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Y E A R 1 1 C A M B R I D G E M a t h e m a t i c s B IL L P E N D E R D A V ID S A D LER JU LIA S H EA D ER EK W A R D E n h a n c e d 3 U n i t E x te n s io n 1 ISBN: 9781107633322 Photocopying is restricted under law and this material must not be transferred to another party © Bill Pender, David Sadler, Julia Shea, Derek Ward 2012 Cambridge University Press

cambridge university press Cambridge, New York, Melbourne, Madrid, Cape Town, Singapore, S˜ ao Paulo, Delhi, Mexico City Cambridge University Press 477 Williamstown Road, Port Melbourne, VIC 3207, Australia www.cambridge.edu.au Information on this title: www.cambridge.org/9781107633322 c⃝ Bill Pender, David Sadler, Julia Shea, Derek Ward 2012 This publication is in copyright. Sub ject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press. First published 1999 Reprinted 2001, 2002, 2008 Reprinted with Student CD 2009, 2010 (twice), 2011 Second edition, Enhanced version 2012 Cover design by Sylvia Witte, revisions by Kane Marevich Typ eset by Aptara Corp Printed in Singapore by C.O.S Printers Pte Ltd ACataloguing-in-Publicationentryisavailablefromthecatalogue of the National Library of Australia at www.nla.gov.au ISBN 978-1-107-63332-2 Paperback Additional resources for this publication at www.cambridge.edu.au/GO Reproduction and Communication for educational purposes The Australian Copyright Act 1968 (the Act) allows a maximum of one chapter or 10% of the pages of this publication, whichever is the greater, to be reproduced and/or communicated by any educational institution for its educational purposes provided that the educational institution (or the body that administers it) has given a remuneration notice to Copyright Agency Limited (CAL) under the Act. For details of the CAL licence for educational institutions contact: Copyright Agency Limited Level 15, 233 Castlereagh Street Sydney NSW 2000 Te l e p h o n e : ( 0 2 ) 9 3 9 4 7 6 0 0 Fa c s i m i l e : ( 0 2 ) 9 3 9 4 7 6 0 1 Email: info@copyright.com.au Reproduction and Communication for other purposes Except as permitted under the Act (for example a fair dealing for the purposes of study, research, criticism or review) no part of this publication may be reproduced, stored in a retrieval system, communicated or transmitted in any form or by any means without prior written permission. All inquiries should be made to the publisher at the address above. Cambridge University Press has no responsibility for the persistence or accuracy of URLS for external or third-party Internet websites referred to in this publication and does not guarantee that any content on such websites is, or will remain, accurate or appropriate. Information regarding prices, travel timetables and other factual information given in this work are correct at the time of ﬁrst printing but Cambridge University Press does not guarantee the accuracy of such information thereafter. ISBN: 9781107633322 Photocopying is restricted under law and this material must not be transferred to another party © Bill Pender, David Sadler, Julia Shea, Derek Ward 2012 Cambridge University Press

Contents Preface .......................................... vii How to Use This Book .................................. x About the Authors ....................................xiv Chapter One — Methods in Algebra .......................... 1 1A Terms, Factors and Indices . . . . . . . . . . . . . . . . . . . . . . . . . 1 1B Expanding Brackets . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 1C Factorisation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 1D Algebraic Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 1E Four Cubic Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 1F Linear Equations and Inequations . . . . . . . . . . . . . . . . . . . . . 13 1G Quadratic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 1H Simultaneous Equations . . . . . . . . . . . . . . . . . . . . . . . . . . 18 1I Completing the Square . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 1J The Language of Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 Chapter Two — Numbers and Functions ........................ 29 2A Cardinals, Integers and Rational Numbers . . . . . . . . . . . . . . . . 29 2B The Real Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 2C Surds and their Arithmetic . . . . . . . . . . . . . . . . . . . . . . . . 39 2D Rationalising the Denominator . . . . . . . . . . . . . . . . . . . . . . 42 2E Equality of Surdic Expressions ...................... 45 2F Relations and Functions . . . . . . . . . . . . . . . . . . . . . . . . . . 48 2G Review of Known Functions and Relations . . . . . . . . . . . . . . . . 52 2H Inverse Relations and Functions . . . . . . . . . . . . . . . . . . . . . . 59 2I Shifting and Reﬂecting Known Graphs . . . . . . . . . . . . . . . . . . 64 2J Further Transformations of Known Graphs . . . . . . . . . . . . . . . . 69 Chapter Three — Graphs and Inequations ....................... 73 3A Inequations and Inequalities . . . . . . . . . . . . . . . . . . . . . . . . 73 3B Intercepts and Sign . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78 3C Domain and Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . 82 3D The Absolute Value Function . . . . . . . . . . . . . . . . . . . . . . . 85 3E Using Graphs to Solve Equations and Inequations . . . . . . . . . . . . 91 3F Regions in the Number Plane . . . . . . . . . . . . . . . . . . . . . . . 96 3G Asymptotes and a Curve Sketching Menu . . . . . . . . . . . . . . . . 100 ISBN: 9781107633322 Photocopying is restricted under law and this material must not be transferred to another party © Bill Pender, David Sadler, Julia Shea, Derek Ward 2012 Cambridge University Press

! ! iv Contents Chapter Four — Tr i g o n o m e t r y .............................107 4A Trigonometry with Right Triangles . . . . . . . . . . . . . . . . . . . . 107 4B Theoretical Exercises on Right Triangles . . . . . . . . . . . . . . . . . 114 4C Trigonometric Functions of a General Angle . . . . . . . . . . . . . . . 117 4D The Quadrant, the Related Angle and the Sign . . . . . . . . . . . . . 121 4E Given One Trigonometric Function, Find Another . . . . . . . . . . . . 127 4F Trigonometric Identities and Elimination . . . . . . . . . . . . . . . . . 129 4G Trigonometric Equations . . . . . . . . . . . . . . . . . . . . . . . . . . 133 4H The Sine Rule and the Area Formula . . . . . . . . . . . . . . . . . . . 139 4I The Cosine Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146 4J Problems Involving General Triangles ...................150 Chapter Five — Coordinate Geometry .........................156 5A Points and Intervals . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156 5B Gradients of Intervals and Lines . . . . . . . . . . . . . . . . . . . . . . 162 5C Equations of Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167 5D Further Equations of Lines . . . . . . . . . . . . . . . . . . . . . . . . . 170 5E Perpendicular Distance . . . . . . . . . . . . . . . . . . . . . . . . . . . 176 5F Lines Through the Intersection of Two Given Lines ...........180 5G Coordinate Methods in Geometry . . . . . . . . . . . . . . . . . . . . . 184 Chapter Six — Sequences and Series .........................188 6A Indices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188 6B Logarithms . ................................192 6C Sequences and How to Specify Them . . . . . . . . . . . . . . . . . . . 196 6D Arithmetic Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200 6E Geometric Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203 6F Arithmetic and Geometric Means . . . . . . . . . . . . . . . . . . . . . 207 6G Sigma Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211 6H Partial Sums of a Sequence . . . . . . . . . . . . . . . . . . . . . . . . 213 6I Summing an Arithmetic Series . . . . . . . . . . . . . . . . . . . . . . 215 6J Summing a Geometric Series . . . . . . . . . . . . . . . . . . . . . . . 219 6K The Limiting Sum of a Geometric Series . . . . . . . . . . . . . . . . . 223 6L Recurring Decimals and Geometric Series . . . . . . . . . . . . . . . . 227 6M Factoring Sums and Di ﬀerences of Powers . . . . . . . . . . . . . . . . 229 6N Proof by Mathematical Induction . . ...................231 Chapter Seven — The Derivative ............................237 7A The Derivative — Geometric Deﬁnition . . . . . . . . . . . . . . . . . 237 7B The Derivative as a Limit . . . . . . . . . . . . . . . . . . . . . . . . . 241 7C A Rule for Di ﬀerentiating Powers of x ..................245 7D The Notation dy dx for the Derivative . . . . . . . . . . . . . . . . . . . . 250 7E The Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254 7F The Product Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 260 7G The Quotient Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262 7H Rates of Change . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 265 7I Limits and Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . 268 7J Di ﬀerentiability . . . ............................273 7K Extension — Implicit Di ﬀerentiation . . . . . . . . . . . . . . . . . . . 276 ISBN: 9781107633322 Photocopying is restricted under law and this material must not be transferred to another party © Bill Pender, David Sadler, Julia Shea, Derek Ward 2012 Cambridge University Press

! ! Contents v Chapter Eight — The Quadratic Function ........................280 8A Factorisation and the Graph . . . . . . . . . . . . . . . . . . . . . . . . 280 8B Completing the Square and the Graph . . . . . . . . . . . . . . . . . . 285 8C The Quadratic Formulae and the Graph . . . . . . . . . . . . . . . . . 289 8D Equations Reducible to Quadratics . . . . . . . . . . . . . . . . . . . . 292 8E Problems on Maximisation and Minimisation . . . . . . . . . . . . . . 294 8F The Theory of the Discriminant . . . . . . . . . . . . . . . . . . . . . . 299 8G Deﬁnite and Indeﬁnite Quadratics . . . . . . . . . . . . . . . . . . . . 304 8H Sum and Product of Roots . . . . . . . . . . . . . . . . . . . . . . . . . 307 8I Quadratic Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . 311 Chapter Nine — The Geometry of the Parabola ....................316 9A A Locus and its Equation . . . . . . . . . . . . . . . . . . . . . . . . . 316 9B The Geometric Deﬁnition of the Parabola . . . . . . . . . . . . . . . . 320 9C Translations of the Parabola . . . . . . . . . . . . . . . . . . . . . . . . 325 9D Parametric Equations of Curves . . . . . . . . . . . . . . . . . . . . . . 327 9E Chords of a Parabola . . . . . . . . . . . . . . . . . . . . . . . . . . . . 330 9F Tangents and Normals: Parametric Approach . . . . . . . . . . . . . . 333 9G Tangents and Normals: Cartesian Approach . . . . . . . . . . . . . . . 338 9H The Chord of Contact . . . . . . . . . . . . . . . . . . . . . . . . . . . 341 9I Geometrical Theorems about the Parabola . . . . . . . . . . . . . . . . 345 9J Locus Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 350 Chapter Ten — The Geometry of the Derivative ....................357 10A Increasing, Decreasing and Stationary at a Point . . . . . . . . . . . . 357 10B Stationary Points and Turning Points . . . . . . . . . . . . . . . . . . . 362 10C Critical Values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 367 10D Second and Higher Derivatives . . . . . . . . . . . . . . . . . . . . . . 371 10E Concavity and Points of Inﬂexion . . . . . . . . . . . . . . . . . . . . . 373 10F Curve Sketching using Calculus . . . . . . . . . . . . . . . . . . . . . . 378 10G Global Maximum and Minimum . . . . . . . . . . . . . . . . . . . . . . 380 10H Applications of Maximisation and Minimisation . . . . . . . . . . . . . 383 10I Maximisation and Minimisation in Geometry . . . . . . . . . . . . . . 388 10J Primitive Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 391 Chapter Eleven — Integration ..............................397 11A Finding Areas by a Limiting Process ...................397 11B The Fundamental Theorem of Calculus ..................402 11C The Deﬁnite Integral and its Properties . . . . . . . . . . . . . . . . . 407 11D The Indeﬁnite Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . 412 11E Finding Area by Integration . . . . . . . . . . . . . . . . . . . . . . . . 415 11F Area of a Compound Region . . . . . . . . . . . . . . . . . . . . . . . . 419 11G Volumes of Solids of Revolution . . . . . . . . . . . . . . . . . . . . . . 423 11H The Reverse Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . 429 11I The Trapezoidal Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . 432 11J Simpson’s Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 434 ISBN: 9781107633322 Photocopying is restricted under law and this material must not be transferred to another party © Bill Pender, David Sadler, Julia Shea, Derek Ward 2012 Cambridge University Press

! ! vi Contents Chapter Twelve — The Logarithmic Function .....................438 12A Review of Logarithmic and Exponential Functions ...........438 12B The Logarithmic Function and its Derivative . . . ...........441 12C Applications of Di ﬀerentiation . . . . . . . . . . . . . . . . . . . . . . . 450 12D Integration of the Reciprocal Function . . . . . . . . . . . . . . . . . . 454 12E Applications of Integration . . . . . . . . . . . . . . . . . . . . . . . . . 459 Chapter Thirteen — The Exponential Function .....................462 13A The Exponential Function and its Derivative . . . . . . . . . . . . . . 462 13B Applications of Di ﬀerentiation . . . . . . . . . . . . . . . . . . . . . . . 467 13C Integration of the Exponential Function . . . . . . . . . . . . . . . . . 472 13D Applications of Integration . . . . . . . . . . . . . . . . . . . . . . . . . 476 13E Natural Growth and Decay . . . . . . . . . . . . . . . . . . . . . . . . 479 Chapter Fourteen — The Trigonometric Functions ..................487 14A Radian Measure of Angle Size . . . . . . . . . . . . . . . . . . . . . . . 487 14B Mensuration of Arcs, Sectors and Segments . . . . . . . . . . . . . . . 491 14C Graphs of the Trigonometric Functions in Radians . . . . . . . . . . . 496 14D Trigonometric Functions of Compound Angles . . . . . . . . . . . . . . 504 14E The Angle Between Two Lines . . . . . . . . . . . . . . . . . . . . . . 509 14F The Behaviour of sin xNear the Origin . . . . . . . . . . . . . . . . . . 513 14G The Derivatives of the Trigonometric Functions . . . . . . . . . . . . . 517 14H Applications of Di ﬀerentiation . . . . . . . . . . . . . . . . . . . . . . . 523 14I Integration of the Trigonometric Functions . . . . . . . . . . . . . . . . 528 14J Applications of Integration . . . . . . . . . . . . . . . . . . . . . . . . . 534 Answers to Exercises ..................................538 Index ...........................................633 ISBN: 9781107633322 Photocopying is restricted under law and this material must not be transferred to another party © Bill Pender, David Sadler, Julia Shea, Derek Ward 2012 Cambridge University Press

Preface This textbook has been written for students in Years 11 and 12 taking the course previously known as ‘3 Unit Mathematics’, but renamed in the new HSC as two courses, ‘Mathematics’ (previously called ‘2 Unit Mathematics’) and ‘Mathemat- ics, Extension 1’. The book develops the content at the level required for the 2 and 3 Unit HSC examinations. There are two volumes — the present volume is roughly intended for Year 11, and the second for Year 12. Schools will, however, diﬀer in their choices of order of topics and in their rates of progress. Although these Syllabuses have not been rewritten for the new HSC, there has been a gradual shift of emphasis in recent examination papers. • The interdependence of the course content has been emphasised. • Graphs have been used much more freely in argument. • Structured problem solving has been expanded. • There has been more stress on explanation and proof. This text addresses these new emphases, and the exercises contain a wide variety of di ﬀerent types of questions. There is an abundance of questions in each exercise — too many for any one student — carefully grouped in three graded sets, so that with proper selection the book can be used at all levels of ability. In particular, those who subse- quently drop to 2 Units of Mathematics, and those who in Year 12 take 4 Units of Mathematics, will both ﬁnd an appropriate level of challenge. We have written a separate book, also in two volumes, for the 2 Unit ‘Mathematics’ course alone. We would like to thank our colleagues at Sydney Grammar School and Newington College for their invaluable help in advising us and commenting on the successive drafts, and for their patience in the face of some di ﬃculties in earlier drafts. We would also like to thank the Head Masters of Sydney Grammar School and Newington College for their encouragement of this pro ject, and Peter Cribb and the team at Cambridge University Press, Melbourne, for their support and help in discussions. Finally, our thanks go to our families for encouraging us, despite the distractions it has caused to family life. Preface to the enhanced version To provide students with practice for the new ob jective response (multiple choice) questions to be included in HSC examinations, online self-marking quizzes have been provided for each chapter, on Cambridge GO (access details can be found in the following pages). In addition, an interactive textbook version is available through the same website. Dr Bill Pender Sub ject Master in Mathematics Sydney Grammar School College Street Darlinghurst NSW 2010 David Sadler Mathematics Sydney Grammar School Julia Shea Head of Mathematics Newington College 200 Stanmore Road Stanmore NSW 2048 Derek Ward Mathematics Sydney Grammar School ISBN: 9781107633322 Photocopying is restricted under law and this material must not be transferred to another party © Bill Pender, David Sadler, Julia Shea, Derek Ward 2012 Cambridge University Press

