8 + . x+2 (x + 1)(x + 2)



Perform synthetic division on –2



3



3x 2 – 5x – 2 . x+2



–5 3(–2)



3



–6



– 11



–2 –11(–2)



22



20



remainder



Step 4: Write the final answer. 8 20 + 3x – 11 + (x + 1)(x + 2) x+2



20(x + 1) + 8 (Combining the remainders from Steps 2 and 4) (x + 1)(x + 2) 20x + 28 = 3x – 11 + (x + 1)(x + 2)



= 3x – 11 +







∴ 3x – 11 +



20x + 28 and the remainder is 20x + 28. (x + 1)(x + 2)



PRACTICE NOW 7



Divide 3x3 – 5x – 1 by x2 – x – 2 using synthetic division and state the remainder.



SIMILAR QUESTIONS



Exercise 2A Questions 10(d)-(f)



Thinking Time When we divide a polynomial by another polynomial, (a) the degree of the dividend must be greater than or equal to the degree of the divisor. Why is this so? (b) the degree of the remainder must be smaller than the degree of the divisor. Why is this so? (c) what is the relationship between the degrees of the dividend, divisor and quotient?



Polynomials



Chapter 2



034



Exercise



2A



BASIC LEVEL



1. Find the sum of each of the following polynomials. (a) x3 + 5x2 + 2, 3x2 – 4x + 7 (b) 2x + 5x3 – 2x2 + 7, 9 + 4x – 5x2 – 2x3 (c) 5x3 + 2x2 + 4, 2x2 + 7x – 5, 7 – 5x2 + 3x – 4x3 (d) 2x4 – 3x2 + 4x3 – 5, 9x2 – 7x3 – 2x4, 7 – x + 3x4 2. Subtract (a) 3x2 – x – 1 from 5x2 + 3x – 7, (b) 2x2 – 7x + 5 from 5 + 8x – 2x3, (c) 4x3 + 2x – 9 from 11 – 3x – 4x2 + 5x3, (d) 3x2 + 5x4 – 6x + 7 from 8x + 5x3 + x5 – 2x4. 3. Find each of the following products. (a) (2x + 3)(7x – 4) (b) (5x + 7)(6x – 1) (c) (2x – 5)(2x2 + 5x – 1) (d) (x + 3y)(2x2 + xy + y) (e) (3x + 5y)(3x2 – 2xy + y2) INTERMEDIATE LEVEL



4. Simplify each of the following. (a) (2x – 1)(3x + 5) + (x + 5)(3x – 7) (b) (5x – 9)(x + 4) – (2x – 3)(7 – x) (c) (x + 3)(x2 – 5x + 2) + (3x2 – 6x + 7)(2x – 1) (d) (x + y)(x2 – xy + y2) – (2x – y)(x2 + 5xy – 3y2) (e) (x + 3)(x3 – 2x + 4) – (2x – 7)(3 + 4x – 2x3) 5. Find the value of A and of B in each of the following. (a) A(3x – 2) + B(5 – x) = 11x – 16 (b) A(x – 3)(x + 1) + B(x – 3) = 7x2 – 17x – 12 (c) 3x3 + 5x2 + 13x + 1 = (3x – 1)(x2 + 2x + A) + B 6. Find the values of A, B and C, given that 3x2 + 5x + C = A(x + 1)2 + B(x + 1) + 4 for all values of x.



035



Chapter 2



Polynomials



7. Find the values of A, B and C in each of the following. (a) 4x2 + 3x – 7 = A(x – 1)(x + 3) + B(x – 1) + C (b) 2x3 + 6x2 – 2x + 8 = (Ax + 2)(x – 1)(x – B) + C (c) 6x3 – 11x2 + 6x + 5 = (Ax – 1)(Bx – 1)(x – 1) + C (d) 2x4 – 13x3 + 19x2 + 5x + 1 = (x – 4)(Ax + 1)(x2 + Bx + 1) + C 8. Find the quotient and remainder for each of the following using long division. (a) (5x2 + 7x – 8) ÷ (x – 1) (b) (–4x2 + 7x + 160) ÷ (7 – x) (c) (3x3 + 4x – 5) ÷ (x + 1) (d) (x3 + 7x2 – 4x) ÷ (x + 3) (e) (5x3 + 11x2 – 5x + 1) ÷ (x + 2) 9. Divide –x3 + 7x2 – 4x – 12 by 2x – 1. Hence, express –x3 + 7x2 – 4x – 12 in terms of 2x – 1. 10. Find the quotient and remainder for each of the following using synthetic division. (a) (9x3 + 6x2 + 4x + 5) ÷ (3x + 1) (b) (12x3 + 12x2 + 7x + 1) ÷ (2x + 1) (c) (3x3 – 7) ÷ (x – 3) (d) (x3 + 4x2 + 3x + 7) ÷ (x2 + 2x + 1) (e) (2x3 + 5x2 + 9) ÷ (x2 + 3x − 4) (f) (5x4 + 2x2 + 7x − 4) ÷ (x2 − 4x + 3) ADVANCED LEVEL



11. The expression (ax + b)(x − 1) + c(x2 + 5) is equal to 18 for all real values of x. By substituting values of x or otherwise, find the values of a, b and c.



2.2



Remainder Theorem



In the previous section, we have learned how to use long division and synthetic division to find the remainder when a polynomial is divided by another polynomial. We will now learn another method to find the remainder when a polynomial is divided by a linear divisor.



Investigation Remainder Theorem



1. Copy and complete the following table. The remainder R can be found by



long division. For (a), the long division has been done in Worked Example 5. No. Dividend P(x) (a)



x + 2x − 7



(b)



4x + 7x + 15



3



x−2 x+1



3



(c) 3 − 8x + 4x + x 3



(d)



2x − 4x + 2 3



2



Linear Divisor x − c Remainder R P(c)



4







P(2) =







P(−1) =



x+3 x−1



2. What is the degree of the remainder? Why? 3. What is the relationship between the remainder and the value of P(c)? 4. What can you say about the divisor when the remainder is zero?



From the investigation, we observe that if a polynomial P(x) is divided by a linear divisor x − c, the remainder is P(c). By considering the Division Algorithm for Polynomials, how do we prove the above result?



Polynomials



Chapter 2



036



Thinking Time When a polynomial P(x) is divided by a linear factor x − c, then Q(x) and R(x) represent the quotient and remainder respectively. Using the Division Algorithm for Polynomials, express P(x) in terms of x − c. 1. State the degree of the polynomial R(x). To find R(x), we need to eliminate Q(x). What value of x can be used to eliminate the term involving Q(x)? Hence, evaluate the remainder. 2. If x − c is a factor of P(x), what can you say about the remainder? Using your result in Question 1, state the value of P(c).



In general, the Remainder Theorem states that if a polynomial P(x) is divided by a linear divisor ax + b, ⎛ b⎞ the remainder is P ⎜− ⎟ . ⎝ a⎠







Worked Example



8



(Using the Remainder Theorem to find the Remainder) Find the remainder when 4x3 − 8x2 + 9x − 5 is divided by (a) x + 2, (b) 2x − 3.



Solution:



(a) R = P(−2) = 4(−2)3 − 8(−2)2 + 9(−2) − 5 = −87 ⎛ 3⎞ (b) R = P ⎜ ⎟ ⎝2⎠



⎛ 3 ⎞2 ⎛ 3 ⎞2 ⎛ 3⎞ = 4⎜ ⎟ – 8⎜ ⎟ + 9⎜ ⎟ – 5 ⎝2⎠ ⎝2⎠ ⎝2⎠ = 4



PRACTICE NOW 8



Find the remainder when 3x3 – 4x2 – 5 is divided by (a) x – 2, (b) x + 5, (c) 2x – 3.



037



Chapter 2



Polynomials



SIMILAR QUESTIONS



Exercise 2B Questions 1(a)-(f)



Worked Example



9



(Finding an Unknown using the Remainder Theorem) Find the value of k if 4x7 + 5x3 – 2kx2 + 7k – 4 has a remainder of 12 when divided by x + 1.



Solution:



Let f (x) = 4x7 + 5x3 – 2kx2 + 7k – 4. Using the Remainder Theorem, f (–1) = 12. i.e. 4(–1)7 + 5(–1)3 – 2k(–1)2 + 7k – 4 = 12 –4 – 5 – 2k + 7k – 4 = 12 5k = 25 k = 5



PRACTICE NOW 9



SIMILAR QUESTIONS



1. Find the value of k if 5x5 + 2kx3 – 6kx2 + 9 has a remainder of 22 when divided by x – 1.



Exercise 2B Questions 2-10



2. The expression 4x3 + Ax2 + Bx + 3 leaves a remainder of 9 when divided by x – 1 and a remainder of −27 when divided by x + 3. Find the value of A and of B.



Exercise



2B



BASIC LEVEL



1. Find the remainder of each of the following. (a) x3 – 5x2 + 3x + 7 is divided by x – 1 (b) −2x3 − 7x2 + 6x − 9 is divided by x + 1 (c) x3 + 2x – 5 is divided by x + 3 (d) −5x2 + 3x − 6 is divided by 2x + 1 (e) 5x3 + 4x2 – 6x + 7 is divided by x – 2 (f ) 2x3 + 4x2 – 6x + 7 is divided by 2x – 1 2. Find the value of k in each of the following. (a) x3 – 3x2 + 8kx + 5 has a remainder of 17 when divided by x – 3. (b) x3 + x2 – kx + 4 has a remainder of 8 when divided by x – 2. (c) 7x9 + kx8 + 3x5 + 5x4 + 7 has a remainder of –3 when divided by x + 1. (d) 5x3 – 4x2 + (k + 1)x – 5k has a remainder of –2 when divided by x + 2.



3. The remainder obtained when 5x3 – 6x2 + kx – 3 is divided by x – 1 is equal to the remainder obtained when the same expression is divided by x – 2. Find the value of k. INTERMEDIATE LEVEL



4.



The expression 2x 3 + hx 2 – 6x + 1 leaves a remainder of 2k when divided by x + 2 and a remainder of k when the same expression is divided by x – 1. Find the value of h and of k.



5. Given that y = 3x3 + 7x2 – 48x + 49 and that y has the same remainder when it is divided by x + k or x – k, find the possible values of k. 6. When x6 + 5x3 – px – q and px2 – qx – 1 are each divided by x + 1, the remainders obtained are 7 and –6 respectively. Find the value of p and of q.



Polynomials



Chapter 2



038



7. When the expression 2x4 + px3 + 5x2 + 7 is divided by (x + 1)(x – 2), the remainder is qx + 18. This result may be expressed as 2x4 + px3 + 5x2 + 7 = (x + 1)(x – 2)Q(x) + (qx + 18), where Q(x) is the quotient. By substituting suitable values of x, find the value of p and of q. 8. Find the value of k if (3x + k)3 + (4x – 7)2 has a remainder of 33 when divided by x – 3.



2.3



ADVANCED LEVEL



9. When the expression 3x3 + px2 + qx + 8 is divided by x2 – 3x + 2, the remainder is 5x + 6. Find the value of p and of q. 10. When x2 + ax + b and x2 + hx + k are divided by x + p, their remainders are equal. Express p in terms of a, b, h and k.



Factor Theorem



From the previous section, we have learned that if a polynomial P(x) is divided by a linear divisor x − c, then the Division Algorithm for Polynomials states that P(x) = (x − c)Q(x) + R and the Remainder Theorem states that the remainder R = P(c). What if the remainder R = 0? Then P(c) = 0 and P(x) = (x − c)Q(x), i.e. x − c is a factor of P(x). In other words, x – c is a linear factor of the polynomial P(x) if and only if P(c) = 0. In general, the Factor Theorem states that ax + b is a linear factor of the polynomial P(x) ⎛ b⎞ if and only if P ⎜ – ⎟ = 0. ⎝ a⎠



The Factor Theorem is a special case of the Remainder Theorem when the remainder = 0.



Worked Example



039



Chapter 2



10 Polynomials



(Finding an Unknown using the Factor Theorem) Find the value of k for which x – 2 is a factor of f (x) = 3x3 – 2x2 + 5x + k. Hence, find the remainder when f (x) is divided by 2x + 3.



Solution:



f (x) = 3x3 – 2x2 + 5x + k x – 2 is a factor of f (x), so f (2) = 0. 3(2)3 – 2(2)2 + 5(2) + k = 0 k = –26



∴ f (x) = 3x3 – 2x2 + 5x – 26 ⎛ 3⎞ When f (x) is divided by 2x + 3, remainder R = f ⎜ – ⎟ . ⎝ 2⎠ ⎛ 3 ⎞3 ⎛ 3 ⎞2 ⎛ 3⎞ R = 3⎜ – ⎟ – 2 ⎜ – ⎟ + 5 ⎜ – ⎟ – 26 ⎝ 2⎠ ⎝ 2⎠ ⎝ 2⎠ 1 = −– 48 8



SIMILAR QUESTIONS



PRACTICE NOW 1 0



Find the value of k for which x – 3 is a factor of f(x) = 4x3 + 6x2 – 9x + 2k. Hence, find the remainder when f(x) is divided by x + 2.



Worked Example



11



Exercise 2C Questions 1-5



(Finding an Unknown using the Factor Theorem) Given that 3x2 − x − 2 is a factor of the polynomial 3x3 + hx2 − 5x + k, find the value of h and of k. Hence, factorize the polynomial completely.



Solution:



Let f(x) = 3x3 + hx2 − 5x + k. Since 3x2 − x − 2 = (3x + 2)(x − 1) is a factor of f(x), then 3x + 2 and x − 1 are also factors of f(x). ⎛ 2⎞ By the Factor Theorem, f ⎜ – ⎟ = 0 and f(1) = 0. ⎝ 3⎠ 3 2 ⎛ 2⎞ ⎛ 2⎞ ⎛ 2⎞ ⎛ 2⎞ f ⎜ – ⎟ = 3⎜ – ⎟ + h ⎜ – ⎟ – 5 ⎜ – ⎟ + k = 0 ⎝ 3⎠ ⎝ 3⎠ ⎝ 3⎠ ⎝ 3⎠ 8 4 10 − + h+ +k=0 9 9 3 −8 + 4h + 30 + 9k = 0 4h + 9k = −22 --- (1) f(1) = 3(1)3 + h(1)2 – 5(1) + k = 0 3 + h – 5 + k = 0 h + k = 2 --- (2) (2) × 4: 4h + 4k = 8 --- (3) (1) – (3): 5k = −30 k = −6 Substitute k = −6 into (2): h – 6 = 2 h=8 ∴ h = 8 and k = −6 Hence, f(x) = 3x3 + 8x2 − 5x − 6 = (3x2 − x − 2)(ax + b) By observation, a = 1 and b = 3. ∴ 3x3 + 8x2 − 5x − 6 = (3x2 − x − 2)(x + 3) = (3x + 2)(x − 1)(x + 3)



AT



TE



NTI



ON



Since 3x2 − x − 2 is a factor of f(x), then f(x) = (3x2 − x − 2)Q(x) = (3x + 2)(x − 1)Q(x). This shows that 3x + 2 and x − 1 are also factors of f(x).



RE



CAL



L



Refer to Worked Example 4 on how to find a and b by observation. You can also use long division.



Polynomials



Chapter 2



040



PRACTICE NOW 1 1



SIMILAR QUESTIONS



If x2 – 3x + 2 is a factor of f(x) = 2x4 + px3 + x2 + qx – 12, find the value of p and of q. Hence, factorize f(x) completely.



Exercise 2C Questions 6-10



Exercise



2C



BASIC LEVEL



1. Find the value of k in each of the following. (a) x – 1 is a factor of 3x3 + 7x2 – kx + 5 (b) x + 2 is a factor of 2x3 + kx2 – 8x – 18 (c) x + 4 is a factor of x3 + 5x2 – kx + 48 (d) 2x – 1 is a factor of 4x4 + 3x3 – 7x2 + 14x + k (e) x – 1 is a factor of k2x4 – 3kx2 + 2 (f ) x + 2 is a factor of (x + 1)5 + (5x + k)3 2. Given that x – 1 is a factor of the expression x 3 – kx + 2, find the value of k and the remainder when the expression is divided by x – 2. INTERMEDIATE LEVEL



3. Given that x2 – x – 2 and x3 + kx2 – 10x + 6 have a common factor, find the possible values of k. 4. Find the value of p and of q for which x – 3 is a common factor of the expressions x2 + ( p + q)x – q and 2x2 + ( p – 1)x + ( p + 2q). 5. Find the value of p and of q for which x + 1 and 2x – 1 are factors of 4x4 + 2x3 + (q – 1)x – q – p. Hence, explain why x – 3 is not a factor of the expression.



041



Chapter 2



Polynomials



6. Given that x3 + 3x2 + hx + k, where h and k are constants, is exactly divisible by x – 2 and leaves a remainder of 30 when divided by x + 1, find the value of h and of k. Hence, factorize the expression completely. 7. If x – k is a factor of the expression kx3 + 5x2 – 7kx – 8, where k is a positive integer, find the value of k. Hence, find the other factors of the expression. 8. If x2 – 2x – 3 is a factor of the expression x4 + px3 + qx – 81, find the value of p and of q. Hence, factorize the expression completely. 9. It is given that x + 2 is a factor of f (x) = a(x – 1)2 + b(x – 1) + c. The remainders when f (x) is divided by x + 1 and x – 1 are –11 and 9 respectively. Find the values of a, b and c. ADVANCED LEVEL



10. Given that x + 1 and x – 3 are factors of the expression x4 + px3 + 5x2 + 5x + q, find the value of p and of q. Hence, find the other two factors of the expression.



2.4



Cubic Expressions and Equations



Cubic Expressions Worked Example



12



(Factorization of Cubic Expressions) Factorize x3 – x2 – 4x + 4 completely.



Solution:



Let f (x) = x3 – x2 – 4x + 4. The positive and negative factors of 4 are ±1, ±2 and ±4. Try x – 1, f (1) = 1 – 1 – 4 + 4 = 0. ∴ x – 1 is a factor of f (x). The other factors can be found by any of the 3 methods shown below. Method 1: Trial and error Try x + 2, f (–2) = –8 – 4 + 8 + 4 = 0. ∴ x + 2 is a factor of f (x). Try x – 2, f (2) = 8 – 4 – 8 + 4 = 0. ∴ x – 2 is a factor of f (x).



Method 2: Long division x2 −4 3 2 x −1 x − x − 4x + 4 −( x 3 − x2 ) − 4x + 4 − (− 4 x + 4 ) 0



)



∴ f (x) = (x – 1)(x2 – 4) ∴ f (x) = (x – 1)(x + 2)(x – 2) = (x – 1)(x + 2)(x – 2) Method 3: Comparing coefficients Since f(x) is of degree 3 and x – 1 is one of the factors, the other factor must be of degree 2 and in the form ax2 + bx + c. x3 – x2 – 4x + 4 = (x – 1)(ax2 + bx + c) = ax3 + (b – a)x2 + (c – b)x – c By observation, a = 1 and c = –4. Equating the coefficients of x : c – b = –4 b=0 ∴ f(x) = (x – 1)(x2 – 4) = (x – 1)(x + 2)(x – 2)



PRACTICE NOW 1 2



1. Factorize x3 + 2x2 – 29x + 42 completely.



SIMILAR QUESTIONS



Exercise 2D Questions 1(a)-(g)



2. Factorize 2x + 9x + 7x − 6 completely. 3



2



Polynomials



Chapter 2



042



Investigation Number of Real Roots of a Cubic Equation



Use a suitable graphing software to plot the graph of y = x3 – 5x2 + 2x + 8. (i) What are the x-coordinates of the points of intersection of the graph of y = x3 – 5x2 + 2x + 8 and the x-axis? (ii) Solve the equation x3 – 5x2 + 2x + 8 = 0. (iii) What is the relationship between the solutions of a cubic equation and the x-coordinates of the points of intersection of the curve with the x-axis? (iv) What are the linear factors of the polynomial? Hence, factorize the polynomial completely. Check your answer using methods you have learned earlier. Repeat the above by drawing the graphs of y = x3 + 4x2 – 3x – 18, y = x3 + 4x2 – 23x + 6, y = x3 – 8x2 + 13x – 6 and y = x3 – x2 + 4x – 12. Using the graph, can you state a linear factor of each cubic polynomial? What can you conclude about the number of real roots of a cubic polynomial and the number of points of intersection of the graph with the x-axis?



Cubic Equations To solve the quadratic equation ax2 + bx + c = 0, we can use the formula x=



−b ± b 2 − 4 ac . 2a



We can also solve the quadratic equation using Rational Root Theorem. In this section, we will learn how to find possible roots given a polynomial equation using Rational Root Theorem. Given a polynomial function with integer or whole number coefficients, a list of possible roots can be determined by listing the factors of the constant over the factors of the coefficient of the leading term.



Worked Example



13



(Solving a Quadratic Equation using Rational Root Theorem) List the possible roots for 4x2 + 3x − 1 = 0.



Solution:



Factors of −1: ±1



Factors of 4: ±1, ±2, ±4 Using Rational Root Theorem, 1 1 factors of −1 possible rational roots = = ±1, ± , ± 2 4 factors of 4



043



Chapter 2



Polynomials



P So roblem lvin g T ip



Any rational root must have a numerator that is a factor of −1 and a denominator that is a factor of 4.



PRACTICE NOW 1 3



SIMILAR QUESTIONS



List the possible roots for 2x2 + 17x − 20 = 0. From Worked Example 13, the possible rational roots are ±1, ±



Exercise 2D Questions 1(a)-(c)



1 1 , ± . However, not 2 4



all possible rational roots are roots of the equation. For example, if we substitute −1 and +1 into the equation: When x = −1,



4(−1)2 + 3(−1) − 1 = 0 When x = 1,



4(1)2 + 3(1) − 1 = 6



Therefore, we can see that x + 1 is a factor of the equation whereas x − 1 gives a remainder of 6.



Worked Example



14



(Solving a Cubic Equation using Rational Root Theorem) Given the equation x3 + 3x2 − 10x − 24 = 0, (a) list the possible roots using the Rational Root Theorem. (b) Hence, determine the roots to the equation.



Solution:



(a) Using Rational Root Theorem, factors of − 24 = ±1, ±2, ±3, ±4, ±6, ±8, ±12 and ±24 possible roots = factors of 1 (b) When x = −2,



(−2)3 + 3(−2)2 − 10(−2) − 24 = 0



When x = 3,



INF



OR



MA



TIO N



Since the roots of the equation are −2, 3 and −4, the factors of the equation are (x + 2), (x − 3) and (x + 4), i.e. (x + 2)(x − 3)(x + 4) = 0.



(3)3 + 3(3)2 − 10(3) − 24 = 0 When x = −4,







(−4)3 + 3(−4)2 − 10(−4) − 24 = 0







∴ The roots of the equation are −2, 3 and −4.



PRACTICE NOW 1 4



Given the equation 6x3 + 8x2 − 7x − 3 = 0,



SIMILAR QUESTIONS



Exercise 2D Questions 1(d)-(g)



(a) list the possible roots using the Rational Root Theorem. (b) Hence, determine the roots to the equation.



Polynomials



Chapter 2



044



In this section, we will also learn how to solve cubic equations of the form ax3 + bx2 + cx + d = 0 (a ≠ 0), with the help of the Factor Theorem to find a linear factor first and by factorization subsequently.



Worked Example



15



(Solving a Cubic Equation using Factor Theorem) Solve the cubic equation 2x3 + 3x2 − 5x − 6 = 0.



Solution:



Let P(x) = 2x3 + 3x2 − 5x − 6. The positive and negative factors of −6 are ±1, ±2, ±3 and ±6. By trial and error, P(−1) = −2 + 3 + 5 − 6 = 0. By the Factor Theorem, x − (−1), i.e. x + 1, is a factor of P(x). ∴ P(x) = 2x3 + 3x2 − 5x − 6 = (x + 1)(ax2 + bx + c) By observation, a = 2 and c = −6. ∴ P(x) = 2x3 + 3x2 − 5x − 6 = (x + 1)(2x2 + bx − 6) Equating coefficients of x2: 3=b+2 b=1



∴ P(x) = 2x3 + 3x2 − 5x − 6 = (x + 1)(2x2 + x − 6) = (x + 1)(2x − 3)(x + 2) Hence, 2x3 + 3x2 − 5x − 6 = 0 (x + 1)(2x − 3)(x + 2) = 0 3 x = −2, −1 or 2



PRACTICE NOW 1 5



INF



OR



MA



TIO N



Since P(1) = 2 + 3 − 5 − 6 ≠ 0, so x − 1 is not a factor of P(x). By trial and error, you may have obtained P(−2) = 0, which implies that x + 2 is a factor of P(x). Then P(x) = (x + 2)(ax2 + bx + c).



RE



CAL



L



We have learned how to equate coefficients in Worked Example 6.



SIMILAR QUESTIONS



1. Solve the equation 2x3 + 11x2 – 7x – 6 = 0.



Exercise 2D Questions 2(a)-(e)



2. Solve the equation 3x + 4x – 49x + 30 = 0. 3



2



Class Discussion Long Division vs Factor Theorem



With reference to Worked Example 15, discuss with your classmates: 1. We can also find the values of a, b and c by dividing 2x3 + 3x2 − 5x − 6 by x + 1 using long division. Which method do you prefer? Explain your choice. 2. What value of x must you substitute into P(x) so that you can use the Factor Theorem to conclude that 2x − 3 is a factor? How is this value of x linked to the factors of the constant term −6 and the coefficient of x3? 3. Can you use the Factor Theorem to obtain all the factors of the cubic polynomial? Explain your answer.



045



Chapter 2



Polynomials



From the discussion, we observe that it may not be possible to find all the factors using the Factor Theorem if there is only one linear factor for the cubic polynomial, e.g. f(x) = 8x3 − 10x2 + x − 6 = (2x − 3)(4x2 + x + 2), where the second factor is quadratic and cannot be factorized into two linear factors. Moreover, to find the linear factor, it is not enough to substitute only the positive and negative factors of the ⎛ 3⎞ constant term −6 into f(x) because in this case f ⎜ ⎟ = 0 . ⎝2⎠ If the coefficient of x3 is not 1, then we have to consider the factors of the coefficient of x3 (i.e. 8) and the constant term (i.e. −6) to find the linear factor.



From the linear factor 2x – 3 given above, we observe that 3 • the numerator 3 of is a factor of −6, the constant term, 2 3 • the denominator 2 of is a factor of 8, the coefficient of x3. 2



From the above, how do you determine the values of x that can be substituted into f(x) = ax3 + bx2 + cx + d, so as to find a linear factor of f(x)?



Worked Example



16



(Solving a Cubic Equation) Solve the cubic equation 8x3 + 10x2 − 5x − 6 = 0.



Solution:



Let f(x) = 8x3 + 10x2 − 5x − 6. The positive and negative factors of −6 are ±1, ±2, ±3 and ±6 and those of 8 are ±1, ±2, ±4 and ±8. By trial and error, ⎛ 3⎞ ⎛ 3 ⎞3 ⎛ 3 ⎞2 ⎛ 3⎞ f ⎜ – ⎟ = 8 ⎜ – ⎟ + 10 ⎜ – ⎟ – 5 ⎜ – ⎟ – 6 = 0. ⎝ 4⎠ ⎝ 4⎠ ⎝ 4⎠ ⎝ 4⎠ 3 By the Factor Theorem, x + 4 , i.e. 4x + 3, is a factor of f(x). ∴ f(x) = 8x3 + 10x2 − 5x − 6 = (4x + 3)(ax2 + bx + c)



By observation, a = 2 and c = −2. ∴ f(x) = 8x3 + 10x2 − 5x − 6 = (4x + 3)(2x2 + bx − 2) Equating coefficients of x: −5 = 3b − 8 b=1 ∴ f(x) = 8x3 + 10x2 − 5x − 6 = (4x + 3)(2x2 + x − 2) Hence, 8x3 + 10x2 − 5x − 6 = 0



(4x + 3)(2x2 + x − 2) = 0



−1 ± 12 − 4 ( 2 )( −2 ) 3 or 2(2 ) 4 −1 ± 17 3 = − or 4 4 3 = − , −1.28 or 0.781 (to 3 s.f.) 4







x = −



Polynomials



Chapter 2



046



SIMILAR QUESTIONS



PRACTICE NOW 1 6



1. Solve the equation x3 + 4x2 – 23x + 6 = 0, giving your answers correct to 2 decimal places where necessary.



Exercise 2D Questions 4(a)-(f)



2. Solve the equation x3 + 5x2 – 19x + 10 = 0, giving your answers correct to 2 decimal places where necessary.



Worked Example



17



(Finding an Unknown Polynomial) The term containing the highest power of x in the



polynomial f(x) is 6x3. One of the roots of the 1 equation f(x) = 0 is 2 . Given that x2 − 2x + 3 is a quadratic factor of f(x), find (i) an expression for f(x) in ascending powers of x,



(ii) the number of real roots of the equation f(x) = 0.



Solution:



1 1 (i) Since 2 is a root of the equation f(x) = 0, by Factor Theorem, x − 2 , i.e. 2x − 1 is a factor of f(x).



Since 2x − 1 and x2 – 2x + 3 are factors of f(x), hence f(x) can be written as f(x) = a(2x – 1)(x2 – 2x + 3).



Since the term containing the highest power of x in f(x) is 6x3, by observation (or equating coefficients of x3), a = 3.



∴ f(x) = 3(2x − 1)(x2 − 2x + 3)







= 6x3 − 15x2 + 24x − 9



(ii) f(x) = 3(2x − 1)(x2 − 2x + 3) = 0



2x − 1 = 0



or



1 x = 2



or



x=



1 = 2



or



=







x2 − 2x + 3 = 0



∴ f(x) = 0 has only one real root.



− ( −2 ) ± ( −2 )2 − 4 (1)( 3) 2 (1) 2 ± −8 (no real solution) 2



PRACTICE NOW 1 7



1. The term containing the highest power of x in the polynomial f(x) is 6x4. Given that x2 – 5x + 1 is a quadratic factor of f(x) and two of the solutions of 1 f(x) = 0 are 2 and 2, find an expression for f(x) in descending powers of x. 2. It is given that x2 – 5x + 3 is a quadratic factor of f(x) and that two of the roots of the equation f(x) = 0 are x = −2 and x = 3. If the term containing the highest power of x in f(x) is 2x4, find an expression for f(x) in ascending powers of x.



047



Chapter 2



Polynomials



SIMILAR QUESTIONS



Exercise 2D Questions 5, 6



Thinking Time How many real roots can a cubic equation have? Can it have 2 real roots? Explain your reasons clearly. Hint: See Worked Examples 15-17 and the investigation earlier in this section 2.4.



Recall the following three algebraic identities: AT



(a + b) = a + 2ab + b (a − b)2 = a2 − 2ab + b2 a2 − b2 = (a + b)(a − b) 2



2



2







TE



NTI



ON



(a + b)2 ≠ a2 + b2 (a − b)2 ≠ a2 − b2 (a + b)3 ≠ a3 + b3 (a − b)3 ≠ a3 − b3



We have also learned that a2 + b2 cannot be factorized into two linear factors. In this section, we will learn how to factorize a3 + b3 and a3 − b3.



Investigation Factorization of a3 ± b3 1.



Factorize each of the following expressions completely. (i) a3 + a2b + ab2 (ii) a2b + ab2 + b3 By subtracting your answer in (ii) from that in (i), factorize a3 – b3.



2.



Factorize each of the following expressions completely. (i) a3 − a2b + ab2 (ii) a2b − ab2 + b3 By adding your answers in (i) and (ii), factorize a3 + b3.



Use the Factor Theorem to find the factors of a3 − b3 and a3 + b3. Do you obtain the same answers as above?



From the above investigation, we have two special algebraic identities: a3 + b3 = (a + b)(a2 − ab + b2)







Sum of Cubes:







Difference of Cubes: a3 − b3 = (a − b)(a2 + ab + b2)



Polynomials



Chapter 2



048



Thinking Time Using Factor Theorem, explain why (i) a + b is a factor of a3 + b3, (ii) a − b is a factor of a3 − b3.



Worked Example



18



(Factorization using the Sum and Difference of Two Cubes) Factorize each of the following expressions completely. (a) 8x3 + 125y3 (b) (3x + 1)3 – 8



Solution:



(a) 8x3 + 125y3 = (2x)3 + (5y)3 = (2x + 5y)[(2x)2 – (2x)(5y) + (5y)2] = (2x + 5y)(4x2 – 10xy + 25y2) (b) (3x + 1)3 – 8 = (3x + 1)3 – (2)3 = [(3x + 1) – 2][(3x + 1)2 + (3x + 1)(2) + 22] = (3x – 1)[9x2 + 6x + 1 + 6x + 2 + 4] = (3x – 1)(9x2 + 12x + 7) SIMILAR QUESTIONS



PRACTICE NOW 1 8



Factorize each of the following expressions completely. (a) 216p3 + 343q6 (b) 64 – (2x + 3)3



Applications of Cubic Equations The time we started to learn mathematics, we also started to learn its vocabulary. We have learned about the definitions of terms such as integer, quadrilateral, polynomial, etc.



Worked Example



19



(Van der Waals‘ Equation) The Van der Waals‘ equation is an equation that relates several properties of real gases, which may be written as v3 –



1 ⎛ 8T ⎞ 2 v 1 ⎜1 + ⎟v + v – = 0 , 3⎝ p ⎠ p p



where v, T and p are the volume, temperature and pressure of the gas respectively. At the critical temperature, i.e. T = 1 K and p = 1 atm, the equation becomes v3 – 3v2 + 3v – 1 = 0, where v is measured in dm3. Solve the equation v3 – 3v2 + 3v – 1 = 0 to find the value of v at the critical temperature. 049



Chapter 2



Polynomials



Exercise 2D Questions 3(a)-(e)



Solution:



Let f(v) = v3 – 3v2 + 3v – 1. The positive and negative factors of –1 are ±1. By trial and error, f(1) = (1)3 – 3(1)2 + 3(1) – 1 = 0. By the Factor Theorem, v – 1 is a factor of f(v). ∴ f(v) = v3 – 3v2 + 3v – 1 = (v – 1)(av2 + bv + c) By observation, a = 1 and c = 1. ∴ f(v) = (v – 1)(v2 + bv + 1)



Equating coefficients of v: 3 = –b + 1 b = −2



∴ f(v) = (v – 1)(v2 – 2v + 1) = (v – 1)3 Hence, v3 – 3v2 + 3v – 1 = 0 (v – 1)3 = 0 v=1 ∴ At the critical temperature, the volume is 1 dm3. PRACTICE NOW 1 9



SIMILAR QUESTIONS



The price of an object, P(x), is related to the supply quantity, x, by the polynomial P(x) = x3 – 4x2 + hx, where h is a constant and x > 3. Given that P(5) = 29, find the value of h.



Exercise 2D Question 7



Exercise



2D



BASIC LEVEL



1. List the possible roots for each of the following equations using the Rational Root Theorem. Hence, determine the roots to each equation.



(a) x2 – 5x + 6



(b) x2 – 11x + 24 (c) 3x2 + x – 30



(d) x3 – 9x2 + 26x – 24 (e) 2x3 + x2 – 13x + 6 (f) x3 – 6x2 – x + 30



(g) 12x3 – 8x2 – 3x + 2 2. Factorize each of the following.



3. Solve each of the following equations. (a) x3 + 2x2 – x – 2 = 0



(b) x3 – 3x2 – 4x + 12 = 0



(c) 4x3 + x2 – 27x + 18 = 0



(d) 3x3 – 16x2 – 37x + 14 = 0 (e) 5x3 – 26x2 + 35x – 6 = 0 INTERMEDIATE LEVEL



4. Solve each of the following equations, giving your answers correct to 2 decimal places where necessary. (a) x3 – 9x2 + 11x + 6 = 0 (b) x3 – 8x2 + 16x – 3 = 0



(a) x3 – 8



(c) x3 – 10x2 + 28x – 16 = 0



(c) 27x6 + 8y9



(e) 3x3 – 23x2 + 7x + 5 = 0



(e) 216 − (3x – 4)3



(g) x4 + 2x3 – 7x2 – 8x + 12 = 0



(b) 125x3 – 64y3



(d) 2x3 – 3x2 – 13x + 7 = 0



(d) 343a3 + 216x6



(f) 4x3 – 16x2 + 15x – 4 = 0



(h) 2x4 – 19x3 + 61x2 – 74x + 24 = 0 Polynomials



Chapter 2



050



5. The term containing the highest power of x in the polynomial f(x) is 3x4. Given that two of the roots of the equation f(x) = 0 are −2 and 4 and that x2 – 7x + 2 is a quadratic factor of f(x), find an expression for f(x) in descending powers of x. 6. It is given that 2x2 – 7x + 1 is a quadratic factor of f(x) and the term containing the highest power of x in f(x) is 4x4. If two of the roots of f(x) = 0 are − 0.5 and 3, find an expression for f(x) in ascending powers of x. 7. The sum of the radii of two spherical balls is 22 cm 2 and the difference of their volumes is 3050 cm3. 3 Find the radii of the balls. 22 (Take π to be .) 7 8. The expression x2n – k has x + 3 as a factor and leaves a remainder of −80 when divided by x + 1. Calculate the value of n and of k. With these values of n and k, factorize x2n – k completely.



2.5



ADVANCED LEVEL



9. Given that f (x) = x2n – ( p + 1)x2 + p, where n and p are positive integers, show that x – 1 is a factor of f (x) for all values of p. When p = 4, find the value of n for which x + 2 is a factor of f (x). Factorize f (x) completely. 8 10. Given that 81x4 – 6x3 – 9a2x2 – 11x – 3 9 is divisible by 3x + a, show that 2a3 + 33a – 35 = 0. Hence, solve the equation for all real values of a.



11. The data in the table lists the annual water consumption of each person in a city. Year Water used (m3)



2000 2002 2004 2006 2008 2010 57.6



59.8



60.0



60.6



64.0



72.6



(i) Using ICT, generate a scatter plot for the data provided. (ii) Suggest a polynomial function that may be used to model the data. (iii) It is believed that a polynomial of the form y = ax3 + bx2 + cx + d models the data. Hence, estimate the water consumption in 2012.



Partial Fractions



Algebraic Fraction (Recap) Three examples of algebraic fractions are x3 + 8 1 7 and ,− 2 . 5−x x+ x 4x − x + 6



The numerator and denominator of the first two algebraic fractions are polynomials while the denominator of the last algebraic fraction is not a polynomial. In this section, we will study only algebraic fractions that are ratios of two P(x) polynomials P(x) and Q(x), i.e. Q ( x ) , where Q(x) ≠ 0. We have learned about proper and improper fractions. 2 For example, is a proper fraction because the numerator < the denominator; 3 7 3 while and are improper fractions because the numerator > the 3 3 denominator. So how do we define a proper and an improper algebraic fraction? 051



Chapter 2



Polynomials



INF



OR



MA



TIO N



P( x ) , where Q( x ) P(x) and Q(x) are polynomials and Q(x) ≠ 0, is also called a rational expression.



An algebraic fraction



Investigation Proper and Improper Algebraic Fractions



The table below shows some examples of proper and improper algebraic fractions No.



Proper Algebraic Fractions



(a)



1 5−x



(b)



8−x x − 2x + 3



x4 − x3 + 2 x 9 − 5x



4 − 2 x + 3x 4 2 x5 + 7



x x+2



(c)



2



−



P(x) . Q( x )



Improper Algebraic Fractions −



x3 + 8 4 x2 − x + 6



Compare the degrees of P(x) and Q(x) in both the proper and improper algebraic fractions. (i) What do you call an algebraic fraction if the degree of P(x) < the degree of Q(x)?



(ii) What do you call an algebraic fraction if the degree of P(x) > the degree of Q(x)?



We have learned in Section 2.1 that we can use the Division Algorithm for Positive Integers to express an 7 improper fraction, such as 3 , as:



7 = 3 × 2 + 1 dividend = divisor × quotient + remainder



7 We can also divide the above equation by the divisor 3, so another way to express 3 is: 7 1 = 2+ 3 3 which is the sum of a positive integer and a proper fraction.



Similarly, we can use the Division Algorithm for Polynomials to express an improper algebraic fraction as the sum of a polynomial and a proper algebraic fraction.



Polynomials



Chapter 2



052



Thinking Time x3 + 3 (i) Is x − 1 a proper or an improper algebraic fraction?



(ii) When x3 + 3 is divided by x − 1, the quotient is x2 + x + 1 and the remainder is 4. Use the Division Algorithm for Polynomials to express x3 + 3 in terms of x − 1. x3 + 3 Hence write x − 1 as the sum of a polynomial and a proper fraction.







In general, another way to write the Division Algorithm is: Dividend Remainder = Quotient + Divisor Divisor



Worked Example



20



(Expressing an Improper Algebraic Fraction as the Sum of a Polynomial and a Proper Algebraic Fraction) Express



x3 + 2 x2 − 3 as the sum of a polynomial and x +1



a proper fraction.



Solution:



By long division, we have x2 + x − 1 x + 1 x3 + 2 x2 −3 3 2 −( x + x ) x2 + 0x −( x2 + x ) −x−3 − ( − x − 1) −2



)



(leave a space for the term in x) (write 0x in the space)



The expression can also be obtained by using synthetic division.



x3 + 2 x2 − 3 2 = x2 + x − 1 − x +1 x +1



∴



SIMILAR QUESTIONS



PRACTICE NOW 20



Express each of the following improper fractions as the sum of a polynomial and a proper fraction. (a)



053



P So roblem lvin g T ip



2 x3 + 7 x + 5 x−2



Chapter 2







Polynomials



6 x2 − 8 x + 9 (b) ( 3 x − 1)( x + 2 )



Exercise 2E Questions 1-8



Exercise



2E



BASIC LEVEL



1. Express each of the following improper fractions as the sum of a polynomial and a proper fraction.



8x + 3 2x − 1 14 x 2 + 3 (c) 2 x2 + 7



(a)



(b)



15 x 3 + 3 3x 2 − 1



6. Find the values of a, b, c and d for which 6 x 3 + 13 x 2 − 2 x − 31 cx + d = ax + b + 2 . 2 x +x−2 x +x−2



2. Express each of the following improper fractions as the sum of a polynomial and a proper fraction. 15 x 2 − 7 x + 3 (a) 2 x − 5x − 2



(b)



4 x3 − 4 x2 + 2 x − 1 ( 2 x − 3)( x + 1)



3 x 3 − x 2 − 24 x + 6 (c) x2 − 9











7. Find the values of a, b, c and d for which cx + d 9 x 4 + 12 x 3 + 3 x 2 + 5 x − 1 = ax + b + 3 . 3 3x + x 3x + x 15 x 3 − 11x 2 − 33 x can be expressed 5x − 2 d in the form ax2 + bx + c + , find the values 5x − 2



8. Given that



of a, b, c and d.



INTERMEDIATE LEVEL



3. By performing long division, express



5. Find the values of a, b, c and d for which 6 x 3 + 4 x 2 − 3x bx 2 + cx + d . a = + 3x 3 + 2 3x 3 + 2



5 x 3 + 3 x 2 − 13 x2 − 2 x + 5



as the sum of a polynomial and a proper fraction. 9 x 3 + 3 x 2 − 13 can be expressed 3x 2 + 5 cx + d in the form ax + b + 2 . 3x + 5



4. It is given that



Find the values of a, b, c and d.



Polynomials



Chapter 2



054



Partial Fractions (Recap) Let us recall how to add and subtract algebraic fractions. For example, (a) 2 distinct linear factors



x−2 3( x + 1) 1 3 + + = ( x + 1)( x − 2) ( x + 1)( x − 2) x +1 x − 2







( x − 2) + 3( x + 1) ( x + 1)( x − 2) 4x + 1 = ( x + 1)( x − 2)



=



(b) 1 linear factor and 1 repeated linear factor 2( x + 1) + 1 2 1 + = x + 1 ( x + 1)2 ( x + 1)2 2x + 2 + 1 = ( x + 1)2 2x + 3 = ( x + 1)2 (c) 1 linear factor and 1 quadratic factor



4 ( x 2 + 3) + ( x − 6 )( 2 x − 1) 6 x 2 − 13 x + 18 2x − 1 4 + 2 = = x−6 x +3 ( x − 6 )( x 2 + 3) ( x − 6 )( x 2 + 3)



In this section, we will learn how to do the reverse, i.e. we will split or decompose a single algebraic fraction into two or more partial fractions.



Case 1: Proper Fraction with Distinct Linear Factors in the Denominator Thinking Time The following shows two examples of combining two algebraic fractions into a single algebraic fraction: (a) (b) 1. 2. 3. 4.



055



4x +1 1 3 + = x + 1 x − 2 ( x + 1)( x − 2)



2 4 22 − =− x + 3 2x − 5 ( x + 3)(2 x − 5) What is the relationship between the denominator of the fraction on the RHS and the denominators of the two fractions on the LHS? What are the degrees of the numerator and the denominator of each of the two fractions on the LHS? In other words, are the fractions on the LHS proper or improper fractions? What are the degrees of the numerator and the denominator of the fraction on the RHS? In other words, is the fraction on the RHS a proper or an improper fraction? From the above observations, to do the reverse, we have px + q A B = + , where A and B are constants. ( ax + b )( cx + d ) ax + b



Chapter 2



Polynomials



Worked Example



21



(Distinct Linear Factors in the Denominator) 10 x − 11 Express in partial fractions. ( x + 1)( 2 x − 5 )



Solution: Let



10 x − 11 A B = , where A and B are constants. + ( x + 1)( 2 x − 5 ) x + 1 2 x − 5



Multiply throughout by (x + 1)(2x − 5), we have 10x − 11 = A(2x − 5) + B(x + 1) Let x = −1. 10(−1) − 11 = A[2(−1) − 5] + B(−1 + 1) −21 = −7A A = 3 Let x =



INF



OR



MA



TIO N



We can also solve for A and B by comparing coefficients.



5 . 2



⎡ ⎛5⎞ ⎤ ⎛5 ⎞ ⎛5⎞ 10 ⎜ ⎟ – 11 = A ⎢2 ⎜ ⎟ – 5⎥ + B ⎜ + 1⎟ 2 2 2 ⎝ ⎠ ⎝ ⎠ ⎣ ⎝ ⎠ ⎦







14 =



7 B 2



B = 4



10 x − 11 3 4 ∴ ( x + 1)( 2 x − 5 ) = + x + 1 2x − 5



PRACTICE NOW 21



9x − 5 1. Express ( x − 3)( 2 x + 5 ) in partial fractions.



SIMILAR QUESTIONS



Exercise 2F Questions 1(a)-(d), 7, 8



2. Factorize 2x3 + 3x2 − 17x + 12 completely. Hence, express 7 x 2 − 25 x + 8 in partial fractions. 2 x 3 + 3 x 2 − 17 x + 12



Case 2: Proper Fraction with Repeated Linear Factors in the Denominator For Case 1 where the two linear factors in the denominator are distinct, px + q A B = + . ( ax + b )( cx + d ) ax + b cx + d We shall now consider the case where the linear factor in the denominator px + q . is repeated, such as in ( ax + b )2



Polynomials



Chapter 2



056



Thinking Time The following shows two examples of combining two algebraic fractions into a single algebraic fraction:



(a) (b)



2x + 3 2 1 + = 2 x + 1 ( x + 1) ( x + 1)2



6 x − 17 3 2 − = 2 2 x − 5 (2 x − 5) (2 x − 5)2



1. What is the relationship between the denominator of the fraction on the RHS and the denominators of



the two fractions on the LHS?



2. From the above observation, to do the reverse, we have constants.



Worked Example



22



(Repeated Linear Factors in the Denominator) Express



3x − 2 in partial fractions. ( x − 1)2



Solution: Let



3x − 2 A B + , where A and B are constants. 2 = x −1 ( x − 1) ( x − 1)2



Multiply throughout by (x − 1)2, we have 3x − 2 = A(x − 1) + B Let x = 1. 3(1) − 2 = A(1 − 1) + B B=1 Let x = 0. 3(0) − 2 = A(0 − 1) + B A=3 ∴



057



3x − 2 3 1 + 2 = x −1 ( x − 1) ( x − 1)2



Chapter 2



Polynomials



px + q A = + ( ax + b )2 ax + b



B



, where A and B are



SIMILAR QUESTIONS



PRACTICE NOW 22



5 − 15 x in partial fractions. ( 2 x + 1)2 5 x − 32 in partial fractions. 2. Express 2 x + 4x + 4



Exercise 2F Questions 2(a)-(b)



1. Express



Worked Example



23



(Repeated Linear Factors in the Denominator) 10 − 3 x − 2 x 2 Express in partial fractions. ( 2 x + 3)( x − 1)2



Solution:



10 − 3 x − 2 x 2 A B C = + + , where A, B and C are constants. 2 x x 2 3 1 + − (2 x + 3)( x − 1) ( x − 1)2 Multiply throughout by (2x + 3)(x − 1)2, 10 − 3x − 2x2 = A(x − 1)2 + B(x − 1)(2x + 3) + C(2x + 3) Let x = 1. 5 = 5C C=1 3 Let x = − . 2 25 10 = A 4 8 A= 5 Let x = 0. 10 = A − 3B + 3C 9 B= − 5



Let



∴



10 − 3 x − 2 x 2 8 9 1 = − + 5 ( 2 x + 3) 5 ( x − 1) ( x − 1)2 ( 2 x + 3)( x − 1)2



PRACTICE NOW 23



1. Express



4 x 2 + 5 x − 32 in partial fractions. ( x + 2 )2 ( x − 11)



SIMILAR QUESTIONS



Exercise 2F Questions 2(c)-(f)



x 2 − 10 x + 36



2. Factorize x3 − 2x2 − 4x + 8 completely. Hence, express 3 in x − 2 x2 − 4 x + 8 partial fractions.



Case 3: Proper Fraction with a Quadratic Factor in the Denominator that Cannot be Factorized We will only consider the case where the quadratic factor in the denominator of a proper algebraic fraction is in the form x2 + c2, where c is a constant. What happens if the quadratic factor in the denominator can be factorized? Polynomials



Chapter 2



058



Worked Example



24



(Quadratic Factor in the Denominator that cannot be Factorized) 7 x 2 + x + 15 in partial fractions. Express x 3 + 3x



Solution:



The denominator x3 + 3x = x(x2 + 3). (factorize the denominator) 2



2



7 x + x + 15 7 x + x + 15 A Bx + C = = + 2 , where A, B and C are constants. Let x x +3 x 3 + 3x x ( x 2 + 3)



Multiply throughout by x(x2 + 3),



7x2 + x + 15 = A(x2 + 3) + x(Bx + C) Let x = 0.



AT



TE



NTI



ON



When the quadratic factor in the denominator cannot be factorized, the numerator of this partial fraction is linear, i.e. Bx + C, where B and C are constants.



15 = A(0 + 3) + 0 A=5



Let x = 1.



7 + 1 + 15 = 5(1 + 3) + 1[B(1) + C] B + C = 3 ------------- (1)



Let x = −1.



7 − 1 + 15 = 5(1 + 3) − 1[B(−1) + C] B − C = 1------------- (2)



Solving (1) and (2) simultaneously, we have B = 2 and C = 1. ∴



7 x 2 + x + 15 5 2 x + 1 = + 2 x x +3 x 3 + 3x



SIMILAR QUESTIONS



PRACTICE NOW 24a



5 x 2 + 12 x + 9 in partial fractions. ( x + 1)( x 2 + 5 ) 1 2. Express in partial fractions. 4 x( x2 + 4 )



Exercise 2F Questions 3(a)-(d)



1. Express



Class Discussion Partial Fraction



Discuss the following with your classmates.



Bx + C , where the denominator is quadratic and x 2 + c2 Bx + C the numerator is linear. Is it possible for the numerator to be a constant? If yes, why do we write 2 x + c2 C and not 2 ? x + c2 C



(a) In Worked Example 24, the second partial fraction is



(b) In Worked Example 23, the last partial fraction is



059



( x − 1)2







numerator is a constant. Why is the numerator not linear?







Hint: See Worked Example 22.



