PEARSON EDEXCEL INTERNATIONAL ALEVEL FURTHER PURE MATHEMATICS 2

PEARSON EDEXCEL INTERNATIONAL A LEVEL STUDENT BOOK Pearson Edexcel International A Level Further Pure Mathematics 2 Student Book provides comprehensive coverage of the Further Pure Mathematics 2 unit. This book is designed to provide students with the best preparation possible for the examination: • • • • • • • Content is fully mapped to the specification to provide comprehensive coverage and easy reference Engaging and relevant international content in a real-world context Exam-style questions at the end of each chapter, and an exam practice paper at the end of the book, provide practice for exam writing technique Signposted transferable skills prepare for further education and employment Reviewed by a language specialist to ensure the book is written in a clear and accessible style Glossary of key Mathematics terminology, and full answers, included at the back of the book Interactive practice activities also included IAL FURTHER PURE MATHS 1 Student Book ISBN: 9781292244648 An Online Teacher Resource Pack (9781292244624) provides further planning, teaching and assessment support. This Student Book supports the following qualifications: Pearson Edexcel International Advanced Subsidiary in Further Mathematics (XFM01) Pearson Edexcel International Advanced Level in Further Mathematics (YFM01) For first teaching September 2018 IAL FURTHER PURE MATHS 3 Student Book ISBN: 9781292244662 www.pearsonglobalschools.com IAL_FPM2_Cover.indd 1-3 PEARSON EDEXCEL INTERNATIONAL A LEVEL FURTHER PURE MATHEMATICS 2 STUDENT BOOK FURTHER PURE MATHEMATICS 2 MATHEMATICS eBook included PEARSON EDEXCEL INTERNATIONAL A LEVEL FURTHER PURE MATHEMATICS 2 STUDENT BOOK 16/04/2019 14:53 PEARSON EDEXCEL INTERNATIONAL A LEVEL FURTHER PURE MATHEMATICS 2 Student Book Series Editors: Joe Skrakowski and Harry Smith Authors: Greg Attwood, Jack Barraclough, Tom Begley, Dave Berry, Ian Bettison, Linnet Bruce, Lee Cope, Charles Garnet Cox, Keith Gallick, Tim Garry, Alistair Macpherson, Bronwen Moran, Johnny Nicholson, Laurence Pateman, Joe Petran, Keith Pledger, Joe Skrakowski, Harry Smith, Geoff Staley, Ibrahim Wazir, Dave Wilkins F01_IAL_FP2_44655_PRE_i-x.indd 1 25/04/2019 08:54 Published by Pearson Education Limited, 80 Strand, London, WC2R 0RL. www.pearsonglobalschools.com Copies of official specifications for all Pearson qualifications may be found on the website: https://qualifications.pearson.com Text © Pearson Education Limited 2019 Edited by Richard Hutchinson Typeset by Tech-Set Ltd, Gateshead, UK Original illustrations © Pearson Education Limited 2019 Illustrated by © Tech-Set Ltd, Gateshead, UK Cover design by © Pearson Education Limited 2019 The rights of Greg Attwood, Jack Barraclough, Tom Begley, Dave Berry, Ian Bettison, Linnet Bruce, Lee Cope, Charles Garnet Cox, Keith Gallick, Tim Garry, Alistair Macpherson, Bronwen Moran, Johnny Nicholson, Laurence Pateman, Joe Petran, Keith Pledger, Joe Skrakowski, Harry Smith, Geoff Staley, Ibrahim Wazir and Dave Wilkins to be identified as the authors of this work have been asserted by them in accordance with the Copyright, Designs and Patents Act 1988. 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Printed in Slovakia by Neografia Picture Credits The authors and publisher would like to thank the following individuals and organisations for permission to reproduce photographs: Shutterstock.com: Tatiana Shepeleva 90, Jag_cz 125; Getty Images: Steve DF 22, Digital Vision. 105, mbbirdy 149; Alamy Stock Photo: Kevin Britland 1, Science Photo Library 46; Science Photo Library: Andrew Brookes, National Physical Laboratory 14 Cover images: Front: Getty Images: Werner Van Steen Inside front cover: Shutterstock.com: Dmitry Lobanov All other images © Pearson Education Limited 2019 All artwork © Pearson Education Limited 2019 F01_IAL_FP2_44655_PRE_i-x.indd 2 25/04/2019 08:54 CONTENTS iii COURSE STRUCTURE iv ABOUT THIS BOOK vi QUALIFICATION AND ASSESSMENT OVERVIEW viii EXTRA ONLINE CONTENT x 1 INEQUALITIES 1 2 SERIES 14 3 COMPLEX NUMBERS 22 4 FURTHER ARGAND DIAGRAMS 46 REVIEW EXERCISE 1 83 5 FIRST-ORDER DIFFERENTIAL EQUATIONS 90 6 SECOND-ORDER DIFFERENTIAL EQUATIONS 105 7 MACLAURIN AND TAYLOR SERIES 125 8 POLAR COORDINATES 149 REVIEW EXERCISE 2 168 EXAM PRACTICE 178 GLOSSARY 180 ANSWERS 183 INDEX 230 F01_IAL_FP2_44655_PRE_i-x.indd 3 25/04/2019 08:54 iv COURSE STRUCTURE CHAPTER 1 INEQUALITIES 1 1.1 ALGEBRAIC METHODS 1.2 USING GRAPHS TO SOLVE INEQUALITIES 1.3 MODULUS INEQUALITIES CHAPTER REVIEW 1 2 CHAPTER 2 SERIES 2.1 THE METHOD OF DIFFERENCES CHAPTER REVIEW 2 CHAPTER 3 COMPLEX NUMBERS 3.1 EXPONENTIAL FORM OF COMPLEX NUMBERS 3.2 MULTIPLYING AND DIVIDING COMPLEX NUMBERS 3.3 DE MOIVRE’S THEOREM 3.4 TRIGONOMETRIC IDENTITIES 3.5 nTH ROOTS OF A COMPLEX NUMBER CHAPTER REVIEW 3 F01_IAL_FP2_44655_PRE_i-x.indd 4 5 8 11 14 15 20 22 23 26 29 32 37 42 CHAPTER 4 FURTHER ARGAND DIAGRAMS 46 4.1 LOCI IN AN ARGAND DIAGRAM 47 4.2 FURTHER LOCI IN AN ARGAND DIAGRAM55 4.3 REGIONS IN AN ARGAND DIAGRAM 63 4.4 FURTHER REGIONS IN AN ARGAND DIAGRAM 65 4.5 TRANSFORMATIONS OF THE COMPLEX PLANE 70 CHAPTER REVIEW 4 78 REVIEW EXERCISE 1 83 CHAPTER 5 FIRST-ORDER DIFFERENTIAL EQUATIONS 90 5.1 SOLVING FIRST-ORDER DIFFERENTIAL EQUATIONS WITH SEPARABLE VARIABLES 91 5.2 FIRST-ORDER LINEAR DIFFERENTIAL EQUATIONS OF THE FORM dy __     + Py = Q WHERE P AND Q dx ARE FUNCTIONS OF x 95 5.3 REDUCIBLE FIRST-ORDER DIFFERENTIAL EQUATIONS 98 CHAPTER REVIEW 5 102 25/04/2019 08:54 COURSE STRUCTURE CHAPTER 6 SECOND-ORDER DIFFERENTIAL EQUATIONS 105 6.1 SECOND-ORDER HOMOGENEOUS DIFFERENTIAL EQUATIONS 6.2 SECOND-ORDER NON-HOMOGENEOUS DIFFERENTIAL EQUATIONS 6.3 USING BOUNDARY CONDITIONS 6.4 REDUCIBLE SECOND-ORDER DIFFERENTIAL EQUATIONS CHAPTER REVIEW 6 106 110 115 118 121 F01_IAL_FP2_44655_PRE_i-x.indd 5 8.1 POLAR COORDINATES AND EQUATIONS 8.2 SKETCHING CURVES 8.3 AREA ENCLOSED BY A POLAR CURVE 8.4 TANGENTS TO POLAR CURVES CHAPTER REVIEW 8 149 150 153 158 162 165 REVIEW EXERCISE 2 168 EXAM PRACTICE 178 126 128 GLOSSARY 180 132 136 ANSWERS 183 INDEX 230 CHAPTER 7 MACLAURIN AND TAYLOR SERIES 125 7.1 HIGHER DERIVATIVES 7.2 MACLAURIN SERIES 7.3 SERIES EXPANSIONS OF COMPOUND FUNCTIONS 7.4 TAYLOR SERIES 7.5 SERIES SOLUTIONS OF DIFFERENTIAL EQUATIONS CHAPTER REVIEW 7 CHAPTER 8 POLAR COORDINATES v 140 144 25/04/2019 08:54 vi ABOUT THIS BOOK ABOUT THIS BOOK The following three themes have been fully integrated throughout the Pearson Edexcel International Advanced Level in Mathematics series, so they can be applied alongside your learning. 1. Mathematical argument, language and proof • Rigorous and consistent approach throughout • Notation boxes explain key mathematical language and symbols 2. Mathematical problem-solving • Hundreds of problem-solving questions, fully integrated into the main exercises • Problem-solving boxes provide tips and strategies • Challenge questions provide extra stretch 3. Transferable skills The Mathematical Problem-Solving Cycle specify the problem interpret results collect information process and represent information • Transferable skills are embedded throughout this book, in the exercises and in some examples • These skills are signposted to show students which skills they are using and developing Finding your way around the book Each chapter is mapped to the specification content for easy reference Each chapter starts with a list of Learning objectives The Prior knowledge check helps make sure you are ready to start the chapter The real world applications of the maths you are about to learn are highlighted at the start of the chapter. Glossary terms will be identified by bold blue text on their first appearance. F01_IAL_FP2_44655_PRE_i-x.indd 6 25/04/2019 08:54 ABOUT THIS BOOK vii Each section begins with explanation and key learning points Exercise questions are carefully graded so they increase in difficulty and gradually bring you up to exam standard Exercises are packed with exam-style questions to ensure you are ready for the exams Exam-style questions are flagged with E Problem-solving questions are flagged with P Problem-solving boxes provide hints, tips and strategies, and Watch out boxes highlight areas where students often lose marks in their exams Challenge boxes give you a chance to tackle some more difficult questions Each chapter ends with a Chapter review and a Summary of key points After every few chapters, a Review exercise helps you consolidate your learning with lots of exam-style questions A full practice paper at the back of the book helps you prepare for the real thing F01_IAL_FP2_44655_PRE_i-x.indd 7 25/04/2019 08:54 viii QUALIFICATION AND ASSESSMENT OVERVIEW QUALIFICATION AND ASSESSMENT OVERVIEW Qualification and content overview Further Pure Mathematics 2 (FP2) is an optional* unit in the following qualifications: International Advanced Subsidiary in Further Mathematics International Advanced Level in Further Mathematics *It is compulsory to study either FP2 or FP3 for the International Advanced Level in Further Mathematics. Assessment overview The following table gives an overview of the assessment for this unit. We recommend that you study this information closely to help ensure that you are fully prepared for this course and know exactly what to expect in the assessment. Unit Percentage _1 FP2: Further Pure Mathematics 2 33   3 % of IAS Paper code WFM02/01 16 _  3 % of IAL Mark Time Availability 75 1 hour 30 mins January and June 2 First assessment June 2020 IAS: International Advanced Subsidiary, IAL: International Advanced A Level. Assessment objectives and weightings Minimum weighting in IAS and IAL AO1 Recall, select and use their knowledge of mathematical facts, concepts and techniques in a variety of contexts. 30% AO2 Construct rigorous mathematical arguments and proofs through use of precise statements, logical deduction and inference and by the manipulation of mathematical expressions, including the construction of extended arguments for handling substantial problems presented in unstructured form. 30% AO3 Recall, select and use their knowledge of standard mathematical models to represent situations in the real world; recognise and understand given representations involving standard models; present and interpret results from such models in terms of the original situation, including discussion of the assumptions made and refinement of such models. 10% AO4 Comprehend translations of common realistic contexts into mathematics; use the results of calculations to make predictions, or comment on the context; and, where appropriate, read critically and comprehend longer mathematical arguments or examples of applications. 5% AO5 Use contemporary calculator technology and other permitted resources (such as formulae booklets or statistical tables) accurately and efficiently; understand when not to use such technology, and its limitations. Give answers to appropriate accuracy. 5% F01_IAL_FP2_44655_PRE_i-x.indd 8 25/04/2019 08:54 QUALIFICATION AND ASSESSMENT OVERVIEW ix Relationship of assessment objectives to units Assessment objective FP2 Marks out of 75 % AO1 AO2 AO3 AO4 AO5 25–30 25–30 0–5 7–12 5–10 _1 _1 1 9  _3 –16 6_  3 –13_  3 33  3 –40 33  3 –40 _2 0–6  3 2 1 Calculators Students may use a calculator in assessments for these qualifications. Centres are responsible for making sure that calculators used by their students meet the requirements given in the table below. Students are expected to have available a calculator with at least the following keys: +, –, ×, ÷, π, x2, __ 1 y x √ x ,  __ x , x , ln x, e , x!, sine, cosine and tangent and their inverses in degrees and decimals of a degree, and in radians; memory. Prohibitions Calculators with any of the following facilities are prohibited in all examinations: • databanks • retrieval of text or formulae • built-in symbolic algebra manipulations • symbolic differentiation and/or integration • language translators • communication with other machines or the internet F01_IAL_FP2_44655_PRE_i-x.indd 9 25/04/2019 08:54 x EXTRA ONLINE CONTENT Extra online content Whenever you see an Online box, it means that there is extra online content available to support you. SolutionBank SolutionBank provides worked solutions for questions in the book. Download the solutions as a PDF or quickly find the solution you need online. Use of technology Explore topics in more detail, visualise problems and consolidate your understanding. Use pre-made GeoGebra activities or Casio resources for a graphic calculator. GeoGebra-powered interactives y x Online Find the point of intersection graphically using technology. Graphic calculator interactives Interact with the maths you are learning using GeoGebra's easy-to-use tools Interact with the maths you are learning using GeoGebra's easy-to-use tools Explore the maths you are learning and gain confidence in using a graphic calculator Calculator tutorials Our helpful video tutorials will guide you through how to use your calculator in the exams. They cover both Casio's scientific and colour graphic calculators. Online Work out each coefficient quickly using the nCr and power functions on your calculator. F01_IAL_FP2_44655_PRE_i-x.indd 10 Step-by-step guide with audio instructions on exactly which buttons to press and what should appear on your calculator's screen 25/04/2019 08:54 INEQUALITIES CHAPTER 1 1 1 INEQUALITIES 1.1 Learning objectives After completing this unit you should be able to: ● Manipulate inequalities involving algebraic fractions → pages 2–5 ● Use graphs to find solutions to inequalities → pages 5–8 ● Solve modulus inequalities → pages 8–11 Prior knowledge check 1 Solve: a 3x2 − 2x − 1 . 0 b x2 + 4x − 2 , 0 ← Pure 1 Section 3.4 2 Solve: a |3x − 1| . 5 b |4x − 8| , 2 ← Pure 3 Section 2 M01_IAL_FP2_44655_U01_001-013.indd 1 Inequalities are used in collisiondetection algorithms in video games. Positions of objects on a screen can be defined by x- and y-coordinates, and the area in which a player or an object is allowed to move can be defined by inequalities. 25/04/2019 08:54 2 CHAPTER 1 INEQUALITIES 1.1 Algebraic methods If you multiply both sides of an inequality by a negative number you reverse the direction of the inequality sign. You need to be more careful if you multiply or divide both sides of an inequality by a variable or expression. If the variable or expression could take either a positive or a negative value then you don’t know which direction is correct for the inequality sign. You can overcome this problem by multiplying by an expression squared. 1 Suppose you want to solve the inequality __  x  . x, x ≠ 0. y y 1 y=x y = x3 y y = x2 y=x y=x y=1 –1 O 1 x –1 The values of x where the 1 graph of y = __   x is above the graph of y = x give you the solution: x , −1 or 0 , x , 1. O 1 x If you multiply both sides of the inequality by x you get 1 . x2. The solution to this inequality is −1 , x , 1, which is not the required solution. –1 O 1 x If you multiply both sides of the inequality by x2 you get x . x 3. The graph of y = x is above the graph of y = x3 for x , −1 and 0 , x , 1, which is the solution to the original inequality. In the third example above, you can solve the inequality x . x3 by algebraically rearranging and factorising. x3 − x , 0 You can add or subtract any term from both sides of an inequality. x(x2 − 1) , 0 x(x − 1)(x + 1) , 0 The critical values are x = 0, x = 1 and x = −1. You can consider a sketch of the graph of y = x(x − 1)(x + 1) to work out which intervals satisfy (i.e. meet the requirements of) the inequality. ■ To solve an inequality involving algebraic fractions: • Step 1: multiply by an expression squared to remove fractions • Step 2: rearrange the inequality to get 0 on one side • Step 3: find critical values • Step 4: use a sketch to identify the correct intervals M01_IAL_FP2_44655_U01_001-013.indd 2 25/04/2019 08:54 INEQUALITIES CHAPTER 1 Example 3 1 x2 Use algebra to solve the inequality  _____  , x + 1, x ≠ 2 x−2 Multiply both sides by (x − 2)2 x2 (x − 2)2 × ______     , (x − 2)2 × (x + 1) x−2 x2 (x − 2)2 × _______     , (x − 2)2 × (x + 1) (x − 2) (x − 2)x2 − (x + 1)(x − 2)2 , 0 (x − 2)(x2 − (x + 1)(x − 2)) , 0 (x − 2)(x2 − x2 + x + 2) , 0 or (x − 2)(x + 2) < 0 Critical values are x = ±2 The sketch of y = (x − 2)(x + 2) is y Problem-solving A natural first step would be to multiply both sides by (x − 2) but we cannot be sure that this is positive. A simple solution is to multiply both sides of the inequality by (x − 2)2 as this will always be positive. Do not aim to multiply out but cancel, collect terms on one side and factorise. This is a quadratic inequality so you can solve it in the usual way. ← Pure 1 Section 3.5 y = (x – 2)(x + 2) Watch out –2 O 2 x When a question says 'Use algebra…' you can still use a sketch to identify which intervals to include in your solution set. However, you should make sure you show algebraic working to find the critical values. The solution to (x − 2)(x + 2) , 0 is −2 , x , 2. When the inequality is not strict you have to be a bit more careful. In the above example, the left-hand side of the inequality is undefined when x = 2, so you cannot include x = 2 in your solution set. Hint Values for which one side of the inequality is undefined will usually be explicitly excluded. In the above example you are given x ≠ 2. ■ When solving an inequality involving ⩽ or ⩾, check whether or not each of your critical values should be included in the solution set. Example 2 x 2 Find all values of x such that  _____  < _____     where x ≠ −1 and x ≠ −3, and express your answer x+1 x+3 using set notation. M01_IAL_FP2_44655_U01_001-013.indd 3 25/04/2019 08:54 4 CHAPTER 1 INEQUALITIES Multiply both sides by (x + 1)2(x + 3)2 So x(x + 1)2(x + 3)2 _______________ 2(x + 1)2(x + 3)2 _______________                <      x+1 x+3 x(x + 1)(x + 3)2 − 2(x + 1)2(x + 3) < 0 (x + 1)(x + 3)(x(x + 3) − 2(x + 1)) < 0 (x + 1)(x + 3)(x2 + x − 2) < 0 (x + 1)(x + 3)(x + 2)(x − 1) < 0 So the critical values are: x = −1, −3, −2 or 1 A sketch of y = (x + 1)(x + 3)(x + 2)(x − 1) is In order to remove the fractions and guarantee that you are not multiplying by a negative quantity, use (x + 1)2(x + 3)2. Cancel terms on each side. Collect terms on LHS. Factorise as much as possible. To find the critical values you need to solve (x + 1)(x + 3)(x + 2)(x − 1) = 0. y –3 –2 –1 O 1 x The curve y = (x + 1)(x + 3)(x + 2)(x − 1) is a quartic graph with positive x4 coefficient, so it starts in top left and ends in top right and passes through (−3, 0), (−2, 0), (−1, 0) and (1, 0). y = (x + 1)(x + 3)(x + 2)(x – 1) The solution to (x + 1)(x + 3)(x + 2)(x − 1) < 0 corresponds to the sections of this graph that are on or below the x-axis. So the solution is {x : −3 , x < −2} ∪ {x : −1 , x < 1} Exercise The inequality is non-strict so you need to check whether the critical values should be included in the solution. The conditions x ≠ −1 and x ≠ −3 are given in the question, so use strict inequalities to exclude these values. 1A 1 Solve the inequalities. a x2 , 5x + 6 b x(x + 1) > 6 x e _____  < 2x x ≠ 1 x−1 2  i _____  , 3 x−4 3 2 f _____  , __     x+1 x 3 1 j _____  . _____     x+2 x−5 2   . 1 c ______ 2 x +1 3     , 1 g ____________ (x + 1)(x − 1) 2   . 1 d ______ 2 x −1 3 2 h __2  > ____________      (x + 1)(x − 2) x 2 Solve the inequalities, giving your answers using set notation. 3x2 + 5 _______   . 1   a x+5 x2 + 7x + 10 ___________ d     . 2x + 7 x+1 M01_IAL_FP2_44655_U01_001-013.indd 4 3x _____ b     . x x−2 x+1 _____ e   2  . 6 x 1+x 2−x c  _____  . _____     1−x 2+x x2 1 f  _____  . __     x+1 6 25/04/2019 08:54 INEQUALITIES E E E/P CHAPTER 1 2x + 1 _____ x+2 3 a Use algebra to find the set of values for which ______     ,     x+5 x+4 x 1 4 a Use algebra to find the set of values for which ______       , _____  giving your answer 2x + 1 x − 3 in set notation. 5 (6 marks) (6 marks) x 1 5 A teacher asks a student to solve the inequality ______         , __ 3x + 4 x The student’s attempt was as follows: 1   , __     x 3x + 4 x ______   x2 , 3x + 4 x2 − 3x − 4 , 0 (x − 4)(x + 1) , 0 −1 , x , 4 E/P a Identify the mistake made by the student and explain why it will produce an incorrect answer. (2 marks) b Solve the inequality correctly. (6 marks) 1 4       , x , ______ 6 Use algebra to solve __  giving your answer using set notation. 2x + 1 x (6 marks) Hint Challenge 1 1 Solve ______   , __   1 − ex ex You probably won’t be able to sketch the graph in this question. Find the critical values, then test values within each interval to determine the solution set. 1.2 Using graphs to solve inequalities ■ If you can sketch the graphs of y = f(x) and y = g(x) then you can solve an inequality such as f(x) , g(x) by observing when one curve is above the other. The critical values will be the solutions to the equation f(x) = g(x). Example Watch out If you are asked to solve an inequality algebraically you should not start by sketching graphs. 3 7x a On the same set of axes, sketch the graphs of the curves with equations y = ______     and y = 4 − x 3x + 1 7x   and y = 4 − x b Find the points of intersection of y = ______   3x + 1 7x   , 4 − x c Solve ______   3x + 1 M01_IAL_FP2_44655_U01_001-013.indd 5 25/04/2019 08:54 6 CHAPTER 1 INEQUALITIES 7x a Sketch y = 4 − x and y = ______    : 3x + 1 y = 4 − x is a straight line crossing the axes at (4, 0) and (0, 4). 7x y = ______    crosses the coordinate axes at (0, 0). 3x + 1 1 There is a vertical asymptote at x = −  __ 3 7 There is a horizontal asymptote at y =  __ 3 So the sketch looks like this y 4 y= 7 3 y= O 7x 3x + 1 x 4 y= 4 – x x = – 31 b Using algebra to find critical values: 7x ______     = 4 − x 3x + 1 7x = 12x + 4 − 3x2 − x Multiply out and collect terms to form a quadratic equation. (3x + 2)(x − 2) = 0 2 So x = −  __ 3 or 2 c Marking these points on the graph: Solve the equation in the usual way: this one factorises. y 4 y= 7 3 y= 2 3 x=– O 1 3 To sketch unfamiliar curves, look for: ● points where the curve meets or crosses the axes ● vertical asymptotes (where the denominators of fractions equal 0) ● behaviour on either side of vertical asymptotes ● behaviour as x gets very large or very small. You can find horizontal asymptotes by rearranging the fraction to see how it behaves as x → ∞. 7x 7 1 7 3x ______         1 − ______     = __     ______   = __   3x + 1 3 (3x + 1 ) 3 ( 3x + 1 ) 1 As x → ∞,  ______  → 0 so the curve has a 3x + 1 horizontal asymptote at y =  _73 Multiply both sides by 3x + 1. This is an equation, not an inequality, so you don't need to multiply by an expression squared. 3x2 − 4x − 4 = 0 – Problem-solving 2 7x 3x + 1 4 y= 4 – x x Look on the sketch for the places where the line is above the curve. These places will give the solution. Watch out Any vertical asymptotes will also be critical values when you are finding your solution set. So the solution is __1 2 x , −  __ 3 or −   3 , x , 2 M01_IAL_FP2_44655_U01_001-013.indd 6 Online Explore the solution to the inequality using GeoGebra. 25/04/2019 08:54 INEQUALITIES CHAPTER 1 Exercise 1B SKILLS 7 PROBLEM-SOLVING 1 Sketch the graphs of the functions. a y = x  2 − 5x + 6 1 c y =  _____  x+1 b y = x  3 + 2x  2 − 3x 4x d y =  ______  1 − 2x 2 Sketch each pair of functions on the same sets of axes. a y = x  2 − 2x + 1and y = 4 − 4 x  2 3 c y = 2x − 1 and y = _____     x−2 E E/P E/P E/P E 1 __ b y = xand y =   x  x d y = 4 − 3xand y = ______     4x − 2 3 Find the points of intersection of the pairs of functions. 3x 1 2   and y =  _____  b y = x − 2and y =  _____  a y = _____   x+1 x−3 x+2 4(x + 2)   c y = x  2 − 4and y = _______   x−2 4   4 a On the same set of axes, sketch the graphs of y = x − 1and y = _____   x−1 4   b Find the points of intersection of y = x − 1 and y = _____   x−1 4   c Write down the solution to the inequality x − 1 . _____   x−1 3 2   x ≠ 3 5 f(x) = __ 2  x ≠ 0 and g(x) = _____ 3−x x a Sketch y = f(x) and y = g(x) on the same set of axes. (3 marks) (2 marks) (2 marks) (3 marks) b Solve f(x) = g(x) (2 marks) c Hence write down the solution to the inequality f(x) . g(x). Give your answer using set notation. (3 marks) 3x 4x 6 a On the same set of axes, sketch the graphs of y = _____       and y = _______   2−x (x − 1)2 4x 3x     and y = _______   b Find the points of intersection of y = _____   2−x (x − 1)2 4x 3x     <_______   c Hence, or otherwise, solve the inequality _____   2 − x (x − 1)2 6(2 − x) 7 a On the same set of axes, sketch the graphs of y = x − 2and y = ____________       (x + 2)(x − 3) 6(2 − x) b Find the points of intersection of y = x − 2and y = ____________       (x + 2)(x − 3) 6(2 − x) c Write down the solution to the inequality x − 2 < ____________       (x + 2)(x − 3) x 1 8 a On the same set of axes, sketch the graphs of y = __     and y = _____     x+2 x x 1     b Find the points of intersection of y = __    and y = _____ x+2 x x 1     c Solve __     . _____ x x+2 M01_IAL_FP2_44655_U01_001-013.indd 7 (4 marks) (2 marks) (2 marks) (4 marks) (3 marks) (2 marks) (3 marks) (2 marks) (2 marks) 25/04/2019 08:54 8 CHAPTER 1 INEQUALITIES Challenge a Sketch the circle with equation (x − 2) 2 + (y − 4) 2 = 10 b Determine the coordinates of all points of intersection between this 4x − 5   circle and the curve with equation y =  ______ x−2 c Sketch this curve on the same set of axes as your answer to part a. d Hence, or otherwise, find the solutions to the inequality 2 4x − 5 (x − 2)2 + (  ______   − 4) , 10 x−2 1.3 Modulus inequalities You need to be able to solve inequalities that include modulus signs. It is often useful to sketch the relevant modulus graph when solving inequalities like this. Example 4 Solve ∣ x2 − 4x∣ , 3 Sketch y = | x2 − 4x| and y = 3: y y = |x2 – 4x| 3 Sketch y = ∣ x2 − 4x∣ and y = 3 on the same set of axes. To sketch y = ∣ x2 − 4x∣ consider the graph of y = x2 − 4x, and reflect any sections of the graph that are below the x-axis in the x-axis. ← Pure 3 Section 2.1 y= 3 Watch out O 4 x To find the critical values, solve |x2 − 4x| = 3 x2 − 4x = 3 ⇒ x2 − 4x − 3 = 0 (x – 2)2 – 4 – 3 = 0 (x – 2)2 = 7 __ x = 2 ± √7  −(x2 − 4x) = 3 ⇒ x2 − 4x + 3 = 0 (x − 3)(x − 1) = 0 x = 1 or 3 M01_IAL_FP2_44655_U01_001-013.indd 8 Solve ∣ x2 − 4x∣ = 3 to find the critical values. You need to consider the two separate cases: when the argument of ∣ x2 − 4x∣ is positive and when it is negative. Use your sketch to determine whether these critical values all correspond to points of intersection. Complete the square or use the quadratic formula. The line y = 3 intersects the graph of y = ∣x2 − 4x∣ at four places, so all of these values of x correspond to points of intersection. Look at example 6 for a situation where this is not the case. 25/04/2019 08:54 INEQUALITIES CHAPTER 1 9 Marking these values on the sketch: y = |x2 – 4x| y 3 y= 3 You need to identify where the points of intersection are on the sketch. O 2– 7 1 3 4 2+ 7 x So the solution is: __ __ √ 2 − 7  , x , 1 or 3 , x , 2 + √7  Example Finally write down the solution to the inequality: the points where the line y = 3 is above the curve. 