This textbook is supported and enhanced by online resources... www.cambridge.edu.au/GO Digital resources and support material for schools. About the additional online resources... Additional resources are available free for users of this textbook online at Cambridge GO and include: • the PDF Textbook – a downloadable version of the student text, with note-taking and bookmarking enabled • extra material and activities • links to other resources. Use the unique 16 character access code found in the front of this textbook to activate these resources. About the Interactive Textbook... The Interactive Textbook is designed to make the online reading experience meaningful, from navigation to display. It also contains a range of extra features that enhance teaching and learning in a digital environment. Access the Interactive Textbook by purchasing a unique 16 character access code from your Educational Bookseller, or you may have already purchased the Interactive Textbook as a bundle with this printed textbook. The access code and instructions for use will be enclosed in a separate sealed pocket. The Interactive Textbook is available on a calendar year subscription. For a limited time only, access to this subscription has been included with the purchase of the enhanced version of the printed student text at no extra cost. You are not automatically entitled to receive any additional interactive content or updates that may be provided on Cambridge GO in the future. Preview online at: ISBN: 9781107633322 Photocopying is restricted under law and this material must not be transferred to another party © Bill Pender, David Sadler, Julia Shea, Derek Ward 2012 Cambridge University Press

Access online resources today at www.cambridge.edu.au/GO Go to the My Resources page on Cambridge GO and access all of your resources anywhere, anytime.* * Technical speciﬁ cations: You must be connected to the internet to activate your account and to use the Interactive Textbook. Some material, including the PDF Textbook, can be downloaded. To use the PDF Textbook you must have the latest version of Adobe Reader installed. 2. 3. 1. Log in to your existing Cambridge GO user account OR Create a new user account by visiting: www.cambridge.edu.au/GO/newuser • All of your Cambridge GO resources can be accessed through this account. • You can log in to your Cambridge GO account anywhere you can access the internet using the email address and password with which you are registered. Activate Cambridge GO resources by entering the unique access code found in the front of this textbook. Activate the Interactive Textbook by entering the unique 16 character access code found in the separate sealed pocket. • Once you have activated your unique code on Cambridge GO , you don’t need to input your code again. Just log in to your account using the email address and password you registered with and you will ﬁ nd all of your resources. Contact us on 03 8671 1400 or help@cambridgego.com.au ISBN: 9781107633322 Photocopying is restricted under law and this material must not be transferred to another party © Bill Pender, David Sadler, Julia Shea, Derek Ward 2012 Cambridge University Press

How to Use This Book This book has been written so that it is suitable for the full range of 3 Unit students, whatever their abilities and ambitions. The book covers the 2 Unit and 3 Unit content without distinction, because 3 Unit students need to study the 2 Unit content in more depth than is possible in a 2 Unit text. Nevertheless, students who subsequently move to the 2 Unit course should ﬁnd plenty of work here at a level appropriate for them. The Exercises: No-one should try to do all the questions! We have written long exercises so that everyone will ﬁnd enough questions of a suitable standard — each student will need to select from them, and there should be plenty left for revision. The book provides a great variety of questions, and representatives of all types should be selected. Each chapter is divided into a number of sections. Each of these sections has its own substantial exercise, subdivided into three groups of questions: Foundation: These questions are intended to drill the new content of the sec- tion at a reasonably straightforward level. There is little point in proceeding without mastery of this group. Development: This group is usually the longest. It contains more substantial questions, questions requiring proof or explanation, problems where the new content can be applied, and problems involving content from other sections and chapters to put the new ideas in a wider context. Later questions here can be very demanding, and Groups 1 and 2 should be su ﬃcient to meet the demands of all but exceptionally di ﬃcult problems in 3 Unit HSC papers. Extension: These questions are quite hard. Some are algebraically challeng- ing, some establish a general result beyond the theory of the course, some make di ﬃcult connections between topics or give an alternative approach, some deal with logical problems unsuitable for the text of a 3 Unit book. Students taking the 4 Unit course should attempt some of these. The Theory and the Worked Exercises: The theory has been developed with as much rigour as is appropriate at school, even for those taking the 4 Unit course. This leaves students and their teachers free to choose how thoroughly the theory is presented in a particular class. It can often be helpful to learn a method ﬁrst and then return to the details of the proof and explanation when the point of it all has become clear. The main formulae, methods, deﬁnitions and results have been boxed and num- bered consecutively through each chapter. They provide a summary only, and represent an absolute minimum of what should be known. The worked examples ISBN: 9781107633322 Photocopying is restricted under law and this material must not be transferred to another party © Bill Pender, David Sadler, Julia Shea, Derek Ward 2012 Cambridge University Press

! ! How to Use This Book xi have been chosen to illustrate the new methods introduced in the section, and should be su ﬃcient preparation for the questions of the following exercise. The Order of the Topics: We have presented the topics in the order we have found most satisfactory in our own teaching. There are, however, many e ﬀective or- derings of the topics, and the book allows all the ﬂexibility needed in the many diﬀerent situations that apply in di ﬀerent schools (apart from the few questions that provide links between topics). The time needed for the algebra in Chapter One will depend on students’ expe- riences in Years 9 and 10. The same applies to other topics in the early chapters —trigonometry,quadraticfunctions,coordinategeometryandparticularlycurve sketching. The Study Notes at the start of each chapter make further speciﬁc remarks about each topic. We h a v e l e f t E u c l i d e a n g e o m e t r y a n d p o l y n o m i a l s u n t i l Ye a r 1 2 f o r t w o r e a s o n s . First, we believe as much calculus as possible should be developed in Year 11, ideally including the logarithmic and exponential functions and the trigonometric functions. These are the fundamental ideas in the course, and it is best if Year 12 is used then to consolidate and extend them (and students subsequently taking the 4 Unit course particularly need this material early). Secondly, the Years 9 and 10 Advanced Course already develops much of the work on polynomials and Euclidean geometry in Options recommended for those proceeding to 3 Unit, so that revisiting them in Year 12 with the extensions and far greater sophistication required seems an ideal arrangement. The Structure of the Course: Recent examination papers have included longer ques- tions combining ideas from di ﬀerent topics, thus making clear the strong inter- connections amongst the various topics. Calculus is the backbone of the course, and the two processes of di ﬀerentiation and integration, inverses of each other, dominate most of the topics. We have introduced both processes using geomet- rical ideas, basing di ﬀerentiation on tangents and integration on areas, but the subsequent discussions, applications and exercises give many other ways of un- derstanding them. For example, questions about rates are prominent from an early stage. Besides linear functions, three groups of functions dominate the course: The Quadratic Functions: These functions are known from earlier years. They are algebraic representations of the parabola, and arise naturally in situations where areas are being considered or where a constant acceleration is being applied. They can be studied without calculus, but calculus provides an alternative and sometimes quicker approach. The Exponential and Logarithmic Functions: Calculus is essential for the study of these functions. We have chosen to introduce the logarithmic function ﬁrst, using deﬁnite integrals of the reciprocal function y=1 /x .This approach is more satisfying because it makes clear the relationship between these functions and the rectangular hyperbola y=1 /x , and because it gives aclearpictureofthenewnumber e. It is also more rigorous. Later, however, one can never overemphasise the fundamental property that the exponential function with base eis its own derivative — this is the reason why these func- tions are essential for the study of natural growth and decay, and therefore occur in almost every application of mathematics. ISBN: 9781107633322 Photocopying is restricted under law and this material must not be transferred to another party © Bill Pender, David Sadler, Julia Shea, Derek Ward 2012 Cambridge University Press

! ! xii How to Use This Book Arithmetic and geometric sequences arise naturally throughout the course. They are the values, respectively, of linear and exponential functions at in- tegers, and these interrelationships need to be developed, particularly in the context of applications to ﬁnance. The Trigonometric Functions: Again, calculus is essential for the study of these functions, whose deﬁnition, like the associated deﬁnition of π,is based on the circle. The graphs of the sine and cosine functions are waves, and they are essential for the study of all periodic phenomena — hence the detailed study of simple harmonic motion in Year 12. Thus the three basic functions of the course — x2,exand sin x—andtherelated numb ers eand π are developed from the three most basic degree 2 curves — the parabola, the rectangular hyperbola and the circle. In this way, everything in the course, whether in calculus, geometry, trigonometry, coordinate geometry or algebra, is easily related to everything else. The geometry of the circle is mostly studied using Euclidean methods, and the highly structured arguments used here contrast with the algebraic arguments used in the coordinate geometry approach to the parabola. In the 4 Unit course, the geometry of the rectangular hyperbola is given special consideration in the context of a coordinate geometry treatment of general conics. Polynomials are a generalisation of quadratics, and move the course a little be- yond the degree 2 phenomena describ ed ab ove. The particular case of the bi- nomial theorem then becomes the bridge from elementary probability using tree diagrams to the binomial distribution with all its practical applications. Unfor- tunately the power series that link polynomials with the exponential and trigono- metric functions are too sophisticated for a school course. Pro jective geometry and calculus with complex numbers are even further removed, so it is not really possible to explain that exponential and trigonometric functions are the same thing, although there are many clues. Algebra, Graphs and Language: One of the chief purposes of the course, stressed in recent examinations, is to encourage arguments that relate a curve to its equation. Being able to predict the behaviour of a curve given only its equation is a constant concern of the exercises. Conversely, the behaviour of a graph can often be used to solve an algebraic problem. We have drawn as many sketches in the book as space allowed, but as a matter of routine, students should draw diagrams for almost every problem they attempt. It is because sketches can so easily be drawn that this type of mathematics is so satisfactory for study at school. This course is intended to develop simultaneously algebraic agility, geometric intuition, and rigorous language and logic. Ideally then, any solution should ISBN: 9781107633322 Photocopying is restricted under law and this material must not be transferred to another party © Bill Pender, David Sadler, Julia Shea, Derek Ward 2012 Cambridge University Press

! ! How to Use This Book xiii display elegant and error-free algebra, diagrams to display the situation, and clarity of language and logic in argument. Theory and Applications: Elegance of argument and perfection of structure are fun- damental in mathematics. We have kept to these values as far as is reasonable in the development of the theory and in the exercises. The application of mathe- matics to the world around us is an equally fundamental, and we have given many examples of the usefulness of everything in the course. Calculus is particularly suitable for presenting this double view of mathematics. We w o u l d t h e r e f o r e u r g e t h e r e a d e r s o m e t i m e s t o p a y a t t e n t i o n t o t h e d e t a i l s o f argument in proofs and to the abstract structures and their interrelationships, and at other times to become involved in the interpretation provided by the applications. Limits, Continuity and the Real Numbers: This is a ﬁrst course in calculus, geometri- cally and intuitively developed. It is not a course in analysis, and any attempt to provide a rigorous treatment of limits, continuity or the real numbers would be quite inappropriate. We believe that the limits required in this course present little di ﬃculty to intuitive understanding — really little more is needed than limx→∞ 1/x = 0 and the occasional use of the sandwich principle in proofs. Char- acterising the tangent as the limit of the secant is a dramatic new idea, clearly marking the beginning of calculus, and quite accessible. Continuity and di ﬀer- entiability need only occasional attention, given the well-behaved functions that occur in the course. The real numbers are deﬁned geometrically as points on the number line, and provided that intuitive ideas about lines are accepted, ev- erything needed about them can be justiﬁed from this deﬁnition. In particular, the intermediate value theorem, which states that a continuous function can only change sign at a zero, is taken to be obvious. These unavoidable gaps concern only very subtle issues of ‘foundations’, and we are fortunate that everything else in the course can be developed rigorously so that students are given that characteristic mathematical experience of certainty and total understanding. This is the great contribution that mathematics brings to all our education. Te c h n o l o g y : There is much discussion, but little agreement yet, about what role tech- nology should play in the mathematics classroom and which calculators or soft- ware may be e ﬀective. This is a time for experimentation and diversity. We have therefore given only a few speciﬁc recommendations about technology, but we encourage such investigation, and to this version we have added some optional technology resources that can be accessed via the Cambridge GO website. The graphs of functions are at the centre of the course, and the more experience and intuitive understanding students have, the better able they are to interpret the mathematics correctly. A warning here is appropriate — any machine drawing of a curve should be accompanied by a clear understanding of why such a curve arises from the particular equation or situation. ISBN: 9781107633322 Photocopying is restricted under law and this material must not be transferred to another party © Bill Pender, David Sadler, Julia Shea, Derek Ward 2012 Cambridge University Press

About the Authors Dr Bill Pender is Sub ject Master in Mathematics at Sydney Grammar School, where he has taught since 1975. He has an MSc and PhD in Pure Mathematics from Sydney University and a BA (Hons) in Early English from Macquarie Uni- versity. In 1973–4, he studied at Bonn University in Germany and he has lectured and tutored at Sydney University and at the University of NSW, where he was aVisitingFellowin1989. HewasamemberoftheNSWSyllabusCommittee in Mathematics for two years and subsequently of the Review Committee for the Years 9–10 Advanced Syllabus. He is a regular presenter of inservice courses for AIS and MANSW, and plays piano and harpsichord. David Sadler is Second Master in Mathematics and Master in Charge of Statistics at Sydney Grammar School, where he has taught since 1980. He has a BSc from the University of NSW and an MA in Pure Mathematics and a DipEd from Sydney University. In 1979, he taught at Sydney Boys’ High School, and he was aVisitingFellowattheUniversityofNSWin1991. Julia Shea is Head of Mathematics at Newington College, with a BSc and DipEd from the University of Tasmania. She taught for six years at Rosny College, a State Senior College in Hobart, and then for ﬁve years at Sydney Grammar School. She was a member of the Executive Committee of the Mathematics Association of Tasmania for ﬁve years. Derek Ward has taught Mathematics at Sydney Grammar School since 1991, and is Master in Charge of Database Administration. He has an MSc in Applied Mathematics and a BScDipEd, both from the University of NSW, where he was subsequently Senior Tutor for three years. He has an AMusA in Flute, and sings in the Choir of Christ Church St Laurence. The mathematician’s patterns, like the painter’s or the poet’s, must be beautiful. The ideas, like the colours or the words, must ﬁt together in a harmonious way. Beauty is the ﬁrst test. — The English mathematician G. H. Hardy (1877–1947) ISBN: 9781107633322 Photocopying is restricted under law and this material must not be transferred to another party © Bill Pender, David Sadler, Julia Shea, Derek Ward 2012 Cambridge University Press

CHAPTER ONE Methods in Algebra Mathematics is the study of structure, pursued using a highly reﬁned form of language in which every word has an exact meaning, and in which the logic is expressed with complete precision. As the structures and the logic of their explanation become more complicated, the language describing them in turn becomes more specialised, and requires systematic study for the meaning to be understood. The symbols and methods of algebra are one aspect of that special language, and ﬂuency in algebra is essential for work in all the various topics of the course. Study Notes: Several topics in this chapter will probably be quite new — the four cubic identities of Section 1E, solving a set of three simultaneous equations in three variables in Section 1G, and the language of sets in Section 1J. The rest of the chapter is a concise review of algebraic work which would normally have been carefully studied in previous years, and needs will therefore vary as to the amount of work required on these exercises. 1A Te r m s , Fa c t o r s a n d I n d i c e s A pronumeral is a symbol that stands for a number. The pronumeral may stand for a known number, or for an unknown number, or it may be a va r i a b l e ,standing for any one of a whole set of possible numbers. Pronumerals, being numbers, can therefore be sub jected to all the operations that are possible with numbers, such as addition, subtraction, multiplication and division (except by zero). Like and Unlike Terms: An algebraic expression is an expression such as x2+2 x+3 x2− 4x− 3, in which pronumerals and numbers and operations are combined. The ﬁve terms in the above expression are x2,2 x,3 x2,−4xand −3. The two like terms x2and 3x2can be combined to give 4 x2, and the like terms 2 xand −4xcan be combined to give −2x. This results in three unlike terms 4x2,−2x and −3, which cannot be combined. WORKED EXERCISE : x2+2 x+3 x2− 4x− 3=4 x2− 2x− 3 Multiplying Terms: To s i m p l i f y a p r o d u c t l i k e 3 xy × (−6x2y)× 12y,itisbesttowork systematically through the signs, the numerals, and the pronumerals. WORKED EXERCISE : (a) 4ab ×7bc =28 ab 2c (b) 3xy ×(−6x2y)× 12y= −9x3y3 ISBN: 9781107633322 Photocopying is restricted under law and this material must not be transferred to another party © Bill Pender, David Sadler, Julia Shea, Derek Ward 2012 Cambridge University Press