Chapter 2



Polynomials



, where the denominator is quadratic but the



Case 4: Improper Fraction If an algebraic fraction is improper, we have to first express it as the sum of a polynomial and a proper fraction (see Worked Example 20), and then decompose the proper fraction into partial fractions. 4 x 3 + 8 x 2 − 50 x − 132 For example, to express in partial fractions, we use long 2 x 2 + x − 28 division to divide 4x3 + 8x2 − 50x − 132 by 2x2 + x − 28 to obtain the quotient 2x + 3 and the remainder 3x − 48. Therefore, 4 x 3 + 8 x 2 − 50 x − 132 3 x − 48 = 2x + 3 + . 2 2 x + x − 28 2 x 2 + x − 28



Next, we decompose



3 x − 48 into partial fractions using Case 1 since 2 x 2 + x − 28



the denominator can be factorized into two distinct linear factors: 3 x − 48 3 x − 48 = 2 ( x + 4)(2 x − 7) 2 x + x − 28 3 x − 48 A B + Let = , where A, B and C are constants. ( x + 4)(2 x − 7) x + 4 2x − 7 Multiply throughout by (x + 4)(2x – 7),



3x – 48 = A(2x – 7) + B(x + 4) Let x = −4. 3(−4) – 48 = A[2(−4) – 7] + B(−4 + 4) −60 = −15A A=4 7 Let x = . 2 ⎡ ⎛7⎞ ⎤ ⎛7 ⎞ ⎛7⎞ 3⎜ ⎟ – 48 = A ⎢2 ⎜ ⎟ – 7⎥ + B ⎜ + 4 ⎟ ⎝2 ⎠ ⎝2⎠ ⎣ ⎝2⎠ ⎦ −



75 15 = B 2 2



B = −5 3 x − 48 4 5 = − 2 2 x + x − 28 x + 4 2 x − 7 ∴



4 x 3 + 8 x 2 − 50 x − 132 4 5 = 2x + 3 + . − 2 x + 4 2 x −7 2 x + x − 28



PRACTICE NOW 24b



1. Express 2. Express



3x 3 − 5 in partial fractions. x2 − 1



6 x 3 − 37 x 2 + 62 x − 14 in partial fractions. 2 x 2 − 11x + 15



Polynomials



Chapter 2



060



Application of Partial Fractions One reason for decomposing an algebraic fraction into partial fractions is the original algebraic fraction is too complicated to manipulate. In this section, we will look at the following example.



Worked Example



25



(Application of Partial Fractions) 1 (i) Express m ( m + 1) in partial fractions.



(ii) Hence, evaluate



1 1 1 1 1 + + + ... + + 1× 2 2 × 3 3× 4 2011 × 2012 2012 × 2013.



Solution:



1 A B (i) Let m ( m + 1) = m + m + 1 , where A, B and C are constants.







Multiply throughout by m(m + 1),







1 = A(m + 1) + Bm







Let m = 0.







1 = A(0 + 1)



A = 1



Let m = −1.







1 = A(−1 + 1) + B(−1)



B = −1



1 1 1 ∴ m ( m + 1) = m − m + 1



1 1 1 1 1 (ii) 1 × 2 + 2 × 3 + 3 × 4 + ... + 2011 × 2012 + 2012 × 2013 ⎛1 1 ⎞ ⎛ 1 1 ⎞ ⎛ 1 1 ⎞ ⎛ 1 1 ⎞ ⎛ 1 1 ⎞ = ⎜ – ⎟ + ⎜ – ⎟ + ⎜ – ⎟ + … + ⎜ – – ⎟+⎜ ⎟ ⎝1 2 ⎠ ⎝ 2 3 ⎠ ⎝ 3 4 ⎠ ⎝ 2011 2012 ⎠ ⎝ 2012 2013 ⎠



= 1 − 1 2013 = 2012 2013



PRACTICE NOW 25



(i) Express



2 in partial fractions. n ( n + 1)( n + 2 )



(ii) Hence, evaluate 2 2 2 2 2 + + + ... + + . 1× 2 × 3 2 × 3× 4 3× 4 × 5 2010 × 2011 × 2012 2011 × 2012 × 2013 061



Chapter 2



Polynomials



SIMILAR QUESTIONS



Exercise 2F Question 10



Exercise



2F



6. Divide 18x3 − 15x2 − 42x + 3 by 6x2 − x − 15.



BASIC LEVEL



1. Express each of the following in partial fractions. 7x + 6 11x − 2 (a) x ( x + 2 ) (b) 3 x ( x − 1) 7 x + 11 15 x + 13 (c) ( x + 3)( x − 2 ) (d) ( x − 1)( 3 x + 1) 2. Express each of the following in partial fractions. x+3 x (a) (b) 2 ( x − 1) ( x + 4 )2 5 x 2 − 3x + 2 8 x 2 + 15 x + 12 (c) (d) 2 x ( x − 1) x ( x + 2 )2 −20 x − 2 39 x 2 − 35 x + 11 (e) (f) ( x + 1)( x − 2 )2 ( 2 x + 3)( 3 x − 1)2



3. Express each of the following in partial fractions.



(a)



2



5 x + 3 x + 20 x( x2 + 4 )



(b)



2



5x + x + 4 ( x + 1)( x 2 + 3)



2 x 2 − 3 x + 27 −8 x 2 − 7 x + 28 (c) (d) ( x + 3)( x 2 + 7 ) ( 3 x − 2 )( 2 x 2 + 9 )



INTERMEDIATE LEVEL



4. Express each of the following improper fractions



in partial fractions.







2 x 2 − 11x + 12 (a) x 2 − 5 x + 6



2 x 3 − 5 x 2 − 14 x − 17 (b) 2 x − 3x − 4



4 x 3 + x 2 − 15 x + 21 (c) ( x + 2 )( x − 1)2







Hence, express







fractions.



18 x 3 − 15 x 2 − 42 x + 3 in partial 6 x 2 − x − 15



7. Factorize 2x3 + 3x2 − 50x − 75 completely.



Hence, express







fractions.



3 x 2 + 19 x + 90 in partial 2 x + 3 x 2 − 50 x − 75 3



8. Factorize 8x3 − 22x2 − 9x + 9 completely. 25 x + 75 Hence, express in partial 8 x 3 − 22 x 2 − 9 x + 9 fractions. ADVANCED LEVEL



9. (i) The cubic polynomial f(x) = 3x3 + hx2 + kx − 8 is exactly divisible by 3x − 2 and leaves a remainder of 5 when it is divided by x − 1. Calculate the value of h and of k. Hence, factorize f(x) completely. (ii) Using the results in (i), express 4 x + 24 in f( x ) partial fractions.



10. (i) Express



20 in partial fractions. n(n + 2 )



(ii) Hence, simplify



20 20 20 20 20 + . + + + ...… + 2 1× 3 2 × 4 3× 5 n − 1 n2 + 2 n



6 x 4 − 5 x 3 + 15 x 2 − 4 x + 5 (d) ( 2 x − 3)( x 2 + 5 ) 2 x 3 − 3x 2 − 2 x − 2 can be x2 − 4 x + 3 c d expressed in the form ax + b + x − 1 + x − 3 .



5. It is given that



Find the values of a, b, c and d.



Polynomials



Chapter 2



062



1. The Division Algorithm for Polynomials states that:



P(x) = D(x) × Q(x) + R(x)







dividend = divisor × quotient



+ remainder



and the degree of R(x) < the degree of D(x).



2. The Remainder Theorem states that: If a polynomial P(x) is divided by a linear divisor x – c, the remainder is P(c). ⎛ b⎞ If a polynomial P(x) is divided by a linear divisor ax + b, the remainder is P ⎜ – ⎟ . ⎝ a⎠ 3. The Factor Theorem states that: ⎛ b⎞ ax + b is a linear factor of the polynomial P(x) if and only if P ⎜ – ⎟ = 0. ⎝ a⎠ The Factor Theorem can be used to solve certain cubic equations. 4. Two special algebraic identities: Sum of Cubes: a3 + b3 = (a + b) (a2 − ab + b2) Difference of Cubes: a3 − b3 = (a − b) (a2 + ab + b2) 5. To decompose



Step 1: If







P(x) , where P(x) and Q(x) are polynomials and Q(x) ≠ 0, into partial fractions: Q( x )



P(x) is improper, express it as the sum of a polynomial and a proper algebraic Q( x )



fraction first.







Step 2: Factorize the denominator of the proper algebraic fraction if possible.







Step 3: Express in partial fractions according to Case 1, 2 or 3. Case Denominator contains







063



Algebraic Fraction px + q ( ax + b )( cx + d )



Partial Fractions A B + ax + b cx + d



1



Distinct linear factors



2



Repeated linear factors



px + q ( ax + b )2



A B + ax + b ( ax + b )2



3



Quadratic factor x2 + c2 (cannot be factorized)



px + q ( ax + b )( x 2 + c 2 )



Bx + C A + ax + b x 2 + c 2



Step 4: Solve for unknown constants by substituting suitable values of x.



Chapter 2



Polynomials



2 1. Find the values of A, B and C in each of the following.



(a) 4x2 – 13x + 5 = A(x – 2)(x – 3) + B(x – 2) + C



(b) 7x2 – 14x + 13 = A(x – 1)(x + 3) + B(x – 1) + C







(c) 2x3 + 3x2 – 14x – 5 = (Ax + B)(x + 3)(x + 1) + C



(d) x4 – 10x2 – 13x + 1



= (x2 + Ax + 3)(x – 1)(x – 3) + Bx + C 2. Find the value of k in each of the following.



(a) x + 2 is a factor of 2x3 – 3x2 + kx + 18



(b) 2x – 1 is a factor of 2x3 + x2 – kx + 30







(c) x – 2 is a factor of 2x3 + kx2 + 11x – 2



(d) x + 3 is a factor of (2x + 5)9 + (3x + k)3







(e) (3x k – 7x2 + 15) ÷ (x – 2) has a remainder of 11



(f) (2kx2 – k2x – 11) ÷ (x – 3) has a remainder of 4



3. The expression 4x3 – 16x2 + 4x + 24 = a(x + 1)(x – 2)(x – b) holds for all values of x. Find the value of a and of b. 4. Find the remainder when



(a) 3x3 + 2x2 – 7 is divided by x – 1,



(b) 5x3 – 4x2 + 7x – 5 is divided by x + 1,



(e) 6x3 – 4x + 5 is divided by 2x – 3,



(f) 2(x + 4)3 + 2(x2 – 3)2 + 2 is divided by 2x + 5.



(c) 2x3 – 8x2 + 5x – 1 is divided by 2x – 1,



(d) 4x3 – 7x2 – 5x + 3 is divided by 3x + 1,



5. Factorize each of the following.



(a) 2x3 – 7x2 + 9 (c) 6x3 + 19x2 – 24x – 16 (e) 8x6 − 343 (g) (2x + 3)3 − (4x − 5)3



(b) x3 – 2x2 – 15x + 36 (d) 12x3 + 28x2 – 7x – 5 (f) 27a3 + 64y6 (h) (3x + 5)3 + (x + 1)3



6. Solve each of the following equations, giving your answers correct to 2 decimal places where necessary.



(a) 2x3 – 3x2 – 17x + 30 = 0 (c) 2x3 – 7x2 – 10x + 24 = 0 (e) x3 – 6x2 + 2x + 12 = 0 (g) 4x3 – 37x2 + 37x – 7 = 0



(b) 6x3 + 11x2 – 4x – 4 = 0 (d) 8x4 – 42x3 + 29x2 + 42x + 8 = 0 (f) 2x3 – 31x2 + 29x – 7 = 0 (h) 6x3 + 5x2 – 26x + 11 = 0



7. When the expression 7x21 – 5x15 + ax6 is divided by x + 1, the remainder is 2. Find the value of a. Hence, find the remainder when the expression is divided by x – 1. 8. Find the value of p and of q for which the expression 12x4 + 16x3 + px2 + qx – 1 is divisible by 4x2 – 1. Hence, find the other factors of the expression.



Polynomials



Chapter 2



064



9. The expression ( px + q)(x – 1) + r(x2 + 2) is equal to 12 for all values of x. By substituting suitable values of x, or otherwise, find the values of p, q and r. 10. Find the value of p and of q if 4x2 – 4x – 3 is a factor of the expression 8x4 + px3 + qx2 + x + 3. Hence, factorize the expression completely. 11. Given that x + 5 is a common factor of x3 + kx2 − Ax + 15 and x3 − x2 − (A + 5)x + 40, find the value of k and of A. Hence, factorize x3 + kx2 − Ax + 15 completely. 12. When the expression f(x) = hx3 – 2x2 + kx + 3 is divided by x – 1, the remainder is 9. When the expression



is divided by x + 2, the remainder is –39. Find the value of h and of k.



13. The expression 2x3 + px2 – qx + 12 has a remainder of –3 and 6 when divided by x + 1 and x – 2 respectively.







Find the value of p and of q.



14. The expression 4x3 – px2 + 3x – q has a remainder of 2 and –5 when divided by 2x – 1 and x – 1 respectively.







Find the value of p and of q.



15. Express each of the following in partial fractions.



(c)



5x ( 2 x + 1)( x 2 + 1)



(d)



9 x 2 + 2 x + 14 ( 2 x + 3)( x − 1)2







(e)



x2 + 4 x − 7 x 2 + x − 12



(f)



4x − 1 x3 − 4 x







(g)



3 x 2 – x  + 1 x ( x –1)2



(h)



4 x 2  + 11x  + 4 x ( x  + 2)2







(i)



3 x 2  + 11x  + 14 ( x  + 3)( x  + 1)2



(j)



9 x 2  + 4 x –4 x 3 –4 x







(k)



5x (2 x  + 1)( x 2 + 1)



(l)



2 x 2 –9 x  + 20 3 x ( x 2 + 4)



16. It is given that



c 2 x3 − 9 x2 + 7 x + 8 can be expressed in the form ax + b + ( 2 x + 1)( x − 2 ) . ( 2 x + 1)( x − 2 )



Find the values of a, b and c.



17. It is given that



065



x 2 + 16 (b) ( 3 − x )( 2 x − 1)2



10 x − 4 (a) ( 2 x − 1)( x + 2 )



Chapter 2



x3 − x2 − 4 x + 1 c = ax + b + . Find the values of a, b and c. ( x − 2 )( x + 2 ) ( x − 2 )( x + 2 )



Polynomials



Challenge −b ± b 2 − 4 ac . There is also a general formula 2a for solving a cubic equation. Find out more about this general formula and use it to solve 2x3 + 3x2 − 5x − 6 = 0.



1. The general formula to solve a quadratic equation is x =



2. Decomposing an algebraic fraction, whose denominator contains only distinct linear factors, into partial fractions can be done easily using Heaviside’s Cover-Up Rule. Find out about this rule and use it to 3x − 1 express ( x − 2 )( x + 3) in partial fractions.



Polynomials



Chapter 2



066



Graphs of Functions and Graphical Solution A newspaper article states that the growth in the number of members of a social network increased exponentially in its first year of operation. What is meant by an exponential increase? How can you represent this information on a graph?



Three LEARNING OBJECTIVES At the end of this chapter, you should be able to: • draw the graphs of simple sums of power functions y = axn, where n = 3, 2, 1, 0, –1 and –2, • draw the graphs of exponential functions y = kax, where a is a positive integer, • estimate the slope of a curve by drawing a tangent, • interpret and analyze data from tables and graphs, including distance-time and speed-time graphs.



3.1



Graphs of Cubic Functions



In Grade 9, we have learned how to draw the graphs of y = ax2 + bx + c, where a, b and c are constants and a ≠ 0. We have also learned how to sketch graphs of the form y = (x – h)(x – k) and y = –(x – h)(x – k), where h and k are constants, and y = (x – p)2 + q and y = –(x – p)2 + q, where p and q are constants. In this section, we will learn how to draw the graphs of cubic functions. In general, cubic functions are of the form y = ax3 + bx2 + cx + d, where a, b, c and d are constants and a ≠ 0.



Investigation Graphs of Cubic Functions



1. Using a graphing software, draw each of the following graphs.



(a) y = x3



(b) y = 2x3



(e) y = –2x3



(f) y = –5x3



(c) y = 5x3



(d) y = –x3



2. For the graph of y = ax3, where a is a constant, how does the value of a affect the shape of the graph? 3. Using a graphing software, draw each of the following graphs. (a) y = x3 – x2 + 1



(c) y = x3 + x



(e) y = –x3 + 2x2 + 1



(b) y = x3 + 4x2 – 3



(d) y = –x3 + x2 – 2



(f) y = –x3 – 0.5x – 1



4. For each of the graphs in Question 3, how does the coefficient of x3 affect the shape of the graph?



069



Chapter 3



Graphs of Functions and Graphical Solution



From the investigation, we observe that for the graph of a cubic function of the form y = ax3, the graph takes the shape in Fig. 3.1(a) and (b) for positive and negative values of the coefficient of x3. y y



x



O



x



O



(a) y = ax3, where a  0 (b) y = ax3, where a  0 Fig. 3.1 In general, the graph of a cubic function, i.e. a function of the form y = ax3 + bx2 + cx + d would take the shape in Fig. 3.2 (but may not pass through the origin if d ≠ 0). y



y



x



O



(a) y = ax3 + bx2 + cx + d, where a  0



y



O







x



O



y



x



O



x



(b) y = ax3 + bx2 + cx + d, where a  0 Fig. 3.2



Graphs of Functions and Graphical Solution



Chapter 3



070



Worked Example



1



(Drawing the Graph of a Cubic Function) Using a scale of 2 cm to represent 1 unit on the x-axis and 2 cm to represent 10 units on the y-axis, draw the graphs of y = x3 + 3 and y = –x3 – 3 for –3  x  3. For each graph, find (i) the value of y when x = 1.5,



(ii) the value of x when y = 20.



Solution: x



y=x +3



–3



–2



–1



0



1



24



5



–2



–3



–4



–24



3



y = –x – 3 3



–5



Scale: x-axis: 2 cm represent 1 unit y-axis: 2 cm represent 10 units



2



3



4



2



3



11



–11



30



–30



y 30



20 y = –x3 – 3



–3



–1



–2



y = x3 + 3



10



0



1



–10



–20



–30



(i) Consider y = x3 + 3.



From the graph, when x = 1.5, y = 6.5.



Consider y = –x3 – 3.



071



From the graph, when x = 1.5, y = –6.5.



Chapter 3



Graphs of Functions and Graphical Solution



2



3



x



(ii) Consider y = x3 + 3.



From the graph, when y = 20, x = 2.6.



Consider y = –x3 – 3.



From the graph, when y = 20, x = –2.85.



PRACTICE NOW 1



SIMILAR QUESTIONS



Using a scale of 2 cm to represent 1 unit on the x-axis and 2 cm to represent 10 units on the y-axis, draw the graphs of y = x3 + 2 and y = –x3 – 2 for –3  x  3.



Exercise 3A Questions 1, 2, 5, 6, 11



For each graph, find (i) the value of y when x = 2.5,



3.2



(ii) the value of x when y = 15.



Graphs of Reciprocal Functions



Graph of y = a x



Investigation a



Graphs of y = x



Using a graphing software, draw the graph of y = a = –3.



a for a = 1, a = 5, a = –1 and x



1. (i) For a > 0, which quadrants do the graphs lie in?



(ii) For a < 0, which quadrants do the graphs lie in?



INF



OR



MA



TIO N



The four quadrants on the Cartesian plane are labeled 1st, 2nd, 3rd and 4th as follows: y 2nd 1st x O 3rd 4th



2. What can you say about the rotational symmetry of each of the graphs? 3. Do the graphs intersect the x-axis and the y-axis? Explain your answer. AT



TE



NTI



ON



The order of rotational symmetry about a particular point is the number of distinct ways in which a figure can map onto itself by rotation in 360°.



Graphs of Functions and Graphical Solution



Chapter 3



072



From the investigation, we observe that for the graph of y = • when x = 0, the function y =



a , x



a is not defined, i.e. there is a break when x = 0, x



• there is rotational symmetry of order 2 about the origin, i.e. it maps onto itself twice by rotation in 360°, • if a > 0, the graph would take the shape in Fig. 3.3(i),



if a < 0, the graph would take the shape in Fig. 3.3(ii). y



x



O



(i) y =



x



O



a , where a > 0 x



(ii) y =



Fig. 3.3



Consider the graph in Fig. 3.3(i) y =



y



a , where a < 0 x



a , where a > 0. The graph consists of two parts x



that lie in the 1st and 3rd quadrants. In the 1st quadrant, we observe that: • as x increases, y decreases;



INF



• as x approaches zero, y becomes very large;



e.g. for a = 1, y =



1 1 , if x = 0.000 001, y = 0.000 001 = 1 000 000; x



• as x becomes very large, y approaches zero; 1 1 e.g. for a = 1, y = , if x = 1 000 000, y = 1 000 000 = 0.000 001; x



• the curve gets very close to the x-axis and y-axis but never touches them. Can you describe the part of the graph that is in the 3rd quadrant? Can you describe the graph of y =



a , where a < 0? x



Thinking Time What are the equations of the lines of symmetry of the graph y = (a) a > 0?



073



Chapter 3



(b) a < 0?



Graphs of Functions and Graphical Solution



a when x



OR



MA



TIO N



As the positive value of x decreases, the value of y increases rapidly and it gets very close to the y-axis. We say that the value of y approaches infinity, written as y → ∞.



Worked Example



2



(Drawing the Graph of y =



a ) x



Using a scale of 1 cm to represent 1 unit on the x-axis and 1 cm to represent 5 units on the y-axis, draw the 6 graph of y = for –5  x  5, x ≠ 0. Find x (i) the value of y when x = 1.4, (ii) the value of x when y = –8.



Solution: x



–5



–4



–3



y –1.2 –1.5 –2



–2



–3



–1 –0.5 –0.3 –6



–12



Scale: x-axis: 1 cm represents 1 unit y-axis: 1 cm represents 5 units



0.3



–20



0.5



20



1



12



6



2



3



3



2



4



1.5



5



1.2



y 25 20 15 10



6 y= x



5 –5



–4



–3



–2



–1



0



1



2



3



4



5



x



–5 –10 –15 AT



–20



(i) From the graph, when x = 1.4, y = 4.5.



(ii) From the graph, when y = –8, x = –0.8.



PRACTICE NOW 2



ON



SIMILAR QUESTIONS



Using a scale of 1 cm to represent 1 unit on both axes, draw the graph of y = (i) the value of y when x = 2.5,



NTI



For (i), although the answer is 4.2857… by calculation, the answer obtained from the graph can only be accurate up to half of a small square grid, which is 0.5. Similarly, for (ii), although x = −0.75 by calculation, the answer obtained from the graph is accurate to half of a small square grid, i.e. 0.1.



–25



for –5  x  5, x ≠ 0. Find



TE



3 x



Exercise 3A Questions 3, 7, 8, 12



(ii) the value of x when y = –1.2.



Graphs of Functions and Graphical Solution



Chapter 3



074



Graph of y = a x



2



Investigation Graphs of y =



a x2



a Using a graphing software, draw the graph of y = 2 for a = 2, a = 4, a = –1 and x a = –3.



1. (i) For a > 0, which quadrants do the graphs lie in?



(ii) For a < 0, which quadrants do the graphs lie in?



2. What can you say about the line symmetry of each of the graphs? 3. Do the graphs intersect the x-axis and the y-axis? Explain your answer.



From the investigation, we observe that for the graph of y = • when x = 0, the function y =



a , x2



a is not defined, i.e. there is a break when x = 0, x2



• if a > 0, the values of y are always positive, i.e. the graph lies entirely above the x-axis; if a < 0, the values of y are always negative, i.e. the graph lies entirely below the x-axis. • the graph is symmetrical about the y-axis, i.e. the y-axis is the line of symmetry, • if a > 0, the graph would take the shape in Fig. 3.4(i), if a < 0, the graph would take the shape in Fig. 3.4(ii). y y



x



O O



(i) y =



x



a , where a > 0 x2



(ii) y =



Fig. 3.4



Consider the graph in Fig. 3.4(i) y =



a , where a < 0 x2



a , where a > 0. The graph consists of two x2 st



parts that lie in the 1st and 2nd quadrants. In the 1 quadrant, we observe that: • as x increases, y decreases;



• as x approaches zero, y becomes very large; 1 1 e.g. for a = 1, y = 2 , if x = 0.001, y = = 1 000 000; x 0.0012 • as x becomes very large, y approaches zero; 1 1 e.g. for a = 1, y = 2 , if x = 1000, y = = 0.000 001; x 1000 2 • the curve gets very close to the x-axis and y-axis but never touches them. Can you describe the part of the graph that is in the 2nd quadrant? a Can you describe the graph of y = 2 , where a < 0? x 075



Chapter 3



Graphs of Functions and Graphical Solution



Worked Example



3



(Drawing the Graph of y =



a ) x2



Using 1 cm to represent 1 unit on the x-axis and 2 cm to represent 1 unit on the y-axis, draw the graph of 1 y = 2 for –4  x  4, x ≠ 0. Find x (i) the value of y when x = 1.6,



(ii) the values of x when y = 3.5.



Solution: x y



–4



0.06



–3



–2



0.11



0.25



–1 1



–0.5 4



0.5



1



4



2



1



3



0.25



0.11



4



0.06



y Scale: x-axis: 1 cm represents 1 unit y-axis: 2 cm represent 1 unit



4



3



2



y = 12 x



1



–4



–3



–2



–1



0



1



2



3



4



x



AT



TE



NTI



ON



The accuracy of the answer can only be accurate up to half of a small square grid.



(i) From the graph, when x = 1.6, y = 0.4.



(ii) From the graph, when y = 3.5, x = 0.6 or x = –0.6.



PRACTICE NOW 3



SIMILAR QUESTIONS



Using 1 cm to represent 1 unit on the x-axis and 2 cm to represent 1 unit on the 2 y-axis, draw the graph of y = − 2 for –4  x  4, x ≠ 0. Find x (i) the value of y when x = 1.5, (ii) the values of x when y = –3.2.



Exercise 3A Questions 4, 9, 10, 13



Graphs of Functions and Graphical Solution



Chapter 3



076



Exercise



3A



1. The table below shows some values of x and the corresponding values of y, where y = x3. x



–3



y



–2



–1



–8



0



1



1



2



3



27



(a) Copy and complete the table.







(b) Using a scale of 2 cm to represent 1 unit, draw a horizontal x-axis for –3  x  3.



Using a scale of 2 cm to represent 5 units, draw a vertical y-axis for –27  y  27.







On your axes, plot the points given in the table and join them with a smooth curve.



(c) Use your graph to find







(i) the value of y when x = 1.5, (ii) the value of x when y = 12.



2. The table below shows some values of x and the corresponding values of y, correct to 1 decimal place, where y = 2x3 + 3. x y



–2



–13



–1.5



–3.8



–1 1



–0.5 p



0



3



0.5



3.3



3. The table below shows some values of x and the 4 corresponding values of y, where y = . x



1



5



1.5



1 4



y



16



1 2



1



2



3



4



2



5



1



(a) Copy and complete the table. (b) Using a scale of 2 cm to represent 1 unit on the x-axis and 1 cm to represent 1 unit on the 4 1 y-axis, draw the graph of y = for  x  5. x 4 (c) Use your graph to find (i) the value of y when x = 3.6, (ii) the value of x when y = 1.5.



4. The table below shows some values of x and the corresponding values of y, correct to 1 decimal 10 place, where y = 2 . x



2



9.8 19



x







x y



1



10



2



2.5



3



a



4



0.6



5



b



(a) Find the value of a and of b.



(b) Using a scale of 4 cm to represent 1 unit, draw a horizontal x-axis for –2  x  2.



(b) Using a scale of 2 cm to represent 1 unit 10 on both axes, draw the graph of y = 2 for x 1  x  5.







Using a scale of 1 cm to represent 5 units, draw a vertical y-axis for –15  y  20.



(i) the value of y when x = 2.8,







On your axes, plot the points given in the table and join them with a smooth curve.











(a) Find the value of p.











(c) Use your graph to find



(ii) the value of x when y = 4.4.



(c) Use your graph to find



(i) the value of y when x = –1.2, (ii) the value of x when y = 14.



5.



Using a scale of 2 cm to represent 1 unit on the x-axis and 2 cm to represent 5 units on the y-axis, draw the graph of y = 3x – x3 for –3  x  3. Use your graph to find







(i) the value of y when x = 1.4,







077



Chapter 3



Graphs of Functions and Graphical Solution



(ii) the values of x when y = –6.6.



6. Using a suitable scale, draw the graph of y = x3 – 6x2 + 13x for 0  x  5. Use your graph to find



(a) the value(s) of y when



(i) x = 1.5,



(ii) x = 3.5,



(iii) x = 4.45.



(i) y = 7,



(ii) y = 15,



(iii) y = 22.



(b) the value of x when



7. Using a scale of 4 cm to represent 1 unit on both 1 2 axes, draw the graph of y = − − 1 for  x  4. 2 x Use your graph to find



(i) the value of y when x = 2.5,



(ii) the value of x when y = –1.6.



8. The table below shows some values of x and the corresponding values of y, correct to 1 decimal 3 place, where y = x − . x







x y



0.5



–5.5



1



–2



2



0.5



3



h



4



3.3



5



4.4



6



(c) Use your graph to find



(i) the value of y when x = 1.6,



(ii) the value of x when y = –2.5. 9. Using a scale of 2 cm to represent 1 unit on the x-axis and 4 cm to represent 1 unit on the y-axis, draw the graph of y = 2 − 32 for 1  x  6. Use x your graph to find



(i) the value of y when x = 1.5, (ii) the value of x when y = 1.5.



10. Using a scale of 2 cm to represent 1 unit on both 2 axes, draw the graph of y = x + 2 for 1  x  6. x Use your graph to find



(b) On the same axes, draw the straight line y = x for –3  x  3.



(i) Write down the x-coordinates of the points at which the line y = x meets the curve y = x3 – 2x – 1. (ii) Hence, state the solutions of the equation x3 – 2x – 1 = x. Explain your answer. 12. The variables x and y are connected by the 1 equation y = x + −1. 2x The table below shows some values of x and the corresponding values of y, correct to 1 decimal place. x 0.1 0.5 0.8 y



(a) Find the value of h and of k.



(b) Using a scale of 2 cm to represent 1 unit on 3 both axes, draw the graph of y = x − for x 0.5  x  6.







(a) Use your graph to find the x-coordinates of the points of intersection of the curve with the x-axis.



k











11. Using a suitable scale, draw the graph of y = x3 – 2x – 1 for –3  x  3.



(i) the value of y when x = 5.4, (ii) the values of x when y = 3.







1



1.5



2



4.1 0.5 0.4 0.5 0.8 1.3



2.5 3.5 p



4



2.6 3.1



(a) Calculate the value of p.



(b) Using a scale of 4 cm to represent 1 unit on 1 both axes, draw the graph of y = x + − 1 for 2x 0.1  x  4.



(c) Use your graph to find the values of x in the 1 range 0.1  x  4 for which x + = 1. 2x



13. Using a scale of 2 cm to represent 1 unit on the x-axis and 1 cm to represent 1 unit on the y-axis, 1 8 draw the graph y = x 2 + − 9 for 0.5  x  7. 4 x (a) Use your graph to find the minimum value of y in the given range. (b) By drawing suitable straight lines to the graph, solve each of the following equations, giving your answers correct to 1 decimal place. 1 2 8 (i) x + =6 4 x 1 2 8 (ii) x + =x+4 4 x 1 2 8 (iii) x + 2 x = 15 − 4 x



Graphs of Functions and Graphical Solution



Chapter 3



078



3.3



Graphs of Exponential Functions



Graphs of y = ax and y = kax



Investigation Graphs of y = ax and y = kax



1. Using a graphing software, draw each of the following graphs.



(a) y = 2x



(b) y = 3x



(c) y = 4x



(d) y = 5x



2. For each of the graphs in Question 1, answer the following questions. (a) Write down the coordinates of the point where the graph intersects the y-axis.



(b) As x increases, what happens to the value of y? (c) Does the graph intersect the x-axis?



3. How does the value of a affect the shape of the graph of y = ax? 4. Using a graphing software, draw each of the following graphs.



(a) y = 2x



(c) y = 5(2x)



(e) y = –4(2x)



(b) y = 3(2x) (d) y = –2x



5. For each of the graphs in Question 4, answer the following questions. (a) Write down the coordinates of the point where the graph intersects the y-axis.



(b) As x increases, what happens to the value of y? (c) Does the graph intersect the x-axis?



6. How does the value of k affect the shape of the graph of y = kax?



079



Chapter 3



Graphs of Functions and Graphical Solution



From the investigation, we observe that for the graph of y = ax, • the values of y are always positive, i.e. the graph lies entirely above the x-axis, • the graph intersects the y-axis at (0, 1). y



1 x



O Fig. 3.5



As the positive value of x increases and tends to the right of the graph, the value of y increases very rapidly and approaches infinity. When x is negative and tends to the left of the graph, y becomes smaller as x becomes smaller. The curve gets very close to the x-axis but never touches it. For the graph of y = kax,



• if k > 0, the values of y are always positive, i.e. the graph lies entirely above the x-axis (see Fig. 3.6(i)),



if k < 0, the values of y are always negative, i.e. the graph lies entirely below the x-axis (see Fig. 3.6(ii)), • the graph intersects the y-axis at (0, k). y



y



k



k O



O



x



x



(i) y = kax, where k > 0



(ii) y = kax, where k < 0



Fig. 3.6



Journal Writing A newspaper article states that the growth in the number of members of a social network increased exponentially in its first year of operation and can be represented by the equation y = 28x, where x is the number of months and y is the number of members. y 1 0



x



(i) Describe how the number of members of the social network changes with time. (ii) Search on the Internet for more real-life applications of exponential graphs.



Graphs of Functions and Graphical Solution



Chapter 3



080



4



Worked Example



(Graph of y = ax)



Using a scale of 4 cm to represent 1 unit on the x-axis and 2 cm to represent 1 unit on the y-axis, draw the graph of y = 2x for –1  x  2.5. Use your graph to find (i) the value of y when x = 1.8, (ii) the value of x when y = 4.6.



Solution: x y



–1



0.5



–0.5



0.71



0



0.5



1



1.41



1



2



1.5



2.83



2



4



2.5



5.66



y Scale: x-axis: 4 cm represent 1 unit y-axis: 2 cm represent 1 unit



y = 2x



5



4



3



2



1



–1



–0.5



0



0.5



1



1.5



2



2.5



x



(i) From the graph, when x = 1.8, y = 3.5. (ii) From the graph, when y = 4.6, x = 2.2.



PRACTICE NOW 4



SIMILAR QUESTIONS



Using a scale of 4 cm to represent 1 unit on the x-axis and 2 cm to represent 1 unit on the y-axis, draw the graph of y = 3x for –2  x  2. Use your graph to find (i) the value of y when x = –1,



081



Chapter 3



(ii) the value of x when y = 0.7.



Graphs of Functions and Graphical Solution



Exercise 3B Questions 1-5



Class Discussion Matching Graphs of Power Functions with the Corresponding Functions Work in pairs.



Match the graphs with their respective functions and justify your answers. If your classmate does not obtain the correct answer, explain to him what he has done wrong. A: y = 2x3 E: y = −



3 x2



B: y = −



6 x



5 2x 2



D: y = 5x



G: y =



1 2x



H: y = –3x3



F: y = –2(6x)



y



y



x



O



C: y =



O



y



x



O



Graph 1



Graph 2



Graph 3



y



y



y



x



O



O



Graph 4



Graph 5



y



y



O



x



O



x



O



x



x



Graph 6



x SIMILAR QUESTIONS



Exercise 3B Question 6



Graph 7



Graph 8



Graphs of Functions and Graphical Solution



Chapter 3



082



3.4



Slope of a Curve



When a straight line touches a curve at a single point A, the line is called the tangent to the curve at the point A. y curve tangent A x



O



When a line l1 touches the curve at P, l1 is called the tangent to the curve at P. Similarly, when a line l2 touches the curve at Q, l2 is called the tangent to the curve at Q. y



20 l1 10 P



–1



0



1



2



3



4



5



x



Q



–10



l2



Fig. 3.7 The slope of the curve at a point is defined as the slope of the tangent to the curve at that point. Hence, the slope of the curve at P in Fig. 3.7 is equal to the slope of the line l1 and the slope of the curve at Q is equal to the slope of the line l2.



083



Chapter 3



Graphs of Functions and Graphical Solution



Worked Example



5



(Slope of a Curve) The variables x and y are connected by the equation 1 y = ( 5x − x 2 ) . 2 The table below shows some values of x and the corresponding values of y. x y



1 2



0



1



2



a



0



2



3



−



1 2



3



4



5



b



3



2



0



2



(a) Find the value of a and of b.



(b) Using a scale of 2 cm to represent 1 unit on both axes, 1 1 2 draw the graph of y = ( 5x − x ) for −  x  5. 2 2 (c) By drawing a tangent, find the slope of the curve at the point (1, 2). (d) The slope of the curve at the point (h, k) is zero.



(i) Draw the tangent at the point (h, k). (ii) Hence, find the value of h and of k.



Solution:



1 (a) When x = − , 2 2 1⎡ 1 1 ⎤ ⎥ y = ⎢5 − − − 2⎣ 2 2 ⎦ 3 = −1 8 3 ∴ a = −1 8 = −1.375



( )( )



P So roblem lvin g T ip



Give the value of a and of b in decimals for easy plotting of points.



1 When x = 2 , 2 2 1⎡ 1 1 ⎤ ⎥ y = ⎢5 2 − 2 2⎣ 2 2 ⎦ = 3 1 8 ∴ b = 31 8 = 3.125



( )( )



Graphs of Functions and Graphical Solution



Chapter 3



084



(b)



Scale: x-axis: 2 cm represent 1 unit y-axis: 2 cm represent 1 unit y 4



3



vertical change = 1.5 y=



2



1 (5x – x2) 2



horizontal change = 1 1



0



1



2



3



4



5



x



–1



–2 (c) A tangent is drawn to the curve at the point (1, 2).



From the graph,







vertical change horizontal change 1.5 = 1



slope =



= 1.5



(d) A line parallel to the x-axis at the maximum point of the curve has a slope 1 equal to zero. From the graph and table, h = 2.5, k = 3 . 8



085



Chapter 3



Graphs of Functions and Graphical Solution



RE



CAL



L



A line parallel to the x-axis has a slope equal to zero.



PRACTICE NOW 5



SIMILAR QUESTIONS



The variables x and y are connected by the equation y = x2 – 4x.



Exercise 3B Questions 7, 8, 9



The table below shows some values of x and the corresponding values of y. x y



–1



0



1



0



a



–3



2



–4



3



4



5



0



b



5



(a) Find the value of a and of b.



(b) Using a scale of 2 cm to represent 1 unit on both axes, draw the graph of y = x2 – 4x for –1  x  5. (c) By drawing a tangent, find the slope of the curve at the point where x = 2.8.



(d) The slope of the curve at the point (h, k) is zero.



(i) Draw the tangent at the point (h, k). (ii) Hence, find the value of h and of k.



Exercise



3B



1. The table below shows some values of x and the corresponding values of y, where y = 4x. x y







–1



0.25



–0.5



0



1



0.5 2



1



4



1.5



2



2.5



(a) Copy and complete the table.



(b) Using a scale of 4 cm to represent 1 unit, draw a horizontal x-axis for –1  x  2.5.



Using a scale of 1 cm to represent 2 units, draw a vertical y-axis for 0  y  32.







On your axes, plot the points given in the table and join them with a smooth curve.







(c) Use your graph to find



(i) the value of y when x = 1.8,



(ii) the value of x when y = 0.4.



2. The variables x and y are connected by the equation y = 3(2x).



The table below shows some values of x and the corresponding values of y correct to 1 decimal place. x



y



–1



1.5



–0.5 2.1



0



0.5



1



6



1.5



2



12



2.5



17.0



(a) Copy and complete the table.



(b) Using a scale of 4 cm to represent 1 unit on the x-axis and 1 cm to represent 1 unit on the y-axis, draw the graph of y = 3(2x) for –1  x  2.5.



(c) Use your graph to find



(i) the values of y when x = 0.7 and x = 2.3, (ii) the values of x when y = 2.5 and y = 7.4.



Graphs of Functions and Graphical Solution



Chapter 3



086



3.



Using a scale of 4 cm to represent 1 unit on the x-axis and 1 cm to represent 2 units on the y-axis, draw the graph of y = –2(3x) for –2  x  2. Use your graph to find







(i) the value of y when x = 1.2,







6. The sketch represents the graph of y = kax, where a > 0. y



(ii) the value of x when y = –6.7.



3 x



O 4. The table below shows some values of x and the corresponding values of y, correct to 1 decimal place, where y = 2 + 2x. x y







–1 a



–0.5 2.7



0



3



1



4



1.5



4.8



2



6



2.5 b



3



10



(c) Use your graph to find



(i)



the values of y when x = –0.7 and x = 2.7,



(ii) the values of x when y = 5.3 and y = 7.5. 5. Using a scale of 4 cm to represent 1 unit on the x-axis and 1 cm to represent 1 unit on the y-axis, draw the graph of y = 3x for –2  x  2. (a) Use your graph to find the value of x when y = 5.8.



(b) On the same axes, draw the graph of 1 1 y = x − , x ≠ 0. 2 x (i) Write down the coordinates of the point 1 1 at which the graph of y = x − meets the 2 x curve y = 3x.



(ii) Hence, state the solution of the equation 1 1 3x + − x = 0. x 2



087



Chapter 3



Graphs of Functions and Graphical Solution



Write down the value of k.



7. The table below shows some values of x and the corresponding values of y, where y = (x + 2)(4 – x). x y



(a) Find the value of a and of b.



(b) Using a scale of 4 cm to represent 1 unit on the x-axis and 2 cm to represent 1 unit on the y-axis, draw the graph of y = 2 + 2x for –1  x  3.











–2 0



–1



0



8



1



9



2



3



4 0



(a) Copy and complete the table.



(b) Using a scale of 2 cm to represent 1 unit on both axes, draw the graph of y = (x + 2)(4 – x) for –2  x  4. (c) By drawing a tangent, find the slope of the curve at the point where x = –1.



(d) The slope of the curve at the point (h, k) is zero. (i) Draw the tangent at the point (h, k).



(ii) Hence, find the value of h and of k.



8. (a) Using a scale of 2 cm to represent 1 unit on the x-axis and 2 cm to represent 5 units on the y-axis, draw the graph of y = 12 + 10x – 3x2 for –2  x  5.



(b) Find the slope of the curve when x = 4.



(c) Find the slope of the curve at the point where the curve intersects the y-axis.



10. Using a scale of 4 cm to represent 1 unit on the x-axis and 1 cm to represent 2 units on the y-axis, draw the graph of y = 2 x + 12 for –2  x  3. x (a) (i) On the same axes, draw the line y = 1 – x. (ii) Hence, solve the equation 1 2 x + 2 − 1+ x = 0. x



(b) Explain why the graph of y = 2 x + 12 will not x lie below the x-axis for all real values of x.



9. (a) Using suitable scale, draw the graph of 1 y = 1+ for 0.5  x  3. x (b) On the same axes, draw the line y = –x. (c) Hence, find the coordinates on the graph of y = 1+ 1 at which the slope of the curve x is –1.



3.5



Applications of Graphs in Real-World Contexts



In this section, we will apply our knowledge of coordinate geometry and graphs to analyze and interpret graphs in various real-world contexts, including distancetime and speed-time graphs. Important features of the graphs such as intercepts, slopes, variables and scale of the x- and y-axes will provide information to help us analyze the graphs. For example, the slope of a line segment in a distance-time graph gives the speed. If the graph is a curve, the slope of the curve at a point will give the speed at that instant.



Graphs of Functions and Graphical Solution



Chapter 3



088



Class Discussion Linear Distance-Time Graphs Work in pairs.



Fig. 3.8 shows the graph of a cyclist’s journey between 0800 and 1200. The graph can be divided into 4 sections – 0800 to 0900, 0900 to 0930, 0930 to 1030 and 1030 to 1200. Distance (km) 50 40 30 20 10 0 08 00



09 00



10 00



11 00



12 00



Time (h)



Fig. 3.8 Since the slope of the graph from 0800 to 0900 =



20 km 1h



= 20 km/h,



the cyclist travels at a constant speed of 20 km/h in the first hour.



1. Consider the section of the graph from 0900 to 0930. Since the graph is a horizontal line, what is its slope? State clearly what this slope represents. 2. Find the slope of the section of the graph from 0930 to 1030. What does this slope tell you about the motion of the cyclist? 3. Find the slope of the section of the graph from 1030 to 1200. What does the negative slope represent? Describe briefly the motion of the cyclist. 4. Explain why the average speed of the cyclist cannot be calculated by using 20 + 0 + 30 + 50 km/h. Hence, find the average speed of the cyclist for the 4 whole journey.



089



Chapter 3



Graphs of Functions and Graphical Solution



RE



CAL



L



The average speed of an object is defined as the total distance traveled by the object per unit time.



Thinking Time Match the scenarios with their respective graphs and justify your answers. A: A few years ago, the exchange rate between Singapore dollars and Hong Kong dollars was S$1 = HK$6.



B: The height of water in a uniform cylindrical container increased at a constant rate from 10 cm to 60 cm.



C: Mr Reyes was driving at a constant speed of 60 km/h when he suddenly applied the brakes and came to a stop.



D: The battery level in a smartphone decreased non-uniformly from 60% to 15%.



E: The temperature of a substance in a freezer decreased uniformly from 60 °C to 15 °C in 20 minutes.



F: A plant grew slowly at a constant rate to a height of 15 cm when it was kept indoors for 4 weeks, then grew more quickly at a constant rate to a height of 60 cm when it was placed outdoors for the next 4 weeks.



60



60



15 0



10 0



Graph 1



60



60



15



15



0



0



Graph 3



60



0



Graph 2



Graph 4



60



Graph 5



10



0



Graph 6



Graphs of Functions and Graphical Solution



Chapter 3



090



Worked Example



6



(Distance-Time Curve) A train started from station A and traveled to station B 8 km from A. The table below shows the readings of the time, in minutes, since leaving station A and the corresponding distance, in km, from A. Time (in minutes) Distance (in km)



1



2



3



4



5



6



7



8



0.3 1.1 2.3 4.8 6.8 7.4 7.8 8.0



(a) Using a scale of 2 cm to represent 2 minutes on the horizontal axis and 2 cm to represent 2 km on the vertical axis, plot the points given in the table and join them with a smooth curve. (b) Use your graph to estimate the time taken to travel the first 4 km of the journey. (c) By drawing a tangent, find the approximate speed of the train 5 minutes after it has left station A. (d) By considering the slope of the graph, compare and describe briefly the motion of the train during the first 4 minutes and the last 4 minutes of the journey.



Solution: (a)



Scale: x-axis: 2 cm represent 2 minutes y-axis: 2 cm represent 2 km



Distance from A (km)



8



6



vertical change = 4.5 km



4 horizontal change = 4 minutes



2 0 091



Chapter 3



2



4 6 8 Time since leaving A (minutes)



Graphs of Functions and Graphical Solution



(b) From the graph, the train takes approximately 3.8 minutes to travel the first 4 km. (c) The slope of the tangent at the point 5 minutes after it left station A gives the speed at that particular point. It is called the instantaneous speed. A tangent is drawn to the curve at the point 5 minutes after it has left station A.



From the graph, slope =







4.5 km 4 minutes



=







=







= 67.5 km/h



TE



NTI



ON



The graphical method of finding the slope of a curve yields only approximate results. A slight change in the drawing may give very different results.



vertical change horizontal change







AT



4.5 km 4 h 60



∴ The speed of the train 5 minutes after it has left station A is approximately 67.5 km/h. (d) During the first 4 minutes, the speed of the train increases as the slope of the curve increases.



During the last 4 minutes, the speed of the train decreases as the slope of the curve decreases.



Distance from A (km)



0



4



8



Time since leaving A (minutes)



Graphs of Functions and Graphical Solution



Chapter 3



092



PRACTICE NOW 6



SIMILAR QUESTIONS



A train started from station P and traveled to station Q, 7.6 km from P.



Exercise 3C Questions 1-3, 6-9



The table below shows the readings of the time, in minutes, since leaving station P and the corresponding distance, in km, from P. Time (in minutes) Distance (in km)



1 0.2



2 0.8



3 2.6



4 5.0



5 6.5



6 7.2



7 7.5



8 7.6



(a) Using a scale of 2 cm to represent 2 minutes on the horizontal axis and 2 cm to represent 2 km on the vertical axis, plot the points given in the table and join them with a smooth curve. (b) Use your graph to estimate the time taken to travel the first 4 km of the journey.



(c) By drawing a tangent, find the approximate speed of the train 6 minutes after it has left station P. (d) By considering the slope of the graph, compare and describe briefly the motion of the train during the first 4 minutes and the last 4 minutes of the journey.



Thinking Time Fig. 3.9 shows three containers, each of which is being filled with liquid at a constant rate from a tap to a height of a cm. The containers are initially empty. The graph of Fig. 3.9(a) shows the height (h cm) of the liquid as the container is being filled in t seconds. Height (h cm)



h



Height (h cm)



a



(a)



t1



Time (t s)



a



h



(b)



t2



Time (t s)



Height (h cm)



h



a



(c)



t3



Time (t s)



Fig. 3.9 1. Explain clearly why the graph of Fig. 3.9(a) is a straight line. 2. Complete the graphs in Fig. 3.9(b) and (c). Explain the shape of each graph obtained.



093



Chapter 3



Graphs of Functions and Graphical Solution



SIMILAR QUESTIONS



Exercise 3C Questions 10, 19



Worked Example



7



(Speed-Time Graph) The graph shows the speed of an object over a period of 10 seconds. Speed (cm/s) 10 8 6 4 2 0



4 6 Time (s)



2



8



10



(i) Find the acceleration in the first 2 seconds, (ii) Given that the distance traveled is given by the area under the speed-time graph, find the average speed during the whole journey. (iii) Find the deceleration in the last 2 seconds.



Solution:



AT



(i) Acceleration =



2 cm/s 2s



TE



NTI



ON



For (i), the slope can be found by vertical change or by using the horizontal change y2 − y1 formula x2 − x1 .



= 1 cm/s2



(ii) Total distance = area under graph = Area of (A + B + C + D) 1 1 1 = × 2 × 2 + (2 × 2) + (2 + 8) × 2 + × 8 × 4 2 2 2 = 2 + 4 + 10 + 16 Speed (cm/s) = 32 cm 8 Total distance Average speed = Total time 32 = 10 2 = 3.2 cm/s A B C D



(



)



(



0



2



AT



)



4 6 Time (s)



TE



NTI



ON



Acceleration of an object = rate of change of speed = slope of a speed-time graph



10







If the slope is negative, we say that the object is decelerating. The unit for acceleration is always of speed per unit time, i.e. if the unit of speed is cm/s, then the unit of acceleration is cm/s2.



(iii) Deceleration in the last 2 seconds = deceleration in the last 4 seconds 8−0 = 10 − 6 = 2 cm/s2 Graphs of Functions and Graphical Solution



Chapter 3



094



PRACTICE NOW 7



SIMILAR QUESTIONS



The graph shows the speed of an object over a period of 9 seconds.



Exercise 3C Questions 4, 5, 11, 12



Speed (m/s) 8 6 4 2 0



1



3



2



4



5 6 Time (s)



7



8



9



(i) Find the acceleration in the first 3 seconds. (ii) Given that the distance traveled is given by the area under the speed-time graph, find the average speed during the whole journey. (iii) Find the deceleration in the last 3 seconds.



Worked Example



8



(Speed-Time Graph) A particle moves along a straight line from A to B so that, t seconds after leaving A, its speed, v m/s, is given by v = 3t2 – 15t + 20. The table below shows some values of t and the corresponding values of v. t



v



0



20



1



8



1.5 a



2



2



2.5 b



3



2



4



8



5



20



(a) Find the value of a and of b. (b) Using a scale of 2 cm to represent 1 second on the horizontal axis and 2 cm to represent 5 m/s on the vertical axis, draw the graph of v = 3t2 – 15t + 20 for 0  t  5. (c) Use your graph to estimate (i) the value of t when the speed is 10 m/s, (ii) the time at which the acceleration is zero, (iii) the slope at t = 4, and explain what this value represents, (iv) the time interval when the speed is less than 15 m/s.



095



Chapter 3



Graphs of Functions and Graphical Solution



Solution:



(a) When t = 1.5,



v = 3(1.5)2 – 15(1.5) + 20 = 4.25



∴ a = 4.25



When t = 2.5,



v = 3(2.5)2 – 15(2.5) + 20 = 1.25



∴ b = 1.25



(b) v (m/s) 20 v = 3t2 – 15t + 20 15



10 5



acceleration = 0



0



2



1



3



4



5



t (s)



(c) (i) From the graph, when v = 10, t = 0.8 or t = 4.2.



(ii) The acceleration is zero when the slope of the curve is zero.







From the graph, the acceleration is zero at t = 2.5.



(iii) A tangent is drawn to the curve at the point t = 4.







From the graph,







vertical change slope = horizontal change 14.5 = 1.65



AT



TE



NTI



ON



In (iii), the unit for acceleration is m/s , i.e. m/s2. s



�9



∴ The acceleration of the particle at t = 4 is approximately 9 m/s2.