5 Solve ∣3x∣ + x < 2 Rearranging gives: |3x| < 2 − x Sketching y = |3x| and y = 2 − x gives y y = |3x| Problem-solving Sketching y = ∣3x∣ + x is quite difficult so it is usually simpler to rearrange and isolate the modulus function. 2 y= 2– x O 2 x Critical values are given by: 3x = 2 − x 4x = 2 x = __   21 or −3x = 2 − x −2 = 2x x = −1 So the line is above |3x| for 1 −1 < x <  __ 2 M01_IAL_FP2_44655_U01_001-013.indd 9 Find the critical values in the usual way. Remember the two cases. By considering (i.e. taking into account) the positions of the critical values, identify the places where the line is above the V-shaped graph. 25/04/2019 08:54 10 CHAPTER 1 INEQUALITIES Sometimes care must be taken to identify the correct roots when solving modulus equations. Example 6 Find all values of x such that ∣x2 − 19∣ < 5(x − 1), expressing your answer in set notation. Sketching both graphs: Online Explore the solution to the inequality using GeoGebra. y y = |x2 – 19| y = 5(x – 1) Sketch the graphs. O 7 x 3 x2 − 19 = 5x − 5 ⇒ x2 − 5x − 14 = 0 (x − 7)(x + 2) = 0 x = 7 or −2 Find the critical values. −(x2 − 19) = 5x − 5 ⇒ x2 + 5x − 24 = 0 (x + 8)(x − 3) = 0 x = −8 or 3 The set of points for which the line is above the curve can be written as {x : 3 < x < 7}. Exercise Watch out Solving the equations gives four values but the graphs only have two crossing points. The valid critical values are x = 3 and x = 7. Write down the solution. 1C 1 Solve the inequalities. a |x − 6| . 6x b |x − 3| . x2 d |2x + 1| > 3 e |2x| + x . 3 c |(x − 2)(x + 6)| , 9 x+3 f ______  , 2 |x| + 1 2 a On the same set of axes, sketch the graphs of y = |3x − 2|and y = 2x + 4 b Solve, giving your answer in set notation, |3x − 2| < 2x + 4 4     2 3 a On the same set of axes, sketch the graphs of y = | x 2 − 4|and y = ______ x −1 4     2 b Solve | x 2 − 4| < ______ x −1 3−x 4 Solve the inequality ______     . 2, giving your answer in set notation. |x | + 1 Problem-solving x x E/P 5 Solve the inequality _____     , 1 − x, To sketch y = _____    rearrange it into the x+2 x+2 B giving your answer in set notation. form y = A +  _____  for constants A and B. E/P | | x+2 M01_IAL_FP2_44655_U01_001-013.indd 10 (5 marks) (5 marks) 25/04/2019 08:54 INEQUALITIES E/P CHAPTER 1 1 6 a On the same set of axes, sketch the graphs of y = _____   x − a  and y = 4∣x − a∣. 1 b Solve, giving your answer in terms of the constant a, _____   x − a , , 4∣x − a∣. E/P 4x 7 Solve ______     , x |x | + 2 E/P 8 A student attempts to solve the inequality | x 2 + x − 8 | , 4x + 2. 11 (5 marks) (3 marks) (6 marks) The working is shown below: x  2 + x − 8 = 4x + 2 ⇒ x  2− 3x − 10 = 0 and − x  2 − x + 8 = 4x + 2 ⇒ x  2 + 5x − 6 = 0 So critical values are x = −6, −2, 1, 5. Solution is: −6 , x , −2and 1 , x , 5 a Identify the mistake in the student’s answer. b Find the correct values of x for which the inequality is satisfied. (1 mark) (3 marks) Challenge f(x) = x 3 + 3 x 2 − 13x − 15 a Show that (x + 1)is a factor of f(x). b Find the other factors and hence sketch the graph of y = f(x). c Hence or otherwise, solve the inequality | x 3 + 3 x 2 − 13x − 15| < x + 5 Chapter review 1 E 1 2 1 Use algebra to solve _____     < __     x−2 x (6 marks) E 2x2 − 2   . 4 2 Use algebra to solve _______   x+2 (4 marks) E 2x2 − 3x + 4 3 Use algebra to solve ___________     , 4x − 2 x−2 (4 marks) E E x+1 1 4 Use algebra to find the set of values of x for which ______     , _____     giving your answer 2x − 3 x − 3 (6 marks) in set notation. (x + 3)(x + 9)      . 3x − 5, giving 5 Use algebra to find the set of values of x for which ____________   x−1 your answer in set notation. M01_IAL_FP2_44655_U01_001-013.indd 11 (4 marks) 25/04/2019 08:54 12 CHAPTER 1 INEQUALITIES 6 a Sketch, on the same axes, the line with equation y = 2x + 2and the graph with 2x + 4 equation y =  ______  x−2 2x +4 _____   b Solve the inequality 2x + 2 .   x−2 P 2x − 4 7 a Sketch, on the same set of axes, the graph with equation y = ______   2  and the line with x −2 equation y = 2 − 4x 2x − 4   b Solve the inequality 2 − 4x ,  ______ x2 − 2 P E/P E/P x−2 2 8 a Sketch, on the same set of axes, the graphs with equations y = ______     and y = _____     x+2 3x − 1 x−2 2 b Solve the inequality  ______  , _____     3x − 1 x + 2 x+1 2x − 1 9 a Sketch, on the same set of axes, the graphs with equations y = _____     and y = ______     x−2 x+4 x + 1 2x − 1 b Solve the inequality  _____  , ______     x−2 x+4 (4 marks) (3 marks) (4 marks) (3 marks) 10 Solve the inequality ∣x2 − 7∣ , 3(x + 1) x2 11 Solve the inequality  ______  , 1 ∣x∣ + 6 E 12 Find the set of values of x for which ∣x − 1∣ . 6x − 1 (3 marks) E 13 Find the complete set of values of x for which ∣x2 − 2∣ . 2x (3 marks) E 14 a Sketch, on the same set of axes, the graph with equation y = ∣2x − 3∣, and the line with equation y = 5x − 1 (3 marks) b Solve the inequality ∣2x − 3∣ , 5x − 1 E 15 a Use algebra to find the exact solution of ∣2x2 + x − 6∣ = 6 − 3x (3 marks) (4 marks) b On the same diagram, sketch the curve with equation y = ∣2x2 + x − 6∣ and the line with equation y = 6 − 3x (3 marks) c Find the set of values of x for which ∣2x2 + x − 6∣ . 6 − 3x E/P (1 mark) 16 a On the same diagram, sketch the graphs of y = ∣x2 − 4∣ and y = ∣2x − 1∣, showing the coordinates of the points where the graphs meet the x-axis. (4 marks) b Solve ∣x2 − 4∣ = ∣2x − 1∣, giving your answers in surd form where appropriate. c Hence, or otherwise, find the set of values of x for which ∣x2 − 4∣ . ∣2x − 1∣ M01_IAL_FP2_44655_U01_001-013.indd 12 (4 marks) (1 mark) 25/04/2019 08:54 INEQUALITIES CHAPTER 1 13 Challenge Solve the inequality | x 2 − 5x + 2 | . |x − 3| Give your answer in set notation, expressing any critical values as surds where appropriate. Summary of key points 1 To solve an inequality involving algebraic fractions: • Step 1: multiply by an expression squared to remove fractions • Step 2: rearrange the inequality to get 0 on one side • Step 3: find critical values • Step 4: use a sketch to identify the correct intervals. 2 When solving an inequality involving < or >, check whether or not each of your critical values should be included in the solution set. 3 If you can sketch the graphs of y = f(x) and y = g(x) then you can solve an inequality such as f(x) , g(x) by observing when one curve is above the other. The critical values will be the solutions to the equation f(x) = g(x). 4 When solving inequalities that include modulus signs, it is often useful to sketch a graph. M01_IAL_FP2_44655_U01_001-013.indd 13 25/04/2019 08:54 2 SERIES 2.1 Learning objectives After completing this unit you should be able to: ● Understand and use the method of differences to sum finite series → pages 15–19 Prior knowledge check 1 Find the sums of the following series. 18 a ∑(99 − 4n) n=1 16 b ∑_12 (3)n−1 n=6 ← Pure 2 Section 5.6 2 a Show that n ∑(r2 + 2r + 3) = _16 n(2n2 + 9n + 25) r=1 30 b Hence find ∑(r2 + 2r + 3) 10 ← Further Pure 1 Section 8.1 M02_IAL_FP2_44655_U02_014-021.indd 14 Series are widely used in the mathematics of disciplines such as physics, computer science, statistics and finance. The method of differences allows the sums of many finite series to be calculated quickly and easily. 25/04/2019 08:55 SERIES CHAPTER 2 15 2.1 The method of differences You can use the method of differences to find the sum of a finite series. ■ If the general term, ur, of a series can be expressed in the form f(r) − f(r + 1) n n r=1 r=1 then ∑  ur = ∑  (f( r) − f(r + 1)) u1 = f(1) − f(2) u2 = f(2) − f(3) u3 = f(3) − f(4) ⋮ un = f(n) − f(n + 1) so n r = f(1) − f(n + 1) Then adding ∑  u r=1 Example You can also start with ur written in the form f(r + 1) − f(r). After adding and cancelling, n you get ∑   u   r = f(n + 1) − f(1) r=1 u 1 + u 2 = f(1) − f(2) + f(2) − f(3) = f(1) − f(3) The f(2) terms cancel. By summing u  1 + u 2 + … + un  all terms cancel except the very first term, f(1), and the very last term, f(n + 1). 1 a Show that 4r 3 ≡ r 2(r + 1)2 − (r − 1)2r 2 b Hence prove, by the method of differences, that n _1 ∑ r 3 =   4 n2(n + 1)2 r=1 a r 2(r + 1)2 − (r − 1)2r 2 ≡ r 2(r 2 + 2r + 1) − (r 2 − 2r + 1)r 2 ≡ r 4 + 2r 3 + r 2 − r 4 + 2r 3 − r 2 ≡ 4r 3 Start with the RHS. Expand and simplify the brackets. n b Consider ∑  (r 2(r + 1)2 − (r − 1)2r 2) r=1 r = 1: r = 2: r = 3: ⋮ r = n: Let Sum of terms 12(2)2 − (0)212 22(3)2 − (1)222 33(4)2 − (2)223 n2(n + 1)2 − (n − 1)2n2 = n2(n + 1)2 n Then 4 ∑  r 3 = n2(n + 1)2 r=1 n So ∑  r 3 = __   41  n2(n + 1)2 r=1 M02_IAL_FP2_44655_U02_014-021.indd 15 All the terms cancel except the first and last. Watch out When using the method of differences, be sure to write out enough terms to make it clear which terms cancel. When you cancel terms, make sure that they can still be clearly read. You could cross them out in pencil. The same result could be proved by mathematical induction. ← Further Pure 1 Section 8.1 25/04/2019 08:55 16 CHAPTER 2 Example SERIES 2 n 1 1 1 1 Verify that _______       _____       __  and hence find ∑  _______  using the method of differences. r(r + 1) ≡ r − r + 1 r(r + 1) r=1 r   − _____   1   ≡ ________   r + 1 −   r(r + 1) r r+1 1 _ ≡ _______   1   r(r + 1) n Write as a single fraction. Simplify. n   1   ≡ ∑  (_  1 − _____   1  ) ∑   _______ r+1 r = 1 r(r + 1) r=1 r Let r = 1: _1  − __   1   1 2 1   − __ r = 2: __   1   2 3 1   − __ r = 3: __   1   3 4 ⋮ All terms cancel except the first and last.   1   r = n: __1 − _____ n n+1 n 1   = 1 − _____   1   So ∑   _______ n+1 r = 1 r(r + 1) n+1−1 =  ________   n+1 Put over a common denominator. n =  _____  n+1 Example 3 n 1 Find ∑   ______ 2 − 1 using the method of differences. 4r r=1 1 _______ 1     ≡ ______________      (2r + 1)(2r − 1)   4r 2 − 1 1 ______________      ≡ ______   A   + ______   B   (2r + 1)(2r − 1) 2r + 1 2r − 1 A(2r − 1) + B(2r + 1) ___________________ ≡           (2r + 1)(2r − 1) so 1 ≡ A(2r − 1) + B(2r + 1) M02_IAL_FP2_44655_U02_014-021.indd 16 Use the difference of two squares to factorise the denominator. Split the fraction into partial fractions. ← Pure 4 Section 2.1 Add the fractions. Set numerators of both sides equal to each other. 25/04/2019 08:55 SERIES CHAPTER 2 Let r = _  21  : 1=0+B×2 Let r = −  _21  : 1 = A × −2 B=_  21   A = −  _21   −  _21   ______ 1 _______ So  17 Put values of r in to find A and B. _1  2  ______   +       ≡   4r 2 − 1 2r + 1 2r − 1   1   − ______ ≡_  21  (______   1   2r − 1 2r + 1) n n 1   = _ ⇒ ∑   _______   1   − ______   1  )  21   ∑  (______ 2 2r + 1 r = 1 4r − 1 r = 1 2r − 1 Let r = 1: _11  − _  31   r = 2: _31   − _  51   _1 _1 r = 3: 5  −  7  ⋮ 1   − ______ r = n: ______   1   2n − 1 2n + 1 All terms cancel except the first and last. 1 1 Substitute the values of r into ______     − ______     2r − 1 2r + 1 only. The  _12 is only required later. n 1   = _   1  ) So ∑   _______  21  (1 − ______ 2 2n + 1 r = 1 4r − 1 2n + 1 − 1 =  _21  ( _________   2n + 1 ) n = ______     2n + 1 If the general term of the series is given in the form f(r) − f(r + 2), you need to adapt the method of differences to consider the terms f(1), f(2), f(n + 1) and f(n + 2). Example 4 2 a Express ____________     in partial fractions. (r + 1)(r + 3) b Hence prove by the method of differences that n n(an + b) 2 ∑   ____________     = ______________      (r + 1)(r + 3) 6(n + 2)(n + 3) r =1 where a and b are constants to be found. 30 2 c Find the value of ∑   ____________    to 5 decimal places. r = 21 (r + 1)(r + 3) M02_IAL_FP2_44655_U02_014-021.indd 17 25/04/2019 08:55 18 CHAPTER 2 SERIES 2 A   + _____ a ____________     ≡  _____   B   (r + 1)(r + 3) r + 1 r + 3 Split into partial fractions. A(r + 3) + B(r + 1)    ≡ _________________        (r + 1)(r + 3) ⇒ Add the fractions. 2 ≡ A(r + 3) + B(r + 1) Compare numerators. Let r = −3: 2 = −2B ⇒ B = −1 2 = 2A ⇒ A = 1 2 1   − _____ Therefore ____________      ≡  _____   1   (r + 1)(r + 3) r + 1 r + 3 Let r = −1: b Using the method of differences,  41   when r = 1: _21   − _ r = 2: _31   − _  51   _1 Cancel terms. __1 r = 3: 4  −  6  Problem-solving ⋮ 1 _____ 1 r = n − 1: __ n −  n + 2  r = n:  1 _____ n ∑  (f(r) − f(r + 2))= r=1 f(1) + f(2) − f(n + 1) − f(n + 2) 1 _____   −     n+1 n+3 n 5 2     =  __  − _____ So ∑   ____________   1   − _____   1   6 n+2 n+3 r = 1 (r + 1)(r + 3) Put these four terms over a common denominator. 5(n + 2)(n + 3) − 6(n + 3) − 6(n + 2)          = _________________________________________________     6(n + 2)(n + 3) 5n2 + 25n + 30 − 6n − 18 − 6n − 12          =  _________________________________________________   6(n + 2)(n + 3) 5n2 + 13n       = ______________     6(n + 2)(n + 3) n(5n + 13)       = ______________     6(n + 2)(n + 3) Factorise. So a = 5 and b = 13. 30 30 20 2 2 2     = ∑   ____________     − ∑    ____________     c ∑   ____________ r = 1 (r + 1)(r + 3) r = 1 (r + 1)(r + 3) r = 21 (r + 1)(r + 3) 20 30 r=1 r=1 Subtract ∑  from ∑  30(5 × 30 + 13) _________________ 20(5 × 20 + 13)          = _________________     −        6(30 + 2)(30 + 3) 6(20 + 2)(20 + 3) M02_IAL_FP2_44655_U02_014-021.indd 18 815 565   = _____     − ____   1056 759 Evaluate. 665 = _______     = 0.02738 to 5 d.p. 24 288 Give answer to 5 d.p. 25/04/2019 08:55 SERIES CHAPTER 2 Exercise 19 2A 1 a Show that r ≡ _ 2 (r(r + 1) − r(r − 1)). 1 n n __ b Hence show that ∑ r =   2 (n + 1) using the method of differences. r=1 E 1 1 1 2 Given  _____________     ≡ ________          − _____________ r(r + 1)(r + 2) 2r(r + 1) 2(r + 1)(r + 2) n 1 _____________    using the method of differences. find ∑   r(r + 1)(r + 2) r=1 E/P 1 3 a Express  _______ in partial fractions. r(r + 2) n 1 _______ b Hence find the sum of the series ∑   r(r + 2) using the method of differences. r=1 E 1 4 a Express ____________     in partial fractions. (r + 2)(r + 3) (5 marks) (1 mark) (5 marks) (1 mark) n 1 ____________ b Hence find the sum of the series ∑  using the method of differences. (5 marks)      (r + 2)(r + 3) r=1 E/P E P 1 1 r     − _______   5 a Show that _______     ≡ __   (r + 1)! r! (r + 1)! n r _______ b Hence find ∑   (r + 1)!  r=1 (2 marks) (5 marks) n 1 _______ 1 2r + 1 2r + 1 __ ________   ≡     −     , find 6 Given that  _________ ∑   r 2(r + 1)2  r 2(r + 1)2 r 2 (r + 1)2 r=1 (6 marks) n n 11 7 a Use the method of differences to prove that ∑   ___________   = ______     , where a and b are an + b r = 1 (2r + 3)(2r + 5) constants to be found. b Prove your result from part a using mathematical induction. E/P n n(an + b) 8 8 Prove that ∑     ___________     = ______________       where a and b are constants to be found. (3n + 1)(3n + 4) r = 1 (3r – 2)(3r + 4) (6 marks) Hint This question can be answered using either the method of differences or proof by mathematical induction. In the exam, either method would be acceptable. If you use proof by induction, you will need to substitute values of n to find the values of a and b. E/P n 9 Prove that ∑ ( r + 1)  2 − ( r − 1)  2 = an(n + 1), where a is a constant to be found. (4 marks) r=1 M02_IAL_FP2_44655_U02_014-021.indd 19 25/04/2019 08:55 20 CHAPTER 2 SERIES Chapter review 2 E/P 2 1 a Express  __________ in partial fractions. (r + 2)( r + 4) n 7n  2 + 25n 2       b Hence show that  ∑   __________  = _____________ r = 1 (r + 2)(r + 4) 12(n + 3)(n + 4) E/P 4 2 a Express ____________      in partial fractions. (4r − 1)( 4r + 3) (1 mark) (5 marks) (2 marks) b Using your answer to part a and the method of differences, show that n 4   ∑   ____________     = r = 1 (4r − 1)( 4r + 3) 4n ________   (   34n + 3) 200 4 c Evaluate   ∑   ____________    giving your answer to 3 significant figures. ( r = 100 4r − 1)( 4r + 3) E 3 a Show that ( r + 1)  3 − ( r − 1)  3 = 6r  2 + 2 (3 marks) (2 marks) (2 marks) b Using the result from part a and the method of differences, show that n   ∑ r  2 =  _6 n(n + 1)(2n + 1) 1 (5 marks) r=1 E/P E/P n n(an + b) 4 4 Prove that   ∑ __________     = ____________       where a and b are constants r = 1 (r + 1)(r + 3) 3(n + 2)(n + 3) to be found. 2n 5 Prove that   ∑ ((r + 1)3 − (r − 1)3)= an  3 + bn  2 + cn + d, where a, b, c and d are r=n constants to be found. E/P (5 marks) (5 marks) n 3 an 6 a Prove that ∑ ____________       = ______     where a, b, and c are constants (3r + 1)( 3r + 4) bn + c r=1 (5 marks) to be found. 3 3(n + 1)     = ______________       b Hence, or otherwise, show that ∑   ___________ ( 23n + 1)( 3n + 2) r = n (3r + 1)(3r + 4) 2n (4 marks) n 2r + 1 1     = 1 − _____     7 Robin claims that ∑  _______ n + 1 r(r + 1) r=1 His workings are shown below. Explain the error that he has made. Using partial fractions: 2r + 1 _______   B A       ≡ __     + _____ r r+1 r(r + 1) Therefore 2r + 1 ≡ A(r + 1) + Br M02_IAL_FP2_44655_U02_014-021.indd 20 25/04/2019 08:55 SERIES CHAPTER 2 21 So A = 1 and B = 1. Using the method of differences, f(1) = 1 + __   21 f(2) = __   21 + __   31 f(3) = __   31 + __   41 ⋮ 1 1 f(n − 1) =  _____  + __     n−1 n 1 1   f(n) = __     + _____   n n+1 n 2r + 1 1  r(r Summing the differences: ∑ _______   = 1 − _____     n+1 + 1) r=1 E/P E/P (2 marks) an + b 3 1 1 1 1 8 Show that _____         where a and b are   + _____     + _____     + … + _______     = __     − _____________ 1×3 2×4 3×5 n(n + 2) 4 2(n + 1 ) (n + 2) constants to be found. (6 marks) 4 9 a Express ____________      in partial fractions. (2r + 1)( 2r + 5) 25 4      to 4 decimal places. b Find the value of ∑  ____________ r = 16 (2r + 1)( 2r + 5) (3 marks) (5 marks) Challenge 30 1 a Given that ∑   ln(1 + _____     = ln k, where k is an integer, find k. r + 2) r=1 n n(an2 + bn + c) 18 b Given that ∑    _______  =   _________________        find a, b, and c. (n + 1)(n + 3)(n + 3) r = 1 r(r + 3) Summary of key points If the general term, ur , of a series can be expressed in the form f(r) − f(r + 1) n n r=1 r=1 then   ∑u  r =   ∑(  f(r) − f(r + 1)) so u1 = f(1) − f(2) u2 = f(2) − f(3) u3 = f(3) − f(4) ⋮ un = f(n) − f(n + 1) n   ur = f(1) − f(n + 1) Then adding ∑ r=1 M02_IAL_FP2_44655_U02_014-021.indd 21 25/04/2019 08:55 3 COMPLEX NUMBERS 3.1 3.2 Learning objectives After completing this unit you should be able to: ● Express a complex number in exponential form → pages 23–26 ● Multiply and divide complex numbers in exponential form → pages 26–29 ● Understand de Moivre’s theorem → pages 29–32 ● Use de Moivre’s theorem to derive trigonometric identities → pages 32–36 ● Know how to solve completely equations of the form − a − ib = 0, giving special attention to cases where a = 1 and b = 0 → pages 37–42 zn Prior knowledge check 1 __ π π z = 4 + 4i√3 and w = 2 (cos __ + i sin __). 6 6 Find: b arg(z) c |zw| d arg(zw) z f arg(__ ← Further Pure 1 Sections 1.5, 1.6 w) a |z| z e __ w | | 2 f(z) = z4 + 4z3 + 9z2 + 4z + 8 Given that z = i is a root of f(z) = 0, show all the roots of f(z) = 0 on an Argand diagram. ← Further Pure 1 Section 1.4 3 Use the binomial expansion to find the n4 term in the expansion of (2 + n)9. ← Pure 2 Section 4.3 M03_IAL_FP2_44655_U03_022-045.indd 22 The relationships between complex numbers and trigonometric functions allow electrical engineers to analyse oscillations of voltage and current in electrical circuits more easily. 25/04/2019 08:56 COMPLEX NUMBERS CHAPTER 3 23 3.1 Exponential form of complex numbers You can use the modulus−argument form of a complex number to express it in the exponential form: z = reiθ. Links The modulus−argument form of a complex number is z = r(cos θ + i sin θ), where r = |z| and θ = arg z. You can write cos θ and sin θ as infinite series of powers of θ: θ 2 θ 4 __ θ6 (−1)r θ 2r     + __     −     + … + _______     + … cos θ = 1 − __ 2! 4! 6! (2r)! ← Further Pure 1 Section 1.6 (1) θ 3 θ 5 __ θ7 (−1)r θ 2r + 1     + __     −     + … +  _________  + … (2) sin θ = θ − __ 3! 5! 7! (2r + 1)! You can also write ex, x ∈ ℝ, as a series expansion in powers of x. Links These are the Maclaurin series expansions of sin θ, cos θ and ex . 2 x3  +  __ x4  +  __ x5  + … +  __ xr  + …  x   +  __ ex = 1 + x + __ r! 2! 3! 4! 5! → Further Pure 2 Section 7.2 You can use this expansion to define the exponential function for complex powers, by replacing x with a complex number. In particular, if you replace x with the imaginary number iθ, you get (iθ )2 (iθ )3 ____ (iθ )4 (iθ )5 ____ (iθ )6     + ____ eiθ = 1 + iθ + ____     +     + ____     +     + … 2! 3! 4! 5! 6! 2 2 3 3 4 4 5 5 6 6   i θ   + ____  i θ   + ____  i θ   + ____  i θ   + … = 1 + iθ + ____   i θ   + ____ 2! 3! 4! 5! 6! 2 3 4 5 6  θ   − ___   iθ   + __  θ   + …  θ   + ___  iθ   − __ = 1 + iθ − __ 2! 3! 4! 5! 6! 2 4 6 3 5  θ   + __  θ   + … + i θ − __  θ   + __   θ   − __   θ   − … = 1 − __ 2! 4! 6! 3! 5! ( ) ( ) By comparing this series expansion with (1) and (2), you can write eiθ as eiθ = cos θ + i sin θ This formula is known as Euler’s relation. It is important for you to remember this result. ■ You can use Euler’s relation, eiθ = cos θ + i sin θ, to write a complex number z in exponential form: z=r where r = |z| and θ = arg z eiθ M03_IAL_FP2_44655_U03_022-045.indd 23 Notation Substituting θ = πinto Euler’s relation yields Euler’s identity: eiπ + 1 = 0 This equation links the five fundamental constants 0, 1, π , e and i, and is considered an example of mathematical beauty. 25/04/2019 08:56 24 CHAPTER 3 Example COMPLEX NUMBERS 1 Express in the form r eiθ, where −π , θ < π. __ π π π π a z=√ 2  cos ___     + i sin ___ b z = 5 cos __     − i sin __          8 8 10 10 ( ( ) ) __ π π a z = √2 cos __ 10 + i sin __ 10 __ π So r = √2 and θ = __ 10 ( ) Compare with r(cos θ + i sin θ). πi __ __ Therefore, z = √2 e10 z = reiθ π π b z = 5 cos __− i sin __ 8 8 ( ) ( ( ) Problem-solving ( − __8π )) π z = 5 cos − __ + i sin 8 Use cos (−θ) = cos θ and sin (−θ) = −sin θ. π So r = 5 and θ = − __ 8 Compare with r(cos θ + i sin θ). πi − __   Therefore, z = 5e 8 Example z = reiθ 2 Express z = 2 − 3i in the form r eiθ, where −π , θ < π. Im O α Sketch the Argand diagram, showing the position of the complex number. 2 Re r 3 z = 2 – 3i ___________ __ r = |z| = √22 + (−3)2  = √ 13  θ = arg z = −arctan (__   32 ) = −0.983 (3 s.f.) __ Therefore, z = √13 e−0.983i M03_IAL_FP2_44655_U03_022-045.indd 24 Here z is in the fourth quadrant so the required argument is −α. Find r and θ. z = reiθ 25/04/2019 08:56 COMPLEX NUMBERS CHAPTER 3 Example 25 3 3πi __ ____     Express z = √ 2 e 4 in the form x + iy, where x, y ∈ ℝ. __ ___ __     3π z=√ 2 e 4 , so r = √2  and θ = ___     4 Compare with reiθ. __ 3π 3π z=√ 2  cos  ___  + i sin  ___   4 4 Write z in modulus–argument form. 3πi ( ) __ 1__  + i  ___ 1__   = √2  − ___ √ 2  √2  ( ) Therefore, z = −1 + i Example Simplify. 4 23πi _____   Express z = 2e 5   in the form r (cos θ + i sin θ), where −π , θ < π. 23πi  _____   5 , so 23π r = 2 and θ = ____     5 23π 13π 13π 3π ____     − 2π = ____    , ____     − 2π = ___     5 5 5 5 3π ___    is in the range −π , θ < π 5 3π 3π So z = 2 cos  ___  + i sin  ___   5 5 Compare with reiθ. z = 2e ( Problem-solving cos θ = cos (θ + 2π) and sin θ = sin (θ + 2π). 23π Subtract multiples of 2π from ____    until you find a 5 value in the range −π , θ < π. ) Write z in the form r (cos θ + i sin θ ). Example 5 Use eiθ = cos θ + i sin θ to show that cos θ = _ 2 (eiθ + e−iθ ). 1 eiθ = cos θ + i sin θ −iθ (1) i(−θ) e = e = cos (−θ) + i sin (−θ) So e = cos θ − i sin θ −iθ (2) e + e = 2 cos θ iθ e iθ + e−iθ ⇒ ________     = cos θ 2 ), as required. Hence, cos θ =  _21   (eiθ + e−iθ M03_IAL_FP2_44655_U03_022-045.indd 25 Use cos (−θ ) = cos θ and sin (−θ ) = −sin θ. −iθ Add (1) and (2). Divide both sides by 2. 25/04/2019 08:56 26 CHAPTER 3 Exercise COMPLEX NUMBERS 3A 1 Express in the form r eiθ , where −π , θ < π. Use exact values of r and θ where possible, or values to 3 significant figures otherwise. __ a −3 b 6i c −2√3  − 2i d −8 + i e 2 − 5i f −2√3  + 2i√3  i 2 (cos  __π5  − i sin  __π5   ) __ π π g√8 (cos  __  + i sin  __   ) 4 4 __ π π h 8cos  __  − i sin  __   6 6 ( ) 2 Express in the form x + iy where x, y ∈ ℝ. πi __     __ b 4eπi ae 3 πi __     πi − __      ge−πi h 3√2 e fe 6 4πi − ____  3 3π i −___  4 __ i 8e 3 Express in the form r (cos θ + i sin θ), where −π , θ < π.     P b 4e 17πi ____       5πi ____ e 3e 2 16πi ____ πi __ c 3√2  e 4 d 8e 6 ae 13 __   5 9πi − ___  8 c 5e 1 4 Use e iθ= cos θ + i sin θ to show that sin θ = __     ( e iθ− e−iθ ) 2i 3.2 Multiplying and dividing complex numbers You can apply the modulus−argument rules for multiplying and dividing complex numbers to numbers written in exponential form. Recall that, for any two complex numbers z  1 and z  2, z z = | z 1||z 2| ●arg (z  1 z  2) = arg (z  1) + arg (z  2) z 1 |z 1| ● __   z    = ___     |z 2| 2 z  1 __ ●arg     = arg (z  1)− arg (z  2) (z 2) ●|   1   2| | | Links These results can be proved by considering the numbers z  1 and z  2 in the form r(cos θ + i sin θ) and using the addition formulae for cos and sin. ← Further Pure 1 Section 1.6 Applying these results to numbers in exponential form gives the following result: ■ If z1 = r1e  iθ1  and z2 = r2e  iθ2  , then: • z1z2 = r1r2e  i(θ1  + θ2  ) z1 __ r1 i(θ  − θ  ) 1 2   =   •  __ z2 r2  e  M03_IAL_FP2_44655_U03_022-045.indd 26 Watch out You cannot automatically assume the laws of indices work the same way with complex numbers as with real numbers. This result only shows that they can be applied in these specific cases. 