! ! 2 CHAPTER 1: Methods in Algebra CAMBRIDGE MATHEMATICS 3U NIT YEAR 11 Index Laws: Here are the standard laws for dealing with indices (see Chapter Six for more detail). 1 INDEX LAWS : axay= ax+y (ab )x= axbx ax ay= ax−y !a b "x = ax bx (ax)n= axn WORKED EXERCISE : (a) 3x4× 4x3=12 x7 (b) (48 x7y3)÷ (16 x5y3)=3 x2 (c) (3a4)3=27 a12 (d) (−5x2)3× (2xy )4= −125 x6× 16 x4y4 = −2000 x10y4 (e) (6x4y)2 3(x2y3)3= 36 x8y2 3x6y9 = 12 x2 y7 Exercise 1A 1. Simplify: (a) 3x− 2y+5 x+6 y (b) 2a2+7 a− 5a2− 3a (c) 9x2− 7x+4 − 14 x2− 5x− 7 (d) 3a− 4b− 2c+4 a+2 b− c+2 a− b− 2c 2. Find the sum of: (a) x+ y+ z,2 x+3 y− 2zand 3 x− 4y+ z (b) 2a− 3b+ c,15 a− 21 b− 8cand 24 b+7 c+3 a (c) 5ab + bc − 3ca ,ab − bc + ca and −ab +2 ca + bc (d) x3− 3x2y+3 xy 2,−2x2y− xy 2− y3and x3+4 y3 3. Subtract: (a) xfrom 3 x (b) −xfrom 3 x (c) 2afrom −4a (d) −bfrom −5b 4. Fr o m : (a) 7x2− 5x+6 take 5 x2− 3x+2 (b) 4a− 8b+ ctake a− 3b+5 c (c) 3a+ b− c− dtake 6 a− b+ c− 3d (d) ab − bc − cd take −ab + bc − 3cd 5. Subtract: (a) x3− x2+ x+1 from x3+ x2− x+1 (b) 3xy 2− 3x2y+ x3− y3from x3+3 x2y+3 xy 2+ y3 (c) b3+ c3− 2abc from a3+ b3− 3abc (d) x4+5+ x− 3x3from 5 x4− 8x3− 2x2+7 6. Multiply: (a) 5aby 2 (b) 6xby −3 (c) −3aby a (d) −2a2by −3ab (e) 4x2by −2x3 (f ) −3p2qby 2 pq 3 7. Simplify: (a) 2a2b4× 3a3b2 (b) −6ab 5× 4a3b3 (c) (−3a3)2 (d) (−2a4b)3 ISBN: 9781107633322 Photocopying is restricted under law and this material must not be transferred to another party © Bill Pender, David Sadler, Julia Shea, Derek Ward 2012 Cambridge University Press

! ! CHAPTER 1: Methods in Algebra 1B Expanding Brackets 3 8. If a= −2, ﬁnd the value of: (a) 3a2− a+4 (b) a4+3 a3+2 a2− a 9. If x=2 and y= −3, ﬁnd the value of: (a) 8x2− y3 (b) x2− 3xy +2 y2 10. Simplify: (a) 5x x (b) −7x3 x (c) −12 a2b −ab (d) −27 x6y7z2 9x3y3z 11. Divide: (a) −2xby x (b) 3x3by x2 (c) x3y2by x2y (d) a6x3by −a2x3 (e) 14 a5b4by −2a4b (f ) −50 a2b5c8by −10 ab 3c2 DEVELOPMENT 12. Simplify: (a) 3a× 3a× 3a 3a+3 a+3 a (b) 3c× 4c2× 5c3 3c2+4 c2+5 c2 (c) ab 2× 2b2c3× 3c3a4 a3b3+2 a3b3+3 a3b3 13. Simplify: (a) (−2x2)3 −4x (b) (3xy 3)3 3x2y4 (c) (−ab )3× (−ab 2)2 −a5b3 (d) (−2a3b2)2× 16 a7b (2a2b)5 14. What must be added to 4 x3− 3x2+2 to give 3 x3+7 x− 6? 15. Ta k e t h e s u m o f 2 a− 3b− 4cand −4a+7 b− 5cfrom the sum of 4 c− 2band 5 b− 2a− 2c. 16. If X =2 b+3 c− 5dand Y =4 d− 7c− b,take X − Y from X + Y. 17. Divide the product of ( −3x7y5)4and ( −2xy 6)3by ( −6x3y8)2. EXTENSION 18. For what values of xis it true that: (a) x× x≤ x+ x? (b) x× x× x≤ x+ x+ x? 1B Expanding Brackets The laws of arithmetic tell us that a(x+ y)= ax + ay , whatever the values of a, x and y.Thisenablesexpressionswithbracketstobe expanded ,meaningthat they can be written in a form without brackets. WORKED EXERCISE : (a) 3x(x− 2xy )=3 x2− 6x2y (b) a2(a− b)− b2(b− a) = a3− a2b− b3+ ab 2 (c) (4x− 2)(4 x− 3) =4 x(4x− 3) − 2(4 x− 3) =16 x2− 12 x− 8x+6 =16 x2− 20 x+6 Special Quadratic Identities: These three identities are so important that they need to be memorised rather than worked out each time. 2 SQUARE OF A SUM :( A + B)2= A2+2 AB + B2 SQUARE OF A DIFFERENCE :( A − B)2= A2− 2AB + B2 DIFFERENCE OF SQUARES :( A + B)(A − B)= A2− B2 WORKED EXERCISE : (a) (4x+5 y)2=16 x2+40 xy +25 y2(square of a sum) (b) ! t− 1 t "2 = t2− 2+ 1 t2(square of a di ﬀerence) (c) (x2+3 y)(x2− 3y)= x4− 9y2(di ﬀerence of squares) ISBN: 9781107633322 Photocopying is restricted under law and this material must not be transferred to another party © Bill Pender, David Sadler, Julia Shea, Derek Ward 2012 Cambridge University Press

! ! 4 CHAPTER 1: Methods in Algebra CAMBRIDGE MATHEMATICS 3U NIT YEAR 11 Exercise 1B 1. Expand: (a) 4(a+2 b) (b) x(x− 7) (c) −3(x− 2y) (d) −a(a+4) (e) 5(a+3 b− 2c) (f ) −3(2 x− 3y+5 z) (g) −2x(x3− 2x2− 3x+1) (h) 3xy (2x2y− 5x3) (i) −2a2b(a2b3− 2a3b) 2. Expand and simplify: (a) 3(x− 2) − 2(x− 5) (b) −7(2 a− 3b+ c)− 6(−a+4 b− 2c) (c) x2(x3− 5x2+6 x− 1) − 2x(x4+10 x3− 2x2− 7x+3) (d) −2x3y(3x2y4− 4xy 5+5 y7)− 3xy 2(x2y6+2 x4y3− 2x3y4) 3. Expand and simplify: (a) (x+2)( x+3) (b) (2a+3)( a+5) (c) (x− 4)( x+2) (d) (2b− 7)( b− 3) (e) (3x+8)(4 x− 5) (f ) (6 − 7x)(5 − 6x) 4. (a) By expanding ( A+B)(A+B), prove the special expansion ( A+B)2= A2+2 AB +B2. (b) Similarly, prove the special expansions: (i) ( A − B)2= A2− 2AB + B2 (ii) ( A − B)(A + B)= A2− B2 5. Expand, using the special expansions: (a) (x− y)2 (b) (a+3) 2 (c) (n− 5)2 (d) (c− 2)( c+2) (e) (2a+1) 2 (f ) (3p− 2)2 (g) (3x+4 y)(3 x− 4y) (h) (4y− 5x)2 6. Multiply: (a) a− 2bby a+2 b (b) 2− 5xby 5 + 4 x (c) 4x+ 7 by itself (d) x2+3 yby x2− 4y (e) a+ b− cby a− b (f ) 9x2− 3x+1 by 3 x+1 7. Expand and simplify: (a) ! t+ 1 t "2 (b) ! t− 1 t "2 (c) ! t+ 1 t "! t− 1 t " DEVELOPMENT 8. (a) Subtract a(b+ c− a)fromthesumof b(c+ a− b)and c(a+ b− c). (b) Subtract the sum of 2 x2− 3(x− 1) and 2 x+3( x2− 2) from the sum of 5 x2− (x− 2) and x2− 2(x+1). 9. Simplify: (a) 14 − #10 − (3x− 7) − 8x$ (b) 4 % a− 2(b− c)− #a− (b− 2)$& 10. Use the special expansions to ﬁnd the value of: (a) 102 2 (b) 999 2 (c) 203 × 197 11. Expand and simplify: (a) (a− b)(a+ b)− a(a− 2b) (b) (x+2) 2− (x+1) 2 (c) (a− 3)2− (a− 3)( a+3) (d) (p+ q)2− (p− q)2 (e) (2x+3)( x− 1) − (x− 2)( x+1) (f ) 3(a− 4)( a− 2) − 2(a− 3)( a− 5) 12. If X = x− aand Y =2 x+ a,ﬁndtheproductof Y − X and X +3 Y in terms of xand a. 13. Expand and simplify: (a) (x− 2)3 (b) (x+ y+ z)2− 2(xy + yz + zx ) (c) (x+ y− z)(x− y+ z) (d) (a+ b+ c)(a2+ b2+ c2− ab − bc − ca ) ISBN: 9781107633322 Photocopying is restricted under law and this material must not be transferred to another party © Bill Pender, David Sadler, Julia Shea, Derek Ward 2012 Cambridge University Press

! ! CHAPTER 1: Methods in Algebra 1C Factorisation 5 14. Prove the identities: (a) (a+ b+ c)(ab + bc + ca )− abc =( a+ b)(b+ c)(c+ a) (b) (ax + by )2+( ay − bx )2+ c2(x2+ y2)=( x2+ y2)(a2+ b2+ c2) EXTENSION 15. If 2 x= a+ b+ c, show that ( x− a)2+( x− b)2+( x− c)2+ x2= a2+ b2+ c2. 16. If ( a+ b)2+( b+ c)2+( c+ d)2=4( ab + bc + cd ), prove that a= b= c= d. 1C Factorisation Factorisation is the reverse process of expanding brackets, and will be needed on a routine basis throughout the course. The various methods of factorisation are listed systematically, but in every situation common factors should always be taken out ﬁrst. 3 METHODS OF FACTORISATION : HIGHEST COMMON FACTOR :Alwaystrythisﬁrst. DIFFERENCE OF SQUARES :Thisinvolvestwoterms. QUADRATICS :Thisinvolvesthreeterms. GROUPING :Thisinvolvesfourormoreterms. Fa c t o r i n g s h o u l d c o n t i n u e u n t i l e a c h f a c t o r i s irreducible , meaning that it cannot be factored further. Factoring by Highest Common Factor and Difference of Squares: In every situation, look for any common factors of all the terms, and then take out the highest common factor. WORKED EXERCISE : Fa c t o r : (a) 18 a2b4− 30 b3 (b) 80 x4− 5y4 SOLUTION : (a) The highest common factor of 18 a2b4and 30 b3is 6 b3, so 18 a2b4− 30 b3=6 b3(3a2b− 5). (b) 80 x4− 5y4=5(16 x4− y4)(highestcommonfactor) =5(4 x2− y2)(4 x2+ y2) (di ﬀerence of squares) =5(2 x− y)(2 x+ y)(4 x2+ y2) (di ﬀerence of squares again) Factoring Monic Quadratics: Aquadraticiscalled monic if the coe ﬃcient of x2is 1. Suppose that we want to factor a monic quadratic expression like x2− 13 x+36. We l o o k f o r t w o n u m b e r s w h o s e s u m i s −13 (the coe ﬃcient of x)andwhose product is 36 (the constant). WORKED EXERCISE : Fa c t o r : (a) x2− 13 x+36 (b) a2+12 ac − 28 c2 SOLUTION : (a) The numbers with sum −13 and product 36 are −9and −4, so x2− 13 x+36 =( x− 9)( x− 4). (b) The numbers with sum 12 and product −28 are 14 and −2, so a2+12 ac − 28 c2 =( a+14 c)(a− 2c). ISBN: 9781107633322 Photocopying is restricted under law and this material must not be transferred to another party © Bill Pender, David Sadler, Julia Shea, Derek Ward 2012 Cambridge University Press

! ! 6 CHAPTER 1: Methods in Algebra CAMBRIDGE MATHEMATICS 3U NIT YEAR 11 Factoring Non-monic Quadratics: In a non-monic quadratic like 2 x2+11 x+ 12, where the coe ﬃcient of x2is not 1, we look for two numbers whose sum is 11 (the coe ﬃcient of x), and whose product is 24 (the product of the constant term and the coe ﬃcient of x2). WORKED EXERCISE : Fa c t o r : (a) 2x2+11 x+12 (b) 6s2− 11 st − 10 t2 SOLUTION : (a) The numbers with sum 11 and product 24 are 8 and 3, so 2 x2+11 x+12 =(2 x2+8 x)+(3 x+12) =2 x(x+4)+3( x+4) =(2 x+3)( x+4) . (b) The numbers with sum −11 and product −60 are −15 and 4, so 6 s2− 11 st − 10 t2 =(6 s2− 15 st)+(4 st − 10 t2) =3 s(2s− 5t)+2 t(2s− 5t) =(3 s+2 t)(2 s− 5t). Factoring by Grouping: When there are four or more terms, it is sometimes possible to split the expression into groups, factor each group in turn, and then factor the whole expression by taking out a common factor or by some other method. WORKED EXERCISE : Fa c t o r : (a) 12 xy − 9x− 16 y+12 (b) s2− t2+ s− t SOLUTION : (a) 12 xy − 9x− 16 y+12 = 3 x(4y− 3) − 4(4 y− 3) =(3 x− 4)(4 y− 3) (b) s2− t2+ s− t=( s+ t)(s− t)+( s− t) =( s− t)(s+ t+1) Exercise 1C 1. Write as a product of two factors: (a) ax − ay (b) x2+3 x (c) 3a2− 6ab (d) 12 x2+18 x (e) 6a3+2 a4+4 a5 (f ) 7x3y− 14 x2y2+21 xy 2 2. Factor by grouping in pairs: (a) ax − ay + bx − by (b) a2+ ab + ac + bc (c) x2− 3x− xy +3 y (d) 2ax − bx − 2ay + by (e) ab + ac − b− c (f ) 2x3− 6x2− ax +3 a 3. Fa c t o r e a c h d i ﬀerence of squares: (a) x2− 9 (b) 1− a2 (c) 4x2− y2 (d) 25 x2− 16 (e) 1− 49 k2 (f ) 81 a2b2− 64 4. Factor each of these quadratic expressions: (a) x2+8 x+15 (b) x2− 4x+3 (c) a2+2 a− 8 (d) y2− 3y− 28 (e) c2− 12 c+27 (f ) p2+9 p− 36 (g) u2− 16 u− 80 (h) x2− 20 x+51 (i) t2+23 t− 50 (j) x2− 9x− 90 (k) x2− 5xy +6 y2 (l) x2+6 xy +8 y2 (m) a2− ab − 6b2 (n) p2+3 pq − 40 q2 (o) c2− 24 cd +143 d2 ISBN: 9781107633322 Photocopying is restricted under law and this material must not be transferred to another party © Bill Pender, David Sadler, Julia Shea, Derek Ward 2012 Cambridge University Press