(iv) From the graph, when v < 15, 0.35 < t < 4.65.



Graphs of Functions and Graphical Solution



Chapter 3



096



PRACTICE NOW 8



SIMILAR QUESTIONS



A particle moves along a straight line from P to Q so that, t seconds after leaving P, its speed, v m/s, is given by v = 2t2 – 9t + 12.



Exercise 3C Questions 13-16, 20-22



The table below shows some values of t and the corresponding values of v. t



v



0



1



12



2



5



3



4



3



a



5



8



b



(a) Find the value of a and of b.



(b) Using a scale of 2 cm to represent 1 second on the horizontal axis and 2 cm to represent 5 m/s on the vertical axis, draw the graph of v = 2t2 – 9t + 12 for 0  t  5. (c) Use your graph to estimate



(i) the values of t when the speed is 7 m/s,







(ii) the time at which the acceleration is zero,







(iii) the slope at t = 4.5, and explain what this value represents,







(iv) the time interval when the speed is less than 10 m/s.



Other Graphs Graphs can also be used in other real-world contexts such as in the calculation of postage rates, parking charges and labor costs.



Worked Example



9



(Graphs involving Rates) The step-function graph below shows the local postage rates for ordinary mails offered by Company A.



Postage (pesos) P So roblem lvin g T ip



60



When the mass m is such that 0  m  20, the postage is 14 pesos. Similarly, for 20  m  50, the postage is 23 pesos.



50 40



AT



30



10



097



Chapter 3



NTI



ON



An empty node indicates that the point is excluded from the graph and a shaded node indicates that the point is included. For example, when the mass is 100 g, the postage is 34 pesos, not 51 pesos.



20



0



TE



50



100



150



Graphs of Functions and Graphical Solution



200



250



Mass (g)



(a) Write down the postage to mail a letter with a mass of 80 g. Company B offers the following postage rates: 25 pesos for the first 50 g and 0.2 pesos for each subsequent gram. (b) Given that Kate wishes to post a letter with a mass of 150 g, insert the graph corresponding to the rates offered by Company B and use your graph to determine which company offers a lower postage.



Solution:



(a) From the graph, the postage is 34 pesos.



(b) For the first 50 g, the postage is 25 pesos.



When the mass is 100 g, the postage is 25 + 0.2 × (100 – 50) = 35 pesos. When the mass is 250 g, the postage is 25 + 0.2 × (250 – 50) = 65 pesos.



Postage (pesos) 70 Company B 60 50 40



Company A



30 20 10 0



50



100



150



200



250



Mass (g)



From the graph, for a mass of 150 g, Company A charges 51 pesos and



Company B charges 45 pesos. Hence, Company B offers a lower postage.



Graphs of Functions and Graphical Solution



Chapter 3



098



PRACTICE NOW 9



SIMILAR QUESTIONS



The step-function graph below shows the parking charges for the first 6 hours at Carpark X. Parking Charges (pesos) 80 60 40 20 0



1



2



3



4



5



6



Duration (h)



Carpark Y has the following charges: Free for the first 12 minutes 25 Centavos per minute thereafter Insert the graph corresponding to the rates offered by Carpark Y and use your graph to determine which carpark Mr Cruz should choose if he has to park for 2 hours.



Worked Example



10



(Other Types of Graphs) The graph below shows the battery level of a smartphone. It had an initial level of 20%, increasing to 60% in half an hour while connected to the power supply. Carlo then removed the smartphone from the power supply to watch a 20-minute long video clip, before connecting the smartphone to the power supply again. Battery level (%) 100



80



60 40 20 0



099



Chapter 3



20 40 60 80 100



Graphs of Functions and Graphical Solution



Time (minutes)



Exercise 3C Question 17



(a) Find the battery level of the smartphone when Carlo was exactly halfway through the video clip. (b) Find the rate of increase in the battery level of the smartphone when it was connected to the power supply again.



Solution:



(a) From the graph, the battery level was 50%. (b) To find the rate of increase in the battery level, we need to calculate the slope of the line from the 50th minute to the 90th minute.



Slope =



= 1.5%/minute







100 − 40 90 − 50



∴ The rate of increase in the battery level is 1.5%/minute.



PRACTICE NOW 10



SIMILAR QUESTIONS



The graph below shows the heart rate, in beats per minute, of an adult who is at the park. He rests at the bench for the first 10 minutes, after which he begins to brisk walk for 10 minutes. He then slows down for 5 minutes, before brisk walking again for a further 5 minutes. He then jogs at a constant speed for 10 minutes, before gradually slowing down.



Exercise 3C Question 18



Heart rate (beats per minute) 140 120 100



80



60 40 20 0



10 20 30 40 50 60



Time (minutes)



(a) Write down his resting heart rate. (b) Find the rate of increase in his heart rate as he brisk walks for the first time. (c) Find the rate of decrease in his heart rate as he slows down in the last



20 minutes.



Graphs of Functions and Graphical Solution



Chapter 3



100



Exercise



3C



4. The graph shows the speed-time graph of a car. 1. A cyclist set out at 0900 for a destination 40 km away. He cycled at a constant speed of 15 km/h until 1030. Then he rested for half an hour before completing his journey at a constant speed of 20 km/h. (i) Draw the distance-time graph to represent the journey. (ii) Hence, find the time at which the cyclist reached his destination, giving your answer to the nearest minute. 2. Carlo starts a 30-km journey at 0900. He maintains a constant speed of 20 km/h for the first 45 minutes and then stops for a rest. He then continues his journey at a constant speed of 30 km/h, finally arriving at his destination at 1120. (i) Find the distance traveled in the first 45 minutes. (ii) Draw the distance-time graph to represent the journey. (iii) Hence, state the duration of his stop, giving your answer in minutes. 3. The figure shows the distance-time graph of a car.



Speed (m/s) 10



0



1



2



3



4



5



Time (s)



6



(i) Find the acceleration in the first 2 seconds. (ii) Given that the distance traveled is given by the area under the speed-time graph, find the average speed during the whole journey. 5. The graph shows the speed, v m/s, of a car after t seconds. Speed (m/s) 14



A



B



Distance (km) 120 100



80



60



0



40 20 0



Chapter 3



30



60



90



120



Time (s)



(i) State what the slope of OA represents. 1



2



3



Time (h)



(i) Find the duration during which the car is not moving. (ii) Find the average speed of the car in the first 2 hours of the journey. (iii) Find the average speed of the car for the whole journey. (iv) Draw the speed-time graph of the car for the whole journey.



101



C



Graphs of Functions and Graphical Solution



(ii) Find the speed of the car when t = 15.



6.



A lift moves from ground level to a height of 60 meters in 10 seconds, stops for 10 seconds and then descends to the ground in 10 seconds. The table shows the height, h m, of the lift on the upward and downward journeys, t seconds after leaving ground level. t (in seconds) h (in m)



0



2



4



6



8



0



3



16 44 57 60



10



t (in seconds) 20 22 24 26 28 30 h (in m)



60 57 44 16



3



0



(i) Using a scale of 2 cm to represent 5 seconds, draw a horizontal t-axis for 0  t  30.



7. A company which manufactures automated vehicles is putting them on a test run. One of the vehicles starts from a point X and travels to a point Y, 3 km away. The table shows the distance, d km, of the vehicle from X, t minutes after leaving X.



Using a scale of 1 cm to represent 5 meters, draw a vertical h-axis for 0  h  60. On your axes, plot the points given in the table and join them with a smooth curve.







(ii) Find the slope of the graph at t = 8 and explain briefly what this slope represents.







A construction worker, waiting at the 40-meters level, starts to walk down at t = 5.



(iii) Assuming that he descends at a steady speed of 0.8 m/s, use your graph to find the time when the worker and the lift are at the same height.



Time (in minutes)



0



Distance (in km)



0



1



2



3



4



5



6



0.2 0.7 1.8 2.5 2.9 3.0



(a) Using a scale of 2 cm to represent 1 minute on the horizontal axis and 4 cm to represent 1 km on the vertical axis, plot the points given in the table and join them with a smooth curve.



(b) Use your graph to find



(i) the approximate time taken to travel the first 1 km, 1 (ii) the slope of the graph when t = 1 and 2 explain briefly what this value represents, (iii) the time taken to travel the last 1 km. 8. Eric and Miguel start moving towards each other at the same time. The initial distance between them is 32 km. (a) Given that Eric is cycling at a constant speed of 20 km/h and Miguel is walking at a constant speed of 7 km/h, draw a distance-time graph to illustrate this information.



(b) Use your graph to find



(i) how long it will take for them to pass each other, (ii) the times when they will be 5 km apart.



Graphs of Functions and Graphical Solution



Chapter 3



102



9. At 0900, Imee travels to meet Kate, who stays 20 km away. Imee travels at a uniform speed of 18 km/h for half an hour. She rests for 20 minutes and then continues her journey at a uniform speed of 8 km/h.



At 0900, Kate sets off from home on the same road to meet Imee and travels at a uniform speed of 7 km/h.



11. The diagram shows the speed-time graph of an object which travels at a constant speed of 36 m/s and then slows down at a rate of 12 m/s2, coming to rest at time t seconds. Speed (m/s) 36



(a) Draw the distance-time graph for the above information. (i) the time at which Imee and Kate meet, (ii) the distance away from Kate’s home when they meet. 10. The figure shows three containers, each with a height of 50 cm and a width of 10 cm. The other dimensions are as shown. The containers are initially empty and it takes 20 seconds to fill each container at a constant rate. 5 cm 20 cm



50 cm



Depth (d cm)







A



B



50 cm



6



Time (s)



t



(i) Find the value of t. (ii) Given that the distance traveled when the object is slowing down is 54 m, find the average speed for the whole journey. 12. The diagram shows the speed-time graph of a train. Speed (m/s) 30



0



20



Time (s)



60



C



(i) Find the acceleration of the train during the first 20 seconds. 10 cm (ii) Given that the train decelerates at a rate of 30 cm 40 cm 20 cm 0.75 m/s2, find the time taken for the whole journey. The diagram below shows the relationship between the depth, d cm, of the liquid and the time, t seconds, 13. A particle moves along a straight line from A to B taken to fill container A. so that, t minutes after leaving A, its speed, v m/min, is given by v = t2 – 7t + 16. 50 30 cm



40



t (minutes)



30



v (m/min)



20 10 0



2



4



6



8 10 12 14 16 18 20 Time (t seconds)



On the same diagram, sketch the graph of the depth of the liquid against time for container B and container C.



103



0



(b) Use your graph to find







Chapter 3



Graphs of Functions and Graphical Solution



0



1



2



3



4



5



6



16



10



6



a



4



6



b



(a) Find the value of a and of b. (b) Using a scale of 2 cm to represent 1 minute on the horizontal axis and 1 cm to represent 1 m/min on the vertical axis, draw the graph of v = t2 – 7t + 16 for 0  t  6. (c) Use your graph to estimate (i) the value(s) of t when the speed is 7 m/min, (ii) the time at which the speed is a minimum, (iii) the slope at t = 2, and explain what this value represents, (iv) the time interval when the speed is not more than 5 m/min.



14. The speed of a body, v m/s, after time t seconds is given in the table. t (s) v (m/s)



0



2



4



6



8



10



12



0



2



7



12



19



28



42



(i) Using a scale of 1 cm to represent 1 second on the horizontal axis and 1 cm to represent 5 m/s on the vertical axis, plot the graph of v against t for 0  t  12.



16.



17. The step-function graph below shows the postage rates to Malaysia for letters offered by Company A. Postage (pesos)



(ii) Use your graph to estimate the speed of the body when t = 5 and when t = 11.



600



(iii) By drawing two tangents, find the acceleration of the body when t = 4 and when t = 10.



500



15. Object P moves along a straight line from A to B so that, t hours after leaving A, its speed, v km/h, is given by v = 3t2 – 17t + 30. t (h) v (km/h)



0



1



2



3



4



5



30



16



h



6



k



20



(a) Find the value of h and of k.



(b) Using a scale of 2 cm to represent 1 hour on the horizontal axis and 2 cm to represent 5 km/h on the vertical axis, draw the graph of v = 3t2 – 17t + 30 for 0  t  5.



(c) Use your graph to estimate



(i)



A taxi starts from rest and accelerates at a uniform rate for 45 seconds to reach a speed of 30 m/s. It then travels at this constant speed. Sketch the speed-time graph and use it to find the speed after 10 seconds.



400 300 200 100 0



50



100



150 200 Mass (g)



250



300



the time at which the speed is a minimum,



(ii) the slope at t = 4.5, and explain what this value represents,



(a) Write down the postage to mail a letter with a mass of 50 g to Malaysia.



(iii) the time interval when the speed does not exceed 10 km/h.



Company B offers the following postage rates to Malaysia: 100 for the first 80 g and 1 peso for each subsequent gram.



(d) Use your graph to determine the value of t at which both objects have the same speed.



(b) Given that Daniel wishes to post a small package with a mass of 220 g to Malaysia, determine which company offers a lower postage. Show your working to support your answer.



Object Q moves along a straight line from A to B with a constant speed of 24 km/h.



Graphs of Functions and Graphical Solution



Chapter 3



104



18. The graph below shows the speed of a coach as it ferried passengers from Blue Town to Summer City via the highway one afternoon. The coach made only one stop during the journey. Speed (km/h)



20. A toy car starts from a point A and moves towards a point B, which it reaches after 7 seconds. The speed, v cm/s, after t seconds, is given in the table. t (s)



v (cm/s)



40



30







20



0



4.5



2



8



3



4



5



6



7



10.5 12 12.5 12 10.5



(b) Use your graph to estimate



(i) the acceleration of the body when t = 2 and when t = 6, 1300



1400



1500



1600



1700



Time (h) (ii) the time interval when the speed is greater than 11 m/s.



(a) Write down the duration of the stop. (b) Find the initial acceleration of the coach. (c) Given that the distance traveled is given by the area under the speed-time graph, explain why the distance between Blue Town and Summer City is less than 250 km. (d) Determine the time when the coach reached Summer City.



(c) Given that this motion can be modeled by the equation v = at2 + bt + c, where a, b and c are constants, find the values of a, b and c. 21. The graph shows the distance-time graph of a body during a period of 6 seconds. Distance (m) 18



19. The diagram below shows three containers of a fixed volume and varying cross-sectional areas, each with a height of 8 cm. The containers are initially empty. A tap is used to fill each of the containers at a constant rate. 2x cm 3z cm 4y cm



8 cm



8 cm



8 cm



4 cm 4x cm



2z cm



4 cm 2y cm



Chapter 3



8



0 4 cm



4z cm



Given that it takes 60 seconds to fill each of the containers, sketch the graph of the height (h cm) of the water level against time (t s) for each of the containers.



105



1



(a) Using a scale of 2 cm to represent 1 second on the horizontal axis and 2 cm to represent 2 cm/s on the vertical axis, plot the graph of v against t for 0  t  7.



50



10 0 1200



0



Graphs of Functions and Graphical Solution







1



2



3



4



5



6



Time (s)



Sketch the speed-time graph for the same journey.



22. The speed of an object, v m/s, at time t seconds, is given by v = 6 + 2t. (a) Sketch the speed-time graph for the motion. (b) Find the speed when t = 3. (c) Sketch the acceleration-time graph for the motion.



1. Graphs of Power Functions y = axn (a) n = 3, a > 0, y = ax3



n = 3, a < 0, y = ax3



y



y



O



x



x



O



(b) n = 2, a > 0, y = ax2



n = 2, a < 0, y = ax2



y







y O



x



O (c) n = 1, a > 0, y = ax y



O



n = 1, a < 0, y = ax y



x



(d) n = 0, a > 0, y = a



O



O



x



n = 0, a < 0, y = a



y







x



y



x



O



x



Graphs of Functions and Graphical Solution



Chapter 3



106



(e) n = –1, a > 0, y =



y







a x



n = –1, a < 0, y = y



x



O



a x



x



O



(f) n = –2, a > 0, y =



y



a x2



O



n = –2, a < 0, y = y



x



a x2



x



O



2. Graph of y = kax, where a > 0



k > 0, y = kax



k < 0, y = kax



y







k



k O



y O



x



3. Slope of a Curve



107



The slope of a curve at a point can be obtained by drawing a tangent to the curve at that point and finding the slope of the tangent.



Chapter 3



Graphs of Functions and Graphical Solution



x



3 1. The table below shows some values of x and the corresponding values of y, where y = x3 – 3x – 10. x y







–3



–28



–2



–1



0



–10



1



2



3



8



4



42







Using a scale of 1 cm to represent 5 units, draw a vertical y-axis for –28  y  42.







On your axes, plot the points given in the table and join them with a smooth curve.



(c) Use your graph to find



(i) the value of y when x = 1.8,



2. The variables x and y are connected by the equation y = x(x – 2)(x + 2).



The table below shows some values of x and the corresponding values of y. x y







–3



–15



–2 0



–1



0



0



1



–3



2



3



(a) Copy and complete the table.



(b) Using a scale of 2 cm to represent 1 unit on the x-axis and 2 cm to represent 5 units on the y-axis, draw the graph of y = x(x – 2)(x + 2) for –3  x  3.



(a) Use your graph to find



(c) Use your graph to find



(ii) the values of x when y = 4.5.



(b) Write down the coordinates of the point on the curve where the tangent to the curve is a horizontal line. 4. Using a suitable scale, draw the graph of y = 3x – 2 for –1.5  x  2. Using your graph,



(ii) the value of x when y = 10.











(i) the value of y when x = –0.75,



(a) Copy and complete the table.



(b) Using a scale of 1 cm to represent 1 unit, draw a horizontal x-axis for –3  x  4.







3. Using a scale of 4 cm to represent 1 unit on the x-axis and 2 cm to represent 1 unit on 1 the y-axis, draw the graph of y = 1 − 2x − for x –4  x  –0.25.



(i) solve the equation 3x = 2,



(ii) find the coordinates of the point on the graph of y = 3 x – 2 where the slope of the tangent is 2.   5. Using a scale of 4 cm to represent 1 unit on 3 both axes, draw the graph of y = x − 2 + for x 0.5  x  4.



(i) State the minimum value of y and the corresponding value of x.



(ii) Find the range of values of x for which y < 2.2.



(iii) By drawing a tangent, find the slope of the graph at the point where x = 3.



(iv) Using your graph, find the value of x for 3 which 2x + = 8. x



(i) the value of y when x = 1.4, (ii) the value of x when y = 4.5, (iii) the solutions to the equation x(x – 2)(x + 2) = 0.



Graphs of Functions and Graphical Solution



Chapter 3



108



6.



A coach traveled from the airport to the hotel in 20 minutes at a constant speed of 45 km/h. After stopping for half an hour, it traveled back to the airport at a constant speed of 60 km/h.



(i) Draw the distance-time graph to represent the journey. (ii) Hence, find the average speed of the coach for the whole journey.



8. The intensity of illumination, I units, at a point on a screen a distance of D cm from the light source k is modeled by the equation I = 2 , where k is D a constant. (i) Using the data in the table, find the value







of k. Distance (D cm) 10 Intensity (I units)



7. The graph shows the distance-time graph of a heavy goods vehicle. Distance (km)



20 25



1250 800



294 200



(ii) Complete the table and use a suitable scale to draw the graph of I against D. (iii) From the graph, find the intensity of illumination when the light source is 30 cm from the screen.



60



(iv) What can you say about the relationship between I and D2? 40



0



1



2



3 3.5



Time (t h)



(a) Find



9. The graph below shows the temperature of soup as it was being heated in an electric cooker. Ten minutes later, Mrs Cruz added some fresh vegetables from the refrigerator, continued to heat the soup until it reached a temperature of 90 °C and switched off the cooker after five minutes.



(i) the time interval during which the vehicle stopped to unload goods,



Temperature (°C)



(ii) the speed when t = 3,



100



(iii) the maximum speed during the journey,



80



(iv) the average speed for the whole journey.



60



(b) Sketch the speed-time graph for the motion.



40 20



0



10



20



30



Time (minutes)



(a) Find the rate of increase in the temperature of the soup in the first ten minutes.



(b) Suggest why there is a drop in the temperature between the 10th and the 11th minute. (c) Find the rate of decrease in the temperature of the soup after the cooker was switched off.



109



Chapter 3



Graphs of Functions and Graphical Solution



Challenge The graph shows an acceleration-time graph of an object traveling in a straight line. Acceleration (m/s2) 2 1 0



2



4



6



Time (s)



–1 –2 –3 Sketch a possible speed-time graph for the motion of the object. Explain your answer.



Graphs of Functions and Graphical Solution



Chapter 3



110



A1



Revision Exercise



1. Consider the sequence 6, 15, 24, 33, 42, … (i) Find, in terms of n, a formula for the nth term of the sequence. (ii) Given that the kth term of the sequence is 159, find the value of k.



7.



2. Given that x + 2 is a factor of F(x) = A(x – 1)2 + B(x – 1) + C and that when F(x) is divided by x – 1 and x + 1, the remainders are 9 and –11 respectively, find the values of the constants A, B and C.



The travel graphs below show the journeys of Antonio and Eric. Antonio starts from the train station at 08 00 and travels towards the airport, 40 km away. Jun Wei starts from the airport at 09 00 and travels towards the train station.



Distance in km



40



3. When x2 – bx + 3 and 2b – x are divided by x – a, the remainders are 1 and 4 respectively. Find the value of a and of b. 4. Solve the equation 3x3 + x2 – 15x + 2 = 0, giving your answers correct to two decimal places where necessary. 5. Express each of the following in partial fractions.



(a)



8 x – 27 ( x – 3)( x – 4 )



5x + 9 (b) ( x + 3)2



6. The diagram shows the speed-time graph of a particle over a period of t seconds. Speed (m/s) V



8 Time (s)



Revision Exercise A1



Travel graph of Antonio



20 10



08 00 09 00 10 00 11 00 12 00 13 00 Time From the graph, find (i) Eric average speed for the whole journey, (ii) the time when Antonio and Eric meet and how far they are from the airport when they meet, (iii) the time interval during which Antonio took a rest, (iv) the distance between Antonio from the airport when Eric reaches the train station.



x



t



(i) Given that the acceleration of the particle during the first 8 seconds of its motion is 1.5 m/s2, find the value of V. (ii) Given that the retardation of the particle is 1 m/s2, find the value of t.



111



30



8. The variables x and y are connected by the 12 +x−6. equation y = x The table below shows some values of x and the corresponding values of y.



4 0



Travel graph of Eric



y



1



7



1.5 h



2



2



3



1



4



1



5



1.4



6



2



7 k



8



3.5



(a) Find the value of h and of k. (b) Using a scale of 2 cm to represent 1 unit 12 on each axis, draw the graph of y = +x−6 x for 1  x  8. (c) Use your graph to find (i) the value of y when x = 2.3, (ii) the minimum value of y. (d) By drawing a tangent, find the slope of the curve at the point where x = 5. (e) Use your graph to obtain one solution of the equation x2 + 12 = 10x.



A2



Revision Exercise



1. Consider the sequence 44, 41, 38, 35, 32, … (i) Find, in terms of n, a formula for the nth term of the sequence. (ii) Given that the kth term of the sequence is –13, find the value of k. 2. Consider the following number pattern: 12 – 2 × 1 = –1 22 – 2 × 2 = 0 32 – 2 × 3 = 3 42 – 2 × 4 = 8 k2 – 2k = 63 (i) Write down the 6th line in the pattern. (ii) Find the value of k. 3. Given that 3x – 11x + 7 = Ax(x – 2) + B(x – 2) + C for all values of x, find the values of A, B and C. 2



4. When x4 + ax3 + bx2 + 6 is divided by x – 1 and x – 2, the remainders are 12 and 54 respectively. Find the value of a and of b. Hence, find the remainder when the expression is divided by x + 1. 5. If x – 3 is a factor of x3 – x2 + ax + 9, find the value of a. Hence, factorize the expression completely. 6. Express each of the following in partial fractions.



(a)



2 x 2 + 6 x + 15 ( x – 2 )( x 2 + 3)



(b)



2x2 – 4 x( x – 4 )



7. A coach leaves Watertown for Sandcity 120 km away at 11 00 and travels at a uniform speed of 50 km/h. An hour later, a car traveling at a uniform speed of 80 km/h leaves Sandcity for Watertown by the same route. (a) Draw the distance-time graph to represent the journey. (b) Use your graph to find (i) the time when the car meets the coach and the distance from Watertown at this instant, (ii) the distance between the coach and the car at 13 00.



8.



The diagram shows the speed-time graph of a car which decelerates uniformly from 45 m/s to 27 m/s in 30 seconds. It then travels at a constant speed of 27 m/s for 30 seconds. Speed (m/s) 45 27



0



20



50



Time (s)



(i) Given that the car begins to decelerate uniformly at 0.6 m/s2 until it comes to rest, find the total time taken for the journey. Give your answer in minutes and seconds. (ii) Sketch the acceleration-time graph for the motion. 9. A string of length 15 cm is used to form a rectangle. Given that the area of the rectangle is y cm2 and that one side of the rectangle is x cm long, 1 show that y = x (15 − 2x ) . 2 The table below shows some values of x and the corresponding values of y. x y



0.5



3.5



1



6.5



2



11



3



13.5



4



14



5



12.5



6



9



6.5 p



(i) Find the value of p. (ii) Using a scale of 2 cm to represent 1 unit, draw a horizontal x-axis for 0  x  7. Using a scale of 1 cm to represent 1 unit, draw a vertical y-axis for 0  y  16. On your axes, plot the points given in the table and join them with a smooth curve. (iii) Use your graph to find the solutions of the equation x(15 – 2x) = 23. (iv) By drawing a tangent, find the slope of the curve at the point (5, 12.5). (v) Use your graph to find the maximum value of y and the corresponding value of x. (vi) Hence, write down the dimensions of the rectangle when the area is a maximum. What can you say about your answer? Revision Exercise A2



112



Arc Length, Area of Sector and Radian Measure The cross section of train tunnels may be in the shape of a major segment of a circle. What is meant by a ‘major segment of a circle’? How do we find the cross-sectional area of such a train tunnel?



Four



LEARNING OBJECTIVES At the end of this chapter, you should be able to: • find the arc length of a circle by expressing the arc length as a fraction of the circumference of the circle, • find the area of the sector of a circle by expressing the area of a sector as a fraction of the area of the circle, • find the area of a segment of a circle, • convert angular measure from radians to degrees and vice versa, 1 • use the formulae s = rθ and A = r 2θ to solve 2 problems involving arc length, area of a sector and area of a segment of a circle.



4.1



Length of Arc



S



S



O



major sector O



P



Q



P



minor sector



S major segment O Q



P



Q



R



R



R



(a)



(b)



(c)



minor segment



Fig. 4.1 Fig. 4.1(a) shows a circle with center O. The line PQ is called a chord. PRQ is part of the circumference which is called an arc. The arc PRQ is called the minor arc and the arc PSQ is called the major arc. The part of a circle enclosed by any two radii of a circle and an arc is called a sector. In Fig. 4.1(b), the region enclosed by the radii OP, OQ and the minor arc PRQ is called a minor sector of the circle. The region enclosed by the radii OP, OQ and the major arc PSQ is called a major sector of the circle. In Fig. 4.1(c), the chord PQ divides the circle into two segments. The region enclosed by the chord PQ and the minor arc PRQ is called a minor segment. The region enclosed by the chord PQ and the major arc PSQ is called the major segment.



Investigation Arc Length



Go to http://www.shinglee.com.sg/StudentResources/and open the geometry software template ‘Arc Length‘ as shown. 1. The template shows a circle with ∠AOB subtended by the (blue) arc AB at the center O.



115



Chapter 4



Arc Length, Area of Sector and Radian Measure



Fig. 4.2 2. Click and drag point B to change the size of the (blue) arc AB. To change the radius of the circle, move point A. Copy and complete Table 4.1. No.



Length of Circumference Length of Blue Arc AB Circumference of Circle ∠AOB Blue Arc AB of Circle



∠AOB 360°



(a) (b) (c) (d) (e) Table 4.1 3. What do you notice about the third last column and the last column of Table 4.1? 4.



Click on the button ‘Show how to do animation’ in the template and it will show you how to add 10 more entries to the table as the points A and B move automatically. What do you notice about the third last column and the last column of the table in the template?



5. Hence, write down a formula for finding the length of an arc of a circle.



From the investigation, we observe that O x° arc length = × circumference 360° r x° r x° = × 2πr , 360° A B where x° is the angle subtended by the arc at the center of the circle of radius r.



Arc Length, Area of Sector and Radian Measure



Chapter 4



116



Worked Example



1



(Finding the Arc Length) In the figure, O is the center of a circle of radius 12 cm and ∠AOB = 124°. X A 124°



O 12 cm B Y Find (i) the length of the minor arc AXB,



(ii) the perimeter of the major sector OAYB.



Solution:



(i) Length of minor arc AXC =



124° × 2π × 12 360°



= 26.0 cm (to 3 s.f.)



(ii) Perimeter of major sector = length of arc AYB + OA + OB 360° − 124° × 2π × 12 + 12 + 12 = 360° = 73.4 cm (to 3 s.f.)



PRACTICE NOW 1



Y 228°



O 25 cm B



A



X



Find (i) the length of the major arc AYB,



(ii) the perimeter of the minor sector OAXB.



Chapter 4



Alternatively, Length of arc AYB = 2πr – length of arc AXB.



SIMILAR QUESTIONS



1. In the figure, O is the center of a circle of radius 25 cm and reflex ∠AOB = 228°.



117



P So roblem lvin g T ip



Arc Length, Area of Sector and Radian Measure



Exercise 4A Questions 1(a)-(d), 2(a)-(c), 3(a)-(b), 4(a)-(d), 5-9



2. In the figure, O is the center of a circle of radius 9 cm and ∠AOB = 150°.



A



O 150°



9 cm B



Find the perimeter of the shaded region, giving your answer in the form a + bπ, where a and b are rational numbers. 3. The figure shows the design of a logo in which a sector has been removed from the circle, center O and radius r cm.



X







O



P



50° r cm Q



Given that the length of the major arc PXQ is 36 cm and ∠POQ = 50°, find the value of r.



Worked Example



2



(Finding the Arc Length) In the figure, O is the center of a circle of radius 8 m. The points A and C lie on the circumference of the circle and OCB is a straight line.



O 8 m 45° A



C



B



Given that ∠AOB = 45° and OA is perpendicular to AB, find the perimeter of the shaded region ABC.



Arc Length, Area of Sector and Radian Measure



Chapter 4



118



Solution:



∠ABC = 180° – 90° – 45°



= 45°



i.e. ΔAOB is isosceles and AB = 8 m. Length of arc AC =



45° × 2π × 8 360°



= 2π m



AT



OB2 = 82 + 82 =



NTI



ON



For accuracy, we should work in terms of π in the intermediate steps.



Using Pythagorean Theorem, OB =



TE



82 + 82



128



= 11.31 m (to 4 s.f.)



BC = OB – OC



= 11.31 – 8



AT



= 3.31 m



∴ Perimeter of shaded region ABC = AB + BC + length of arc AC = 8 + 3.31 + 2π







ON



In order for the final answer to be accurate to three significant figures, any intermediate working must be correct to four significant figures.



SIMILAR QUESTIONS



1. In the figure, O is the center of a circle of radius 8 m. The points P and R lie on the circumference of the circle and OPQ is a straight line. O 8m R



36.9° P Q



Given that ∠QOR = 36.9° and OR is perpendicular to QR, find the perimeter of the shaded region PQR.



2. The figure shows a sector of a circle of radius 10 cm. Given that the angle at the center of the circle is 80°, find the perimeter of the sector. 10 cm 80° 10 cm



119



NTI



= 17.6 m (to 3 s.f.)



PRACTICE NOW 2







TE



Chapter 4



Arc Length, Area of Sector and Radian Measure



Exercise 4A Questions 10-12, 14



Worked Example



3



(Finding the Arc Length) In the figure, AQB is the minor arc of a circle with center O and radius 17 cm. ARB is a semicircle with AB as its diameter and P as its center. ΔOPB is a right-angled triangle with ∠OPB = 90° and ∠POB = 59°. R Q P



A



59° O



B 17 cm



Find (i) the length of AB, (ii) the perimeter of the shaded region.



Solution: (i) In ΔOPB,



PB OB PB sin 59° = 17



AT



sin ∠POB =







NTI



ON



In ∆OPB, PB is the side opposite ∠POB and OB is the hypotenuse.



PB = 17 sin 59°







TE



= 14.57 cm (to 4 s.f.)



∴ AB = 2PB







= 2(14.57)



= 29.1 cm (to 3 s.f.)



(ii) ∠AOB = 2∠POB







118° × 2π × 17 360°



= 35.01 cm (to 4 s.f.)



Length of arc ARB =







= 118°



Length of arc AQB =







= 2(59°)



1 × 2π × 14.57 2



= 45.77 cm (to 4 s.f.)



∴ Perimeter of shaded region = 35.01 + 45.77







= 80.8 cm (to 3 s.f.)



Arc Length, Area of Sector and Radian Measure



Chapter 4



120



PRACTICE NOW 3



SIMILAR QUESTIONS



In the figure, PAQ is the minor arc of a circle with center O and radius 35 cm. PBQ is a semicircle with PQ as its diameter and R as its center. ΔORQ is a right-angled triangle with ∠ORQ = 90° and ∠RQO = 36°.



Exercise 4A Questions 13, 15



B A R



P



36°



Q



35 cm O



Find (i) the length of PQ,



(ii) the perimeter of the shaded region.



Exercise



4A



2. For each of the following circles, find 1. Find the length of each of the following arcs AXB.



A



(a) O



82°



X



(b) O



14 cm B 134°



A



8 cm B







(i) the length of the minor arc AXB,



(ii) the perimeter of the major sector OAYB.



(a)



9 cm



Y



X



O



214° O 17 cm A



X



(d)



(b)



Y O



O



B A



X



76°



B



(c) X



A



46° 9.8 cm B



A



16 cm 112° B X



(c)



Y







215° O A 17.6 cm X



121



Chapter 4



Arc Length, Area of Sector and Radian Measure



B



3. Find the radius of each of the following circles.



(a)







7. Find the radius of each of the following circles.



(a) O



O A



95°







B



Length of minor arc = 26.53 cm



(b)



214° A



O







B



148°



Perimeter of minor sector = 77.91 cm



(b)







O



B A







A



Length of major arc = 104.6 cm







44°



B Perimeter of major sector = 278.1 cm



  4. The radius of a circle is 14 m. Find the angle at the center of the circle subtended by an arc of length



(a) 12 m,



(c) 64.2 m,



(b) 19.5 m,



8. The figure shows two sectors OAB and OPQ with O as the common center. The lengths of OA and OQ are 8 cm and 17 cm respectively.



(d) 84.6 m,







giving your answers correct to the nearest degree.



5.



The hour hand of a large clock mounted on a clock tower travels through an angle of 45°. If the hour hand is 1.5 m long, how far does the tip of the hour hand travel?



6. A piece of wire 32 cm long is bent to form a sector of a circle of radius 6 cm. Find the angle subtended by the wire at the center of the circle.



O



8 cm



A



P



60°



B 17 cm



Q



Given that ∠AOB = ∠POQ = 60°, find the perimeter of the shaded region, giving your answer in the form a + bπ, where a and b are rational numbers. 7 of 9. In the figure, the length of the minor arc is 24 the circumference of the circle.



O A



B







(i) Find ∠AOB.







(ii) Given that the diameter of the circle is 14 cm, find the length of the minor arc.



Arc Length, Area of Sector and Radian Measure



Chapter 4



122



10. In the figure, O is the center of a circle of radius 7.5 cm. The points A and B lie on the circumference of the circle and OBP is a straight line.



13. In the figure, ABD is the minor arc of a circle with center O and radius 9 cm. ACD is the minor arc of a circle with center P and radius 16 cm. The point X lies on AP such that AX = XP. A



O 7.5 cm



B P



14 cm



B



A



X P O 9 cm 16 cm



C







Given that PA = 14 cm and OA is perpendicular to AP, find







(i) ∠POA,







(ii) the perimeter of the shaded region PBA.







(i) ∠APB,







(ii) ∠AOB,



11. The figure shows a circle with center O and radius 26 cm. Triangles OPR and OQR are congruent and ∠OPR = ∠OQR = 90°.







(iii) the perimeter of the shaded region.



P 26 cm 138° O



R



D Find



14. In the figure, O is the center of a circle. The points A and B lie on the circumference of the circle 15 such that AB = 3 cm and ∠OAB = 30°. 2



Q



O



Given that ∠POQ = 138°, find







(i) the length of QR,







(ii) the perimeter of the shaded region.



12. The figure shows a sector of a circle with center O and radius 13 cm. O 13 cm A







22 cm



B



A







B



Find the perimeter of the shaded region.



15. The figure shows a semicircle with center O and diameter SR. QR is an arc of another circle with center T and T lies on RS produced. Q



Given that the length of the chord AB = 22 cm,







(i) show that ∠AOB is approximately 115.6°,







(ii) find the perimeter of the shaded region.



P



T



123



30° 15 3 cm 2



Chapter 4



Arc Length, Area of Sector and Radian Measure



36°



S 14 cm



O



R



Given that TO = 14 cm, ∠OTP = 36° and OP is perpendicular to PT, find the perimeter of the shaded region.



4.2



Area of Sector



Investigation Area of Sector



Go to http://www.shinglee.com.sg/StudentResources/ and open the geometry software template ‘Sector Area‘ as shown below. 1. The template shows a circle with center O and a (yellow) sector OAB.



Fig. 4.3 2. Click and drag point B to change the size of the (yellow) sector OAB. To change the radius of the circle, move point A. Copy and complete Table 4.2. No.



Area of Shaded Sector OAB



Area of Circle



Area of Sector OAB ∠AOB Area of Circle



∠AOB 360°



(a) (b) (c) (d) (e) Table 4.2



Arc Length, Area of Sector and Radian Measure



Chapter 4



124



3. What do you notice about the third last column and the last column of Table 4.2? 4.



Click on the button ‘Show how to do animation’ in the template and it will show you how to add 10 more entries to the table as the points A and B move automatically. What do you notice about the third last column and the last column of the table in the template?



5. Hence, write down a formula for finding the area of a sector of a circle.



From the investigation, we observe that



x° × area of the circle 360° x° × πr 2 , = 360°



O



area of a sector =



A



r x° r Sector



B



where x° is the angle of the sector subtended at the center of the circle of radius r.



Worked Example



4



(Finding the Area of a Sector) In the figure, O is the center of a circle of radius 26 cm. The length of minor arc AQB is 52 cm. P O A



26 cm B



Q (i) Show that ∠AOB is approximately 114.6°.



(ii) Hence, find the area of the major sector OAPB.



Solution:



(i) Since the length of minor arc AQB is 52 cm, ∠AOB x° × 2π × 26 = 52 (use arc length = × 2πr ) 360° 360° ∠AOB 52 = 360° 52π 1 = π 1 ∠AOB = × 360° π = 114.6° (to 1 d.p.)



125



Chapter 4



Arc Length, Area of Sector and Radian Measure



(ii) Reflex ∠AOB = 360° – 114.59° (∠s at a point)



= 245.4°



Area of major sector OAPB =







245.4° × π × 26 2 360°



= 1450 cm2 (to 3 s.f.)



PRACTICE NOW 4



SIMILAR QUESTIONS



In the figure, O is the center of a circle of radius 15 cm. The length of minor arc AQB is 33 cm. P O A



Exercise 4B Questions 1(a)-(f), 2(a)-(c), 3(a)-(d), 5(a)-(b), 6(a)(d), 7-9



15 cm B



Q



(i) Show that ∠AOB is approximately 126.1°.



(ii) Hence, find the area of the major sector OAPB.



Worked Example



5



(Finding the Area of a Sector) In the figure, ABCD is a trapezoid in which AD is parallel to BC, AD = 7 cm, BC = 10 cm and ∠ADC = ∠BCD = 90°. PXD is an arc of a circle center A and PYC is an arc of a circle center B. A 7 cm D



P X Y



B 10 cm C



(i) Show that ∠BAD is approximately 100.2°. (ii) Hence, find the area of the shaded region.



Arc Length, Area of Sector and Radian Measure



Chapter 4



126



Solution:



(i) Draw a line AT such that T lies on BC and AT is perpendicular to BC.



A



7 cm P



7 cm D



X



10 cm



Y



B 3 cm T 7 cm C



BT = BC – TC



= 10 – 7 = 3 cm



AB = 7 + 10 (AB = AP + PB) = 17 cm In ΔATB,



sin ∠BAT =



3 17



∠BAT = sin –1



AT



3 17



= 10.16° (to 2 d.p.)



∠BAD = 90° + 10.16°







= 100.16°







= 100.2° (to 1 d.p.)



(ii) Using Pythagorean Theorem,



AT2 = 172 – 32 AT =



=



17 2 − 32 280



= 16.73 cm (to 4 s.f.)







∠ABT = 180° – 90° – 10.16°







Area of shaded region







= 79.84° (to 2 d.p.)



= Area of trapezoid ABCD – area of sector APD – area of sector BCP 1 100.2° 79.84° × π × 72 − × π × 10 2 = (7 + 10)(16.73) − 2 360° 360° = 29.7 cm2 (to 3 s.f.)







127



Chapter 4



Arc Length, Area of Sector and Radian Measure



TE



NTI



ON



In ∆ATB, BT is the side opposite ∠BAT and AB is the hypotenuse.



PRACTICE NOW 5



SIMILAR QUESTIONS



In the figure, ABCD is a trapezoid in which AB is parallel to DC, AB = 19 m, DC = 15 m and ∠ABC = ∠DCB = 90°. PXB is an arc of a circle center A and PYC is an arc of a circle center D. A 19 m B



P X Y



Exercise 4B Questions 10-13



D 15 m C



(i) Show that ∠ADC is approximately 96.8°. (ii) Hence, find the area of the shaded region.



Exercise



4B



2. For each of the following circles, find 1. Copy and complete the table for sectors of a circle. Angle Arc Area Perimeter Radius at length center 72° (a) 7 cm 136 mm (b) 35 mm 2 270° 1848 mm (c) 150° 220 cm (d) 55 m (e) 14 m 75° 154 cm2 (f)







(i) the perimeter,







(ii) the area,







of the minor sector.







(a)







(b)



7 cm 30° O







O







3.5 cm



340° (c) 140°



O 6 cm



Arc Length, Area of Sector and Radian Measure



Chapter 4



128



3. The figure shows a circle with center O and ∠AOB = θ°. The circumference of the circle is 88 cm.







B C



θ°



O



6. The diameter of a circle is 18 cm. Find the angle subtended by the arc of a sector with each of the following areas. (a) 42.6 cm2



(b) 117.2 cm2



(c) 214.5 cm2



(d) 18.9 cm2



A



Find the length of arc ACB and the area of sector OACB for each of the following values of θ. (a) 60



7. The figure shows two sectors OAB and ODC with O as the common center.



(b) 99



(c) 126



(d) 216



r cm



4. The figure shows a circle with center O and ∠POQ = x°. The area of the circle is 3850 cm2. S



Q



x°



O



R cm



Find the area of sector OPSQ and the length of arc PSQ for each of the following values of x.



(c) 108



(d) 198



(a)







(iii) r = 35, R = 49, m = 160



8. In the figure, the area of the shaded sector POQ 5 is of the area of the whole circle. 18 Q O P



Area of minor sector = 114 cm2







(i) Find ∠POQ.



(ii) Given that the area of the shaded sector is 385 cm2, find the diameter of the circle.



66°



129







(ii) r = 5, R = 8, m = 120



150°



(b)











(i) r = 10, R = 20, m = 45



O







C



(b) 84



5. Find the radius of each of the following circles.



B



Given that OA = r cm, OC = R cm and ∠AOB = m°, find the perimeter and the area of the shaded region ABCD for each of the following cases.



(a) 36



m°



P



O







D



A



O



9.



Area of major sector = 369 cm2



Chapter 4



Arc Length, Area of Sector and Radian Measure



During an Art lesson, students are required to make a shape in the form of a sector of a circle of radius 12 cm. If the perimeter of the shape is 38 cm, find the area of the paper used in making the shape.



10. The figure shows two circular discs of radii 11 cm and 7 cm touching each other at R and lying on a straight line MPQN.



A 11 cm M



P



R



B 7 cm Q



P N



Q







(i) Show that ∠PAB is approximately 77.2°.







(ii) Hence, find the area of the shaded region.



11. Two circular discs of radii 4p cm and p cm touch each other externally and lie on a straight line AB as shown. 4p cm



A



X



p cm



12. In the figure, O is the center of a circle with radius 16 cm. PYB is the minor arc of a circle center A and radius 32 cm. OX divides ΔOAQ into two congruent triangles.



Y B



Find an expression, in terms of p, for the area enclosed by the two discs and the line AB.



Y



X A 66° B 16 cm O 16 cm







Given that ∠OAQ = 66°, find



(i) ∠BOQ,



(ii) the length of AQ,







(iii) the perimeter of the shaded region,







(iv) the area of the shaded region.



13. The figure shows a quadrant of a circle of radius 12 cm. A B 12 cm



C



D



O



Given that B is the midpoint of the arc AC, find







(i) the length of BD,







(ii) the perimeter of the shaded region,







(iii) the area of the shaded region.



Arc Length, Area of Sector and Radian Measure



Chapter 4



130



4.3



Radian Measure



Conversion Between Degrees and Radians



So far we have been using the measurement of 360° to denote the angle for one complete revolution. However, this value is arbitrary and in some branches of mathematics, angular measurement cannot be conveniently done in degrees. Hence, a new unit, called the radian, is introduced to describe the magnitude of an angle. Consider a circle with center O and radius r units (see Fig. 4.4(a)). Suppose the arc AB has a length of r units.



Then the measure of the angle θ, subtended by the arc AB at the center O, is defined to be one radian. In other words, a radian is the size of the angle subtended at the center of a circle by an arc equal in length to the radius of the circle. B r



θ O r



s



Q



r



r



A



O



(a)



θ



r



P



(b) Fig. 4.4



In general, the angle θ, in radians, subtended at the center of a circle with radius r units by an arc PQ of length s units, is equal to the ratio of the arc length to the radius: θ (in radians) = Since



s r



s is a ratio, θ has no units, i.e. radian is not a unit. r



The abbreviation of ‘radian’ is ‘rad’.



For example, if the radius of the circle is 2 cm and arc PQ = 2 cm, then θ = 1 rad; if the radius of the circle is 2 cm and arc PQ = 4 cm, then θ = 2 rad.



131



Chapter 4



Arc Length, Area of Sector and Radian Measure



INF



OR



MA



TIO N



Th i s s y s t e m o f a n g u l a r measurement involving radians, is applied especially to those branches of mathematics which involve the differentiation and integration of trigonometric ratios.



Investigation Visualize the Size of an Angle of 1 radian



For this activity, you will require a sheet of paper, a pair of compasses and a piece of string (5 cm long).



1. On a sheet of paper, use a rule and a pair of compasses to draw a circle, center O and with radius 5 cm. 2. Mark a point A on the circumference, as shown in Fig. 4.5.



3. Using a piece of string of length 5 cm, mark a point B on the circumference such that the length of arc AB is 5 cm.



4. Consider the minor sector AOB. Notice that the radius is equal to the arc length. What is the value of θ in radians? C 5 cm B 5 cm 5 cm 5 cm D A O 5 cm



RE



CAL



L



θ (in radians) =



s r



Fig. 4.5 5. Repeat Step 3 to mark out arcs BC, CD, DE, … of length 5 cm each around the circle. 6. (a) Estimate the size of the angle, in radians, subtended at the center by (i) a semicircle, (ii) a circle. (b) Using this estimate, what is the approximate size of ∠AOB (1 radian) in degrees?



(c) Hence, what is an angle of 1 radian approximately equal to in degrees?



7. (a) Using the formula for the circumference of a circle, find the exact value of the angle at the center of the circle in radians. (b) Hence, find the exact value of an angle of 1 radian in degrees.



In general, for a circle of radius r units,



if arc PQ = r, then θ = 1 rad;



INF



if arc PQ = 2r, then θ = 2 rad;



When arc PQ = 2πr, it means that the arm OP has made one complete revolution, i.e. OM has moved through an angle of 360°, i.e. 2π rad = 360°.



180° ≈ 57.3° π π 1° = ≈ 0.017 46 rad 180°



1 rad =



MA



TIO N



The table gives a conversion table for some special angles.



if arc PQ = 2πr, then θ = 2π rad.



π rad = 180°



OR



Angle (°) 30 45 60 90



Arc Length, Area of Sector and Radian Measure



Angle (rad) π 6 π 4 π 3 π 2



Chapter 4



132



Class Discussion Estimate the Size of Angles in Radians Work in pairs.



Look at the figures below.



θ θ







(a)



(b)



θ θ







(c)



(d)



θ



θ



(e)



133



(f)



Identify which figure shows an angle θ of







(i) 1 radian,



(ii) 2 radians,



(iii) 3 radians,







(iv) 6 radians,



(v) 0.5 radians,



(vi) 1.5 radians.



Chapter 4



Arc Length, Area of Sector and Radian Measure



Worked Example



6



(Converting from Radians to Degrees) Convert each of the following angles from radians to degrees. π (a) (b) 2.8 12



Solution:



(a) Since π radians = 180°, π 180° rad = 12 12 = 15°



(b) Since π radians = 180°,



1 radian =



180° . π



2.8 radians = 2.8 × 180° π = 160.4° (to 1 d.p.)



PRACTICE NOW 6



7



Gottfried Wilhelm von Leibnitz (1646-1716) was a German genius who won recognition in law, philosophy, religion, literature, metaphysics and mathematics. He developed a method of calculating π without reference to π a circle. He proved that could 4 be determined to any desired degree of accuracy by the formula: π = 1 − 1 + 1 − 1 + 1 − 1 + ... 4 3 5 7 9 11



SIMILAR QUESTIONS



Convert each of the following angles from radians to degrees. π 3π (a) (b) 15 2 (c) 3.04 (d) 8



Worked Example



un



Just For F



Exercise 4C Questions 1(a)-(d)



(Converting from Degrees to Radians) Convert each of the following angles from degrees to radians. (a) 27°



(b) 315°



Solution:



(a) Since 180° = π radians, π 27° = × 27° 180° = 0.471 radians (to 3 s.f.) (b) Since 180° = π radians, π 315° = × 315° 180° 7π = 4 = 5.50 radians (to 3 s.f.)



PRACTICE NOW 7



SIMILAR QUESTIONS



Convert each of the following angles from degrees to radians. (a) 36°



(b) 288°



(c) 197.5°



(d) 400°



Exercise 4C Questions 2(a)-(d), 3(a)-(d)



Arc Length, Area of Sector and Radian Measure



Chapter 4



134



Use of Calculators Worked Example



8



(Use of a Calculator)



INF



Find the value of each of the following. (a) sin 1.2 (c) tan 1.012



(b) cos 0.879



Solution:



OR



MA



TIO N



Most scientific calculators enable us to find the value of a trigonometric function in both radians and degrees by setting the mode to either ‘radian’ or ‘degree’.



(a) To find the value of sin 1.2, first set the calculator to the ‘radian’ mode. Press sin 1 . 2 = and the display shows 0.932 039 086, i.e. sin 1.2 = 0.932 (to 3 s.f.)



(b) To find the value of cos 0.879, press cos 0 . 8 7 9 = and the display shows 0.637 921 564, i.e. cos 0.879 = 0.638 (to 3 s.f.)



(c) Press tan 1 . 0 1 2 = and the display shows 1.599 298 86, i.e. tan 1.012 = 1.60 (to 3 s.f.)



PRACTICE NOW 8



SIMILAR QUESTIONS



Find the value of each of the following. (a) sin 0.65 (b) cos 0.235



Worked Example



9



Exercise 4C Questions 4(a)-(f)



(c) tan 1.23



(Use of a Calculator) For each of the following, find the value of x in the range π 0 0. Then



CP = r



(a) Consider ∆CPQ. Using Pythagorean Theorem, write down an equation in terms of x, y, a, b and r. (b) If a = 0 and b = 0, (i) state the coordinates of the center of the circle, (ii) write down the equation of the circle for this particular case.



In summary, the standard form of the equation of a circle with center C(a, b) and radius r is given by: (x − a)2 + (y − b)2 = r2, where r > 0.



215



Chapter 6



Coordinate Geometry



Class Discussion Plotting Graphs using Equation of Circle in Standard Form



1. Use a suitable graphing software to plot the circles whose equations are given below. Copy and complete the table by finding the coordinates of the center of each circle and its radius. No.