25/04/2019 08:56 COMPLEX NUMBERS Example CHAPTER 3 27 6 __ __ πi πi __ a Express 2e 6 × √ 3 e 3 in the form x + iy. b z = 2 + 2i, Im(zw) = 0 and |zw| = 3|z| Use geometrical reasoning to find the two possibilities for w, giving your answers in exponential form. __ πi __ __ __ πi a 2e 6 × √ 3  e 3 = (2 × √ 3 ) ei ( 6   +  3 ) __ __ π __ π z 1z 2 = r 1r  2e  i(θ1  +θ2  ) __ πi = 2√3  e 2 __ π π = 2√3 (cos __  + i sin __ ) 2 2 = __ 2√3 (0 __ Convert the complex number to modulus− argument form. + i) = 2i√3  b |zw| = 3|z| ⇒ |w| = 3 2 π    ) = __     arg z = arctan (__ 2 4 |zw| = |z||w| = 3|z|. Im(zw) = 0 so arg (zw) = 0 or π 3π π     or −  __  So arg w = ___ 4 4 Im 3π wz lies on the real axis, so z is rotated ___     4 π __ clockwise or    anticlockwise when multiplied by w. 4 z = 2 + 2i 3π 4 zw2 π zw1 π 4 4 O ___ πi Re ___ 3πi w1 = 3e− 4  and w2 = 3e    4 Example 7 π π 2(cos  ___  + i sin  ___ ) 12 12 __________________ Express          in the form reiθ. __ 5π 5π ___ ___ √  2 (cos     + i sin    ) 6 6 π π __ πi 2(cos  ___  + i sin  ___ ) 12 12 2e 12 ___________________ ______           =   __ ___   5πi __ 5π 5π ___ ___ √2 e 6 √ 2 (cos     + i sin    ) 6 6 5π 2 i __π −__ = ___   __  e( 12   6 ) √ 2  __ r   1  r    e  i(θ1  −θ2  )  z    = __ 2 2 z  1 __ 3πi − ___ = √2  e   4 M03_IAL_FP2_44655_U03_022-045.indd 27 Convert the numerator and denominator to exponential form. Simplify. 25/04/2019 08:56 28 CHAPTER 3 COMPLEX NUMBERS 3B Exercise 1 Express in the form x + iy, where x, y ∈ ℝ. πi __ __ ___ 2πi __ πi __ ae   3 × e   4 πi __ 7πi ___ c√2 e   3 × e −   3 × 3e   6 √5 e  iθ × 3e  3iθ b 2 Express in the form x + iy where x, y ∈ ℝ. __ ___ 3πi   2 ___ a  ___   9πi 8e   2 __ √ 3 e   7 _____ 7πi 2e ____ 15πi ___ √ 2 e −  6 ______ c    b  ___ 2πi 4e −  7 3 Express in the form reiθ __ ___ 19πi   × √ 2  e   3 2e   3 πi __ )( ( ) 8π 8π 3π 3π bcos  ___  + i sin  ___   cos  ___  + i sin  ___   11 11 11 11 a (cos 2θ + i sin 2θ)(cos 3θ + i sin 3θ) π π c 3   + i sin  ___   ) (cos  __π4  + i sin  __π4   ) × 2(cos  ___ 12 12 __ __ π π π π d√6  cos (− ___   ) + i sin (−  ___   )  × √3  cos  __  + i sin  __   3 3 12 12 ( ) 4 Express in the form reiθ ( ) π π √ 2 (cos  __  + i sin  __ ) 2 2 ________________ b       π π   _1 __ __ __ cos 5θ + i sin 5θ a  _____________       cos 2θ + i sin 2θ   2 (cos     + i sin    ) 4 4 π π 3(cos  __  + i sin  __ ) 3 3        c  _________________ 5π 5π 4(cos  ___  + i sin  ___ ) 6 6 __ __ 7π     5 z and w are two complex numbers where z = −9 + 3i √3 , |w | = √ 3  and arg w = ___ 12 Express in the form reiθ, where −π , θ < π. z a z b w c zw d __ w P 6 Use the exponential form for a complex number to show that (cos 9θ + i sin 9θ)(cos 4θ + i sin 4θ) ____________________________         ≡ cos 6θ + i sin 6θ cos 7θ + i sin 7θ E/P __ z   2 z   2 7 z = 1 + i√3 , Re(__   w ) = 0 and __   w  = |z| | | Use geometrical reasoning to find the two possibilities for w, giving your answers in exponential form. (4 marks) E/P 8 a Evaluate (1 + i)2, giving your answer in exponential form. nπi _n ___ b Use mathematical induction to prove that (1 + i)  n = 2 2 e 4 for n ∈ ℤ+. c Hence find (1 + i)16. P (2 marks) (4 marks) (1 mark) 9 Use Euler’s relation for eiθ and e−iθ to verify that cos2 θ + sin2 θ ≡ 1. M03_IAL_FP2_44655_U03_022-045.indd 28 25/04/2019 08:56 COMPLEX NUMBERS CHAPTER 3 29 Watch out Challenge You cannot assume that the laws of indices will apply to complex numbers. Prove these results using only the properties a Given that n is a positive integer, prove by induction that (re  iθ)  n= r  ne  inθ 1 b Given further that z  −n = __   n  for all z ∈ ℂ, show that z   ( re  iθ)  −n= r  −ne  −inθ z 1z  2 = r 1r  2e  i(θ1  +θ2  ) r  1 __ i(θ1  −θ2  ) z    2  = r    2  e  z   1 __ 3.3 De Moivre’s theorem You can use Euler’s relation to find powers of complex numbers given in modulus−argument form. ( r(cos θ + i sin θ)) 2 = ( re  iθ)  2 = re  iθ × re  iθ = r  2 e  i2θ = r  2( cos 2θ + i sin 2θ) Similarly, ( r(cos θ + i sin θ))  3 = r  3(cos 3θ + i sin 3θ), and so on. The generalisation of this result is known as de Moivre’s theorem: ■ For any integer n, n (r(cos θ + i sin θ))  n= r   (cos nθ + i sin nθ) You can prove de Moivre’s theorem quickly using Euler’s relation. ( r(cos θ + i sin θ))  n= ( re  iθ)  n = r  n e  inθ = r  n(cos nθ + i sin nθ) You can also prove de Moivre’s theorem for positive integer exponents directly from the modulus− argument form of a complex number using the addition formulae for sin and cos. This step is valid for any integer exponent n. ← Exercise 3B Challenge Links This proof uses the method of proof by induction. ← Further Pure 1 Section 8.1 1. Basis step n = 1; LHS = (r(cos θ + i sin θ))1 = r(cos θ + i sin θ) RHS = r1(cos 1θ + i sin 1θ) = r(cos θ + i sin θ) As LHS = RHS, de Moivre’s theorem is true for n = 1. 2. Assumption step Assume that de Moivre’s theorem is true for n = k, k ∈ ℤ + : (r(cos θ + i sin θ))k = rk (cos kθ + i sin kθ) M03_IAL_FP2_44655_U03_022-045.indd 29 25/04/2019 08:56 30 CHAPTER 3 COMPLEX NUMBERS 3. Inductive step When n = k + 1, (r(cos θ + i sin θ))k + 1 = (r(cos θ + i sin θ))k × r(cos θ + i sin θ) = r k(cos kθ + i sin kθ) × r(cos θ + i sin θ) By assumption step = r k + 1(cos kθ + i sin kθ)(cos θ + i sin θ) = r k + 1((cos kθ cos θ − sin kθ sin θ) + i (sin kθ cos θ + cos kθ sin θ)) = r k + 1(cos(kθ + θ) + i sin(kθ + θ)) By addition formulae = r k + 1(cos((k + 1)θ) + i sin((k + 1)θ)) Therefore, de Moivre’s theorem is true when n = k + 1. 4. Conclusion step Links The corresponding proof for negative integer exponents is left as an exercise. If de Moivre’s theorem is true for n = k, then it has been shown to be true for n = k + 1. As de Moivre’s theorem is true for n = 1, it is now proven to be true for all n ∈ ℤ+ by mathematical induction. Example → Exercise 3C Challenge 8 9π 9π     + i sin  ___   ) cos ___ (_________________ 17 17 Simplify     2π 2π (cos ___     − i sin ___      ) 17 17 5 3 ( ( Problem-solving ) ) 9π 9π 5 cos  ___  + i sin  ___   17 17 __________________        2π 2π 3 ___ ___ cos     − i sin      17 17 You could also show this result by writing both numbers in exponential form: 9π 9π 5 cos  ___  + i sin  ___   17 17 _______________________     =         2π 2π 3 cos−___      + i sin −___        17 17 ( e ) ____ e i(___ − − __ ______     =   ___   = e   17 (   17 )) = e  3πi = e  πi = −1 6πi 3 45π 45π cos  ____  + i sin  ____  17 17 ____________________     =        6π 6π ___ ___ cos (−θ) = cos θ and sin (−θ) = −sin θ ( ( ( ) ( cos−  9πi 5 ___ ) 17 )  17 (    + i sin −  )    17 45π 45π 6π 6π     − (− ___ )  + i sin ____     − ( − ___ )  = cos____ 17 17 17 17 ( 51π ____ = cos   ) (   17 45π 6π (e   17 ) e 2πi − ___ )) ( 45πi ___ ) −   17 Apply de Moivre’s theorem to both the numerator and the denominator. z1 __     = cos (θ1 − θ2) + i sin (θ1 − θ2) z2 51π ____   + i sin     17 17 = cos 3π + i sin 3π Simplify. = cos π + i sin π Subtract 2π from the argument. = −1 + i(0) 9π 9π   + i sin  ___   ) ( cos  ___ 17 17 __________________ So        = −1 2π 2π ___ ___ (cos     − i sin      ) 17 17 5 3 M03_IAL_FP2_44655_U03_022-045.indd 30 25/04/2019 08:56 COMPLEX NUMBERS CHAPTER 3 Example 31 9 __ 7 Express ( 1 + i√ 3 ) in the form x + iy where x, y ∈ ℝ. First, you need to find the modulus and argument __ of 1 + i√3  . You may want to draw an Argand diagram to help you. Im 3 θ O 1 _________ __ Re __ r = √ 12 + (√ 3  )2   = √ 4  = 2 Find r and θ. __ ( ) π θ = arctan      = __     1 3 √3  ___ __ π π So 1 + i√3  = 2(cos  __  + i sin  __   ) 3 3 __ π π 7 (1 + i√3  )7 = ( 2(cos  __  + i sin  __ ))  3 3 7π 7π = 27 cos  ___  + i sin  ___   3 3 ( __ Write 1 + i√3 in modulus–argument form. ) Apply de Moivre’s theorem. π π = 128(cos  __  + i sin  __   ) 3 3 ( 1 __ __ ( )) √3  ___ = 128    + i       2 2 __ Subtract 2π from the argument. __ Therefore, ( 1 + i√3  ) = 64 + 64i√3  Exercise 7 3C 1 Use de Moivre’s theorem to express in the form x + iy, where x, y ∈ ℝ. π π 5 c cos  __  + i sin  __   6 6 ( ) a (cos θ + i sin θ)6 b (cos 3θ + i sin 3θ)4 π π 8 d cos  __  + i sin  __   3 3 2π 2π 5 e cos  ___  + i sin  ___   5 5 π π 15 f cos  ___  − i sin  ___   10 10 cos 5θ + i sin 5θ a __________________      2  (cos 2θ + i sin 2θ) (cos 2θ + i sin 2θ)7 b __________________   3  (cos 4θ + i sin 4θ) 1 c __________________   3  (cos 2θ + i sin 2θ) (cos 2θ + i sin 2θ)4   3  d __________________ (cos 3θ + i sin 3θ) cos 5θ + i sin 5θ e __________________        (cos 3θ − i sin 3θ)2 cos θ − i sin θ f __________________        (cos 2θ − i sin 2θ)3 ( ) ( ) ( ) 2 Express in the form einθ M03_IAL_FP2_44655_U03_022-045.indd 31 25/04/2019 08:56 32 CHAPTER 3 COMPLEX NUMBERS 3 Evaluate, giving your answers in the form x + iy, where x, y ∈ ℝ. ( cos  13  − i sin  13 ) _________________ 7π ___ 7π ___ ( cos   7  – i sin   7  ) _________________ 4 3π ___ a   6  4π 4π (cos  ___  + i sin  ___ ) 13 13 11π ____ ( cos   3  – i sin   3 ) __________________ 3      b    15π π 2 ____ __ (cos     + i sin    ) 7 7 4π ___ 2π ___ 7      c    10π 4π 4 ____ ___ (cos     + i sin    ) 3 3 4 Express in the form x + iy where x, y ∈ ℝ. a (1 + i)5 b (−2 + 2i)8 c (1 − i)6 d (1 − i√ ) 3 6 e( _2  − _  2  i√3   ) f(−2√3  − 2i ) __ E E E 3 __ 5 1 __ 9 __ 5 __ 5 Express (3 + i√  3 ) in the form a + bi√3  where a and b are integers. (2 marks) π π 6 w = 2( cos  __  + i sin  __ ) 6 6 Find the exact value of w4, giving your answer in the form a + ib where a, b ∈ ℝ. (2 marks) __ 3π 3π 7 z = √ 3  (cos  ___  – i sin  ___ ) 4 4 Find the exact value of z6, giving your answer in the form a + ib where a, b ∈ ℝ. (3 marks) __ 1 + i √ 3 __ in the form reiθ where r > 0 and −π , θ < π. E/P 8 a Express _______   1 − i √ 3 (3 marks) __ n 1 + i √3  __    is real b Hence find the smallest positive integer value of n for which ( _______   1 − i √3  ) (2 marks) 9 Use de Moivre’s theorem to show that (a + bi)n + (a − bi)n is real for all integers n. (5 marks) and positive. E/P Challenge Problem-solving Without using Euler’s relation, prove that if n is a positive integer, (r(cos θ + i sin θ))  −n= r  −n(cos (−nθ) + i sin (−nθ)) You may assume de Moivre’s theorem for positive integer exponents, but do not write any complex numbers in exponential form. 3.4 Trigonometric identities You can use de Moivre’s theorem to derive trigonometric identities. Applying the binomial expansion to (cos θ + i sin θ)n allows you to express cos nθ in terms of powers of cos θ, and sin nθ in terms of powers of sin θ. Links (a + b)  n= a  n+ n   C  1 a  n−1b + n   C  2 a  n−2 b  2 + … + n   C  r a  n−r b  r+ … + b  n, n ∈ ℕ n n! where n   C  r = ( r )  =  ________   ← Pure 2 Section 4.3 r!(n − r)! M03_IAL_FP2_44655_U03_022-045.indd 32 25/04/2019 08:56 COMPLEX NUMBERS CHAPTER 3 33 Example 10 Use de Moivre’s theorem to show that cos 6θ = 32 cos6 θ − 48 cos4 θ + 18 cos2 θ − 1 (cos θ + i sin θ)6 = cos 6θ + i sin 6θ = cos6 θ cos5 θ(i sin θ) Apply de Moivre’s theorem. cos4 θ(i sin θ)2 + + 2 1 6 3 3 + C3 cos θ(i sin θ) + 6C 4 cos2 θ(i sin θ)4 + 6C5 cos θ(i sin θ)5 + (i sin θ)6 6C 6C = cos6 θ + 6i cos5 θ sin θ + 15i2 cos4 θ sin2 θ + 20i3 cos3 θ sin3 θ + 15i4 cos2 θ sin4 θ + 6i5 cos θ sin5 θ + i6 sin6 θ = cos6 θ + 6i cos5 θ sin θ − 15 cos4 θ sin2 θ − 20i cos3 θ sin3 θ + 15 cos2 θ sin4 θ + 6i cos θ sin5 θ − sin6 θ Equating the real parts gives cos 6θ = cos6 θ − 15 cos4 θ sin2 θ + 15 cos2 θ sin4 θ − sin6 θ = cos6 θ − 15 cos4 θ(1 − cos2 θ) + 15 cos2 θ(1 − cos2 θ)2 − (1 − cos2 θ)3 = cos6 θ − 15 cos4 θ(1 − cos2 θ) + 15 cos2 θ(1 − 2 cos2 θ + cos4 θ) − (1 − 3 cos2 θ + 3 cos4 θ − cos6 θ) = cos6 θ − 15 cos4 θ + 15 cos6 θ + 15 cos2 θ − 30 cos4 θ + 15 cos6 θ − 1 + 3 cos2 θ − 3 cos4 θ + cos6 θ = 32 cos6 θ − 48 cos4 θ + 18 cos2 θ − 1 Apply the binomial expansion to (cos θ + i sin θ)6. Simplify. Simplify the powers of i. The real part of cos 6θ + i sin 6θ is cos 6θ. Apply sin2 θ ; 1 − cos2 θ, sin4 θ ; (sin2 θ)2 and sin6 θ ; (sin2 θ)3 Multiply out the brackets. Apply a cubic binomial expansion. Expand the brackets. Simplify. Therefore, cos 6θ = 32 cos6 θ − 48 cos4 θ + 18 cos2 θ − 1 You can also find trigonometric identities for sinn θ and cosn θ where n is a positive integer. If z = cos θ + i sin θ, then __   1z = z−1 = (cos θ + i sin θ)−1 = (cos(−θ) + i sin(−θ)) = cos θ − i sin θ Apply de Moivre’s theorem. Use cos θ = cos (−θ) and −sin θ = sin (−θ). It follows that z + __   1z = cos θ + i sin θ + cos θ − i sin θ = 2 cos θ z − __   1z = cos θ + i sin θ − (cos θ − i sin θ) = 2i sin θ M03_IAL_FP2_44655_U03_022-045.indd 33 25/04/2019 08:56 34 CHAPTER 3 COMPLEX NUMBERS Also, z n = (cos θ + i sin θ)n = cos nθ + i sin nθ By de Moivre’s theorem. 1   = z–n = (cos θ + i sin θ)–n  __ zn = (cos(−nθ) + i sin(−nθ)) Apply de Moivre’s theorem. Use cos θ = cos (−θ) and sin (−θ) = −sin θ. = cos nθ − i sin nθ It follows that   1n  = cos nθ + i sin nθ + cos nθ − i sin nθ = 2 cos nθ zn + __ z zn − __   1n  = cos nθ + i sin nθ − (cos nθ − i sin nθ) = 2i sin nθ z It is important that you remember and are able to apply these results: 1 ■ z + __   z = 2 cos θ 1 ■ zn + __   n = 2 cos nθ Notation In exponential form, these results are equivalent to: z 1 ■ zn − __   n = 2i sin nθ z 1 ■ z − __   z = 2i sin θ 1 1 cos nθ = __    (e  inθ + e  −inθ) sin nθ = __    (e  inθ − e  −inθ) 2 2i Example 11 Express cos5 θ in the form a cos 5θ + b cos 3θ + c cos θ, where a, b and c are constants. Let z = cos θ + i sin θ 5 (z + _ z1   ) = (2 cos θ)5 = 32 cos5 θ 2 3 = z5 + 5C1 z4( _z1   ) + 5C2 z3( _z1   ) + 5C3 z2( _z1   ) 5 4 + 5C 4 z( _z1   ) + (  _z1   ) 1    + 10z2 __ 1    = z5 + 5z4 _1   + 10z3 __ z2 z3 z 1    +  __ 1    + 5z __ z5 z4 ( ) () ( ) ( ) ( ) 1 Use z + __   z  = 2 cos θ. Apply the binomial expansion to 1 5 (z + __   z ) . Simplify. 10 5 __   + __ = z5 + 5z3 + 10z + __   3  +   15  z z z ( ) ( ) ( ) = z5 + __   15   + 5z3 + __   13   + 10z + _  1   z z z = 2 cos 5θ + 5(2 cos 3θ) + 10(2 cos θ) So, 32 cos5 θ = 2 cos 5θ + 10 cos 3θ + 20 cos θ ⇒ 5 cos5 θ = __  161   cos 5θ + __  16   cos 3θ + __  58  cos θ M03_IAL_FP2_44655_U03_022-045.indd 34 1 Group zn and __   n  terms. z 1 Use zn + __   n  = 2 cos nθ z 1 This is in the required form with a = __   16 , 5 5 __ _ b =   16 and c =   8 25/04/2019 08:56 COMPLEX NUMBERS CHAPTER 3 35 Example 12 a Express sin4 θ in the form d cos 4θ + e cos 2θ + f, where d, e and f are constants. ∫ π __     2 b Hence find the exact value of   sin4 θ dθ. 0 a Let z = cos θ + i sin θ 4 ( z − _ z1   ) = (2i sin θ)4 = 16i4 sin4 θ = 16 sin4 θ 2 = z4 + 4C1 z3(− _z1   ) + 4C2 z2 (− _z1   ) + 4C3 z1(− _z1   ) + ( − _z1   ) 4 3 ( ) ( ) 1    +  __ 1    + 4z(− __ z ) (z ) 1 Use z − __   z  = 2i sin θ, noting that i4 = 1 Apply the binomial expansion to 1 4 (z − __   z ) 1    = z4 + 4z3 − _1   + 6z2 __ z2 z 4 3 Simplify. 4 = z4 − 4z2 + 6 − __   2  + __   14  z z ( ) ( ) = z4 + __   14   − 4z2 + __   12   + 6 z z = 2 cos 4θ − 4(2 cos 2θ) + 6 So, 16 sin4 θ = 2 cos 4θ − 8 cos 2θ + 6 sin4 θ = __  81   cos 4θ − _   21   cos 2θ + __  38  ⇒ ∫ b   π __     2 0 sin4 θ dθ ∫ π __     2 1 =   (__   8  cos 4θ − __   21   cos 2θ + __   38 ) dθ 0 __ π 1 = [___   32 sin 4θ − __   41 sin 2θ + __   38 θ] 02 π 1 = ( ___   32   sin 2π − __   41   sin π + __   83 (__     − 0 2 )) 3π = 0 − 0 + ___     16 3π = ___     16 Exercise 1 Group zn and __   n  terms. z 1 Use zn + __   n  = 2 cos nθ z This is in the required form with d = _  18 , e = −  _12 and f = _  38  Use the answer from part a. 1 cos kθ integrates to __     sin kθ. k 3D Use de Moivre’s theorem to prove the trigonometric identities: P 1 a sin 3θ ≡ 3 sin θ − 4 sin3 θ c cos 7θ ≡ 64 cos7 θ − 112 cos5 θ + 56 cos3 θ − 7 cos θ e sin5 θ ≡ __  16  (sin 5θ − 5 sin 3θ + 10 sin θ) 1 M03_IAL_FP2_44655_U03_022-045.indd 35 b sin 5θ ≡ 16 sin5 θ − 20 sin3 θ + 5 sin θ d cos4 θ ≡ _ 8  (cos 4θ + 4 cos 2θ + 3) 1 25/04/2019 08:56 36 CHAPTER 3 COMPLEX NUMBERS E/P 2 a Use de Moivre’s theorem to show that cos 5θ ≡ 16 cos5 θ − 20 cos3 θ + 5 cos θ (5 marks) b Hence, given also that cos 3θ = 4 cos3 θ − 3 cos θ, find all the solutions of cos 5θ + 5 cos 3θ = 0 (6 marks) in the interval 0 < θ , π. Give your answers to 3 decimal places. E/P 3 a Show that 32 cos6 θ ≡ cos 6θ + 6 cos 4θ + 15 cos 2θ + 10. ∫ π __ (6 marks) __     6 b Hence find   cos6 θ dθ in the form aπ + b√ 3  where a and b are rational constants to be 0 (3 marks) found. E/P 4 a Show that 32 cos  2 θ sin  4 θ ≡ cos 6θ − 2 cos 4θ − cos 2θ + 2. ∫ (6 marks) π _   3  b Hence find the exact value of   cos  2 θ sin  4 θ dθ. (3 marks) 0 P 5 By using de Moivre’s theorem, or otherwise, compute the integrals. ∫ π _   2  a   sin6 θ dθ 0 E/P ∫ π _   4  b   sin  2 θ cos  4 θ dθ 0 ∫ π _   6  c   sin  3 θ cos  5 θ dθ 0 6 a Use de Moivre’s theorem to show that cos 6θ ≡ 32 cos6 θ − 48 cos4 θ + 18 cos2 θ − 1 b Hence find the six distinct solutions of the equation 32x6 − 48x4 + 18x2 − _  2 = 0 3 giving your answers to 3 decimal places where necessary. (5 marks) E/P (5 marks) Problem-solving Use the substitution x = cos θto reduce the equation to the form cos 6θ = k. Find as many values of θas you need to find six distinct values of x. 7 a Use de Moivre’s theorem to show that sin 4θ ≡ 4 cos3 θ sin θ − 4 cos θ sin3 θ 4 tan θ − 4 tan3 θ    b Hence, or otherwise, show that tan 4θ ≡ ________________      1 − 6 tan2 θ + tan4 θ (4 marks) (4 marks) c Use your answer to part b to find, to 2 decimal places, the four solutions of the equation (5 marks) x4 + 4x3 − 6x2 − 4x + 1 = 0 M03_IAL_FP2_44655_U03_022-045.indd 36 25/04/2019 08:56 COMPLEX NUMBERS CHAPTER 3 37 3.5 nth roots of a complex number You can use de Moivre’s theorem to solve an equation of the form zn = w, where z, w ∈ ℂ. This is equivalent to finding the nth roots of w. __ __ Just as a real number, x, has two square roots, √ x  and −√ x , any complex number has n distinct nth roots. ■ If z and w are non-zero complex numbers and n is a positive integer, then the equation zn = w has n distinct solutions. You can find the solutions to zn = w using de Moivre’s theorem, and by considering the fact that the argument of a complex number is not unique. Notation cos (θ + 2kπ) = cos θ and sin (θ + 2kπ) = sin θfor integer values of k. ■ For any complex number z = r(cos θ + i sin θ), you can write z = r(cos (θ + 2kπ) + i sin (θ + 2kπ)), where k is any integer. Example 13 a Solve the equation z3 = 1 b Represent your solutions to part a on an Argand diagram. c Show that the three cube roots of 1 can be written as 1, ω and ω2 where 1 + ω + ω2 = 0 az  3 = 1 z3 = cos 0 + i sin 0 (r(cos θ + i sin θ))3 = cos (0 + 2kπ) + i sin (0 + 2kπ), k ∈ ℤ r3(cos 3θ + i sin 3θ) = cos (0 + 2kπ) + i sin (0 + 2kπ), k ∈ ℤ So r = 1 3θ = 2kπ k = 0 ⇒ θ = 0, so z1 = cos 0 + i sin 0 = 1 2π k = 1 ⇒ θ = ___     3 __ √ 3  2π 2π ___ ___ 1    ) = −  __ + i   so z  2 = cos (   ) + i sin (___   2 2 3 3 2π k = −1 ⇒ θ = −  ___  3 __ √ 3  2π 2π ___ ___ 1 − i   so z 3 = cos (−    ) + i sin (−  ___ ) = −  __   2 2 3 3 Therefore, __ __ √ √ 3  3  ___ __1 ___ 1 + i   − i   z = 1, z = −  __   or z = −     2 2 2 2 M03_IAL_FP2_44655_U03_022-045.indd 37 Start by writing 1 in modulus−argument form. Write z in modulus−argument form, and write the general form of the argument on the right-hand side by adding integer multiples of 2π. Apply de Moivre’s theorem to the left-hand side of the equation. Compare the modulus on both sides to get r = 1. Compare the arguments on both sides. Problem-solving Choose values of k to find the three distinct roots. By choosing values on either side of k = 0 you can find three different arguments in the interval [ −π, π]. These are the cube roots of unity. 25/04/2019 08:56 38 CHAPTER 3 b z2 = – COMPLEX NUMBERS Im 3 1 +i 2 2 __ 2π 2π 3 3 2π 3 z3 = – z1 = 1 Re   √3 Plot the points z1 = 1, z2 = −  __12  + i  ___  and __ 2 3 √  z3 = −  __12  − i  ___ on an Argand diagram: 2 The points z1, z2 and z3 lie on a circle of radius 1 unit. The angles between each of the vectors z1, z2 and 2π z3 are ___     , as shown on the Argand diagram. 3 3 1 –i 2 2 1 __ __ √ 3  ___ ___ 2πi c Let ω = z 2 = −     + i     = e   3 2 2 Then, ω  2 = (e   3 ) = e   3 __ ___ 2πi √ 3  1 = e −  3 = −  __  − i  ___  = z 3 2 2 ___ 2πi 2 1 + ω + ω2 = __ 4πi ___ __ 1 + ( −     + i    ) + ( −     − i    ) = 0 2 2 2 2 1 __ √ 3  ___ 1 __ √ 3  ___ Notice that ω* = ω2. Notation It can be proved that the sum of the nth roots of unity is zero, for any positive integer n > 2. 2πk 2πk ____ ____   ■ In general, the solutions to z  n= 1 are z = cos (____   n  ) + i sin(  n  ) = e  n for k = 1, 2, … , n and are 2πik known as the nth roots of unity. If n is a positive integer, then there is an nth root of unity ω = e    n  such that: 2πi __ • the nth roots of unity are 1, ω, ω  2, … , ω  n−1 • 1, ω, ω  2, … , ω  n−1form the vertices of a regular n-gon • 1 + ω + ω  2 + … + ω  n−1 = 0 M03_IAL_FP2_44655_U03_022-045.indd 38 25/04/2019 08:56 COMPLEX NUMBERS CHAPTER 3 39 Example 14 __ Solve the equation z4 = 2 + 2i√3  Im 2 + 2i 3 2 3 O θ 2 Re ___________ __ 2 + (2 √ 3 )2  2    modulus = √ __ To solve an equation of the form zn = w, start by writing w in modulus−argument form. _______ = √ 4 + 12  = 4 2√ 3  argument = arctan (____     = 2 ) π π So z4 = 4(cos  __  + i sin  __ ) 3 3 π __     3 Now let z = r(cos θ + i sin θ), and write the general form of the argument on the RHS by adding integer multiples of 2π. (r(cos θ + i sin θ))4 π π = 4(cos (__     + 2kπ) + i sin (__     + 2kπ)), k ∈ ℤ 3 3 Apply de Moivre’s theorem to the LHS. r4(cos 4θ + i sin 4θ) π π = 4(cos (__     + 2kπ) + i sin (__     + 2kπ)), k ∈ ℤ 3 3 4 __ __ So r4 = 4 ⇒ r = √ 4  = √ 2  π 4θ =  __  + 2kπ 3 __ π π π k = 0 ⇒ θ =  ___ , so z1 = √ 2 (cos  ___  + i sin  ___ ) 12 12 12 7π k = 1 ⇒ θ =  ___ , 12 7π 7π so z2 = √ 2 (cos  ___  + i sin  ___ ) 12 12 __ __ 5π 5π 5π k = −1 ⇒ θ = −  ___ , so z3 = √ 2 (cos (−  ___ ) + i sin (−  ___ )) 12 12 12 11π 11π 11π k = −2 ⇒ θ = −  ___ , so z4 = √ 2 (cos (−  ___ ) + i sin (−  ___ )) 12 12 12 __ __ __ πi __ ___ 7πi __ 5πi ___ __ ___ 11πi or z = √ 2  e 12, z = √ 2  e 12 , z = √ 2  e −   12 or z = √ 2  e −  12 Compare the modulus on both __ sides to get r = √ 2 . Compare the arguments on both sides. π When k = 1, 4θ =  __  + 2π 3 π ___ 2π ___ 7π ___ ⇒ θ =     +     =     12 4 12 Watch out Make sure you choose n consecutive values of k to get n distinct roots. If an argument is not in the interval [ −π, π]you can add or subtract a multiple of 2π. These are the solutions in the form reiθ. M03_IAL_FP2_44655_U03_022-045.indd 39 25/04/2019 08:56 40 CHAPTER 3 COMPLEX NUMBERS You can also use the exponential form of a complete number when solving equations. Example 15 __ __ Solve the equation z3 + 4√2  + 4i√2  = 0 __ __ z3 + 4√2  + 4i√2  = 0 __ __ z3 = −4√2  − 4i√2  Im 4 2 O Re θ 4 2 –4 2 – 4i 2 ___________________ __ __ ________ ___ 2 2 −4√2   ) + ( −4√2   )   = √ 32 + 32  = √64  = 8 modulus = √(   ( __ 4√__2  _____ ) π __ 3π ___ argument = −π + arctan      = −π +     = −   4 4 4√2  i (reiθ)3 = 8e (−   4 + 2kπ) __ 3π r3e3iθ = 8e (−   4 + 2kπ)i __ 3π 3 __ So r3 = 8 ⇒ r = √ 8  = 2 3π 3θ = −  ___  + 2kπ 4 ___ −πi π k = 0 ⇒ θ = −  __ , so z1 = 2e    4  4 5πi ___ 5π k = 1 ⇒ θ =  ___ , so z2 = 2e    12  12 ____ −11πi 11π k = −1 ⇒ θ = −  ____ , so z3 = 2e    12  12 5π 5π π π or z = 2(cos(−  __ ) + i sin(−  __ )), z = 2(cos  ___  + i sin  ___ ) 4 4 12 12 11π 11π or z = 2(cos(−  ____ ) + i sin(−  ____ )). 12 12 M03_IAL_FP2_44655_U03_022-045.indd 40 Find the modulus__ and __ argument of –4√2  – 4i√2 . Write z = reiθ and use ( reiθ)n = rneinθ. Remember to write the general form of the argument on the righthand side by adding integer multiples of 2π. Compare the modulus on both sides to get r = 2. Compare the arguments on both sides. Choose values of k to find three distinct roots. Either choose values that produce arguments in the interval −π , θ < π, or add or subtract multiples of 2π as necessary. 25/04/2019 08:56 COMPLEX NUMBERS Exercise CHAPTER 3 41 3E 1 Solve the equations, expressing your answers for z in the form x + iy, where x , y ∈ ℝ. a z4 − 1 = 0 b z3 − i = 0 c z3 = 27 d z4 + 64 = 0 e z4 + 4 = 0 f z3 + 8i = 0 2 Solve the equations, expressing the roots in the form r (cos θ + i sin θ), where −π , θ < π. a z7 = 1 b z4 + 16i = 0 c z5 + 32 = 0 d z3 = 2 + 2i e z4 + 2i√3  = 2 f z3 + 32√3  + 32i = 0 __ __ 3 Solve the equations, expressing the roots in the form r e iθ, where r . 0 and −π , θ < π. Give θ to 2 decimal places. ___ a z4 = 3 + 4i P b z3 = √ 11  − 4i __ c z4 = −√ 7  + 3i 4 a Find the three roots of the equation (z + 1)3 = −1 Give your answers in the form x + iy, where x, y ∈ ℝ. b Plot the points representing these three roots on an Argand diagram. c Given that these three points lie on a circle, find its centre and radius. P 5 a Find the five roots of the equation z5 − 1 = 0 Give your answers in the form r (cos θ + i sin θ), where −π , θ < π. b Hence or otherwise, show that 2π 4π 1 cos  ___   + cos  ___   = − _2 5 5 ( ) E ( ) __ 6 a Find the modulus and argument of −2 − 2i√3  __ b Hence find all the solutions of the equation + 2 + =0 Give your answers in the form r eiθ , where r . 0 and −π , θ < π and illustrate the roots on an Argand diagram. z4 E Problem-solving Use the fact that the sum of the five roots of unity is zero. (2 marks) 2i√3  (4 marks) __ 7 Find the four distinct roots of the equation z4 = 2(1 – i√3 ) in exponential form, and show these roots on an Argand diagram. (7 marks) M03_IAL_FP2_44655_U03_022-045.indd 41 25/04/2019 08:56 42 CHAPTER 3 E/P 8 z = √ 6  + i√2  __ COMPLEX NUMBERS __ a Find the modulus and argument of z. b Find the values of w such that and −π , θ < π. P w3 = z4, (2 marks) giving your answers in the form 9 a Solve the equation b Hence deduce that 1+z+ z2 + z3 + z4 + 1) and + z5 + z6 (z4 + where r . 0 (4 marks) Problem-solving 1 + z + z2 + z3 + z4 + z5 + z6 + z7 = 0 (z2 reiθ, + 1) are factors of 1 + z + z2 + z3 + … + z7 is the sum of a geometric series. z7. Challenge a Find the six roots of the equation z6 = 1 in the form eiθ, where −π , θ < π. b Hence show that the solutions to (z + 1)6 = z6 are kπ z = −  _12 + _  12 i cot (___    ), k = 1, 2, 3, 4, 5. 