! ! CHAPTER 1: Methods in Algebra 1D Algebraic Fractions 7 5. Wr i t e e a c h q u a d r a t i c e x p r e s s i o n a s a p r o d u c t o f t w o f a c t o r s : (a) 2x2+5 x+2 (b) 3x2+8 x+4 (c) 6x2− 11 x+3 (d) 3x2+14 x− 5 (e) 9x2− 6x− 8 (f ) 6x2− 7x− 3 (g) 6x2− 5x+1 (h) 3x2+13 x− 30 (i) 12 x2− 7x− 12 (j) 12 x2+31 x− 15 (k) 24 x2− 50 x+25 (l) 2x2+ xy − y2 (m) 4a2− 8ab +3 b2 (n) 6p2+5 pq − 4q2 (o) 18 u2− 19 uv − 12 v2 6. Write each expression as a product of three factors: (a) 3a2− 12 (b) x4− y4 (c) x3− x (d) 5x2− 5x− 30 (e) 25 y− y3 (f ) 16 − a4 (g) 4x2+14 x− 30 (h) x3− 8x2+7 x (i) x4− 3x2− 4 (j) ax 2− a− 2x2+2 (k) 16 m3− mn 2 (l) ax 2− a2x− 20 a3 DEVELOPMENT 7. Factor as fully as possible: (a) 72 + x− x2 (b) (a− b)2− c2 (c) a3− 10 a2b+24 ab 2 (d) a2− b2− a+ b (e) x4− 256 (f ) 4p2− (q+ r)2 (g) 6x4− x3− 2x2 (h) a2− bc − b+ a2c (i) 9x2+36 x− 45 (j) 4x4− 37 x2+9 (k) x2y2− 13 xy − 48 (l) x(x− y)2− xz 2 (m) 20 − 9x− 20 x2 (n) 4x3− 12 x2− x+3 (o) 12 x2− 8xy − 15 y2 (p) x2+2 ax + a2− b2 (q) 9x2− 18 x− 315 (r) x4− x2− 2x− 1 (s) 10 x3− 13 x2y− 9xy 2 (t) x2+4 xy +4 y2− a2+2 ab − b2 (u) (x+ y)2− (x− y)2 EXTENSION 8. Fa c t o r f u l l y : (a) a2+ b(b+1) a+ b3 (b) a(b+ c− d)− c(a− b+ d) (c) (a2− b2)2− (a− b)4 (d) 4x4− 2x3y− 3xy 3− 9y4 (e) (x2+ xy )2− (xy + y2)2 (f ) (a2− b2− c2)2− 4b2c2 (g) (ax + by )2+( ay − bx )2+ c2(x2+ y2) (h) x2+( a− b)xy − aby 2 (i) a4+ a2b2+ b4 (j) a4+4 b4 1D Algebraic Fractions An algebraic fraction is a fraction containing pronumerals. They are manipulated in the same way as arithmetic fractions, and factorisation plays a ma jor role. Addition and Subtraction of Algebraic Fractions: Acommondenominatorisrequired, but ﬁnding the lowest common denominator can involve factoring all the denom- inators. 4 ADDITION AND SUBTRACTION OF ALGEBRAIC FRACTIONS : First factor all denominators. Then work with the lowest common denominator. ISBN: 9781107633322 Photocopying is restricted under law and this material must not be transferred to another party © Bill Pender, David Sadler, Julia Shea, Derek Ward 2012 Cambridge University Press

! ! 8 CHAPTER 1: Methods in Algebra CAMBRIDGE MATHEMATICS 3U NIT YEAR 11 WORKED EXERCISE : (a) 1 x− 4− 1 x = x− (x− 4) x(x− 4) = 4 x(x− 4) (b) 2 x2− x− 5 x2− 1= 2 x(x− 1) − 5 (x− 1)( x+1) = 2(x+1) − 5x x(x− 1)( x+1) = 2− 3x x(x− 1)( x+1) Multiplication and Division of Algebraic Fractions: The key step here is to factor all numerators and denominators completely before cancelling factors. 5 MULTIPLICATION AND DIVISION OF ALGEBRAIC FRACTIONS : First factor all numerators and denominators completely. Then cancel common factors. To d i v i d e b y a n a l g e b r a i c f r a c t i o n , m u l t i p l y b y i t s r e c i p r o c a l i n t h e u s u a l w a y. WORKED EXERCISE : (a) 2a 9− a2× a− 3 a3+ a= 2a (3 − a)(3 + a)× a− 3 a(a2+1) = − 2 (a+3)( a2+1) (b) 6abc ab + bc ÷ 6ac a2+2 ac + c2= 6abc b(a+ c)× (a+ c)2 6ac = a+ c Simplifying Compound Fractions: A compound fraction is a fraction in which either the numerator or the denominator is itself a fraction. 6 SIMPLIFYING COMPOUND FRACTIONS : Multiply top and bottom by something that will clear fractions from numerator and denominator together. WORKED EXERCISE : (a) 12− 13 14+ 16 = 12− 13 14+ 16 × 12 12 = 6− 4 3+2 = 25 (b) 1 t+ 1 t+1 1 t− 1 t+1 = 1 t+ 1 t+1 1 t− 1 t+1 × t(t+1) t(t+1) = (t+1)+ t (t+1) − t =2 t+1 Exercise 1D 1. Simplify: (a) x 2x (b) a a2 (c) 3x2 9xy (d) 12 ab 4a2b (e) 12 xy 2z 15 x2yz 2 (f ) uvw 2 u3v2w ISBN: 9781107633322 Photocopying is restricted under law and this material must not be transferred to another party © Bill Pender, David Sadler, Julia Shea, Derek Ward 2012 Cambridge University Press

! ! CHAPTER 1: Methods in Algebra 1D Algebraic Fractions 9 2. Simplify: (a) x 3× 3 x (b) a 4÷ a 2 (c) x× 3 x2 (d) a2 2b× b2 a2 (e) 3x2 4y2× 2y x (f ) x2 3ay 3÷ x2 3ay 3 (g) 5 a÷ 10 (h) 2ab 3c × c2 ab 2 (i) 8a3b 5 ÷ 4ab 15 (j) 2a 3b× 5c2 2a2b× 3b2 2c (k) 12 x2yz 8xy 3 × 24 xy 2 36 yz 2 (l) 3a2b 4b3c× 2c2 8a3÷ 6ac 16 b2 3. Wr i t e a s a s i n g l e f r a c t i o n : (a) x 2+ x 5 (b) a 3− a 6 (c) x 8− y 12 (d) 2a 3 + 3a 2 (e) 7b 10 − 19 b 30 (f ) xy 30 − xy 18 (g) 1 x+ 1 2x (h) 3 4x+ 4 3x (i) 1 a− 1 b (j) x+ 1 x (k) a+ b a (l) 1 2x− 1 x2 4. Simplify: (a) x+1 2 + x+2 3 (b) 2x− 1 5 − x+3 2 (c) 2x+1 3 − x− 5 6 + x+4 4 (d) 3x− 7 5 + 4x+3 2 − 2x− 5 10 (e) x− 5 3x − x− 3 5x (f ) 1 x− 1 x+1 (g) 1 x+1 − 1 x+1 (h) 2 x− 3+ 3 x− 2 (i) 2 x+3 − 2 x− 2 (j) x x+ y+ y x− y (k) a x+ a− b x+ b (l) x x− 1− x x+1 5. Factor where possible and then simplify: (a) a ax + ay (b) 3a2− 6ab 2a2b− 4ab 2 (c) x2+2 x x2− 4 (d) a2− 9 a2+ a− 12 (e) x2+2 xy + y2 x2− y2 (f ) x2+10 x+25 x2+9 x+20 (g) ac + ad + bc + bd a2+ ab (h) y2− 8y+15 2y2− 5y− 3 (i) 9ax +6 bx − 6ay − 4by 9x2− 4y2 6. Simplify: (a) 3x+3 2x × x2 x2− 1 (b) a2+ a− 2 a+2 × a2− 3a a2− 4a+3 (c) c2+5 c+6 c2− 16 ÷ c+3 c− 4 (d) x2− x− 20 x2− 25 × x2− x− 2 x2+2 x− 8÷ x+1 x2+5 x (e) ax + bx − 2a− 2b 3x2− 5x− 2 × 9x2− 1 a2+2 ab + b2 (f ) 2x2+ x− 15 x2+3 x− 28 ÷ x2+6 x+9 x2− 4x ÷ 6x2− 15 x x2− 49 7. Simplify: (a) 1 x2+ x+ 1 x2− x (b) 1 x2− 4+ 1 x2− 4x+4 (c) 1 x− y+ 2x− y x2− y2 (d) 3 x2+2 x− 8− 2 x2+ x− 6 (e) x a2− b2− x a2+ ab (f ) 1 x2− 4x+3 + 1 x2− 5x+6 − 1 x2− 3x+2 ISBN: 9781107633322 Photocopying is restricted under law and this material must not be transferred to another party © Bill Pender, David Sadler, Julia Shea, Derek Ward 2012 Cambridge University Press

! ! 10 CHAPTER 1: Methods in Algebra CAMBRIDGE MATHEMATICS 3U NIT YEAR 11 8. Simplify: (a) b− a a− b (b) v2− u2 u− v (c) x2− 5x+6 2− x (d) 1 a− b− 1 b− a (e) m m − n + n n− m (f ) x− y y2+ xy − 2x2 DEVELOPMENT 9. Study the worked exercise on compound fractions and then simplify: (a) 1− 12 1+ 12 (b) 2+ 13 5− 23 (c) 12− 15 1+ 110 (d) 1720 − 34 45− 310 (e) 1x 1+ 2x (f ) t− 1t t+ 1t (g) 1 1b+ 1a (h) xy+ yx xy− yx (i) 1− 1x+1 1x+ 1x+1 (j) 3x+2 − 2x+1 5x+2 − 4x+1 10. If x= 1 λ and y= 1 1− x and z= y y− 1, show that z= λ. 11. Simplify: (a) x4− y4 x2− 2xy + y2÷ x2+ y2 x− y (b) 8x2+14 x+3 8x2− 10 x+3 × 12 x2− 6x 4x2+5 x+1 ÷ 18 x2− 6x 4x2+ x− 3 (c) (a− b)2− c2 ab − b2− bc × c a2+ ab − ac ÷ ac − bc + c2 a2− (b− c)2 (d) x− y x + x3+ y3 xy 2 − x2+ y2 x2 (e) x+4 x− 4− x− 4 x+4 (f ) 4y x2+2 xy − 3x xy +2 y2+ 3x− 2y xy (g) 8x x2+5 x+6 − 5x x2+3 x+2 − 3x x2+4 x+3 (h) 1 x− 1+ 2 x+1 − 3x− 2 x2− 1− 1 x2+2 x+1 12. (a) Expand ! x+ 1 x "2 . (b) Suppose that x+ 1 x = 3. Use part (a) to evaluate x2+ 1 x2without attempting to ﬁnd the value of x. EXTENSION 13. Simplify these algebraic fractions: (a) 1 (a− b)(a− c)+ 1 (b− c)(b− a)+ 1 (c− a)(c− b) (b) ! 1+ 45 x− 8− 26 x− 6 "! 3− 65 x+7 + 8 x− 2 " (c) ! 2− 3n m + 9n2− 2m2 m2+2 mn " ÷ ⎛ ⎜⎜⎝ 1 m − 1 m − 2n− 4n2 m + n ⎞ ⎟⎟⎠ (d) 1 x+ 1 x+2 × 1 x+ 1 x− 2 ÷ x− 4 x x2− 2+ 1 x2 ISBN: 9781107633322 Photocopying is restricted under law and this material must not be transferred to another party © Bill Pender, David Sadler, Julia Shea, Derek Ward 2012 Cambridge University Press

! ! CHAPTER 1: Methods in Algebra 1E Four Cubic Identities 11 1E Fo u r C u b i c I d e n t i t i e s The three special quadratic identities will be generalised later to any degree. For now, here are the cubic versions of them. They will be new to most people. 7 CUBE OF A SUM :( A + B)3= A3+3 A2B +3 AB 2+ B3 CUBE OF A DIFFERENCE :( A − B)3= A3− 3A2B +3 AB 2− B3 DIFFERENCE OF CUBES : A3− B3=( A − B)(A2+ AB + B2) SUM OF CUBES : A3+ B3=( A + B)(A2− AB + B2) The proofs of these identities are left to the ﬁrst two questions in the following exercise. WORKED EXERCISE : Here is an example of each identity. (a) (x+5) 3= x3+15 x2+75 x+125 (b) (2x− 3y)3=8 x3− 36 x2y+54 xy 2− 27 y3 (c) x3− 8= ( x− 2)( x2+2 x+4) (d) 43+5 3=(4+5)(16 − 20 + 25)= 9 × 21 = 3 3× 7 WORKED EXERCISE : (a) Simplify a3+1 a+1 . (b) Fa c t o r a3− b3+ a− b. SOLUTION : (a) a3+1 a+1 = (a+1)( a2− a+1) a+1 = a2− a+1 (b) a3− b3+ a− b =( a− b)(a2+ ab + b2)+( a− b) =( a− b)(a2+ ab + b2+1) Exercise 1E 1. (a) Prove the factorisation A3− B3=( A − B)(A2+ AB + B2)byexpandingtheRHS. (b) Similarly, prove the factorisation A3+ B3=( A + B)(A2− AB + B2). 2. (a) Prove the identity ( A + B)3= A3+3 A2B +3 AB 2+ B3by writing (A + B)3=( A + B)(A2+2 AB + B2)andexpanding. (b) Similarly, prove the identity ( A − B)3= A3− 3A2B +3 AB 2− B3. 3. Expand: (a) (a+ b)3 (b) (x− y)3 (c) (b− 1)3 (d) (p+2) 3 (e) (1 − c)3 (f ) (t− 3)3 (g) (2x+5 y)3 (h) (3a− 4b)3 4. Fa c t o r : (a) x3+ y3 (b) a3− b3 (c) y3+1 (d) g3− 1 (e) b3− 8 (f ) 8c3+1 (g) 27 − t3 (h) 125 + a3 (i) 27 h3− 1 (j) u3− 64 v3 (k) a3b3c3+1000 (l) 216 x3+125 y3 5. Write as a product of three factors: (a) 2x3+16 (b) a4− ab 3 (c) 24 t3+81 (d) x3y− 125 y (e) 250 p3− 432 q3 (f ) 27 x4+1000 xy 3 (g) 5x3y3− 5 (h) x6+ x3y3 ISBN: 9781107633322 Photocopying is restricted under law and this material must not be transferred to another party © Bill Pender, David Sadler, Julia Shea, Derek Ward 2012 Cambridge University Press

! ! 12 CHAPTER 1: Methods in Algebra CAMBRIDGE MATHEMATICS 3U NIT YEAR 11 6. Simplify: (a) x3− 1 x2− 1 (b) a2− 3a− 10 a3+8 (c) a3+1 6a2 × 3a a2+ a (d) x2− 9 x4− 27 x÷ x+3 x2+3 x+9 7. Simplify: (a) 3 a− 2− 3a a2+2 a+4 (b) 1 x3− 1+ x+1 x2+ x+1 (c) 1 x2− 2x− 8− 1 x3+8 (d) a2 a3+ b3+ a− b a2− ab + b2+ 1 a+ b DEVELOPMENT 8. Factor as fully as possible: (a) a3+ b3+ a+ b (b) x6− 64 (c) 2a4− 3a3+16 a− 24 (d) (x+ y)3− (x− y)3 (e) s3− t3+ s2− t2 (f ) (t− 2)3+( t+2) 3 (g) (a− 2b)3+(2 a− b)3 (h) x6− 7x3− 8 (i) u7+ u6+ u+1 (j) 2+ x3− 3x6 (k) x7− x3+8 x4− 8 (l) a5+ a4+ a3+ a2+ a+1 9. Simplify: (a) 6a2+6 a2+ a+1 × a3− 1 a3− 3a2× a3+ a2 a4− 1 (b) x4− 8x x2− 4x− 5× x2+2 x+1 x3− x2− 2x÷ x2+2 x+4 x− 5 (c) (a+1) 3− (a− 1)3 3a3+ a (d) 1 x− 3− 8x x3− 27 − x− 3 x2+3 x+9 (e) 3x2+2 x+4 x3− 1 − x+1 x2+ x+1 − 2 x− 1 (f ) 1+ x+ x2 1− x3 + x− x2 (1 − x)3 EXTENSION 10. Find the four quartic identities that correspond to the cubic identities in this exercise. That is, ﬁnd the expansions of ( A + B)4and ( A − B)4and ﬁnd factorisations of A4+ B4 and A4− B4. 11. Factor as fully as possible: (a) x7+ x (b) x12 − y12 12. If x+ y=1 and x3+ y3= 19, ﬁnd the value of x2+ y2. 13. Simplify ( x− y)3+( x+ y)3+3( x− y)2(x+ y)+3( x+ y)2(x− y). 14. If a+ b+ c= 0, show that (2 a− b)3+(2 b− c)3+(2 c− a)3=3(2 a− b)(2 b− c)(2 c− a). 15. Simplify a4− b4 a2− 2ab + b2÷ a2b+ b3 a3− b3 × a2b− ab 2+ b3 a4+ a2b2+ b4. 16. Simplify (1 + a)2÷ ⎛ ⎜⎜⎝1+ a 1− a+ a 1+ a+ a2 ⎞ ⎟⎟⎠. ISBN: 9781107633322 Photocopying is restricted under law and this material must not be transferred to another party © Bill Pender, David Sadler, Julia Shea, Derek Ward 2012 Cambridge University Press