Equation of Circle



(a)



x2 + y2 = 9



(b)



(x − 1)2 + (y − 3)2 = 16



(c)



(x + 2)2 + (y + 4)2 = 25



(d)



x2 + (y − 5)2 = 1



(e)



⎛ 1⎞ 2 ⎜x – ⎟ + (y + 1.5) = 4 4 ⎝ ⎠



Coordinates of Center of Circle



Radius of Circle



2



2. By observing the second and the last columns of the table, how do you find the radius of a circle just by looking at its equation? 3. By observing the second and the third columns of the table, how do you find the coordinates of the center of a circle just by looking at its equation? 4. The standard form of the equation of a circle is (x − a)2 + (y − b)2 = r2, where



r > 0.



(i) Discuss with your classmate the effect of increasing the value of a on the circle. Explain your answer. (ii) Discuss with your classmate the effect of decreasing the value of b on the circle. Explain your answer.



Coordinate Geometry



Chapter 6



216



Worked Example



6



(Finding Coordinates of the Center and the Radius of a Circle) Write down the coordinates of the center and the radius of each of the following circles. (b) (x – 2)2 + (y + 3)2 = 25 (a) x2 + y2 = 36 2 ⎛ 1⎞ (c) ⎜x – ⎟ + y2 – 49 = 0 2⎠ ⎝



Solution:



(a) Center (0, 0)



Radius =



36 = 6



(b) Center (2, −3)



Radius =



25 = 5



2 ⎛ 1⎞ (c) ⎜x – ⎟ + y2 – 49 = 0 2⎠ ⎝







2



⎛ 1⎞ 2 ⎜x – ⎟ + y = 49 2⎠ ⎝



⎛ 1 ⎞ Center ⎜– , 0 ⎟ ⎝ 2 ⎠







Radius =



49 = 7



PRACTICE NOW 6



Write down the coordinates of the center and the radius of each of the following circles. (b) (x + 4)2 + (y – 6)2 = 100 (a) x2 + y2 = 81 2 ⎛ 1⎞ (c) x2 + ⎜y + ⎟ – 16 = 0 3⎠ ⎝



Thinking Time A satellite is an object that orbits the earth. (a) Given that a weather satellite orbits the earth such that its distance from the center of the earth is always 9000 km, write down an equation that it satisfies, in terms of x and y, where x and y are the longitudinal and latitudinal distances from the center of the earth respectively in kilometers, as shown on a map. (b) Your classmate says that the equation of the orbit of another satellite around the earth can be written as x2 + y2 = 3240. Explain why he is incorrect.



217



Chapter 6



Coordinate Geometry



SIMILAR QUESTIONS



Exercise 6C Questions 1(a)-(c)



Journal Writing Search on the Internet for the radius of each of the following structures. (ii) The London Eye (i) The Singapore Flyer (iii) The Star of Nanchang (iv) The Great Beijing Wheel (v) The Great Berlin Wheel Assuming that the coordinates of the center of each structure are (0, 0), write down the equation of each of them.



Worked Example



7



(Finding the Equation of a Circle given the Center and the Radius) Find the equation of the circle with center C(2, 3) and radius 5, leaving your answer in standard form.



Solution:



Equation of the circle is (x – 2)2 + (y – 3)2 = 52 (x – 2)2 + (y – 3)2 = 25



PRACTICE NOW 7



Find the equation of each of the following circles. (a) (−1, 3), radius 4 (b) (4, −2), radius 7



SIMILAR QUESTIONS



Exercise 6C Questions 2(a)-(e)



Coordinate Geometry



Chapter 6



218



Worked Example



8



(Finding the Equation of a Circle given the End Points of a Diameter) Find the equation of a circle with diameter AB, where the coordinates of A and B are (4, −3) and (2, 1) respectively.



Solution: y



P So roblem lvin g T ip



B(2, 1)



It is advisable to sketch the circle to better visualize the position of its center.



x



O C



A(4, –3)



RE



The center of the circle, C, is the midpoint of AB. ⎛ 4 + 2 –3 + 1⎞ , Coordinates of C = ⎜ ⎟ 2 ⎠ ⎝ 2 = (3, −1) Radius of circle, AC = =



CAL



L



Midpoint of A(x1, y1) and ⎛x + x y + y ⎞ B(x2, y2) is ⎜ 11 22 ,, 11 22 ⎟ . 2 ⎠ ⎝ 2



( 4 − 3)2 + ( −3 + 1)2 5



Equation of the circle is (x – 3)2 + (y + 1)2 =



( 5)



2



(x – 3)2 + (y + 1)2 = 5



PRACTICE NOW 8



1. Find the equation of a circle with diameter PQ, where the coordinates of P and Q are (−2, 1) and (4, 3) respectively.



2. Find the equation of a circle which touches the x-axis and whose center is C(−3, 5).



219



Chapter 6



Coordinate Geometry



SIMILAR QUESTIONS



Exercise 6C Questions 3, 4, 6, 7



Equation of a Circle in General Form If you expand (x + 3)2 + (y − 7)2 = 36, you will obtain x2 + y2 + 6x − 14y + 22 = 0. This is called the general form of the equation of the circle: x2 + y2 + 2gx + 2fy + c = 0, where g = 3, f = −7 and c = 22.



Class Discussion Plotting Graphs using Equation of Circle in General Form Work in pairs.



1. Use a suitable graphing software to plot the circles whose equations are given below. Then complete the table by finding the values of g, f, c and g2 + f 2 − c, the coordinates of the center of each circle and its radius. No.



Equation of Circle



(a)



x2 + y2 + 8x + 10y + 16 = 0



(b)



x2 + y2 + 3x − 4y − 6 = 0



(c)



x 2 + y 2 − 5x − 9y +



(d)



x2 + y2 − 7x − 3.75 = 0



(e)



x 2 + y2 +



g



f



c



g2 + f 2 − c



Coordinates of Center of Circle



Radius of Circle



51 =0 2



9 7 y+ =0 2 2



2. By observing the various columns of the table, write down the coordinates of the center of a circle in



terms of g, f and/or c.



3. By observing the various columns of the table, write down the radius of a circle in terms of g, f and/or c.



Coordinate Geometry



Chapter 6



220



4. Without using any graphing software, write down the radius and the coordinates of the center of the following circles.



No.



g



Equation of Circle



(a)



x2 + y2 + 12x − 2y + 1 = 0



(b)



11yy == = xx22 ++ yy22 −− 11



(c)



4x2 + 4y2 − 24x + 20y + 25 = 0



f



c



Coordinates of Center of Circle



Radius of Circle, g2 + f 2 − c



33 22



5. Express g, f and c in terms of a, b and/or r. 6. Without using any graphing software, write down the radius of the circle whose equation is



x2 + y2 + 6x − 8y + 25 = 0. Now use the graphing software to graph the equation. What do you obtain?



7. Without using any graphing software, find the value of g2 + f 2 − c for the equation x2 + y2 + 6x − 8y + 26 = 0. Is this a circle? Explain your answer. Now use the graphing software to graph the equation. What do you obtain?



8. From Questions 6 and 7, how does the value of g2 + f 2 − c affect whether x2 + y2 + 2gx + 2fy + c = 0 is the equation of a circle?



From the discussion, the equation (x – a)2 + (y – b)2 = r2 can be written in the form x2 + y2 + 2gx + 2fy + c = 0, where g2 + f 2 − c > 0, such that (−g, −f) is the center and



g 2 + f 2 − c is the radius.



This is the equation of the circle in the general form.



Thinking Time By using the method of completing the square, show that the equation x2 + y2 + 2gx + 2fy + c = 0 can be expressed as (x – a)2 + (y – b)2 = r2. Write down the values of a, b and r in terms of g, f and c respectively.



221



Chapter 6



Coordinate Geometry



Worked Example



9



(Finding the Center and the Radius of a Circle) Find the center and the radius of the circle 2x2 + 2y2 – 3x + 4y = –1.



Solution:



AT



Method 1:



Rewriting the equation, we have 2x2 + 2y2 – 3x + 4y + 1 = 0 3 1 x 2 + y 2 − x + 2 y + == 00 2 2 Comparing this with the general form of a circle x2 + y2 + 2gx + 2fy + c = 0,



2g = −







TE



NTI



ON



When using the general form, always ensure that the right-hand side of the equation is zero.



3 3 i.e. –g = , 2 4



2f = 2 i.e. –f = –1, 1 and c = . 2 ⎛3 ⎞ ∴ Center of circle = (–g, –f) = ⎜ , – 1⎟ ⎝4



Radius of circle =



g2 + f 2 − c



=



⎛ 3⎞ 1 2 ⎜– ⎟ + (1) – 2 ⎝ 4⎠



=



17 4



⎠



2



Method 2: 3 1 x 2 + y 2 − x + 2 y + == 00 2 2 By completing the square, x2 –



2



2



⎛ 3⎞ ⎛ 3⎞ 3 1 x + ⎜– ⎟ + y 2 + 2y + (1)2 – ⎜– ⎟ – (1)2 + = 0 2 4 4 2 ⎝ ⎠ ⎝ ⎠ 2



⎛ 3⎞ 9 1 2 – 1+ = 0 ⎜x – ⎟ + (y + 1) – 4⎠ 16 2 ⎝ 2



⎛ 3⎞ 17 2 ⎜x – ⎟ + (y + 1) = 4⎠ 16 ⎝



Comparing this with the standard form (x − a)2 + (y − b)2 = r2,



17 ⎛3 ⎞ 17 = 4 . center of circle = ⎜ , – 1⎟ and radius = 16 ⎝4 ⎠



Coordinate Geometry



Chapter 6



222



Class Discussion Finding the Center and the Radius of a Circle



Discuss with your classmates which of the two methods in Worked Example 9 you prefer, explaining your reasons clearly.



PRACTICE NOW 9



SIMILAR QUESTIONS



Find the center and the radius of each of the following circles. (a) x2 + y2 – 6x + 8y + 9 = 0 (b) 2x2 + 2y2 + 4x – 3y = –2



Worked Example



10



Exercise 6C Questions 1(d)-(f)



(Finding the Equation of a Circle given Two Points and Another Condition) Find the equation of the circle which passes



through the points A(5, 0), B(3, 4) and has its center lying on the line 2y + x = 4.



Solution: y B(3, 4) P So roblem lvin g T ip



2y + x = 4



The center of the circle lies on the intersection of the line 2y + x = 4 and the perpendicular bisector of the chord AB.



C O



223



Chapter 6



Coordinate Geometry



A(5, 0)



x



⎛3 + 5 0 + 4⎞ , Midpoint of AB = ⎜ ⎟ = (4, 2) 2 ⎠ ⎝ 2 4−0 Slope of AB = = = −–22 3− 5



RE



Equation of perpendicular bisector of AB: 1 y – 2 = (x – 4) 2 1 y = x ----------------- (1) 2 The center also lies on the line



AT



2y + x = 4 ----------------- (2)



The perpendicular bisector of the chord AB will pass through the center of the circle. Solving (1) and (2), x = 2, y = 1 ∴ Center of circle = (2, 1)



(x – 2)2 + (y – 1)2 =



10



)



NTI



ON



OR



MA



TIO N



You can give the answer either in the standard form or in the general form since the question does not specify.



( 2 − 5 )2 + (1 − 0 )2



(



TE



The center of a circle lies on the perpendicular bisector of any chord. INF



= 10 ∴ Equation of the circle:



L



If two lines l1 and l2 are perpendicular, then m1m2 = −1.



1 Slope of perpendicular bisector of AB = 2



Radius of circle, r =



CAL



2



(x – 2)2 + (y – 1)2 = 10 (in the standard form)



x2 + y2 – 4x – 2y – 5 = 0 (in the general form)



SIMILAR QUESTIONS



PRACTICE NOW 1 0



1. Find the equation of the circle which passes through the points A(2, 6) and B(−2, −2) and has its center lying on the line y = x + 1.



Exercise 6C Questions 12-14



2. The line y = 2x + 5 cuts the circle x2 + y2 = 10 at two points A and B.



(i) Find the coordinates of A and of B.



(ii) Find the equation of the perpendicular bisector of AB and show that it passes through the center of the circle.



(iii) Given that the perpendicular bisector cuts the circle at P and Q, show that the x-coordinates of P and Q are k 2 and −k 2 respectively, where k is an integer to be found.



Thinking Time 1. Given a line and a circle, describe possible cases using diagrams. 2. If we have a line and a parabola, how can we find the coordinates of the points of intersection? Describe the method to find the coordinates of the points of intersection of a line and a curve.



Coordinate Geometry



Chapter 6



224



Exercise



6C



BASIC LEVEL



1. Find the center and the radius of each of the following circles. (a) x2 + y2 = 49







A(4, −7) and its center is at the origin, find



(i) the equation of the circle,



(b) 2x2 + 2y2 = 5



(ii) the length of the chord passing through A, with slope 2.



(d) x2 + y2 = 2x + 8



5 11. Given a circle with radius 2 2 and center ⎛ 1 1⎞ ⎜−1 , ⎟ , find ⎝ 2 2⎠



(c) (x + 2)2 + (y – 3)2 = 16 (e) x2 + y2 – x – 5y + 4 = 0 (f) 4 x 2 + 4 y 2 − 6 x + 10 y = =1



1 2



2. Find the equation of each of the following circles. (a) center (0, 0), radius 3



(b) center (−2, 3), radius 4 ⎛1 2 ⎞ (c) center ⎜ , – ⎟ , radius 1 1 ⎝2 5 ⎠ 2 (d) center (4, −1), passing through (−2, 0)



(e) center (3, −4), passing through (9, −4) 3. A diameter of a circle has its end points at A(0, −1) and B(2, 3). Find the equation of the circle.



4. Find the equation of the circle with the points (−3, 2) and (3, −2) as the end points of its diameter. 5. Find the points of intersection of the line y = x + 4 and the circle (x + 1)2 + (y – 4)2 = 25. INTERMEDIATE LEVEL



6. A diameter of a circle x2 + y2 – 4x – 2y + 3 = 0



passes through (1, 3). Find the equation of this



diameter.







(i) the equation of the circle, (ii) the length of the chord 4x – 3y – 5 = 0 of the circle. 12. Given that a circle which passes through the points P(3, 5) and Q(−1, 3) has radius



10 , find



(i) the equation of the circle, (ii) the equation of the line which passes through the center and the midpoint of PQ. 13. Find the equation of the circle which passes through the points A(0, 1) and B(3, −2) and has its center lying on the line y + 3x = 2. 14. Find the equation of the circle which passes through the points P(−6, 5) and Q(2, 1) and has its center lying on the line y – x = 4. 15. The line y + x = 0 cuts the circle x2 + y2 – 8x + 8y + 10 = 0 at A and B. Show that the x-coordinates of A and B are a + b 11 and a − b 11 respectively, where a and b are integers to be found. ADVANCED LEVEL



7. Find the equation of a circle which touches the



225



10. Given that a circle passes through the point



x-axis and whose center is C(5, 4).



16. The line x + 7y = 25 cuts the circle x2 + y2 = 25 at two points A and B. (i) Find the coordinates of A and of B.



8. Find the equation of the circle which passes through the point (4, 2) and has a center at (2, −2).



(ii) Find the equation of the perpendicular bisector of AB. Does it pass through the center of the circle? Show your working clearly.



9. The line x – y – 3 = 0 intersects the circle x 2 + y 2 – 2x – 2y – 7 = 0 at two points A and B. Find the length of AB.



(iii) Find the coordinates of the points where the perpendicular bisector cuts the circle.



Chapter 6



Coordinate Geometry



1. Length







The length of any line segment AB, where the coordinates of the points A and B are (x1, y1) and (x2, y2) respectively is 2



y B(x2, y2)



2



( x2 − x1 ) + ( y2 − y1 ) . 2. Midpoint If A(x1, y1) and B(x2, y2) are two distinct points,



⎛x + x y + y ⎞ Midpoint of AB = ⎜ 1 2 , 1 2 ⎟ 2 ⎠ ⎝ 2



A(x1, y1)



x 3. Equation of a Circle • Standard form The equation of a circle in the standard form is (x – a)2 + (y – b)2 = r2, where r > 0, (a, b) is the center and r is the radius. • General form The equation of a circle in the general form is x2 + y2 + 2gx + 2fy + c = 0, where g2 + f 2 − c > 0, (−g, −f)







O



is the center and g 2 + f 2 − c is the radius.



6 1. The line l has equation 3x – 4y = 24. It intersects the x-axis at A and the y-axis at B. Given that M is the point (4, –3), find



(i) the slope of l,



(ii) the length of AB,



(iii) the equation of the line passing through B and having the same slope as OM, where O is the origin.



2. The coordinates of the points A and B are (0, 6) and (8, 0) respectively. (i) Find the equation of the line passing through A and B.



Given that the line y = x + 1 cuts the line AB at the point M, find



(ii) the coordinates of M,



(iii) the equation of the line which passes through M and is parallel to the x-axis, (iv) the equation of the line which passes through M and is parallel to the y-axis.



Coordinate Geometry



Chapter 6



226



3. The diagram shows the line l passing through the points A(–1, 1) and B(5, 5). y







B(5, 5)



5







4 3 2



A(–1, 1)







0







C(4, 1)



1



–2 –1



1



2



3



4



5



6



x



(i) the slope of l,



(ii) the equation of l,







(iii) the area of ΔABC,



(iv) the length of BC, giving your answer correct to 2 decimal places. 4. The diagram shows the points A(–1, 0) and B(0, 3).



y 4 3



B



2 1 A –2



–1



0



1



2



3



x



(i) Find the equation of the line passing through A and B. (ii) Given that the length of AB is the value of h.



h units, find



The point (–5, k) lies on BA produced.



(iii) Find the value of k.



(iv) Given that y = x + 1 is the line of symmetry of ΔABC, find the coordinates of C.



227



Chapter 6



Coordinate Geometry



(i) Write down the coordinates of P and of Q. (ii) Find the length of PQ.



Another line l2 has the same slope as l1 and passes through the point (0, –2).



(iii) Find the equation of the line l2.



(iv) Given that the y-axis is the line of symmetry of ΔPQR, find the coordinates of R. 6. The straight line y = 2x + 3 meets the curve xy + 20 = 5y at the points A and B. Find the coordinates of the midpoint of AB.



Given that C is the point (4, 1), find











5. The equation of a straight line l1 is 5y + 12x – 60 = 0. It intersects the x-axis at P and the y-axis at Q.



7.



Find the equation of the circle, where P is the center and PQ is the radius. (a) P(2, 7), Q(5, 11) (b) P(−2, 9), Q(4, 6)



8. A weather satellite orbits planet P such that the equation of its path can be modeled by the equation x2 + y2 – 16x – 12y + 75 = 0, where x and y are the longitudinal and latitudinal distances from the center of P respectively in kilometers, as shown on a map. State the coordinates of the center and the radius of the orbit. 9. Find the center and the radius of each of the following circles. (a) x2 + y2 – 4x – 10y + 20 = 0 (b) x2 + y2 + 4x – 4y – 17 = 0 (c) 3x2 + 3y2 = 16 (d) (x + 1)(x – 5) + (y – 2)(y – 4) = 0 10. Find the equation of the circle with center (0, 2) and radius 4. Hence, find the coordinates of the points where the circle cuts the x- and y-axes. 11. A diameter of a circle has end points at A(−5, 0) and B(9, 0). (i) Find the equation of the circle. (ii) If C(2, k) lies on the circle, find the value of k. Hence, determine whether ∆ABC is an isosceles right-angled triangle, showing your working clearly.



12. Find the equation of the circle which passes through the points A(2, 1) and B(3, −2) and has its center lying on the line y + x = 0. 13. The coordinates of P, Q and R are (x, y), (9, 0) and (1, 0) respectively. Given that PQ = 3PR, (i) find a relationship between x and y, (ii) determine whether this relationship represents a circle, showing your working clearly. 14. (i) (ii)



Given two points P(1, 4) and Q(−1, −2), find the equation of the circle with PQ as its diameter. Does the point R(−3, 2) lie on the circle? Hence, determine whether ∠PRQ is 90˚, explaining your working clearly.



15. The line y + x = 6 cuts the circle x2 + y2 – 2x – 6y + 6 = 0 at two points A and B. (i) Find the midpoint of AB. (ii) Find the equation of the perpendicular bisector of AB. Hence, show that it passes through the center of the circle. (iii) Given that the perpendicular bisector cuts the circle at P and Q, show that the x-coordinates of P and Q are a + b 2 and a − b 2 respectively. Find the value of a and of b.



Challenge 1. Given that the y-axis is a tangent to the circle with radius 5 units and the circle passes through the point (1, −4), find the possible equations of the circle. 2.



Given three circles, C1: (x + 1)2 + (y – 3)2 = 1 C2: 2x2 + 2y2 + 3x + y – 9 = 0 C3: 4x2 + 4y2 + 7x – 11y = 0, show that the centers of C1 and C2 are the ends of a diameter of C3.



3. The circle x2 + y2 – 6x – 9y + 8 = 0 cuts the x-axis at A and B and the y-axis at C and D. Prove that OA × OB = OC × OD, where O is the origin.



Coordinate Geometry



Chapter 6



228



B1



Revision Exercise



1. In the figure, OAB is a A sector of a circle of radius 8 cm 8 cm and the length of arc AB = 15.2 cm. O Find (i) ∠AOB in radians, (ii) the area of the sector. B



15.2 cm



3. PA and PB are tangents to the circle with center O. Given that ∠APB = 48° and ∠OBC = 18°, find (i) ∠BAC, 48° (ii) ∠ABC. P 4.



A, B, C, D and E are points on a circle. Given that AB = AE, ∠ABE = 26° and ∠AED = 118°, find (i) ∠BAE, (ii) ∠BCD.







P



B



y



4



O



z



D 118° C



B



w y



x D



B u vE 229



40°



Find the angles u, v, w, x, y and z. Revision Exercise B1



A –2



5. In the figure, AB is a diameter of the circle with center O. BEF is a tangent to the circle at B, ∠BAE = 20° and ∠AFB = 40°.



20°



2



B



A



F



C



6



C



26°



Q



7. Find the equation of the line joining the points A(5, 7) and B(8, 12).



18°







b



Given that ∠QTR = a and ∠QPR = b, express each of the following in terms of a and b. (i) ∠RQS (ii) ∠PRS (iii) ∠PST (iv) ∠QRS (v) ∠PSQ



O



A



P



8. The figure shows the points A(–2, 1), B(2, 1) and C(4, 6).



A



E



a



T



2. The figure shows a semicircle AOBP where O is the center and ΔAPB is 0.6 rad right-angled at P. A O Given that AB = 16 cm and ∠PAB = 0.6 radians, find (i) the length of AP, (ii) the area of ΔPAB, (iii) the area of the shaded region.



R



6. In the figure, TPQ and TSR are straight lines and SR = QR. S







0



B 2



4



x



Find (i) the area of ΔABC, (ii) the coordinates of the point D such that ABCD is a parallelogram, (iii) the coordinates of the point P such that ACPB is a parallelogram.



9. Find the equation of the circle with center (–2, 5) and radius 5 units. The line y = 2x + 4 cuts the circle at points P and Q. Find the length of the chord PQ. 10. The line y + 7x = 34 cuts the circle x2 + y2 – 2x – 4y – 20 = 0 at two points, A and B. Find (i) the coordinates of A and of B, (ii) the equation of the perpendicular bisector of AB and show that it passes through the center of the circle, (iii) the coordinates of the points where the perpendicular bisector cuts the circle, giving your answers in radical form.



B2



Revision Exercise



1. In the figure, the radius of the circle and of each arc is 4 cm.



6. The line x – 2y = –4 cuts the x-axis and the y-axis at P and Q respectively. M is a point on PQ such that it is equidistant from the coordinate axes. y Q x – 2y = –4







P



Find the area of the shaded region.



2. In the figure, EB is the tangent to the circle with center O at B. C



O B



E Given that ∠ABE = 50°, find (i) ∠AOB, (ii) ∠ACB.



C(a, 4)



14 cm B







Given that the radius of the circle is 14 cm and ∠APB = 56°, find the area of the shaded region.



4. The tangents to a circle at P, Q and R intersect at A, B and C as shown. C R



A



42°



Q 56° P



O A



O



56°



Find (i) the coordinates of P and of Q, (ii) the coordinates of M, (iii) the area of ΔPMO.



D(0, 6)



3. In the figure, PA and PB are tangents to the circle, center O. A P



x



O



7. ABCD is a trapezoid in which DC = 13 units. The coordinates of B, C and D are (a, –1), (a, 4) and 2 (0, 6) respectively and the slope of AB is . 3 y



A



50°







M



B



Given that ∠BAC = 42° and ∠PQB = 56°, find (i) ∠ACB, (ii) ∠PQR, (iii) ∠RPQ. 5. Find the equation of the line passing through the point (2, –5) and parallel to the line 5x + 7y = 46.







B(a, –1)



x



Find (i) the coordinates of the point C, (ii) the coordinates of the point A, (iii) the area of the trapezoid ABCD, (iv) the length of AB, (v) the equation of AB.



8. A straight line through the point (1, 18) intersects the curve xy = 30 at the point (2, 15). Find the coordinates of the point at which the line meets the curve again. 9. Given that a circle with center (2, 5) touches the x-axis, find the equation of the circle. 10. A diameter of a circle has end points at P(– 4, 3) and Q(4, –3). (i) Find the equation of the circle. (ii) If R(k, 4) lies on the circle, find the value of k, where k < 0. Hence, prove that nPQR is a right-angled triangle.



Revision Exercise B2



230



Medical history was made in Singapore in 1998 when an Indonesian lady gave birth to a set of healthy sextuplets in a local hospital. If the mother is to arrange the six lovely babies for a group photograph, how many different ways of arranging the six babies are possible?



LEARNING OBJECTIVES At the end of this chapter, you should be able to: • illustrate the permutation of objects and derive the formula for finding the number of permutations of n objects taken r at a time, • illustrate the combination of objects and derive the formula for finding the number of combinations of n objects taken r at a time, • differentiate permutation from combinationof n objects taken r at a time, • solve problems involving permutations and combinations.



Seven



Permutations and Combinations



7.1



Addition and Multiplication Principles



Addition Principle Consider the case of a traveler who wishes to visit Manila in the Philippines starting from Singapore. If he has a choice of taking any of three different sea cruise routes and any of four air package routes, in how many ways can he travel from Singapore to Manila? In this case, he will have a total of seven choices of traveling from Singapore to Manila since he can travel by any one of the sea or air routes. This example is an illustration of the Addition Principle which states



if there exist two non-overlapping categories of ways to perform an operation, and there are m ways in the first category and n ways in the second category, then there are (m + n) ways to perform the operation.



Worked Example



1



(Solving Problems involving Addition Principle) Mrs Torres visits the community club in her constituency and finds many interesting courses offered. There are three different dance classes, two different computer classes and four martial arts classes. How many choices does Mrs Torres have?



Solution:



Mrs Torres can choose any of the 3 dance classes, any of the 2 computer classes or any of the 4 martial art classes. Using the Addition Principle, number of choices = 3 + 2 + 4 =9 ∴ Mrs Torres has 9 choices.



PRACTICE NOW 1



Kristel patronizes a hawker center during her lunch time and finds that there are many choices of food available. There are four different types of noodles, three different types of congee and two different types of rice. How many choices does Kristel have?



233



Chapter 7



Permutations and Combinations



SIMILAR QUESTIONS



Exercise 7A Questions 1-2



Multiplication Principle Daniel who lives in Area A travels by bus to his school in Area B. He must take a bus through Point C before boarding a different bus to reach his school. There are two bus services plying between his home and Point C, Buses 170 and 171. There are 3 bus services between Point C and his school, Buses 30, 31 and 32. In how many different ways can Daniel travel to school on these buses? In Grade 8, we have learned to solve problem using a tree diagram. We can use it to solve this problem. Buses to Point C Buses to Area B 30 170



31 32



Area A



30 171



31 32



Starting from Area A, two lines are drawn where each line denotes one possible way of traveling to Point C. From each of these, another three lines are drawn to represent the possible ways of traveling from Point C to Area B. Therefore, Daniel can travel to school in six different ways. For example, he can take Bus 170 followed by Bus 30, Bus 170 followed by Bus 31, etc. This example is an illustration of the Multiplication Principle, which can be generalized as follows. If an operation A can be performed in m ways and a second operation B can be performed in n ways, then the two successive operations can be performed in m × n ways. The principle can be extended to include more than two successive operations.



Extended Multiplication Principle



AT



If an operation A can be performed in m ways, a second operation B can be performed in n ways, a third operation C can be performed in p ways, and so on and the last operation in k ways, the successive operations can be performed in m × n × p … × k ways.



TE



NTI



ON



The Multiplication Principle is true only when the operations are independent, i.e. the choice made for one operation will not affect the choice made for the other operations.



Thinking Time There is also an Extended Addition Principle. Can you generalize it?



Permutations and Combinations



Chapter 7



234



Worked Example



2



(Solving Problems involving Extended Multiplication Principle) There are two roads joining Town A to Town B and five roads joining Town B to Town C. Find how many different routes there are from Town A to Town C using these roads.



Solution:



There are two operations to perform in succession: Town A to Town B 2 ways Town B to Town C 5 ways Using the Multiplication Principle, number of routes = 2 × 5 = 10 ∴ There are 10 routes from Town A to Town C. SIMILAR QUESTIONS



PRACTICE NOW 2



There are three paths joining a restaurant to a hotel and five paths joining the hotel to a shopping mall. Find how many different routes there are from the restaurant to the shopping mall using these paths.



Worked Example



3



Exercise 7A Questions 3-4, 8(a)



(Solving Problems involving Extended Multiplication Principle) Assume during the Asia-Pacific Economic Cooperation (APEC) meeting held in Pasay, Philippines in 2015, five journalists arrived from Singapore and there were five hotels to accommodate them. If each hotel could only accommodate one journalist, how many different ways could the five journalists be accommodated?



Solution:



When the first journalist arrived, he had five choices of hotels. The next journalist would have only 4 choices as the first journalist who arrive earlier occupied one of the hotels, and so on. Therefore, there were 5 operations to perform in succession. Using the Multiplication Principle, number of ways = 5 × 4 × 3 × 2 × 1 = 120 ∴ There are 120 different ways the five journalists could be accommodated.



PRACTICE NOW 3



How many 3-letter code words are possible using the first 8 letters of the alphabet if no letter can be repeated? 235



Chapter 7



Permutations and Combinations



SIMILAR QUESTIONS



Exercise 7A Questions 5-7



Sometimes, both Addition and Multiplication Principles are used to solve a problem.



Worked Example



4



(Solving Problems involving the Combination of Addition and Multiplication Principles) A student wanting to go to town by bus can take any of the four AB Liner Service or any of the five ALPS The Bus. How many choices does he have for a round trip if he must return by a different bus service from his outward journey?



Solution:



The student has a choice of 9 different bus services on his outward journey. If the outward journey is by AB Liner Service, he will return by ALPS The Bus. ∴ Number of choices = 4 × 5 = 20 If the outward journey is by ALPS The Bus, he will return by AB Liner Service. ∴ Number of choices = 5 × 4 = 20 Using the Addition Principle, total number of choices for making two-way trips = 20 + 20 = 40 ∴ He has 40 choices for a round trip. SIMILAR QUESTIONS



PRACTICE NOW 4



Ann wishes to have noodles or rice for her lunch. If she has noodles for lunch, then she will have rice for dinner, and vice versa. How many choices does she have for her meals if there are five types of noodles and six types of rice?



Exercise 7A Question 8(b)



Exercise



7A



BASIC LEVEL



1. A teacher wants to choose a student representative from three classes. If there are 15 students eligible in the first class, 18 students eligible in the second class and 12 students eligible in the third class, how many choices does the teacher have? 2. Kate’s father needs to buy a smartphone manufactured by A, B or C. If there are six models made by either A, seven by B and five by C available in the store he visits, how many choices does he have?



3. There are four bus services plying between Pasay and Manila, and three bus services plying between Manila and Quezon. How many possible bus routes are there from Pasay to Quezon via Manila? 4. Assume there are three roads joining Town P to Town Q, two roads joining Town Q to Town R and four roads joining Town R to Town S. How many different routes are there from Town P to Town S, passing through Town Q and Town R?



Permutations and Combinations



Chapter 7



236



5. Eric has four choices of traveling to school every day. He can walk, cycle, take a bus or the MRT. In how many ways can he arrange his forms of transport for the next three days if he does not want to repeat any form of transportation? 6. The front doors of a row of six terrace houses in an estate are to be painted in white, blue, green, yellow, brown and red. In how many ways can the painting be done if no two doors are to be of the same color?



7. In how many ways can you arrange the set of sextuplets born in Singapore in 1998 for a photograph in a row as it is mentioned at the beginning of this chapter?



INTERMEDIATE LEVEL



8. Three roads run from Town A to Town B and five roads run from Town B to Town C. (a) How many possible routes are there from Town A to Town C via Town B? (b) How many routes are possible for a return trip from Town A via Town B if it is permitted to return by the same route?



7.2



Permutations



Let us consider an example. A waiting room contains a row of seven chairs. In how many different ways can three of these chairs be occupied by three people? The first person can occupy any of the seven chairs, i.e. he has 7 choices. Once he is seated, the second person can occupy any of the six remaining chairs, i.e. he has 6 choices. Similarly, the last person will have 5 choices left. According to the Multiplication Principle in Section 7.1, the total number of ways of the three people can be seated in 7 × 6 × 5 = 210 ways. The example above illustrates the concept of permutation in Mathematics. In general, a permutation is an arrangement of objects taken either all at the same time or a few at a time. The example above shows the permutation of 3 objects out of 7 objects, or, in other words, 7 objects taken 3 at a time. Generally, the number of permutations of r objects taken from n subjects is denoted by nPr, where 1 < r < n. From the above example, we know that the number of permutations of 7 objects taken 3 at a time is 7 P3 = 7 × 6 × 5. Similarly, nP3 = n(n – 1)(n – 2), (assume n > 3)



237



n



P4 = n(n – 1)(n – 2)(n – 3), etc.



Chapter 7



Permutations and Combinations



(assume n > 4)



un



Just For F



How many arrangements of the letters in the word PERMUTING start with a vowel?



In general, the number of permutations of n objects taken r at a time is n



Pr = n(n – 1)(n – 2)(n – 3) … (n – r + 1)



When r = n, nPr = n(n – 1)(n – 2) … 3 . 2 . 1. We define n! = n(n – 1)(n – 2)(n – 3) … 3 . 2 . 1.



un



Just For F



Therefore, nPn = n!. Using the n! symbol, we will have nPr =



Can you prove that



n! (n – r)!



n



Pr =



n!



(n – r)!



?







Thinking Time We define 0! = 1 so the formula nPr =



n! holds true for r = n. n ( – r)!



Can you explain why?



Worked Example



5



(Solving Problems involving Permutation) How many different four-letter words can be formed using the letters in the word OLYMPICS?



Solution: 8



8! P4 = 8 × 7 × 6 × 5 or 8P4 = 8 – ( 4)! = 1680 words = 1680 words



PRACTICE NOW 5



SIMILAR QUESTIONS



How many different three-digit numbers can be formed using the numbers 1 to 9?



Exercise 7B Questions 1-3



Permutations and Combinations



Chapter 7



238



Worked Example



6



(Solving Problems involving Permutation) In how many ways can the 1st, 2nd and 3rd prizes in a lucky draw be awarded to a group of 40 people? Assume each person can be awarded only one prize. INF



Solution: 40



Press 4



40! 37!



0 SHIFT



!



In how many ways can 5 hamburgers be given to a group of 15 students? Assume each student can be given only one hamburger.



7



Fill the number from the right-hand side. Since there are only 4 even numbers (2, 4, 6, 8) in the digits 1 to 9, there are 4 ways to fill the 3rd box. 2nd



Exercise 7B Questions 4-5



(Solving Problems involving Permutation with one Given Condition) How many three-digit even numbers can be formed using the digits 1 to 9 without repetition?



Solution:



P So roblem lvin g T ip



Fill the even number from the right-hand side since the numbers formed are even numbers. Can we fill the numbers starting from the left-hand side?



3rd



The 2nd box can be filled with the remaining 8 digits and the 3rd box can be filled with the remaining 7 digits. 1st



2nd



3rd



∴ 8 × 7 × 4, i.e. 8P2 × 4 = 224 even numbers which can be formed using the digits 1 to 9 without repetition.



PRACTICE NOW 7



How many three-letter words ending with a vowel can be formed using all the letters of the alphabet without repetition?



Worked Example 7 illustrates the general method of treating arrangement problems with given conditions. Worked Example 8 illustrates another similar problem with more possible restrictions. 239



Chapter 7



÷



SIMILAR QUESTIONS



PRACTICE NOW 6



1st



TIO N



3 7 SHIFT ! = to obtain the answer. Some calculators also have the nPr function. Press 4 0 nPr 3 = to get 59 280.



= 59 280 ways



Worked Example



MA



Many scientific calculators come with the factorial functions ‘!’.



40! P3 = (40 – 3)!



=



OR



Permutations and Combinations



SIMILAR QUESTIONS



Exercise 7B Questions 6-9



Worked Example



8



(Solving Problems involving Permutation with more than one Given Condition) At the passing out parade ceremony for the newly appointed officer cadets of the National Armed Forces, 3 army officers, 3 navy officers, 2 civil defense officers and 7 air force officers were asked to take a group photograph. In how many ways can the photographs be taken if the officers are to stand (a) in a row? (b) in a row so that no 2 air force officers are next to each other? (c) in a row so that the 3 army officers are next to each other? (d) in a row so that the 3 navy officers are not next to each other? (e) in a row so that the 2 civil defense officers are at the two ends of the row? (f) in two rows with the air force officers at the back row and the other eight officers at the front?



Solution:



(a) 15! or 15P15 (Permutation of 15 taken 15 at a time)



(b) Since there are 7 air force officers out of 15 officers, and the air force officers are to be arranged in alternate positions between 8 other officers in a row, i.e. 9 available positions = 9P7 ways.



8 other officers are arranged in 8! possible ways. ∴ Total number of ways = 8! × 9P7







= 7 315 660 800



P So roblem lvin g T ip



Note that there are 9 available positions for the arrangement of the 7 air force officers in (b) because of two additional positions at each end of the row.



(c) Since there are 3 army officers out of 15 officers and they are to be next to each other, hence they are considered as one ‘object’, i.e. 13 possible ways of arrangement = 13!



Number of ways for the arrangement of the 3 army officers = 3!



∴ Total number of ways = 13! × 3!







= 37 362 124 800



(d) Since there are 3 navy officers out of 15 officers, and the navy officers are to be arranged in an alternate positions between 12 other officers in a row, i.e. 13 available positions = 13P3 ways.



12 other officers are arranged in 12! possible ways. ∴ Total number of ways = 12! × 13P3







= 821 966 745 600



(e) For the two civil defense officers to be at the two ends of the row, number of ways = 2! 13 other officers will take the remaining positions at any arrangement = 13!



∴ Total number of ways = 2! × 13!







= 12 454 041 600



(f) The possible arrangement for the air force officer to be at the back row = 7!







The remaining officers are arranged at the front row = 8!



∴ Total number of ways = 7! × 8!







= 203 212 800 Permutations and Combinations



Chapter 7



240



SIMILAR QUESTIONS



PRACTICE NOW 8



A fruit basket contains 9 apples, 2 oranges, 5 pears and 4 peaches. The fruits are to be displayed on a shelf. In how many ways can the fruits be arranged if they are to be displayed (a) in a row? (b) in a row so that there no two apples are next to each other? (c) in a row so that the four peaches are next to each other? (d) in a row so that the five pears are not next to each other? (e) in a row so that the two oranges are at the two ends of the row? (f) in two rows with the apples at the back row and the other fruits are the front?



9



Worked Example



(Solving Problems involving Permutation with more than one Given Condition) How many three- and four-digit numbers can be formed using the digits 2, 3, 4, 5 and 6 without repetition? (a) How many of these numbers will be greater than 400? (b) How many will be less than 400?



Solution:



Number of possible ways to form three-digit number = 5P3 Number of possible ways to form four-digit number = 5P4



∴ Total number of three- and four- digit numbers to be formed = 5P3 + 5P4 = 180



(a) Let us consider only three-digit numbers greater than 400 since all four



digit numbers are greater than 400.







The first digit of the number must be 4, 5 or 6, i.e. number of ways = 3P1.







The remaining digits for the number can be filled with the remaining numbers







= 4P1 × 3P1.







Total number of ways to form three-digit number so it will be greater than 400



= 3P1 × 4P1 × 3P1







∴ Total number that is greater than 400



= (3P1 × 4P1 × 3P1) + (5P1 × 4P1 × 3P1 × 2P1) = (3 × 4 × 3) + (5 × 4 × 3× 2)



= 156



(b) Since there are 180 numbers in total in which 156 of them which are greater than 400, and none of them is equal to 400,



241



total number that is less than 400



= 180 – 156 = 24



Chapter 7



Permutations and Combinations



Exercise 7B Questions 10-12



PRACTICE NOW 9



SIMILAR QUESTIONS



How many three- and four-letter words can be formed using the letters A, C, G,



Exercise 7B Questions 13-16



O and N without repetition? (a) How many of these words will end with a vowel? (b) How many end with a consonant?



Exercise



7B



BASIC LEVEL



1. The principal of a school plans to observe three classes, A, B and C, in his school in a morning. In how many possible orders of the classes can he observe? List all the possible orders. 2. An examination paper has ten questions. The students need only to answer eight of them. In how many different ways can the students answer the eight questions? 3. Miguel received an achievement award for his outstanding academic performance in school. He was asked to take a photograph together with two other award recipients and the principal of the school. How many different arrangements were possible for a group photograph of the four? 4. The first and second prizes for a mathematics quiz are to be awarded to a class of 40. How many possibilities are there for the result if no one can be awarded more than one prize?



8. Six married couples are to sit on a bench. In how many ways can they be arranged if



(a) there is no restriction?







(b) each husband must sit beside his wife?



9. Six boys and five girls are to stand in a line. In how many ways can this be done if no two boys are to stand beside each other? 10. Three girls and four boys are to sit on seven chairs in a row. Calculate the number of different possible arrangements if



(a) two of the boys are to sit at the ends,







(b) the three girls are to sit together.



11. Five boys and three girls are to sit in a row of eight chairs. Calculate the number of different possible arrangements if



(a) the boys are to sit on adjacent seats,







(b) the girls are to sit on adjacent seats. INTERMEDIATE LEVEL



5. In how many ways can the first four places in a race be filled by six competitors running in it if there is no tie for any place? 6. How many integers less than 400 can be formed using the digits 2, 3, 4, 5 and 6 without repetition? 7. Ten history books, five geography books and three mathematics books are to be arranged on a shelf. Assume all the books are different. In how many ways can this be done if books of the same subject must be placed together?



12. Four girls, three boys and two old ladies are to be arranged in a straight line. In how many ways can this be done if



(a) the girls must be together?







(b) the boys must be together?



(c) the girls must be together while old ladies cannot stand next to each other?



(d) there is no restriction?



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13. How many numbers between 2000 and 8999 (both inclusive) do not have four different digits? 14. How many four-digit numbers can be made using the digits 1, 2, 3 and 5? (a) How many of these digits are greater than 4000?



(b) How many are less than 4000?



(c) How many of these are odd numbers greater than 4000?



ADVANCED LEVEL



15. How many numbers greater than 5000 can be formed using the digits 3, 4, 5, 6 and 7 without repetition of digits? How many of these are even numbers? 16. How many numbers greater than 250 can be formed using the digits 1, 2, 3, 4 and 5 without repetition of digits?



(d) How many are even numbers greater than 4000?



7.3 Combinations In permutations, attention must be given to the order of arrangement of the objects. In many instances, however, the order of arrangement does not matter. For example, if 3 representatives are to be selected from 5 people from a committee, usually we only need to know which 3 out of the 5 are selected rather than in which order they are selected. If we use P, Q and R to stand for three people selected, PQR, PRQ, QRP are different permutations, but they are the same committee, i.e. the same combination of people. In Mathematics, we define a combination of n different objects taken r at a time as a selection of r objects from n objects without considering the order of objects being selected. Generally, the number of combinations of n different objects taken r at a time is denoted by nCr. A formula of nCr can be derived from the formula for nPr. Each combination of r objects can be arranged in r! ways. Hence nPr = r! × nCr



n



Cr =







=







=



Pr r!



n



n! (n – r)!r!



n (n – 1) (n – 2)…(n – r + 1) r!



Worked Example 10 explains the difference between permutation and combination.



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Chapter 7



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Worked Example



10



(Identifying Difference between Permutation and Combination) (a) Ten restaurants are finalists for three prizes – one for 250 000, one for 200 000 and one for 20 000. In how many different ways can the prizes be awarded? (b) Ten restaurants are finalists for three 20 000 prizes. In how many different ways can the prizes be awarded?



Solution:



(a) Since the prizes are different, the order of winning restaurants is important.



The number of ways of awarding the 3 different prizes to 3 out of the 10 restaurants is 10P3.



10



P3 = 10 × 9 × 8 = 720







The above is a case of permutation.



(b) Since the prizes are equal, the order of winning restaurants is not important. Therefore, the number of combinations of 10 restaurants taken 3 at a time, denoted by 10C3 is determined.



To find 10C3, note for each combination of 3 restaurants. For example, if we take them as A, B, C it can be permutated in 3! = 6 possible ways, i.e. ABC, ACB, BAC, BCA, CAB or CBA.



Hence 10P3 = 3! × 10C3 C3 =



10



P3 3!



10



10 × 9 × 8 3× 2 ×1







=







= 120



∴ There are 120 different ways to award the prizes.



SIMILAR QUESTIONS



PRACTICE NOW 1 0



(a) Fifteen students are finalists for five different scholarship awards. In how many different ways can the awards be given?



Exercise 7C Questions 1(a)-(b)



(b) Fifteen students are finalists for three scholarship awards worth 100 000. In how many different ways can the awards be given?



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11



Worked Example



(Solving Problems involving Combination) (a) In how many ways can two prefects be selected from a group of 20 prefects? (b) Seven points are marked on a piece of paper such that no three of them lie on a straight line. How many straight lines can possibly be drawn to pass through any two points?



Solution:



P2 2! 20 × 19 = 2 ×1



C2 =



20



(a)



20



= 190



(b) Since two points are required to form a straight line, 7







P2 2! 7×6 = 2 ×1



C2 =







7



= 21



∴ There are 21 ways the straight lines can be drawn.







PRACTICE NOW 1 1



SIMILAR QUESTIONS



(a) In how many ways can four science books be selected from 18 science books? (b) Four pupils are to be selected to form a team for competition. If eight pupils are under consideration for selection, find the number of ways in which this team may be selected.



Worked Example



12



Exercise 7C Questions 2(a)-(e), 3-6, 9-11, 13-15



(Proving Equation involving Combination) Show that nCr = nCn – r.



Solution: Cr =



n



n



Cn – r =



n!



(n – r)!r!



n! ⎡⎣n – (n – r)⎤⎦!(n – r)!



n! = r!(n – r)!



∴ nCr = nCn – r PRACTICE NOW 1 2



Show that nCn = nC0 = 1. 245



Chapter 7



Permutations and Combinations



SIMILAR QUESTIONS



Exercise 7C Question 7



Worked Example



13



Solution:



(Solving Problems involving Combination) The number of combinations of two objects selected from n objects is equal to the number of combinations of three objects selected from n objects. Determine n.



No. of combinations of 2 objects from n = nC2.



No. of combinations of 3 objects from n = nC3. Since nC2 = nC3, n n P2 P3 i.e. = 2! 3! n (n – 1) n (n – 1) (n – 2) = 2 ×1 3× 2 ×1 n–2=3 n=5



PRACTICE NOW 13



SIMILAR QUESTIONS



Determine n if the number of combinations of two objects selected from n objects is equal to 10.



Exercise 7C Question 8



Exercise



7C



BASIC LEVEL



1. (a)



A committee consisting of a chairman and a secretary is to be selected from a class of 40 pupils. How many possibilities are there to form a committee?



(b) If the same class is to select two representatives to seek clarification from the teachers regarding certain class matters, how many different groups of representatives can be possibly formed? 2. (a) In the game of Toto a punter selects six numbers from the 49 available numbers. How many different selections are possible? (b)



(c) The Philippines hosted the Asia-Pacific Economic Cooperation (APEC) meeting in 2015. If we are to select 5 ministers from 14 ministers to pose for a photograph, how many different groups of ministers can we select? (d) A committee of three men and four women is to be selected from six men and seven women. Find the number of ways in which such a committee can be formed. (e) The old system of the game of Toto requires the punter to select six numbers from a total of 45 numbers. How many different selections are possible?



An inspector from the Ministry of Environment selects five kinds of vegetables from eight different kinds of vegetables from the whole sale market for analysis. In how many ways can this be done?



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3. In a memorial service to mourn the deaths of 114 victims of an airplane crash, representatives of the nine religious groups were present to give memorial service at the National Indoor Stadium. (a) How many different ways were there to arrange the representatives to stand in a straight line? (b) If a group of four representatives was to be selected to come forward at a time, how many different selections were possible? 4. 12 friends enter a restaurant where there are only three tables left. One table can seat three people, the other can seat four and the largest can seat five. In how many ways can the waiter seat 12 friends assuming that the order of seating at the tables does not matter? 5. A librarian has to select four newspapers and seven magazines from the seven regional newspapers and 12 international magazines available. In how many ways can the librarian make the choices? 6. A soccer team is to be selected for a tour. The final selection is made by selecting eight forwards out of 10, four center-forwards out of seven, three full-backs out of five and two goalkeepers out of four. Calculate the total number of possibilities of selecting the touring team. 7. Show that 15C5 = 15C10. 8. Determine n if the number of permutations of three objects selected from n objects is equals to 36 times the number of combinations of two objects selected from n objects. INTERMEDIATE LEVEL



11. Three different cars are available to take five women and a man on a particular journey. In how many different ways can the six people be allocated to the cars so that the man is alone? 12. Five boys and five girls are selected from ten boys and 11 girls to form five mixed pairs for squash. Find the total number of different mixed pairs which can be formed. ADVANCED LEVEL



13. A committee of seven is to be selected from six men and five women. Calculate the number of different possible ways of forming the committee if



(a) there is no restriction,



(b) the number of women must be greater than the number of men. 14. A committee of three doctors is to be selected from a class of seven male doctors and four female doctors. Calculate



(a) the total number of possible committees,



(b) the number of possible committees if at least one member of the committee is a male doctor. 15. From a group of six boys and five girls, a committee of four children is to be formed. Calculate



(a) the total number of ways of forming a



9. A committee of six is to be formed from five single men, four single women and two married couples.



committee,



(a) Find the total number of ways of forming the committee.







(b) If the committee is to consist of a chairperson, who must be a single woman, 3 other men and 2 women, in how many different ways can the committee be formed?



247



10. A committee of nine is to be chosen from ten men and seven women. How many ways of selection are there if there should not be more than five men in the committee?



Chapter 7



Permutations and Combinations







(b) the number of possible committees if at least one member of the group must be a girl,



(c) the number of possible committees if at least







one member of the group must be a boy.



1.



Addition Principle If there exist two non-overlapping categories of ways to perform an operation, and there are m ways in the first category and n ways in the second category, then there are (m + n) ways to perform the operation.



2.



Multiplication Principle If an operation A can be performed in m ways, a second operation B can be performed in n ways, a third operation C can be performed in p ways, and so on and the last operation in k ways, the successive operations can be performed in (m × n × p … × k) ways.



3. A permutation is an arrangement of objects taken either all at the same time or a few at a time. The number of permutations of n objects taken r at a time n! is nPr = , where 1 < r < n and n! = n(n – 1)(n – 2) … 3 . 2 . 1. n ( – r)! When r = n, it is a permutation of n objects taken all at a time, nPr = n!



4. A combination of n different objects taken r at a time is defined as a selection of the r objects from n objects without considering the order of the objects being selected. The number of combinations of n different objects taken n n (n – 1) (n – 2)…(n – r + 1) P n! r at a time is nCr = r = = where 1 < r < n. r! r! n – r !r! ( )



7 1. There are three bus services plying between Area A and Area B, and there are five bus services plying between Area B and Area C. How many different bus routes can a commuter take from Area A to Area C via Area B? 2. In how many ways can seven school prefects be arranged in a line?



3. In how many ways can all the letters in the word JURONG be arranged? How many of these words will begin with the letter O? How many of these words will end with the letter R? 4. The final year mathematics paper consists of 11 questions and each pupil is required to answer any five questions. How many choices are there for the students to select the five questions?



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248



5.