6 Chapter review 3 P E/P 1 a Use eiθ = cos θ + i sin θ to show that cos θ = _  2 (e  iθ + e  −iθ) cos (A + B) + cos (A − B)        b Hence prove that cos A cos B ≡ ______________________     2 1 2 Given that z = r(cos θ + i sin θ), r ∈ ℝ, prove by induction that zn = rn(cos nθ + i sin nθ), n ∈ ℤ  +. (5 marks) 2 x + i sin 3 x ) (cos 3 ________________    3 Express       in the form cos nx + i sin nx where n is an integer to be determined. cos x − i sin x 4 Use de Moivre’s theorem to evaluate: 1 a (−1 + i)8 b _________   1 1 16  _ (  2  − _ 2 i ) E/P 1   n  = 2 cos nθ. 5 a Given z = cos θ + i sin θ, use de Moivre’s theorem to show that zn + __ z 1 3   2   in terms of cos 6θ and cos 2θ. b Express z2 + __ z ( ) (4 marks) (3 marks) c Hence, or otherwise, find constants a and b such that cos3 2θ = a cos 6θ + b cos 2θ. (3 marks) ∫ π __     6 __ d Hence, or otherwise, show that   cos3 2θ dθ = k√ 3 , where k is a rational constant. (4 marks) 0 M03_IAL_FP2_44655_U03_022-045.indd 42 25/04/2019 08:56 COMPLEX NUMBERS E/P CHAPTER 3 6 a Show that cos  5 θ ≡ __   16 (cos 5θ + 5 cos 3θ + 10 cos θ) 43 1 π __ (5 marks) π __ The diagram shows the curve with equation y = cos5 x, –     < x <    . The finite region R is 2 2 bounded by the curve and the x-axis. y 1 R y = cos5 x O x b Calculate the exact area of R. E/P (6 marks) 7 a Show that ( ) sin6 θ ≡ −  __ (5 marks) 32 cos 6θ − 6 cos 4θ + 15 cos 2θ − 10 π     − θ), or otherwise, find a similar identity for cos6 θ. (3 marks) b Using the substitution α = ( __ 2 1 a 5π     , find the exact value of a. c Given that ∫   (cos  6θ + sin  6 θ ) dθ = ___ 32 0 E/P (5 marks) 8 Use de Moivre’s theorem to show that sin 6θ ≡ sin 2θ (16 cos4 θ − 16 cos2 θ + 3) E/P (5 marks) 9 a Use de Moivre’s theorem to show that cos 5θ ≡ 16 cos5 θ − 20 cos3 θ + 5 cos θ (5 marks) b Hence find all solutions to the equation 16x5 − 20x3 + 5x + 1 = 0 (5 marks) giving your answers to 3 decimal places where necessary. E/P 10 a Show that sin5 θ ≡ __   16 (sin 5θ − 5 sin 3θ + 10 sin θ) 1 (5 marks) b Hence solve the equation sin 5θ − 5 sin 3θ + 9 sin θ = 0 for 0 < θ , π E/P (4 marks) 11 a Use de Moivre’s theorem to show that cos 5θ ≡ cos θ (16 cos4 θ − 20 cos2 θ + 5) __ 5 + √ 5  π     b By solving the equation cos 5θ = 0, deduce that cos2  ___   = ______ 10 8 ( ) ( ) (4 marks) ( ) 3π 7π c Hence, or otherwise, write down the exact values of cos2  ___   , cos2  ___   and 10 10 9π ___ 2 cos      . 10 ( ) M03_IAL_FP2_44655_U03_022-045.indd 43 (5 marks) (3 marks) 25/04/2019 08:56 44 CHAPTER 3 COMPLEX NUMBERS E/P 12 a Use de Moivre’s theorem to find an expression for tan 3θ in terms of tan θ. cot  3 θ − 3 cot θ      b Deduce that cot 3θ = _____________      3 cot  2 θ − 1 E (4 marks) (2 marks) 13 a Express 4 − 4i in the form r (cos θ + i sin θ), where r . 0, −π , θ < π, where r and θ are exact values. (2 marks) b Hence, or otherwise, solve the equation z5 = 4 − 4i, leaving your answers in the form z = Reikπ , (4 marks) where R is the modulus of z and k is a rational number such that −1 < k < 1. c Show on an Argand diagram the points representing the roots. E/P 14 a Find the cube roots of 2 − 2i in the form reiθ where r . 0 and −π , θ < π. (2 marks) (5 marks) These cube roots are represented by points A, B and C in the Argand diagram, with A in the fourth quadrant and ABC going anticlockwise. The midpoint of AB is M, and M represents the complex number w. b Draw an Argand diagram, showing the points A, B, C and M. (2 marks) c Find the modulus and argument of w. (2 marks) d Find w6 in the form a + bi. (3 marks) Challenge Show that the points on an Argand diagram that represent the roots z+1 6 of (_____   z  )  = 1lie on a straight line. Summary of key points 1 You can use Euler’s relation, eiθ = cos θ + i sin θ, to write a complex number z in exponential form: z = reiθ where r = |z| and θ = arg z. 2 2 For any two complex numbers z1 = r1eiθ1  and z2 = r2eiθ • z1z2 = r1r2ei(θ1 + θ2) z1 ___ r1 i(θ − θ ) 1 2 •  __ z2  =   r2 e 3 De Moivre’s theorem: For any integer n, (r(cos θ + i sin θ))n = rn(cos nθ + i sin nθ) 1 4 • z + __   z  = 2 cos θ 1 • z − __   z  = 2i sin θ M03_IAL_FP2_44655_U03_022-045.indd 44 1   n  = 2 cos nθ • zn + __ z 1   n  = 2i sin nθ • zn − __ z 25/04/2019 08:56 COMPLEX NUMBERS CHAPTER 3 45 5 If z and w are non-zero complex numbers and n is a positive integer, then the equation zn = w has n distinct solutions. 6 For any complex number z = r(cosθ + i sinθ), you can write z = r(cos (θ + 2kπ) + i sin (θ + 2kπ)) where k is any integer. 2πk 2πk ___ ____   2πik    n  7 In general, the solutions to z  n = 1 are z = cos (____ ) + i sin (  n  ) = e  n for k = 1, 2, … , n and are known as the nth roots of unity. If n is a positive integer, then there is an nth root of unity ω = e    n  such that: 2πi __ • The nth roots of unity are 1, ω, ω2, … , ω n−1 • 1, ω, ω2, … , ωn−1 form the vertices of a regular n-gon • 1 + ω + ω2 + … + ω n−1 = 0 8 The nth roots of any complex number s lie on the vertices of a regular n-gon with its centre at the origin. M03_IAL_FP2_44655_U03_022-045.indd 45 25/04/2019 08:56 46 CHAPTER 4 FURTHER ARGAND DIAGRAMS 4 FURTHER ARGAND DIAGRAMS 3.3 3.4 Learning objectives After completing this chapter you should be able to: ● Represent loci on an Argand diagram → pages 47–55 ● Determine the loci of sets of points, z, in an Argand diagram given z−a in the forms |z − a| = k|z − b| and arg(_) = β, where k, β ∈ ℝ, z−b k . 0, k ≠ 1 and a, b ∈ ℂ → pages 55–63 ● Represent regions on an Argand diagram → pages 63–65 ● Represent regions on an Argand diagram of the forms α < arg (z − z1) < β and p < Re(z) < q, where α, β, p, q ∈ ℝ and z1 ∈ ℂ → pages 65–69 ● Apply elementary transformations that map points from the z-plane to the w-plane, including those of the forms w = z2 and az + b w = _ where a, b, c, d ∈ ℂ. → pages 70–78 cz + d Prior knowledge check 1 Show the complex numbers z1 = −2 + 3i, z2 = 4 + i and z3 = 1 − 3i on an Argand diagram. ← Further Pure 1 Section 1.4 2 Draw the roots of the quadratic equation z2 + 10z + 26 = 0 on an Argand diagram. ← Further Pure 1 Section 1.7 3 Draw the roots of the quadratic equation z2 + 2z + 4 = 0 on an Argand diagram. ← Further Pure 1 Section 1.7 M04A_IAL_FP2_44655_U04_046-082.indd 46 This is an image of a Julia set. Sets such as these are generated by examining the behaviour of points under the repeated application of mappings in the complex plane. 25/04/2019 08:57 FURTHER ARGAND DIAGRAMS CHAPTER 4 47 4.1 Loci in an Argand diagram Complex numbers can be used to represent a locus of points on an Argand diagram.  1 + i y 1 and z  2 = x  2 + i y 2, |z2 − z1| ■ For two complex numbers z  1 = x represents the distance between the points z1 and z2 on an Argand diagram. Im z2 = x2 + iy2 z2 – z1 z2 z1 = x1 + iy1 z1 Using the above result, you can replace z 2with the general point z. O The locus of points described by |z − z1| = r is a circle with centre (x1, y1) and radius r. Online Explore the locus of z, when Im Re |z − z1| = r, using GeoGebra. z = x + iy z – z1 Locus of points. Every point z, on the circumference of the circle, is a distance of r from the centre of the circle. z1 = x1 + iy1 O Re ■ Given z  1 = x  1 + i y 1, the locus of points z on an Argand diagram such that |z − z1| = r, or |z − (x1 + i y1)| = r, is a circle with centre (x1, y1) and radius r. You can derive a Cartesian form of the equation of a circle from this form by squaring both sides: |z − z1| = r Links |(x − x1) + i(y − y1)| = r _____ (x − x1)2 + (y − y1)2 = r2 Since | p + qi| = √ p  2 + q  2  The Cartesian equation of a circle with centre (a, b) and radius r is (x − a)2 + ( y − b)2 = r2 ← Pure 2 Section 2.5 The locus of points that are an equal distance from two different points z1 and z2 is the perpendicular bisector of the line segment joining the two points. Im z = x + iy Locus of points. Every point z on the line is an equal distance from points z1 and z2. z2 = x2 + iy2 z1 = x1 + iy1 Online O Re Explore the locus of z, when |z − z1| = |z − z2|, using GeoGebra. ■ Given z  1 = x  1 + i y 1 and z  2 = x  2 + i y 2, the locus of points z on an Argand diagram such that |z − z1| = |z − z2| is the perpendicular bisector of the line segment joining z1 and z2. M04A_IAL_FP2_44655_U04_046-082.indd 47 25/04/2019 08:57 48 CHAPTER 4 Example FURTHER ARGAND DIAGRAMS 1 Given that z satisfies |z − 4| = 5, a sketch the locus of z on an Argand diagram. b Find the values of z that satisfy: i both |z − 4| = 5 and Im(z) = 0 ii both |z − 4| = 5 and Re(z) = 0 a |z − 4| = 5 is a circle with centre (4, 0) and radius 5. |z − (x1 + i y1)| = r is represented by a circle with centre (x1, y1) and radius r. Im Sketch a circle with centre (4, 0) and radius 5 on an Argand diagram. 5 O 4 Re b i Im(z) = 0 represents the real axis. The points where the circle cuts the real axis are (−1, 0) and (9, 0). The values of z at these points are z = −1 and z = 9. ii |z − 4| = 5 ⇒ (x − 4)2 + y2 = 52 Centre of circle is (4, 0) and radius is 5. So consider 4 + 5 = 9 and 4 − 5 = −1. Watch out Give your answers as complex numbers, not as coordinates. This is the Cartesian equation of a circle with centre (4, 0) and radius 5. (0 − 4)2 + y2 = 52 16 + y2 = 25 y2 = 9 y = ±3 Re(z) = 0 for all points on the imaginary axis, so set x = 0. The points where the circle cuts the real axis are (0, 3) and (0, −3). The values of z are z = 3i and z = −3i. Example 2 A complex number z is represented by the point P in the Argand diagram. Given that |z − 5 − 3i| = 3, a sketch the locus of P b find the Cartesian equation of this locus c find the maximum value of arg z in the interval (−π, π). M04A_IAL_FP2_44655_U04_046-082.indd 48 25/04/2019 08:57 FURTHER ARGAND DIAGRAMS CHAPTER 4 a Im |z − 5 − 3i| can be written as |z − (5 + 3i)|. As this distance is always equal to 3, the locus of P is a circle centre (5, 3), radius 3. P 3 3 49 (5, 3) O 5 Re The standard Cartesian equation of a circle is (x − a)2 + ( y − b)2 = r2 b The Cartesian equation of the locus is (x − 5)2 + (y − 3)2 = 9 c Im The maximum value of arg z is the angle OA makes with the positive real axis. A The line OC bisects the angle AOB. C 5 + 3i θ 2 O Problem-solving 3 θ 5 2 B Re Using triangle OBC: 3 θ     tan(__    ) = __ 5 2 3    ) = 1.08 rad (3 s.f.) θ = 2arctan(__ 5 Example When solving geometrical problems like this one, it is helpful to draw an Argand diagram. The maximum value of arg(z) occurs when the line between the origin and P is a tangent to the circle. Use circle properties. OB is perpendicular to BC, and triangles OBC and OAC are congruent. 3 Given that the complex number z = x + iy satisfies the equation |z − 12 − 5i| = 3, find the minimum value of |z| and maximum value of |z|. The locus of z is a circle centre C(12, 5), radius 3. Im 5 O | z | represents the distance from the origin to any point on this locus. 3 Y 3 C (12, 5) X 12 Re |z|min = OC − CX = 13 − 3 = 10. |z|max = OC + CY = 13 + 3 = 16. The minimum value of |z| is 10 and the maximum value of |z| is 16. M04A_IAL_FP2_44655_U04_046-082.indd 49 | z |min and | z |m axare represented by the distances OX and OY respectively. ________ The distance OC = √ 122 + 52  = 13. The radius r = CX = CY = 3. 25/04/2019 08:57 50 CHAPTER 4 FURTHER ARGAND DIAGRAMS Example 4 Given that |z − 3| = |z + i|, a sketch the locus of z and find the Cartesian equation of this locus b find the least possible value of |z|. a | z − 3| = | z + i|is the perpendicular bisector of the line segment joining the points (3, 0) and (0, −1). The gradient of the line joining (0, −1) and 1 (3, 0) is  __ 3 So, the gradient of the perpendicular bisector is −3. The midpoint of the line joining (0, −1) and (3, 0) is (  2  , −   2 ). __ 3 __1 The perpendicular bisector will pass through the midpoint. Substitute (x1, y1) = ( _  32 , − _  12 ) and m = −3 into the equation of a straight line. y − y  1 = m(x − x 1) Problem-solving y + __   21 = − 3(x − __   32 ) __1 The locus of points z satisfying |z − z1| = |z − z2| is the perpendicular bisector of the line segment joining z1 to z2. You could also square both sides of |z − 3| = |z + i|: |x + iy − 3| = |x + iy + i| |(x − 3) + iy| = |x + i(y + 1)| (x − 3)  2 + y  2 = x  2 + ( y + 1)  2 x2 − 6x + 9 + y2 = x2 + y2 + 2y + 1 y = − 3x + 4 __ 9 y +   2 = − 3x +   2 y = − 3x + 4 Im 4 Problem-solving dmin 4 3 O (3, 0) ( (0, –1) The minimum distance is the perpendicular distance from O to the perpendicular bisector. 3 2, ) – 21 Re b The gradient of the line labelled dmin is __   31 The line is parallel to the line joining (0, −1) and (3, 0). The line passes through the origin. __1 The equation of this line is y =   3 x __ 1   3 x = − 3x + 4 10  __ 3 x = 4 6 __ 2 x =  __ 5 ⇒ y =   5 __________ dmin = √ (__   65 )2 + (__   25 )2   ___ Find the point where this line intersects the perpendicular bisector. Solve to find x and substitute into y = _  13  xto find y. Use Pythagoras’ theorem. 2√10  _____ =     5 M04A_IAL_FP2_44655_U04_046-082.indd 50 25/04/2019 08:57 FURTHER ARGAND DIAGRAMS CHAPTER 4 51 Locus questions can also make use of the geometric property of the argument. ■ Given z  1 = x  1 + i y 1, the locus of points z on an Argand diagram such that arg (z − z1) = θ is a half-line from, but not including, the fixed point z1 making an angle θ with a line from the fixed point z1 parallel to the real axis. Notation A half-line is a straight line extending from a point infinitely in one direction only. Im z Online arg (z – z1) z1 Explore the locus of z, when arg (z − z1) = θ, using GeoGebra. θ O Re You can find the Cartesian equation of the half-line corresponding to arg(z − z  1) = θ by considering how the argument is calculated: arg (z − z1) = θ arg ((x − x1) + i ( y − y1)) = θ y − y1 ______ x  − x1  = tan θ y − y1 = tan θ (x − x1) Example θ is a fixed angle so tan θ is a constant. This is the equation of a straight line with gradient tan θ passing through the point (x1, y1). 5 3π Given that arg (z + 3 + 2i) = ___     4 a sketch the locus of z on an Argand diagram b find the Cartesian equation of the locus 3π c find the complex number z that satisfies both |z + 3 + 2i| = 10 and arg (z + 3 + 2i) = ___     4 a arg(z + 3 + 2i) = 3π 4 3π 4 (–3, –2) M04A_IAL_FP2_44655_U04_046-082.indd 51 Im O Re z + 3 + 2i can be written as z − (−3 − 2i). As 3π arg(z + 3 + 2i) = __    , the locus of z is the half-line 4 3π     in an from (−3, −2) making an angle of ___ 4 anticlockwise sense from a line in the same direction as the positive real axis. 25/04/2019 08:57 52 CHAPTER 4 FURTHER ARGAND DIAGRAMS 3π b arg(z + 3 + 2i) = ___     4 3π ___ arg(x + iy + 3 + 2i) =     4 3π ___ arg((x + 3) + i(y + 2)) =     4 y+2 3π ______       = tan ___ 4 x+3 y + 2 = − (x + 3) Hence the Cartesian equation of the locus is y = −x − 5, x , −3 c |z + 3 + 2i| = 10 is a circle with centre (−3, −2) and radius 10. Im z 10 3π 4 4 a (–3, –2) π O Re 2 ⇒ 2a2 = 100 a2 + a2 = 10 ___ __ √2  a = √50  = ± 5 __ __ z = (− 3 − 5 √2 ) + i(− 2 + 5 √2 ) P Group the real and imaginary parts. Remove the argument. 3π tan  ___  = −1 4 Watch out The locus is the half-line so you need to give a suitable range of values for x. Use a geometric approach to find z. Draw part of a circle with centre (−3, −2) and radius 10. 3π π Angle inside the new triangle is π − ___     = __     4 4 a Exercise z can be rewritten as z = x + iy π As the angle is __    , the triangle is isosceles. 4 So the two shorter sides have the same length. Problem-solving An alternative algebraic approach would be to substitute the equation for the half-line, y = −x − 5, into the equation of the circle, ( x + 3)  2 + ( y + 2)  2 = 10  2, and then solve for x and y. You would need to choose the solution which lies on the correct half-line. 4A 1 Sketch the locus of z and give the Cartesian equation of the locus of z when: a |z| = 6 b |z| = 10 c |z − 3| = 2 d |z + 3i| = 3 e |z − 4i| = 5 f |z + 1| = 1 g |z − 1 − i| = 5 h |z + 3 + 4i| = 4 i |z − 5 + 6i| = 5 Hint You may choose a geometric or an algebraic approach to answer these questions. 2 Given that z satisfies |z − 5 − 4i| = 8, a sketch the locus of z on an Argand diagram b find the exact values of z that satisfy: i both |z − 5 − 4i| = 8 and Re(z) = 0 M04A_IAL_FP2_44655_U04_046-082.indd 52 ii both |z − 5 − 4i| = 8 and Im(z) = 0 25/04/2019 08:57 FURTHER ARGAND DIAGRAMS P CHAPTER 4 53 3 A complex number z is represented by the point P on the Argand diagram. Given that |z − 5 + 7i| = 5, a sketch the locus of P b find the Cartesian equation of this locus c find the maximum value of arg z in the interval (−π, π). E/P 4 On an Argand diagram the point P represents the complex number z. Given that |z − 4 − 3i| = 8, E/P 5 6 a find the Cartesian equation for the locus of P (2 marks) b sketch the locus of P (2 marks) c find the maximum and minimum values of |z| for points on this locus. (2 marks) The point P represents a complex number z on an Argand diagram. __ Given that |z + 2 − 2√3 i| = 2, a sketch the locus of P on an Argand diagram b write down the minimum value of arg z c find the maximum value of arg z. Sketch the locus of z and give the Cartesian equation of the locus of z when: a |z − 6| = |z − 2| b |z + 8| = |z − 4| c |z| = |z + 6i| d |z + 3i| = |z − 8i| e |z − 2 − 2i| = |z + 2 + 2i| f |z + 4 + i| = |z + 4 + 6i| g |z + 3 − 5i| = |z − 7 − 5i| h |z + 4 − 2i| = |z − 8 + 2i| |z + 6 − i| __________ j     = 1 |z − 10 − 5i| |z + 3| i  ______  = 1 |z − 6i| E/P E/P (2 marks) (2 marks) (2 marks) 7 Given that |z − 3| = |z − 6i|, a sketch the locus of z b find the exact least possible value of |z|. (3 marks) (4 marks) 8 Given that |z + 3 + 3i| = |z − 9 − 5i|, a sketch the locus of z b find the Cartesian equation of this locus c find the exact least possible value of |z|. (3 marks) (3 marks) (3 marks) 9 Sketch the locus of z and give the Cartesian equation of the locus of z when: a |2 − z| = 3 10 Sketch the locus of z when: π a arg z = __     3 π d arg(z + 2 + 2i) = − __  4 2π ___ g arg(z − 1 + 3i) =     3 M04A_IAL_FP2_44655_U04_046-082.indd 53 b |5i − z| = 4 c |3 − 2i − z| = 3 π b arg(z + 3) = __     4 3π e arg(z − 1 − i) = ___     4 π h arg(z − 3 + 4i) = − __  2 π c arg(z − 2) =  __  2 f arg(z + 3i) = π 3π i arg(z − 4i) = − ___  4 25/04/2019 08:57 54 P E/P E/P E/P CHAPTER 4 FURTHER ARGAND DIAGRAMS 11 Given that z satisfies |z + 2i| = 3, a sketch the locus of z on an Argand diagram π b find |z| that satisfies both |z + 2i| = 3 and arg z =  __  6 12 Given that the complex number z satisfies the equation |z + 6 + 6i| = 4, a find the exact maximum and minimum value of |z| b find the range of values for θ, −π , θ , π, for which arg(z − 4 + 2i) = θ and |z + 6 + 6i| = 4 have no common solutions. (4 marks) (2 marks) (3 marks) (2 marks) 16 Sketch on the same Argand diagram the locus of points satisfying: a |z − 3 + 2i| = 4 π b arg(z − 1) = −  __  4 π The complex number z satisfies both |z − 3 + 2i| = 4 and arg(z − 1) = −  __  4 Given that z = a + ib, where a, b ∈ ℝ, E/P (2 marks) 15 Sketch on the same Argand diagram the locus of points satisfying: a |z − 2i| =| z − 8i| π b arg(z − 2 − i) = __     4 π The complex number z satisfies both |z − 2i| = |z − 8i| and arg(z − 2 − i) =  __  4 c Use your answers to parts a and b to find the value of z. E/P (4 marks) 13 The point P represents a complex number z on an Argand diagram such that |z| = 5. π The point Q represents a complex number z on an Argand diagram such that arg(z + 4) =  __  2 a Sketch, on the same Argand diagram, the locus of P and the locus of Q as z varies. (2 marks) π b Find the complex number for which both |z| = 5 and arg(z + 4) = __     (2 marks) 2 14 Given that the complex number z satisfies |z − 2 − 2i| = 2, a sketch, on an Argand diagram, the locus of z π Given further that arg(z − 2 − 2i) =  __ , 6 b find the value of z in the form a + ib, where a, b ∈ ℝ. E/P (3 marks) c find the exact value of a and the exact value of b. π π 17 If the complex number z satisfies both arg z = __     and arg(z − 4) = __     3 2 a find the value of z in the form a + ib, where a, b ∈ ℝ. b Hence, find arg(z − 8). π E/P 18 Given that arg(z + 4) =  __  3 a sketch the locus of z on an Argand diagram b find the minimum value of |z| for points on this locus. M04A_IAL_FP2_44655_U04_046-082.indd 54 (2 marks) (3 marks) (3 marks) (3 marks) (2 marks) (3 marks) (2 marks) 25/04/2019 08:57 FURTHER ARGAND DIAGRAMS E/P CHAPTER 4 55 19 A complex number z is represented by the point P on the Argand diagram. Given |z + 8 − 4i| = 2, a sketch the locus of P (2 marks) b show that the maximum value of arg(z + 15 − 2i) in the interval (−π, π) 2 is 2 arcsin ____   ___  (√ 53 ) c find the exact values of the complex numbers that satisfy both |z + 8 − 4i| = 2 3π and arg(z + 4i) = ___     4 (3 marks) (3 marks) Challenge The complex number z satisfies both |z + i| = 5 and arg(z − 2i) = θ, where θ is a real constant such that −π , θ < π. Given that |z − 4i| , 3, find the range of possible values of θ. 4.2 Further loci in an Argand diagram You need to be able to determine the locus of a set of points whose distances from two fixed points are in a constant ratio. P Consider a circle with centre O and radius r. The fixed point A lies inside the circle, and the fixed point B lies on the straight line through OA and is such that OA × OB = r2. B r O A For any point P on the circumference of the circle: OB OP OA × OB = OP 2 so ____     = ____     OP OA This means that triangle OPA and triangle OBP are similar, since they have two corresponding sides BP OB     = ____ in the same ratio with an equal included angle (SAS). Hence ___     which is constant for all points AP OP P on the circumference of the circle. Hence BP = kAP for some constant k, and the locus of points which satisfies this relationship is a circle. y For example, the set of points that are exactly twice the distance from (0, 1) as from the point (3, −2). It is not intuitive, but this locus of points is a circle with its centre at (4, −3). (0, 1) 2d x O If you replaced the coordinate axes above with an Argand diagram, this would be equivalent to the set of points that were twice as far from i as from 3 − 2i. You could write this locus as the set of points z that satisfy |z − i| = 2|z − (3 − 2i)|. d (3, 22) (4, 23) 1 The locus of points z that satisfy |z − a| = k|z − b|, where a, b ∈ ℂ and k ∈ ℝ, k > 0, k ≠ is a circle. You can find the centre and radius of the circle by finding its Cartesian equation. M04A_IAL_FP2_44655_U04_046-082.indd 55 25/04/2019 08:57 56 CHAPTER 4 Example FURTHER ARGAND DIAGRAMS 6 Given that |z − 6| = 2|z + 6 − 9i|, a use algebra to show that the locus of z is a circle, stating its centre and its radius b sketch the locus of z on an Argand diagram. Online Explore the locus of z when |z − a| = k|z − b| using GeoGebra. a |z − 6| = 2|z + 6 − 9i| ⇒ |x + iy – 6| = 2|x + iy + 6 − 9i| ⇒ |(x − 6) + iy| = 2|(x + 6) + i(y − 9)| ⇒ |(x − 6) + iy|2 = 22|(x + 6) + i(y − 9)|2 ⇒ (x − 6)2 + y2 = 4((x + 6)2 + (y − 9)2) ⇒ x2 − 12x + 36 + y2 = 4(x2 + 12x + 36 + y2 − 18y + 81) ⇒ x2 − 12x + 36 + y2 = 4x2 + 48x + 144 + 4y2 − 72y + 324 ⇒ 3x2 + 60x + 3y2 − 72y + 432 = 0 ⇒ x2 + 20x + y2 − 24y + 144 = 0 ⇒ (x + 10)2 − 100 + (y − 12)2 − 144 + 144 = 0 ⇒ (x + 10)2 + (y − 12)2 = 100 So the locus of z is a circle with centre (−10, 12) and radius 10. b Group the real and imaginary parts. Square both sides. Remove the moduli. Complete the square twice for x and for y. Circle (x − a)2 + (y − b)2 = r2 with (a, b) = (−10, 12) and r = 10. Im Locus of z (–10, 12) z can be written as z = x + iy. 12 Locus of z as required. B P A –10 O Re Problem-solving |z − 6| represents the distance from the point A(6, 0) to P. |z + 6 − 9i| = |z − (−6 + 9i)| represents the distance from the point B(−6, 9) to P. |z − 6| = 2|z + 6 − 9i| gives AP = 2BP. This means that P is the locus of points such that the distance AP is twice the distance BP. One of the points will always be inside the circle and the other will always be outside the circle. Another previous result for loci in an Argand diagram makes use of the geometric property of the argument of a complex number. M04A_IAL_FP2_44655_U04_046-082.indd 56 25/04/2019 08:57 FURTHER ARGAND DIAGRAMS CHAPTER 4 57 ■ Given z1 = x1 + iy1, the locus of points z on an Argand diagram such that arg(z − z1) = θ is a half-line from, but not including, the fixed point z1, making an angle θ with a line from the fixed point z1 parallel to the real axis. Im Watch out The endpoint z1 is not included in the locus. You show this by drawing it with an open circle. z z1 arg (z – z1) θ O Re You can make use of the following circle properties to determine more complicated loci given in terms of arguments. Angles subtended at an arc in the same segment are equal. ● ● P The angle in a semicircle is a right angle. ● The angle subtended at the centre of the circle is twice the angle at the circumference. P A Q P O O B A A B π ∠ APB =  _  2 z−a ■ The locus of points z that satisfy arg  _ = θ, ( z − b) where θ ∈ ℝ, θ > 0 and a, b ∈ ℂ, is an arc of a circle with endpoints A and B representing the complex numbers a and b, respectively. ∠APB = ∠AQB B ∠AOB = 2∠APB Watch out The endpoints of the arc, A and B, are not included in the locus. You can see why this locus is the arc of a circle by drawing points A and B on an Argand diagram, and drawing a point P such that ∠APB = θ, where θ is a positive, constant angle. Im β θ α P θ=α–β The solution shown for Example 2 below illustrates the same approach developed here. β B α A O M04A_IAL_FP2_44655_U04_046-082.indd 57 Re 25/04/2019 08:57 58 CHAPTER 4 FURTHER ARGAND DIAGRAMS From knowing the locus for an equation of the form arg(z − z1) = θ, you can conclude that arg(z − a) = α and arg(z − b) = β. It follows that ∠APB = α − β This is due to the properties of parallel lines. θ = α − β = arg(z − a) − arg(z − b) z−a = arg(_   z − b) z 1 arg(_   z    ) = arg z 1 − arg z 2 ← Further Pure 1 Section 2.3 2 As P moves, ∠APBis always equal to the constant θ. By the converse of the first circle property on the previous page, ∠APBmust be the angle subtended in the arc of a circle. The locus of P is the arc of a circle that is drawn anticlockwise from A to B. Im Problem-solving P P9 To prove the converse of θ the first circle property, B θ Q suppose P9 did not lie on the circle through A, B and A P. Let Q be the intersection of this circle with the line through A and P9. Then ∠AQB = θ and ∠AQB ≠ ∠AP9B. This is a contradiction since ∠AP9B = θ, so P9 must lie on the circle. θ θ P9 B A O Re π If θ ,  __ , then the locus is a major arc of the circle. 