! ! CHAPTER 1: Methods in Algebra 1F Linear Equations and Inequations 13 1F Linear Equations and Inequations The rules for solving equations and for solving inequations are the same, except for a qualiﬁcation about multiplying or dividing an inequation by a negative: 8 LINEAR EQUATIONS : Any number can be added to or subtracted from both sides. Both sides can be multiplied or divided by any nonzero number. LINEAR INEQUATIONS : The rules for inequations are the same as those for equations, except that when both sides are multiplied or divided by a negative number, the inequality sign is reversed. WORKED EXERCISE : Solve: (a) 4− 7x 4x− 7=1 (b) x− 12 < 5+3 x SOLUTION : (a) 4− 7x 4x− 7=1 × (4x− 7) 4− 7x=4 x− 7 +7 x 4=11 x− 7 +7 11 = 11 x ÷ 11 x=1 (b) x− 12 < 5+3 x − 3x −2x− 12 < 5 +12 −2x< 17 ÷ (−2) x> −812 Because of the division by the neg- ative, the inequality was reversed. Changing the Subject of a Formula: Similar sequences of operations allow the sub ject of a formula to be changed from one pronumeral to another. WORKED EXERCISE : Given the formula y= x+1 x+ a: (a) change the sub ject to a, (b) change the sub ject to x. SOLUTION : (a) y= x+1 x+ a × (x+ a) xy + ay = x+1 − xy ay = x+1 − xy ÷ y a= x+1 − xy y (b) y= x+1 x+ a × (x+ a) xy + ay = x+1 xy − x=1 − ay x(y− 1) = 1 − ay ÷ (y− 1) x= 1− ay y− 1 Exercise 1F 1. Solve: (a) −2x= −20 (b) 3x> 2 (c) −a=5 (d) x −4≤− 1 (e) −1− x=0 (f ) 0·1y=5 (g) 2t 4+ t 5 (g) 19 = 3 − 7y (h) 23 − u 3 ≥ 7 ISBN: 9781107633322 Photocopying is restricted under law and this material must not be transferred to another party © Bill Pender, David Sadler, Julia Shea, Derek Ward 2012 Cambridge University Press

! ! 14 CHAPTER 1: Methods in Algebra CAMBRIDGE MATHEMATICS 3U NIT YEAR 11 3. Solve: (a) 5x− 2< 2x+10 (b) 5− x=27+ x (c) 16 + 9 a> 10 − 3a (d) 13 y− 21 ≤ 20 y− 35 (e) 13 − 12 x≥ 6− 3x (f ) 3(x+7) = −2(x− 9) (g) 8+4(2 − x)> 3− 2(5 − x) (h) 7x− (3x+11) = 6 − (15 − 9x) (i) 4(x+2) = 4 x+9 (j) 3(x− 1) < 2(x+1)+ x (k) (x− 3)( x+6) ≤ (x− 4)( x− 5) (l) (1 + 2x)(4 + 3 x)=(2 − x)(5 − 6x) (m) (x+3) 2> (x− 1)2 (n) (2x− 5)(2 x+5) = (2 x− 3)2 4. Solve: (a) x 8 = 1 2 (b) a 12 = 2 3 (c) y 20 < 4 5 (d) 1 x =3 (e) 2 a =5 (f ) 3= 9 2y (g) 2x+1 5 ≥− 3 (h) 5a 3 − 1≥ 3a 5 +1 (i) 7− 4x 6 < 1 (j) 5+ a a = −3 (k) 9− 2t t =13 (l) 6− c 3>c (m) 1 a+4 = 1 − 2 a (n) 4 x− 1= −5 (o) 3x 1− 2x =7 (p) 11 t 8t+13 = −2 5. Solve: (a) x 3− x 5 ≥ 2 (b) a 10 − a 6< 1 (c) x 6+ 2 3= x 2− 5 6 (d) 1 x− 3= 1 2x (e) 1 2x− 2 3=1 − 1 3x (f ) x 3− 2< x 2− 3 (g) x− 2 3 > x+4 4 (h) 3 x− 2= 4 2x+5 (i) x+1 x+2 = x− 3 x+1 (j) (3x− 2)(3 x+2) (3x− 1)2 =1 (k) a+5 2 − a− 1 3 > 1 (l) 3 4− x+1 12 ≤ 2 3− x− 1 6 (m) 2x 5 + 2− 3x 4 < 3 10 − 3− 5x 2 (n) 34(x− 1) − 12(3x+2) = 0 (o) 4x+1 6 − 2x− 1 15 = 3x− 5 5 − 6x+1 10 (p) 7(1 − x) 12 − 3+2 x 9 ≥ 5(2 + x) 6 − 4− 5x 18 6. (a) If v= u+ at,ﬁnd awhen t=4, v=20 and u=8. (b) Given that v2= u2+2 as , ﬁnd the value of swhen u=6, v=10 and a=2. (c) Suppose that 1 u+ 1 v= 1 t.Find v,giventhat u= −1and t=2. (d) If S = −15, n=10 and a= −24, ﬁnd ℓ,giventhat S = n2(a+ ℓ). (e) Temperatures in degrees Fahrenheit and degrees Celsius are related by the formula F = 95C + 32. Find the value of C that corresponds to F =95. (f ) Suppose that the variables cand dare related by the formula 3 c+1 = 5 d− 1.Find c when d= −2. ISBN: 9781107633322 Photocopying is restricted under law and this material must not be transferred to another party © Bill Pender, David Sadler, Julia Shea, Derek Ward 2012 Cambridge University Press

! ! CHAPTER 1: Methods in Algebra 1F Linear Equations and Inequations 15 DEVELOPMENT 7. Solve each of the following inequations for the given domain of the variable, and graph each solution on the real number line: (a) 2x− 3< 5, where xis a positive integer. (b) 1− 3x≤ 16, where xis a negative integer. (c) 4x+5 > 2x− 3, where xis a real number. (d) 7− 2x≥ x+1, where xis a real number. (e) 4≤ 2x< 14, where xis an integer. (f ) −12 < 3x< 9, where xis an integer. (g) 1< 2x+1 ≤ 11, where xis a real number. (h) −10 ≤ 2− 3x≤− 1, where xis a real number. 8. Solve each of these problems by constructing and then solving a linear equation: (a) Five more than twice a certain number is one more than the number itself. What is the number? (b) Ihave$175inmywallet,consistingof$10and$5notes. IfIhavetwiceasmany$10 notes as $5 notes, how many $5 notes do I have? (c) My father is 24 years older than me, and 12 years ago he was double my age. How old am I now? (d) The fuel tank in my new car was 40% full. I added 28 litres and then found that it was 75% full. How much fuel does the tank hold? (e) Acertaintankhasaninletvalveandanoutletvalve. Thetankcanbeﬁlledviathe inlet valve in 6 minutes and emptied (from full) via the outlet valve in 10 minutes. If both valves are operating, how long would it take to ﬁll the tank if it was empty to begin with? (f ) A basketball player has scored 312 points in 15 games. How many points must he average per game in his next 3 games to take his overall average to 20 points per game? (g) A cyclist rides for 5 hours at a certain speed and then for 4 hours at a speed 6 km/h greater than her original speed. If she rides 294 km altogether, what was her initial speed? (h) Two trains travel at speeds of 72 km/h and 48 km/h respectively. If they start at the same time and travel towards each other from two places 600 km apart, how long will it be before they meet? 9. Rearrange each formula so that the pronumeral written in the brackets is the sub ject: (a) a= bc − d [b] (b) t= a+( n− 1)d [n] (c) p q+ r= t [r] (d) u=1+ 3 v [v] (e) a 2− b 3= a [a] (f ) 1 f + 2 g= 5 h [g] (g) x= y y+2 [y] (h) a= b+5 b− 4 [b] (i) c= 7+2 d 5− 3d [d] (j) u= v+ w − 1 v− w +1 [v] 10. Solve: (a) x x− 2+ 3 x− 4=1 (b) 3a− 2 2a− 3− a+17 a+10 = 1 2 EXTENSION 11. (a) Show that x− 1 x− 3=1+ 2 x− 3. (b) Hence solve x− 1 x− 3− x− 3 x− 5= x− 5 x− 7− x− 7 x− 9. ISBN: 9781107633322 Photocopying is restricted under law and this material must not be transferred to another party © Bill Pender, David Sadler, Julia Shea, Derek Ward 2012 Cambridge University Press

! ! 16 CHAPTER 1: Methods in Algebra CAMBRIDGE MATHEMATICS 3U NIT YEAR 11 1G Quadratic Equations This section reviews the solution of quadratic equations by factorisation and by the quadratic formula. The third method, completing the square, will be reviewed in Section 1I. Solving a Quadratic by Factorisation: This method is the simplest, but it normally only works when the roots are rational numbers. 9 SOLVING A QUADRATIC BY FACTORING : 1. Get all the terms on the left, then factor the left-hand side. 2. Use the principle that if AB =0, then A =0 or B =0. WORKED EXERCISE : Solve 5 x2+34 x− 7=0. SOLUTION : 5x2+34 x− 7=0 (5x− 1)( x+7) = 0 (factoring the LHS) 5x− 1=0 or x+7 = 0 (one of the factors must be zero) x= 15 or x= −7 Solving a Quadratic by the Formula: This method works whether the solutions are rational numbers or involve surds. 10 THE QUADRATIC FORMULA : The solution of ax 2+ bx + c=0 is x= −b+ √b2− 4ac 2a or x= −b− √b2− 4ac 2a . Always calculate b2− 4ac ﬁrst. The formula is proven by completing the square, as discussed in Chapter Eight. WORKED EXERCISE : Use the quadratic formula to solve: (a) 5x2+2 x− 7=0 (b) 3x2+4 x− 1=0 SOLUTION : (a) 5x2+2 x− 7=0 Here b2− 4ac =2 2+140 =144 =12 2, so x= −2+12 10 or −2− 12 10 =1 or −125. (b) 3x2+4 x− 1=0 Here b2− 4ac =4 2+12 =28 =4 × 7, so x= −4+2 √7 6 or −4− 2√7 6 = 13(−2+ √7) or 13(−2− √7) . Exercise 1G 1. Solve: (a) x2=9 (b) a2− 4=0 (c) 1− t2=0 (d) x2= 94 (e) 4x2− 1=0 (f ) 25 y2=16 2. Solve by factoring: (a) x2− 5x=0 (b) c2+2 c=0 (c) t2= t (d) 3a= a2 (e) 2b2− b=0 (f ) 3u2+ u=0 (g) 3y2=2 y (h) 12 u+5 u2=0 ISBN: 9781107633322 Photocopying is restricted under law and this material must not be transferred to another party © Bill Pender, David Sadler, Julia Shea, Derek Ward 2012 Cambridge University Press

! ! CHAPTER 1: Methods in Algebra 1G Quadratic Equations 17 3. Solve by factoring: (a) x2− 3x+2 = 0 (b) x2+6 x+8 = 0 (c) a2+2 a− 15 = 0 (d) y2+4 y=5 (e) p2= p+6 (f ) a2= a+132 (g) c2+18 = 9 c (h) 8t+20 = t2 (i) u2+ u=56 (j) 50 + 27 h+ h2=0 (k) k2=60+11 k (l) α2+20 α =44 4. Solve by factoring: (a) 3a2− 7a+2 = 0 (b) 2x2+11 x+5 = 0 (c) 3b2− 4b− 4=0 (d) 2y2+5 y=12 (e) 5x2− 26 x+5 = 0 (f ) 4t2+9 = 15 t (g) t+15 = 2 t2 (h) 10 u2+3 u− 4=0 (i) 25 x2+9 = 30 x (j) 6x2+13 x+6 = 0 (k) 12 b2+3+20 b=0 (l) 6k2+13 k=8 5. Solve using the quadratic formula, giving exact answers followed by approximations to four signiﬁcant ﬁgures where appropriate: (a) x2− x− 1=0 (b) y2+ y=3 (c) a2+12 = 7 a (d) u2+2 u− 2=0 (e) c2− 6c+2 = 0 (f ) 4x2+4 x+1 = 0 (g) 2a2+1 = 4 a (h) 5x2+13 x− 6=0 (i) 2b2+3 b=1 (j) 3c2=4 c+3 (k) 4t2=2 t+1 (l) x2+ x+1 = 0 6. Solve by factoring: (a) x= x+2 x (b) a+ 10 a =7 (c) y+ 2 y= 9 2 (d) (5b− 3)(3 b+1) = 1 (e) 5k+7 k− 1 =3 k+2 (f ) u+3 2u− 7= 2u− 1 u− 3 7. Find the exact solutions of: (a) x= 1 x+2 (b) 4x− 1 x = x (c) a= a+4 a− 1 (d) 5m 2 =2+ 1 m (e) y+1 y+2 = 3− y y− 4 (f ) 2(k− 1) = 4− 5k k+1 8. (a) If y= px − ap 2,ﬁnd p,giventhat a=2, x=3 and y=1. (b) Given that ( x− a)(x− b)= c,ﬁnd xwhen a= −2, b=4 and c=7. (c) Suppose that S = n 2 #2a+( n− 1)d$.Findthepositivevalueof nifS =80, a=4and d=6. 9. Find ain terms of bif: (a) a2− 5ab +6 b2=0 (b) 3a2+5 ab − 2b2=0 10. Find yin terms of xif: (a) 4x2− y2=0 (b) x2− 9xy − 22 y2=0 DEVELOPMENT 11. Solve each problem by forming and solving a suitable quadratic equation: ( − 7) x cm ( + 2)x cm xcm (a) Find the value of xin the diagram opposite. (b) Find a positive integer which when increased by 30 is 12 less than its square. (c) Two positive numbers di ﬀer by 3 and the sum of their squares is 117. Find the numbers. (d) A rectangular area can be completely tiled with 200 square tiles. If the side length of each tile was increased by 1 cm, it would only take 128 tiles to tile the area. Find the side length of each tile. ISBN: 9781107633322 Photocopying is restricted under law and this material must not be transferred to another party © Bill Pender, David Sadler, Julia Shea, Derek Ward 2012 Cambridge University Press