In how many ways can five boys and four girls be seated in a row of nine chairs when (a) the boys and girls can sit in any chair? (b) the boys and girls must sit in alternate chairs? (c) the girls must sit on adjacent chairs? (d) the boys must sit on adjacent chairs?



6. How many three-digit even numbers greater than 400 can be formed from the digits 1, 2, 3, 4 and 5 if repetition of digits is not allowed? 7. A table has nine seats, four on one side facing the main door and five on the opposite side facing the garden. In how many ways can nine people be seated at the table if (a) there is no restriction? (b) Antonio and Chris must sit on the same side of the table? (c) Miguel, Eric and Sam must sit on the side facing the garden? (d) Kate and Tricia must sit on opposite sides? (e) Kate and Tricia must sit on the same side? 8. Ten people are to be seated in a row. Calculate the total number of ways in which this can be done if two of them, Eric and Carlo (a) are always sitting next to each other, (b) always have exactly one of the other people sitting between them. 9. How many four- and five-digit even numbers can be formed from the digits 1, 2, 4, 6 and 9 if repetition of digits is not allowed? 10. Mr Reyes is to select a team of 13 players for the school’s basketball team from 16 qualifying players. How many different possible ways can he form the team? 11. A committee of four men and five women is to be chosen from seven married couples. How many different possibilities are there of forming such a committee? 12. A tennis club has to select two mixed-double pairs from a group of six men and four women. How many different ways can this be done?



249



Chapter 7



Permutations and Combinations



13. In how many ways can a committee of seven people be selected from six couples if it must contain (a) more men than women? (b) at least two women? 14. A committee of seven is to be chosen from a group of six boys and five girls. In how many ways can this be done if (a) there is no restriction? (b) the number of boys must be greater than the number of girls? (c) the number of girls must be greater than the number of boys? 15. An examination paper consists of two parts, Part A and Part B. Part A consists of ten questions and Part B has seven questions. A pupil is required to answer any eight questions from Part A and any four questions from Part B. In how many ways can the questions be chosen? 16. A school parent-teacher committee consisting of six members is to be formed from seven parents, three teachers, the vice-principal and the principal. In how many ways can the committee be formed in order to include (a) the principal and the vice-principal? (b) only teacher and parents? (c) exactly two teachers? (d) exactly four parents? (e) not more than four parents?



Challenge 1. Find the total number of different permutations, taking all the letters at the same time, of the letters in the word MATHEMATICAL. 2. Calculate the number of different permutations, taking all the letters at the same time, of the letters in the word CONCLUDE. How many of these permutations (a) have all the consonants together? (b) begin with a consonant and end with a vowel? (c) have the vowels together? 3. Every regular telephone number in the Philippines consists of seven numbers from 0 to 9. If a telephone number cannot start with 0, how many telephone numbers can be made up using seven of the ten digits?



Permutations and Combinations



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250



Probability of Combined Events An insurance premium is an amount of money charged by insurance companies for coverage during accidents, sickness, disability or death. The computation of this premium is actually related to probability. As the probability of an older person having major illnesses or dying is higher than that of a younger person, the premium will also be higher. The science of this branch of mathematics is called actuarial science and the person specialized in this field is called the actuary. In this chapter, we shall learn more about probability.



Eight



LEARNING OBJECTIVES At the end of this chapter, you should be able to: • use the Addition Law of Probability to solve problems involving mutually exclusive events, and • use the Multiplication Law of Probability to solve problems involving independent and dependent events.



8.1



Addition Law of Probability and Mutually Exclusive Events



In this section, we will learn the conditions for adding probabilities.



Investigation Mutually Exclusive and Non-Mutually Exclusive Events Eight cards numbered 1 to 8 are placed in a box. A card is drawn at random. Let A be the event of drawing a card with a prime number. Let B be the event of drawing a card with a multiple of 4. Let C be the event of drawing a card with an odd number. 1. List the sample space. Part 1: Mutually Exclusive Events 2. List the favorable outcomes for event A and find the probability that A will occur, i.e. P(A). 3. List the favorable outcomes for event B and find the probability that B will occur, i.e. P(B). 4. Is there any overlap between the favorable outcomes for event A and the favorable outcomes for event B? That is, are there any outcomes that favor the occurrence of both event A and event B? These two events are said to be mutually exclusive. 5. List the favorable outcomes for event A or event B, and find the probability that the combined event A or B, or A  B will occur, i.e. P(A or B) or P(A  B). 6. Is P(A  B) = P(A) + P(B) in this case? Can you explain why? Part 2: Non-Mutually Exclusive Events 7. List the favorable outcomes for event C and find the probability that C will occur, i.e. P(C). 8. Is there any overlap between the favorable outcomes for event A and the favorable outcomes for event C? That is, are there any outcomes that favor the occurrence of both event A and event C? These two events are said to be non-mutually exclusive. 9. List the favorable outcomes for event A or event C, and find the probability that the combined event A or C, or A  C will occur, i.e. P(A or C) or P(A  C). 10. Is P(A  C) = P(A) + P(C) in this case? Can you explain why?



253



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Probability of Combined Events



SIMILAR QUESTIONS



Exercise 8A Questions 5, 10



From the investigation, we can conclude that if two events A and B cannot occur at the same time (i.e. the events are mutually exclusive), then P(A or B) or P(A  B) = P(A) + P(B). On the other hand, if two events A and C can occur at the same time (i.e. the events are non-mutually exclusive), then P(A or C) or P(A  C) ≠ P(A) + P(C). In general, the Addition Law of Probability states that if A and B are mutually exclusive events, P(A or B) or P(A  B) = P(A) + P(B).



Worked Example



1



RE



(Probability involving Mutually Exclusive Events) A card is drawn at random from a standard pack of 52 playing cards. Find the probability that the card is (i) an Ace or a King, (ii) a heart or a diamond, (iii) neither a King nor a Queen.



Solution:



(i) P(Ace or King) = P(Ace) + P(King) 4 4 = + 52 52 1 1 = + 13 13 2 = 13 (ii) P(heart or diamond) = P(heart) + P(diamond) 13 13 + = 52 52 1 1 = + 4 4 1 = 2 (iii) P(King or Queen) = P(King) + P(Queen) 4 4 + = 52 52 1 1 = + 13 13 2 = 13 ∴ P(neither King nor Queen) = 1 – P(King or Queen) 2 = 1− 13 11 = 13



CAL



L



There are 4 suits in a standard pack of 52 playing cards, i.e. club , diamond , heart and spade . Each suit has 13 cards, i.e. Ace, 2, 3, …, 10, Jack, Queen and King. All the clubs and spades are black in color. All the diamonds and hearts are red in color. All the Jack, Queen and King cards are picture cards.



P So roblem lvin g T ip



Since only one card is drawn, the events of drawing an ace and a king cannot occur at the same time, i.e. the events are mutually exclusive. By identifying that the two events are mutually exclusive, we apply the Addition Law of Probability to obtain the answer, i.e. P(Ace or King) = P(Ace) + P(King).



P So roblem lvin g T ip



Alternative solution for (iii): Number of cards excluding King and Queen cards = 52 – 8 = 44 44 52 11 = 13



P(neither King nor Queen) =



Probability of Combined Events







Chapter 8



254



PRACTICE NOW 1



SIMILAR QUESTIONS



A card is drawn at random from a standard pack of 52 playing cards. Find the probability of drawing (i) a picture card or an Ace, (ii) an Ace or a card bearing a number which is divisible by 3, (iii) a King or a Queen, (iv) neither a Jack nor an Ace.



Worked Example



2



Exercise 8A Questions 1-3, 6



(Probability involving Mutually Exclusive Events) The probabilities of three teams, L, M and N winning a 1 1 1 football competition are , and respectively. 4 8 10 Assuming only one team can win, calculate the probability that (i) either L or M wins, (ii) neither L nor N wins. AT



Solution:



(i) P(L or M wins) = P(L wins) + P(M wins) 1 1 = + 4 8 3 = 8 (ii) P(L or N wins) = P(L wins) + P(N wins) 1 1 = + 4 10 7 = 20



NTI



ON



Since only one team can win, the events of each of the teams, L, M and N winning are mutually exclusive. How do we deduce that there are more than three teams in the competition? From (ii), Is P(neither L nor N wins) = P(L not winning) + P(N not winning)? Explain.



P(neither L nor N wins) = 1 – P(L or N wins) 7 = 1− 20 13 = 20



PRACTICE NOW 2



The probabilities of four teams, P, Q, R and S winning the National Football 1 1 1 1 and respectively. Assuming only one team can Championship are , , 5 6 7 8 win the championship, find the probability that (i) either P or Q wins the championship, (ii) Q or R or S wins the championship, (iii) none of these teams win the championship.



255



TE



Chapter 8



Probability of Combined Events



SIMILAR QUESTIONS



Exercise 8A Questions 4, 7-9



Exercise



8A



1. Eleven cards numbered 11, 12, 13, 14,…, 21 are placed in a box. A card is removed at random from the box. Find the probability that the number on the card is (i) even, (ii) prime, (iii) either even or prime, (iv) divisible by 3, (v) neither even nor prime. 2. A bag contains 7 red, 5 green and 3 blue marbles. A marble is selected at random from the bag. Find the probability of selecting (i) a red marble, (ii) a green marble, (iii) either a red or a green marble, (iv) neither a red nor a green marble. 3. The letters of the word ‘MUTUALLY’ and the word ‘EXCLUSIVE’ are written on individual cards and the cards are put into a box. A card is picked at random. What is the probability of picking (i) the letter ‘U’, (ii) the letter ‘E’, (iii) the letter ‘U’ or ‘E’, (iv) a consonant, (v) the letter ‘U’ or a consonant, (vi) the letter ‘U’ or ‘E’ or ‘L’? 4. The probability of a football team winning any 7 match is and the probability of losing any 10 2 match is . What is the probability that 15 (i) the team wins or loses a particular match, (ii) the team neither wins nor loses a match?



5. In a probability experiment, three fair coins are tossed, one after another. (a) Display all the possible outcomes of the experiment using a tree diagram. (b) For the experiment, the events A, B, C and D are defined as follows: A: All three coins show heads. B: At least two coins show tails. C: Exactly one coin shows a head. D: The sides appear alternately. For each part, identify if the following events are mutually exclusive. (i) A, B (ii) C, D (iii) B, C (iv) A, C (v) B, D (vi) A, B, C 6. A card is drawn at random from a standard pack of 52 playing cards. Find the probability of drawing (i) a King or a Jack, (ii) a Queen or a card bearing a prime number, (iii) a card bearing a number that is divisible by 3 or by 5, (iv) neither a King nor a Jack. 7. When a golfer plays any hole, the probabilities 1 2 3 that he will take 4, 5 or 6 strokes are , and 14 7 7 respectively. He never takes less than 4 strokes. Calculate the probability that in playing a hole, he will take (i) 4 or 5 strokes, (ii) 4, 5 or 6 strokes, (iii) more than 6 strokes.



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256



8. In a basketball tournament, three of the participating teams are Alpha, Beta and Gamma. The probabilities of each of these three teams 4 1 1 winning the tournament are , and 15 10 5 respectively. Find the probability that (i) Alpha or Gamma will win the tournament, (ii) Alpha, Beta or Gamma will win the tournament, (iii) neither Alpha nor Gamma will win the tournament, (iv) none of these three teams will win the tournament. 9. Every year, only one student can win the Student of the Year Award. The probabilities of Ann, 1 1 Chris and Kate winning the award are , 3 8 1 and respectively. What is the probability that 20 (i) one of them will win the award, (ii) none of them will win the award, (iii) Ann and Chris will not win the award?



8.2



10. In a game, Miguel attempts to score a penalty kick against a goalkeeper who will try to save his shot. There is an equal chance that he will score or miss his penalty kick. Miguel has three chances to score, and the game ends once Miguel scores a penalty kick. (a) Draw a tree diagram to show all the possible outcomes. What is the total number of outcomes? (b) Events A and B are defined as follows: A: exactly two penalty kicks are attempted. B: at most two penalty kicks are attempted. Are A and B mutually exclusive events? Explain your answer.



Multiplication Law of Probability and Independent Events



In this section, we will learn another type of diagram to represent the sample space, and the conditions for multiplying probabilities.



Class Discussion Choosing a Diagram to Represent the Sample Space Discuss in pairs.



There are 3 blue balls and 2 red balls in a bag. The balls are identical except for their color. A ball is drawn at random from the bag and is replaced. A second ball is then drawn at random from the bag. 1. Try representing the sample space for this probability experiment using (a) a possibility diagram, and (b) a tree diagram. 2. Is it easy or tedious to represent the sample space in each diagram?



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From the class discussion, we observe that it is still possible to draw a possibility diagram as shown in Fig. 8.1(a). But what happens if there is a third draw, or if there are 8 blue balls and 2 red balls? Then it is not possible to draw a possibility diagram for the former case, and it will be very tedious to draw a 10-by-10 tree diagram for the latter case. Similarly, it is very tedious to draw a tree diagram with 5  5 = 25 branches for the investigation.



Second Draw



Therefore, there is a need to simplify the tree diagram to represent the sample space. Fig 8.1(b) shows the use of a simplified tree diagram to represent the sample space for the above experiment, where B represents ‘blue’ and R represents ‘red’, and the probability on each branch represents the probability for the occurrence of the outcome at the end of the branch. First Draw



R R B B B B B B R R First Draw (a)



3 5



B



2 5



R



Second Draw B



3 5



(b)



2 5 3 5



R



2 5



R



B: Blue R: Red



B



Fig. 8.1



Second Draw



Figure 8.2 shows how we could use the possibility diagram in (a) and the tree diagram in (b) to find P(BR), the probability that the first ball drawn is blue and the 6 second ball drawn is red. The answer is in both cases. 25 First Draw



R R B B B



3 5



B B B R R First Draw (a) number of blue balls 3× 2 5×5 6 = 25



P(BR) =



number of red balls



B



2 5



R (b)



3 5



3 2 × 5 5 6 = 25



total number of balls in each draw



Fig. 8.2



B



2 5 3 5



R



2 5



R



probability of drawing first blue ball P(BR) =



Second Draw



B: Blue R: Red



B



probability of drawing red ball after first blue ball



In other words, we can multiply along the connected branches of a probability tree to find the above answer.



Probability of Combined Events



Chapter 8



258



Worked Example



3



(Use of Tree Diagram) There are 7 green marbles and 3 yellow marbles in a bag. The marbles are identical except for their color. A marble is drawn at random from the bag, and is replaced. A second marble is then drawn at random from the bag. Find the probability that (i) the first marble drawn is yellow, (ii) the second marble drawn is yellow given that the first marble drawn is green, (iii) the first marble drawn is green and the second marble drawn is yellow, (iv) the second marble drawn is yellow.



Solution:



First Draw 7 10



3 10



(i) P(first marble is Y) =



3 (see 10



Second Draw 7 10



G



Y



3 10 7 10



G: Green Y: Yellow



G Y



G



This is P(second marble is Y, given that first marble is G).



3 Y 10 Number of yellow marbles in bag for first draw — Total number of marbles in bag for first draw )



(ii) P(second marble is Y, given that first marble is G) Number of yellow marbles in bag for second draw, given first marble is green 3 = (see — ) Total number of marbles in bag for second draw 10 7 3  10 10 21 = 100



(iii) P(first marble is G and second marble is Y) or P(GY) =



(iv) P(second marble is Y) = P(GY) + P(YY) ⎛7 3⎞ ⎛3 3⎞ = ⎜ × ⎟+⎜ × ⎟ ⎝ 10 10 ⎠ ⎝ 10 10 ⎠ 21 9 + 100 100 30 = 100 3 = 10



=



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Probability of Combined Events



PRACTICE NOW 3



SIMILAR QUESTIONS



A box contains 5 blue pens and 7 red pens. The pens are identical except for their color. A pen is selected at random from the box and its color is noted. The pen is replaced back into the box. A second pen is then selected at random from the box. Find the probability that



Exercise 8B Questions 1-3, 8-9



(i) the first pen selected is red, (ii) the second pen selected is blue, given that the first pen selected is blue, (iii) the first pen selected is blue and the second pen selected is blue, (iv) the second pen selected is blue, (v) no blue pen was selected.



Independent Events Two events are independent events if the chance of one of them occurring does not affect the chance of the other event occurring. From Worked Example 3, we observe that: 3 , and 10 3 P(second marble is Y, given that first marble is Y) = . 10



P(second marble is Y, given that first marble is G) =



These two probabilities are equal, regardless of whether the first marble drawn is green or yellow, because the first marble is replaced in the bag before drawing the second marble. In other words, the first event of drawing a green or yellow marble does not affect the second event of drawing a yellow marble. We say that the second event is independent of the first event.



AT



TE



NTI



ON



Independent events are not the same as mutually exclusive events.



Let A be the event that the first marble drawn is green and B be the event that the second marble drawn is yellow. 7 3 From Worked Example 3(iii), we observe that P(A and B) =  , where 10 10 3 is the probability that the second marble is yellow, given that the first 10 3 marble is green. Since P(B) is also equal to from (iv), we can write 10 P(A and B) = P(A)  P(B) in this case.



In general, the Multiplication Law of Probability states that if A and B are independent events, P(A and B) or P(A  B) = P(A)  P(B).



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260



Worked Example



4



(Probability involving Independent Events) There are 25 boys and 15 girls in a class. 12 of the boys and 5 of the girls wear spectacles. A class monitor and a class monitress are selected at random from the 25 boys and the 15 girls respectively. What is the probability that both the class monitor and monitress wear spectacles?



Solution:



12 25 5 P(monitress wears spectacles) = 15 1 = 3



P(monitor wears spectacles) =



Since the selections of the monitor and the monitress are independent, 12 1 P(monitor and monitress wear spectacles) =  25 3 4 = 25



PRACTICE NOW 4



1. Workers from a company work in either the ‘Administrative’ Department or the ‘Technical’ Department. There are 18 men and 12 women in the company. 12 men and 4 women are from the ‘Technical’ Department. A chairman and a chairwoman are selected at random from the 18 men and the 12 women respectively. Find the probability that (i) both the chairman and chairwoman are from the ‘Technical’ Department, (ii) the chairman is from the ‘Administrative’ Department and the chairwoman is from the ‘Technical’ Department. 2. Eric has two laptops, Laptop X and Laptop Y. In any one year, the probability of Laptop X breaking down is 0.1 and the probability of Laptop Y breaking down is 0.35. In any one year, what is the probability that (i) both laptops break down, (ii) Laptop X breaks down but Laptop Y does not, (iii) exactly one of the laptops breaks down?



261



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Probability of Combined Events



SIMILAR QUESTIONS



Exercise 8B Questions 4, 10-11, 17



Dependent Events



Investigation Dependent Events There are 7 green marbles and 3 yellow marbles in a bag. The marbles are identical except for their color. Two marbles are drawn at random from the bag (i.e. without any replacement). 1. Copy and complete the probabilities on the probability tree in Fig. 8.3. First Draw 7 10      



G



Y



     



3 ⎛ 1⎞ ⎜= ⎟ 9 ⎝ 3⎠



Second Draw G Y G: Green Y: Yellow G



2 9



Y



Fig. 8.3 2. Find the probability that (i) the second marble drawn is yellow, given that the first marble drawn is green, (ii) the second marble drawn is yellow, given that the first marble drawn is yellow. 3. Are the probabilities in Question 2(i) and (ii) equal? Does the probability of drawing a yellow marble in the second draw depend on the outcome in the first draw? Explain. 4. Find the probability that the second marble drawn is yellow. Is this probability equal to the probabilities in Question 2(i) and (ii)? 5. Let A be the event that the first marble drawn is green and B be the event that the second marble drawn is yellow. (i) Is event B independent or dependent on Event A? Explain. (ii) Does the Multiplication Law of Probability, P(A and B) = P(A)  P(B) apply in this case?



Probability of Combined Events



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262



From the investigation, we observe that the Multiplication Law of Probability, P(A and B) = P(A)  P(B), does not apply if A and B are dependent events. However, when we are finding the probability for P(A and B) in Question 5(ii) in the investigation, we still multiply the probabilities across two connected branches, 7 3 7 3 i.e. P(A and B) = P(GY) =  = , where ≠ P(B). We can still multiply across two 10 9 30 9 connected branches because the second probability on the branch is not the probability of the second event B.



Worked Example



5



(Probability involving Dependent Events) Out of 33 students in a class, 21 study Geography and 12 study History. No student studies both subjects. Two students are picked at random from the class. Find the probability that (i) the first student studies History and the second student studies Geography, (ii) one student studies History while the other student studies Geography.



Solution: 21 ⎛ 7 ⎞ ⎜= ⎟ 33 ⎝ 11 ⎠



12 ⎛ 4 ⎞ ⎜= ⎟ 33 ⎝ 11 ⎠



First Draw



Second Draw



20 ⎛ 5 ⎞ ⎜= ⎟ 32 ⎝ 8 ⎠ G 12 ⎛ 3 ⎞ ⎜= ⎟ 32 ⎝ 8 ⎠ 21 32 H 11 32



4 21 × 11 32 21 = 88



(i) P(HG) =



(ii) P(H and G) = P(HG) + P(GH) ⎛ 4 21 ⎞ ⎛ 7 3 ⎞ = ⎜ × ⎟+⎜ × ⎟ ⎝ 11 32 ⎠ ⎝ 11 8 ⎠ 21 21 = + 88 88 21 = 44



263



Chapter 8



P So roblem lvin g T ip



Probability of Combined Events



Although the question does not say that the first student is not replaced in the class before picking the second student, this is assumed.



G H G: Geography H: History G H



This is because if the two students are picked together, it is the same as picking at random (same as the no replacement case).



PRACTICE NOW 5



SIMILAR QUESTIONS



1. Mr Sam, a Science teacher, needs two students to assist him with a Science demonstration. Two students are picked at random from his class of 16 boys and 12 girls. Using a tree diagram, find the probability that (i) the first student is a boy and the second student is a girl, (ii) one student is a boy while the other student is a girl, (iii) at least one of the students is a girl.



Exercise 8B Questions 5-7, 12-16, 18-19



2. A bag contains 8 red balls, 7 blue balls and 1 white ball. Two balls are drawn from the bag at random, one after another, without replacement. Find the probability that (i) the first ball is red and the second ball is blue, (ii) one ball is red while the other ball is blue, (iii) the two balls are of the same colour.



Performance Task The mathematical constant pi, π is used to find the circumference or the area of a circle. A commonly used value of pi is 3.142. Since π is a irrational number, it has an infinite, non-repeating number of decimal digits, i.e. its value is 3.141 592.... We can approximate the value of π by simulation modeling, which involves the use of a computer program to generate a scenario based on a set of rules in order to study the outcomes of the interactions of the variables in the model. For this question, we will use a spreadsheet to generate random points that fall inside a square of length 2 units, as shown in Fig. 8.4.



Fig. 8.4 We are interested to find out the number of points that fall inside the unit circle, i.e. a circle of radius 1 unit. This will help us to estimate the area of the unit circle.



Probability of Combined Events



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264



(a) What is the area of the unit circle? Set up the following in a spreadsheet, as shown in Fig. 8.5. The formula in each of the cells in Columns A and B, from A2 to A51 and from B2 to B51 is =2*RAND()-1. Enter the given formula in the cell A2 to generate a random number. Click on the small black square in the lower-right corner of the cell A2, hold and drag it down to the cell A51. This will generate 50 random numbers, from the cell A2 to A51. Repeat this process for Column B. Hence, we will generate 50 points with coordinates (x, y), which fall within a square of length 2 units.



Fig. 8.5 The formula for the cell C2 in Column C is =IF((A2^2+B2^2)=0”). This will count all the points inside the square. The cell F2 gives an estimate of the area of the unit circle, or an estimate of the value of π. (c) What formula should you use in F2? Select the cells from A2 to B51. Insert a scatter plot with markers only. It should look like the diagram in (a), but with fewer points and without the circle.



265



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Probability of Combined Events



Record the value in F2. For most spreadsheets, you can press the 'F9' button to re-generate another 50 random points. Record the new value in F2, and take the average of the two values to find the mean area of the unit circle based on 100 random points. Continue this process until you get 1000 random points. (d) What is the mean area of the unit circle based on 1000 random points? Is it close enough to the value of π?



Exercise



8B



1. Eric has two bags each containing 5 black marbles and 4 red marbles. He takes one marble at random from each bag. (a) Copy and complete the tree diagram shown below. First Bag



2. A bag contains 6 red balls and 4 yellow balls. A ball is chosen at random and then put back into the bag. The process is carried out twice. (a) Copy and complete the tree diagram shown below. First Draw ⎛ ⎞ ⎜ ⎟ ⎝ ⎠



Second Bag



⎛ ⎞ ⎜ ⎟ ⎝ ⎠



Black Black







Red



Red



Black Red



(b) Find the probability that Eric draws (i) a black marble from the first bag,



Second Draw



⎛ ⎞ ⎜ ⎟ ⎝ ⎠



Red



Yellow



⎛ ⎞ ⎜ ⎟ ⎝ ⎠ ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ ⎛ ⎞ ⎜ ⎟ ⎝ ⎠







Red Yellow



Red Yellow



(b) Find the probability of choosing (i) two red balls, (ii) one ball of each color, (iii) a yellow ball on the second draw.



(ii) a red marble from the second bag, given that he draws a black marble from the first bag, (iii) a black marble from the first bag and a red marble from the second bag, (iv) a red marble from the second bag.



Probability of Combined Events



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266



3. The diagram below shows two discs, each with four equal sectors. Each disc has a pointer which, when spun, is equally likely to come to rest in any of the four equal sectors. In a game, the player spins each pointer once. His score is the sum of the numbers shown by the pointers.



4. Eric takes either Bus A or Bus B to school every day. Bus A and B either arrive punctually or late. The probabilities of Bus A and B arriving 2 7 punctually are and respectively. Find the 3 8 probability that (i) both buses are punctual, (ii) Bus A is late while Bus B is punctual,



10



20



0



30



20



30



30



30



(iii) exactly one of the buses is late.



First Disc Second Disc (a) Copy and complete the tree diagram shown. First Disc Second Disc Sum 10



0 ⎛1⎞ ⎜ ⎟ ⎝4⎠



10



)







0



(



)



(b) Copy and complete the tree diagram shown below.



0 30







( (



First Boy



)



(ii) the second number obtained is zero. (c) If the player’s score is between 10 and 50 but excluding 10 and 50, he receives 20. If his score is more than 40, he receives 50. Otherwise, he receives nothing. What is the probability that he receives (iii) 20 or 50,



267



Chapter 8



(ii) 50, (iv) nothing?



Probability of Combined Events



⎛ 3⎞ ⎜ ⎟ ⎝8⎠



⎛ ⎞ ⎜ ⎟ ⎝ ⎠ 







Second Boy



⎛ ⎞ ⎜ ⎟ ⎝ ⎠



)



(i) the first number obtained is less than or equal to the second number obtained,



20,



On another occasion, 2 boys are chosen at random from the same group of 8 boys.



50



(b) With the help of the diagram, calculate the probability that



(i)



A second boy is then chosen at random from the remaining 7 boys. What is the probability that the second boy chosen is also left-handed, given that the first boy chosen is left-handed?



(



30



30



(a)



30



20



⎜ ⎟ ⎝ ⎠



5. In a group of 8 boys, 3 are left-handed. The remaining 5 boys are right-handed. If a boy is chosen at random from the group, state the probability that the boy chosen is left-handed.



L



R



L



⎞ ⎟ ⎠ ⎞ ⎟ ⎠



R



⎛ ⎞ ⎜ ⎟ ⎝ ⎠



R



⎛ ⎜ ⎝ ⎛ ⎜ ⎝



L: Left-handed R: Right-handed L



(c) From the tree diagram in (b), find the probability that (i) the first boy chosen is right-handed and the second boy chosen is left-handed, (ii) both boys chosen are left-handed, (iii) the second boy chosen is left-handed.



6. A class has 30 girls and 15 boys. Two representatives are to be selected at random from the class. Find the probability that (i) the first representative is a girl, (ii) the second representative is a girl, given that the first representative is a boy, (iii) the first representative is a boy and the second representative is a girl, (iv) a boy and a girl are selected as representatives. 7. A bag contains 6 green and 4 blue cards. (a) A card is drawn at random. Find the probability that it is green. (b) The card drawn is returned to the bag and after mixing the cards thoroughly, Megan takes two cards at random from the bag, one after another. Using a tree diagram, calculate the probability that Megan has taken out



(b)



Miguel wants to find out the probability of selecting two rotten potatoes. He multiplies the probabilities along the ‘RR’ branch (highlighted in red) and he says that he is using the ‘Multiplication Law of Probability’.







Do you agree with what Miguel says? Explain your answer clearly.



9.



A red die has the number 1 on one face, the number 2 on two faces and the number 3 on three faces. Two green dice each has the number 6 on one face and the number 5 on five faces. The three dice are rolled together.



(a) Copy and complete the tree diagram shown below by writing the probabilities on the ‘branches’. Red Die



First Second Green Die Green Die



5 5



6 5



1



(i) two green cards,



6



(ii) one card of each color, (iii) at least one blue card.



6 5



5



6 5



2 6 8. Bag A contains 20 potatoes, 4 of which are rotten. Bag B contains 12 potatoes, 3 of which are rotten. Miguel selects one potato at random from each bag. (a) Complete the tree diagram below to show the possible outcomes of Miguel's selections.



Bag A



Bag B



G



 



R G



6 5



6



6







(i) 2 on the red die, 5 on the first green die and 6 on the second green die,



R: Rotten G: Good G



5 3



(b) Using the tree diagram, calculate the probability of obtaining



R R



6 5



(ii) 3 on the red die and 6 on each of the two green dice,



(iii) exactly two sixes,







(iv) a sum of 12,







(v) a sum which is divisible by 3.



Probability of Combined Events



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268



10. A woman goes to the supermarket once a month for grocery shopping. 4 The probability that she buys a sack of rice is . 9 Find the probability that (i) she will not buy a sack of rice in a particular month, (ii) she will not buy a sack of rice in two particular consecutive months, (iii) she buys a sack of rice in just one of two particular months. 11. The table below shows the number of male and female employees working in the front office, middle office and back office of an investment bank. Department Front Office Gender







Middle Office



Back Office



Male



40



55



38



Female



36



35



52



Three representatives, one from each department, are selected at random to attend a seminar. What is the probability that



(i) all three representatives are females, (ii) the representative from the front office is a male while the others are females,



(iii) exactly one of the representatives is a male?



12. In a wardrobe, there are 16 shirts, of which 8 are black, 6 are white and 2 are blue. The shirts are identical except for their color. (a) If two shirts are taken out of the wardrobe, find the probability that (i) both are black, (ii) one shirt is black and the other is white, (iii) the two shirts are of the same color.



269



13. Ten cards are marked with the letters P, R, O, P, O, R, T, I, O and N respectively. These cards are placed in a box. Two cards are drawn at random, without replacement. Calculate the probability that (i) the first card bears the letter ‘O’, (ii) the two cards bear the letters ‘P’ and ‘O’ in that order, (iii) the two cards bear the letters ‘P’ and ‘O’ in any order, (iv) the two cards bear the same letter. 14. Five balls numbered 1, 2, 5, 8 and 9 are put in a bag. (a) One ball is selected at random from the bag. Write down the probability that it is numbered ‘8’. (b) On another occasion, two balls are selected at random from the bag. Find the probability that



(i) the number on each ball is even,



(ii) the sum of the numbers on the balls is more than 10, (iii) the number on each ball is not a prime number, (iv) only one ball bears an odd number. 15. Box A contains 7 blue balls and 5 yellow balls. Box B contains 3 blue balls and 7 yellow balls. One ball is removed at random from Box A and placed into Box B. After thoroughly mixing the balls, a ball is drawn at random from Box B and placed back into Box A. (a) Draw a tree diagram to illustrate this experiment. (b) Find the probability that at the end of the experiment, Box A has



(b) If a third shirt is taken out from the wardrobe, calculate the probability that all three shirts are black.



(i) more yellow balls than blue balls,



(c) Can we use the Multiplication Law of Probability to obtain the answer in (b)? Explain your answer.



balls.



Chapter 8



Probability of Combined Events



(ii) exactly 7 blue and 5 yellow balls, (iii) twice as many blue balls as yellow



16. Class A has 18 boys and 17 girls and Class B has 14 boys and 22 girls. A student from Class A is transferred to Class B. The teacher selects a student at random from the extended Class B. Find the probability that the student selected is (i) the student who was initially from Class A, (ii) a boy.



17. In any given year, the probabilities of a volcanic eruption in each of the countries A, B and C are 0.03, 0.12 and 0.3 respectively. For any given year, find the probability that (i) volcanic eruptions will occur in all three countries,



18. A bag contains 10 red balls, 9 blue balls and 7 yellow balls. Three balls are drawn in succession without replacement. By drawing a tree diagram or otherwise, find the probability of obtaining (i) a red and two blue balls in that order, (ii) a red, a yellow and a blue ball in that order, (iii) three balls of different colours. 19. A game is such that a fair die is rolled repeatedly until a ‘6’ is obtained. Find the probability that



(a) (i) the game ends on the third roll,



(ii) the game ends on the fourth roll, (iii) the game ends by the fourth roll.



(ii) no volcanic eruptions will occur,



(b) Suppose now that the game is such that the same die is rolled repeatedly until two ‘6’s are obtained. Find the probability that



(iii) there is at least one volcanic eruption,



(i) the game ends on the third roll,



(iv) there are exactly two volcanic eruptions.



(ii) the game ends on the third roll and the sum of the scores is odd.



1. The Addition Law of Probability states that if A and B are mutually exclusive events, then P(A or B) or P(A  B) = P(A) + P(B).



2. The Multiplication Law of Probability states that if A and B are independent events, then P(A and B) or P(A  B) = P(A)  P(B).



Probability of Combined Events



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270



8 1. Assuming that the birthdays of people are equally likely to occur in any month, find the probability that (i) two people selected at random are born in the same month, (ii) three people selected at random are not born in the same month, (iii) four people selected at random are born in the same month. 2. On any day, the probability that Imee will miss 1 her bus is . Find the probability that 7 (i) she will catch her bus on a particular day, (ii) she will miss her bus on two particular consecutive days, (iii) she will miss her bus on just one of two particular consecutive days, (iv) she will catch her bus on three particular consecutive days. 3. The probabilities of Antonio, Eric and Robert winning the gold medal for the 100-meter 1 1 1 freestyle swimming competition are , and 2 6 8 respectively. Find the probability that



A box contains x white chocolates and y dark chocolates. Eric selects a chocolate from the box, followed by Kristel who also selects a chocolate. Find, in terms of x and y, the probability that



(i) Eric selects a dark chocolate, (ii) Eric selects a white chocolate while Kristel selects a dark chocolate, (iii) the chocolates selected by them are of different types. 6. In a city, the probability that it will rain on any 1 particular day is . The probability of a traffic jam 4 2 1 is when it rains and when it does not rain. 5 5 What is the probability that there will be a traffic jam in the city on a particular day? 7. In a class of 30 students, 20 are boys and 10 are girls. Of the 10 girls, 6 travel to school by bus and 4 travel by car. (a) If two students are selected at random, calculate the probability that (i) one is a girl and one is a boy, (ii) no girls are selected.



(i) one of them wins the gold medal,



(b) If two of the 10 girls are selected at random, calculate the probability that



(ii) none of them win the gold medal,



(i) both travel to school by bus,



(iii) Antonio fails to win the gold medal.



(ii) both travel to school by different means of transportation,



4.



The letters of the word ‘FOLLOW’ are written on six individual cards. The cards are placed face down on the table and their positions are rearranged randomly. The cards are turned over one at a time. For each of the following cases, find the probability that



(i) the first two cards turned over will each have the letter ‘O’ written on them, (ii) the second card turned over will have the letter ‘F’ written on it, (iii) the word ‘FOLLOW’ is obtained, in that order. 271



5.



Chapter 8



Probability of Combined Events



(iii) at least one travels to school by bus. 8. A weather forecast station describes the weather for the day as either fine or wet. If the weather is fine today, the probability that it will be fine the next day is 0.8. If the weather is wet today, the probability that it will be wet the next day is 0.6. Given that Monday is wet, find the probability that (i) the next two days will also be wet, (ii) Tuesday will be wet and Wednesday will be fine,



10. Three airplanes are scheduled to land at either Terminal 1 or Terminal 2 of the airport. The probabilities of each airplane landing at Terminal 1 3 2 5 are , and respectively. Find the probability 4 3 6 that



(iii) there will be one fine day and one wet day for the next two days, (iv) two of the next three days will be wet. 9. A bowl of sweets contains 2 fruit gums, 3 mints and 5 toffees. Three sweets are to be chosen at random, without replacement, from the bowl. Calculate the probability that



(i) all three airplanes land at Terminal 2, (ii) exactly two airplanes land at Terminal 1, (iii) exactly one airplane lands at Terminal 1.



(i) the first two sweets chosen will be different,



(ii) the three sweets chosen will be the same, (iii) of the three sweets chosen, the first two will be the same and the third will be a toffee.



Challenge 1. During a game show, the host picks you to take part in a contest and you are given the following scenario.



There are three closed doors and you are invited to pick one of them. There is a brand new car behind one of the doors, and a goat behind each of the other two doors. The host knows what is behind the doors.







Suppose that you pick Door 1 and the host opens Door 3 because he knows that the car is not behind it, as shown below. 1



2



3



You are then given the option to switch to Door 2. Should you switch doors to increase your probability of winning the car? Hint: This is counter-intuitive. Search on the Internet for an explanation to this famous probability puzzle and compare it with your own reasoning.



2. (a) There are 367 students in the school hall. What is the probability that at least two of the students have their birthday falling on the same day of the year (i.e. they do not have to be born in the same year)? (b) There are 40 students in the classroom. What is the probability that at least two of the students have their birthday falling on the same day of the year? Is the probability very high? (c) What is the least number of students in a classroom for the probability that at least two of them have their birthday falling on the same day of the year to be greater than 0.5?



Probability of Combined Events



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272



Statistical Data Analysis



In Singapore, the daily mean maximum temperature was 31.2 °C in 2012 and 31.3 °C in 2013. Based on these statistics, can we conclude that the temperatures for both years were about the same? To have a better understanding of the temperatures for both years, we should also know how the temperatures were spread throughout the year. In this chapter, we will learn how to measure the spread of a set of data using statistics such as the interquartile range and standard deviation.



Nine



LEARNING OBJECTIVES At the end of this chapter, you should be able to: • construct and interpret data from dot diagrams and stem-and-leaf diagrams, • state the features of cumulative frequency curves, • interpret and analyze cumulative frequency curves, • estimate the median, quartiles, percentiles and deciles from cumulative frequency curves, • calculate the quartiles for a set of discrete data, • interpret and analyze box-and-whisker plots, • calculate the standard deviation, • use the mean and standard deviation to compare two sets of data.



9.1



Statistical Diagrams



Recap (Pictograms, Bar Graphs, Pie Charts and Line Graphs) The choice of an appropriate statistical diagram depends on the type of data collected and the purpose of collecting the data. In Grade 7, we have learned the following statistical diagrams as shown in Table 9.1.



Students’ Favorite Fruit



INF



OR



MA



TIO N



‘Data’ is the plural of ‘datum’.



Students’ Favorite Fruit 160



Number of students



Apples Honeydew Pears Watermelons



140 120 100 80 60 40 20 0



Oranges



Apples Honeydew Pears Watermelons Oranges



Type of Fruit Pictogram



Bar graph



Students’ Favorite Fruit



Rubber Consumption



Oranges



Apples 72˚ 54˚ 108˚



Honeydew



36˚



Watermelons



Consumption (tonnes)



250 000



200 000



150 000



100 000



50 000



Pears



0



January February



March



Month Pie chart



Line graph Table 9.1



We have also learned to display information by constructing a histogram. In this chapter, we will learn more statistical diagrams and how to analyze them.



275



Chapter 9



Statistical Data Analysis



April



May



9.2 Dot Diagrams A dot diagram consists of a horizontal number line and dots placed above the number line. The dots represent the values in a set of data.



Worked Example



1



(Dot Diagram) The table shows the time taken, in minutes, for 12 employees of a company to travel from their homes to the office. 20



22



21



21



18



18



30



20



22



23



22



20



(i) Represent the data on a dot diagram. (ii) Briefly describe the distribution of the data.



Solution:



AT



NTI



ON



The values on the number line should always be at equal intervals. For example, the dot diagram in Worked Example 1 should not be drawn like this:



18



(i) Step 1: Identify the range of values, i.e. 18 minutes to 30 minutes. Step 2: Draw a horizontal number line with equal intervals to represent the range of values identified in Step 1. Step 3: Using the data from the table, plot each datum with a dot over its value on the number line.



TE



20



AT



TE



21



NTI



22



23



30



ON



To describe a distribution, we may consider the following. • Range (variation of data) • Clusters (data that is visually grouped about certain values)



18 19 20 21 22 23 24 25 26 27 28 29 30 (ii) The time taken for the 12 employees to travel from their homes to the office ranges from 18 minutes to 30 minutes. The time taken is clustered around 20 to 23 minutes. There is an extreme value of 30 minutes.



• Extreme data (data that deviates significantly from the other data) • Symmetry (equal number of data on each side of a value)



SIMILAR QUESTIONS



PRACTICE NOW 1



The table shows the marks obtained by 16 students in a class for a Mathematics test. 3



18



20



19



14



9



8



13



16



19



14



10



14



10



12



10



Exercise 9A Questions 1, 4, 12



(i) Represent the data on a dot diagram. (ii) If 50% of the students passed the test, find the passing mark for the test. (iii) Briefly describe the distribution of the data.



Statistical Data Analysis



Chapter 9



276



Worked Example



2



(Dot Diagram) The dot diagram represents the masses, in kg, of 24 students.



25



30



35



40



(i) What is the most common mass? (ii) Calculate the percentage of students who have a mass of more than or equal to 32 kg. (iii) A student is selected at random. The probability 1 that the student has a mass of at least x kg is . 3 Find the value of x. (iv) Briefly describe the distribution of the data.



Solution:



(i) The most common mass is 32 kg. (ii) Percentage of students who have a mass of more than or equal to 32 kg 15 = × 100% 24



= 62.5%



(iii) Number of students who have a mass of at least x kg 1 = × 24 (since probability = ) 3 = 8



Since 4 + 2 + 1 + 1 = 8,







25



30



35



40



∴ Value of x = 34



(iv) The masses of the 24 students range from 25 kg to 40 kg. The masses are symmetrical about 32 kg. The extreme masses of 25 kg and 40 kg deviate considerably from the other masses recorded.



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AT



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An advantage of a dot diagram is that it is an easy way to display small sets of data that do not contain many distinct values. A disadvantage is that if a set of data is too large, the diagram may appear packed.



SIMILAR QUESTIONS



PRACTICE NOW 2



The dot diagram represents the ages, in years, of contestants who join a karaoke contest.



10



15



20



25



Exercise 9A Questions 2, 5-6



30



(i) How old is the youngest contestant? (ii) Find the total number of contestants who join the karaoke contest. (iii) A contestant is selected at random. The probability that the age of the contestant 1 is less than or equal to x years is . Find the value of x. 4 (iv) Briefly describe the distribution of the data.



9.3



Stem-and-Leaf Diagrams



A stem-and-leaf diagram consists of a few stems, each with a different number of leaves. Fig. 9.1 shows an example of a stem-and-leaf diagram. Stem 2 3 4



0 0 0



4 0 4



4 0 4



5 6 8



5 6



8 7



9 7



Leaf 7



7



7



7



8



8



8



9



9



Key: 2 | 4 means 24 Fig. 9.1 In Fig. 9.1, each stem represents a digit in the tens place and each leaf represents a digit in the ones places (refer to the key). For example, all the numbers 20, 24, 24, 25, 25, 28 and 29 have a common stem 2. What is the common stem of 40, 44, 44 and 48?



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• In a stem-and-leaf diagram, the stems must be arranged in numerical order. • The leaves must be recorded in ascending order. Do not use commas to separate the leaves. • The number of leaves must tally with the total number of data collected. • An advantage of a stem-and-leaf diagram is that the individual data values are retained; and if the appropriate units are used, the shape of the data distribution can be easily observed. • A disadvantage is that it is not useful if there are extreme values in a set of data as many stems with no leaves will be present.



Statistical Data Analysis



Chapter 9



278



Worked Example



3



(Stem-and-Leaf Diagram)



The table shows the masses, in g, of 10 salt containers. 56



58



67



72



61



63



76



50



64



79



Represent the data using a stem-and-leaf diagram.



Solution:



Stem Leaf 5 0 6 8 6 1 3 4 7 7 2 6 9 Key: 5 | 6 means 56 g



arrange the stems in numerical order



arrange the leaves in ascending order



SIMILAR QUESTIONS



PRACTICE NOW 3



The table shows the number of text messages sent by a group of students on a typical weekend. 20



33



30



59



14



41



56



12



37



44



23



35



Represent the data using a stem-and-leaf diagram.



Sometimes, a stem-and-leaf diagram may have more leaves on some stems, such as the leaves corresponding to the stem 3 in Fig. 9.1. When this happens,



we can break each stem into two halves – one for the leaves numbered 0 to 4 and the other for the leaves numbered 5 to 9. We will then obtain a stem-and-leaf diagram with split stems (see Fig. 9.2). Stem 2 2 3 3 4 4



0 5 0 6 0 8



4 5 0 6 4



4 8 0 7 4



Leaf 9 7



7



7



7



Key: 2 | 4 means 24 Fig. 9.2



279



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7



Statistical Data Analysis



8



8



8



9



9



Exercise 9A Questions 3, 13



Worked Example



4



(Stem-and-Leaf Diagram with Split Stems) The stem-and-leaf diagram represents the amount of money, in pesos, collected by a group of students during a charity run. Stem 7 8 9



0 2 0



4 3 0



8 4 0



9 4 5



Leaf 7 6



9 7



9 9



9



9



9



Key: 7 | 8 means 78



(i) Construct a stem-and-leaf diagram with split stems. (ii) What is the amount of money that was most commonly collected? (iii) Find the ratio of the number of students who collected more than 70 but less than 85 to the total number of students.



Solution:



(i) To construct a stem-and-leaf diagram with split stems, we separate the stems into smaller number of equal-sized units. Stem Leaf 7 0 4 7 8 9 8 2 3 4 4 8 7 9 9 9 9 9 9 0 0 0 9 5 6 7 9 Key: 7 | 8 means 78



(consists of values from 40 – 44) (consists of values from 45 – 49) (consists of values from 50 – 54) (consists of values from 55 – 59) (consists of values from 60 – 64) (consists of values from 65 – 69)



(ii) The amount of money that was most commonly collected is 89. (iii) Required ratio = 7 : 21 =1:3



PRACTICE NOW 4



SIMILAR QUESTIONS



The stem-and-leaf diagram represents the masses, in kg, of some boxes. Stem 8 9 10



0 0 0



3 0 1



4 0 2



6 0 4



Leaf 6 6 1 1 4 6



7 2 8



8 2 8



8 5



9 6



9 7



7



Exercise 9A Questions 7-9



9



Key: 8 | 3 means 83 kg (i) Construct a stem-and-leaf diagram with split stems. (ii) What is the most common mass? (iii) Find the percentage of boxes with a mass more than 85 kg but less than 90 kg. Statistical Data Analysis



Chapter 9



280



When two sets of data are given, we can use a stem-and-leaf diagram with a common stem to represent the data. This is known as a back-to-back stem-and-leaf diagram.



Worked Example



5



(Back-to-Back Stem-and-Leaf Diagram) The lengths, in mm, of terrapins from two different habitats are recorded in the table. Habitat A Habitat B



55



50



60



64



60



71



55



50



62



68



58



63



69



59



75



72



70



64



77



70



69



53



73



69



65



55



(i) Represent these two sets of data by a back-to-back stem-and-leaf diagram. (ii) Which habitat is more suitable for the growth of terrapins? Explain your answer.



Solution:



(i) A back-to-back stem-and-leaf diagram consists of a stem and leaves on both sides of the stem. Step 1: Construct a stem-and-leaf diagram for Habitat A. Step 2: U sing the same stem, construct the stem-and-leaf diagram for Habitat B on the left side of the stem. Leaves for Habitat B 9



9



7



8



9



4



5



5



3



3



2



Stem 5



0



6



0



7



0



0



0



Key: 5 | 5 means 55 mm



Leaves for Habitat A 0



2



1



3



4



5



5



5



9



8 AT



(ii) Habitat B is more suitable for the growth of terrapins as there are more terrapins with greater lengths (between 70 and 77 mm) in Habitat B than in Habitat A.



The times, in minutes, Miguel and Carlo spent using their tablet computers for the past 20 days are recorded in the table.



Carlo



83



83



41



43



59



47



79



42



83



70



52



92



89



65



90



49



77



65



85



82



72



84



78



46



81



88



82



52



91



81



53



46



98



89



84



95



58



99



49



78



(i) Represent these two sets of data by a back-to-back stem-and-leaf diagram. (ii) Who used his tablet computer for the longest time in a day? (iii) Who used his tablet computer for the shortest time in a day? (iv) Who used his tablet computer for a longer duration of time? Explain your answer. 281



Chapter 9



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The leaves corresponding to Habitat B are arranged in ascending order from the right to the left.



SIMILAR QUESTIONS



PRACTICE NOW 5



Miguel



TE



Statistical Data Analysis



Exercise 9A Questions 10 -11



Exercise



9A



BASIC LEVEL



1. The table shows the masses, in kg, of 16 bags of rice. 43



25



30



26



36



30



25



20



36



25



15



20



36



15



36



23



(i) Represent the data on a dot diagram. (ii) What is the mass of the lightest bag of rice? (iii) What is the most common mass? (iv) Bags of rice with a mass of less than 30 kg have to be refilled. How many bags have to be refilled? 2. The dot diagram represents the time taken, in minutes, for a group of people to complete a questionnaire.



3. The table shows the time taken, in hours, for 20 students to complete their Social Studies project. 3.5 7.0 4.8 2.8 5.5 6.8 6.6 5.5 4.5 2.5 6.5 3.6 3.3 4.0 3.5 7.5 4.8 6.5 6.4 2.8 (i) Represent the data using a stem-and-leaf diagram. (ii) Find the percentage of students who spent less than 3 hours to complete their project. (iii) Students who spent more than 6 hours to complete their project had 2 marks deducted. What fraction of the students had 2 marks deducted? INTERMEDIATE LEVEL



4. The average daily temperature (in °C) of a city in autumn is recorded in the table. 22 20 22 19 23 17 21 18 20 21



20 20 18 21 19 21 19 19 21 19



0







5



10



15



(i) What was the most common time taken? (ii) Find the difference between the shortest and the longest time taken to complete the questionnaire. (iii) Find the total number of people who completed the questionnaire. (iv) Find the ratio of the number of people who took 10 minutes or longer to complete the questionnaire to the total number of people who completed the questionnaire.



(i) Represent the data on a dot diagram. (ii) Find the percentage of temperatures recorded that are at most 20 °C. (iii) Comment on the distribution of the data. 5. The dot diagram represents the attention span, in minutes, of 22 pre-schoolers.



4.0



5.0



7.0



6.0



(i) What is the most common attention span? (ii) A pre-schooler is chosen at random. Find the probability that the attention span of the pre-schooler is not more than 5.5 minutes. (iii) Briefly describe the distribution of the data.



Statistical Data Analysis



Chapter 9



282



8. The times, in minutes, for 20 sportsmen to complete a running test are recorded in the table.



6. The dot diagrams represent the results of a Mathematics test by gender. 0



5



10



15



20



25



7.9 8.3 7.8 8.5 7.4 8.3 7.0 8.1 8.3 8.5 8.0 9.4 7.4 8.5 7.5 9.8 8.1 7.7 9.4 8.3



30



Results of the girls



(i) Construct a stem-and-leaf diagram with split stems. (ii) What were the fastest and the slowest time taken? 0 5 10 15 20 25 30 (iii) Find the percentage of sportsmen who took Results of the boys at least 8.5 minutes to complete the test. (iv) A sportsman is selected at random. Find the (a) If the passing mark for the test is 15, how many probability that the sportsman failed the test if boys failed the test? the passing time is 8.4 minutes. (b) Distinctions are awarded to students who scored 25 marks and above in the test. 9. The table shows the unemployment rates in a (i) Express the number of girls who scored country from 2002 to 2012. distinctions as a percentage of the total number of girls in the class. Year Unemployment Rate (%) (ii) Express the number of boys who scored distinctions as a percentage of the total 2002 3.5 number of boys in the class. 2003 4.0 (c) Which gender performed better in the test? 2004 3.5 Explain your answer. 2005 3.2 7. The stem-and-leaf diagram represents the number of chocolates in some packets.