2 π __ If θ .    , then the locus is a minor arc of the circle. 2 π __ If θ =    , then the locus is a semicircle. 2 In these two examples, a = 2i and b = −3. The arcs are drawn anticlockwise from A to B. ( ) arg z – 2i = π 3 z +3 Im Im ( ) arg z – 2i = 2π z +3 3 P π 3 A B O P A 2π 3 O B Re Re In the following two examples the values of a and b are reversed. Im ( ) arg z + 3 = π 3 z – 2i A Im ( B B 2π 3 π 3 A O Re P M04A_IAL_FP2_44655_U04_046-082.indd 58 ) arg z + 3 = 2π z – 2i 3 P O Re 25/04/2019 08:57 FURTHER ARGAND DIAGRAMS CHAPTER 4 59 Finding the centre of the circle on which the major or minor arc is located requires algebraic and/or geometric working. This is illustrated in Example 2. Example 7 Online Explore the locus of z when z−6 π (z − b) Given that arg(_         = _ z − 2) 4 a sketch the locus of P(x, y) which is represented by z on an Argand diagram z−a arg _____     = θ using GeoGebra. b find the Cartesian equation of this locus. You must show your reasoning clearly. z−6 π aarg(_     = arg (z − 6) − arg (z − 2) =  _  z − 2) 4 Let L1 be the half-line satisfying arg(z − 6) = α z 1 Use arg(_   z    ) = arg z 1 − arg z 2. 2 and let L2 be the half-line satisfying arg(z − 2) = β. π It follows that α − β =  _  (1) 4 Im L1 P(x, y) O B(2, 0) α A(6, 0) D Re From the diagram, ∠PBA = β and ∠PAD = α. π Use α − β = _     4 4 From the circle theorems, angles in the same segment of a circle are equal. P(x, y) 4 M04A_IAL_FP2_44655_U04_046-082.indd 59 B(2, 0) A(6, 0) D (1) π P can vary but ∠BPA must always be _     4 π O Therefore the point P is found lying on π both L1 and L2 where α − β = _     4 As P lies on L1 and L2, it is found where L1 and L2 intersect. The exterior angle of a triangle is the sum of the two opposite interior angles. From △ABP, it follows that ∠BPA + ∠PBA = ∠PAD ⇒ ∠BPA + β = α ⇒ ∠BPA = α − β π ⇒ ∠BPA = _     4 π As α and β vary, ∠BPA is constant and is _     4 Im Locus of P π All points on L1 satisfy arg(z − 6) = α All points on L2 satisfy arg(z − 2) = β L2 β π Use arg(z − 6) − arg(z − 2) = _     4 Re Therefore as P varies, ∠BPA will always π be equal to _     4 π π So, since _     , _    it follows that P must 4 2 lie on the major arc starting at (6, 0) and finishing at (2, 0), but not including the points (6, 0) and (2, 0). 25/04/2019 08:57 60 CHAPTER 4 FURTHER ARGAND DIAGRAMS b Method 1: Geometric Im π π ∠BPA = _     ⇒ ∠ACB = _    , as the angle 4 2 subtended at the centre of the circle is twice the angle at the circumference. P π Locus of P 4 As CA and CB are both radii, then the radius is r = CA = CB. C This implies that △CAB is isosceles π and ∠CAB = ∠CBA = _     4 π 2 O B(2, 0) Re A(6, 0) C π 4 π Since the locus is the major arc of the circle which lies above the real axis, then the Cartesian equation for the locus must include the condition that y . 0. π 4 B(2, 0) 2 4 2 X(4, 0) A(6, 0) _ _ AX = CX = 2 ⇒ AC = √ 2  2 + 2  2 = 2√ 2  and C is the point (4, 2). So the Cartesian equation of the locus of P is (x − 4)2 + (y − 2)2 = 8, where y . 0. Method 2: Algebraic x − 6 + iy z−6 ______     = __________       z−2 x − 2 + iy (x − 6 + iy ) (x − 2 − iy)      =  ___________________ (x − 2 + iy ) (x − 2 − iy) x  2− 8x + 12 + y  2+ 4iy    = __________________        (x − 2)  2+ y  2 x  2− 8x + 12 + y  2 4y        + ___________       i = ______________   ( (x − 2)  2+ y  2 ) ( (x − 2)  2+ y  2) x  2 − 8x + 12 + y  2 4y π So arg ____________        + ___________              i = _ 2 2 (( (x − 2)  + y  ) ( (x − 2)  2 + y  2) ) 4 x  2 − 8x + 12 + y  2 ___________ 4y       ⇒  ____________   =        (x − 2)  2 + y  2 (x − 2)  2 + y  2 x ⇒   2 − 8x + 12 + y  2 = 4y ⇒ (x − 4)  2 + ( y − 2)  2 = 8, where y . 0 M04A_IAL_FP2_44655_U04_046-082.indd 60 Let X be the midpoint of AB. Hence π π ∠CXA = _     and ∠XCA = ∠CAX = _     4 2 So △CAX is isosceles and AX = CX = 2. Watch out The locus is only a part of a circle (an arc), so you need to give a suitable range of values for x and/ or y to indicate which part of the circle is included. Problem-solving z−6 In order to deal with arg(_     z − 2) algebraically, you need to identify its real and imaginary parts. Write z = x + iy then multiply the numerator and denominator by (z − 2)*. Im (w) If arg w = θ, then _     = tan θ. In this Re(w) π π case, θ = _     and tan  _  = 1, so the real 4 4 and imaginary parts are equal. Watch out If you use an algebraic method to find the equation of the circle, you still need to use geometric considerations to work out which arc of the circle satisfies the given condition. In this case y . 0. 25/04/2019 08:57 FURTHER ARGAND DIAGRAMS Example CHAPTER 4 61 8 π z Given the equation arg(_        ) = _ 2 z − 4i a sketch the locus of points z that satisfy the equation on an Argand diagram. b Hence write down the range of possible values of Re(z). π z aarg(_         = argz − arg(z − 4i) = α − β. So α − β = _ z − 4i ) 2 Im 4 –β Locus of P –β α O b π Since the constant angle at P is _    , the 2 locus of P is a semicircle from (0, 0) anticlockwise to (0, 4), not including (0, 0) and (0, 4). P α Re Problem-solving Im The point on the locus furthest to the right is 2 + 2i, so the largest possible value of Re(z) is 2. The endpoints of the semicircle are at 0 and 4i. These points are not included in the locus of z, so use a strict inequality to show the smallest possible value of Re(z). 4 2 2 + 2i O 2 arg z = αand arg (z − 4i) = β π z arg( _ ) = _     2 z − 4i π π _ α − β =     and α , _     ⇒ β , 0. 2 2 Re The range of possible values for Re(z) are 0 , Re(z) < 2. Exercise 4B 1 Sketch the locus of z and give the Cartesian equation of the locus of z when: a |z + 3| = 3|z − 5| b |z − 3| = 4|z + 1| c |z − i| = 2|z + i| d |z + 2 − 7i| = 2|z − 10 + 2i| e |z + 4 − 2i| = 2|z − 2 − 5i| f |z| = 2|2 − z| 2 Sketch the locus of z when: π z aarg(_         = _ z + 3) 4 π z        ) = _ carg(_ 3 z−2 π earg z − arg (z − 2 + 3i) = _     3 M04A_IAL_FP2_44655_U04_046-082.indd 61 z − 3i π barg(_         = _ z + 4) 6 z − 3i π darg(_        ) = _ z−5 4 π z − 4i _ _ farg(    =     z + 4) 2 25/04/2019 08:57 62 E CHAPTER 4 FURTHER ARGAND DIAGRAMS 3 The complex number z = x + iy satisfies the equation |z + 1 + i| = 2|z + 4 − 2i| The complex number z is represented by the point P on the Argand diagram. (4 marks) a Show that the locus of P is a circle with centre (−5, 3). b Find the exact radius of this circle. E/P (1 mark) 4 The point P represents a complex number z in an Argand diagram. π Given that arg z − arg(z + 4) =  _ is a locus of points P lying on an arc of a circle C, 4 a sketch the locus of points P b find the coordinates of the centre of C (3 marks) c find the radius of C (2 marks) d find a Cartesian equation for the circle C (1 mark) e find the finite area bounded by the locus of P and the x-axis. E/P E/P E/P (3 marks) 5 A curve F is described by the equation |z| = 2|z + 4| a Show that F is a circle, and find its centre and radius. (5 marks) b Sketch F on an Argand diagram. (2 marks) c Given that z lies on F, find the range of possible values of Im(z). (3 marks) 6 The set of points z lie on the curve defined by |z − 8| = 2|z − 2 − 6i|. Find the range of (7 marks) possible values of arg(z). w − 8i π 7 A curve S is described by the equation arg(_       , w ∈ ℂ.  ) = _ 2 w+6 a Sketch S on an Argand diagram. (2 marks) b Find the Cartesian equation for S. (3 marks) c Given that z lies on S, find the largest value of a and the smallest value of b that satisfy a , arg(z) , b. (2 marks) d State the range of possible values of Re(z). E/P E/P (2 marks) (1 mark) 8 The point P represents the complex number z that satisfies the equation 3π arg(z − 1) − arg(z + 3) = _     , z ≠ − 3 4 Use a geometric approach to find the Cartesian equation of the locus of P. (5 marks) 9 Each of the three Argand diagrams below shows an arc of a circle drawn from point A to point B that is the locus of a set of complex numbers z. Write down a complex (6 marks) equation for each locus. a b Im c Im Im B(0, 4) π 6 π 4 B(–5, 0) A(–2, 0) O Re O M04A_IAL_FP2_44655_U04_046-082.indd 62 B(1, 2) A(0, 1) 2π 3 A(6, 1) Re O Re 25/04/2019 08:57 FURTHER ARGAND DIAGRAMS E CHAPTER 4 10 The curve C has equation |z + 3| = 3|z − 5|, z ∈ ℂ. a Show that C is a circle with equation x  2 + y  2 − 12x + 27 = 0 (2 marks) b Sketch C on an Argand diagram. (2 marks) π c The point z1 lies on C such that arg z 1 = _    . Express z 1in the form r (cos θ + i sin θ). 6 E/P 63 (3 marks) 11 In an Argand diagram, points A and B represent the numbers 6i and 3 respectively. As z varies, the locus of points P satisfying the equation |z − z1| = k|z − z2|, where z1, z2 ∈ ℂ and k ∈ ℝ, is the circle C such that each point P on the circle is twice the distance from point A than it is from point B. a Write down the complex numbers z  and z  , and the value of k. (2 marks) Hint AP = 2BP 1 2 b Show that the Cartesian equation of circle C is x  2 + y  2 − 8x + 4y = 0 (2 marks) The locus of points w satisfying the equation arg(w − 6) = αwhere α ∈ ℝ passes through the centre of circle C and intersects it at point Q. c Find the value of α. (3 marks) (3 marks) d Find the exact coordinates of Q. Challenge Fully describe the locus of points z that satisfy the equation |z − a| + |z + a| = b, where a and b are real constants and b . 2a. 4.3 Regions in an Argand diagram You can use complex numbers to represent regions on an Argand diagram. Example 9 a On separate Argand diagrams, shade in the regions represented by: π iii 0 < arg(z − 2 − 2i) <  __  4 b Hence, on the same Argand diagram, shade the region which satisfies i |z − 4 − 2i| < 2 ii |z − 4| , |z − 6| π {z ∈ ℂ : |z − 4 − 2i| < 2} ∩ {z ∈ ℂ : |z − 4| , |z − 6|} ∩ { z ∈ ℂ : 0 < arg (z − 2 − 2i) < __   }    4 a i |z − 4 − 2i| < 2 |z − 4 − 2i| = 2 represents a circle centre (4, 2), radius 2. Im 2 2 O M04A_IAL_FP2_44655_U04_046-082.indd 63 |z − 4 − 2i| , 2 represents the region on the inside of this circle. (4, 2) 4 Re |z − 4 − 2i| < 2 represents the boundary inside of this circle. 25/04/2019 08:57 64 CHAPTER 4 FURTHER ARGAND DIAGRAMS ii |z − 4| , |z − 6| |z − 4| = |z − 6| is represented by the line x = 5. This line is the perpendicular bisector of the line segment joining (4, 0) to (6, 0). Im |z − 4| , |z − 6| represents the region x , 5. All points in this region are closer to (4, 0) than to (6, 0). O 4 6 Re 5 x=5 π iii 0 < arg(z − 2 − 2i) <  __  4 π arg(z − 2 − 2i) =  __ is the half-line from the point 4 π (2, 2) at angle  __ to the horizontal. 4 Im 2 (2, 2) O Note this region does not include the line x = 5. So x = 5 is represented by a dashed line. arg(z − 2 − 2i) = 0 is the other half-line shown from the point (2, 2). π 4 Re 2 π 0 < arg(z − 2 − 2i) <  __ is represented by the region 4 in between and including these two half-lines. Notation b |z − 4 − 2i| < 2, |z − 4| , |z − 6| π and 0 < arg(z − 2 − 2i) < __     4 Im 4 2 Exercise 2 4 π The line arg(z − 2 − 2i) =  __ and the circle 4 |z − 4 − 2i| = 2 both go through the point (4, 4). The region shaded is satisfied by all three of |z − 4 − 2i| < 2 |z − 4| , |z − 6| π 0 < arg(z − 2 − 2i) <  __  4 (4, 2) O The symbol ∩ is the symbol for the intersection of two sets. You need to find the region of points that lie in all three sets. 5 6 Re Online Explore this region using GeoGebra. 4C 1 On an Argand diagram, shade in the regions represented by the inequalities: a |z| , 3 b |z − 2i| . 2 c |z + 7| > |z − 1| e 2 < |z| < 3 f 1 < |z + 4i| < 4 g 3 < |z − 3 + 5i| < 5 M04A_IAL_FP2_44655_U04_046-082.indd 64 d |z + 6| . |z + 2 + 8i| 10/09/19 6:27 PM FURTHER ARGAND DIAGRAMS E/P E/P E/P CHAPTER 4 65 2 The complex number z is represented by a point P on an Argand diagram. 3π Given that |z + 1 − i| < 1 and 0 < arg z < ___    , shade the locus of P. 4 3 Shade on an Argand diagram the region satisfied by π {z ∈ ℂ : |z| < 3} ∩ {z ∈ ℂ : __   < arg (z + 3) < π} 4 (6 marks) 4 a Sketch on the same Argand diagram: i (2 marks) the locus of points representing |z − 2| = |z − 6 − 8i| ii the locus of points representing arg(z − 4 − 2i) = 0 π iii the locus of points representing arg(z − 4 − 2i) = __     2 b Shade on an Argand diagram the set of points (2 marks) (2 marks) π 2 {z ∈ ℂ : |z − 2| < |z − 6 − 8i|} ∩ { z ∈ ℂ : 0 < arg (z − 4 − 2i) < __   }    E/P (6 marks) 5 a Find the Cartesian equations of: (2 marks) __ the locus of points representing |z + 10| = |z − 6 − 4i√ 2 | i ii the locus of points representing |z + 1| = 3 __ (6 marks) b Find the two values of z that satisfy both |z + 10| = |z − 6 − 4i√ 2 | and |z + 1| = 3 (2 marks) R on an Argand diagram which satisfies both c Hence shade in the region __ |z + 10| < |z − 6 − 4i√ 2 | and |z + 1| < 3 (4 marks) Challenge The sets A, B and C are defined as: A = {z ∈ ℂ: |z + 5 + 8i| < 5} B = {z ∈ ℂ: |z + 8 + 4i| < |z + 2 + 12i|} π C={ z ∈ ℂ: 0 < arg(z + 10 + 8i) < __   }    4 Shade the set of points A ∩ B ∩ C9   , that are in set A and in set B, but not in set C. 4.4 Further regions in an Argand diagram You can use inequalities to represent regions in the Argand diagram. ■ The inequality θ 1 < arg(z − z  1) < θ2  describes a region in an Argand diagram that is enclosed by the two half-lines arg(z − z  1) = θ1  and arg(z − z  1) = θ2  , and also includes the two halflines, but does not include the point represented by z1. Im z1 O M04A_IAL_FP2_44655_U04_046-082.indd 65 Imagine that the enclosed region in the diagram, represented by θ 1 < arg(z − z 1) < θ2  , is formed by rotating the half-line with argument θ 1 anticlockwise by the angle θ 2 − θ1  about the point z1. θ2 – θ1 θ2 θ1 Watch out Re The region described by θ 1  , arg(z − z 1) , θ2  would not include the two half-lines. You would use dotted lines to represent them. 10/09/19 6:27 PM 66 CHAPTER 4 FURTHER ARGAND DIAGRAMS Example 10 Describe algebraically, in terms of z, the region shown in each Argand diagram. a b Im Im 4 1 1 –3 c O Re O Re 3 d Im Im O 1 5 3 Re (3, –2) 1 O 3 Re a The region is enclosed by the two half-lines π arg(z − (−3 + i)) = 0and arg(z − (−3 + i)) = _     4 The region is described by the inequality π     0 , arg(z + 3 − i) < _ 4 b The region_ is enclosed by the two _ half-lines 5π π _ √ √ arg(z − ( 3  + i)) =     and arg(z − ( 3  + i)) = −  _  2 6 The region is described by the inequality _ 7π π      _  < arg(z − √ 3  − i) < _ 2 6 _ 5π c The initial half-line is arg(z − √ 3  − i) = −  _  6 _ π     and the terminal half-line is arg(z − √ 3  − i) = _ 2 The region is described by the inequality _ 5π π     −  _  < arg(z − √ 3  − i) < _ 6 2 These are the same half-lines as part b. You can consider this region as being formed by _ 5π rotating the half-line arg(z − √ 3   − i) = −  _  6 _   , 1) from an anticlockwise about the point (√ 3 5π π angle of −  _ to an angle of _     2 6 M04A_IAL_FP2_44655_U04_046-082.indd 66 The initial half-line is horizontal, so θ1 = 0. The gradient of the terminal half-line is 1 since it extends from (−3, 1) through ( 0, 4). so π θ 2 = _     4 Since the initial half-line is dashed it is not included (,) in the region. The terminal halfline is solid so it is included (<) in the region. Watch out The argument θ of any complex number is usually given in the range − π , θ < π. This is called the principal argument. However, you could also give the _ 7π second half-line as arg(z − (√   3 + i)) = _     6 It makes more sense to use this value in the final inequality so that the second upper value is greater than the lower value. ← Further Pure 2 Section 3 25/04/2019 08:57 FURTHER ARGAND DIAGRAMS CHAPTER 4 d The shaded region in the diagram is the intersection of a circle and its interior with the region between two half-lines. The circle and its interior is given by | z − 3 + 2i| < 2 The equation for the initial half-line is 3π arg(z − 5 + 2i) = _    and the equation for the 4 terminal half-line is arg(z − 5 + 2i) = π. So the region between the two half-lines is described 3π by the inequality _     < arg(z − 5 + 2i) < π. 4 The shaded region is given by {z ∈ ℂ : |z − 3 + 2i| < 2} 3π ∩{ z ∈ ℂ : ___     < arg(z − 5 + 2i) < π} 4 67 The set of points z satisfying the inequality |z − z1| < r is a circle and its interior with radius r and centre at the point representing z1. The initial half-line extends from (5, −2) 3π through (3, 0), so θ1 = _     4 The terminal half-line extends horizontally to the left, so θ2 = π. Notation Use set notation, with the symbol ⋂ denoting the intersection of the two sets. Example 11 On separate Argand diagrams, shade the region satisfied by each set of points: 2π a{z ∈ ℂ :  _  < arg z < π} ⋂ { z ∈ ℂ : |z + 3 − 4i| < 5} 3 b{z ∈ ℂ : 2|z − 4| < |z|} ⋂ { z ∈ ℂ : 4 < Re(z) < 6} a 2π 3 Im The region described by the inequality 2π _     < arg z < πis between the two half-lines 3 2π arg z = _     and arg z = π. 3 2π < arg z < π 3 Problem-solving π O and |z + 3 – 4i| < 5 Re If you have to sketch a union or intersection of regions on an Argand diagram, it is helpful to sketch each region separately first. Im 8 (–3, 4) –6 M04A_IAL_FP2_44655_U04_046-082.indd 67 O Re 25/04/2019 08:57 68 CHAPTER 4 FURTHER ARGAND DIAGRAMS Therefore the intersection is z [ ℂ:2π < arg z < π ù {z [ ℂ :|z + 3 – 4i| < 5} 3 2π Im 3 { } 8 Online π –6 Re O b Let z = x + iy. 2| x − 4 + iy| < | x + iy| 2 2 |(x − 4) + iy|2 < | x + iy|2 2 4 (x  − 8x + 16 + y  2) < x  2 + y  2 3x  2 + 3y  2 − 32x + 64 < 0 x  2 + y   2 − ___   32 + ___   64 3x 3 < 0 64 256 16 __ ___ ___ ( x −   3 )  + y  2 <   9 −   3 2 Explore this region using GeoGebra. Problem-solving The equation 2|z − 4| = |z| represents a circle, so the inequality 2|z − 4| < |z| represents a region consisting of either a circle and its interior, or a circle and the region outside it. You need to use an algebraic approach to find the centre and radius of the circle. 64 16 __ ___ ( x −   3 )  + y  2 <   9 2 Therefore 2|z − 4| < |z| describes the region consisting of the circle with centre __ 8 (__   16 3 , 0) and radius   3 and its interior. The region described by 4 < Re(z) < 6is the region between, and including, the vertical lines x = 4and x = 6. So {z ∈ ℂ : 2|z − 4| < |z|} ∩ { z ∈ ℂ : 4 < Re(z) < 6} describes the region shaded below. Im x=4 x=6 Complete the square. The locus of points satisfying the equation Re(z) = 4 is the vertical line x = 4, and the locus of points satisfying the equation Re(z) = 6 is the vertical line x = 6. (163 , 0) O 8 3 M04A_IAL_FP2_44655_U04_046-082.indd 68 8 Re 25/04/2019 08:57 FURTHER ARGAND DIAGRAMS CHAPTER 4 69 4D Exercise 1 On separate Argand diagrams, shade the regions, R, described by: π a0 < arg(z − 4 − i) < _     b − 1 < Im(z) < 2 2 π π 1 c _  < |z| , 1 d −  _  < arg(z + i) < _     2 3 4 2 The region R in an Argand diagram is satisfied by the inequalities |z| < 5 and |z| < |z − 6i|. Draw an Argand diagram and shade in the region R. 3 Shade on an Argand diagram the region satisfied by the set of points P(x, y), where |z + 1 − i| < 1 3π and 0 < arg z , _     4 4 Shade on an Argand diagram the region, R, satisfied by the set of points P(x, y), where |z| , 3 π and _     < arg(z + 3) < π 4 E E/P 5 On separate Argand diagrams, shade the regions, R, defined by the sets of points: π π a{z ∈ ℂ : − _     < arg(z + 1 + i) < −  _}    ⋂ { z ∈ ℂ : |z + 1 + 2i| < 1} 2 4 {z ∈ ℂ : 2|z − 6| < |z − 3|}  ⋂ { z ∈ ℂ : Re(z) < 7} b (4 marks) 6 a Shade on an Argand diagram the region defined by |z + 6| < 3 (2 marks) (4 marks) b The complex number z satisfies |z + 6| < 3. Find the range of possible values of arg z. (4 marks) E/P 7 a Indicate on an Argand diagram the region consisting of the set of points satisfying 3π both _     < arg(z − 8) < π and Im(z) < Re(z). (3 marks) 4 b Find the exact area of this region. (3 marks) E/P 8 a Shade on an Argand diagram the region R defined by π {  z ∈ ℂ : |z − 3 + 2i| > √ 2  |z − 1|}   ⋂ { z ∈ ℂ : 0 < arg(z + 1 + 2i) < _   }    _ b Find the exact area of region R. 3 c The complex number z lies in region R. Find the maximum value of Im(z). (4 marks) (3 marks) (5 marks) Challenge On an Argand diagram, shade the set of points {  z ∈ ℂ : 6 < Re((2 − 3i)z) , 12  } ⋂ {  z ∈ ℂ : (Rez)(Im z) > 0  } M04A_IAL_FP2_44655_U04_046-082.indd 69 10/09/19 6:27 PM 70 CHAPTER 4 FURTHER ARGAND DIAGRAMS 4.5 Transformations of the complex plane You need to be able to transform simple loci, such as lines and circles, from one complex plane (the z-plane) to another complex plane (the w-plane). Transformations will map points in the z-plane to points in the w-plane by applying a formula relating z = x + iyto w = u + iv. Notation The convention is to use u for the real part and v for the imaginary part of a complex number in the w-plane. It is helpful to be able to recognise the type of transformation – translation, enlargement or rotation – from the formula for some simple transformations. Example 12 The point P represents the complex number z on an Argand diagram, where |z| = 2. T1, T2 and T3 represent transformations from the z-plane, where z = x + iy, to the w-plane where w = u + iv. Describe the locus of the image of P under the transformations: 1 b T2: w = 3z c T3: w = _  2 z + i a T1: w = z − 2 + 4i The locus of P in the z-plane is a circle with centre (0, 0) and radius 2. This is the locus of P in the z-plane before any transformations have been applied. y 2 |z| = 2 2 O Rearrange to make z the subject. x Apply the modulus to both sides of the equation. Use |z| = 2 a T1: w = z − 2 + 4i ⇒ w + 2 − 4i = z ⇒ |w + 2 − 4i| = |z| ⇒ |w + 2 − 4i| = 2 v (–2, 4) P(u, v) –2 O b T2: w = 3z ⇒ |w| = |3z| ⇒ |w| = |3||z| ⇒ |w| = 3(2) = 6 M04A_IAL_FP2_44655_U04_046-082.indd 70 The image of the locus of P under T1 is |w + 2 − 4i| = 2. This is a circle with centre (−2, 4) and radius 2. 6 Problem-solving 4 The transformation T1: w = z − 2 + 4i represents a −2 translation of z by the vector ( )  4 2 Apply the modulus to both sides of the equation. u Use |z1z2| = |z1||z2| Use |z| = 2 Online Explore these transformations using GeoGebra. 25/04/2019 08:57 FURTHER ARGAND DIAGRAMS CHAPTER 4 71 The image of the locus of P under T2 is |w| = 6. This is a circle with centre (0, 0) and radius 6. v 6 Problem-solving 6 u O The transformation T2: w = 3z represents an enlargement of z by scale factor 3 with centre (0, 0). Rearrange to make _ 12 z the subject. P(u, v) Apply the modulus to both sides of the equation. c T3: w = __   21 z + i 1 ⇒ w − i =  __ 2 z Use |z1z2| = |z1||z2| 1 ⇒ |w − i| = | __ 2 ||z| Use |z| = 2 1 ⇒ |w − i| = | __ 2 z| __1 ⇒ |w − i| =   2 (2) = 1 The image of the locus of P under T3 is |w − i| = 1. This is a circle with centre (0, 1) and radius 1. v P(u, v) Problem-solving 1 –1 O 1 u The transformation T3: w = _  12 z + i represents an enlargement of z by scale factor  _12 about the point (0, 0), followed by a translation by the 0 vector (  ) 1 Example 13 π For the transformation w = iz − 1, find the locus of w when z lies on the half-line arg(z + 2) = _     4 = iz − 1 w ⇒ iz = w + 1 w 1 ⇒ z = _     + _    i i ⇒ z = − iw − i π arg(z + 2) = _     4 π     ⇒ arg(− iw − i + 2) = _ 4 π     ⇒ arg(− iw + 2 − i) = _ 4 π     ⇒ arg(− i(w + 2i + 1)) = _ 4 ⇒ arg(− i) + arg(w + 1 + 2i) = π π     ⇒ −  _  + arg(w + 1 + 2i) = _ 2 4 3π     ⇒ arg(w + 1 + 2i) = _ 4 M04A_IAL_FP2_44655_U04_046-082.indd 71 Rearrange the transformation formula w = iz − 1 to make zthe subject. Substitute − iw − ifor z. π _     4 Use arg(z 1 z 2) = arg(z 1) + arg(z 2) 25/04/2019 08:57 72 CHAPTER 4 FURTHER ARGAND DIAGRAMS The locus of points in the w-plane is the half‑line, 3π arg(w + 1 + 2i) = _    , that extends from the point 4 3π    to the horizontal ( − 1, −2)at an angle of _ 4 extending to the left of (− 1, −2). v O arg (w + 1 + 2i) = 3π 4 u 3π 4 (–1, –2) Problem-solving The transformation w = iz − 1represents an π anticlockwise rotation through _    about the origin 2 −1 followed by a translation by the vector ( )  . 0 Im arg (z + 2) = π 4 –2 O Re arg (w + 1 + 2i) = 3π 4 (–1, –2) Examples 6 and 7 lead to the following general results: a ■ w = z + a + ib represents a translation by the vector (  ), where a, b ∈ ℝ. b ■ w = kz, where k ∈ ℝ, represents an enlargement by scale factor k with centre (0, 0), where k ∈ ℝ. π ■ w = iz represents an anticlockwise rotation through  _ about the origin. 2 Compound transformations, such as the one in Example 7, are represented by transformation formulae which combine more than one of the characteristics listed above. For example, the transformation formula w = kz + a + ibrepresents an enlargement by scale factor kwith centre (0, 0) a followed by a translation by the vector (  ), where a, b, k ∈ ℝ. b Example 14 A transformation from the z-plane to the w-plane is given by w = z  2, where z = x + iy and w = u + iv. Describe the locus of w and give its Cartesian equation when z lies on: a a circle with equation x  2 + y  2 = 16 b the line with equation x = 1 M04A_IAL_FP2_44655_U04_046-082.indd 72 Notation A Cartesian equation for a locus in the z-plane will be in terms of xand ybecause z = x + iy. However, a Cartesian equation for a locus in the w-plane will be in terms of uand v because w = u + iv. 25/04/2019 08:57 FURTHER ARGAND DIAGRAMS CHAPTER 4 a |z| = 4 w = z2 ⇒ |w| = |z2| ⇒ |w| = |z||z| ⇒ |w| = 4 × 4 ⇒|w| = 16 Hence the locus of w is a circle with centre (0, 0) and radius 16, and the Cartesian equation for the locus of w is u  2 + v  2 = 16  2 = 256 b Let z = 1 + iy w = z  2⇒ w = ( 1 + iy)  2 ⇒ w = (1 − y  2) + 2yi So u = 1 − y  2 and v = 2y − 4u = −4 + 4y  2 and v  2 = 4y  2 The Cartesian equation for w is v  2 = −4u + 4 The locus of w is a parabola that is symmetric about the real axis, with vertex at (1, 0), as shown in the diagram. 