! ! 18 CHAPTER 1: Methods in Algebra CAMBRIDGE MATHEMATICS 3U NIT YEAR 11 (e) The numerator of a certain fraction is 3 less than its denominator. If 6 is added to the numerator and 5 to the denominator, the value of the fraction is doubled. Find the fraction. (f ) A photograph is 18 cm by 12 cm. It is to be surrounded by a frame of uniform width whose area is equal to that of the photograph. Find the width of the frame. (g) A certain tank can be ﬁlled by two pipes in 80 minutes. The larger pipe by itself can ﬁll the tank in 2 hours less than the smaller pipe by itself. How long does each pipe take to ﬁll the tank on its own? (h) Two trains each make a journey of 330 km. One of the trains travels 5 km/h faster than the other and takes 30 minutes less time. Find the speeds of the trains. 12. Solve each of these equations: (a) 2 a+3 + a+3 2 = 10 3 (b) k+10 k− 5 − 10 k = 11 6 (c) 3t t2− 6= √3 (d) 3m +1 3m − 1− 3m − 1 3m +1 =2 EXTENSION 13. (a) Find xin terms of c,giventhat 2 3x− 2c+ 3 2x− 3c= 7 2c. (b) Find xin terms of aand bif a2b x2 + ! 1+ b x " a=2 b+ a2 x. 1H Simultaneous Equations This section will review the two algebraic approaches to simultaneous equations — substitution and elimination (graphical interpretations will be discussed in Chapters Two and Three). Both linear and non-linear simultaneous equations will be reviewed, and the methods extended to systems of three equations in three unknowns. Solution by Substitution: This method can be applied whenever one of the equations can be solved for one of the variables. 11 SIMULTANEOUS EQUATIONS BY SUBSTITUTION : Solve one of the equations for one of the variables, then substitute it into the other equation. WORKED EXERCISE : Solve these simultaneous equations by substitution: (a) 3x− 2y=29 (1) 4x+ y=24 (2) (b) y= x2 (1) y= x+2 (2) SOLUTION : (a) Solving (2) for y, y=24 − 4x. (2A) Substituting (2A) into (1), 3 x− 2(24 − 4x)=29 x=7 . Substituting x= 7 into (1), 21 − 2y=29 y= −4. So x=7 and y= −4. ISBN: 9781107633322 Photocopying is restricted under law and this material must not be transferred to another party © Bill Pender, David Sadler, Julia Shea, Derek Ward 2012 Cambridge University Press

! ! CHAPTER 1: Methods in Algebra 1H Simultaneous Equations 19 (b) Substituting (1) into (2), x2= x+2 x2− x− 2=0 (x− 2)( x+1) = 0 x=2 or −1. Fr o m ( 1 ) , w h e n x=2, y=4, and when x= −1, y=1. So x=2 and y=4, or x= −1and y=1. Solution by Elimination: This method, when it can be used, is more elegant, and can involve less algebraic manipulation with fractions. 12 SIMULTANEOUS EQUATIONS BY ELIMINATION :Takesuitablemultiplesoftheequations so that one variable is eliminated when the equations are added or subtracted. WORKED EXERCISE : Solve these simultaneous equations by elimination: (a) 3x− 2y=29 (1) 4x+5 y=8 (2) (b) x2+ y2=53 (1) x2− y2=45 (2) SOLUTION : (a) Ta k i n g 4 × (1) and 3 × (2), 12 x− 8y=116 (1A) 12 x+15 y=24 . (2A) Subtracting (1A) from (2A), 23 y= −92 y= −4. Substituting into (1), 3x+8 = 29 x=7 . So x=7 and y= −4. (b) Adding (1) and (2), 2x2=98 x2=49 . Subtracting (2) from (1), 2y2=8 y2=4 . So x=7 and y=2, or x=7 and y= −2, or x= −7and y=2, or x= −7and y= −2. Systems of Three Equations in Three Variables: The key step here is to reduce the system to two equations in two variables. 13 SOLVING THREE SIMULTANEOUS EQUATIONS : Using either substitution or elimination, produce two simultaneous equations in two of the variables. WORKED EXERCISE : Solve simultaneously: 3 x− 2y− z= −8(1) 5x+ y+3 z=23 (2) 4x+ y− 5z= −18 (3) SOLUTION : Subtracting (3) from (2), x+8 z=41 . (4) Doubling (3), 8 x+2 y− 10 z= −36 (3A) and adding (1) and (3A), 11 x− 11 z= −44 x− z= −4. (5) Equations (4) and (5) are now two equations in two unknowns. Subtracting (5) from (4), 9z=45 z=5 . ISBN: 9781107633322 Photocopying is restricted under law and this material must not be transferred to another party © Bill Pender, David Sadler, Julia Shea, Derek Ward 2012 Cambridge University Press

! ! 20 CHAPTER 1: Methods in Algebra CAMBRIDGE MATHEMATICS 3U NIT YEAR 11 Substituting z=5 into (5), x=1 and substituting into (2), y=3 . So x=1, y=3 and z= 5 (which should be checked in the original equations). Exercise 1H 1. Solve by substitution: (a) y=2 xand 3 x+2 y=14 (b) y= −3xand 2 x+5 y=13 (c) y=4 − xand x+3 y=8 (d) x=5 y+4 and 3 x− y=26 (e) 2x+ y=10 and 7 x+8 y=53 (f ) 2x− y=9 and 3 x− 7y=19 (g) 4x− 5y=2 and x+10 y=41 (h) 2x+3 y=47 and 4 x− y=45 2. Solve by elimination: (a) 2x+ y=1 and x− y= −4 (b) 2x+3 y=1 6and2 x+7 y=2 4 (c) 3x+2 y= −6and x− 2y= −10 (d) 5x− 3y=28 and 2 x− 3y=22 (e) 3x+2 y=7 and 5 x+ y=7 (f ) 3x+2 y=0 and 2 x− y=56 (g) 15 x+2 y=27 and 3 x+7 y=45 (h) 7x− 3y=41 and 3 x− y=17 (i) 2x+3 y=28 and 3 x+2 y=27 (j) 3x− 2y=11 and 4 x+3 y=43 (k) 4x+6 y=11 and 17 x− 5y=1 (l) 8x=5 yand 13 x=8 y+1 3. Solv ebysubstitution: (a) y=2 − xand y= x2 (b) y=2 x− 3and y= x2− 4x+5 (c) y=3 x2and y=4 x− x2 (d) x− y=5 and y= x2− 11 (e) x− y=2 and xy =15 (f ) 3x+ y=9 and xy =6 (g) x2− y2=16 and x2+ y2=34 (h) x2+ y2= 117 and 2 x2− 3y2=54 DEVELOPMENT 4. Solve each of these problems by constructing and then solving a pair of simultaneous equations: (a) If 7 apples and 2 oranges cost $4, while 5 apples and 4 oranges cost $4 ·40, ﬁnd the cost of each apple and orange. (b) Twice as many adults as children attended a certain concert. If adult tickets cost $8 each, child tickets cost $3 each and the total takings were $418, ﬁnd the numbers of adults and children who attended. (c) Amanis3timesasoldashisson. In12yearstimehewillbetwiceasoldashisson. How old is each of them now? (d) At a meeting of the members of a certain club, a proposal was voted on. If 357 members voted and the proposal was carried by a ma jority of 21, how many voted for and how many voted against the proposal? (e) The value of a certain fraction becomes 15if one is added to its numerator. If one is taken from its denominator, its value becomes 17.Findthefraction. (f ) Kathy paid $320 in cash for a CD player. If she paid in $20 notes and $10 notes and there were 23 notes altogether, how many of each type were there? (g) Two people are 16 km apart on a straight road. They start walking at the same time. If they walk towards each other, they will meet in 2 hours, but if they walk in the same direction (so that the distance between them is decreasing), they will meet in 8 hours. Find their walking speeds. ISBN: 9781107633322 Photocopying is restricted under law and this material must not be transferred to another party © Bill Pender, David Sadler, Julia Shea, Derek Ward 2012 Cambridge University Press

! ! CHAPTER 1: Methods in Algebra 1I Completing the Square 21 (h) A certain integer is between 10 and 100. Its value is 8 times the sum of its digits and if it is reduced by 45, its digits are reversed. Find the integer. 5. Solve simultaneously: (a) y 4− x 3 =1 and x 2+ y 5=10 (b) 4x+ y− 2 3 =12 and 3 y− x− 3 5 =6 6. Solve simultaneously: (a) x=2 y y=3 z x+ y+ z=10 (b) x+2 y− z= −3 3x− 4y+ z=13 2x+5 y= −1 (c) 2a− b+ c=10 a− b+2 c=9 3a− 4c=1 (d) p+ q+ r=6 2p− q+ r=1 p+ q− 2r= −9 (e) 2x− y− z=17 x+3 y+4 z= −20 5x− 2y+3 z=1 9 (f)3u+ v− 4w = −4 u− 2v+7 w = −7 4u+3 v− w =9 7. Solve simultaneously: (a) x+ y=15 and x2+ y2=125 (b) x− y=3 and x2+ y2=185 (c) 2x+ y=5 and 4 x2+ y2=17 (d) x+ y= 9 and x2+ xy + y2=61 (e) x+2 y=5 and 2 xy − x2=3 (f ) 3x+2 y=16 and xy =10 EXTENSION 8. Solve simultaneously: (a) 7 x− 5 y=3 and 2 x+ 25 2y=12 (b) 9x2+ y2=52 and xy =8 9. Consider the equations 12 x2− 4xy +11 y2=64 and 16 x2− 9xy +11 y2=78. (a) By letting y= mx , show that 7 m2+12 m − 4=0. (b) Hence, or otherwise, solve the two equations simultaneously. 1I Completing the Square We w i l l s e e i n C h a p t e r E i g h t t h a t c o m p l e t i n g t h e s q u a r e , b e c a u s e i t c a n b e d o n e i n all situations, is more important for the investigation of quadratics than factoring. Fo r e x a m p l e , t h e q u a d r a t i c f o r m u l a r e v i e w e d e a r l i e r i s p r o v e n b y c o m p l e t i n g t h e square. The review in this section will be restricted to monic quadratics, in which the coe ﬃcient of x2is 1. Perfect Squares: When the quadratic ( x+ α)2is expanded, (x+ α)2= x2+2 αx+ α2, the coe ﬃcient of xis twice α and the constant is the square of α. Reversing the process, the constant term in a perfect square can be found by taking half the coe ﬃcient of xand squaring it. 14 COMPLETING THE SQUARE IN AN EXPRESSION :Tocompletethesquareinagiven expression x2+ bx + ··· ,halvethecoe ﬃcient bof xand square it. ISBN: 9781107633322 Photocopying is restricted under law and this material must not be transferred to another party © Bill Pender, David Sadler, Julia Shea, Derek Ward 2012 Cambridge University Press

! ! 22 CHAPTER 1: Methods in Algebra CAMBRIDGE MATHEMATICS 3U NIT YEAR 11 WORKED EXERCISE : Complete the square in: (a) a2+16 a+··· (b) x2−3x+··· SOLUTION : (a) The coe ﬃcient of ais 16, half of 16 is 8, and 8 2=64, so a2+16 a+64 = ( a+8) 2. (b) The coe ﬃcient of xis −3, half of −3is −112,and( −112)2=2 14, so x2− 3x+2 14=( x− 112)2. Solving Quadratic Equations by Completing the Square: This is the process underlying the quadratic formula. 15 SOLVING QUADRATIC EQUATIONS BY COMPLETING THE SQUARE : Complete the square in the quadratic by adding the same to both sides. WORKED EXERCISE : Solve: (a) t2+8 t=20 (b) x2−x−1=0 (c) x2+x+1 = 0 SOLUTION : (a) t2+8 t=20 t2+8 t+16 = 36 (t+4) 2=36 t+4 = 6 or t+4 = −6 t=2 or −10 (b) x2− x− 1=0 x2− x+ 14=1 14 (x− 12)2= 54 x− 12= 12 √5or −12 √5 x= 12+ 12 √5or 12− 12 √5 (c) x2+ x+1 = 0 x2+ x+ 14= −34 (x+ 12)2= −34 This is impossible, because a square can’t be negative, so the equation has no solutions. Exercise 1I 1. Wr i t e d o w n t h e c o n s t a n t w h i c h m u s t b e a d d e d t o e a c h e x p r e s s i o n i n o r d e r t o c r e a t e a perfect square: (a) x2+2 x (b) y2− 6y (c) a2+10 a (d) m2− 18 m (e) c2+3 c (f ) x2− x (g) b2+5 b (h) t2− 9t 2. Fa c t o r : (a) x2+4 x+4 (b) y2+2 y+1 (c) p2+14 p+49 (d) m2− 12 m +36 (e) t2− 16 t+64 (f ) 400 − 40 u+ u2 (g) x2+20 xy +100 y2 (h) a2b2− 24 ab +144 3. Copy and complete: (a) x2+6 x+ ··· =( x+ ··· )2 (b) y2+8 y+ ··· =( y+ ··· )2 (c) a2− 20 a+ ··· =( a+ ··· )2 (d) b2− 100 b+ ··· =( b+ ··· )2 (e) u2+ u+ ··· =( u+ ··· )2 (f ) t2− 7t+ ··· =( t+ ··· )2 (g) m2+50 m + ··· =( m + ··· )2 (h) c2− 13 c+ ··· =( c+ ··· )2 4. Solve each of the following quadratic equations by completing the square: (a) x2− 2x=3 (b) x2− 6x=0 (c) a2+6 a+8 = 0 (d) y2+3 y=10 (e) b2− 5b− 14 = 0 (f ) x2+4 x+1 = 0 (g) x2− 10 x+20 = 0 (h) y2− y+2 = 0 (i) a2+7 a+7 = 0 ISBN: 9781107633322 Photocopying is restricted under law and this material must not be transferred to another party © Bill Pender, David Sadler, Julia Shea, Derek Ward 2012 Cambridge University Press

! ! CHAPTER 1: Methods in Algebra 1J The Language of Sets 23 5. Complete the square for each of the given expressions: (a) p2− 2pq + ··· (b) a2+4 ab + ··· (c) x2− 6xy + ··· (d) c2+40 cd + ··· (e) u2− uv + ··· (f ) m2+11 mn + ··· DEVELOPMENT 6. Solve by dividing both sides by the coe ﬃcient of x2and then completing the square: (a) 3x2− 15 x+18 = 0 (b) 2x2− 4x− 1=0 (c) 3x2+6 x+5 = 0 (d) 2x2+8 x+3 = 0 (e) 4x2+4 x− 3=0 (f ) 4x2− 2x− 1=0 (g) 3x2− 8x− 3=0 (h) 2x2+ x− 15 = 0 (i) 2x2− 10 x+7 = 0 7. (a) If x2+ y2+4 x− 2y+ 1 = 0, show that ( x+2) 2+( y− 1)2=4. (b) Show that the equation x2+ y2− 6x− 8y=0can be written in the form ( x− a)2+ (y− b)2= c,where a,band care constants. Hence write down the values of a,b and c. (c) If x2+1 = 10 x+12 y, show that ( x− 5)2=12( y+2). (d) Find values for A,B and C if y2− 6x+16 y+94 = ( y+ C)2− B(x+ A). EXTENSION 8. (a) Write down the expansion of ( x+ α)3and hence complete the cube in x3+12 x2+ ··· =( x+ ··· )3. (b) Hence use a suitable substitution to change the equation x3+12 x2+30 x+4 = 0 into acubicequationoftheform u3+ cu + d=0. 1J The Language of Sets We will often want to speak about collections of things such as numbers, points and lines. In mathematics, these collections are called sets ,andthissectionwill introduce or review some of the language associated with sets. Logic is very close to the surface when talking about sets, and particular attention should be given to the words ‘if ’, ‘if and only if ’, ‘and’, ‘or’ and ‘not’. Listing Sets and Describing Sets: A set is a collection of things. When a set is speciﬁed, it needs to be made absolutely clear what things are its members. This can be done by listing the members inside curly brackets: S = {1, 3, 5, 7, 9 }, read as ‘ S is the set whose members are 1, 3, 5, 7 and 9’. It can also be done by writing a description of its members inside curly brackets, for example, T = {odd integers from 0 to 10 }, read as ‘ T is the set of odd integers from 0 to 10’. Equal Sets: Two sets are called equal if they have exactly the same members. Hence the sets S and T in the previous paragraph are equal, which is written as S = T. The order in which the members are written doesn’t matter at all, neither does repetition, so, for example, {1, 3, 5, 7, 9 }= {3, 9, 7, 5, 1 }= {5, 9, 1, 3, 7 }= {1, 3, 1, 5, 1, 7, 9 }. ISBN: 9781107633322 Photocopying is restricted under law and this material must not be transferred to another party © Bill Pender, David Sadler, Julia Shea, Derek Ward 2012 Cambridge University Press