2006



Stem Leaf 2 0 2 4 4 4 5 6 6 6 6 6 7 8 9 9 9 3 1 1 1 2 2 3 4 4 5 5 7 7 8 8 4 0 1 2 3 3 3 5 5 6 8



2008



2007 2009 2010 2011



Key: 2 | 2 means 22 (i) Construct a stem-and-leaf diagram with split stems. (ii) What is the most common number of chocolates? (iii) Find the percentage of packets which have more than or equal to 25 chocolates but fewer than 32 chocolates.



283



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Statistical Data Analysis



2012



2.7 2.0 2.2 2.9 2.2 2.1 2.0



(a) If we say that a year is ‘good’ when the unemployment rate is less than 2.5%, which years are considered ‘good’? (b) Construct (i) a dot diagram, (ii) a stem-and-leaf diagram with split stems. (c) State an advantage of using (i) the dot diagram, (ii) the stem-and-leaf diagram with split stems, to represent the employment rates in the country from 2002 to 2012.







10. The back-to-back stem-and-leaf diagram represents the scores of the students in two different schools for a common Geography quiz. In each school, 29 students sat for the quiz.



Leaves for School P Stem Leaves for School Q 5 2 6 8 9 6 2 5 8 8 9 9 7 4 6 7 8 8 9 8 0 3 4 4 6 7 7 9 0 2 7 8 10 0 0 Key: 6 | 9 means 69 marks







9 9 8 5 3 9 9 9 7 7 6 6 9 8 8 7 6 6 3



4 9 2 4 2



0 6 0 2 0



(i) Which school had the highest scorer? (ii) Which school had the lowest scorer? (iii) Which school performed better in the quiz? Explain your answer. 11. The results of a Mathematics examination of two Grade 10 classes are recorded in the table. Class A Class B



77 69 68 55



70 50 90 94



31 68 42 79



45 98 80 92



39 66 82 66



45 85 88 72



59 60 69 78



63 55 67 70



47 50 69 87



68 70 75 65



(i) Represent these two sets of data by a back-to-back stem-and-leaf diagram. (ii) Find the percentage of students who scored distinctions, i.e. 70 marks and above, in each class. (iii) Which class performed better in the examination? Explain your answer.



ADVANCED LEVEL



12. Some plants are grown by a group of students for an experiment. The table shows the heights, in cm, of the plants after 8 weeks. 4.5



8.1







8.6



4.7



5.6



7.2



5.5



6.0



8.1



5.3



4.4



7.2



6.5



13. The table shows the total number of goals scored at full-time in each match from the round of 16 to the final in the FIFA World Cup hosted by South Africa in 2010. Match



Number of goals scored



Round of 16



Uruguay : Korea Republic USA : Ghana Germany : England Argentina : Mexico Netherlands : Slovakia Brazil : Chile Paraguay : Japan Spain : Portugal



3 2 5 4 3 3 0 1



Quarterfinals



Netherlands : Brazil Uruguay : Ghana Argentina : Germany Paraguay : Spain



3 2 4 1



Semifinals



Uruguay : Netherlands Germany : Spain



5 1



Match for third place







5



Final



Uruguay : Germany



Netherlands : Spain



0



(i) Represent the data on a dot diagram. (ii) What are the most common number of goals scored? (iii) Express the number of matches that have at least 3 goals scored as a percentage of the total number of matches from the round of 16 to the final. (iv) Consider the matches from the round of 16 to the final of the FIFA World Cup. From the World Cup in 2006 to that in 2010, there was a decrease of 37.5% in the number of matches that have not more than one goal scored at full time (before penalty shootouts). Find the number of matches with more than one goal scored (excluding penalty shootouts) in the 2006 FIFA World Cup.



7.8



Explain clearly whether you would use a dot diagram or a stem-and-leaf diagram to represent the data.



Statistical Data Analysis



Chapter 9



284



9.4



Cumulative Frequency Table and Curve



Cumulative Frequency Table In this section, we will learn how to present a set of data by constructing a table of cumulative frequencies.



Class Discussion Constructing a Table of Cumulative Frequencies Discuss in pairs.



Table 9.2(a) shows the frequency table for the number of hours spent surfing the Internet by 40 students on a particular day, while Table 9.2(b) shows the corresponding table of cumulative frequencies (or cumulative frequency table). To find the cumulative frequency for a particular hour k, we must add up the frequencies which are less than or equal to k, i.e. t  k. For example, the cumulative frequency for 4 hours, i.e. t  4 is 3 + 5 = 8. Number of Hours Spent Surfing the Internet by 40 Students Number of hours, t



Frequency



Number of hours, t



Cumulative Frequency



0t2



3



t2



3



2t4



5



t4



3+5=8



4t6



16



6t8



12



8  t  10



4 (a)



(b) Table 9.2



1. Using the information from Table 9.2(a), copy and complete Table 9.2(b). 2. Using your answers in Table 9.2(b), find the number of students who surf the Internet for (i) 6 hours or less, (ii) more than 8 hours, (iii) more than 4 hours but not more than 10 hours. 3. What does the last entry under ‘Cumulative Frequency’ of Table 9.2(b) represent? Explain your answer.



285



Chapter 9



Statistical Data Analysis



From the class discussion, we have learned that the cumulative frequency for a particular value can be obtained by adding up the frequencies which are less than or equal to that value. In other words, the cumulative frequency is a ‘running total’ of frequencies. The cumulative frequency table allows us to gather information such as the number of students whose score is below a passing mark or the number of animal species shorter than or equal to a certain length, as shown below.



SIMILAR QUESTIONS



PRACTICE NOW



The lengths of 40 insects of a certain species were measured, to the nearest millimeter. The frequency distribution is given in the table below. Length (x mm)



Frequency



25  x  30



1



30  x  35



3



35  x  40



6



40  x  45



12



45  x  50



10



50  x  55



6



55  x  60



2



Exercise 9B Questions 1-2, 8



(a) Using the table given, construct a cumulative frequency table. (b) Using the cumulative frequency table which you have constructed, find the number of insects which are



(i) 50 mm or less in length,







(ii) more than 45 mm in length,







(iii) more than 35 mm but less than or equal to 50 mm in length.



Cumulative Frequency Curve In this section, we will learn how to draw and interpret a cumulative frequency curve. In Worked Example 6, we will make use of the cumulative frequencies from Table 9.2(b) to learn how to draw a cumulative frequency curve.



Statistical Data Analysis



Chapter 9



286



6



Worked Example



(Drawing and Interpretation of a Cumulative Frequency Curve) The table below shows the cumulative frequencies for the number of hours (t) spent by 40 students surfing the Internet, on a particular day. Number of hours, t Cumulative Frequency (a)



t2



t4



t6



t8



t  10



3



8



24



36



40



Using a scale of 1 cm to represent 1 hour on the horizontal axis and 1 cm to represent 5 students on the vertical axis, draw a cumulative frequency curve for the data given in the table.



(b) Using the cumulative frequency curve, estimate (i) the number of students who surf the Internet for 5 hours or less, (ii) the percentage of students who surf the Internet for more than 6.5 hours, (iii) the value of t, such that 80% of the students surf the Internet for t hours or less.



Solution: (a)



Cumulative Frequency Curve for the Number of Hours Spent Surfing the Internet 40 32



Cumulative Frequency



35







P So roblem lvin g T ip



To plot a cumulative frequency curve:



29



30



Step 1: Label the vertical axis, i.e. ‘Cumulative Frequency’.



25



Step 2: Label the horizontal axis, i.e. ‘Number of hours (t)’.



20



Step 3: Plot the points on the graph paper, i.e. (2, 3), (4, 8), (6, 24), (8, 36) and (10, 40).



14



15



Step 4: Join all the points with a smooth curve.



10 5



5 0



6.5 7.0 AT



1



2



3



4



5



6



7



8



9



10



Number of hours (t) (b) (i) From the curve, the number of students who surf the Internet for 5 hours or less is 14.



287



Chapter 9



Statistical Data Analysis



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The reading of ‘14’ indicates that 14 students surf the Internet for less than or equal to 5 hours.



(ii) From the curve, the number of students who surf the Internet for 6.5 hours or less is 29. ∴ 40 – 29 = 11 students surf the Internet for more than 6.5 hours. ∴ The percentage of students who surf the Internet for more than 6.5 hours 11 is × 100% % = 27.5%. 40 80 (iii) 80% of the students means × 40 = 32, i.e. 32 students surf the Internet 100 for t hours or less.



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The answer can only be accurate up to half of a small square grid.



From the curve, t = 7.0.



PRACTICE NOW 6



SIMILAR QUESTIONS



The table below shows the amount of milk (in liters) produced by each of the 70 cows of a dairy farm, on a particular day. Amount of milk (x liters)



Exercise 9B Questions 3-5, 9



Number of cows



0  x  4



7



4  x  6



11



6  x  8



17



8  x  10



20



10  x  12



10



12  x  14



5



(a) Copy and complete the following cumulative distribution table for the data given. Amount of milk (x liters)



Number of cows



x  4



7



x  6



18



x  8 x  10 x  12 x  14 (b) Using a scale of 1 cm to represent 1 liter on the horizontal axis and 1 cm to represent 5 cows on the vertical axis, draw a cumulative frequency curve for the data given. (c) Using the curve in (b), estimate (i) the number of cows that produce less than or equal to 9.4 liters of milk, (ii) the fraction of the 70 cows that produce more than 7.4 liters of milk, (iii) the value of x, if 70% of the cows produce more than x liters of milk. In Worked Example 6, the upper-end points of the cumulative frequency classes are given as ‘less than or equal to’, i.e. t  k. The cumulative frequencies can also be computed by having the upper-end points as ‘less than’, i.e. t < k. In Worked Example 7, a ‘less-than’ cumulative frequency curve is used.



Statistical Data Analysis



Chapter 9



288



Worked Example



7



(Interpretation of a ‘Less-than’ Cumulative Frequency Curve) The ‘less-than’ cumulative frequency curve shows the distribution of the masses (g) of 120 apples. Cumulative Frequency Curve for the Masses of Apples



Cumulative Frequency



120 100 80 60 40 20 0



60



70



80



90



100 110



120



130



Mass (g) Estimate from the curve (i) the number of apples having masses less than 98 g, (ii) the fraction of the total number of apples having masses 117 g or more, (iii) the value of k, given that 20% of the apples have masses k g or more.



Solution:



Cumulative Frequency Curve for the Masses of Apples



Cumulative Frequency



120 100 80 60 40 20 0



60



70



80



90



100 110



Mass (g)



289



Chapter 9



Statistical Data Analysis



120



130



(i) From the curve, it is estimated that 48 apples have masses less than 98 g. AT



(ii) From the curve, 104 apples have masses less than 117 g.







∴ 120 – 104 = 16 apples have masses 117 g or more. 16 2 ∴ The required fraction is . = 120 15



(iii) 20% of 120 =



TE



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ON



For (i), the reading of 48 indicates that 48 apples have masses less than 98 g.



20 × 120 100



= 24 ∴ 24 apples have masses k g or more, i.e. 120 – 24 = 96 apples have masses less than k g.



From the curve, 96 apples have masses less than 114 g. ∴ k = 114



PRACTICE NOW 7



SIMILAR QUESTIONS



The Vitamin C content of 200 oranges is measured. The cumulative frequency curve below shows the Vitamin C content, x mg, and the number of oranges having Vitamin C content less than x mg.



Exercise 9B Questions 6-7, 10



Cumulative Frequency Curve for the Vitamin C content of Oranges 200



Cumulative Frequency



175 150 125 100 75 50 25 0



20



22



24



26



28



30



32



34



Amount of Vitamin C (x mg) Use the curve to estimate (i) the number of oranges having Vitamin C content less than 32 mg, (ii) the fraction of the total number of oranges having Vitamin C content of 26 mg or more, (iii) the value of p, given that 40% of the oranges have Vitamin C content of p mg or more.



Statistical Data Analysis



Chapter 9



290



Exercise



9B



1. 120 students took a Mathematics examination and their results are shown in the table below.







2.



230 students took part in a physical fitness test and were required to do pull-ups. The number of pull-ups done by the students is shown in the frequency table below.



Marks (m)



Number of students



0  m  10



3



0  x  6



69



10  m  20



12



6  x  8



63



20  m  30



9



8  x  10



28



30  m  40



11



10  x  12



24



40  m  50



17



12  x  16



19



50  m  60



19



14



60  m  70



20



16  x  20  20  x  25



70  m  80



14



80  m  90



10



90  m  100



5



Number of Pull-ups (x)



Frequency



13



(a) Copy and complete the following cumulative frequency table. Number of Pull-ups (x)



(a) Construct a table of cumulative frequencies for the given data. (b) Using the table in (a), find the number of students who (i) scored less than or equal to 30 marks, (ii) scored more than 80 marks, (iii) scored more than 40 marks but not more than 90 marks.







x6







x8







x  10







x  12







x  16







x  20







x  25



Cumulative Frequency 69



(b) Students have to do at least 12 pull-ups to qualify for the Gold Award, 8 pull-ups to qualify for the Silver Award and 6 pull-ups for the Bronze Award. Using the table in (a), find the number of students who achieved for the (i) Gold Award, (iii) Bronze Award.



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Chapter 9



Statistical Data Analysis



(ii) Silver Award,



3.



The masses, in kg, of 100 students are measured. The cumulative frequency curve shows the mass, x kg, and the number of students with masses less than or equal to x kg. Cumulative Frequency Curve for the Masses of Students 100



4. The masses of 50 loaves of bread from a bakery are measured. The cumulative frequency curve below shows the mass y g, and the number of loaves of bread which are less than or equal to y g. Cumulative Frequency Curve for the Masses of Loaves of Bread



90 50



70



45



60



40



50



35 Cumulative Frequency



Cumulative Frequency



80



40 30 20 10 0



60



62



64



66



68



70



72



74



25 20 15 10



Mass (x kg)



30



5



Use the curve to estimate



(i) the number of students whose masses are less than or equal to 65 kg,



0



445 447 449 451 453 455 Mass ( y g)



(ii) the number of students whose masses are more than 68.6 kg,







(iii) the percentage of the total number of students whose masses are more than 64.4 kg.



(i) the number of loaves of bread having masses less than or equal to 450.4 g,



From the graph, estimate



(ii) the number of loaves of bread which are rejected because they are underweight or overweight, given that a loaf is underweight if its mass is 446.2 g or less, and overweight if its mass is more than 453.6 g, 3 (iii) the value of x, if of the loaves of bread have 10 masses more than x g.



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5. 500 earthworms were collected from a sample of Soil A and 500 earthworms from Soil B, and their lengths were measured. The cumulative frequency curve below shows the length x mm, and the number of earthworms which have lengths less than or equal to x mm.



6. 120 students took a music examination. The cumulative frequency curve below shows the results (x marks) and the number of students who obtained less than x marks. The highest possible mark is 80. Cumulative Frequency Curve for the Results of a Music Examination



Cumulative Frequency Curve for the Lengths of Earthworms 120



400 Cumulative Frequency



Cumulative Frequency



500



300



200 Soil A



100



Soil B



100



80



60



40 0



20



40



60



80



100



Length (x mm) (a) For both Soil A and Soil B, use the graphs to estimate (i) the number of earthworms having lengths less than or equal to 46 mm, (ii) the percentage of earthworms having lengths greater than 76 mm, (iii) the value of a, if 18% of the earthworms have lengths a mm or less. (b) Which soil produced the longest earthworm among the 1000 earthworms? (c) Earthworms which grew more than 60 mm are said to be ‘satisfactory‘. From the graph, estimate the percentage of ‘satisfactory‘ earthworms from (i) Soil A, (ii) Soil B.



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20



0



20



40



60



80



Marks (x) From the graph, estimate (i) the number of students who scored less than 45 marks, (ii) the fraction of the total number of students who failed the music examination, given that 34 is the lowest mark to pass the examination, (iii) the value of a, if 27.5% of the students obtained at least a marks in the music examination.



7.



The speeds of 100 bicycles on a cycling lane at East Coast Park are recorded. The cumulative frequency curve below shows the speed u km/h, and the number of bicycles which traveled at a speed less than u km/h.



8. The table below shows the cumulative frequencies for the annual income of 200 households in a certain district. Annual Income ( x, in ten thousands)



Cumulative Frequency



x=0



0



x  20



41



100



x  40



78



90



x  60



99



80



x  80



118



70



x  100



164



60



x  120



200



Cumulative Frequency Curve for the Speeds of Bicycles at East Coast Park



Cumulative Frequency







50



(i) On a sheet of graph paper, draw a histogram to represent the frequency distribution.



40



(ii) If a household is selected at random from this district, what is the range of annual income the household is most likely to earn?



30 20 10 0



5



10



15



20



25



30



35



40



Speed (u km/h)



Use the curve to estimate



(i) the number of bicycles that traveled at a speed less than 18 km/h, (ii) the fraction of the total number of bicycles that traveled at a speed greater than or equal to 29 km/h, (iii) the value of v, if 40% of the bicycles have a speed less than v km/h.



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80



0



60



40 30 20



0



0 40



45



50



55



60



65



Mass (g)







(a) (i) Tomatoes with masses more than 56 g are rated as grade A tomatoes. Find the percentage of grade A tomatoes. (ii) Estimate the value of y if 15% of the tomatoes are y g or less. These are rated as grade C tomatoes. (iii) Find the number of grade B tomatoes which are between grades A and C. (b) (i) From the curve, the data are transformed into the frequency distribution table below. Copy and complete the table. Masses (x g)



Frequency



40  x  45 45  x  50 50  x  55 55  x  60



60  x  65



(ii) Using the table, find an estimate of the mean mass of a tomato produced at the nursery.



295



0



Score Examination B



50



10



Score Examination A Cumulative Frequency



Cumulative Frequency



70



Cumulative Frequency



Cumulative Frequency for the Masses of Tomatoes



Cumulative Frequency



9. The cumulative frequency curve shows the distribution of the masses (in grams) of 80 tomatoes produced at a nursery.



10. The cumulative frequency curves for the results of three different Mathematics examinations are shown below. All three examinations are attempted by the same 1000 students.



Chapter 9



Statistical Data Analysis







Score Examination C



Out of the three Mathematics examinations, explain clearly which one is likely to be



(i) the most challenging, (ii) the least challenging.



Explain your answers.



9.5



Median, Quartiles, Percentiles, Deciles, Range and Interquartile Range



In Grade 7, we have learned how to find the median of a set of data. The median is a measure of the average and is the ‘middle value’ when the data are arranged in an ascending order. In this section, we will learn how to find the quartiles, range and interquartile range for both discrete and continuous data.



Discrete Data Discrete data refers to a set of data which only takes on distinct values. For example, a data set showing the number of phone calls received in a day 2 can only take on distinct values such as 1, 5, 12, etc., but not 1.5, 4 , etc. 3 Consider the following set of distinct data arranged in ascending order: Set A: 2, 5, 6, 7, 8, 12, 14, 16, 20, 21, 30 The total number of data values is 11, i.e. n = 11. In Grade 7, we have learned that for a set of discrete data, the median is the value of the data in the middle position, i.e. the 6th position in the case for Fig. 9.3.



2 5 6 7



8



12 14 16 20 21 30 median Fig. 9.3



From Fig. 9.3, we see that the median 12 divides the data in 2 equal halves, with 5 values on each side of the median. In Fig. 9.4, we consider the 5 values on the left of the median. The middle value of these 5 values is 6 and it is called the lower quartile or the first quartile Q1. lower half 2 5 6 7 lower quartile Q1



8



12 14 16 20 21 30 median Q2 Fig. 9.4



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The first quartile can be considered as the ‘first-quarter value’. 25% (or one quarter) of the data is less than or equal to this value. Since the median is the middle value or ‘second-quarter value’, the median is also called the second quartile Q2. 50% (or half) of the data is less than or equal to this value. Similarly, in Fig. 9.5, we consider the 5 values on the right of the median. The middle value of the 5 values is 20, and it is called the upper quartile or the third quartile Q3. 75% (or three quarters) of the data is less than or equal to this value.



Internet Resources There are other formulae and methods to find the lower quartile, Q1 and the upper quartile, Q3. Search on the Internet for more information. But we will use the method shown in the textbook.



upper half 2 5 6 7



8



12 14 16 20 21 30 median Q2



upper quartile Q3



Fig. 9.5 From Fig. 9.4 and 9.5, we see that the quartiles obtained by the above method divide the data which is arranged in ascending order into 4 roughly equal parts. Now that we have learned how to find the median and quartiles for a given set of data, we shall learn how to measure the spread of the data by using the range and the interquartile range. Fig. 9.6 shows the range and interquartile for the data values in Set A. The median, Q1, Q3, range and the interquartile range are indicated in the dot diagram as shown. range 0 2 6 12 20  Q1



median



30



Q3



interquartile range Fig. 9.6 These measures of spread show the degree of variation or how ‘spread out’ the data values are. For Set A, Range = Largest value – Smallest value = 30 – 2 = 28 Interquartile range = Q3 – Q1 = 20 – 6 = 14



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AT



TE



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The interquartile range is the range of the middle 50% of the data.



The interquartile range is a better measure of the spread of the data than the range because it tells us how the middle 50% of the data are distributed. The range only consists the difference between the largest and the smallest values of the set of data. The interquartile range is not affected by extreme values as it does not consider the behavior of the lower 25% or upper 25% of the data. A statistical measure that takes into account the behavior of every value of a data set will be discussed in Section 9.7.



Worked Example



8



(Finding and Interpreting the Range and Interquartile Range for a Set of Discrete Data with an Even Number of Data Values) The data below shows the marks for a multiple choice quiz with 20 questions, taken by 8 students. 10, 12, 12, 13, 9, 17, 11, 14 (i) For the given set of data, find Q1, Q2 and Q3. (ii) Find the range. (iii) Find the interquartile range.



Solution:



(i) Arranging the given data in ascending order: lower half 9



upper half



10 11 12 12 13 14 17 Q1



Q2 median



Q3



For the given data, n = 8. 12 + 12 ∴ Q2 = = 12 (When n is even, the median is the average of the two 2 middle values.) 10 + 11 Q1 = = 10.5 (When the number of data in the lower half is even, Q1 is 2 the average of the two middle values.) 13 + 14 Q3 = = 13.5 (When the number of data in the upper half is even, Q3 is 2 the average of the two middle values.)



(ii) Range = 17 – 9 =8 (iii) Interquartile Range = Q3 – Q1 = 13.5 – 10.5 =3



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PRACTICE NOW 8



SIMILAR QUESTIONS



1. The following set of data shows the number of sit-ups done in 1 minute by 10 students during a physical fitness test. 12, 22, 36, 10, 14, 45, 59, 44, 38, 25







(i) For the given set of data, find Q1, Q2 and Q3. (ii) Find the range. (iii) Find the interquartile range. 2. Another physical fitness test is conducted one month later. However, only 9 of the students took the test as one of the students is sick. The following set of data shows the number of sit-ups done in 1 minute by these 9 students. 23, 54, 15, 32, 16, 26, 47, 9, 35



(i) For the given set of data, find Q1, Q2 and Q3.







(ii) Find the range.







(iii) Find the interquartile range.



Continuous Data Continuous data refers to data which can take on any value within a range of numbers. For example, a data set showing the height (cm) of 30 girls in a class can take on values such as 150.4, 169.34, 150, etc. For continuous data, we can estimate the quartiles from the cumulative frequency curve. Let us look at the cumulative frequency curve from Worked Example 7 on page 289, where n = 120.



299



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Exercise 9C Questions 1-3



Cumulative Frequency Curve for the Masses of Apples



100 80 60



0



60



70



80



90



100 110



P90



Q1



20



Q3



40 Median



Cumulative Frequency



120



120



130



Mass (g) Interquartile range: 111 – 92 = 19 Fig. 9.7



For continuous data, we can obtain the quartiles by dividing the data set into 4 equal parts. n 120 When the cumulative frequency is = = 60, from Fig. 9.7, the mass of the 2 2 apples is 101.



Since the median is the middle value, then the median is 101 g. Similarly,



n 120 3 3 = = 60 and n = × 120 = 90. 4 4 4 4



∴ From Fig. 9.7, Q1 = 92 g and Q3 = 111 g.



Thus, the interquartile range is Q3 – Q1 = 111 – 92 = 19 g (as shown in Fig 9.7).



The range is the difference between the largest end-point and smallest end-point, i.e. 130 – 60 = 70 g.



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Percentiles and Deciles For continuous data, we can also find a measure called percentiles and deciles. Percentiles are values that divide the data set into 100 equal parts. 90 For example, in Fig. 9.7, 90% of the distribution (i.e. × 120 = 108 apples) 100 have masses less than 119 g. We say that the 90th percentile, P90 = 119 g. Since the median value of 101 means that 50% of the distribution, i.e. 60 apples have masses less than 101 g, the median is also called the 50th percentile, P50. Similarly, Q1 = P25 and Q3 = P75. Deciles are values that divide the data set into 10 equal parts. Based on the example above, the 90th percentile can be represented as the 9th decile, D9 while the median is also called the 5th decile, D5.



Worked Example



9



(Estimating the Quartiles, Interquartile, Percentiles and Deciles from a Cumulative Frequency Curve) The cumulative frequency curve represents the instantaneous speeds of 100 motor vehicles taken at a particular point on a street. Cumulative Frequency Curve for the Speeds of Motor Vehicles 100 90 Cumulative Frequency



80 70 60 50 40 30 20 10 0



10



20



30



40



50



60



70



80



Speed (km/h) Estimate (i) the median, the lower and the upper quartiles, (ii) the interquartile range, (iii) the range of the speed, (iv) the 1st decile, (v) the value of v, if 85% of motor vehicles have speeds less than or equal to v km/h. 301



Chapter 9



Statistical Data Analysis



Solution:



Cumulative Frequency Curve for the Speeds of Motor Vehicles 100 90



Cumulative Frequency



80 70 60 50 40 30 20 10 0



10



20



30



40



50



60



70



80



Speed (km/h) (i) For this set of data, n = 100. n n 3n ∴ = 50, = 25 and = 75 2 4 4 From the graph, median speed = 49 km/h, lower quartile = 42 km/h, upper quartile = 57 km/h. (ii) Interquartile range = 57 – 42 = 15 km/h (iii) Range = 80 – 10 = 70 km/h (iv) 10% of the total frequency =



10 × 100 100



= 10 st From the graph, the 1 decile = 32 km/h.



(v) 85% of drivers =



equal to v km/h.



85 × 100 = 85, i.e. 85 drivers have speeds less than or 100



From the graph, v = 62.



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302



PRACTICE NOW 9



SIMILAR QUESTIONS



120 students take a Science test. The cumulative frequency curve shows the test marks (m) and the number of students scoring less than m marks.



Cumulative Frequency



Cumulative Frequency Curve for the Science Test Marks of Students 120 100 80 60 40 20 0



10 20 30 40 50 60 70 80 90 100 Marks (m)



From the graph, estimate (i) the median, the lower quartile and the upper quartile, (ii) the interquartile range, (iii) the 1st and 8th deciles, (iv) the passing mark if 60% of the students passed the test.



Worked Example



10



(Comparing and Analyzing Two Cumulative Frequency Curves) The diagram below shows the cumulative frequency curves for the annual incomes (in thousands of pesos) of 60 households in two towns, A and B.



Cumulative Frequency Curves for the Annual Household Incomes of Households



Cumulative Frequency



60 50 40



Town A



Town B



30 20 10 0



100 200 300 400 500 600 700 800



Annual Household Income ( , in thousands) (a) For Town A, find (i) the median income level, (ii) interquartile range. 303



Chapter 9



Statistical Data Analysis



Exercise 9C Questions 4-7



(b) For Town B, find (i) the median income level, (ii) the interquartile range. (c) ‘Households in Town B generally have higher income levels than households in Town A’. Do you agree? Explain your answer. (d) Which town is more likely to have an ‘income-gap’ problem? Justify your answer.



Solution:



(a) For each set of data, n = 60. n n 3n ∴ = 30, = 15 and = 45 2 4 4



Cumulative Frequency Curve for the Annual Household Incomes of Households



Cumulative Frequency



60 50



Town A



40



Town B



30 20 10 0



100 200 300 400 500 600 700 800



Annual Household Income ( , in thousands) (i) From the graph, median income level of Town A = 380 000. (ii) From the graph, lower quartile = 320 000 upper quartile = 440 000 ∴ Interquartile range of Town A = 440 000 – 320 000 = 120 000



AT



(b) (i) From the graph, median income level of Town B = 500 000.



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For (d), in Town B, the interquartile range of 180 000 indicates that the difference (or gap) in income levels between the bottom 25% of households and top 25% of households is 180 000.



(ii) From the graph, lower quartile = 400 000 upper quartile = 580 000 ∴ Interquartile range of Town B = 580 000 – 400 000 = 180 000



(c) Agree. The median annual income level of Town B is higher by 500 000 – 380 000 = 120 000. (d) Town B. The interquartile range of 180 000 for Town B is much higher than the interquartile range of 120 000 for Town A.



Therefore, it is more likely to have an ‘income-gap’ problem than Town A, whose interquartile range of 120 000 is much smaller.



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PRACTICE NOW 1 0



SIMILAR QUESTIONS



300 students, each from School A and School B, participated in an IQ quiz. The maximum marks for the quiz is 100. The cumulative frequency curves below show the distribution of the marks scored by the students from each of the two schools.



Exercise 9C Questions 8-12



Cumulative Frequency Curve for the IQ Quiz Marks of Students



Cumulative Frequency



300 250 200



School B



School A



150 100 50 0



10



20



30



40



50



60



70



80



90 100



IQ Quiz Marks (a) For School A, estimate



(i) the median,



(ii) the interquartile range.



(b) For School B, estimate the



(i) the median,



(ii) the interquartile range.



(c) State, with a reason, if School A or School B performed better overall. (d) In which school were the quiz marks more consistent? Justify your answer.



Exercise



9C



1. Find the range, lower quartile, median, upper quartile and interquartile range for the following sets of data.



0, 1, 6, 9, 24, 0, 27, 6, 9, 29



(a) 7, 6, 4, 8, 2, 5, 10







(b) 63, 80, 54, 70, 51, 72, 64, 66



(i) For the given data, find the median, the lower and upper quartiles. (ii) Find the range and the interquartile range.



(c) 14, 18, 22, 10, 27, 32, 40, 16, 9 (d) 138, 164, 250, 184, 102, 244, 168, 207, 98, 86 (e) 10.4, 8.5, 13.1, 11.8, 6.7, 22.4, 4.9, 2.7, 15.1



305



2. The following set of data shows the number of distinctions scored by 10 classes for a particular examination. Each class has 40 students.



Chapter 9



Statistical Data Analysis



3. The stem-and-leaf diagram below represents the Mathematics quiz marks of 20 students.











9 2 1 0 0 0 2 7 5



Leaf



2



8



1



1



3 7



Cumulative Frequency Curve for the Heights of Plants



2



8



60



9



Cumulative Frequency



Stem 0 1 2 3 4 6 7 8 9



5. The following diagram shows the cumulative frequency curve for the heights, in cm, of 56 plants grown under experimental conditions.



9 8



Key: 0 | 9 means 9 marks



Find



50 40 30 20 10



(i) the median mark, 0



(ii) the range,



(iii) the interquartile range.



4. The graph shows the cumulative frequency curve for the daily earnings of 300 employees in a company. Cumulative Frequency Curve for the Daily Earnings of 300 Employees



Cumulative Frequency



300







10



20



30



40



50



60



70



Height (cm) Use the curve to estimate



(i) the median height,



(ii) the upper quartile,



(iii) the lower quartile, (iv) the number of plants having heights greater than 57 cm.



250 200 150 100 50 0



600 700 800 900 1000 1100 1200 Daily Earnings (pesos)







(a) Use the graph to estimate (i) the median, the lower and upper quartiles, (ii) the interquartile range. (b) Find the (i) 2nd decile, (ii) 90th percentile,



of the daily earnings of the employees.



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306



(a) Estimate from the graph 6. The following diagram shows the cumulative curve for the lengths of 600 leaves from a tree.



(iv) the number of participants who scored more than or equal to 26 marks but less than 30 marks.



Cumulative Frequency



600



(b) Given that 37.5% of the students passed the quiz, use the graph to find the passing mark.



500 400



8. The graph shows the cumulative frequency curves of the daily traveling expenses of 800 students in two schools, A and B.



300 200



Cumulative Frequency Curves for the Daily Traveling Expenses of Students



100 0



20







25



30



35



40



45



50



800



(b) Given that 65% of the leaves are considered healthy if their length is longer than h mm, use the graph to find the value of h. 7. 80 students participated in a multiple choice quiz which consists of 40 questions. The cumulative curve shows the marks scored. Cumulative Frequency Curve for the Quiz Marks of 80 Students



Cumulative Frequency



Length (x mm)



(a) Use the graph to find (i) the median length, (ii) the interquartile range.



Cumulative Frequency



(ii) the upper quartile, (iii) the interquartile range,



Cumulative Frequency Curve for the Lengths of Leaves



600 400



School A



School B



200 0



20



40



60



80



100



Traveling Expenses (centavo)



80



(a) Use the graph to estimate the median traveling expenses of the students from (i) School A, (ii) School B.



60



(b) Find the interquartile range of the traveling expenses of the students from (i) School A, (ii) School B. (c) Find the 80th percentile of the traveling expenses of the students from School B.



40



(d) State, with a reason, whether the students from School A or School B spend more on daily traveling expenses.



20



0 



307



(i) the median mark,



10



20



30



Marks



Chapter 9



Statistical Data Analysis



40



9.



All the students from two classes, A and B, took the same Mathematics Olympiad examination paper. The cumulative frequency curves below show the scores for the two classes.



10. The cumulative frequency curves show the distribution of marks scored by 500 cadets in a physical fitness test from each of the two military schools, A and B. Cumulative Frequency Curves for Marks Scored by Cadets



40



500



30



400 Cumulative Frequency



Cumulative Frequency



Cumulative Frequency Curves for the Mathematics Olympiad Scores of Students



Class B 20 Class A 10



0



10



20



30



40



School B



300 School A 200



100



50



Scores (i) Estimate the lower quartile, median, and upper quartile in Class A.



0



20



40



60



80



Marks



(ii) How many students are there in Class B?



(a) For School A, estimate from the graph,



(iii) Find the interquartile range of Class B.



(i) the median mark,



(iv) Estimate the percentage of the students from Class B who received a gold award, given that the qualifying mark for a gold award is more than 38.



(ii) the 70th percentile,



(v) Do you agree with the statement that ‘Class A generally performed better and their results are more consistent’? Justify your answer.



100



(iii) the interquartile range, (iv) the number of cadets who scored less than 43 marks, (v) the passing mark given that 60% of the cadets passed the physical fitness test. (b) It is given that a distinction grade is equivalent to 70 marks and above. Find the percentage of cadets who scored distinctions in each school. (c) ‘Cadets from School B performed better in general, than School A’. Do you agree? Give two reasons to support your answer.



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308







City X



City Y



80 65 21 81 16 23 37 50 53 100



103 79 99 121 200 308 114 171 198 235



(a) For the PSI data given for City X, find (i) the range, (ii) the median, (iii) interquartile range. (b) For the PSI data given for City Y, find (i) the range, (ii) the median, (iii) the interquartile range.



(c) Which city's data show a greater spread?



(d) Compare and comment on the air quality of the two cities. Give two reasons to support your answer.



Cumulative Frequency Curve for the Waiting Times of Clients in a Bank 60 Cumulative Frequency



11. The table below shows the Pollutant Standards Index (PSI) of City X and City Y, measured in the same period of 10 days. A higher PSI reading indicates worse air quality, and vice versa.



12. The waiting times (in minutes) of 60 clients at a bank, on a particular day were measured. The cumulative frequency curve shows the waiting times (t), and the number of clients with waiting times more than t minutes.



50 40 30 20 10 0



2.5



5



7.5



10 12.5 15 17.5 20



Waiting Time (t minutes) (a) (i) Estimate the lower quartile, median and upper quartile of the waiting times in the bank. (ii) Find the interquartile range. (b) Find the percentage of clients who waited for not more than 15 minutes at the bank. (c) For the same 60 clients, a second cumulative frequency curve is plotted to show the waiting times (t), and the number of clients with waiting times less than or equal to t minutes. What does the intersection of the two cumulative curves represent? Explain your answer clearly.



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9.6



Box-and-Whisker Plots



In this section, we will learn how to draw and interpret a box-and-whisker plot, which is another way to show the distribution of a set of data. Let us look at Worked Example 9 on page 301 again, where estimates were obtained from the cumulative frequency curve for the speeds of 100 motor vehicles. In the example, • the maximum speed is 80 km/h, • the minimum speed is 10 km/h, • the median speed is 49 km/h, • the lower quartile (Q1) is 42 km/h and



AT



TE



• the upper quartile (Q3) is 57 km/h. We can present this information on a box-and-whisker plot. To begin, we draw a horizontal number line using a suitable scale. The number line must be long enough to contain all the data points. On top of the number line, the positions of the MIN (minimum speed), the MAX (maximum speed) and the quartiles are indicated, as shown in Fig. 9.8. MIN



10



Q1 Median Q3



20



30



40 50 Speed (km/h)



60



70



80



Above the number line, the MIN and the MAX are marked. Two line segments are then drawn to connect the MIN and MAX to the sides of the box. These two line segments represent the whiskers of a box-and-whisker plot.



10



20



30



49



40 50 Speed (km/h)



The number line must be drawn with an arrow pointing to the right.



AT



As shown in Fig. 9.9, a rectangular box is drawn above the number line, with the left side at the lower quartile and the right side at the upper quartile. A vertical line is then drawn inside the box to indicate the median. This rectangular box represents the box of a box-and-whisker plot.



42



ON



MAX



Fig. 9.8



10



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It is important to label the values of the MIN, MAX, the quartiles and the median on the box-and-whisker plot if there is no graph grid (e.g. Fig. 9.9). With or without the grid, any units used for the number line, e.g. speed (km/h), must also be indicated.



80



57 60



TE



70



80



Fig. 9.9



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310



Therefore, the final figure in Fig. 9.9 is called a box-and-whisker plot. From the box-and-whisker plot, Range = MAX – MIN and Interquartile Range = Q3 – Q1 = 80 – 10 = 57 – 42 = 70 km/h, = 15 km/h. A box-and-whisker plot is a way of summarizing a set of data. If we are interested in only the five values (i.e. min, max, Q1, Q2 and Q3), then we use the box-and-whisker plot. But if we need to find the cumulative frequencies or percentiles, then we use the cumulative frequency curve. When comparing two sets of data, it is easier to use the box-and-whisker plot than the cumulative frequency curve because we will usually compare only the medians and the interquartile ranges.



Worked Example



11



(Drawing a Box-and-Whisker Plot) Draw a box-and-whisker plot for the given set of data. 10, 4, 3, 16, 14, 13, 4, 7, 11, 5, 17, 14



Solution:



Arranging the given data in ascending order: lower half



upper half



3 4 4 5 7 10 11 13 14 14 16 17 Q1



Q3



median



For the given data, n = 12, MIN = 3 and MAX = 17. ∴ Median =



10 + 11 = 10.5 2



4+5 = 4.5 (When the number of data in the lower half is even, Q1 is the average Q1 = 2 of the two middle values.) 14 + 14 Q3 = = 14 (When the number of data in the upper half is even, Q3 is the 2 average of the two middle values.)



The box-and-whisker plot is drawn below.



3 2



311



Chapter 9



4.5 4



10.5 6



8



Statistical Data Analysis



10



17



14 12



14



16



18



PRACTICE NOW 1 1



SIMILAR QUESTIONS



Draw a box-and-whisker plot for the given set of data.



Exercise 9D Question 1(a)-(d)



20, 14, 23, 9, 7, 13, 29, 9, 16



Class Discussion Vertical Box-and-Whisker Plots



Box-and-whisker plots can also be drawn vertically. Table 9.3 shows the summary statistics for two sets of data, A and B. Set A 20 120 36 50 70



MIN MAX Q1 Median Q3



Set B 10 110 80 90 100



Table 9.3 Fig 9.10 shows the box-and-whisker plot, which is drawn vertically for the data in Set A.



120 100 80 60 40 20 0



Set A



Set B



Fig. 9.10 1. On the square grid and scale given in Fig. 9.10, draw a vertical box-and-whisker plot for the data in Set B. 2. What do the heights of the rectangular boxes represent? Compare the heights of the two rectangular boxes corresponding to the data in Set A and Set B. 3. From the height of the rectangular boxes, what can we infer about the spread of the data in Set A and Set B?



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From the class discussion, we have learned that box-and-whisker plots can also be drawn vertically. Box-and-whisker plots give us a visualization of the spread of a set of data and also facilitate comparisons between two or more sets of data.



Worked Example



12



(Interpreting a Box-and-Whisker Plot) A class of students took an English proficiency test. The results are represented by a box-and-whisker plot, as shown below.



20



30



40



50



60 Marks



70



80



(i) State the median mark. (ii) Find the range of the marks of the class. (iii) Find the interquartile range of the mark.



Solution:



20



30



40



50



60 Marks



70



80



90



100



(i) From the box-and whisker plot, the median score is 60 marks. (ii) Range = MAX – MIN = 100 – 20 = 80 (iii) Interquartile range = Q3 – Q1 = 82 – 31 = 51



313



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90



100



PRACTICE NOW 1 2



SIMILAR QUESTIONS



A class of 50 students took a Geography test. The results are represented by a box-and-whisker plot, as shown below. The maximum mark of the test is 80.



0



20



40 Marks



60



Exercise 9D Questions 2-5



80



(i) the median mark, (ii) the range, (iii) the interquartile range.



Worked Example



13



(Interpreting and Comparing Two Box-and-Whisker Plots) The box-and-whisker plots show the distribution of the battery life (hours) of two brands of smartphones, Smartphone A and Smartphone B. 150 smartphones of each type were fully charged and tested for their battery lives.



Smartphone A Smartphone B 10



11



12 13 14 15 Battery Life (hours)



16



(a) For Smartphone A, use the diagram to find,



(i) the range,







(ii) the median,







(iii) the interquartile range.



(b) For Smartphone B, use the diagram to find



(i) the range,







(ii) the median,







(iii) the interquartile range.



(c) Which brand of smartphone has a longer battery life on average? State a reason.



Statistical Data Analysis



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Solution: Smartphone A Smartphone B 10



11



12 13 14 15 Battery Life (hours)



16



(a) For Smartphone A,



(i) range = MAX – MIN = 15.2 – 11 = 4.2 hours







(ii) median = 12.6 hours



(iii) interquartile range = Q3 – Q1 = 13.5 – 11.5 = 2 hours (b) For Smartphone B,



(i) range = MAX – MIN = 15.5 – 10 = 5.5 hours







(ii) median = 13 hours



(iii) interquartile range = Q3 – Q1 = 14 – 12.1 = 1.9 hours (c) Smartphone B. Smartphone B has a longer median battery life.



315



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SIMILAR QUESTIONS



PRACTICE NOW 1 3



A developer can choose between two different types of bricks for the construction of a new shopping complex. The box-and-whisker plots show the results of tests on the compressive strength of 200 bricks, measured in pounds per square inch (psi) of the two types of bricks. The higher the value of the psi, the stronger the brick.



Exercise 9D Questions 6-12



Brick A Brick B



2000



3000 4000 5000 6000 Compressive Strength (psi)



7000



(a) For Brick A, find



(i) the range,







(ii) the median,







(iii) the interquartile range.



(b) For Brick B, find



(i) the range,







(ii) the median,







(iii) the interquartile range.



(c) On average, which type of brick is stronger? State a reason to support your answer.



Internet Resources In the study of Statistics, an outlier is an observation that is ‘far away’ from other observations in a data set. Search on the Internet to find out more about outliers and how quartiles and the interquartile range are used to determine if a certain data point is an outlier. In addition, box-and-whisker plots with outliers are also drawn differently. Find out also how a box-and-whisker plot is drawn for a data set with outliers.



Statistical Data Analysis



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316



Exercise



9D



1. Draw a box-and-whisker plot for each of the following sets of data.



(a) 1, 14, 9, 8, 20, 11, 5







(b) 45, 51, 57, 43, 45, 60, 58, 54







(c) 3, 6, 11, 2, 17, 22, 15, 8, 21, 3, 15, 12



4.



The following diagram shows the box-and whisker plot for the alcohol content (grams per deciliter of blood) in the blood of drivers who were given breathalyzer tests.



(d) 79, 87, 66, 96, 98, 87, 82, 77, 93 2. The following diagram shows the box-and whisker plot for the daily temperature (°C) from 1st June to 30th June in a city.



0



0.02 0.04 0.06 0.08 0.10 0.12 0.14 Blood Alcohol Content (g/dL)



(i) State the lower quartile, median and upper quartile of the alcohol content of the drivers. (ii) Compare the spread of the alcohol content between the highest 25% and the lowest 25% of the drivers.



15 16 17 18 19 20 21 22 23 24 25 26 27 28 Temperature (°C) (i) State the lower quartile, median and upper quartile of the temperature. (ii) Find the range of the temperature in June. 3. The box-and-whisker plot below shows the blood pressure level (in mm of mercury) of patients who have taken a certain prescription drug.



130



317



140



150 160 170 180 mm of mercury



190



5. The heights of basketball players (cm) in a NBA team are given below.







168, 180, 185, 192, 192, 195, 195, 196, 198, 200, 205, 213 The data can be represented in the box-andwhisker plot below.



a



200



c d b Height (cm)



e



(i) State the median blood pressure level of the patients.



(i) Find the values of a, b, c, d and e.



(ii) Find the interquartile range.



(iii) Calculate e – a. What does it represent?



Chapter 9



Statistical Data Analysis



(ii) Calculate d – b. What does it represent?



6. The following box-and-whisker plots show the masses (g) of three types of apples. Type A Type B



8.



The box-and-whisker plots show the marks obtained by some students in the History and Geography examinations. The maximum mark for both examinations is 100.



Geography History



Type C 100



200 300 Mass (g)







400



0



(a) Which type of apples has (i) the highest median mass, (ii) the lowest median mass? (b) Which type of apples has masses which are more evenly distributed? (c) Which type of apples has masses which have a greater spread?



20



40 60 Marks



80



100



(a) For the Geography examination, find (i) the range, (ii) the median, (iii) the interquartile range. (b) For the History examination, find (i) the range, (ii) the median, (iii) the interquartile range.



7. The box-and-whisker plots show the masses (kg) of Grade 10 students from School A and School B.



(c) Imee said that the Geography examination is easier than the History examination. Do you agree with Imee? Give two reasons for your answer.



School A



(d) Which examination has a wider spread of marks? Give a reason for your answer.



School B



40 



50



60 70 Mass (kg)



80



90



(a) For School A, find (i) the range, (ii) the median, (iii) the interquartile range. (b) For School B, find (i) the range, (ii) the median, (iii) the interquartile range. (c) ‘Students from School B are generally heavier than students from School A.’ Do you agree with this statement? Give a reason for your answer.



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318



9.



64 adults were asked to indicate the weekly number of hours they spent watching television. The cumulative frequency curve below shows the information obtained. Cumulative Frequency Curve for the Weekly Number of Hours Adults Spent on Watching Television



10. The box-and-whisker plots show the distribution of the ages (in years) of 60 members from Prestige Country Club and Luxury Country Club.



Prestige Luxury



Cumulative Frequency



70 60



30



50 40



35



40 45 50 Age (years)



55



60



(a) For Prestige Country Club, find



30



(i) the median age,



20



(ii) the interquartile range.



10



(b) For Luxury Country Club, find (i) median age, 0



10



20



30



40



Number of Hours (a) Use the graph to estimate (i) the median, (ii) the interquartile range, (iii) the number of adults who spent more than 25 hours per week watching television.







The box-and-whisker plot below shows the number of hours that a group of 64 teenagers spent watching television.



0



20



40 60 Number of Hours



80



(b) Find (i) the median, (ii) the interquartile range. (c) ‘Teenagers spent more time watching television in general.’ Do you agree? Give a reason to support your answer. (d) Compare and comment on the spread of the time spent watching television of these two groups of people.



319



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Statistical Data Analysis



(ii) the interquartile range, (c) For the box-and-whisker plot for Luxury Country Club, the left whisker is much longer than the right whisker. Explain what this means. (d) Which country club shows a greater spread of ages? (e) Comment briefly on the distribution of ages between the members in Prestige Country Club and Luxury Country Club. 11. The following diagrams show the box-and whisker plots for two sets of data, X and Y.



Set X Set Y







0



10



20



30



(a) For each set of data, find (i) the median, (ii) the range, (iii) the interquartile range.



40



50



x



(b) Which set of data has a more balanced spread?



12. The box-and-whisker diagrams for three sets of data, X, Y and Z, are shown below.



(c) Which set of data has a greater spread?



x



(d) Which set of data has a lower median? (e) Which one of the cumulative curves (A, B or C) shown below best represents the set X? Cumulative Frequency Curves for Data in Set X x



0 B



0



10



20



C



30



40



50



x



Y







The histograms below (A, B and C) show the frequency distributions for the three sets of data.







Match each of the three data sets to their respective histograms. Justify your answers.



A



B



x



x



x



Frequency



Frequency



Q



Z



Frequency



Frequency



Frequency



(f) The histograms below (P, Q and R) show the frequency distributions for the three cumulative frequency curves for data in Set X (A, B and C in part (e)). Match each of the three curves to their respective histograms. Justify your answers.



P



X



Frequency



Cumulative Frequency



A



R



x



C



x



(g) Describe a context for each of the histograms P, Q and R.



Statistical Data Analysis



Chapter 9



320



9.7



Standard Deviation



In Grade 7, we have learned a statistical measure, standard deviation, to describe the distribution of a set of data.



Investigation Are Averages Adequate for Comparing Distributions?



Fig. 9.11(a) and (b) show the dot diagrams for two sets of data, Set A and Set B, both with size n = 6, mode = 3, median = 3 and mean = 3. Although the three averages (mode, median and mean) are all equal to 3 for Set A and Set B, the two distributions are different.



1



2



3 4 (a)



5



1



2



3 4 (b)



5



Fig. 9.11 1. Draw another two dot diagrams with distributions such that n = 6, and with the mode, median and mean all equal to 3. 2. Are the three averages (mode, median and mean) adequate for comparing two sets of data? Explain.



From the above investigation, we have learned that two sets of data can have the same averages (mode, median and mean), but the distributions can still be different. Therefore, there is a need for another method to measure the spread of the data or distribution. In Section 9.5, we have learned how to find the interquartile range for both discrete and continuous data. The interquartile range is a measure of the spread of the data about the median. It tells us about the range of the middle 50% of the distribution. It is often used when the median is the appropriate measure of the average of the data, and we have learned in Grade 7 when we should use the median.



321



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Statistical Data Analysis



As mentioned in Section 9.5, we will revise in this section on standard deviation which describes how the data are spread about the mean and which also takes into account all the values of the data set. It is often used when the mean is the appropriate measure of the average of the data. We have learned a formula for calculating the mean, x of a set of data, x=



∑ fx , ∑f



where f is the frequency of each data value x. If f = 1 for each data value x, then x=



∑x , n



where n = ∑ f is the size of the data.



Investigation Obtaining a Formula for Standard Deviation



Table 9.4 shows the temperatures, in degree Celsius (°C) of two cities, City A and City B on a particular day, taken at 4-hour intervals. Time



Temperature of City A (°C)



Temperature of City B (°C)



0000



25



21



0400



24



15



0800



26



23



1200



33



36



1600



31



41



2000



29



32 Table 9.4



Part 1: Mean Temperatures 1. Find the mean temperature of City A and of City B. 2. Are the mean temperatures of both cities equal? 3. By looking at Table 9.4 closely, what can you say about the spread of the temperatures of City A as compared to the spread of the temperatures of City B in relation to the respective mean temperatures?



Statistical Data Analysis



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322



Part 2: Spread of the Temperatures 4. In order to find a better measure of the spread of the temperatures, copy and complete Table 9.5 for City A. The first row has been done for you. x



x−x



25



25 – 28 = –3



24 26 33 31 29 ∑ (x − x ) =



Sum



Table 9.5 5. Fig. 9.12 shows the graphs of the temperatures of both cities. Compare the graphs and decide which set of data is more spread out.



xA



Temperature of City B Temperature (°C)



Temperature (°C)



Temperature of City A 45 40 35 x−x



30



xB



25



45 40



30 25



20



20



15



15



0



x−x



35



0 0000 0400 0800 1200 1600 2000



0000 0400 0800 1200 1600 2000



Time



Time



(a)



(b) Fig. 9.12



6.