73 This has Cartesian equation x  2 + y  2 = 16 Take the modulus of each side of the equation. Use |z1z2| = |z1||z2|, where z1 = z2 = z Use |z| = 4 The line x = 1 in the z-plane is the locus of Re(z) = 1 This is a parametric equation of a curve in the w-plane with y as the parameter. Problem-solving A Cartesian equation in the w-plane should be in terms of u and v. You need to eliminate y from the equations. v v2 = –4u + 4 O 1 u az + b ■ You need to be able to apply transformation formulae of the form w =  _ cz + d where a, b, c, d ∈ ℂ, that map points in the z-plane to points in the w-plane. Example 15 The transformation T from the z-plane, where z = x + iy, to the w-plane, where w = u + iv, is given 5iz + i by w = _     , z ≠ −1. z+1 a Show that the image, under T, of the circle |z| = 1 in the z-plane is a line l in the w-plane. b Sketch l on an Argand diagram. M04A_IAL_FP2_44655_U04_046-082.indd 73 25/04/2019 08:57 74 CHAPTER 4 FURTHER ARGAND DIAGRAMS 5iz + i a w = _     z+1 ⇒ w(z + 1) = 5iz + i ⇒ wz + w = 5iz + i ⇒ wz − 5iz = i − w ⇒ z(w – 5i) = i − w i−w ⇒z=_     w − 5i i−w   So |z| = _   w − 5i |i − w| ⇒ 1 =  ________  |w − 5i| ⇒ |w − 5i| = |i − w| ⇒ |w − 5i| = |(−1)(w − i)| ⇒ |w – 5i| = |−1||w − i| ⇒ |w − 5i| = |w − i| v b | Rearrange the transformation equation to make z the subject of the equation. | Take the modulus of each side of the equation. z 1 |z 1| Use |z| = 1 and _   z    = _   z    | | 2 Take out a factor of −1 on the RHS. Use |z1z2| = |z1||z2| (0, 5) v=3 l (0, 1) O | 2| u As you are working in the w-plane, plot v against u. |w − 5i| = |w − i| is in the form |w − w1| = |w − w2| so represents points on the perpendicular bisector of the line segment joining (0, 1) and (0, 5). Therefore the line l has equation v = 3 Therefore the image of |z| = 1, under T, is the line l with equation v = 3. Example 16 The transformation T from the z-plane, where z = x + iy, to the w-plane, where w = u + iv, is given 3z − 2 by w = _     , z ≠ − 1. z+1 Show that the image, under T, of the circle with equation x2 + y2 = 4 in the z-plane is a circle C in the w-plane. State the centre and radius of C. 3z − 2 w = _     z+1 ⇒ w(z + 1) = 3z − 2 ⇒ wz + w = 3z − 2 ⇒ w + 2 = 3z − wz ⇒ w + 2 = z(3 − w) w+2 ⇒  _ = z 3−w M04A_IAL_FP2_44655_U04_046-082.indd 74 3z − 2 Rearrange the transformation equation w = _     z+1 to make z the subject of the equation. 25/04/2019 08:57 FURTHER ARGAND DIAGRAMS CHAPTER 4 x2 + y2 = 4 can also be written as |z| = 2. w+2 _   = |z| = 2 3−w |w + 2| ⇒  _  = 2 | 3 − w| ⇒ |w + 2| = 2|3 − w| ⇒ |w + 2| = 2|−1||w − 3| ⇒ |w + 2| = 2|w − 3| ⇒ |u + iv + 2| = 2|u + iv − 3| ⇒ |(u + 2) + iv| = 2|(u − 3) + iv| ⇒ |(u + 2) + iv|2 = 22|(u − 3) + iv|2 ⇒ (u + 2)2 + v2 = 4((u − 3)2 + v2) ⇒ u2 + 4u + 4 + v2 = 4(u2 − 6u + 9 + v2) ⇒ u2 + 4u + 4 + v2 = 4u2 − 24u + 36 + 4v2 ⇒ 3u2 − 28u + 3v2 + 32 = 0 ___ 32 2 ⇒ u2 − ___   28 3 u + v +   3 = 0 | | 2 196 ___ ___ 32 2 ⇒ ( u − __   14 3 )  −   9 + v  +   3 = 0 2 ⇒ ( u − __   14   2 = ___   100 9 3 )  + v 75 x2 + y2 = 4 is the equation of a circle with centre (0, 0) and radius 2. Take the modulus of each side of the equation. z 1 |z 1| Use _   z    = _   z    and |z| = 2. | | 2 | 2| Write w as u + iv. Group the real and imaginary parts. Square both sides. Remove the moduli. Complete the square for u. Therefore the image of x2 + y2 = 4, under T, __ 10 is a circle C with centre (__   14 3 , 0) and radius   3 Example 17 iz − 2 A transformation T of the z-plane to the w-plane is given by w = _     , z ≠ 1. 1−z Show that as z lies on the real axis in the z-plane, then w lies on a line l in the w-plane. Sketch l on an Argand diagram. iz − 2 w = _   1−z ⇒ w(1 − z) = iz − 2 ⇒ w − wz = iz − 2 ⇒ w + 2 = wz + iz ⇒ w + 2 = z(w + i) w+2 ⇒  _  = z w+i u + iv + 2 So z = _     u + iv + i (u + 2) + iv     ⇒ z =  __________ u + i(v + 1) (u + 2) + iv u − i(v + 1)    ⇒ z = __________       × ___________     u + i(v + 1) u − i(v + 1) u(u + 2) − i(u + 2)(v + 1) + iuv + v(v + 1)            ⇒ z =  ____________________________________ u2 + (v + 1)2 M04A_IAL_FP2_44655_U04_046-082.indd 75 iz − 2 Rearrange the transformation equation w = _   1−z to make z the subject of the equation. Write w as u + iv. Group the real and imaginary parts. Multiply the numerator and denominator by the complex conjugate of u + i(v + 1). Use the difference of two squares. 25/04/2019 08:57 76 CHAPTER 4 FURTHER ARGAND DIAGRAMS u(u + 2) + v(v + 1) uv − (u + 2)(v + 1) ⇒ z = ________________          + i ________________          2 2 ( u + (v + 1) u2 + (v + 1)2 ) u(u + 2) + v(v + 1) uv − (u + 2)(v + 1) So x + iy =  ________________         + i ________________          ( u2 + (v + 1)2 ) u2 + (v + 1)2 Since z lies on the real axis, y = 0. u(u + 2) + v(v + 1) uv − (u + 2)(v + 1) So x + 0i =  ________________         + i ________________          ( u2 + (v + 1)2 ) u2 + (v + 1)2 uv − (u + 2)(v + 1) Hence, 0 =  ________________         u2 + (v + 1)2 ⇒ uv − (u + 2)(v + 1) = 0 ⇒ uv − (uv + u + 2v + 2) = 0 ⇒ uv − uv − u − 2v − 2 = 0 ⇒ 2v = −u − 2 1 So w lies on the line with equation v = −  __ 2 u − 1. v O –1 Exercise Write z as x + iy. Equate the imaginary parts. Multiply both sides by u2 + (v + 1)2 Rearrange to make v the subject. As you are working in the w-plane, plot v against u. v = – 21 u – 1 –2 Group the real and imaginary parts. u The line l has equation v = − _12 u − 1 and cuts the coordinate axes at (−2, 0) and (0, −1). 4E y 1 Consider the triangle shown on the right in the z-plane. For each of the 3 transformation formulae: z3 2 i sketch the image of the triangle by plotting the images of 1 z z2 z 1, z 2 and z 3, in the w-plane 1 ii give a geometrical description of the mapping from the z-plane O 1 2 3 x to the w-plane. a w = z − 3 + 2i b w = 2z c w = iz − 2 + i d w = 3z − 2i −2 2 A transformation T from the z-plane to the w-plane is a translation by the vector ( )  followed 3 by an enlargement with scale factor 4 and centre O. Write down the transformation T in the form w = az + b, where a, b ∈ ℂ. 3 Determine the formula for a transformation from the z-plane to the w-plane in such a way that the locus of w points is the image of the locus of z points rotated 90° anticlockwise and enlarged by a scale factor of 4, both about the point (0, 0). 4 For the transformation w = 2z − 5 + 3i, find the Cartesian equation of the locus of w as z moves on the circle |z − 2| = 4. 5 For the transformation w = z − 1 + 2i, sketch on separate Argand diagrams the locus of w when z lies on: π a the circle |z − 1| = 3 b the half-line arg(z − 1 + i) = _     4 c the line y = 2x M04A_IAL_FP2_44655_U04_046-082.indd 76 25/04/2019 08:57 FURTHER ARGAND DIAGRAMS CHAPTER 4 77 1 6 For the transformation w =  _ z  , z ≠ 0, describe the locus of w when z lies on: π b the half-line with equation arg z = _     a the circle |z| = 2 4 c the line with equation y = 2x + 1 E E E/P E 7 For the transformation w = z2, a show that as z moves once round a circle with centre (0, 0) and radius 3, w moves twice round a circle with centre (0, 0) and radius 9 (6 marks) b find the locus of w when z lies on the real axis (2 marks) (2 marks) c find the locus of w when z lies on the imaginary axis. i 2   , z ≠ _     8 The transformation T from the z-plane to the w-plane is given by w = _   2 i − 2z The circle with equation |z| = 1 is mapped by T onto the curve C. a i Show that C is a circle. ii Find the centre and radius of C. The region |z| < 1 in the z-plane is mapped by T onto the region R in the w-plane. b Shade the region R on an Argand diagram. 1   , z ≠ 2, show that the image, under T, of the circle 9 For the transformation w = _   2−z with centre O, and radius 2 in the z-plane is a line l in the w-plane. Sketch l on an Argand diagram. (8 marks) (2 marks) (6 marks) z−i   , z ≠ − i. 10 A transformation from the z-plane to the w-plane is given by w = _   z+i a Show that the circle with equation |z − i| = 1 in the z-plane is mapped to a circle in the w-plane, giving an equation for this circle. (5 marks) b Sketch the new circle on an Argand diagram. (1 mark) 11 The transformation T from the z-plane, where z = x + iy, to the w-plane where w = u + iv, is 3   , z ≠ 2. given by w = _   2−z Show that, under T, the straight line with _ equation 2y = x is transformed to a circle in the √ 5  3 3 3 w-plane with centre ( _  4 ,  _2 ) and radius  _   (7 marks) 4 E/P 12 The transformation T from the z-plane, where z = x + iy, to the w-plane, where w = u + iv, is − iz + i given by w = _     , z ≠ − 1. z+1 a The transformation T maps the points on the circle with equation x2 + y2 = 1 in the z-plane, to points on a line l in the w-plane. Find the Cartesian equation of l. (4 marks) b Hence, or otherwise, shade and label on an Argand diagram the region R of the (2 marks) w-plane which is the image of |z| < 1 under T. c Show that the image, under T, of the circle with equation x2 + y2 = 4 in the z-plane is a circle C in the w-plane. Find the equation of C. (4 marks) E/P E 13 The transformation T from the z-plane, where z = x + iy, to the w-plane, where w = u + iv, is 4z − 3i given by w =  _  , z ≠ 1. z−1 Show that the circle |z| = 3 is mapped by T onto a circle C, and state the centre and radius of C. (6 marks) M04A_IAL_FP2_44655_U04_046-082.indd 77 25/04/2019 08:57 78 CHAPTER 4 FURTHER ARGAND DIAGRAMS E 14 The transformation T from the z-plane, where z = x + iy, to the w-plane, where w = u + iv, is 1 given by w =  _  , z ≠ − i. z+i a Show that the image, under T, of the real axis in the z-plane is a circle C1 in the (5 marks) w-plane and find the equation of C1. b Show that the image, under T, of the line x = 4 in the z-plane is a circle C2 in the (5 marks) w-plane, and find the equation of C2. E/P 15 The transformation T from the z-plane, where z = x + iy, to the w-plane where w = u + iv, is 4 given by w = z +  _ z  , z ≠ 0. Show that the transformation T maps the points on a circle |z| = 2 to points in the interval (7 marks) [−k, k] on the real axis. State the value of the constant k. E/P 16 The transformation T from the z-plane, where z = x + iy, to the w-plane, where w = u + iv, is 1 given by w = _     , z ≠ − 3. z+3 Show that T maps the line with equation 2x − 2y + 7 = 0 onto a circle C, and state the centre (6 marks) and the exact radius of C. Challenge A transformation T : w = az + b, a, b ∈ ℂmaps the complex numbers 0, 1 and 1 + i in the z-plane to the points 2i, 3i and − 1 + 3i, respectively, in the w-plane. Find a and b. Chapter review 4 E/P 1 The point P represents a complex number z in an Argand diagram. Given that |z + 1 − i| = 1 P a find a Cartesian equation for the locus of P (2 marks) b sketch the locus of P on an Argand diagram (2 marks) c find the greatest and least possible values of |z| (2 marks) d find the greatest and least possible values of |z − 1|. π 2 Given that arg(z − 2 + 4i) =  __  4 a sketch the locus of P(x, y) which represents z on an Argand diagram (2 marks) b find the minimum value of |z| for points on this locus. E/P 3 The complex number z satisfies |z + 3 − 6i| = 3. Show that the exact maximum value of π 1 arg zin the interval (−π, π) is  __  + 2 arcsin ___   __  (4 marks) (√ 2 5  ) E/P 4 A complex number z is represented by the point P on the Argand diagram. Given that |z − 5| = 4, a sketch the locus of P. π b Find the complex numbers that satisfy both |z − 5| = 4 and arg(z + 3i) = __     3 giving your answers in radians to 2 decimal places. (2 marks) (6 marks) c Given that arg (z + 5) = θ and |z − 5| = 4 have no common solutions, find the range (3 marks) of possible values of θ, −π , θ , π. M04A_IAL_FP2_44655_U04_046-082.indd 78 25/04/2019 08:57 FURTHER ARGAND DIAGRAMS E/P E/P E/P CHAPTER 4 79 5 Given that |z + 5 − 5i| = |z − 6 − 3i| a sketch the locus of z (3 marks) b find the Cartesian equation of this locus (3 marks) c find the least possible value of |z|. (3 marks) 6 a Find the Cartesian equation of the locus of points that satisfies |z − 4| = |z − 8i| π b Find the value of z that satisfies both |z − 2| = |z − 4i| and arg z = __     4 c Shade on an Argand diagram the set of points π {z ∈ ℂ : |z − 4| < |z − 8i|} ∩ { z ∈ ℂ : __     < arg z < π} 4 (3 marks) (3 marks) (3 marks) 7 a Find the Cartesian equations of: i the locus of points representing |z − 3 + i| = |z − 1 − i| __ ii the locus of points representing |z − 2| = 2√2  __ b Find the two values of z that satisfy both |z − 3 + i| = |z − 1 − i| and |z − 2| = 2√2  __ (6 marks) (2 marks) The region R is defined by the inequalities |z − 3 + i| > |z − 1 − i| and |z − 2| < 2√2  c Show the region R on an Argand diagram. (4 marks) 8 For each equation: i use an algebraic approach to determine a Cartesian equation for the locus of z on an Argand diagram ii describe the locus geometrically. a |z| = |z − 4| b |z| = 2|z − 4| E/P E E E/P _ (3 marks) 9 a Sketch the locus of points that satisfies the equation |z − 2 + i| = √ 3  The half-line L with equation y = mx − 1, x ≥ 0, m > 0is tangent to the locus from part a at point A. (3 marks) b Find the value of m. (2 marks) c Write an equation for L in the form arg(z − z 1) = θ, z 1 ∈ ℂ, − π , θ < π. d Find the complex number a represented by point A. (3 marks) 10 a Find the Cartesian equation of the locus of points representing |z + 2| = |2z − 1| (3 marks) π _ b Find the value of z which satisfies both |z + 2| = |2z − 1| and arg z =     (3 marks) 4 c Hence shade in the region R on an Argand diagram which satisfies both |z + 2| >|2z − 1| π and  _  < arg z < π (2 marks) 4 π z − 4 − 2i   =_       11 Given that arg(_ ) 2 z − 6i a sketch the locus of P(x, y) which represents z on an Argand diagram (4 marks) b deduce the exact value of |z − 2 − 4i|. (2 marks) 12 A curve has equation 2|z + 3| = |z − 3|, where z ∈ ℂ. a Show that the curve is a circle with equation x  2 + y  2 + 10x + 9 = 0 b Sketch the curve on an Argand diagram. The line L has equation bz * + b * z = 0, where b ∈ ℂ and z ∈ ℂ. c Given that the line L is a tangent to the curve and that arg b = θ, find the possible values of tan θ. M04A_IAL_FP2_44655_U04_046-082.indd 79 (2 marks) (2 marks) (5 marks) 25/04/2019 08:57 80 E/P E CHAPTER 4 FURTHER ARGAND DIAGRAMS z − 5 − 2i π 13 A curve S is described by the equation arg(_         = _ z − 1 − 6i ) 2 a Show that S is a semicircle, and find its centre and radius. b Find the maximum value of |z|, and express it exactly. (5 marks) (4 marks) 14 a Indicate on an Argand diagram the region, R, consisting of the set of points satisfying the inequality 2 < |z − 2 − 3i| < 3 b Find the exact area of region R. c Determine whether or not the point represented by 4 + i lies inside R. (3 marks) (2 marks) (3 marks) E 15 The transformation T from the z-plane, where z = x + iy, to the w-plane where w = u + iv, 1 is given by w = _   z  , z ≠ 0. 1 a Show that the image, under T, of the line with equation x = _  2 in the z-plane is a (4 marks) circle C in the w-plane. Find the equation of C. b Hence, or otherwise, shade and label on an Argand diagram the region R of the 1 (3 marks) w-plane which is the image of x > _  2 under T. E 16 The point P represents the complex number z on an Argand diagram. Given that |z + 4i| = 2 a sketch the locus of P on an Argand diagram. b Hence find the maximum value of |z|. T1, T2, T3 and T4 represent transformations from the z-plane to the w-plane. Describe the locus of the image of P under the transformations: c i T1: w = 2z ii T2: w = iz iii T3: w = −iz iv T4: w = z* (2 marks) (3 marks) (8 marks) E 17 The transformation T from the z-plane, where z = x + iy, to the w-plane where w = u + iv, is z+2 given by w =  _  , z ≠ − i. z+i a Show that the image, under T, of the imaginary axis in the z-plane is a line l in the w-plane. Find the equation of l. (4 marks) b Show that the image, under T, of the line y = x in the z-plane is a circle C in the w-plane. _ √ 10  _ (5 marks) Find the centre of C and show that the radius of C is     2 E/P 18 The transformation T from the z-plane, where z = x + iy, to the w-plane where w = u + iv, 4−z is given by w =  _  , z ≠ − i. z+i The circle |z| = 1 is mapped by T onto a line l. Show that l can be written in the form au + bv + c = 0, where a, b and c are integers to be determined. (5 marks) E/P 19 The transformation T from the z-plane, where z = x + iy, to the w-plane where w = u + iv, 3iz + 6 is given by w = _     , z ≠ 1. 1−z Show that the circle |z| = 2 is mapped by T onto a circle C. State the centre of C and show _ that the radius of C can be expressed in the form k √ 5  where k is an integer to be determined. (5 marks) M04A_IAL_FP2_44655_U04_046-082.indd 80 25/04/2019 08:57 FURTHER ARGAND DIAGRAMS E/P CHAPTER 4 az + b 20 The mapping from the z-plane to the w-plane given by w =  ______ z + c  , z, w ∈ ℂ, a, b, c ∈ ℝ maps the origin onto itself, and reflects the point 1 + 2i in the real axis. a Find the values of a, b and c. A second complex number ω is also mapped to itself. b Find ω. E/P 81 (5 marks) (5 marks) az + b 21 A transformation from the z-plane to the w-plane is defined by w = _   z + c , where a, b, c ∈ ℝ. Given that w = 1 when z = 0 and that w = 3 − 2i when z = 2 + 3i, a find the values of a, b and c (5 marks) b find the exact values of the two points in the complex plane which remain invariant under the transformation. (5 marks) 22 The transformation T from the z-plane, where z = x + iy, to the w-plane where w = u + iv, z+i is given by w = _   z  , z ≠ 0. a The transformation T maps the points on the line with equation y = x in the z-plane, other than (0, 0), to points on the line l in the w-plane. Find an equation of l. (4 marks) b Show that the image, under T, of the line with equation x + y + 1 = 0 in the z-plane (4 marks) is a circle in the w-plane, where C has equation u2 + v2 − u + v = 0. c On the same Argand diagram, sketch l and C. (3 marks) Challenge π 1 The complex number z satisfies arg(z − 3 + 3i) = – __  4 The complex number w is such that |w − z| = 3. a Sketch the locus of w. b State the exact minimum value of |w|. 2 The complex function f maps any point in an Argand diagram represented by z = x + iyto its reflection in the line x + y = 1. Express f in the form f(z) = az* + b, where a, b ∈ ℂ. Summary of key points 1 You can represent complex numbers on an Argand diagram. The x-axis on an Argand diagram is called the real axis and the y-axis is called the imaginary axis. The complex number z = x + iy is represented on the diagram by the point P(x, y), where x and y are Cartesian coordinates. x 2 The complex number z = x + iy can be represented as the vector ( y )  on an Argand diagram. 3 For two complex numbers z1 = x1 + iy1 and z2 = x2 + iy2, |z2 − z1| represents the distance between the points z1 and z2 on an Argand diagram. 4 Given z1 = x1 + iy1, the locus of points z on an Argand diagram such that |z − z1| = r, or |z − (x1 + iy1)| = r, is a circle with centre (x1, y1) and radius r. M04A_IAL_FP2_44655_U04_046-082.indd 81 25/04/2019 08:57 82 CHAPTER 4 FURTHER ARGAND DIAGRAMS 5 Given z1 = x1 + iy1 and z2 = x2 + iy2, the locus of points z on an Argand diagram such that |z − z1| = |z − z2| is the perpendicular bisector of the line segment joining z1 and z2. 6 Given z1 = x1 + iy1, the locus of points z on an Argand diagram such that arg (z − z1) = θ is a half-line from, but not including, the fixed point z1 making an angle θ with a line from the fixed point z1 parallel to the real axis. 7 The locus of points z that satisfy |z – a| = k|z – b|, where a, b ∈ ℂ and k ∈ ℝ, k > 0, k ≠1 is a circle. z−a 8 The locus of points z that satisfy arg(  _ ) = θ, where θ ∈ ℝ, θ . 0 and a, b ∈ ℂ, is an arc of a z−b circle with endpoints A and B representing the complex numbers a and b, respectively. The endpoints of the arc are not included in the locus. π • If θ ,  __ , then the locus is a major arc of the circle. 2 π __ • If θ .    , then the locus is a minor arc of the circle. 2 π __ • If θ =    , then the locus is a semicircle. 2 9 The inequality θ  1 < arg(z − z 1) < θ  2describes a region in an Argand diagram that is enclosed by the two half-lines arg(z − z 1) = θ  1 and arg(z − z 1) = θ  2, and also includes the two half-lines, but does not include the point represented by z1. a 10 • w = z + a + ibrepresents a translation by the vector (   )  , where a, b ∈ ℝ. b • w = kz, where k ∈ ℝ, represents an enlargement by scale factor k with centre (0, 0), where k ∈ ℝ. π • w = iz represents an anticlockwise rotation through  __ about the origin. 2 az + b 11 You need to be able to apply transformation formulae of the form w =   _  , where cz + d a, b, c, d ∈ ℂ, that map points in the z-plane to points in the w-plane. M04A_IAL_FP2_44655_U04_046-082.indd 82 25/04/2019 08:57 REVIEW EXERCISE 1 1 Review exercise E/P 2 1 1 Use algebra to solve  _____  , _____     x−2 x+1 c Hence, or otherwise, solve the inequality 4x 2x  _____  < _______     2 − x (x + 1)  2 giving your answer using set notation. (2) (6) ← Further Pure 2 Section 1.1 E 2 Find the set of values of x for which x2   _____  . 2x x−2 (5) ← Further Pure 2 Section 1.2 ← Further Pure 2 Section 1.1 E 3 Find the set of values of x for which x2 − 12   _______ x  . 1 E (5) 4 Find the set of values of x for which 3 2x − 5 . __  x  giving your answer using set notation. (5) c Write down the solution to the inequality |x − 5| , |3x − 2| 5 Given that k is a constant and that k . 0, find, in terms of k, the set of values of x+k k x for which ______       (7) . __ x + 4k x E 6 a On the same set of axes, sketch the graphs of 2 y = 2 − x and y = −  _____ (3) x−1 b Find the points of intersection of 2 y = 2 − x and y = −  _____ (2) x−1 c Write down the solution to the inequality 2 2 − x . −  _____  (2) x−1 ← Further Pure 2 Section 1.2 E 7 a On the same set of axes sketch the 4x 2x graphs of y =  _____  and y = _______   (4)   2−x (x + 1)  2 b Find the points of intersection of 4x 2x y = _____     (2)   and y = _______   2−x (x + 1)  2 M04B_IASL_FP2_44655_RE1_083-089.indd 83 9 a Sketch the graph of y = |x + 2| (2) b Use algebra to solve the inequality 2x . |x + 2| (4) ← Further Pure 2 Section 1.3 ← Further Pure 2 Section 1.1 E (2) ← Further Pure 2 Section 1.3 ← Further Pure 2 Section 1.1 E/P 8 a On the same set of axes, sketch the graphs of (3) y = |x − 5| and y = |3x − 2| b Finds the coordinates of the points of intersection of y = |x − 5| and (3) y = |3x − 2| ← Further Pure 2 Section 1.1 E 83 E 10 a Sketch the graph of y = |x − 2a|, given that a . 0 (2) b Solve |x − 2a| . 2x + a, where a . 0 (4) ← Further Pure 2 Section 1.3 E/P | | x 11 Solve the inequality _____     , 8 − x, x−3 giving your answer in set notation. (6) ← Further Pure 2 Section 1.3 E 12 a On the same set of axes, sketch the (3) graphs of y = x and y = |2x − 1| b Use algebra to find the coordinates of the points of intersection of the two graphs. (2) c Hence, or otherwise, find the set of values of x for which |2x − 1| . x (4) ← Further Pure 2 Section 1.3 25/04/2019 08:56 84 1 REVIEW EXERCISE E/P 13 Use algebra to find the set of real values (5) of x for which |x − 3| . 2|x + 1| a Use algebra to solve the equation x2 − 1   ______ = 3(1 − x) (6) |x + 2| b Hence, or otherwise, find the set of values of x for which x2 − 1   ______ , 3(1 − x) |x + 2| Give your answer using set notation. (2) ← Further Pure 2 Section 1.3 E/P 14 Solve, for x, the inequality |5x + a| < |2x|, where a . 0 (6) ← Further Pure 2 Section 1.3 E/P 15 a Using the same set of axes, sketch the curve with equation y = |x2 − 6x + 8| and the line with equation 2y = 3x − 9 State the coordinates of the points where the curve and the line meet the x-axis. (4) b Use algebra to find the coordinates of the points where the curve and the line intersect and, hence, solve the inequality 2|x2 − 6x + 8| . 3x − 9 (5) ← Further Pure 2 Section 1.3 E/P n n 2  ∑ ___________      = _____     n+2 r = 1 (r + 1)(r + 2) 16 a Sketch, on the same set of axes, the graph of y = |(x − 2)(x − 4)|, and the (3) line with equation y = 6 − 2x b Find the exact values of x for which (3) |(x − 2)(x − 4)| = 6 − 2x c Hence solve the inequality |(x − 2)(x − 4)| , 6 − 2x E/P E/P n n(an + b) 2 ∑     ____________     = _____________       c(n + 2)(n + 3) r = 1 (r + 1)(r + 3) where a, b and c are constants to be found. (5) 20 a Show that r+1 1 r  _____  − _____     , r  ℤ+      ___________ r + 2 r + 1 (r + 1)(r + 2) (2) b Hence, or otherwise, find n 1  ∑  ___________    giving your answer r = 1 (r + 1)(r + 2) as a single fraction in terms of n. (3) (2) y 17 19 Prove that ← Further Pure 2 Section 2.1 ← Further Pure 2 Section 1.3 E/P (5) ← Further Pure 2 Section 2.1 ← Further Pure 2 Section 1.3 E 18 Prove that ← Further Pure 2 Section 2.1 E/P 2 y= x –1 |x + 2| 2 21 f(x) =  __________________     (x + 1)(x + 2)(x + 3) a Express f(x) in partial fractions. (2) b Hence find   ∑  f(r). (3) n r=1 –2 –1 O 1 x E/P The diagram above shows a sketch of the curve with equation x2 − 1 y =   ______, x ≠ −2 |x + 2| The curve crosses the x-axis at x = 1 and x = −1 and the line x = −2 is an asymptote of the curve. M04B_IASL_FP2_44655_RE1_083-089.indd 84 ← Further Pure 2 Section 2.1 22 a Express as a simplified single fraction 1 1  _______2  − __     (2) (r − 1) r 2 b Hence prove, by the method of differences, that n 2r − 1 1 __  ∑  ________   2 (3) 2  = 1 −   n2  r = 2 r (r − 1) ← Further Pure 2 Section 2.1 25/04/2019 08:56 REVIEW EXERCISE E/P 1 23 a Prove that n n(an + b) 4 ∑     _______  = ____________       (n + 1)(n + 2) r=1 r(r + 2) where a and b are constants to be found. 100 4 b Find the value of  ∑_______    , to r = 50 r(r + 2) 4 decimal places. 