! ! 24 CHAPTER 1: Methods in Algebra CAMBRIDGE MATHEMATICS 3U NIT YEAR 11 Members and Non-members: The symbol ∈ means ‘is a member of ’, and the symbol /∈ means ‘is not a member of ’, so if A = {3, 4, 5, 6 },then 3∈ A and 2 /∈ A and 6 ∈ A and 9 /∈ A, which is read as ‘3 is a member of A’ and ‘2 is not a member of A’, and so on. Set-builder Notation: Athirdwaytospecifyasetistowritedown,usingacolon(:), all the conditions something must fulﬁl to be a member of the set. For example, {n:n is a positive integer, n< 5}= {1, 2, 3, 4 }, which is read as ‘The set of all n such that n is a positive integer and n is less than 5’. If the type of number is not speciﬁed, real numbers are normally intended. Here are some sets of real numbers, in set-builder notation, with their graphs on the number line: x 30 {x:x< 3} x −20 {x:x≥− 2} x −11 0 {x:−1≤ x< 1} x −11 0 {x:x< −1or x≥ 1} The Size of a Set: Asetmaybe ﬁnite or inﬁnite .Ifaset S is ﬁnite, then |S|is the symbol for the number of members in S.Forexample, {positive even numbers }is inﬁnite and |{ a, e, i, o, u }| =5 . Some sets have only one member, for example {3}is ‘the set whose only member is 3’. The set {3}is a di ﬀerent ob ject from the number 3: 3∈ {3} and 3 ̸= {3} and |{ 3}| =1 . The Empty Set: The symbol ∅ represents the empty set , which is the set with no members: |∅|=0 and x/∈∅ , whatever xis. There is only one empty set, because any two empty sets have the same members (that is, none at all) and so are equal. Subsets of Sets: Aset A is called a subset of a set B if every member of A is a member of B.Thisrelationiswrittenas A ⊂ B.Forexample, {men in Australia }⊂ {people in Australia } {2, 3, 4 }̸⊂ {3, 4, 5 }. Because of the way subsets have been deﬁned, every set is a subset of itself. Also the empty set is a subset of every set. For example, {1, 3, 5 }⊂ {1, 3, 5 }, ∅⊂ {1, 3, 5 } and {3}⊂ {1, 3, 5 }. ‘If’ means Subset, ‘If and Only If’ means Equality: The word ‘if ’ and the phrase ‘if and only if ’ are fundamental to mathematical language. They have an important interpretation in the language of sets, the ﬁrst in terms of subsets of sets, the second in terms of equality of sets: A ⊂ B means ‘If x∈ A,then x∈ B’. A = B means ‘ x∈ A if and only if x∈ B’. ISBN: 9781107633322 Photocopying is restricted under law and this material must not be transferred to another party © Bill Pender, David Sadler, Julia Shea, Derek Ward 2012 Cambridge University Press

! ! CHAPTER 1: Methods in Algebra 1J The Language of Sets 25 Union and Intersection: The union A ∪B of two sets A and B is the set of everything belonging to A or to B or to both. Their intersection A∩B is the set of everything belonging to both A and B.Forexample, if A = {0, 1, 2, 3 }and B = {1, 3, 6 }, then A ∪B = {0, 1, 2, 3, 6 }and A ∩B = {1, 3 }. Two sets A and B are called disjoint if they have no elements in common, that is, if A ∩B = ∅.Forexample,thesets {2, 4, 6, 8 }and {1, 3, 5, 7 }are disjoint. ‘Or’ means Union, ‘And’ means Intersection: The deﬁnitions of union and intersection can be written in set-builder notation using the words ‘and’ and ‘or’: A ∪B = {x:x∈ A or x∈ B} A ∩B = {x:x∈ A and x∈ B}. This connection between the words ‘and’ and ‘or’ and set notation should be carefully considered. The word ‘or’ in mathematics always means ‘and/or’, and never means ‘either, but not both’. The Universal Set and the Complement of a Set: A universal set is the set of everything under discussion in a particular situation. For example, if A = {1, 3, 5, 7, 9 }, then possible universal sets are the set of all positive integers less than 11, or the set of all real numbers. Once a universal set E is ﬁxed, then the complement A of any set A is the set of all members of that universal set which are not in A.Forexample, if A = {1, 3, 5, 7, 9 }and E = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10 }, then A = {2, 4, 6, 8, 10 }. Notice that every member of the universal set is either in A or in A, but never in both A and A: A ∪A = E and A ∩A = ∅. ‘Not’ means Complement: There is an important connection between the word ‘not’ and the complement of a set. If the deﬁnition of the complementary set is written in set-builder notation, A = {x∈ E :xis not a member of A}. Venn Diagrams: A Ve n n d i a g r a m is a diagram used to represent the relationship be- tween sets. For example, the four diagrams below represent the four di ﬀerent possible relationships between two sets A and B.Ineachcase,theuniversalset is again E = {1, 2, 3, ... ,10 }. A B 5 7 1 32 4 68 9 10 E A = {1, 3, 5, 7 } B = {1, 2, 3, 4 } AB 55 77 1133 2244 668 8 99 1010 E A = {1, 3, 5, 7 } B = {2, 4, 6, 8 } A B 5 7 1 3 2 4 68 9 10 E A = {1, 2, 3 } B = {1, 2, 3, 4, 5 } A B 5 7 1 3 2 4 6 8 9 10 E A = {1, 3, 5, 7 ,9 } B = {1, 3 } ISBN: 9781107633322 Photocopying is restricted under law and this material must not be transferred to another party © Bill Pender, David Sadler, Julia Shea, Derek Ward 2012 Cambridge University Press

! ! 26 CHAPTER 1: Methods in Algebra CAMBRIDGE MATHEMATICS 3U NIT YEAR 11 Compound sets such as A ∪ B,A ∩ B and A ∩ B can be visualised by shading regions the of Venn diagram, as is done in several questions in the following exercise. The Counting Rule for Sets: To calculate the size of the union A∪B of two sets, adding the sizes of A and of B will not do, because the members of the intersection A∩B would be counted twice. Hence |A ∩ B|needs to be subtracted again, and the rule is |A ∪B|= |A|+ |B|− |A ∩B|. Fo r e x a m p l e , t h e Ve n n d i a g r a m o n t h e r i g h t s h o w s t h e t w o sets: A = {1, 3, 4, 5, 9 } and B = {2, 4, 6, 7, 8, 9 }. A B 5 7 1 3 2 4 6 8 9 E From the diagram, |A ∪ B|=9, |A|=5, |B|=6and |A ∩B|= 2, and the formula works because 9 = 5 + 6 − 2. When two sets are disjoint, there is no overlap between A and B to cause any double counting. With A ∩B = ∅and |A ∩B|= 0, the counting rule becomes |A ∪B|= |A|+ |B|. Problem Solving Using Venn Diagrams: A Venn diagram is often the most convenient way to sort out problems involving overlapping sets of things. In the following exercise, the number of members of each region is written inside the region, rather than the members themselves. WORKED EXERCISE : 100 Sydneysiders were surveyed to ﬁnd out how many of them had visited the cities of Melbourne and Brisbane. 31 people had visited Mel- bourne, 26 people had visited Brisbane and 12 people had visited both cities. Find how many people had visited: (a) Melbourne or Brisbane, (b) Brisbane but not Melbourne, (c) only one of the two cities, (d) neither city. SOLUTION : Let M be the set of people who have visited Mel- bourne, let B be the set of people who have visited Brisbane, and let E be the universal set of all people surveyed. Calcu- lations should begin with the 12 people in the intersection of the two regions. Then the numbers shown in the other three regions of the Venn diagram can easily be found, and so: M B 14 19 12 55 E (a) |{ visited Melbourne or Brisbane }| =19+14+12=45 (b) |{ visited Melbourne only }| =19 (c) |{ visited only one city }| = 19 + 14 = 33 (d) |{ visited neither city }| =100 − 45 = 55 ISBN: 9781107633322 Photocopying is restricted under law and this material must not be transferred to another party © Bill Pender, David Sadler, Julia Shea, Derek Ward 2012 Cambridge University Press

! ! CHAPTER 1: Methods in Algebra 1J The Language of Sets 27 Exercise 1J 1. State whether each set is ﬁnite or inﬁnite. If it is ﬁnite, state its number of members: (a) {1, 3, 5, ... } (b) {0, 1, 2, ... ,9 } (c) ∅ (d) {points on a line } (e) {n:n is a positive integer and 1

! ! 28 CHAPTER 1: Methods in Algebra CAMBRIDGE MATHEMATICS 3U NIT YEAR 11 (a) A = {positive integers },B = {positive real numbers } (b) A = {7, 1, 4, 8, 3, 5 },B = {2, 9, 0, 7 } (c) A = {multiples of 3 },B = {multiples of 5 } (d) A = {l, e, a, r, n },B = {s, t, u, d, y } (e) A = {politicians in Australia },B = {politicians in NSW }. 11. In each of the following, A and B represent sets of real numbers. For each part, graph on separate number lines: (i) A,(ii) B,(iii) A ∪B,(iv) A ∩B. (a) A = {x:x> 0},B = {x:x≤ 3} (b) A = {x:x≤− 1},B = {x:x> 2} (c) A = {x:−3≤ x< 1},B = {x:−1≤ x≤ 4} A B 12. (a) Explain the counting rule |A ∪B|= |A|+ |B|− |A ∩B|by making reference to the Venn diagram opposite. (b) If |A ∪B|=17, |A|=12 and |B|= 10, ﬁnd |A ∩B|. (c) Show that the relationship in part (a) is satisﬁed when A = {3, 5, 6, 8, 9 }and B = {2, 3, 5, 6, 7, 8 }. 13. Use a Venn diagram to solve each of these problems: (a) In a group of 20 people, there are 8 who play the piano, 5 who play the violin and 3whoplayboth. Howmanypeopleplayneither? (b) Each person in a group of 30 plays either tennis or golf. 17 play tennis, while 9 play both. How many play golf ? (c) In a class of 28 students, there are 19 who like geometry and 16 who like trigonometry. How many like both if there are 5 students who don’t like either? DEVELOPMENT P Q R 14. Shade each of the following regions on the given diagram (use a separate diagram for each part). (a) P ∩Q ∩R (b) (P ∩R)∪(Q ∩R) (c) P ∪Q ∪R (where P denotes the complement of P) 15. Agroupof80peoplewassurveyedabouttheirapproachestokeepingﬁt. Itwasfound that 20 jog, 22 swim and 18 go to the gym on a regular basis. If 10 people both jog and swim, 11 people both jog and go to the gym, 6 people both swim and go to the gym and 43 people do none of these activities on a regular basis, how many people do all three? EXTENSION 16. (a) Explain why a ﬁve-member set has twice as many subsets as a four-member set. (b) Hence ﬁnd a formula for the number of subsets of an n-member set. 17. How many di ﬀerent possibilities for shading are there, given a Venn diagram with three overlapping sets within a universal set? 18. Express in words: ‘ {∅}̸= ∅because ∅∈ {∅}’. Is the statement true or false? 19. Decide whether or not the following statement is true: A ⊂ B if and only if, if x̸∈ B then x̸∈ A. 20. Simplify #A ∩(A ∩B)$∪#(A ∩B)∪(B ∩A)$. 21. The deﬁnition A = {sets that are not members of themselves } is impossible. Explain why, by considering whether or not A is a member of itself. Online Multiple Choice Quiz ISBN: 9781107633322 Photocopying is restricted under law and this material must not be transferred to another party © Bill Pender, David Sadler, Julia Shea, Derek Ward 2012 Cambridge University Press

CHAPTER TWO Numbers and Functions The principal purpose of this course is the study of functions of real numbers, and the ﬁrst task is therefore to make it clear what numbers are and what functions are. The ﬁrst ﬁve sections of this chapter review the four number systems, with particular attention to the arithmetic of surds. The last ﬁve sections develop the idea of functions and relations and their graphs, with a review of known graphs, and a discussion of various ways in which the graph of a known function or relation can be transformed, allowing a wide variety of new graphs to be obtained. Study Notes: Although much of the detail here may be familiar, the sys- tematic exposition of numbers and functions in this chapter will be new and demanding for most pupils. Understanding is vital, and the few proofs that do occur are worth emphasising. In the work on surds, the exact value of the num- ber must constantly be distinguished from its decimal approximation produced on the calculator. In the work on functions, computer sketching can make routine the understanding that a function has a graph — an understanding fundamental for the whole course but surprisingly elusive — and computers are particularly helpful in understanding transformations of graphs and how they can be e ﬀected algebraically, because a large number of similar examples can be examined in a short time. Nevertheless, pupils must eventually be able to construct a graph from its equation on their own. 2A Cardinals, Integers and Rational Numbers Our experience of numbers arises from the two quite distinct ﬁelds of counting and geometry, and we shall need to organise these contrasting insights into a uniﬁed view. This section concerns the cardinal numbers, the integers and the rational numbers, which are all based on counting. The Cardinal Numbers: Counting things requires the numbers 0, 1, 2, 3, ... .These numb ers are called the cardinal numbers ,andthesymbol N is conventionally used for the set of all cardinal numbers. 1 DEFINITION : N = {cardinal numbers }= {0, 1, 2, 3, ... } This is an inﬁnite set, because no matter how many cardinal numbers are listed, there will always be more. The number 0 is the smallest cardinal, but there is no largest cardinal, because given any cardinal n,thecardinal n+1 is bigger. Closure of N: If two cardinals aand bare added, the sum a+ band the product ab are still cardinals. We therefore say that the set N of cardinals is closed under ISBN: 9781107633322 Photocopying is restricted under law and this material must not be transferred to another party © Bill Pender, David Sadler, Julia Shea, Derek Ward 2012 Cambridge University Press

! ! 30 CHAPTER 2: Numbers and Functions CAMBRIDGE MATHEMATICS 3U NIT YEAR 11 addition and multiplication. But the set of cardinals is not closed under either subtraction or division. For example, 5− 7 is not a cardinal and 6 ÷ 10 is not a cardinal. Divisibility — HCF and LCM: Division of cardinals sometimes does result in a cardinal. Acardinal ais called a divisor of the cardinal bif the quotient b÷ ais a cardinal. Fo r e x a m p l e , {divisors of 24 }= {1, 2, 3, 4, 6, 8, 12, 24 } {divisors of 30 }= {1, 2, 3, 5, 6, 10, 15, 30 } The highest common factor or HCF of two or more cardinals is the largest cardinal that is a divisor of each of them, so the HCF of 24 and 30 is 6. The key to cancelling a fraction down to its lowest terms is dividing the numerator and denominator by their HCF: 24 30 = 24 ÷ 6 30 ÷ 6= 4 5 If ais a divisor of b,then bis a multiple of a.Forexample, {multiples of 24 }= {24, 48, 72, 96, 120, 144, ... } {multiples of 30 }= {30, 60, 90, 120, 150, ... } The lowest common multiple or LCM of two or more cardinals is the smallest positive cardinal that is a multiple of each of them, so the LCM of 24 and 30 is 120. The key to adding and subtracting fractions is ﬁnding the LCM of their denominators, called the lowest common denominator : 5 24 + 7 30 = 5× 5 120 + 7× 4 120 = 53 120 Prime Numbers: A prime number is a cardinal number greater than 1 whose only divisors are itself and 1. The primes form a sequence whose distinctive pattern has confused every mathematician since Greek times: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, ... Acardinalgreaterthan1thathasfactorsotherthanitselfand1iscalleda composite number , and without giving its rather di ﬃcult proof, we shall assume the ‘unique factorisation theorem’: 2 THEOREM :Everypositivecardinalnumbercanbewrittenasaproductofprime numb ers in one and only one way, apart from the order of the factors. So, for example, 24 = 2 3× 3, and 30 = 2 × 3× 5. This theorem means that as far as multiplication is concerned, the prime numbers are the building blocks for all cardinal numbers, no matter how big or complicated they might be. (No primes divide 1, and so the factorisation of 1 into primes requires the qualiﬁcation that aproductofnofactorsis1.) The Greeks were able to prove that there are inﬁnitely many prime numbers, and the proof of this interesting result is given here because it is a clear example of ‘proof by contradiction’, where one assumes the theorem to be false and then works towards a contradiction. ISBN: 9781107633322 Photocopying is restricted under law and this material must not be transferred to another party © Bill Pender, David Sadler, Julia Shea, Derek Ward 2012 Cambridge University Press