Instead of a graph, we need to obtain a formula for measuring spread. Consider the value of ∑ (x − x ) . You have obtained this value for City A in Table 9.5. Use a similar method to obtain the value of ∑ (x − x ) for City B. Compare the values obtained for City A and City B. Is this a good measure of spread? Why?



7. Now consider ∑ (x − x )2 . Find the value of ∑ (x − x )2 for City A and City B and compare these values. Do you think it is a good measure of spread? Why?



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8.



What happens to ∑ (x − x )2 if the temperatures are taken at 2-hour intervals instead of 4-hour intervals, i.e. what happens to ∑ (x − x )2 if we have 12 data values instead of 6 data values? Does this mean that the spread will increase when there are more data values?



9. Find the value of



∑ (x − x )2 for City A and City B. Do you think it is a good n



measure of spread? Why? ∑ (x − x )2 is (°C)2 because n ∑ (x − x )2 we have squared (x − x ) . Hence we need to take the square root of n



10. The unit for temperature is °C. However, the unit for



to make the unit consistent. Find the value of



∑ (x − x )2 . n



∑ (x − x )2 is called the standard deviation. It measures how the temperatures n are spread about the mean x .



11. Calculate the standard deviation



∑ (x − x )2 for City B. n



12. Compare the standard deviation for both cities. Which standard deviation is larger? What does it mean when the standard deviation is larger?



From the above investigation, we have learned how the formula for the standard deviation comes about: Standard Deviation =



∑ (x − x )2 n



INF



OR



MA



TIO N



Some students may have observed that, instead of squaring x − x as in the investigation, the absolute value of x − x i.e. | x − x | can be used in order to eliminate the negative signs. There is actually such a measure of spread called the mean absolute deviation (MAD), ∑|x−x | , but statisticians i.e. n have found that the standard deviation is more useful for higher-level Statistics.



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324



Internet Resources



Alternative Formula for Standard Deviation There is an alternative formula for standard deviation: Standard Deviation =



∑ x2 − x2 n



It is easier to compute the standard deviation using this formula than the first formula, i.e.



Worked Example



∑ (x − x )2 . n



14



i.e.



(Finding the Standard Deviation Using the Alternative Formula) The data for the temperature of City A is shown below. Find the standard deviation using the alternative formula



x 25 24 26 33 31 29 ∑ x =168



∑x n 168 = 6 = 28



Mean x =



Standard deviation =



∑ x2 − x2 n



=



4768 − 282 6



= 3.27 (to 3 s.f.)



Chapter 9



∑ x2 − x2 . n



Temperature of City A 25 24 26 33 31 29 AT



Solution:



325



∑ x2 − x 2 , from n ∑ (x − x )2 . n



Time 0000 0400 0800 1200 1600 2000







Go to http://www.shinglee.com.sg/ StudentResources/ and open the worksheet ‘Standard Deviation‘ to find out how to obtain the alternative formula for standard deviation,



Statistical Data Analysis



x2 625 576 676 1089 961 841 ∑ x2 = 4768



TE



NTI



ON



Compare the solution in this Worked Example with Table 9.5. Which formula is easier to use?







SIMILAR QUESTIONS



PRACTICE NOW 1 4



The table below shows the number of grammatical errors made by Kate in eight English essays submitted this semester. Find the standard deviation of the number of errors made. Show your working clearly. Essay Number



1



2



3



4



5



6



7



8



Number of Errors



6



9



15



26



10



14



21



3



Exercise 9E Questions 1(a)-(c)



Use of Calculator to find Standard Deviation for Ungrouped Data In Worked Example 14, we can also make use of the statistical functions of scientific calculators to find the standard deviation directly. The data presented in Worked Example 14 is an example of ungrouped data. We use the same data points (temperature of City A) as in Worked Example 14, i.e. 25, 24, 26, 31, 33, 29



AT



TE



NTI



ON



The buttons on calculators vary with different models. Refer to the instruction manual of your calculator.



Before we start, we must always remember to clear all the data currently stored in the calculator memory. To do so, press SHIFT 9 (CLR) 2



=



AC . Follow the steps below to obtain the standard deviation. AT



STEPS 1.



MODE



2.



2 (STAT)



3.



1 (1-VAR)



(this changes the calculator to ‘Statistics’ mode)



Step 9. =



2



5



=



2



4



=



2



6



=







3



1



=



3



3



=



2



9



=



We can also use the calculator to find the mean, i.e. x , for ungrouped data first.



5. AC



Steps 1 to 7 are the same. Step 8. 2 ( x )



6. SHIFT 1



8.



3 (xσn)



ON



Step 8. 2 ( x )







4 (VAR)



NTI



To obtain the mean after step 9, repeat step 5 and 7 and continue with



4. Enter the data one at a time, i.e.



7.



TE



Step 9. =



(the screen displays xσn)



9. = (the screen displays the value of the standard deviation, i.e. 3.265…)



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326



PRACTICE NOW



SIMILAR QUESTIONS



The ages of 7 people are 16, 21, 22, 18, 20, 12 and 24 years. With the help of a calculator, find the standard deviation of their ages.



Exercise 9E Questions 2(a)-(c), 8-10, 14-15



Standard Deviation for Grouped Data For grouped data, the formula for finding standard deviation is essentially the same: Standard Deviation = where the mean, x =



2



∑ f (x − x ) ∑f



computation of the standard deviation.



15



∑ fx − x2 , ∑f



or



∑ fx . ∑f



Similarly, the second formula, i.e.



Worked Example



2



∑ fx 2 − x 2 , is easier to use for the ∑f



(Finding the Standard Deviation for Grouped Data) 100 Grade 10 students, each from School A and School B, were asked for the amount of time they spent watching television each week. The results are given in the table below. School A Number of Hours 10  x  15



Number of Students 3



15  x  20



12



20  x  25



19



25  x  30



36



30  x  35



22



35  x  40



8 School B



(i)



Mean



26.3 hours



Standard Deviation



5.12 hours



Find an estimate of the mean and standard deviation of the number of hours spent watching television by the 100 students from School A, showing your working clearly.



(ii) Compare and comment briefly on the results of the two schools.



327



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INF



OR



MA



TIO N



For grouped data, both the mean and the standard deviation computed are estimates as the mid-value is used to represent the data in each group.



Solution:



RE



(i)  Number of Hours



Frequency



Mid-value (x)



fx



fx2



10  x  15



3



12.5



37.5



468.75



15  x  20



12



17.5



210



3675



20  x  25



19



22.5



427.5



9618.75



25  x  30



36



27.5



990



27 225



30  x  35



22



32.5



715



23 237.5



35  x  40



8



37.5



300



11 250



Sum



∑ f = 100



∑ fx = 2680



∑ fx2 = 75 475







Mean x =



= 26.8 hours







Standard deviation =



=



As the calculation of the standard deviation involves the mean, the mid-value of each class interval needs to be calculated. The calculator can also be used to obtain the values of ∑ fx and ∑ fx 2 .



AT







L



In order to obtain the mean for a set of grouped data, we need to calculate the mid-value of each class interval.



∑ fx ∑f



2680 = 100



CAL



TE



NTI



ON



For Worked Example 15, use the first formula



∑ fx 2 − x2 ∑f



i.e.



∑ f (x − x )2 ∑f



to compute



the standard deviation and compare which formula is easier to use.



75 475 − 26.82 100



= 6.04 hours (to 3 s.f.)







∴ For School A, the mean is 26.8 hours and the standard deviation is 6.04 hours.



(ii)



The students in both schools spent approximately the same number of hours, on average, watching television. However, School A has a higher standard deviation, which indicates that there is a greater spread in the number of hours spent watching television, i.e. some students spent long hours while some spent very little time watching television.



Statistical Data Analysis



Chapter 9



328



PRACTICE NOW 1 5



SIMILAR QUESTIONS



30 students, each from Class A and Class B took the same examination. Information on the examination results is shown in the tables below. Class A Marks



0x4



4x8



8  x  12



Frequency



3



8



14



12  x  16 16  x  20 2



3



Class B Mean



9.7



Standard Deviation



3.1



(a) For Class A, find an estimate of



(i) the mean mark,







(ii) the standard deviation,







showing your working clearly.



(b) Compare and comment briefly on the results for the two classes.



Use of Calculator to find Standard Deviation for Grouped Data Similarly, we can use the calculator to find the standard deviation directly for grouped data. The data for School A in Worked Example 15 is an example of grouped data, as the data was grouped according to the given class intervals. The data for School A is shown again, in Table 9.6 below. Hours



Mid-value (x)



Frequency



10  x  15



12.5



3



15  x  20



17.5



12



20  x  25



22.5



19



25  x  30



27.5



36



30  x  35



32.5



22



35  x  40



37.5



8



Table 9.6 Similarly, we press SHIFT 9 (CLR) 2 memory.



= AC to clear the calculator



In addition, we need to switch on the ‘FREQ’ column in order to input the frequency values. To do so, key in SHIFT MODE



329



Chapter 9



Statistical Data Analysis



↓



4 (STAT) 1 (ON).



Exercise 9E Questions 3-6, 11



Follow the steps below to obtain the standard deviation. STEPS 1.



MODE



2.



2 (STAT)



(this changes the calculator to ‘Statistics’ mode)



3.



1 (1-VAR)



(we will see the column ‘FREQ’)



4. Enter the data (mid-value) and its corresponding frequency, one at a time. Use the arrow keys to move to the position where you want to input the values.



X FREQ







1



2



.



5



=







3



=







1



7



.



5



=







1



2



=







2



2



.



5



=



1



9



=







2



7



.



5



=



3



6



=







3



2



.



5



=



2



2



=







3



7



.



5



=



8



=







5. AC 6. SHIFT 1 7.



4 (VAR)



8.



3 (xσn) (the screen displays xσn)



9.



=



(the screen displays the value of the standard deviation, i.e. 6.042…)



Statistical Data Analysis



Chapter 9



330



PRACTICE NOW



SIMILAR QUESTIONS



The table shows the frequency distribution of the masses of 60 snails in grams. Mass (g)



Number of Snails



0  x  10



1



10  x  20



2



20  x  30



10



30  x  40



18



40  x  50



20



50  x  60



6



60  x  70



3



With the help of a calculator, find an estimate of the standard deviation for the given data.



Thinking Time A meteorologist has calculated the mean temperature and standard deviation of a particular day, with temperatures measured at hourly intervals. The calculation was done using statistical software. Mean: 29.5 °C Standard Deviation: 3.2 °C Due to a systematic error in the software, the hourly measurements taken were all overestimated by 1.5 °C. Explain clearly how the measured mean and standard deviation have been affected by this error.



331



Chapter 9



Statistical Data Analysis



Exercise 9E Questions 7(a)-(b), 12-13, 16



Class Discussion Matching Histograms with Data Sets Discuss in groups of three. Table 9.7 shows the mean, median and standard deviation of several data sets. Each data set has the same sample size. Data set



Mean



Median



Standard Deviation



I



70



60



13



II



90



90



30



III



85



60



25



IV



110



120



23



V



83



100



17



VI



81



80



19



Table 9.7



Frequency



Frequency



Fig. 9.13 shows six histograms A to F.



x



B



x



Frequency



Frequency



A



x



D



x



Frequency



Frequency



C



E



x



F



x



Fig. 9.13 Match the histograms to the data sets. Explain your answers.



Statistical Data Analysis



Chapter 9



332



Class Discussion Can We Always Trust the Statistics We Read? Discuss in groups of four. Part 1: The Choice of Averages Read the news report on an employment survey, as shown in Fig. 9.14.



NEWS REPORT Fresh graduates of Statville National University (SNU) have the highest starting salary. 86% of the 2014 SNU cohort who graduated participated in an employment survey. The survey revealed that the average monthly starting salary of SNU fresh graduates in full-time employment is 24 895, the highest among all public universities in the country. Fig. 9.14 Table 9.8 shows the actual summary statistics from the employment survey. However, this was not published in the news report. Summary Statistics of Employment Survey Mean Starting Salary



24 895



Lower Quartile



19 367



Median Starting Salary



20 583



Upper Quartile



25 674



Modal Starting Salary



19 819



th



90 Percentile



35 604



Table 9.8 1. From Table 9.8, how do we interpret (i) the median starting salary, (ii) the 90th percentile? 2. Find the numerical difference between the mean starting salary and median starting salary. What does the difference suggest about the distribution of starting salaries among SNU’s fresh graduates, and what could have caused this difference? 3.



Combining your analysis from Questions 1 and 2 and Table 9.8, do you think that the ‘average monthly starting salary’ used in the news report gives an accurate description of the starting salaries of fresh graduates from SNU? Explain your answer.



4. What are some points that you can learn about the choice of averages presented in everyday statistics?



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Part 2: The Collection of Statistical Data Fig. 9.15 shows a printed advertisement by a toothpaste company, Superclean. Study Fig. 9.15 and answer the questions.



90% of respondents in a recent survey recommend Superclean Toothpaste!



an



Supercle



Fig. 9.15 5. In Fig. 9.15, do you think that the statistic ‘90% of respondents in a recent survey recommend Superclean Toothpaste’ is credible? Give two reasons to support your answer. 6. The Statville Advertising Standards Authority has banned this advertisement as it believes that it is ‘misleading’ consumers. With reference to Fig. 9.16, explain why the advertisement in Fig. 9.15 is considered to be misleading.



SURVEY QUESTIONNAIRE Name 3 toothpaste brands you will recommend. 1. _______________________ 2. _______________________ 3. _______________________



Fig. 9.16 7. What are some points that you can learn about the collection of statistical data?



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334



Projected Growth in Dividend Payment



Part 3: The Display of Statistical Data Statistical graphs and diagrams are often used to summarize or highlight data findings. Study each of the diagrams given.



pesos/unit of stock 4.5



An investment company displayed the bar graph in Fig. 9.17 to its prospective clients, to demonstrate the positive outlook in investing in one of its financial products.



4.0



8. Explain what is misleading about the bar graph.



3.0



3.5



2.5 0



2015



2016



2017



Year



Fig. 9.17 Global Smartphone Market Share The Chief Executive Officer of a big smartphone company used the following 3D pie chart to present his company's market share in the global smartphone market during a company presentation.



Other Brands 25% Company B 30%



Company C 19%



9. (i) Which company do you think used the 3D pie chart in Fig. 9.18? Explain your answer. (ii) Explain what is misleading about this 3D pie chart.



Fig. 9.18



Company A 26%



10. What are some points that you can learn about the display of statistical data?



Performance Task We have learned various methods to handle statistical data and recognized the importance of statistic in representing data and information meaningfully. Conduct a mini-research on your classmates’ performance in the final examination in Mathematics and Science last year. Work in groups of four to evaluate and interpret your results using the following methods: •



Cumulative frequency curve



•



Box-and-whisker plot



•



Stem-and-leaf diagram



Your evaluation should include interpretation of results using appropriate measures of positions to formulate meaningful decisions.



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Chapter 9



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Exercise



9E



1. Calculate the standard deviation of each set of data. Show your working clearly. (a) 3, 4, 5, 7, 8, 10, 13 (b) 28, 25, 32, 20, 30, 19, 22, 24, 27, 23 (c) –5, –4, 0, 1, 4, –2







2. Use your calculator to find the standard deviation of each set of data. (a) 128, 135, 156, 123, 144, 130 (b) 0, 1, 25, 14, 2, 16, 22, 4 (c) 39.6, 12, 13.5, 22.6, 31.3, 8.4, 5.5, 4.7 3. The distribution of marks scored by students for a class quiz is shown in the table below. Marks



2



3



4



5



6



7



8



Number of Students



5



7



6



4



9



3



6







Calculate the standard deviation for the marks. Show your working clearly.



4. The number of goals scored per match by Spurs United during the soccer league season is shown in the frequency table below. Number of Goals Scored per Match







6. The weekly salaries, in pesos, of 60 workers in a company are shown in the table below.



0



1



2



3



4



5



6



Number of Matches 10



8



7



6



2



3



1



Calculate the standard deviation. Show your working clearly.



5. Find an estimate of the standard deviation for the following set of data. Show your working clearly. x 0x5 5  x  10 10  x  15 15  x  20 20  x  25 25  x  30



Frequency 4 12 20 24 16 4



Salary ( )



Frequency



5000  x  5200



8



5200  x  5400



23



5400  x  5600



16



5600  x  5800



3



5800  x  6000



10



Find an estimate of the standard deviation of the weekly salary of the workers. Show your working clearly. 7. Use your calculator to find an estimate of the standard deviation of each of the following sets of data. (a) x Frequency 16 30  x  40



(b)



40  x  50



25



50  x  60



35



60  x  70



14



70  x  80



10



70  y  75



Frequency 4



75  y  80



11



80  y  85



15



85  y  90



24



90  y  95



18



95  y  100



9



100  y  105



3



y



Statistical Data Analysis



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336



8. The results for an IQ quiz taken by 8 students from Class A and Class B are shown below. The maximum score for the quiz is 25.







Mrs Santos brought Kate on a holiday to New Zealand for a week. The time taken for Kate to fall asleep on each of the nights during the week is shown in the table below.



Class A: 4, 6, 6, 7, 8, 10, 11, 12 Class B: 0, 1, 1, 2, 3, 14, 17, 25







(i) Calculate the mean and standard deviation for Class A and Class B. Show your working clearly. (ii) Compare and comment briefly on the results of Class A and Class B. 9. Kate scored x marks in a Mathematics quiz and her friends’ scores were 5, 16, 6, 10 and 4. The mean mark of these six students is 10. (i) Find the value of x.



Time Taken to Fall Asleep (Minutes)



Day Monday



20



Tuesday



12



Wednesday



5



Thursday



10



Friday



25



Saturday



3



Sunday



12



(ii) Find the standard deviation of the marks of the six students.



(ii) Calculate the mean and standard deviation of the time taken for Kate to fall asleep during the week in New Zealand.



(iii) How did Kate perform for the quiz, relative to her friends?



(iii) Compare and comment on the answers in parts (i) and (ii).



10. Mrs Santos reads bedtime stories to her daughter, Kate every night. The time taken for Kate to fall asleep on each night in a particular week is shown in the table below.



11. Two trains, A and B, are scheduled to arrive at a station at certain time. The times (in minutes) by which the trains arrived after the scheduled time were recorded in the table below.



Day



Time Taken to Fall Asleep (Minutes)



Time Number of days (minutes) for Train A



Number of days for Train B



Monday



23



2



3



4



Tuesday



15



3



2



3



Wednesday



8



4



5



9



Thursday



13



5



12



9



Friday



28



6



10



7



Saturday



6



7



6



5



8



1



3



Sunday



15



9



1



0



(i) Calculate the mean and standard deviation of the time taken for Kate to fall asleep.







(i) For each train, calculate the mean of the data and standard deviation of the data. (ii) Which train is more consistently arriving late? Briefly explain your answer. (iii) Which train is more punctual on the whole? Briefly explain your answer.



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12. The waiting time, in minutes, for 60 patients at two hospitals are given in the tables below.







Stamford Hospital Time (minutes)



Number of Patients



20  t  22



5



22  t  24



11



24  t  26



27



26  t  28



13



28  t  30



4











14. It is given that the six numbers, 10, 6, 18, x, 15 and y have a mean of 9 and a standard deviation of 6. Find the value of x and of y. 15. A set Q contains n numbers, which has a mean of 5 and a standard deviation of 1.8. Two additional numbers (from sets A, B or C) are to be added to the set Q. A: {–2, 12}







Hillview Hospital Mean



Standard Deviation



25



3.2



(a) For Stamford Hospital, find an estimate of the



B: {5, 16}



C: {4, 6}



(i) Which of the sets, when added to set Q, will result in a new mean which is unchanged from the original mean? (ii) Using your answer in (i) or otherwise, state which of the sets, when added to set Q, will result in a new standard deviation which is closest to the original standard deviation.



(i) mean waiting time, (ii) standard deviation. (b) Compare, briefly, the waiting time for the two hospitals.



16. The table below shows the masses of 100 students from Brighthill School and 100 students from Hogwarts School. (All masses are corrected to the nearest 5 kg.) Brighthill School



13. The table shows the daily temperatures of two cities in Sahara Desert over a period of 50 days. Temperature (°C)







Number of Days



Mass (x kg)



45



50



55



60



65



70



Number of Students



5



36



28



22



7



2



City A



City B



35  x  40



1



2



Hogwarts School



40  x  45



4



14



Mass (y kg) 40 45 50 55 60 65 70 75 80



45  x  50



12



16



50  x  55



23



10



55  x  60



7



5







60  x  65



3



3







(a) For each city, calculate an estimate of the (i) mean temperature, (ii) standard deviation. (b) Which city is warmer on the whole? State a reason. (c) Which city’s daily temperature is more consistent? Explain your answer.



Number of Students



7



21 24



6



3



26



8



1



4



The National Statistics Division requires the combined statistics (mean and standard deviation) of both schools.



(i) Can we use x + y to find the combined mean? Explain your answer. (ii) Can we add the standard deviations of the masses for both schools to find the combined standard deviation? Explain your answer. (iii) Find an estimate of the combined mean and standard deviation of all 200 students.



Statistical Data Analysis



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338



1. Data in a dot diagram are represented by dots above a horizontal number line. 2. The values in stem-and-leaf diagrams are split into two parts, the stems and the leaves. There are variations of the stem-and-leaf diagrams such as those with split stems and the back-to-back stem-and-leaf diagrams. 3. The choice of an appropriate statistical diagram depends on the type of data collected and the purpose of collecting the data. • A dot diagram is most suitable when we want to display a small set of data that does not contain many distinct values. • A stem-and-leaf diagram is most suitable when we need to know the exact values of the original data. 4. A table of cumulative frequencies is a way of presenting a set of data. It can be obtained from a frequency table. The cumulative distribution can be displayed graphically by a cumulative frequency curve. 5. The range of a set of data is the difference between the largest value and the smallest value. 6. The interquartile range is the difference between the upper quartile (Q3) and the lower quartile (Q1). It measures the spread of the middle 50% of the data. 7. A cumulative frequency curve can be used to estimate the median, quartiles, percentiles and deciles of a distribution. It can also be used to obtain estimates such as how many students scored less than a certain mark, etc. 8. A box-and-whisker plot illustrates the range, lower quartile (Q1), median and upper quartile (Q3) of a frequency distribution. MIN



Q1 Whisker



Median



MAX



Q3 Whisker



Box



9. The standard deviation measures the spread of a set of data about the mean. 10. For ungrouped data, the standard deviation is ∑ (x − x )2 n







339



where the mean, x =



Chapter 9



∑x . n



Statistical Data Analysis



or



∑ x2 − x2 , n



11. For grouped data, the standard deviation is ∑ f (x − x )2 ∑f







where the mean, x =



or



∑ fx 2 − x2 , ∑f



∑ fx . ∑f



9



1.



The amount of time spent by 750 students of Hillcrest School to travel from home to school on a particular morning is shown in the cumulative frequency curve. Cumulative Frequency Curve for the Time Spent Traveling to School 800



(a) Use the graph to estimate (i) the number of students who take less than 17.5 minutes to travel to school, (ii) the fraction of the 750 students who take at least 27 minutes to travel to school, (iii) the value of x, given that 40% of the 750 students take at least x minutes to travel to school. (b) Estimate the 90th percentile.



700



2. A survey was done to find out the number of children (aged 13 or below) in a family.



Cumulative Frequency



600 500 0



400



1 2 3 4 5 Number of Children



300



(a) Find the standard deviation of the number of children in a family.



200



(b) A box-and-whisker plot is drawn to represent the data.



100 0



0 10



20



30



Time (minutes)



40



50







x1



x2



x3



5



Number of Children



(i) Find the values of x1, x2 and x3. (ii) Find the interquartile range of the data.



Statistical Data Analysis



Chapter 9



340



3. The daily amounts of money (in pesos) spent by Kate over a period of 14 days are recorded as follows: 200, 220, 300, 150, 190, 220, 400 250, 230, 330, 290, 300, 940, 190



(i) greater than 55 mm,



(ii) the interquartile range, (iii) the mean and standard deviation of the data.



4. In an agricultural experiment, the lengths (mm) of 124 ears of barley from Australia were measured. The cumulative frequency curve shows the length x mm, and the number of ears of barley with lengths less than or equal to x mm.



5.



There are 160 students taking the same examination paper in each of the two schools. The cumulative frequency curves show the marks scored by the students. Cumulative Frequency Curve for the Marks of an Examination



Cumulative Frequency



130 120 110 100



140 120 100 80 60



School A School B



40 20 0



90 Cumulative Frequency



(ii) not greater than 25 mm or greater than 64 mm.



160



Cumulative Frequency Curve for the Lengths of Ears of Barley



10 20 30 40 50 60 70 80 90 100 Marks



80



(a) For the students in School A, use the graph to find (i) the median, (ii) interquartile range. (b) For the students in School B, use the graph to find (i) the median, (ii) interquartile range. (c) Estimate the percentage of students from School B who scored more than 80 marks. (d) Make two comparisons between the scores of the students from School A and School B.



70 60 50 40 30 20 10 0 10



20



30



40



50



Length (mm)



Chapter 9



(ii) lower and upper quartiles.



(c) Using the graph, find the number of ears of barley with lengths



(i) the median of the data,



341



(i) the median length, (b) Find the interquartile range.



Find







(a) Using the graph, estimate



Statistical Data Analysis



60



70



6.



In a rifle range, Sam and Carlo fired 6 shots each at a target. The table shows the distance, in millimeters, of each shot from the center of the target. Shooter



7. The following table shows the lifespans, to the nearest hour, for 100 light bulbs produced by two companies, Brightworks and Lumina. Lifespans (hours)



Distance from center of target (mm)



Number of light bulbs Brightworks



Lumina



600  t  700



2



8



4



700  t  800



9



10



(a) For each shooter, calculate (i) the mean distance from the center of the target, (ii) the standard deviation.



800  t  900



16



12



900  t  1000



21



16



1000  t  1100



29



r



1100  t  1200



18



18



1200  t  1300



5



12



Mean



p



989.5



Standard Deviation



q



t



Sam’s shots



47



16



32



1



19



35



Carlo’s shots



20



9



16



43



13



(b) Make two comparisons between the shots fired by Sam and Carlo.



(i) Find the values of p, q, r and t. (ii) Make two comparisons between the lifespans of the light bulbs produced by Brightworks and Lumina. 8. The table shows the time taken, measured to the nearest minute, for 30 students to read a passage. 2



1



4



2



7



1



1



5



3



4



2



1



5



4



1



3



2



2



1



4



1







2



1



4



2



3



3



2



2



1



(i) Represent the data on a dot diagram. (ii) What were the most common time taken? (iii) Find the percentage of students who took at most 3 minutes to read the passage. (iv) Briefly describe the distribution of the data.



9. The dot diagram represents the results of a survey conducted on a group of primary school children to find out the average number of sweets they eat in a week.











0



1



2



3



4



5



6



(i) How many sweets do most children eat in a week? (ii) What is the greatest number of sweets the children eat in a week? (iii) Find the fraction of children who do not eat sweets in a week.



Statistical Data Analysis



Chapter 9



342



10. The table shows the lengths, in cm, of some metal rods in a factory. 146



130



141



158



150



165



130



133



145



153



152



145



147



151



165



132



131



145



146



141



(i) Represent the data using a stem-and-leaf diagram. (ii) What is the length of the longest metal rod? (iii) What is the most common length? (iv) Find the ratio of the number of metal rods that are shorter than 146 cm to the number of metal rods that are at least 152 cm long. 11. The stem-and-leaf diagram with split stems represents the speeds, in km/h, of some vehicles as they drove past a particular section of an expressway. The speeds of the vehicles were recorded by a speed camera located at that section. Stem 5 5 6 6 7 7 8 8 9 9



3 5 0 6 0 5 0 5 0 9



Leaf 7 1 8 2 5 1 5 0



9 2 8 2 5 2 5 1



3 9 3 6 2 6 2



4 6 2 6



4 8 2 6



8 2 7



9 2 7



9 2 7



3 7



3 7



3 8



3 9



4



4



Key: 5 | 2 means 52 km/h (i) What was the most common speed? (ii) The highest speed limit for vehicles on an expressway is 90 km/h. Find the percentage of vehicles that exceeded the speed limit. 3 (iii) A vehicle is selected at random. The probability that the speed of the vehicle exceeds x km/h is . 10 Find the value of x. 12. The average amounts of money, in pesos, saved by two groups of students in a day are recorded in the table. Group A Group B



73 28 39 62



70 64 80 38



30 96 42 45



36 74 27 61



68 68 60 46



55 85 59 32



42 51 35 73



66 30 59 36



(i) Represent these two sets of data by a back-to-back stem-and-leaf diagram. (ii) What is the greatest average amount of money saved by the students in each group? (iii) Which group of students saves more money in a month? Explain your answer.



343



Chapter 9



Statistical Data Analysis



Challenge 1. A set of data, Set A is given below. Set A: 9, 11, 16, 24, 34



Another four sets of data, Sets W, X, Y and Z are shown below. Set W: –9, –11, –16, –24, –34 Set X: 14, 16, 21, 29, 39 Set Y: 18, 22, 32, 48, 68 Set Z: 16, 26, 34, 39, 41



Without calculating the standard deviation for any of the above sets, explain clearly which of the sets W, X, Y or Z has the same standard deviation as that of Set A. Hint: The standard deviation describes how the data are spread about the mean. 2. Two sets, M and N, have the same mean, standard deviation and data size, i.e. n. Give an example of such a pair of sets.



Statistical Data Analysis



Chapter 9



344



C1



Revision Exercise



1. A waiting room in a private medical clinic has a row of eight chairs. (a) In how many ways can four of these chairs be occupied by a boy, an old lady and two young sisters? (b) In how many of these ways can (i) the two young sisters be seated on adjoining chairs? (ii) the two young sisters be seated at the ends of the row of chairs? (iii) all four be seated on adjoining chairs? 2.



In how many ways can a group of three men be selected from seven men? How many ways of selection are there if one of two particular men must not be included?



3. A bag contains 5 red, 7 yellow and 1 white disc. Two discs are taken out in succession without replacement. By drawing a tree diagram or otherwise, find the probability of getting (i) two red discs, (ii) a red and a yellow disc in that order, (iii) two white discs, (iv) two discs of different colours. 4. The table shows the number of times the students in a class fell ill last year. 8 7 5 2 5



5 10 4 4 5



7 9 6 5 5



5 6 2 9 3



3 8 6 8 2



6 10 0 6 3



7 2 5 6 7



(i) Represent the data on a dot diagram. (ii) What was the most common number of times the students in the class fell ill? (iii) Find the percentage of students who did not fall ill last year.



345



Revision Exercise C1



5. The stem-and-leaf diagram below represents the Science quiz marks of 24 students.











Stem 0 1 2 3 4 5 6 7 8



7 2 4 0 3 0 2 8 3



Leaf 8 4 5 1 2 5



5 6



9 6



6



7



9



6 7



Key: 0 | 7 means 7 marks



Find (i) the median mark, (ii) the range, (iii) the interquartile range.



6. The box-and-whisker plots below show the distributions of the starting monthly salaries of fresh graduates from two universities, A and B. A B



2 2.2 2.4 2.6 2.8 3.0 3.2 3.4 3.6 3.8 4.0 4.2 4.4 Starting Monthly Salary ( ’0000) (i) Find the interquartile range for both universities. (ii) ‘On average, fresh graduates from university A get a higher starting salary than fresh graduates from university B’. Do you agree? Give a reason for your answer. (iii) Which university has a bigger proportion of fresh graduates getting more than 3500 for their starting salary? Give a reason for your answer.



C1



Revision Exercise



7. The cumulative frequency curve below represents the standing broad jump distance (cm) of 80 Grade 10 boys.



8. The period of time 120 cars stayed at a car park was measured and recorded to the nearest minute. The results are shown in the table below. Period of time (t min)



Number of cars



15  x  30



12



30  x  45



18



45  x  60



27



60  x  75



39



60



75  x  90



16



50



90  x  105



8



Cumulative Frequency Curve for the Standing Broad Jump Distance of Grade 10 Boys







Cumulative Frequency



80 70



40



(a) Construct a table of cumulative frequencies for the given data. (b) Using the table in (a), find the number of cars which stayed (i) less than or equal to 60 minutes, (ii) more than 75 minutes, 180 190 200 210 220 230 240 250 260 270 280 (iii) more than 45 minutes but not more than 90 minutes. Distance (cm)



30 20 10 0



(a) Copy and complete the grouped frequency table of the standing broad jump distance of each boy. Distance (cm)



Frequency



180  x  200



9.



Find the range, lower quartile, median, upper quartile and interquartile range of the following sets of data. (a) 10, 7, 4, 3, 6, 9, 2 (b) 95, 60, 66, 48, 76, 87, 82, 55 (c) 170, 219, 152, 208, 140, 167, 96, 223, 181



200  x  220 220  x  240 240  x  260



260  x  280



(b) Using the grouped frequency table, find an estimate of (i) the mean standing broad jump distance, (ii) the standard deviation. (c)



Another batch of 80 students who have taken the standing broad jump test have the same median but a larger standard deviation. Describe how its cumulative frequency curve will differ from the given curve.



Revision Exercise C2



346



C2



Revision Exercise



1. Six girls are to be chosen from a group of 12 girls to form a debating team. How many different ways can we choose the debating team? If two of the girls are sisters and not more than one of them can be selected, how many different ways can the team be formed now? 2. A neighborhood watch group of five members is to be selected from five men and nine women. In how many ways can such a committee be formed if (a) there is no restriction, (b) the committee must include at least a man, (c) the committee must include at least two women?



3. Three national servicemen Chris, Angelo and Daniel took part in a rifle shooting competition. The probabilities that Chris, Angelo and Daniel 2 3 4 will each hit the target are , and 7 3 5 respectively. (a) Three of them fire one shot each simultaneously at the target. Find the probability that (i) all three men hit the target, (ii) all three men miss the target, (iii) exactly two of them hit the target, (iv) at least one man hits the target. (b) In a game, Chris, Angelo and Daniel fire one shot each at the target, in that order. Once the target is hit, the game ends and the winner is the one who hits the target first. Find the probability that (i) the game ends after two shots, (ii) the game ends after three shots, (iii) the game ends by the third shot.



4. The times, in seconds, for 30 students to complete a 40-meter shuttle run are recorded in the table. 11.8 10.7 12.2



10.5 11.3 11.8



10.9 10.6 10.4



14.0 11.2 11.2



10.8 10.9 10.3



11.2 11.4 12.6



11.2 11.5 10.9



10.8 10.8 10.7



10.7 12.3 10.8



11.6 11.9 11.1



(i) Construct a stem-and-leaf diagram with split stems. (ii) To obtain a grade A for the shuttle run, a student must take less than 11.5 seconds to complete the run. Find the percentage of students who scored an A for the run. 5. Draw a box-and-whisker plot for each of the following sets of data.







347



(a) 16, 21, 17, 26, 28, 19, 23, 9, 31



(b) 0.06, 0.06, 0.02, 0.13, 0.09, 0.08, 0.04, 0.08



(c) 95, 92, 113, 58, 85, 80, 92, 98, 95, 100, 96, 105



Revision Exercise C1



C2



Revision Exercise



6. The cumulative frequency curve below shows the distribution of the marks scored by 600 students in a Mathematics examination in Euler High School.



7. The masses of 200 eggs from Rainbow Farm were measured and the results are illustrated by the cumulative frequency curve below.



Cumulative Frequency Curve for Marks Scored by 600 Pupils



Cumulative Frequency Curve for the Masses of Eggs from Rainbow Farm



600



200 Cumulative Frequency



Cumulative Frequency



500



400



300



150



100



50



200 0



100



0



45



20



40



60



80



100



Marks







The eggs are graded according to their masses in grams: Grade 1: 62 g  m  75 g Grade 2: 51 g  m  62 g Grade 3: 40 g  m  51 g



(i) the median mark,



(b) Indicate clearly the upper and lower quartiles on the graph and find the interquartile range.



The box-and-whisker plot gives the information on the marks scored by 600 students in the same examination in Fermat High School.



0 



20



40 60 Marks



80



100



(c) Find the median mark and the interquartile range. Hence, comment briefly on the performance of the students in the two schools.



75



65



Mass (g)



(a) Use the graph to estimate (ii) the passing mark such that 60% of the students will pass the examination.



55



(a) Using the curve, estimate (i) the median mass, (ii) the interquartile range, (iii) the percentage of eggs in each grade. (b) The masses of 200 eggs from Skyhi Farm were also measured and the results are represented by the box-and-whisker plot below.



40



45



50



55 60 Mass (g)



(i) State the median interquartile range.



65



70



mass



75 and



the



(ii) 5% of the eggs produced by Skyhi Farm are Grade 1. Make two comparisons between the quality of eggs from the two farms. Revision Exercise C2



348



Problems in Real-World Contexts PROBLEM 1: The Broken Plate An archaeologist found a part of a broken circular plate, drawn to scale of 1 cm to represent 4 cm, as shown. (i) Explain how you can help to find the diameter of the plate using geometrical properties of circles. (ii) Using the method you have explained in (i), find the diameter of the original circular plate. (iii) Hence, find the area of the original circular plate.



PROBLEM 2: Cooling of a Cup of Tea Very hot coffee, tea and water may cause serious burn injury. It had been reported that hot water at 56 °C could cause serious burns in 15 seconds. An experiment was done by a group of students to study the cooling of a cup of hot tea. The plotted points in the graph below represent the temperature of the tea over time. The students initially used a straight line graph to model the cooling of the tea. Temperature (T °C)



70



60



50



40



30



20



10







0



5



10



15



20



25



Time (t minutes)



(a) Use the graph to find the equation of the straight line. (b) Use the equation obtained in (a) to estimate



349







(i) the time taken for the temperature of the tea to decrease to 56 °C,







(ii) the temperature of the tea after 25 minutes. Explain why this is not a good estimate of the temperature of the tea. Problems in Real-World Contexts



Problems in Real-World Contexts (c) After further research and collection of more data, the students found that the best-fit curve to model the cooling of the tea is given by the equation T = a(2.72)–0.09t + 21, where a is a constant. Temperature (T °C)



70



60



50



40



30



20



10



0



5



10



15



20



25



Time (t minutes)



(i) Find the value of a. (ii) Hence, use the equation to estimate the temperature of the tea after 50 minutes. What do you think this temperature represents?



Problems in Real-World Contexts



350



Problems in Real-World Contexts PROBLEM 3: The Presentation of Data Statville Technologies is a leading producer of tablet computers. Its flagship product, the ‘Statville Tablet’ was released in 2011. In its latest media event, the Chief Executive Officer of Statville Technologies proudly announced the sale of its 120 millionth tablet, and that ‘quarterly sales performances have been very positive’. The following diagram was shown during the media event. AT



Media Release Cumulative Tablet Sales 120 million 100 80 60 40 20 0 Q3



2011



Q4



Q1



Q2



Q3



Q4



2012



Q1



2013



Q2



Q3



Q4



Q1



Q2



Q3



Q4



2014



Study the diagram and answer the questions below. (i) Suppose that for a particular quarter, zero sales were made. How will this be reflected on the curve shown in the diagram? (ii) Does the curve in the diagram support the claim that ‘quarterly sales performances have been very positive’? Explain your answer. (iii) Name one feature of the curve which you think may not reflect the actual quarterly sales figures accurately. (iv) Why do you think that Statville Technologies used a ‘cumulative sales curve’ to present its sales performance data?



351



Problems in Real-World Contexts



TE



NTI



ON



A quarter is a 3-month period on a financial calendar and it is typically expressed as ‘Q’. Company earnings are usually reported quarterly.



Problems in Real-World Contexts Problem 4: Box Office Earning The table below shows the top five worldwide highest grossing movies of all time. The amounts shown are in US dollars and the figures are not adjusted for inflation. The worldwide box office earnings of ‘$2 781 606 847’ for Blue Planet is calculated using the price of movie tickets in 2009.



ALL-TIME WORLDWIDE BOX OFFICE Rank



Title



Year of Release



Worldwide Box Office Earnings



1



Blue Planet



2009



$2 781 606 847



2



The Shipwreck



1997



$2 185 372 302



3



Superheroes



2012



$1 515 679 547



4



Platform 9.75



2011



$1 327 655 619



5



Freeze



2013



$1 259 136 600



(a) (i) How much more money did Blue Planet earn than The Shipwreck at the box office? (ii) Is the comparison between Blue Planet and The Shipwreck fair? Why or why not? Explain your answer. (b) We can also calculate and compare the earnings of the movies after adjusting for inflation. The inflation rate is the rate of increase of the ticket prices. Assuming a yearly inflation rate of 3.5% in movie ticket prices, the box office earnings of Superheroes would increase by 3.5% to $1 515 679 547  1.035 = $1 568 728 331 in 2013.



Calculate the box office earnings of







(i) Blue Planet,







in 2013.



(c)



Hence, compare the box office earnings of the two movies, Blue Planet and The Shipwreck in 2013, taking into account a yearly inflation rate of 3.5%. Which movie made more money and by how much? Why is this comparison a fairer one?



















(ii) The Shipwreck,



(d) The movie, The President, was released in 1991 and made $205 400 000 at the box office. Give an equation that allows you to calculate the box office earnings of The President in 2013, taking into account a yearly inflation rate of 3.5%.



Problems in Real-World Contexts



352



Problems in Real-World Contexts PROBLEM 5: 4-D Draw Study the news report and answer the questions.



NEWS REPORT 6904 Strikes Twice! ‘The four-digit number, 6904, has won both the first and second prizes in the 4-D draw on June 27, 2007. The probability of a particular number winning any prize in a 4-D draw is one out of 10 000 while the probability of any number appearing twice in the same draw is one out of 10 000 times 10 000, or one in 100 million.’



(a) A person can bet on any number from 0000 to 9999 in a 4-D draw. Find the probability of a particular number, e.g. 6904, winning the first prize in the draw. (b) There are 23 prizes to be won in a 4-D draw: First prize, second prize, third prize, 10 starter prizes and 10 consolation prizes. Find the probability of a particular number, e.g. 6904, winning any prize in a 4-D draw, leaving your answer correct to 4 significant figures. (c) The news report wrote, “The probability of a particular number winning any prize in a 4-D draw is one out of 10 000 …” By comparing your solutions for parts (a) and (b), do you agree with this statement in the news report? Explain. (d) Find the probability of a particular number, e.g. 6904, winning both the first and second prizes in the same 4-D draw. (e) Find the probability of any number winning both the first and second prizes in the same 4-D draw. (f) The news report wrote, “… the probability of any number appearing twice in the same draw is one out of 10 000 times 10 000, or one in 100 million.” By comparing your solutions for parts (d) and (e), do you agree with this statement in the news report? Explain.



353



Problems in Real-World Contexts



INF



OR



MA



TIO N



Personal gambling habits, if not controlled, leads to problem gambling, which brings hardship to self and family. Stay away from gambling.



P So roblem lvin g T ip



For part (b), find the probability that a particular number will not appear in the draw first. For part (e), this is similar to tossing two dice and finding the probability that both dice will show the same number.



Problems in Real-World Contexts Problem 6: Car Park Design A school is holding an evening concert. The parade square, which is a rectangular plot of land with dimensions 40 m by 30 m, will be converted into a car park to provide additional parking space. You are required to calculate the maximum number of additional car park labels that can be issued. Parking lots may be arranged to allow for parallel or angled parking. The figure below shows the arrangement of parking lots in parallel parking and the minimum width of the parking aisle. The proposed dimensions of a parking lot for parallel parking are 5.4 m by 2.4 m.



Arrangement of parking lots



3.6 m



3.6 m



6m



Parking lots



one side



both sides



one or both sides



Traffic flow



one-way



one-way



two-way



For angled parking, the figure below shows the arrangement of parking lots arranged at a parking angle of 30° to the line of traffic flow and the minimum width of the parking aisle. The proposed dimensions of a parking lot for angled parking are 4.8 m by 2.4 m.



30˚



30˚



3.6 m



4.2 m 30˚



6.3 m



30˚



Arrangement of parking lots



30˚



Parking lots



one side



both sides



one or both sides



Traffic flow



one-way



one-way



two-way



Problems in Real-World Contexts



354



Problems in Real-World Contexts Your teacher suggests the following guidelines for the minimum width of the parking aisle for the different parking angles, as shown in the table below. Parking angle



Parking lots on one Parking lots on both Parking lots on one or both sides, sides, side, one-way traffic flow one-way traffic flow two-way traffic flow



0° (parallel)



3.6 m



3.6 m



6.0 m



30°



3.6 m



4.2 m



6.3 m



45°



4.2 m



4.8 m



6.3 m



60°



4.8 m



4.8 m



6.6 m



90°



6.0 m



6.0 m



6.6 m



Showing all relevant calculations, suggest the arrangement of car park lots in the parade square so as to maximise the number of lots. What is the maximum number of additional car park labels that can be issued? Guiding Questions: (a) Should the traffic flow be restricted to one way or two ways? (b) Do the entrance and exit of the car park affect your model? (c) Does your model include handicap parking lots? If so, how will this affect the design of the car park?



355



Problems in Real-World Contexts



Practice Now Answers Chapter 1 Practice Now 1 1. (a) 28, 33 (b) –50, –56 (c) 1215, 3645 (d) –18, 6 2. (a) 22, 29 (b) 15, 11 Practice Now 2 (i) 23 (ii) 58 Practice Now 3 1. (a) 4n + 1 (b) 5n + 2 (c) 6n − 4 (d) 3n − 2 2. (i) 23, 27 (ii) 4n – 1 (iii) 199 Practice Now 4 (i) 3 (ii) 38 (iii) 248 Practice Now 5 (i) 7 (ii) 2 882 400.5 (iii) 3 362 800.5 Practice Now 6 2 (i) 5 2 (ii) 416 3 Practice Now 7 (i) 42.2 (ii) 5.21 Practice Now 8 24 and 96 Practice Now 9 (i) 6, 12; 7, 14; n + 1, 2n + 2 (ii) 54; 110 (iii) 59; 60



Chapter 2 Practice Now 1 (i) 8x3 + 3x2 – 7x + 11; 3 (ii) 3x2 – 3x – 1; 2 Practice Now 2 (i) 5x5 − 15x4 + 32x3 − 9x2 + 21x − 18 (ii) −100 (iii) Degree of P(x) × Q(x) = Degree of P(x) + Degree of Q(x) Practice Now 3 A = 3, B = 5, C = 7



Practice Now 13 1 5 ± , ±1, ±2, ± , ±4, ±5, ±10, ±20 2 2 Practice Now 14 1 1 1 3 (a) ± , ± , ± , ±1, ± , ±3 6 3 2 2 (b) −



1 3



Practice Now 15 1 1. −6, − , 1 2 2 2. −5, , 3 3



Practice Now 4 1. A = 3, B = 3, C = −5 2. A = 3, B = 12, C = 17



Practice Now 16 1. −7.27, 0.27, 3 2. −7.65, 0.65, 2



Practice Now 5 (x + 2)(x – 2)(4x + 3)



Practice Now 17 1. 6x4 − 45x3 + 87x2 − 45x + 6 2. −36 + 54x + 4x2 − 12x3 + 2x4



Practice Now 6 2 4x 2 + 6x + 16 – , the remainder is −2. 3 x− 2 Practice Now 7 4x+5 and the remainder 3x + 3 + (x +1) (x − 2) is 4x + 5.