85 E/P 24 a Prove that n 2 1  ∑  ______   2   = 1 − ______     2n + 1 r = 1 4r − 1 b Hence find the exact value of 20 2   2   ∑______   r = 11 4r − 1 E (5) (2) 29 Show that cos 2x + i sin 2x ______________       cos 9x − i sin 9x can be expressed in the form cos nx + i sin nx, where n is an integer to be found. (4) ← Further Pure 2 Section 3.2 (5) E/P 30 a Use de Moivre’s theorem to show that cos 5θ = 16 cos5 θ − 20 cos3 θ + 5 cos θ (4) b Hence find 3 distinct solutions of the equation 16x5 − 20x3 + 5x + 1 = 0, giving your answers to 3 decimal places where appropriate. (5) (2) ← Further Pure 2 Section 2.1 E 25 Given that for all real values of r, ← Further Pure 2 Section 3.4 (2r + 1)3 − (2r − 1)3 = Ar 2 + B where A and B are constants, a find the value of A and the value of B. (2) 31 a Use de Moivre’s theorem to show that sin 5θ = sin θ (16 cos4 θ − 12 cos2 θ + 1) (4) (3) b Hence, or otherwise, solve, for 0 < θ , π, sin 5θ + cos θ sin 2θ = 0(5) E/P b Hence show that n  ∑  r 2 = _ 6 n(n + 1)(2n + 1) 1 r=1 40 c Calculate  ∑(3r − 1)2. r=1 E/P 26 Prove that E/P ← Further Pure 2 Section 2.1 27 a Show that r3 − r + 1 1 1   ________   r − 1 + __     r  − _____   r+1 r(r + 1) for r ≠ 0, −1. (2) n 3 r −r+1 b Find  ∑    ________ , expressing your r = 1 r(r + 1) answer as a single fraction in its simplest form. (3) 32 a Use de Moivre’s theorem to show that 1 sin5 θ = __  16(sin 5θ − 5 sin 3θ + 10 sin θ) (4) b Hence, or otherwise, show that n(an + b) ______________       =       c(n + 1)(2n + 1) r = 1 r(r + 1)(r + 2) where a, b and c are constants to be found. (6) E ← Further Pure 2 Section 3.4 (2) ← Further Pure 2 Section 2.1 1 ∑    _____________ 2n (5) ← Further Pure 2 Section 2.1 ← Further Pure 2 Section 2.1 E/P n 2r + 3 28 Find  ∑  ________   r   r = 1 3 (r + 1) π _   2 8 ∫   sin5 θ dθ = __  15  0 E/P (6) ← Further Pure 2 Section 3.4 33 a Given that z = cos θ + i sin θ, show that (2) zn + z−n = 2 cos nθ b Express cos6 θ in terms of cosines of multiples of θ. (4) c Hence show that π _   2 5π ∫   cos6 θ dθ = ___     32 0 (6) ← Further Pure 2 Section 3.4 ← Further Pure 2 Section 2.1 M04B_IASL_FP2_44655_RE1_083-089.indd 85 25/04/2019 08:56 86 E 1 REVIEW EXERCISE 34 a Solve the equation E/P 5 z = 4 + 4i giving your answers in the form z = reikπ, where r is the modulus of z and k is a rational number such that (6) 0 < k < 2. E/P b Show on an Argand diagram the points representing your solutions. (2) ← Further Pure 2 Section 3.5 E giving your answers in the form reiθ, (6) where r . 0, −π , θ < π. b Show that your solutions satisfy the equation E 37 a Find, in the form re  iθ, the solutions to the equation 38 The point P represents the complex number z in an Argand diagram. Given that |z − 2 + i| = 3, a sketch the locus of P in an Argand diagram (2) 42 The complex number z satisfies the equation |z + 3 + i| = |z − 2 + i|. a Sketch the locus of z. (2) b Find the minimum value of |z|. (1) c Find the value of z that also satisfies 3π arg z = −  ___ 4 (5) ← Further Pure 2 Section 3.5 (3) ← Further Pure 2 Section 4.1 (6) b State the name of the polygon formed. (1) 41 Sketch, on an Argand diagram, the locus of the point P representing a complex number z such that π E/P The solutions form the vertices of a polygon in the Argand diagram. (2) (3) arg(z + 3 + i) =  __ 2 ← Further Pure 2 Section 3.5 E/P Given that |z − 3i| = 3, ← Further Pure 2 Section 4.1 ← Further Pure 2 Section 3.5 z   5 − 16 − 16i √  3 = 0 40 A complex number z is represented by the point P in an Argand diagram. 3π for an integer k, the value of which should be stated. (3) _ (2) ← Further Pure 2 Section 4.1 arg(z − 3i) =  ___ 4 z9 + 2k = 0 E b find the maximum value of |z|. b find the complex number z which satisfies both |z − 3i| = 3 and __ 36 Solve the equation z5 = i, giving your answers in the form cos θ + i sin θ. a sketch the locus of z on an Argand (2) diagram a sketch the locus of P 35 a Solve the equation z3 = 32 + 32√3 i E 39 Given that z stisfies |z − 2i| = 2, E/P (2) ← Further Pure 2 Section 4.1 43 A complex number z is represented by the point P on an Argand diagram. π z+i      , Given that arg(____   = __ z − i) 4 a without calculation, explain why the locus of P forms a major arc. (1) b determine the location of the centre of the circle containing this arc. (4) ← Further Pure 2 Section 4.2 b find the exact values of the maximum and minimum of |z|. (2) ← Further Pure 2 Section 4.1 M04B_IASL_FP2_44655_RE1_083-089.indd 86 25/04/2019 08:56 REVIEW EXERCISE E/P 1 44 The diagram shows the sector of a circle drawn on an Argand diagram. 87 E A B ← Further Pure 2 Section 4.2 E/P X The centre of the circle, X, represents the complex number −1 − 2i, and the arc AB is the locus of points z ∈ ℂ that satisfy z − 3i + 1    = θ, the equation arg(________ z−b ) where b ∈ ℂ, θ ∈ ℝ. ← Further Pure 2 Section 4.2 Find the exact value of p. E E/P E/P (2) 47 A curve L is _ defined in the complex plane 5 + 2i| for z ∈ ℂ. by |z − 4| = √  |z A curve M is defined in the complex _ plane by |z − 6| = √  |z 7 + 6i| for z ∈ ℂ. a Explain why L and M are similar. (2) b Find the exact scale factor of enlargement from L to M. (2) ← Further Pure 2 Section 4.2 M04B_IASL_FP2_44655_RE1_083-089.indd 87 51 Shade on an Argand diagram the set of points 3π π    < arg(z − 3 − 3i)  <  ___ } {z ∈ ℂ: − __ 2 4 (6) ∩ { z ∈ ℂ : |z − 3i|  < 3} 52 Using an Argand diagram shade the region satisfied by π {z ∈ ℂ :  __  < arg(z − 5) < π} 3 5π    } ∩   {z ∈ ℂ : 0 < arg(z − 10) < ___ 6 (4) ← Further Pure 2 Section 4.4 ← Further Pure 2 Section 4.2 E (3) ← Further Pure 2 Section 4.3 ← Further Pure 2 Section 4.2 b Deduce the value of |z + 1 − i| 50 Sketch, on an Argand diagram, the region which satisfies the following condition. 2π π  __  < arg (z − 1) <  ___  4 3 ← Further Pure 2 Section 4.3 45 On an Argand diagram a circle is defined __ 2 − i| for z ∈ ℂ. by |z − 1| = √  |z 46 A curve P is described by the equation π z − 2i arg(_____      , z ∈ ℂ.   = __ z + 2) 2 a Sketch the locus of P (4) (7) ← Further Pure 2 Section 4.2 Determine the radius and centre of this circle. (4) E 49 A circle with circumference of 24π is plotted on an Argand diagram. This circle is known to be defined by the __ equation |z − i| = √p |z + 1|, where p . 1, p ∈ ℝ and z ∈ ℂ. a Write down the complex number represented by the point A. (1) 25π    , b Given that the sector has area ____ 12 find the values of b and θ. (6) E 48 A curve P is described by the equation π z+1 __ arg(_____   z  )=   4  , z ∈ ℂ. Find the exact length of this curve. (6) E/P 53 Drawn on an Argand diagram, a shaded semicircle is defined by {z ∈ ℂ : |z − 6i| < 2|z − 3|} ∩  { z ∈ ℂ : Re(z) < k} where k ∈ ℝ. a Find k. (4) b Find the exact area of the semicircle. (2) ← Further Pure 2 Section 4.4 25/04/2019 08:56 88 1 REVIEW EXERCISE E/P 54 On an Argand diagram a triangular region is defined by π {z ∈ ℂ : 0 < arg(z − p) < __    } 4 ∩   {z ∈ ℂ : |z − p| < |z − q|} E/P where p, q ∈ ℝ. The region has an area of x, x . 0. _ Prove that q = p + √ 8x . (6) b Find the image under T in the w-plane of the circle |z| = 1 in the z-plane.(4) ← Further Pure 2 Section 4.4 E/P b Define, as a locus, the line of symmetry of the image of triangle ABC under T in the w-plane. (3) ← Further Pure 2 Section 4.5 E/P 56 A transformation from the z-plane to the w-plane is given by 2z − 1 w =  ______  z−2 Show that the circle |z| = 1 is mapped onto the circle |w| = 1. (5) ← Further Pure 2 Section 4.5 E/P c Sketch, on separate diagrams, the circle |z| = 1 in the z-plane and its image under T in the w-plane. (2) 55 Three points in the z-plane form the vertices A, B and C of an isosceles triangle. This triangle has area 8 and a line of symmetry defined by Im(z) = 4. A transformation T from the z-plane to the w-plane is defined by w = 3z + 4 − 2i a Find the area of the image of triangle ABC under T in the w-plane. (2) 57 A transformation from the z-plane to the w-plane is given by z−i w =  ____ z  a Show that under this transformation 1 the line Im z = _  2 is mapped to the circle with equation |w| = 1 (5) b Hence, or otherwise, find, in the form az + b w = ______    , where a, b, c and d ∈ ℂ, the cz + d transformation that maps the line 1 Im z = _  2 to the circle with centre 3 − i and radius 2. (4) 58 The transformation T from the z-plane to the w-plane is defined by z+1 w =  _____ , z ≠ i z+i a Show that T maps points on the halfπ line arg z = __    in the z-plane onto points 4 on the circle |w| = 1 in the w-plane. (4) d Mark on your sketches the point P where z = i and its image Q under T in the w-plane. (3) ← Further Pure 2 Section 4.5 E/P 59 A transformation of the z-plane to the w-plane, T, is given by 1 w = az +  __ z , z ∈ ℂ, z ≠ 0, a ∈ ℤ, a . 1 where z = x + iy and w = u + iv The locus of the points in the z-plane that 1   a  is mapped satisfy the equation |z| = __ under T onto a curve C in the w-plane. 1 a Given that |z| = __   a , express z in exponential form. (1) b Hence prove that C may be defined by a Cartesian equation in the w-plane as ( 1 − a)  2 u  2 + ( 1 + a)  2 v  2 = ( 1 − a  2)   2 (6) c T produces an image in the w-plane which forms an ellipse with equation u  2 v2 ___     +   __  = 1 16 4 Sketch the locus of the points on the z-plane which have been transformed under T to create this image. (3) ← Further Pure 2 Section 4.5 ← Further Pure 2 Section 4.5 M04B_IASL_FP2_44655_RE1_083-089.indd 88 25/04/2019 08:56 REVIEW EXERCISE 1 89 Challenge 1 1   1 Solve in the range 0 , x , 2π, ________   ,  _____  1 − sin x sin x 2 a Show that if ω = e    3 , then ← Further Pure 2 Section 1.2 2πi _ { 1n + ωn + (ω 2)n 1 if n is zero or a multiple of 3   _____________      = 0 otherwise 3 Let f(x) be a finite polynomial whose largest power of x is a multiple of 3, so that f(x) = a 0 + a1x + a2x2 + … + a3kx 3k where ai ∈ ℝ, k ∈ ℕ. The sum S is given by k S = a 0 + a3 + a6 + … + a3k = ∑   a  3r r=0 b f(1) + f(ω) + f(ω  2)      By considering a general term of f(x), show that S = ______________   3 15 45 2  45 − 2     = ______ Hence, by considering the binomial expansion of (1 + x)45, show that ∑  (___ c     3 r=0 3r ) M04B_IASL_FP2_44655_RE1_083-089.indd 89 ← Further Pure 2 Section 2.1 25/04/2019 08:56 90 CHAPTER 5 FIRST-ORDER DIFFERENTIAL EQUATIONS 5 FIRST-ORDER DIFFERENTIAL EQUATIONS 4.1 4.2 4.3 Learning objectives After completing this chapter you should be able to: ● Solve first-order differential equations by separation of variables and sketching members of the family of solution curves → pages 91–94 ● Solve first-order differential equations using an integrating factor → pages 95–98 ● Use a given substitution to transform a differential equation into one that can be solved → pages 98–102 Prior knowledge check 1 Find the general solution to the differential dy equation ___ = xex ← Pure 4 Section 6.6 dx 2 Find the particular solution to the dy x differential equation ___ = − _y_ given that dx y = 0 when x = 2 ← Pure 4 Section 6.6 3 Find: a 3 dt ∫ ______ 50 − 2t b ∫∫tan 4x dx M05_IAL_FP2_44655_U05_090-104.indd 90 ← Pure 3 Section 7.2 Population growth can be modelled by a differential equation. For example, the rate of change of the population of bacteria in a petri dish is proportional to the number of bacteria present, subject to the limiting factor of the amount of space on the dish. 25/04/2019 08:55 FIRST-ORDER DIFFERENTIAL EQUATIONS CHAPTER 5 91 5.1 First-order differential equations with separable variables If a first-order differential equation can be dy written in the form ___     = f(x)g(y)then you can dx 1 solve it by writing    ____     dy = ∫    f(x)dx g(y) ∫ Links This process is called separating the variables. ← Pure 4 Section 6.6 When you have integrated and found the general solution, you can let the arbitrary constant take different numerical values, thus generating particular solutions. You can then sketch a graph for each of these solutions. The curves that are sketched are called a family of solution curves. In some questions you will be given a boundary condition, such as y = 1 when x = 0. You can use this to find the arbitrary constant. Different boundary conditions will give rise to different particular solutions. The graph of each solution belongs to the family of solution curves. Example 1 Find the general solution of the differential equation dy ___     = 2 dx and sketch members of the family of solution curves represented by the general solution. Integrating gives y = 2x + C which is the general solution. The solution ‘curves’ corresponding to C = −2, −1, 0, 1, and 2 are shown below. C=2 C=0 y C = –2 1 The graphs of y = 2x + C form the family of solution ‘curves’ for this differential equation. 0.5 –1 –0.5 O –0.5 This is a straight line equation. 0.5 1 1.5 2 x These are a set of straight lines with gradient 2 and intercept C. –1 C=1 M05_IAL_FP2_44655_U05_090-104.indd 91 C = –1 25/04/2019 08:55 92 CHAPTER 5 FIRST-ORDER DIFFERENTIAL EQUATIONS Example 2 Find the general solution of the differential equation dy x ___ y      = −__ dx and sketch members of the family of solution curves represented by the general solution. ∫  ydy = − ∫  xdx Separate the variables and integrate. 1 2 ∴ __   21 y2 = −  __ 2 x + C which is the general solution. This can be written as x2 + y2 = r2, where r2 = 2C. This is a circle equation. The solution curves corresponding to C = 0.5, 1.125, 2, 4.5 are shown below. y The graphs of x2 + y2 = 2C form the family of solution curves for this differential equation. 3 2 1 –3 –2 O –1 –1 –2 1 2 3 x These are a set of circles with centre at the origin and with radius r, where r 2 = 2C. –3 Example 3 Find the general solution to the differential equation y dy  ___ = −  __ x dx and sketch members of the family of solution curves represented by the general solution. ∫ 1 __ ∫ 1 __      y dy = −      x dx ln |y| = −ln |x| + c ln |y| + ln |x| = c ln |xy| = c |xy| = ec A c y = ±__ x , where A = e M05_IAL_FP2_44655_U05_090-104.indd 92 Separate the variables and integrate. Collect the ln terms together and combine using the laws of logarithms. Problem-solving If c is a constant of integration, then A = ec can also be used as a constant. Writing the equation in this form helps you determine the family of solution curves. 25/04/2019 08:55 FIRST-ORDER DIFFERENTIAL EQUATIONS CHAPTER 5 93 Some solution curves corresponding to different values of A are shown below. y A The graphs of y = ±  __ x  form the family of solution A y=± x O curves for this differential equation. x Online Explore families of solution curves using GeoGebra. Example 4 a Find the general solution of the differential equation __ dx  ___  = √x , t > 0 dt b Find the particular solutions which satisfy the initial conditions i x = 0 when t = 0 ii x = 1 when t = 0 iii x = 4 when t = 0 iv x = 9 when t = 0 c Sketch the members of the family of solution curves represented by these particular solutions. 1 a∫  ___   √__  dx = ∫  dt x  __1 2x  2 = t + c t+c 2 x = (_____       , t > 0 2 ) b i Substituting x = 0 when t = 0 gives c = 0 t2  x = __    , t > 0 4 ii Substituting x = 1 when t = 0 gives c = 2 (t + 2)2  x = ________    , t > 0 4 iii Substituting x = 4 when t = 0 gives c = 4 (t  4)2  x = ________    , t > 0 4 iv Substituting x = 9 when t = 0 gives c = 6 (t  6)2  x = ________    , t > 0 4 M05_IAL_FP2_44655_U05_090-104.indd 93 Separate the variables, which in this question are x and t. t usually denotes time. Integrate and make x the subject of the formula. Substitute the initial conditions, i.e. the values of x when t = 0, to find c. Write the equations of the particular solutions. 25/04/2019 08:55 94 CHAPTER 5 c FIRST-ORDER DIFFERENTIAL EQUATIONS x 10 The graphs of t+c 2 x = (  ____    , t > 0, form the 2 ) family of solution curves for this differential equation. 9 8 7 6 5 4 3 These are parts of parabolae. 2 1 O Exercise 1 2 3 4 5 6 t 5A In questions 1–8 find the general solution of the differential equation and sketch the family of solution curves represented by the general solution. dy dy 2 ___  = y 1 ___  = 2x dx dx dy dy 1 3 ___  = x2 4 ___  = __    , x . 0 dx dx x dy 2y dy x 5 ___  = ___         6 ___  = __ dx x dx y dy dy y 7 ___  = e y 8 ___  = ________    , x > 0 dx dx x(x 1 1) dy dy 9  ___  = cos x 10 ___  = y cot x, 0 < x < π dx dx dy dy π π 11 ___  = sec2 t, −  __  < t < __         5 x(1 − x), 0 < x < 1 12 ___ 2 2 dt dx 13 Given that a is an arbitrary constant, show that y2 = 4ax is the general solution of the dy y differential equation  ___  = ___    . dx 2x 1    , 1 and 4. a Sketch the members of the family of solution curves for which a 5 __ 4 b Find also the particular solution which passes through the point (1, 3), and add this curve to your diagram of solution curves. 14 Given that k is an arbitrary positive constant, show that y2 1 kx2 5 9k is the general solution 2xy dy     ∣x∣ < 3. of the differential equation ___     5 ______ dx 9 2 x2 a Find the particular solution, which passes through the point (2, 5). 1 4 b Sketch the family of solution curves for k 5 __    , __    , 1 and include your particular solution 9 9 in the diagram. M05_IAL_FP2_44655_U05_090-104.indd 94 25/04/2019 08:55 FIRST-ORDER DIFFERENTIAL EQUATIONS CHAPTER 5 95 5.2 First-order linear differential equations of the form dy   + Py = Q where P and Q are functions of x ___ dx Example 5 Find the general solution to the differential equation dy x3 ___ + 3x2y = sin x dx You can use the product rule dy x3___   + 3x2y = sin x dx d So ___   (x3y) = sin x dx ⇒ x3y = ∫ sin x dx So = −cos x + c c 1 y = −  ___3 cos x + ___   3  x x dv du d u___    + v___    = ___   ( uv), with u = x3 and v = y, to dx dx dx dy d recognise that x3___   + 3x2y = ___   ( x3y). dx dx Use integration as the inverse process of differentiation. Integrate each side of the equation, including an arbitrary constant on the right-hand side. Make y the subject by dividing each of the terms One side of the differential equation in the on the right-hand side by x3. example above is an exact derivative of a product in the form dy d    (f(x)y) f(x) ___  + f ′(x)y = ___ dx dx You can solve some first-order differential equations by turning them into equations of this form. Example 6 Find the general solution of the equation dy 3y sin x ___       =   _____     + ___ 3 x dx x dy ___ 3y sin x     5  _____     1 ___ x x3 dx Multiply this equation by x3 dy x3 ___     1 3x2 y 5 sin x dx c 1 The solution is y 5 2___3  cos x 1 ___   3  x x   M05_IAL_FP2_44655_U05_090-104.indd 95 You can multiply this equation by x3 to make it into an exact equation. x3 is called an integrating factor. This is an exact equation which was solved as Example 5. 25/04/2019 08:55 96 CHAPTER 5 Example FIRST-ORDER DIFFERENTIAL EQUATIONS 7 dy Solve the general equation ___     1 Py 5 Q, where P and Q are functions of x. dx Multiply the equation by the integrating factor f(x). dy Then f(x) ___     + f(x) Py = f(x)Q (1) dx The equation is now exact and so the left-hand side is of the form dy f(x) ___     + f9(x)y dx dy dy So f(x) ___     + f(x)Py = f(x) ___     + f9(x)y dx dx ∴ f9(x) = f(x)P Dividing by f(x) and integrating f9(x) ∫   ____  dx = ∫  P dx f(x) ∴ ln|f(x)| = ∫  P dx ∴ f(x) = e  ∫  P dx Equation (1) becomes dy e  ∫   Pdx ___     + e  ∫  Pdx Py = e  ∫  Pdx Q dx d ∴ ___  (e  ∫  Pdx y) = e  ∫  PdxQ dx ∴e  ∫  Pdx y = ∫   e  ∫  PdxQdx + C You do this to make the equation exact. Compare the left-hand side of your differential equation with the format for an exact differential equation. Compare the coefficients of y and put them equal. This is a ln integral as the numerator is the derivative of the denominator. You need to learn this formula for the integrating factor. This will lead to a solution provided that these integrals can be found. The left-hand side will always be y 3 integrating factor. This is the solution to the differential equation. dy ■■ For the general equation ___     + Py = Q, where P and Q are functions of x, you obtain the dx integrating factor by finding e  ∫  Pdx ■■ You obtain the general solution to the differential equation by using e  ∫  Pdxy = ∫   e  ∫   PdxQ dx + C Example 8 Find the general solution of the differential equation dy  ___ − 4y = ex Find the integrating factor. dx The integrating factor is e∫P(x) = e∫(−4) dx = e−4x dy e−4x ___   − 4e−4x y = exe−4x dx d ⇒ ___ (e−4x y) = e−3x dx ⇒ e−4x y = ∫e−3x dx 1 −3x = −__ +c 3 e So M05_IAL_FP2_44655_U05_090-104.indd 96 1 x 4x y = −  __ 3 e + ce Multiply each term by the integrating factor. Express the LHS as the derivative of a product. Integrate to get the general solution. Divide every term, including the constant, by the integrating factor to make y the subject. 25/04/2019 08:55 FIRST-ORDER DIFFERENTIAL EQUATIONS Example CHAPTER 5 97 9 dy Find the general solution of the differential equation cos x  ___  + 2y sin x = cos4 x dx Divide through by cos x: dy ___   + 2y tan x = cos3 x (1) dx The integrating factor is 2 e∫P(x)dx = e∫2 tan x dx = e2 ln sec x = eln sec x = sec2 x dy sec2 x ___   + 2y sec2 x tan x = sec2 x cos3 x dx d So  ___ (y sec2 x) = cos x dx ⇒ y sec2 x = ∫cos x dx ⇒ y sec2 x = sin x + c y = cos2 x(sin x + c) ⇒ Exercise Divide by cos x so that equation is in the form dy ___     + P(x) = Q(x) dx Use properties of ln to simplify the integrating factor. Multiply equation (1) by the integrating factor and simplify the right-hand side. Integrate to get the general solution and multiply through by cos2 x. 5B 1 Find the general solutions to these differential equations. dy dy __ __ a x     + y = cos x b e−x     − e−x y =xe x dx dx dy __ dy 1 __ 1 __ x e x2e y     + 2xe y = x d __ x  d  x  −  x2  y = e dx E dy __ c sin x     + y cos x = 3 dx dy __ f 4xy     + 2y2 = x2 dx dy 2 a Find the general solution to the differential equation ___     + 2xy = e  −x  2 dx b Describe the behaviour of y as x → ∞. (4 marks) (1 mark) 3 a Find the general solution to the differential equation dy x2 ___   + 2xy = 2x +1 dx b Find the three particular solutions which pass through the points with coordinates 1 1 1 (−__  , 0), (−__  , 3) and (−__  , 19) respectively and sketch their solution curves for x , 0. 2 2 2 4 a Find the general solution to the differential equation dy y ____________ 1 ln x   ___ + __   =       x . 1 dx x (x + 1)(x + 2) b Find the particular solution which passes through the point (2, 2). 5 Find the general solutions to these differential equations by using an integrating factor. dy dy ___ ___ b   + y cot x = 1 a   + 2y = ex dx dx M05_IAL_FP2_44655_U05_090-104.indd 97 25/04/2019 08:55 98 CHAPTER 5 dy ___ c   + y sin x = ecos x dx dy ___ e   + y tan x = x cos x dx dy x3 g x2  ___ − xy =   _____  , x . −2 x+2 dx d y i (x + 2) ___   − y = x + 2 dx FIRST-ORDER DIFFERENTIAL EQUATIONS dy ___ d   − y = e2x dx dy y __ 1 f  ___ + __   x =   2  dx x d y h 3x  ___ + y = x dx d y ex j x  ___ + 4y =   __2  dx x dy 6 Find y in terms of x given that x  ___ + 2y = ex and that y = 1 when x = 1. dx (8 marks) dy 7 Solve the differential equation, giving y in terms of x, where x3  ___ − x2y = 1 dx and y = 1 at x = 1. (8 marks) 8 a Find the general solution to the differential equation 1 dy (x + __   x ) ___   + 2y = 2(x2 + 1)2 dx giving y in terms of x. (6 marks) b Find the particular solution which satisfies the condition that y = 1 at x = 1. (2 marks) 9 a Find the general solution to the differential equation dy π π cos x ___   + y = 1, −__  , x , __     2 2 dx b Find the particular solution which satisfies the condition that y = 2 at x = 0. E E/P 10 a Find the general solution to the differential equation dy cos x  ___  + y sin x = 1 dx b Find the particular solution such that y = 3 when x = π. 3π π    , 1) and ( ___    , −1)lie on all possible solution curves. c Show that the points ( __ 2 2 dy 11 Find a general solution to the equation a  ___  + by = 0in terms of a and b. dx (6 marks) (2 marks) (6 marks) (2 marks) (3 marks) (6 marks) 5.3 Reducible first-order differential equations You can use a substitution to reduce a first-order differential equation into a form that you know how to solve, either by separating the variables, or by using an integrating factor. M05_IAL_FP2_44655_U05_090-104.indd 98 25/04/2019 08:55 FIRST-ORDER DIFFERENTIAL EQUATIONS CHAPTER 5 99 Example 10 a Show that the substitution y = xz transforms the differential equation dy x   2 + 3y  2 ___     = ________     2xy dx into dz 1 + z  2     x ___     = ______ 2z dx b Hence find the general solution to the original equation, giving y2 in terms of x. a y = xz dy dz   ___   = x  ___   + z dx dx (1) (2) dy x2 + 3y2     = ________ Substituting into ___     gives dx 2xy x2 + 3x2z2 dz      x  ___  + z = __________   dx 2x2z x2(1 + 3z2) dz      x  ___  + z = __________   dx 2x2z dz 1 + 3z2     − z x  ___  = _______ dx 2z 1 + z2     as required. = ______ 2z _____   2z   dz = __   1   dx b x 1 + z2 ∫ ∫ ln(1 + z2) = ln x + c 1 + z2 = Ax, where A is a positive constant y2 ___ (1 + (   2 )) = Ax x y2 = x2(Ax − 1) Watch out Using the substitution, differentiate dy dz to get   ___   in terms of   ___  . Note that z is a function dx dx of x and y, not a constant, so you must use the product rule. Substitute into the differential equation using equations (1) and (2). Rearrange and simplify your equation. Separate the variables, then integrate including a constant of integration. ← Further Pure 2 Section 5.1 Take exponentials and let A = ec. Use the original substitution to transform the general solution in z back into a general solution in x and y. y y 2 y = xz, so z = __   x  and z2 = ( __   x )  . Example 11 dy a Use the substitution z = y−1 to transform the differential equation ___     + xy = xy2 into a dx differential equation in z and x. b Solve the new equation, using an integrating factor. c Find the general solution to the original equation, giving y in terms of x. M05_IAL_FP2_44655_U05_090-104.indd 99 25/04/2019 08:55 100 CHAPTER 5 FIRST-ORDER DIFFERENTIAL EQUATIONS a As z = y−1, y = z−1 dy 1 dz  ___  = − __2   ___  dx z dx dy Substituting into ___     + xy = xy2 gives dx 1 dz − __2   ___  + xz−1 = xz−2 z dx dz ⇒  ___  − xz = −x dx Rearrange the substitution to make y the subject. dy dz Differentiate to give ___     in terms of ___     dx dx Rearrange and simplify your equation. x2 __ −    2 b The integrating factor is e∫−x dx = e x2 dz x2 x2 −  __ −  __ −  __       e 2  ___  − xe 2 z = −xe 2 dx x2 x2 −  __ d −  __      ___  (e 2 z) = −x e 2 dx ∫ x −  __   2 z = − xe x −  __   2 z = e 2 e 2 x −  __   2 2 2 e x −  __   2 dx To solve a differential equation in the form dy ___     + P(x)y = Q(x) , multiply every term in the dx equation by the integrating factor e∫P(x)dx. ← Further Pure 2 Section 5.2 +c x2 __     z = 1 + ce 2 c As y = z−1, 1   y = _______   1+ x2   __  ce 2 Integrate to give result then divide each term by the integrating factor. Use the original substitution to write y in terms of x. Example 12 dy y − x + 2 a Use the substitution u = y − x to transform the differential equation ___   = ________     dx y − x + 3 into a differential equation in u and x. b By first solving this new equation, show that the general solution to the original equation may be written in the form (y − x)2 + 6y − 4x − 2c = 0, where c is an arbitrary constant. a Let u = y − x du dy Then  ___  = ___     − 1 dx dx dy y − x + 2 Substituting into ___     = _________     gives dx y − x + 3 du u+2    ___  + 1 = ______   dx u+3 du u + 2   ⇒  ___  = ______   − 1 dx u + 3 du −1  ___  = ______     dx u + 3 b ∫(u + 3)du = −∫dx 1  __ u2 + 3u = −x + c 2 1  __ (y − x)2 + 3(y − x) = −x + c 2 ∴ (y − x)2 + 6y − 4x − 2c = 0 M05_IAL_FP2_44655_U05_090-104.indd 100 dy du Differentiate to give ___     in terms of  ___  dx dx dy Make ___     the subject and substitute. dx Rearrange and simplify your equation. Separate the variables and integrate. Substitute back to give your result in terms of x and y. 25/04/2019 08:55 FIRST-ORDER DIFFERENTIAL EQUATIONS Exercise CHAPTER 5 101 5C y 1 Use the substitution z = __   x to transform each differential equation into a differential equation in z and x. By first solving the transformed equation, find the general solution to the original equation, giving y in terms of x. x2 dy y __ dy y __ x     +    , x . 