! ! CHAPTER 2: Numbers and Functions 2A Cardinals, Integers and Rational Numbers 31 3 THEOREM :Thereareinﬁnitelymanyprimenumbers. Proof: Suppose, by way of contradiction, that the theorem were false. Then there would be a ﬁnite list p1,p2,p3,...p nof all the primes. Fo r m t h e p r o d u c t N = p1p2p3...p n of all of them. Then N + 1 has remainder 1 after division by each of the primes p1,p2,p3,...p n. So N +1 is not divisible by any of the primes p1,p2,p3,...p n. So the prime factorisation of N + 1 must involve primes other than p1,p2,p3,...p n. But p1,p2,p3,...p nis supposed to be a complete list of primes. This is a contradiction, so the theorem is true. In case you are tempted to think that all theorems about primes are so easily proven, here is the beginning of the list of prime pairs ,whicharepairsofprime numb ers di ﬀering by 2: 3, 5; 5, 7; 11, 13; 17, 19; 29, 31; 41, 43; 59, 61; 71, 73; ... . No-one has yet been able to prove either that this list of prime pairs is ﬁnite, or that it is inﬁnite. Computers cannot answer this question, because no computer search for prime pairs could possibly establish whether this list of prime pairs terminates or not. The Integers: The desire to give a meaning to calculations like 5 − 7leadsto negative numb ers −1, −2, −3, −4, ... , and the positive and negative numbers together with zero are called the integers ,from‘integral’meaning‘whole’. Thesymbol Z (from the German word zahlen meaning numb ers ) is conventionally used for the set of integers. 4 DEFINITION : Z = {integers }= {0, 1, −1, 2, −2, 3, −3, ... } This set Z is another inﬁnite set containing the set N of cardinal numbers. There is neither a greatest nor a least integer, because given any integer n, the integer n+ 1 is greater than n,andtheinteger n− 1islessthan n. Closure of Z: The set Z of integers is closed not only under addition and multiplication, but also under subtraction. For example, 7+( −11) = −4( −8) × 3= −24 ( −16) − (−13) = −3 but the set is still not closed under division. For example, 12 ÷ 10 is not an integer. The Rational Numbers: The desire to give meaning to a calculation like ‘divide 7 into 3 equal parts’ leads naturally to fractions and the system of rational numbers . Positive rational numbers were highly developed by the Greeks, for whom ratio was central to their mathematical ideas. 5 DEFINITION :A rational number is a number that can be written as a ratio or fraction a/b ,where aand bare integers and b̸=0: Q = {rational numbers } (Q stands for quotient.) ISBN: 9781107633322 Photocopying is restricted under law and this material must not be transferred to another party © Bill Pender, David Sadler, Julia Shea, Derek Ward 2012 Cambridge University Press

! ! 32 CHAPTER 2: Numbers and Functions CAMBRIDGE MATHEMATICS 3U NIT YEAR 11 Fo r e x a m p l e , 2 12 = 5 2, − 1 3 = −1 3 ,30 ÷ 24 = 5 4,4= 4 1 and −7= −7 1 . Because every integer a is also a fraction a/ 1, the set Q of rational numbers contains the set Z of integers. So we now have three successively larger systems of numbers, N ⊂ Z ⊂ Q. Lowest Terms: Multiplication or division of the numerator and denominator by the same nonzero number doesn’t change the value of a fraction. So division of the numerator and denominator by their HCF always cancels a fraction down to lowest terms, in which the numerator and denominator have no common factor greater than 1. Multiplying the numerator and denominator by −1reversesboth their signs, so the denominator can always be made positive. 6 ASTANDARD FORM :Everyrationalnumbercanbewrittenintheform a/b ,where aand bare integers with highest common factor 1, and b≥ 1. Closure of Q: The rational numbers are closed under all four operations of addition, multiplication, subtraction and division (except by 0). The opposite of a ratio- nal number a/b is obtained by taking the opposite of either the numerator or denominator. The sum of a number and its opposite is zero: − a b= −a b = a −b and a b+ !−a b " =0 . The reciprocal (inverse is not the correct word) of a nonzero rational number a/b is obtained by exchanging the numerator and denominator. The product of a number and its reciprocal is 1: !a b "−1 = b a ! or 1 a/b = b a " and a b× b a=1 . The reciprocal x−1of a rational number xis analogous to its opposite −x: x× x−1=1 and x+( −x)=0 . Terminating and Recurring Decimals: A terminating decimal is an alternative notation for a rational number that can be written as a fraction with a power of 10 as the denominator: 125= 1410 =1 ·43 18= 3125 1000 =3 ·125 578 350 =578+ 6100 =578 ·06 Other than this rather narrow purpose, decimals are useful for approximating numbers so that they can be compared or placed roughly on a number line. For this reason, decimals are used when a quantity like distance or time is being physically measured — the act of measuring can never produce an exact answer. A rational number that cannot be written with a power of 10 as its denominator can, however, be written as a recurring decimal ,inwhichthedigitstotheright of a certain point cycle endlessly. In the division process, the cycling begins when a remainder occurs which has occurred before. 23=0 ·666 666 ... =0 ·˙6 13 1011 =13 ·909 09 ... =13 ·˙9˙0 637=6 ·428 571 428 571 ... =6 ·˙42857 ˙1 24 3544 =24 ·795 454 545 45 ... =24 ·79 ˙5˙4 Conversely, every recurring decimal can be written as a fraction, by the following method. ISBN: 9781107633322 Photocopying is restricted under law and this material must not be transferred to another party © Bill Pender, David Sadler, Julia Shea, Derek Ward 2012 Cambridge University Press

! ! CHAPTER 2: Numbers and Functions 2A Cardinals, Integers and Rational Numbers 33 WORKED EXERCISE : Write each recurring decimal as a fraction in lowest terms: (a) 0·˙5˙1 (b) 7·3˙48 ˙6 SOLUTION : (a) Let x=0 ·˙5˙1. Then x=0 ·515 151 ... × 100 100 x=51 ·515 151 ... Subtracting the last two lines, 99 x=51 x= 1733. So 0 ·˙5˙1= 1733. (b) Let x=7 ·3˙48 ˙6. Then x=7 ·348 648 ... × 1000 1000 x=7348 ·648 648 ... Subtracting the last two lines, 999 x=7341 ·3 x= 734139990 . So 7 ·3˙48 ˙6= 2719370 . 7 METHOD :Ifthecyclelengthis n,multiplyby10 nand subtract. Note: Some examples in the exercises below show that every terminating dec- imal has an alternative representation as a recurring decimal with endlessly cy- cling 9s. For example, 1=0 ·˙97=6 ·˙95 ·2=5 ·1˙911 ·372 = 11 ·371 ˙9 Exercise 2A 1. Find all primes: (a) less than 100, (b) between 150 and 200. 2. Find the prime factorisations of: (a) 24 (b) 60 (c) 72 (d) 126 (e) 104 (f ) 135 (g) 189 (h) 294 (i) 315 (j) 605 3. Find the HCF of the numerator and denominator of each fraction, then express the fraction in lowest terms: (a) 72 64 (b) 84 90 (c) 78 104 (d) 112 144 (e) 168 216 (f ) 294 315 4. Find the LCM of the two denominators, and hence express as a single fraction: (a) 1 8+ 1 12 (b) 5 18 − 2 15 (c) 3 8− 13 36 (d) 37 42 + 23 30 (e) 55 72 − 75 108 (f ) 7 60 + 31 78 5. Express each number as a recurring or terminating decimal. Do not use a calculator. (a) 58 (b) 23 (c) 716 (d) 59 (e) 320 (f ) 712 (g) 41625 (h) 5411 (i) 238 (j) 176 6. Express each decimal as a rational number in lowest terms: (a) 0·15 (b) 0·˙7 (c) 0·108 (d) 0·˙1˙8 (e) 3·12 (f ) 5·˙4˙5 (g) 1·˙6 (h) 1·˙2˙1 (i) 0·2˙1 (j) 6·5˙3 ISBN: 9781107633322 Photocopying is restricted under law and this material must not be transferred to another party © Bill Pender, David Sadler, Julia Shea, Derek Ward 2012 Cambridge University Press

! ! 34 CHAPTER 2: Numbers and Functions CAMBRIDGE MATHEMATICS 3U NIT YEAR 11 DEVELOPMENT 7. Express each of the following as a recurring or terminating decimal: (a) 116 (b) 1312 (c) 715 (d) 1637 (e) 227 (f ) 1324 (g) 317 (h) 1314 (i) 2713 (j) 1521 8. Express each decimal as a rational number in lowest terms: (a) 0·˙7˙5 (b) 1·˙03 ˙7 (c) 4·˙56 ˙7 (d) 0·˙435 ˙6 (e) 1·˙9 (f ) 2·4˙9 (g) 1·5˙2 (h) 2·3˙4˙5 (i) 7·13 ˙8 (j) 0·11 ˙3˙6 9. Wr i t e d o w n t h e r e c u r r i n g d e c i m a l s f o r 17,27,37,47,57and 67. What is the pattern? 10. Find the prime factorisation of the following numbers and hence determine the square root of each by halving the indices: (a) 256 (b) 576 (c) 1225 (d) 1936 11. Write the HCF and LCM of each pair of numbers in prime factor form: (a) 792 and 1188 (b) 1183 and 1456 (c) 2646 and 3087 (d) 3150 and 5600 12. (a) In order to determine whether a given number is prime, it must be tested for divisibility by smaller primes. Given a numb er b etween 200 and 250, which primes need to b e tested? (b) A student ﬁnds that none of the primes less than 22 is a factor of 457. What can be said about the number 457? (c) Which of 247, 329, 451, 503, 727 and 1001 are primes? EXTENSION 13. Prove that if xis the HCF of aand b,then xmust b e a factor of a− b. 14. The factors of a perfect number, other than itself, add up to that number. (a) Show that 28 is a perfect number. (b) Euclid knew that if 2 n− 1isaprimenumber, then 2 n−1(2n− 1) is a perfect number. Test this proposition for n=2, 3, 4 and 5. 15. (a) Evaluate 13as a decimal on your calculator. (b) Subtract 0 ·333 333 33 from this, multiply the result by 10 8and then take the reciprocal. (c) Show arithmetically that the ﬁnal answer in part (b) is 3. Is the answer on your calculator also equal to 3? What does this tell you about the way fractions are stored on a calculator? 16. Two numbers m and n are called relatively prime if the HCF of m and n is 1. The Euler function φ(n)of n is the number of integers less than or equal to n that are relatively prime to n. (a) Conﬁrm the following by listing the integers that are relatively prime to the given numb er: (i) φ(9) = 6 (ii) φ(25) = 20 (iii) φ(32) = 16 (iv) φ(45) = 24 (b) It is known that φ(pk)= pk− pk−1for a prime pand a positive integer k. Show that this is true for p=2 and k=1 ,2,3,4. (c) Prove that φ(3n)=2 × 3n−1. Generalise this result to φ(pn), where pis prime and n is a positive integer. ISBN: 9781107633322 Photocopying is restricted under law and this material must not be transferred to another party © Bill Pender, David Sadler, Julia Shea, Derek Ward 2012 Cambridge University Press

! ! CHAPTER 2: Numbers and Functions 2B The Real Numbers 35 2B The Real Numbers Whereas the integers are regularly spaced 1 unit apart, the rational numbers are packed inﬁnitely closely together. For example, between 0 and 1 there are 9 rationals with denominator 10: 01 110 210 310 410 510 610 810 910 710 and there are 99 rationals with denominator 100 between 0 and 1: 01 50100 and so on, until the rationals are spread ‘as ﬁnely as we like’ along the whole number line. 8 THE RATIONAL NUMBERS ARE DENSE :Thismeansthatwithinanysmallintervalon the number line, there are inﬁnitely many rational numbers. 1 1 2 There are Numbers that are Not Rational: Although the rational numb ers are dense, there are many more numb ers on the number line. In fact ‘most numbers’ are not rational. In par- ticular, some of the most important numbers you will meet in this course are irrational ,like √2, π and the number e, which we will deﬁne later. The proof by contradiction that√2 is irrational was found by the Greeks, and the result is particularly important, since √2 arises so easily in geometry from the very simple and important process of constructing the diagonal of the unit square. It is a surprising result, and shows very clearly how fractions cannot form a number system su ﬃcient for studying geometry. 9 THEOREM :Thenumber √2isirrational. Proof: Suppose, by way of contradiction, that √2wererational. Then √2couldbewrittenas √2= a b, where aand bare integers with no common factors, and b≥ 1. Multiplying both sides by band then squaring both sides gives a2=2 b2. Since 2 b2is even, then the left-hand side a2must also b e even. Hence amust b e even, b ecause if awere o dd, then a2would b e o dd. So a=2 kfor some integer k,andso a2=4 k2is divisible by 4. So the right-hand side 2 b2is divisible by 4, and so b2is even. Hence bmust b e even, b ecause if bwere o dd, then b2would b e o dd. But now both aand bare even, and so have a common factor 2. This is a contradiction, so the theorem is true. It now follows immediately that every multiple of √2byarationalnumbermust be irrational. The exercises ask for similar proofs by contradiction that numbers like √3, 3√2andlog 23areirrational,andthefollowingworkedexampleshows that log 25isirrational. Unfortunatelytheproofsthat π and eare irrational are considerably more di ﬃcult. ISBN: 9781107633322 Photocopying is restricted under law and this material must not be transferred to another party © Bill Pender, David Sadler, Julia Shea, Derek Ward 2012 Cambridge University Press

! ! 36 CHAPTER 2: Numbers and Functions CAMBRIDGE MATHEMATICS 3U NIT YEAR 11 WORKED EXERCISE : Show that log 25 is irrational (remember that x =log 25 means that 2 x=5). SOLUTION : Suppose, by way of contradiction, that log 25wererational. Then log 25 could be written as log 25= a b, where aand bare integers with no common factors, and b≥ 1. Wr i t i n g t h i s u s i n g p o w e r s , 2ab=5 , and taking the bth power of both sides, 2 a=5 b. Now the LHS is even, being a positive integer power of 2, and the RHS is odd, being a positive integer power of 5. This is a contradiction, so the theorem is true. The Real Numbers and the Number Line: The existence of irrational numbers means that the arithmetic of Section 2A, based on the cardinals and their successive extension to the integers and the rationals, is inadequate, and we require a more general idea of number. It is at this point that we turn away from counting and make an appeal to geometry to deﬁne a still larger system called the real numbers as the points on the number line. Take a line ℓand turn it into a number line by cho osing two p oints on it called 0 and 1: 01 ! The exercises review the standard methods of using ruler and compasses to con- struct a point on ℓcorresponding to any rational number, and the construction of further points on ℓcorresponding to the square roots √2, √3, √5, √6, ... . Even the number π can notionally be placed on the line by rolling a circle of di- ameter 1 unit along the line. It seems reasonable therefore to make the following deﬁnition. 10 DEFINITION :The real numbers are the points on the number line: R = {real numbers } The real numbers are often referred to as the continuum ,becausetherationals, despite being dense, are in a sense scattered along the number line like specks of dust, but do not ‘join up’. For example, the rational multiples of √2, which are all irrational, are just as dense on the number line as the rational numbers. It is only the real line itself which is completely joined up, to be the continuous line of geometry rather than falling apart into an inﬁnitude of discrete points. Note: There is, as one might expect, a great deal more to be done here. First, the operations of addition, subtraction, multiplication and division need to be deﬁned, and shown to be consistent with these operations in the rationals. Sec- ondly, one needs to explain why all other irrationals, like log 25and 3√2, really do have their place amongst the real numbers. And thirdly and most fundamentally, our present notion of a line is far too naive and undeveloped as yet to carry the rigorous development of our deﬁnitions. These are very di ﬃcult questions, which were resolved only towards the end of the 19th century, and then incompletely. Interested readers may like to pursue these questions in a more advanced text. ISBN: 9781107633322 Photocopying is restricted under law and this material must not be transferred to another party © Bill Pender, David Sadler, Julia Shea, Derek Ward 2012 Cambridge University Press

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