Practice Now 8 (a) 3 (b) −480 31 (c) − 8 Practice Now 9 1. −2 2. A = 7, B = −5 Practice Now 10 1 k = −67 ; −125 2 Practice Now 11 p = −7, q = 16; (2x + 3)(x – 1)(x – 2)2 Practice Now 12 1. (x – 2)(x – 3)(x + 7) 2. (2x – 1)(x + 2)(x + 3)



Practice Now 18 (a) (6p + 7q2)(36p2 − 42pq2 + 49q4) (b) (1 – 2x)(37 + 20x + 4x2) Practice Now 19 4 5 Practice Now 20 (a) 2x2 + 54x + 15 + (b) 2 +



13−18 x



35 x−2



(3x −1) (x + 2)



Practice Now 21 2 5 1. + x−3 2x + 5 2. (x – 1)(x + 4)(2x – 3); 2 4 5 + − x −1 x + 4 2 x − 3 Practice Now 22 15 25 1. – + 2(2 x +1) 2(2 x +1)2 5 42 2. − x+2 (x + 2)2



Practice Now Answers



356



Practice Now 23 1 2 3 1. + + x+2 x −11 (x + 2)2



2. (x + 2)(x – 2)2,



5 15 11 − + 4(x + 2) 4(x − 2) (x − 2)2



Practice Now 24a 14 x + 22 1 1. + 3(x +1) 3(x 2 + 5) x 1 2. − 16 x 16(x 2 + 4) Practice Now 24b 4 1 1. 3x + − x +1 x −1 1 7 2. 3x – 2 + − x−3 2x − 5 Practice Now 25 1 2 1 (i) − + n n +1 n + 2 2 025 077 (ii) 4  050 156 Chapter 3 Practice Now 1 (i) 17.5, −17.5 (ii) 2.35, −2.55 Practice Now 2 (i) 1.2 (ii) −2.5 Practice Now 3 (i) −0.9 (ii) 0.8 or −0.8 Practice Now 4 (i) 0.3 (ii) −0.3 Practice Now 5 (a) a = 5, b = −3 (c) 1.61 (d) (ii) h = 2, k = −4



357



Practice Now 8 (a) a = 2, b = 17 (c) (i) 0.7 or 3.9 (ii) t = 2.25 (iii) 9 m/s2 (iv) 0.25 , t , 4.3 Practice Now 9 Carpark X Practice Now 10 (a) 60 beats/minute (b) 6 beats/minute2 (c) 1 beat/minute2 Chapter 4 Practice Now 1 1. (i) 99.5 cm (ii) 108 cm ⎛21π ⎞ 2. ⎜ +18⎟ cm ⎝ 2 ⎠ 3. 6.65 Practice Now 2 1. 13.2 m 2. 34.0 cm Practice Now 3 (i) 56.6 cm (ii) 155 cm Practice Now 4 (ii) 459 cm2 Practice Now 5 (ii) 122 m2 Practice Now 6 (a) 12° (b) 270° (c) 174.2° (d) 458.4°



Practice Now 6 (b) 3.5 minutes (c) 27 km/h



Practice Now 7 (a) 0.628 rad (b) 5.03 rad (c) 3.45 rad (d) 6.98 rad



Practice Now 7 2 (i) 2 m/s2 3 (ii) 4.89 m/s (iii) 2 m/s2



Practice Now 8 (a) 0.605 (b) 0.973 (c) 2.82



Practice Now Answers



Practice Now 9 (a) 1.06 rad (b) 1.22 rad (c) 0.722 rad Practice Now 10 (i) 7.00 cm (ii) 11.9 cm Practice Now 11 (i) 0.772 rad (ii) 3.23 cm (iii) 5.30 cm Practice Now 12 4.71 cm Practice Now 13 1. (i) 0.75 rad (ii) 54 m2 2. 22 cm Practice Now 14 (ii) 15.8 m (iii) 93.4 m2 Practice Now 15 (i) 12.96 cm (ii) 43.4 cm2 Practice Now 16 (ii) (9.6 − 11.5) cm (iii) 52.5 cm2



Chapter 5 Practice Now 1 1. 21.35 cm 2. 5.05 cm Practice Now 2 32.6 cm or 6.13 cm Practice Now 3 1. (i) 23.2° (ii) 11.4 cm (iii) 23.625 cm2 2. (i) 3.9 (ii) 64.0° (iii) 7.10 cm2



Practice Now 4 1. (i) 28° (ii) 59° (iii) 26.3 cm (iv) 369 cm2 2. 32° Practice Now 5 1. (i) 56° (ii) 118° 2. x = 110°, y = 55° 3. 146° Practice Now 6 (i) 25° (ii) 50° (iii) 105° Practice Now 7 1. (i) 44° (ii) 25° (iii) 69° 2. x = 50°, y = 25° 3. 20° Practice Now 8 1. (i) (159 − x)° (ii) (149 − x)° (iii) 64° (iv) 85° 2. 34° Practice Now 9 48° Practice Now 10 (ii) 7.14 cm



Chapter 6 Practice Now 1 (a) 5 units (b) 11.4 units (c) 6 units Practice Now 2 ⎛ 1⎞ (a) ⎜0, 2 ⎟ 4⎠ ⎝ (b) (−3, 0) 3 (c) 3 units2 8 Practice Now 3 1. DEF 2. No



Practice Now 4 (a) (i) (4, 3) (ii) (1, 1) (iii) (2, 5) (b) x = 4, y = 3



Practice Now 7 3000 Practice Now 8



Practice Now 5 ⎛1 11⎞ (i) ⎜ , ⎟ ⎝2 2 ⎠ (ii) (−1, 4) (iii) AC = 5 2 units, BD = 3 2 units



(a) 20! or 20P20 (b) 3 186 701 844 000 000 (c) 8 536 498 274 000 000 (d) 685 430 596 700 000 000 (e) 12 804 747 410 000 000 (f) 14 485 008 380 000



Practice Now 6 (a) (0, 0), 9 (b) (−4, 6), 10 ⎛ ⎞ (c) ⎜0, − 1⎟ , 4 3 ⎝ ⎠



Practice Now 9 480 (a) 160 (b) 320



Practice Now 7 (a) (x + 1)2 + (y − 3)2 = 16 (b) (x − 4)2 + (y + 2)2 = 49 Practice Now 8 1. (x − 1)2 + (y − 2)2 = 10 2. (x + 3)2 + (y − 5)2 = 25



Practice Now 10 (a) 360 360 (b) 455 Practice Now 11 (a) 3060 (b) 70



Practice Now 9 (a) (3, −4); 4



Practice Now 13 n=5



⎛ 3⎞ 3 (b) ⎜−1, ⎟ ; ⎝ 4⎠ 4



Practice Now 10 1. (3x − 2)2 + (3y − 5)2 = 185 2. (i) A(−1, 3), B(−3, −1) 1 (ii) y = − x 2 (iii) 2 Chapter 7 Practice Now 1 9



Practice Now 2 11 (i) 30 73 (ii) 168 307 (iii) 840



Practice Now 2 15 Practice Now 3 336 Practice Now 4 60 Practice Now 5 504 words Practice Now 6 2730



Chapter 8 Practice Now 1 4 (i) 13 4 (ii) 13 2 (iii) 13 11 (iv) 13







Practice Now 3 7 (i) 12 5 (ii) 12 25 (iii) 144 5 (iv) 12 49 (v) 144



Practice Now Answers



358



Practice Now 4 2 1. (i) 9 1 (ii) 9 2. (i) 0.035 (ii) 0.065 (iii) 0.38 Practice Now 5 16 1. (i) 63 32 (ii) 63 43 (iii) 63 7 2. (i) 30 7 (ii) 15 49 (iii) 120 Chapter 9 Practice Now 1 (ii) 14 Practice Now 2 (i) 11 years old (ii) 20 (iii) 20 Practice Now 4 (ii) 90 kg (iii) 25% Practice Now 5 (ii) Miguel (iii) Carlo (iv) Miguel Practice Now (Page 286) (b) (i) 32 (ii) 18 (iii) 28 Practice Now 6 (a) 35, 55, 65, 70 (c) (i) 50 4 (ii) 7 (iii) 6.4



Practice Now 7 (i) 180 4 (ii) 5 (iii) 29.8 Practice Now 8 1. (i) 14, 30.5, 44 (ii) 49 (iii) 30 2. (i) Q1 = 15.5, Q2 = 26, Q3 = 41 (ii) 45 (iii) 25.5 Practice Now 9 (i) 50, 17, 81 (ii) 64 (iii) 1st decile = 8 8th decile = 84 (iv) 34 Practice Now 10 (a) (i) 62 (ii) 18 (b) (i) 70 (ii) 24 (c) School B (d) School A Practice Now 12 (a) (i) 28 (ii) 62 (iii) 14 Practice Now 13 (a) (i) 2500 (ii) 4700 (iii) 1500 (b) (i) 5000 (ii) 5500 (iii) 2200 (c) Brick B Practice Now 14 7.18 Practice Now (Page 327) 3.74 Practice Now 15 (a) (i) 9.2 (ii) 4.18 Practice Now (Page 331) 12.1



359



Practice Now Answers



Answers Chapter 1



Exercise 1A 1. (a) 39, 44



(b) 40, 32







(c) 384, 768







(d) 50, 25







(e) 16, –4







(f) –288, 576







(g) –87, –94







(h) –50, –40



2. (a) 9, 15



(b) 12, 8







(c) –33, –32







(d) 88, 85



3. (a) –67, –131



(b) 8, 13



(c) 144, 196



(d) –216, 343



Exercise 1B 1. (i) 15



(ii) 21







(iii) 105



2. (a) 6n + 1



(b) 3n − 7







(c) 7n + 53







(d) −3n + 17



3. (i) 18, 21



(ii) 3n







(iii) 315



4. (i) 4



(ii) 806



5. (a) (i) −1 (ii) 1 (iii) 55 1 (b) (i) 1 2



1 (ii) 14 2 1 (iii) 77 2 (c) (i) −2







(d) (i) 5



14. x = 80 6 (a) 7 44 (b) 52 49 (c) 686



(ii) 20 (iii) −25 1 (e) (i) − 8 1 (ii) −1 4 7 (iii) −6 8 1 6. (a) (i) 2 1 (ii) 128



15. (i) 7, 12; 8, 14; n + 2, 2n + 2







(iii) 8



(b) (i) −2



(ii) 1024 (iii) –21 846



(c) (i) 4



(ii) 1 310 720 (iii) 1 789 569 705 1 (d) (i) − 3 1 (ii) − 59 049 (iii) 0.250 3 (e) (i) 2 11 (ii) 34 64 (iii) 777 2 7. (a) 3 1 (b) 20 4 3 (c) 60 4 8. (a) 12 3 (b) 16 (c) 0.8



9. 2 3 and 6 3 10. (a) 3, 9, 19, 33



(b) (i) 2n2 − 1



(ii) 301 087 14 9 11. (a) T1= 265 , d = −6 31 31 (b) 44 1 1 12. d = (y − x), T3 = (3x − y) 4 2 13. (a) 8



(b) 80







(ii) 23; 48







(iii) 31; 33



16. (i) 5



(iii) 8







(iv) 89



Review Exercise 1 1. (a) 53, 44



(b) 28, 40 1 1 (c) , 27 81 (d) 121, 169



2. (i) 64, 81



(ii) (n + 2)2







(iii) 729



3. a = 2, b = 10; a = 10, b = 2 4. (a) (i) 12 (ii) 16 2 (b) (i) 3 (ii) 121.5 (iii) 121.22, 99.77 5 × 6 1 5. (ii) 15 = ; n(n + 1) 2 2 (iii) 3003



(iv) 11



6. (i) 13 + 23 + 33 + 43 + 53 + 63 + 73 = 784



= (1 + 2 + 3 + 4 +5 + 6 + 7)2







(ii) 14 400







(iii) 8



7. (ii) 5 × 5 + 1 = 26, 5n + 1



(iii) 281







(iv) It is not possible.



8. (i) 37



(ii) 2n + 2







(iii) 362



(ii) 2 (iii) 110



Answers



360



8. (a) 5x + 12; 4



1. 7 1 2. n(n – 1) 2 3. (i) 4, 9







(b) 4x + 21; 13



6. h = –16, k = 12; (x – 2)(x – 1)(x + 6)







(c) 3x – 3x + 7; –12



7. k = 2; (2x + 1), (x + 4)







(d) x + 4x –16; 48



8. p = 10, q = –90,







(ii) For n > 3, Tn = Tn – 1 + Tn







(iii) Lucas Numbers (which is



2



2



(e) 5x + x –7; 15 ⎛ 1 99 13 3⎞ 9. (2x – 1) ⎜− x 2 + x −   ⎟ − 8 4 8⎠ ⎝ 2 10. (a) 9x2 + 3x + 3, the remainder is 4.



4. (i) 11, 18 .



–2







(ii) For n > 4, Tn = Tn – 2 + Tn







(iii) Perrin Numbers (or Perrin



.



–3



(c) 3x2 + 9x + 27, the remainder is 74. (d) x + 2, the remainder is −2x + 5.



(e) 2x – 1, the remainder is 11x + 5. (f) 5x2 + 20x + 67, the remainder is



Sequence) Chapter 2



2



(b) 12x2 + 6x + 4, the remainder is −1.



different from Lucas Sequence)



5. (i) 10, 12



215x − 205. 11. a = −3, b = −3, c = 3



Exercise 2A 1. (a) x3 + 8x2 – 4x + 9



Exercise 2B







(b) 3x – 7x + 6x + 16



1. (a) 6







(c) x – x + 10x + 6











(d) 3x – 3x + 6x – x + 2







3



2



3



2



4



3



2



2. (a) 2x + 4x – 6 2







(b) −2x – 2x + 15x







(c) x – 4x – 5x + 20







(d) x5 – 7x4 + 5x3 – 3x2 + 14x – 7



3



3



2



2



3. (a) 14x2 + 13x – 12



(b) 30x2 + 37x – 7







(c) 4x3 – 27x + 5







(d) 2x3 + 7x2y + xy + 3xy2 + 3y2







(e) 9x3 + 9x2y – 7xy2 + 5y3



4. (a) 9x2 + 15x – 40



(b) 7x – 6x – 15







(c) 7x – 17x + 7x – 1







(d) –x3 – 9x2y + 11xy2 – 2y3







(e) 5x4 – 11x3 – 10x2 + 20x + 33



2 3



2



5. (a) A = 3, B = −2



(b) A = 7, B = −3







(c) A = 5, B = 6



6. A = 3, B = −1, C = 6 7. (a) A = 4, B = −5, C = 0



(b) A = 2, B = −3, C = 14







(c) A = 3, B = 2, C = 6 or A = 2,



B = 3, C = 6



(d) A = 2, B = −3, C = 5



(b) –20



(c) –38 3 (d) –8 4 (e) 51 1 (f) 5 4 1 2. (a) 2 (b) 4



(c) –5







(d) –8



3. k = –17 1 1 4. h = –1 , k = –4 2 2 5. k = 0, 4 or –4 6. p = 3, q = –8 1 1 7. p = –5 , q = –1 2 2 8. –7 9. p = –8, q = 8 10. p =



k −b h−a



Exercise 2C



Answers



(x – 3)(x + 1)(x + 3)(x + 9) 9. a = 7, b = 24, c = 9 10. p = –5, q = –6, (x – 1), (x – 2) Exercise 2D 1. (a) x = 2 or 3



(b) x = 3 or 8







(c) x = 3 or −







(b) (5x – 4y)(25x2 + 20xy + 16y2)







(c) (3x2 + 2y3)(9x4 – 6x2y3 + 4y6)







(d) (7a + 6x2)(49a2 – 42ax2 + 36x4)







(e) (10 – 3x)(9x2 – 6x + 28)



10 3 (d) x = 2, 3 or 4 1 or 2 (e) x = −3, 2 (f) x = −2, 3 or 5 1 1 2 or (g) x = − , 2 2 3 2. (a) (x – 2)(x2 + 2x + 4)



3. (a) 1, –1 or –2 (b) 2, 3 or –2 3 (c) 2, or –3 4 1 (d) , –2 or 7 3 1 (e) , 2 or 3 5 4. (a) 2, 7.41 or –0.41







(b) 3, 4.79 or 0.21



(c) 4, 5.24 or 0.76 1 (d) , 3.19 or –2.19 2 1 (e) – , 7.32 or 0.68 3 1 (f) , 2.78 or 0.72 2 (g) 1, 2, –2 or –3 1 (h) , 2, 3 or 4 2 5. f(x) = 3x4 – 27x3 + 24x2 + 156x – 48



1. (a) 15 1 (b) 4 2 (c) –16 7 (d) –5 8 (e) 1 or 2



6. f(x) = −3 + 16x + 31x2 – 24x3 + 4x4



(f) 11



11. (ii) y = 0.05x3 – 0.55x2 + 2x + 57.6







2. k = 3; 4 1 3. k = 1  or −15 2 1 4. p = –6, q = 4   2 361



5. p = –1, q = 2



Challenge Yourself − Chapter 1



7. 10 cm, 12 cm 8. n = 2, k = 81; (x2 + 9)(x + 3)(x – 3) 9. n = 2; (x − 1)(x + 1)(x – 2)(x + 2) 10. a = 1



(iii) 88.8 m3



Exercise 2E



7 2 x −1 5x + 3 (b) 5x + 3x 2 −1 46 (c) 7 − 2 x2 + 7 68 x + 33 2. (a) 15 + 2 x − 5x − 2 7x − 4 (b) 2x − 1 + (2 x − 3) (x +1) 1. (a) 4 +



3x − 3 x2 − 9 x − 78 3. 5x + 3 + 2 x − 2x + 5 4. a = 3, b = 1, c = −15, d = −18



(c) 3x − 1 +



5. a = 2, b = 4, c = −3, d = −4 6. a = 6, b = 7, c = 3, d = −17 7. a = 3, b = 4, c = 1, d = −1 8. a = 3, b = −1, c = −7, d = −14 Exercise 2F 3 4 1. (a) + x x+2 2 3 (b) + 3x x −1 2 5 (c) + x+3 x−2 7 6 (d) − x −1 3x +1 1 1 2. (a) + x −1 (x −1)2 1 1 − (b) x+4 (x + 4)2 2 3 4 (c) + + x x −1 (x −1)2 3 5 7 (d) + − x x+2 (x + 2)2 2 2 14 − (e) − x +1 x−2 (x − 2)2 5 1 1 − + (f) 2x + 3 3x −1 (3x −1)2 5 3 3. (a) + 2 x x +4 2 3x − 2 (b) + 2 x +1 x +3 27 9 −11x (c) + 8(x + 3) 8(x 2 + 7) 4x+5 2 (d) − 3x − 2 2 x2 + 9 2 3 4. (a) 2 + − x−2 x−3 2 5 (b) 2x + 1 + − x +1 x−4 23 14 (c) 4 + − + 9(x + 2) 9(x −1) 11 3(x −1)2



185 223x + 30 − 29(2 x − 3) 29(x 2 + 5) 5 19 5. a = 2, b = 5, c = , d = 2 2 x − 27 ; 6. 3x − 2 + 6 x 2 − x −15 3 4 3x − 2 + − 2x + 3 3x − 5 7. (x + 5)(x – 5)(2x + 3); 1 2 3 + − x+5 x−5 2x + 3 8. (2x − 1)(4x + 3)(x − 3); 2 6 7 + − x−3 4x+3 2 x −1 9. (i) h = −2, k = 12,



(d) 3x + 2 +



6. (a) 2, 2



1 , −3 2



2 1 , −2, − 3 2 1 (c) 1 , −2, 4 2 1 1 (d) 2, 4, − , − 4 2 (e) 2, 5.16, −1.16 1 (f) , 0.48, 14.52 2 1 (g) , 8.14, 0.86 4 1 (h) , 1.36, −2.69 2 7. a = 4; 6



(b)



f(x) = (x2 + 4)(3x – 2) 4 x + 24 6 2x (ii) = − 2 3x − 2 f(x) x +4 10 10 10. (i) − n n+2 10 10 (ii) 15 – − n +1 n + 2



8. p = 1, q = −4, (3x + 1), (x + 1)



Review Exercise 2



13. p = –8, q = –5



1. (a) A = 4, B = 7, C = −5



14. p = 16, q = – 4 2 24 15. (a) + 5(2 x −1) 5(x + 2) 13 1 3 (b) + + 3− x 2(2 x −1) 2(2 x −1)2 x+2 2 (c) 2 − 2 x +1 x +1 5 2 5 (d) + + 2x + 3 x −1 (x −1)2 2 1 (e) 1 +   + x−3 x+4 1 9 7 (f) − + 4x 8(x + 2) 8(x − 2) 1 2 3 (g) + + 2 x x  −1 (x −1)







(b) A = 7, B = −28, C = 6







(c) A = 2, B = −5, C = 10







(d) A = 4, B = −13, C = −8



2. (a) −5



(b) 61







(c) −9







(d) 10







(e) 3







(f) 1 or 5



3. a = 4, b = 3 4. (a) –2 (b) –21 1 (c) – 4 20 (d) 3 27 1 (e) 19 4 7 (f) 29 8 5. (a) (x + 1)(2x − 3)(x − 3)







(b) (x − 3)2(x + 4)







(c) (2x + 1)(x + 4)(3x − 4)







(d) (2x − 1)(2x + 5)(3x + 1)







(e) (2x2 − 7)(4x4 + 14x2 + 49)







(f) (3a + 4y2)(9a2 − 12ay2 + 16y4)







(g) 2(4 – x)(28x2 – 26x + 19)







(h) 2(2x + 3)(7x2 + 24x + 21)



9. p = 1, q = −4, r = 4 10. p = −4, q = −14, (x + 1)(2x − 3)(2x + 1)(2x − 1) 11. k = 1, A = 17; (x + 5)(x − 1)(x − 3) 12. h = 3, k = 5







(h)



1 3 3 + + 2 x x+2 (x + 2)



2 1 3 + + 2 x+3 x +1 (x +1) 1 3 5 + (j) + x x+2 x−2 x+2 2 (k) 2 − 2 x +1 x +1 5 x+3 (l) − 2 3x x +4 16. a = 1, b = −3, c = 2



(i)



17. a = 1, b = −1, r = −3 Challenge Yourself 3 1. −2, −1 or 2 1 2 2. + x−2 x+3



Answers



362



Chapter 3



Exercise 3A 1. (a) −27, −1, 0, 8



(c) (i) 3.5



(ii) 2.3 2. (a) 2.75



(c) (i) −0.5



(ii) 1.75 3. (a) 8, 4, 1.3, 0.8



(c) (i) 1.1



(ii) 2.65 4. (a) a = 1.1, b = 0.4



(c) (i) 1.3



(ii) 1.5 5. (i) 1.5



(ii) 2.4



6. (a) (i) 9.5 (ii) 15 (iii) 27



(b) (i) 0.8



(ii) 3.5 (iii) 4.15 7. (i) −1.8



(ii) 3.25



Exercise 3B



12. (i) 1.5 m/s2



1. (a) 0.5, 8, 16, 32











(c) (i) 12



(ii) −0.65 2. (a) 3, 4.2, 8.5



(c) (i) 4.9, 14.8



(ii) −0.25, 1.30 3. (i) −7.4



(ii) 1.1



4. (a) a = 2.5, b = 7.7



(c) (i) 2.6, 8.5



(ii) 1.7, 2.45 5. (a) 1.6



(b) (i) (−1.15, 0.3)



(ii) −1.15 6. 3 7. (a) 5, 8, 5



(c) 4







(d) (ii) h = 1, k = 9



8. (b) −14



(c) 10



9. (c) (1, 2) 10. (a) (ii) −0.85



8. (a) h = 2, k = 5.5



Exercise 3C



(c) (i) −0.3



1. (ii) 1153







(ii) 0.9



2. (i) 15 km



9. (i) 0.65











(ii) 2.45



10. (i) 5.5



(ii) 1, 2.75



11. (a) −1, −0.6, 1.6



(c) (i) −1.5, −0.35, 1.9



(ii) −1.5, −0.35, 1.9 12. (a) 1.7



(c) no solution



13. (a) −4.2



(c) (i) 1.5 or 4



(iii) 65 minutes



3. (i) 1 h



(ii) 30 km/h







(iii) 34.3 km/h



4. (i) 5 m/s2



(ii) 7.5 m/s



5. (ii) 7 m/s 6. (ii) 3.54 m/s



(iii) t = 25.5



7. (b) (i) 2.3 minutes (ii) 0.44 km/minute



(ii) 1.6 or 5.8



(iii) 2.75 minutes



(iii) 0.6 or 4.3



8. (b) (i) 1 h 11 minutes (ii) 1 h and 1 h 22 minutes 9. (b) (i) 1011 (ii) 8.3 km 11. (i) 9



363



Answers



(ii) 30 m/s



(ii) 100 s



13. (a) a = 4, b = 10



(c) (i) 1.7, 5.3



(ii) 3.5 s (iii) −3 (iv) 2.4 , t , 4.6 14. (ii) 9.5 m/s, 34 m/s



(iii) 2.5 m/s2, 5 m/s2



15. (a) h = 8, k = 10



(c) (i) 2.8 s



(ii) 10 km/h2 (iii) 1.65 , t , 4 (d) 0.4 2 16. 6 m/s 3 17. (a) 70 pesos







(b) Company B



18. (a) 30 minutes



(b) 20 km/h2







(d) 1648



20. (b) (i) 3 cm/s2, −1 cm/s2 (ii) 3.3 , t , 6.7



(c) a = −0.5, b = 5, c = 0



22. (b) 12 m/s Review Exercise 3 1. (a) −12, −8, −12, −8



(c) (i) −9.5



(ii) 3.1 2. (a) 3, 0, 15



(c) (i) −2.75



(ii) 2.4 (iii) −2, 0, 2 3. (a) (i) 3.85 (ii) −0.375 (b) (−0.7, 3.8) 4. (i) 0.625



(ii) (0.55, −0.2)



5. (i) y = 1.45, x = 1.75



(ii) 0.9 , x , 3.275







(iii) 0.67







(iv) 3.575



6. (ii) 27.7 km/h



7. (a) (i) 1 < t < 2 1 (ii) 13 km/h 3 (iii) 40 km/h



8. (i) 1 minute 35 seconds



Exercise 4B



9. (i) p = 6.5



(iii) 2.15 and 5.35



1. (a) 8.80 cm, 30.8 cm2, 22.8 cm



(iv) 17.1 km/h







(iv) –2.5



8. (i) 320 000







(v) the maximum value of







(ii) 16, 33, 40; 3200, 512







(iii) 350 units



9. (a) 5 °C/minute



(c) 2 °C/minute



Revision Exercise A1 1. (i) 9n – 3



(ii) 18



2. A = 7, B = 24, C = 9 3. a = 2, b = 3 4. 2, 0.14 or –2.47 3 5 5. (a) + x−3 x−4 6 5 (b) − x+3 (x + 3)2 6. (i) 16



(ii) 24



1 7. (i) 13 km/h 3 (ii) 1030; 20 km



(iii) 1 h







(iv) 10 km



8. (a) h = 3.5, k = 2.7



(c) (i) x = 2.3, y = 1.5.



(ii) the minimum value of y = 0.9.



(d) 0.5







(e) 1.4



Revision Exercise A2 1. (i) –3n + 47



(ii) 20



2. (i) 24



(ii) k = 9



3. A = 3, B = –5, C = –3 4. 6 5. (x – 1)(x + 3)(x – 3) 5 3x 6. (a) − 2 x−2 x +  3 1 7 (b) 2 + + x x−4 7. (b) (i) 1233, 77 km



y = 14.1 x = 3.75. CHAPTER 4 Exercise 4A



1. (a) 11.4 cm



(b) 32.7 cm







(c) 63.5 cm







(d) 53.7 cm



2. (a) (i) 11.9 cm (ii) 62.6 cm



(b) (i) 31.3 cm



(ii) 101 cm



(c) (i) 44.5 cm



(ii) 101 cm 3. (a) 16.0 cm



(b) 28.0 cm



4. (a) 49°



(b) 80°







(c) 263°







(d) 346°



5. 1.18 m 6. 191.0° 7. (a) 17.0 cm (b) 37.0 cm ⎛ 25π ⎞ 8. ⎜8 +   ⎟ cm 3 ⎠ ⎝



9. (i) 105°



(ii) 12.8 cm



10. (i) 61.8°



(ii) 30.5 cm



11. (i) 67.7 cm



(ii) 198 cm



12. (ii) 48.2 cm 13. (i) 27.3°



(ii) 54.5°







(iii) 32.4 cm



14. 44.4 cm 15. 43.0 cm







(b) 108.0°, 66 mm, 1150 mm2







(c) 28.0 mm, 132 mm, 188 mm







(d) 84.0 cm, 9240 cm2, 388 cm







(e) 225.1°, 385 m2, 83 m







(f) 15.3 cm, 20.1 cm, 50.8 cm



2. (a) (i) 17.7 cm



(ii) 12.8 cm2



(b) (i) 8.22 cm



(ii) 2.14 cm2



(c) (i) 26.7 cm



(ii) 44.0 cm2 3. (a) 14.7 cm, 103 cm2



(b) 24.2 cm, 169 cm2







(c) 30.8 cm, 216 cm2







(d) 52.8 cm, 370 cm2



4. (a) 385 cm2, 22.0 cm



(b) 898 cm2, 51.3 cm







(c) 1150 cm2, 66.0 cm







(d) 2120 cm2, 121 cm



5. (a) 9.33 cm



(b) 12.0 cm



6. (a) 60.3°



(b) 165.8°







(c) 303.5°







(d) 26.7°



7. (i) 43.6 cm, 118 cm2



(ii) 33.2 cm, 40.8 cm2







(iii) 263 cm, 1640 cm2



8. (i) 100°



(ii) 42.0 cm



9. 84 cm2 10. (ii) 32.5 cm2 11. 1.47p2 12. (i) 132°



(ii) 13.0 cm







(iii) 92.7 cm







(iv) 200 cm2



13. (i) 8.49 cm



(ii) 21.4 cm







(iii) 20.5 cm2



(ii) 60 km



Answers



364



Exercise 4C 1. (a) 150°



(b) 25.7°







(c) 183.3°







(d) 146.7°







4. (a) 5 cm, 10 cm2



2. (a) 0.653 rad



(b) 1.38 rad







(c) 2.48 rad







(c) 14.1







(d) 0.383







(e) 0.156







(f) 3.08







(c) 12 m, 57.6 m2







(d) 10 m, 12 m







(ii) 243 cm2



6. 32 cm 7. 117 m 8. (i) 22.5 cm2



(ii) (2π − 1.8) rad



(iii) (10π − 9) cm 1 2 9. (i) r θ = 8, 2r + rθ = 18 2 1 (ii) r = 8, θ = 4 10. (i) 15.6 cm







(ii) 100 cm2







(b) 1.40



11. 292 cm2 π 12. (i) 3 (ii) 12.6 cm







(c) 0.448











(d) 0.148



13. (i) 18 cm







(e) 0.694











(f) 0.898



14. (i) 6.4 cm



5. (a) 0.833







6. (i) 11.7 cm







9. (i) 0.159 rad (ii) 11.2 m







(iii) 3.34 m



(ii) 3.80 cm2 (ii) 774 cm2







(ii) 10.9 m







(iii) 85.6 m2



18. (i) 4.80



10. 13.8 m







Exercise 4D







(ii) 73.0 cm2



19. (i) 2.09 rad



1. (a) 9.6 cm



(ii) 22.1 cm2



20. 76.8 cm







(b) 3.5 cm







(c) 43.75 m







(d) 9 mm



21. (i) 1.32



(ii) 6.30 cm2



22. (i) 1.04



2. (a) 70.4 cm2



(b) 66.47 cm2







(c) 108.9 m







(d) 2650.8 mm2



2



Answers



(ii) 19.7 cm2



17. (i) 4 rad



(ii) 14.0 cm







(ii) 72.1 cm2



16. (i) 800 cm2



(ii) 14.4 m



8. (i) 9.36 cm



(iii) 43.5 cm2



15. (i) 6



(ii) 14.6 cm



7. (i) 6.79 m



365



(b) 1.5 rad, 27 cm2



(e) 2 rad, 16 mm 2 (f) 9 mm, rad 3 5. (i) 81.2 cm



(d) 5.38 rad π 3. (a) rad 12 π (b) rad 10 5π (c) rad 12 5π (d) rad 4 4. (a) 0.717 (b) 0.856



















1 rad 8 (ii) 144 cm2



3. (i) 1







(ii) 44.7 cm2



23. (i) 0.284 rad



(ii) 2.93 cm2



24. (i) 14.4 cm



(ii) 19.3 cm2







(iii) 13.6 cm



(iv) 35.8 cm2 3π 25. (i) 14 (ii) 164 cm2



Review Exercise 4 1. (i) 50.3 cm



(ii) 302 cm2



2. (i) 60.8 cm2



(ii) (2π − 1.9) rad







(iii) (16π − 15.2) cm



3. (i) 0.453



(ii) 3.80 cm2



4. (i) 1.55 rad



(ii) 27.6 cm2



5. 22% 6. (ii) 2.91 × 10−3 7. (i) 23.9 cm



(ii) 77.7 cm2







(iii) 98.6 cm2







(iv) 17.1 cm



8. (i) 4.73 cm, 12.5 cm



(ii) 54.1 cm2



9. 0.433r2 10. 50(π − 1) cm2 11. (a) (i) 17.6 cm (ii) 141 cm2 (iii) 107 cm2



(b) 80.0 cm2



Challenge Yourself 1. (a) (7π + 14) cm



(b) (7π + 14) cm







(c) (7π + 14) cm







(d) (7π + 14) cm



2. (i) 3



(ii) 0.644 rad







(iii) 6.99 cm2



3. (ii) 22.1 cm2 4. (i) 16



(ii) 175 cm2



CHAPTER 5 Exercise 5A



1. (a) a = 12, b = 67.4



(b) c = 11.0, d = 61.9







(c) e = 6, f = 50.2



2. 15 cm 3. 13 m 4. 13.7 cm 5. (a) a = 12, b = 90



(b) x = 11, y = 90



6. 18.0 cm



2



7. 17 cm 8. (i) 5.66 cm 9.



(ii) 14.2 cm 1 cm or 7 cm



10. 8.39 cm 11. 63.3 cm Exercise 5B 1. 24° 2. (i) 26° (ii) 122° x 3. 45° + 2 4. (a) a = 49, b = 14







(b) c = 58, d = 15







(c) e = 34, f = 14.8







(d) g = 35, h = 55







(e) i = 8, j = 67.4







(f) k = 12.6, l = 50.0



5. (i) 44°



(ii) 25°



6. (i) 7.5 cm



(ii) 67.4°







(iii) 34.4 cm2



7. 138° 8. 9 cm 9. 7 m 10. 64° 11. 51° 12. 45 cm



Exercise 5C



Review Exercise 5



1. (a) 80



1. (a) x = 50, y = 25







(b) 30







(b) x = 34, y = 114







(c) 40







(c) x = 28.5, y = 16.5







(d) 115







(d) x = 26, y = 38







(e) 125







(e) x = 26, y = 148







(f) 50







(f) x = 62, y = 118







(g) 35



2. (a) x = 70, y = 35







(h) 28







(b) x = 34, y = 56



2. (a) 50







(c) x = 132, y = 114







(b) 45







(d) x = 105, y = 30







(c) 30







(e) x = 6.43, y = 25







(d) 60







(f) x = 54, y = 72



3. (a) 50



3. (a) x = 62, y = 118



(b) x = 116, y = 46



4. 60°







(c) x = 115, y = 57.5



5. 65°







(d) x = 50



6. (a) 40







(e) x = 72, y = 28







(b) 36







(f) x = 48, y = 22







(c) 47



4. (a) x = 41







(d) 130







(b) x = 78, y = 30



7. (i) 70°







(c) x = 108, y = 144



(ii) 70°







(d) x = 24, y = 42



8. 270°







(e) x = 29, y = 59



9. 37°







(f) x = 42, y = 90



10. (i) 62°







(g) x = 103, y = 45



(ii) 47°







(h) x = 22.5, y = 135



11. (i) 64°



5. x° + y°















(b) 12



(ii) 64°



12. 70



6. 20° 7. (i) 24°



13. 78.5 cm







14. 31°



8. x = 74, y = 103



15. 32°



9. (i) 90°



16. 45°







17. 125°



10. (i) 1 cm 1 (ii) y° 2 11. (i) 4 cm



2



18. (i) 90°



(ii) 55°



(ii) 49°



(ii) 110°



19. 40°







20. (i) 35°



12. (i) 46°







(ii) 131°







(ii) 13.6 cm (ii) 134°



21. (b) (i) 8 cm



13. 61° or 119°



(ii) 9 cm



14. (i) 78°



22. (ii) PCB







23. 18°



15. (i) 62°



24. P = 65°, Q = 55°, R = 60°







1 25. (ii) 10 cm 6



(ii) 102° (ii) 126°



16. Yes



Answers



366



Challenge Yourself



4.



(a) 5



⎛ 5 5⎞ 5. (i) ⎜2 ,  2 ⎟ 6⎠ ⎝ 6



(b) 3 (c) 2



26 units 3 6. P(−11, −3), Q(7, 9), R(3, −11)







(d) 2 Chapter 6



1. (a) (0, 0), 7



1. (a) 8.06 units



(b) 8.54 units







(c) 11.4 units







(d) 10.8 units



⎛ ⎞ 3. (a) ⎜0,   − 8 1⎟ 3⎠ ⎝



⎛1 5⎞ (e) ⎜ ,   ⎟ , ⎝2 2⎠



10 2







⎛ 3 5⎞ 3 (f) ⎜ ,− ⎟ , ⎝4 4⎠ 2



⎛ 21⎞ ⎜0,   − 3 ⎟ 26 ⎠ ⎝



(b) x2 + y2 + 4x − 6y − 3 = 0







(c) 25x2 + 25y2 − 25x + 20y − 46 = 0







(d) x2 + y2 − 8x + 2y − 20 = 0







(e) x2 + y2 − 6x + 8y − 11 = 0



(ii) 9.6 units (ii) 5.83 units







(iii) (0, 4)







(iv) 11 or −5



3. x2 + y2 − 2x − 2y − 3 = 0 4. x2 + y2 = 13



5. (−4, 0), (3, 7)



7. (i) 3 units, 4.12 units, 4.47 units



(ii) 6 units2







(iii) 6 or −8



10. (i) x2 + y2 = 65



10. 2.57 units



(ii) 4 5 units



11. (i) x2 + y2 + 3x − y − 10 = 0



11. 3.48 units







(ii) 5 units



12. (i) x2 + y2 − 4x − 4y − 2 = 0



Exercise 6B







1. (a) (6, 4)



(b) (0, 2)







(c) (4, 2)



14. x + y + 6x − 2y − 15 = 0 2







⎛1 ⎞ 1 (e) ⎜ a + b,   a + b⎟ 2 ⎝2 ⎠



⎛1 ⎞ (f) ⎜ a h 2 +  k 2 , ah + 2ak⎟ ⎝2 ⎠ 2. (i) (4, –1)



(



(ii) (4, –11)



(ii) y + 2x = 6



13. x2 + y2 − 2x + 2y − 3 = 0



⎛1 1⎞ (d) ⎜ ,  2 ⎟ 2⎠ ⎝2



)



⎛ 1⎞ ⎜4,   − ⎟ , (5, –6) 2⎠ ⎝



Answers



7. x2 + y2 − 10x − 8y + 25 = 0 9. 3 2 units



9. (ii) 9 units2







6. y + 2x = 5 8. x2 + y2 − 4x + 4y − 12 = 0



8. −2 or 1



367















3.



(d) (1, 0), 3



2. (a) x2 + y2 = 9



6. (i) 4.5 units2











(b) (0, 0),



⎛ ⎞ (b) ⎜2 3 ,  0⎟ ⎝ 11 ⎠



5. (i) 32 units, 48 units2







10 2 (c) (−2, 3), 4







2. 7.07



4.



(ii)



Exercise 6C



Exercise 6A







⎛ 1 7⎞ ⎜− ,  2 ⎟ 6 12 ⎝ ⎠



2



15. a = 4, b = 1 16. (i) A(–3, 4), B(4, 3)



(ii) y = 7x







⎛ 2 ⎛ 2 7 2⎞ 7 2⎞ ,  − (iii) ⎜ ,   ⎟ ⎟ and ⎜− 2 2 ⎠ 2 2 ⎝ ⎠ ⎝



Review Exercise 6 3 1. (i) 4 (ii) 10 units 3 (iii) y = − x − 6 4 3 2. (i) y = − x + 6 4 ⎛ 6 6⎞ (ii) ⎜2 ,   3 ⎟ 7⎠ ⎝ 7 6 (iii) y = 3 7 6 (iv) x = 2 7 2 3. (i) 3 2 5 (ii) y = x + 3 3 (iii) 10 units2



(iv) 4.12 units



4. (i) y = 3x + 3



(ii) 10







(iii) −12







(iv) (2, 1)



5. (i) P(5, 0), Q(0, 12)



(ii) 13 units 12 (iii) y = − x−2 5 (iv) (−5, 0)



⎛ 3 1⎞ ⎜1 ,  6 ⎟ 2⎠ ⎝ 4 2 7. (a) x + y2 − 4x − 14y + 28 = 0 6.



(b) x2 + y2 + 4x − 18y + 40 = 0



8. (8, 6), 5 km 9. (a) (2, 5), 3



(b) (–2, 2), 5







(c) (0, 0),







4 3 3 (d) (2, 3), 10



10. x2 + y2 − 4y − 12 = 0, (0, 6),



(0, −2), (±2 3, 0)



11. (i) x2 + y2 − 4x − 45 = 0



(ii) 7, yes



12. x2 + y2 − 2x + 2y − 3 = 0 13. (i) x2 + y2 = 9



(ii) Yes



14. (i) x2 + y2 − 2y − 9 = 0



(ii) Yes, yes



15. (i) (2, 4)



(ii) y = x + 2







(iii) a = 1, b = 1



Challenge Yourself



8. (5, 6)



1. x2 + y2 − 10x + 14y + 49 = 0 or



9. x + y – 4x – 10y + 4 = 0







x + y − 10x + 2y + 1 = 0 2



2



Revision Exercise B1 1. (i) 1.9 rad



(ii) 60.8 cm







(iii) 8.57 cm



2 2



3. (i) 72°



y = 130° and z = 80°



6. (i) b (ii) b – a







(iii) 2b – a







(iv) 180° – 2b







(v) 180° + a – 3b



(b) 225



Exercise 7B B, C, A; B, A, C; C, A, B; C, B, A



7. 3y = 5x – 4 8. (i) 10 units2



(ii) D(0, 6)







(iii) (D(8, 6)



9. 4 5 units 10. (i) (4, 6) and (5, –1) Revision Exercise B2 1. 17.4 cm (to 3 s.f.) 2



2. (i) 100°



2. 1 814 400 ways 3. 24 4. 1560 5. 360 6. 49 7. 15 676 416 000 8. (a) 479 001 600



3. 157 cm (to 3 s.f.) 2



4. (i) 70°



5. 5x + 7y + 25 = 0 6. (i) (– 4, 0) and (0, 2) ⎛ 1 1⎞ (ii) ⎜−1 ,  1 ⎟ ⎝ 3 3⎠ 2 (iii) 2 units2 3 7. (i) (3, 4)







(ii) (0, –3)







(iii) 21 units







(iv) 3.61 units







(v) 3y = 2x – 9



2



(b) 46 080



1. (a) 1560 (b) 780







2. (a) 13 983 816



(b) 56







(c) 2002







(d) 700







(e) 8 145 060



3. (a) 362 880 (b) 126







4. 27 720 5. 27 720 6. 94 500 8. n = 20



9. (a) 1716 (b) 1400







10. 14 115 11. 90 12. 13 970 880 13. (a) 330



(b) 115



14. (a) 165



(b) 161



15. (a) 330



(b) 315







(c) 325



9. 86 400



Review Exercise 7



10. (a) 1440



1. 15







(b) 720



11. (a) 2880



(ii) 50°



(iii) 55°







1. 6 possible orders; A, B, C; A, C, B;











5. 24



8. (a) 15



5. u = 20°, v = 70°, w = 30°, x = 70°,



(ii) 69°



4. 24



7. 720



(ii) 88°







Exercise 7A



6. 720



(ii) 42°



4. (i) 128°







Chapter 7



3. 12



(ii) 59.7 cm







10. (i) x2 + y2 = 25



2. 18











Exercise 7C 2



1. 45



2



2. (i) 13.2 cm







2



(b) 4320



12. (a) 17 280



(b) 30 240







(c) 11 520







(d) 362 880



13. 3472 14. 24



(a) 6







(b) 18







(c) 4







(d) 2



15. 192, 78 16. 279



2. 5040 3. 720, 120, 120 4. 462 5. (a) 362 880



(b) 2880







(c) 17 280







(d) 14 400



6. 2 7. (a) 362 880



(b) 161 280







(c) 43 200







(d) 201 600







(d) 161 280



8. (a) 725 760



(b) 645 120



Answers



368



9. 144 10. 560 11. 735 12. 180 13. (a) 396



(b) 786



14. (a) 330



(b) 215







(c) 115



15. 1575



16. (a) 210



(b) 210







(c) 378







(d) 350







(e) 812



Challenge Yourself 1. 479 001 600 2. (a) 2880 ways



(b) 10 800 ways



3. 9 000 000 ways Chapter 8



Exercise 8A 5 1. (i) 11 4 (ii) 11 9 (iii) 11 4 (iv) 11 2 (v) 11 7 2. (i) 15 1 (ii) 3 44 (iii) 55 1 (iv) 5 3 3. (i) 17 2 (ii) 17 5 (iii) 17 10 (iv) 17 13 (v) 17 8 (vi) 17 5 4. (i) 6 1 (ii) 6 369



Answers



5. (b) (i) Mutually exclusive (ii) Not mutually exclusive (iii) Not mutually exclusive (iv) Mutually exclusive (v) Not mutually exclusive (vi) Not mutually exclusive 2 6. (i) 13 5 (ii) 13 5 (iii) 13 11 (iv) 13 5 7. (i) 14 11 (ii) 14 3 (iii) 14 7 8. (i) 15 17 (ii) 30 8 (iii) 15 13 (iv) 30 61 9. (i) 120 59 (ii) 120 13 (iii) 24 10. (a) 4



(b) Not mutually exclusive



Exercise 8B 5 4 5 4 5 1. (a) , ; , ,  , 9 9 9 9 9 5 (b) (i) 9 4 (ii) 9 20 (iii) 81 4 (iv) 9 3 2 3 2 3 2. (a) , ; , ,  , 5 5 5 5 5 9 (b) (i) 25 12 (ii) 25 2 (iii) 5 1 1 1 3 1 3. (a) , ; , , , 2 4 4 4 4 Sum: 40, 20, 30, 60 3 (b) (i) 4 1 (ii) 4



4 9



3 8 9 (ii) 16 15 (iii) 16 1 (iv) 16 7 4. (i) 12 7 (ii) 24 3 (iii) 8 3 5. 8 2 (a) 7 5 2 5 3 4 (b) ; , , , 8 7 7 7 7 15 (c) (i) 56 3 (ii) 28 3 (iii) 8 2 6. (i) 3 15 (ii) 22 5 (iii) 22 5 (iv) 11 3 7. (a) 5 1 (b) (i) 3 8 (ii) 15 2 (iii) 3 1 4 1 3 1 3 8. (a) , , , ,   , 5 5 4 4 4 4 (b) Yes



(c) (i)



1 1 1 5 1 , , ; , , 6 3 2 6 6 1 5 1 5 1 ; , , , , 6 6 6 6 6 1 5 1 5 1 , , , , 6 6 6 6 6 5 (b) (i) 108 1 (ii) 72 1 (iii) 36 5 (iv) 18 7 (v) 24



9. (a)



2 5



3 1 3 , , 4 4 4



5 1 5 , , , 6 6 6 5 1 5 , , , 6 6 6



5 9 25 (ii) 81 40 (iii) 81 91 11. (i) 855 182 (ii) 1539 5591 (iii) 15  390 7 12. (a) (i) 30 2 (ii) 5 11 (iii) 30 1 (b) 10 (c) No 3 13. (i) 10 1 (ii) 15 2 (iii) 15 1 (iv) 9 1 14. (a) 5 1 (b) (i) 10 2 (ii) 5 3 (iii) 10 3 (iv) 5 10. (i)



15. (b) (i) 0 17 33 5 (iii) 44 1 16. (i) 37 508 (ii) 1295 (ii)



17. (i) 0.001 08



(ii) 0.598







(iii) 0.402







(iv) 0.0454



3 65 21 (ii) 520 63 (iii) 260 18. (i)



25 216 125 (ii) 1296 671 (iii) 1296 5 (b) (i) 108 1 (ii) 36 Review Exercise 8 1 1. (i) 12 143 (ii) 144 1 (iii) 1728 6 2. (i) 7 1 (ii) 49 12 (iii) 49 216 (iv) 343 19 3. (i) 24 5 (ii) 24 1 (iii) 2 1 4. (i) 15 1 (ii) 6 1 (iii) 180 y 5. (i) x+y xy (ii) (x + y) (x + y −1) 6.



(iii) 1 4



2 xy



(x + y) (x + y −1)



40 87 38 (ii) 87 1 (b) (i) 3 8 (ii) 15 13 (iii) 15 8. (i) 0.36 7. (a) (i)







(ii) 0.24







(iii) 0.32







(iv) 0.24



31 45 11 (ii) 120 5 (iii) 36 1 10. (i) 72 31 (ii) 72 5 (iii) 36 9. (i)



19. (a) (i)







Challenge Yourself 1. Yes 2. (a) 1



(b) 0.891; Yes



(c) 22 Chapter 9



Exercise 9A 1. (ii) 15 kg



(iii) 36 kg







(iv) 9



2. (i) 1 minute



(ii) 13 minutes







(iii) 31



3. (ii) 15% 7 (iii) 20 4. (ii) 60% 5. (i) 5.4 minutes 6 (ii) 11 6. (a) 2 2 (b) (i) 41 % 3 2 (ii) 16 % 3 (c) Girls 7. (ii) 26



(iii) 35%



8. (ii) 7 minutes, 9.8 minutes (iii) 30% 3 (iv) 10 9. (a) 2007, 2008, 2010, 2011, 2012



10. (i) School Q



(ii) School P







(iii) School P



11. (ii) 25%, 60%



(iii) Class B



Answers



370



12. (ii) 3



2. (a) 7.5, 1, 24















1 (iii) 56 % 4 (iv) 8



13. Stem-and-leaf diagram Exercise 9B 1. (b) (i) 24 (ii) 15 (iii) 80 2. (a) 132, 160, 184, 203, 217, 230



(b) (i) 46



(ii) 52 (iii) 63 3. (i) 26



(ii) 24







(iii) 80%



4. (i) 33



(ii) 6







(iii) 450.6



5. (a) (i) Soil A: 130; Soil B: 90 (ii) Soil A: 9%; Soil B: 3% (iii) Soil A: 41; Soil B: 46



(b) (i) Soil A







(c) (i) 36%



(ii) 24% 6. (i) 57 13 (ii) 60 (iii) 52 7. (i) 13 1 (ii) 5 (iii) 23 8. (ii) 80 < x , 100 9. (a) (i) 55% (ii) 52 (iii) 24



(b) (i) 2, 6, 18, 44, 10



(ii) 55.9 g 10. (i) Examination B



(ii) Examination C



Exercise 9C 1. (a) 8, 4, 6, 8, 4



371







(b) 29, 58.5, 65, 71, 12.5







(c) 31, 12, 18, 29.5, 17.5







(d) 164, 102, 166, 207, 105







(e) 19.7, 5.8, 10.4, 14.1, 8.3



Answers



(b) 29, 23



12. (a) (i) 10, 13, 15.25 (ii) 5.25



3. (i) 45







(b) 73.3%







(ii) 86







(c) Median position and median







(iii) 47



4. (a) (i) 970, 880, 1050 (ii) 170



(b) (i) 85



(ii) 110 5. (i) 50



(ii) 57







(iii) 39







(iv) 14



6. (a) (i) 35.5 (ii) 6



(b) 34



7. (a) (i) 23.5 (ii) 26.5 (iii) 6.5 (iv) 16







waiting time.



Exercise 9D 2. (i) 19, 21, 25 (ii) 11







3. (i) 169 (ii) 23







4. (i) 0.05, 0.07, 0.086 5. (i) a = 168, b = 188.5, c = 195, d = 199, e = 213



(ii) 10.5; Interquartile range







(iii) 45; Range



6. (a) (i) Type A (ii) Type C



(b) Type B







(c) Type B



(b) 25



7. (a) (i) 34



(ii) 58



(iii) 8







8. (a) (i) 42



(b) (i) 26



(ii) 24



(c) 75







(d) School B



9. (i) 21, 28, 34



(ii) 38







(iii) 8.5







(iv) 2.63%



10. (a) (i) 50 (ii) 60 (iii) 29 (iv) 195



(v) 44







(b) School A: 14%; School B: 29%







(c) Agree



11. (a) (i) 84 (ii) 51.5 (iii) 57



(b) (i) 229



(ii) 146 (iii) 97



(c) City Y



(ii) 63 (b) (i) 41







(ii) 68 (iii) 14 (c) Agree







8. (a) (i) 64 (ii) 56 (iii) 26



(b) (i) 84



(ii) 42 (iii) 36



(c) Agree







(d) History examination



9. (a) (i) 15 (ii) 7 (iii) 6 (b) (i) 34







(ii) 24 (c) Agree







10. (a) (i) 40.5 (ii) 5



(b) (i) 50



(ii) 9.5



(d) Luxury



11. (a) (i) Set X: 30; Set Y: 21



14. x = 1, y = 4 or x = 4, y = 1



Challenge Yourself



(ii) Set X: 40; Set Y: 25



15. (i) A, C



(iii) Set X: 20; Set Y: 7







2. {−2, −1, 0, 1, 2} and



(ii) C







{− 5 , 0, 0, 0, 5 }







(b) Set X



16. (i) Yes







(c) Set X







(ii) No



Revision Exercise C1







(d) Set Y







(iii) 55.3, 8.58







(e) Curve B



1. (a) 1680







(f) Curve A → P; Curve B → R; Curve C → Q



12. X to B; Y to A; Z to C Exercise 9E 1. (a) 3.27



Review Exercise 9 1. (a) (i) 200 1 (ii) 5 (iii) 23



(b) 31.5



2. (a) 1.45







(b) 4.02











(c) 3.06



(ii) 2.5



2. (a) 11.1



(b) (i) 0.5, 1, 3



3. (i)



240







(b) 9.35







(ii) 100







(c) 11.9







(iii) 300.71, 188.51



3. 1.96



(ii) 37, 50



4. 1.68 5. 6.20 6. 250 7. (a) 11.7



(b) 7.23



8. (i) Class A: Mean = 8



Standard deviation = 2.60



Class B: Mean = 7.875



Standard deviation = 8.87



9. (i) 19



(ii) 5.69



10. (i) 15.4, 7.23



(ii) 12.4, 7.23



11. (i) Class A: Mean = 5.28



Standard deviation = 1.55



Class B: Mean = 4.98



4. (a) (i) 44



Standard deviation = 1.67







(ii) A







(iii) B



12. (a) (i) 25 (ii) 2 13. (a) (i) City A: 51.5 City B: 48.6 (ii) City A: 5.20 City B: 6.19



(b) City A







(c) City A







(b) 13







(c) (i) 15



(ii) 6 5. (a) (i) 48 (ii) 13



(b) (i) 57



(ii) 33



(c) 12.5%



6. (a) (i) Sam: 25; Carlo: 17.5 (ii) Sam: 14.9; Carlo: 12.5 7. (i) p = 990, q = 141, r = 24,







(b) (i) 420



(ii) 60 (iii) 120 2. 35, 30 5 3. (a) 39 35 (b) 156 (c) 0 47 (d) 78 4. (ii) 5 times 6 (iii) 2 % 7 5. (i) 46



(ii) 80







(iii) 27



6. (i) University A: 7600 University B: 5200



(ii) Agree







(iii) A



7. (a) 18, 42, 15, 4, 1



(b) (i) 212



(ii) 16.9 8. (a) 12, 30, 57, 96, 112, 120



(b) (i) 57



(ii) 24 (iii) 82



t = 177



9. (a) 8, 3, 6, 9, 6



8. (ii) 1 minute and 2 minutes 1 (iii) 73 % 3 9. (i) 4







(b) 47, 57.5, 71, 84.5, 27







(c) 127, 146, 170, 213.5, 67.5







(ii) 6







(iii)







(iii) 145 cm







(iv) 2 : 1



4 25 10. (ii) 165 cm



11. (i) 82 km/h



(ii) 5%







(iii) 84



12. (ii) 96, 80



(iii) Group A



Answers



372



Revision Exercise C2 1. 924, 714 2. (a) 2002



(b) 1876







(c) 1956



8 35 2 (ii) 35 46 (iii) 105 4. (ii) 70% 3. (a) (i)



6. (a) (i) 49 (ii) 44



(b) Q1 = 36, Q3 = 64,







Interquartile range = 28



(c) 68, 24



7. (a) (i) 54.5 (ii) 7.5 (iii) Grade 1: 17.5%;



Grade 2: 60%;







Grade 3: 22.5%







(b) (i) 50, 7



Problems in Real-World Contexts 1. (ii) 16 cm



(iii) 201 cm2



2. (a) T = −3.6t + 76.5



(b) (i) 5.69 minutes



(ii) −13.5 °C



(c) (i) 57







(ii) 21.6 °C



4. (a) (i) $596 234 545



(b) (i) $3 191 957 836



(ii) $3 789 405 063



(c) The Shipwreck



(d) 205 400 000(1.035)22 1 5. (a) 10 000 (b) 0.002 297







(c) No







(d)







373



1 100 000 000 1 (e) 10 000 (f) No



Answers