0, y . 0 b ___ = __     +   2 , x . 0 a ___ = __ dx x y dx x y 2 y y d dy x3 + 4y3 y c ___ = __     + __   2 , x . 0 d ___ = _______    , x . 0 dx x x dx 3xy2 E 2 a Use the substitution z = y−2 to transform the differential equation dy 1 p p   ___ + (__     tan x) y = −(2 sec x)y3, − __  , x ,   __  2 2 2 dx dz ___ into the differential equation     − z tan x = 4 sec x. (5 marks) dx b By first solving the transformed equation, find the general solution to the original equation, (6 marks) giving y in terms of x. E 3 a Use the substitution z = x   2to transform the differential equation dx _1 ___     + t 2x = t 2x  2 dt dz 1 1 into the differential equation  ___  +  __ t  2z = __    t  2. (4 marks) 2 dt 2 b By first solving the transformed equation, find the general solution to the original equation, (6 marks) giving x in terms of t. E 4 a Use the substitution z = y−1 to transform the differential equation dy 1 (x + 1)3 ___ 2   x y =  _______   − __ x y dx (x + 1)3 dz 1 into the differential equation ___     + __   (4 marks)     z = −  _______ x dx x b By first solving the transformed equation, find the general solution to the original equation, (6 marks) giving y in terms of x. P 5 a Use the substitution z = y2 to transform the differential equation dy 1 2(1 + x2) ___     + 2xy = __     y dx into a differential equation in z and x. _1 By first solving the transformed equation, b find the general solution to the original equation, giving y in terms of x c find the particular solution for which y = 2 when x = 0. E/P 6 Show that the substitution z = y−(n − 1) transforms the general equation dy ___   + P(x)y = Q(x)yn, dx dz into the linear equation  ___ − P(x)(n − 1)z = − Q(x)(n − 1). dx M05_IAL_FP2_44655_U05_090-104.indd 101 (5 marks) 25/04/2019 08:55 102 CHAPTER 5 E/P FIRST-ORDER DIFFERENTIAL EQUATIONS 7 a Use the substitution u = y + 2x to transform the differential equation dy −(1 + 2y + 4x)      ___ = _____________ 1 + y + 2x dx into a differential equation in u and x. (3 marks) b By first solving this new equation, show that the general solution to the original equation may be written as 4x2 + 4xy + y2 + 2y + 2x = k, where k is a constant. (6 marks) Challenge dy x  2  ___  − xy = y  2 dx By means of a suitable substitution, show that the general solution to the differential equation is given by x y = −  _______  ln x + C where C is a constant of integration. Chapter review 5 E E E E E E/P 1 Find the general solution to the differential equation dy ___     + y tan x = 2 secx dx giving your answer in the form y = f(x). (7 marks) 2 Find the general solution to the differential equation dy (1 − x2)  ___  + xy = 5x, −1 , x , 1 dx giving your answer in the form y = f(x). (7 marks) 3 Find the general solution to the differential equation dy x ___     + x + y = 0 dx giving your answer in the form y = f(x). (7 marks) 4 y satisfies the differential equation __ dy y ___     + __     = √ x  dx x Find y as a function of x. (7 marks) 5 y satisfies the differential equation dy ___     + 2xy = x dx Find y in terms of x. (7 marks) 6 Find the general solution to the differential equation dy x(1 − x2)  ___  + (2x2 − 1)y = 2x3, 0 , x , 1 dx giving your answer in the form y = f(x). (7 marks) M05_IAL_FP2_44655_U05_090-104.indd 102 25/04/2019 08:55 FIRST-ORDER DIFFERENTIAL EQUATIONS E/P CHAPTER 5 103 dy 7 Find the general solution to the equation  ___  − ay = Q(x), where a is a constant, dx giving your answer in terms of a, when a Q(x) = ke  λx (k and λ are constants). (6 marks) Given that Q(x) = kx  ne  ax, where k and n are constants, b find the general solution to the differential equation. E/P E E 8 Find, in the form y = f(x), the general solution to the differential equation dy π tan x  ___  + y = 2 cos x tan x, 0 , x , __     2 dx 9 a Show that the transformation z = y−1 transforms the differential equation dy (1) x ___     + y = y  2 ln x dx into the differential equation dz z ln x ___     − __ (2)   x  = −   ____ x  dx b By solving differential equation (1), find the general solution to differential equation (2). 10 a Show that the substitution z = y2 transforms the differential equation dy (1) 2 cos x  ___  − y sin x + y  −1 = 0 dx into the differential equation dz     − z sin x = − 1 cos x ___ (2) dx b Solve differential equation (2) to find z as a function of x. c Hence write down the general solution to differential equation (1) in the form y2 = f(x). E E y 11 a Show that the substitution z = __   x  transforms the differential equation dy     − xy = 0 (1) (x  2 − y  2)___ dx into the differential equation dz z  3   (2) x  ___  = ______   dx 1 − z  2 b Solve equation (2) and hence obtain the general solution to equation (1). y 12 a Show that the transformation z = __   x  transforms the differential equation dy y(x + y) ___ (1)         = _______ dx x(y − x) into the differential equation dz 2z x ___     = _____     (2) dx z − 1 b Solve equation (2) and hence obtain the general solution to equation (1). M05_IAL_FP2_44655_U05_090-104.indd 103 (7 marks) (6 marks) (4 marks) (6 marks) (4 marks) (6 marks) (1 mark) (4 marks) (6 marks) (4 marks) (6 marks) 25/04/2019 08:55 104 CHAPTER 5 E E FIRST-ORDER DIFFERENTIAL EQUATIONS y 13 a Show that the substitution z =  __ x  transforms the differential equation − 3xy dy ___ ________   (1)     =   2 dx y  − 3x  2 into the equation dy z   3   (2) x  ___  = −  ______ dx z  2 − 3 b By solving equation (2), find the general solution to equation (1). 14 a Use the substitution u = x + y to show that the differential equation dy ___     = (x + y + 1)(x + y − 1) dx can be written as du ___     = u  2 dx b Hence find the general solution to the original differential equation. (4 marks) (6 marks) (3 marks) (4 marks) E 15 a Show that the transformation u = y − x − 2 can be used to transform the differential equation dy (1)  ___  = (y − x − 2)  2 dx into the differential equation du ___ (2) (3 marks)     = u  2 − 1 dx b Solve equation (2) and hence find the general solution to equation (1). (4 marks) E/P 16 A particle is moving with velocity v at time t such that __ dv t  ___  + v = 2t  3v  3, 0 , t , √ 3  (1) dt a Use the substitution u = v−2 to show that the differential equation can be transformed du 2u 2 into  ___  −  ___ (5 marks) t = − 4t  dt 1 b Given that v = _  2 when t = 1, show that the solution to differential equation (1) can ________ 1      2 be written as v = ________ t (c − 4t) where c is a constant to be found. (8 marks) √ Summary of key points 1 2 dy You can solve a first-order differential equation of the form ___    + P(x)y = Q(x)by multiplying dx every term by the integrating factor e  ∫   P(x)dx. You can use a substitution to reduce a first-order differential equation into a form that you know how to solve, either by separating the variables, or by using an integrating factor. M05_IAL_FP2_44655_U05_090-104.indd 104 25/04/2019 08:55 SECOND-ORDER DIFFERENTIAL EQUATIONS CHAPTER 6 105 6 SECOND-ORDER DIFFERENTIAL EQUATIONS 5.1 5.2 Learning objectives After completing this chapter you should be able to: ● Solve second-order homogeneous differential equations using the auxiliary equation → pages 106–110 ● Solve second-order non-homogeneous differential equations using the complementary function and the particular integral → pages 110–115 ● Use boundary conditions to find a particular solution to a second-order differential equation → pages 115–118 ● Use a given substitution to transform a second-order differential equation into one that can be solved → pages 118–121 Prior knowledge check 1 Find the general solutions of these differential equations: dy a x ___ = 2(y 2( − 1) dx dy y b ___ + __ = 2x dx x 2 ← Further Pure 2 Section 5.1 Find the particular solution to the differential equations dy a ___ + 3xy = ex when x = 0, y = 2 dx dy b x___ − y = x3 when x = 1, y = 3 dx ← Pure 4 Section 5.2 M06_IAL_FP2_44655_U06_105-124.indd 105 Many real-life situations can be modelled using differential equations: for example, the displacement of a point on a vibrating spring from a fixed point, or the distance fallen by a parachutist. 25/04/2019 08:55 106 CHAPTER 6 SECOND-ORDER DIFFERENTIAL EQUATIONS 6.1 Second-order homogeneous differential equations A second-order differential equation contains second derivatives. Example 1 Find the general solution to the differential equation d  2y ____   2  = 12x dx  Watch out dy ___     = 6x2 + A dx y = 2x3 + Ax + B The general solution to this secondorder differential equation needs two arbitrary constants. If you wanted to find a particular solution you would need to know two boundary conditions. In this section you will look at techniques for solving linear differential equations that are of the form d  2y dy a  ____2  + b  ___  + cy = 0 dx dx  Notation You sometimes see differential where a, b and c are real constants. Equations equations of this type written as of this form (with 0 on the right-hand side) are d  2y dy ay″ + by′ + cy = 0where y″ = ____   2  and y′ = ___     called second-order homogeneous differential dx dx  equations with constant coefficients. dy Using the techniques from the previous section, the general solution of a  ___  + by = 0is of the form dx b   . Notice that k is the solution to the equation ak + b = 0. y = Ae  kx, where k = −  __ a This suggests that an equation of the form y = Aekx might also be a solution of the second-order dy d   2y differential equation a  ____2  + b  ___  + cy = 0. But it cannot be the general solution, as it only contains dx dx  one arbitrary constant. Since two constants are necessary for a second-order differential equation, you can try a solution of the form y = Ae  λx + Be  μx, where A and B are arbitrary constants and λ and μ are constants to be determined. dy  ___  = Aλe  λx + Bμe  μx dx ____ d  2y   2  = Aλ  2e  λx + Bμ2e  μx dx  Substituting these into the differential equation gives a(Aλ2eλx + Bμ2eμx) + b(Aλeλx + Bμeμx) + c(Aeλx + Beμx) = 0 aAλ  2 e  λx + aBμ  2e  μx + bAλ e  λx + bBμe  μx + cAe  λx + cBe  μx = 0 Ae  λx(aλ  2 + bλ + c) + Be  μx(aμ2 + bμ + c) = 0 This shows that the equation y = Ae  λx + Be  μxwill satisfy the original differential equation if both λ and μ are solutions to the quadratic equation am  2 + bm + c = 0. The equation am  2 + bm + c = 0is called the auxiliary equation. M06_IAL_FP2_44655_U06_105-124.indd 106 25/04/2019 08:55 SECOND-ORDER DIFFERENTIAL EQUATIONS CHAPTER 6 107 The natures of the roots α and β of the auxiliary equation, am2 + bm + c = 0 determine the d  2y dy general solution to the differential equation a  ____2  + b  ___  + cy = 0. dx dx  You need to consider three different cases: •• Case 1: b  2> 4ac The auxiliary equation has two distinct real roots α and β. The general solution will be of the form y = Ae  αx + B e  βx where A and B are arbitrary constants. •• Case 2: b  2 = 4ac The auxiliary equation has one repeated root α. The general solution will be of the form y = (A + Bx) e  αx where A and B are arbitrary constants. •• Case 3: b  2< 4ac Links Case 3 is equivalent to y = Aeαx + Beβx The auxiliary equation has two complex with complex α and β. conjugate roots α and β equal to p ± q i. The general solution will be of the form y = e  px (A cos qx + B sin qx) where A and B Notation If the roots are purely imaginary are arbitrary constants. ( p = 0), the general solution reduces to y = A cos qx + B sin qx Example 2 dy d  2y a Find the general solution to the equation 2  ____2  + 5  ___  + 3y = 0. dx dx  b Verify that your answer to part a satisfies the equation. a 2m  2 + 5m + 3 = 0 3 (2m + 3)(m + 1) = 0 ⇒ m = −  _ 2  or m = − 1 So the general solution is 3 _ y = Ae  − 2 x + Be  −x where A and B are arbitrary constants Write down the auxiliary equation. Solve the auxiliary equation. Write down the general solution. 3 _ b y = Ae  − 2 x + Be  −x dy 3 3 − _  x −x  ___  = −  __ 2 Ae  2 − Be  dx d2y  9 − _3 x  ____2  = _     Ae  2 + Be −x dx  4 3 3 __ 3 9 − _  x −x − _  x −x 2(  __ 4 Ae  2 + Be  ) + 5(−  2 Ae  2 − Be  ) 3 _ + 3(Ae  − 2 x + Be  −x) 3 3 3 _ _ _ 15 __ __ 9 −    x −  =   2 Ae  2 −   2 Ae  2 x + 3Ae  − 2 x + 2Be  −x − 5Be  −x + 3Be  −x = 0 as required. M06_IAL_FP2_44655_U06_105-124.indd 107 Write down expressions for the first and second derivatives. d  2y dy Substitute your expressions for ____   2 , ___     and y into dx  dx the differential equation, expand and simplify. 25/04/2019 08:55 108 CHAPTER 6 Example SECOND-ORDER DIFFERENTIAL EQUATIONS 3 Show that y = (A + Bx )e3x dy satisfies the differential equation   2  − 6   ___ + 9y = 0 dx dx Let y = Ae3x + Bxe3x, then dy ___   = 3Ae3x + 3Bxe3x + Be3x dx 2y d ____   2  = 9Ae3x + 9Bxe3x + 3Be3x + 3Be3x dx = 9Ae3x + 9Bxe3x + 6Be3x d2y dy ____   2  − 6  ___ + 9y = 9Ae3x + 9Bxe3x + 6Be3x dx dx − 6(3Ae3x + 3Bxe3x + Be3x) + 9(Ae3x + Bxe3x) = 0 So y = (A + Bx)e3x is a solution to the equation. Example dy ____ 2 Differentiate the expression for y twice. Substitute into the left-hand side of the differential equation and simplify to show that the result is zero. Notation The auxiliary equation − 6m + 9 = 0has a repeated root at m = 3 so the general solution is in the form you expect. m  2 4 d2y dy Find the general solution to the differential equation ____   2  + 8  ___ + 16y = 0 dx dx m 2   + 8m + 16 = 0 (m + 4)  2 = 0 ⇒ m = −4 So the general solution is y = (A + Bx)e  −4x. Example Write down the auxiliary equation. Solve the auxiliary equation. In this case there is a repeated root. 5 d2y dy Find the general solution to the differential equation ____   2  − 6  ___ + 34y = 0 dx dx   2 − 6m + 34 = 0 ⇒ m = 3 ± 5i m So the general solution is y = e  3x(A cos 5x + B sin 5x) Example Write down the auxiliary equation and solve it using the quadratic formula or by completing the square. In this case there are two complex conjugate roots. 6 d2y Find the general solution to the differential equation ____   2  + 16y = 0 dx m  2 + 16 = 0 ⇒ m = ± 4i So the general solution is y = A cos 4x + B sin 4x M06_IAL_FP2_44655_U06_105-124.indd 108 Write down the auxiliary equation and solve it. In this case there are two purely imaginary roots. This is in the form y = e  px(A cos qx + B sin qx) with p = 0. 25/04/2019 08:55 SECOND-ORDER DIFFERENTIAL EQUATIONS Exercise CHAPTER 6 109 6A 1 Find the general solution to each differential equation. dy d2y ____ a   2  + 5 ___ + 6y = 0 dx dx 2 dy dy c ____2  + 2 ___ − 15y = 0 dx dx 2 dy dy e ____2  + 5 ___ = 0 dx dx 2 dy dy g 4 ____2  − 7 ___ − 2y = 0 dx dx d2y dy b  ____2  − 8 ___ + 12y = 0 dx dx 2 dy dy d ____2  − 3 ___ − 28y = 0 dx dx 2 dy dy f 3 ____2  + 7 ___ + 2y = 0 dx dx 2 dy dy h 15 ____2  − 7 ___ − 2y = 0 dx dx 2 Find the general solution to each differential equation. dy d2y a ____2  + 10  ___ + 25y = 0 dx dx d2y dy c ____2  + 2 ___ + y = 0 dx dx d2y dy e 16 ____2  + 8 ___ + y = 0 dx dx d2y dy g 4 ____2  + 20  ___ + 25y = 0 dx dx d2y dy b ____2  − 18 ___ + 81y = 0 dx dx d2y dy d ____2  − 8 ___ + 16y = 0 dx dx d2y dy f 4 ____2  − 4 ___ + y = 0 dx dx __ dy d2y h ____2  + 2√3  ___ + 3y = 0 dx dx 3 Find the general solution to each differential equation. d2y a ____2  + 25y = 0 dx d2y c ____2  + y = 0 dx d2y dy e ____2  + 8 ___ + 17y = 0 dx dx 2 dy dy g ____2  + 20 ___ + 109y = 0 dx dx d2y b ____2  + 81y = 0 dx d2y d 9  ____2  + 16y = 0 dx 2 dy dy f ____2  − 4 ___ + 5y = 0 dx dx 2 d y __ dy h ____2  + √ 3   ___ + 3y = 0 dx dx 4 Find the general solution to each differential equation. dy d2y a ____2  + 14 ___  + 49y = 0 dx dx d2y dy c ____2  + 4 ___  + 13y = 0 dx dx d2y dy e 9 ____2  − 6 ___  + 5y = 0 dx dx M06_IAL_FP2_44655_U06_105-124.indd 109 d2y dy b ____2  + ___     − 12y = 0 dx dx d2y dy d 16 ____2  − 24 ___  + 9y = 0 dx dx d2y dy f 6 ____2  − ___     − 2y = 0 dx dx 25/04/2019 08:55 110 CHAPTER 6 E/P SECOND-ORDER DIFFERENTIAL EQUATIONS dx d  2x 5 Given the differential equation ____   2  + 2k  ___  + 9x = 0, where k is a real constant, dt dt  a find the general solution to the differential equation when: i |k| . 3 ii |k| , 3 iii |k| = 3 (8 marks) b In the case where k = 2, i find the general solution ii describe what happens to x as t → ∞. (4 marks) P 6 Given that am2 + bm + c = 0 has equal roots m = a, prove that y = (A + Bx)eax is a solution d 2y dy to the differential equation a ___2  + b ___  + c = 0. dx dx P Notation This result is known as 7 Given that y = f(x)and y = g(x)are both solutions to the 2 dy   y d the principle of superposition. second-order differential equation a  ____2  + b  ___  + cy = 0, dx dx  prove that y = Af(x) + Bg(x), where A and B are real constants, is also a solution. Challenge Let α and β be the roots of a real-valued quadratic equation, so that a = p + iq and β = p − iq, p, q [ R. Show that it is possible to choose A, B [ C such that Aeax + Beax can be written in the form epx(Ccosqx + Dsinqx) where C and D are arbitrary real constants. 6.2 Second-order non-homogeneous differential equations A second-order differential equation of the form d  2y dy a  ____2  + b  ___  + cy = f(x) dx dx  is called a non-homogeneous differential equation. Notation You sometimes see differential equations of this type written as d  2y dy ay″ + by′ + cy = f(x)where y ″ =  ____2  and y ′ =  ___  dx dx  To solve an equation of this type you first find the general solution of the corresponding homogeneous d  2y dy differential equation, a  ____2  + b  ___  + cy = 0. This is called the complementary function (C.F.). dx dx  You then need to find a particular integral (P.I.), which is a function that satisfies the differential equation. The form of the particular integral depends on the form of f (x). M06_IAL_FP2_44655_U06_105-124.indd 110 25/04/2019 08:55 SECOND-ORDER DIFFERENTIAL EQUATIONS CHAPTER 6 111 This table provides some particular integrals to try. Form of f(x) p p + qx p + qx + rx2 pekx p cos ωx + q sin ωx Form of particular integral λ λ + μx λ + μx + vx2 λekx λ cos ωx + μ sin ωx Use this form of the P.I. for functions such as 4x2 or 1 – x2. Use this form of the P.I. for functions such as sin 2x or 5 cos x. ■■ A particular integral is a function which satisfies the original differential equation. Example 7 d2y dy Find a particular integral of the differential equation ____   2  − 5 ___  + 6y = f(x) when f(x) is: dx dx x 2 c 3x d e e 13 sin 3x a 3 b 2x dy d2y a Let y = λ, then ___     = 0 and ____   2  = 0 dx dx d2y dy ____ ___ Substitute into   2  − 5    + 6y = 3: dx dx 0 − 5 × 0 + 6λ = 3 ⇒ λ = __   21 So a particular integral is __  21 dy ___ b Let y = λx + μ, then   d y ____ 2 d y ____   = λ and   dx 2   = 0 dx2 dy ___   − 5    + 6y = 2x: dx dx2 0 − 5 × λ + 6(λx + μ) = 2x ⇒ (6μ − 5λ) + 6λx = 2x ⇒ 6μ − 5λ = 0 and 6λ = 2 5   18 ⇒ λ = __   31 and μ = __ 5 So a particular integral is __  31 x + __   18 Substitute into   c Let y = λx2 + μx + ν dy d2y Then ___     = 2λx + μ and ____   2  = 2λ dx dx 2y d dy Substitute into ____   2  − 5 ___  + 6y = 3x2: dx dx 2 2λ − 5(2λx + μ) + 6(λx + μx + ν) = 3x2 ⇒ (2λ − 5μ + 6ν) + (6μ − 10λ)x + 6λx2 = 3x2 ⇒ 2λ − 5μ + 6ν = 0, 6μ − 10λ = 0 and 6λ = 3 19 ⇒ λ = __   21 , μ = __   56 and ν = ___   36 19 So a particular integral is __  21 x2 + __   56 x + ___   36 M06_IAL_FP2_44655_U06_105-124.indd 111 When f(x) = 3, which is constant, choose P.I. = λ, which is also constant. Differentiate twice and substitute the derivatives into the differential equation. Solve equation to give the value of λ. When f(x) = 2x, which is a linear function of x, choose P.I. = λx + μ. Differentiate twice and substitute the derivatives into the differential equation. Equate the constant terms and the coefficients of x to give simultaneous equations, which you can solve to find λ and μ. As f(x) = 3x2, which is a quadratic function of x let P.I. = λx2 + μx + ν. Equate the constant terms, the coefficients of x and the coefficients of x2 to give simultaneous equations, which you can solve to find λ, μ and ν. 25/04/2019 08:55 112 CHAPTER 6 SECOND-ORDER DIFFERENTIAL EQUATIONS dy d2y d Let y = λex, then ___     = λex and ____   2  = λex dx dx 2y d dy Substitute into ____   2  − 5 ___  + 6y = ex: dx dx λex − 5λex + 6λex = ex ⇒ 2λex = ex ⇒ λ = __   21 So a particular integral is __  21 ex e Let y = λ sin 3x + μ cos 3x dy Then ___     = 3λ cos 3x − 3μ sin 3x dx d2y ____ and   2  = −9λ sin 3x − 9μ cos 3x dx d2y dy Substitute into ____   2  − 5 ___  + 6y = 13 sin 3x: dx dx −9λ sin 3x − 9μ cos 3x − 5(3λ cos 3x − 3μ sin 3x) + 6(λ sin 3x + μ cos 3x) = 13 sin 3x ⇒ (−9λ + 15μ + 6λ) sin 3x + (−9μ − 15λ + 6μ) cos 3x = 13 sin 3x ⇒ −9λ + 15μ + 6λ = 13 and −9μ − 15λ + 6μ = 0 5 __ 1 ⇒ λ = −__ 6 and μ =   6 5 __ 1 So a particular integral is −__ 6 sin 3x +   6 cos 3x As f(x) = ex, which is an exponential function of x let P.I. = λex. Equate coefficients of ex to find the value of λ. As f(x) = 13 sin 3x, which is a trigonometric function of x let P.I. = λ sin 3x + μ cos 3x, also a similar trigonometric function. Problem-solving Equate coefficients of sin 3x and of cos 3x and solve simultaneous equations. dy d  2y ■■ To find the general solution to the differential equation a  ____2  + b  ___  + cy = f(x) dx dx  dy d  2y ____ •• Solve the corresponding homogeneous equation a   2  + b  ___  + cy = 0 to find the dx dx  complementary function (C.F.) •• Choose an appropriate form for the particular integral (P.I.) and substitute into the original equation to find the values of any coefficients. •• The general solution is y = C.F. + P.I. Example 8 d2y dy Find the general solution to the differential equation ____   2  − 5 ___  + 6y = f(x) when f(x) is: dx dx x 2 c 3x d e e 13 sin 3x a 3 b 2x m2 − 5m + 6 = 0 (m − 3)(m − 2) = 0 ⇒ m = 3 or m = 2 Hence the complementary function is y = Ae  3x + Be  2x where A and B are arbitrary constants. M06_IAL_FP2_44655_U06_105-124.indd 112 Solve the auxiliary equation to find the values of m. 25/04/2019 08:55 SECOND-ORDER DIFFERENTIAL EQUATIONS CHAPTER 6 113 The particular integrals were already found in Example 7 so the general solutions are: a y = Ae3x + Be2x + __   21 5 b y = Ae3x + Be2x + __   31 x + __   18 19 c y = Ae3x + Be2x + __   21 x2 + __   56 x + ___   36 The general solution is y = C.F. + P.I. __1 d y = Ae3x + Be2x +   2 ex e y = Ae3x + Be2x − __   61 sin 3x + __   56 cos 3x You need to be careful if the standard form of the particular integral contains terms which form part of the complementary function. If this is the case, you need to modify your particular integral so that no two terms in the general solution have the same form. For example, this situation occurs when f(x) is of the form pekx, and k is one of the roots of the auxiliary equation. In this case you can try a particular integral of the form,λxekx. Example 9 d2y dy Find the general solution to the differential equation ____   2  − 5 ___  + 6y = e2x dx dx As in Example 8, the complementary function is y = Ae3x + Be2x. The particular integral cannot be λe2x, as this is part of the complementary function. So let y = λxe2x dy Then ___     = 2λxe2x + λe2x dx and d y ____ The function λe2x is part of the C.F. and satisfies the differential equation d2y dy ____   2  − 5 ___  + 6y = 0, so it cannot also satisfy dx dx 2y dy d ____   2  − 5 ___  + 6y = e2x dx dx 2   = 4λxe2x + 2λe2x + 2λe2x = 4λxe2x + 4λe2x dx2 d2y dy Substitute into ____   2  − 5 ___  + 6y = e2x: dx dx 4λxe2x + 4λe2x − 5(2λxe2x + λe2x) + 6λxe2x = e2x ⇒ −λe2x = e2x ⇒ λ = −1 So a particular integral is −xe2x. The general solution is y = Ae3x + Be2x − xe2x.   Watch out Let the P.I. be λxe2x and differentiate, substitute and solve to find λ. The general solution is y = C.F. + P.I. When one of the roots of the auxiliary equation is 0, the complementary function will contain a constant term. If f(x) is a polynomial, you will need to multiply its particular integral by x to make sure the P.I. does not also contain a constant term. M06_IAL_FP2_44655_U06_105-124.indd 113 25/04/2019 08:55 114 CHAPTER 6 SECOND-ORDER DIFFERENTIAL EQUATIONS Example 10 Find the general solution to the differential equation dy d2y ____   2  − 2 ___  = 3 dx dx d2y dy First consider the equation  ____2  − 2 ___  = 0 dx dx m2 − 2m = 0 m(m − 2) = 0 ⇒ m = 0 or m = 2 Write down and solve the auxiliary equation. So the complementary function is y = A + Be2x. The particular integral cannot be a constant, as this is part of the complementary function, so let y = λx. dy d2y Then  ___  = λ and ____   2  = 0 dx dx 2y d dy Substitute into ____   2  − 2 ___  = 3: dx dx 0 − 2λ = 3 3 ⇒ λ = − __ 2  3 So a particular integral is −  __ 2 x 3 The general solution is y = A + Be2x −  __ 2 x. Exercise Find the complementary function by putting the right-hand side of the differential equation equal to zero, and solving this new equation. Try to find a particular integral. The righthand side of the original equation was 3, which was a constant and usually this would imply a constant P.I. As the C.F. includes a constant term ‘A’, the P.I. cannot also be constant. A value of λ d 2y dy would satisfy  ____2  − 2 ___  = 0 rather than dx dx d2y dy ____ ___   2  − 2    = 3 dx dx Multiply the ‘expected’ P.I. by x and try λx instead. The general solution is y = C.F. + P.I. 6B 1 Solve each differential equation, giving the general solution. d2y dy d2y dy b ____2  − 8 ___  + 12y = 36x a ____2  + 6 ___  + 5y = 10 dx dx dx dx E d2y dy c ____2  + ___     − 12y = 12e2x dx dx d2y dy d ____2  + 2 ___  − 15y = 5 dx dx d2y dy e ____2  − 8 ___  + 16y = 8x + 12 dx dx d2y dy f ____2  + 2 ___  + y = 25 cos 2x dx dx d2y g ____2  + 81y = 15e3x dx d2y h ____2  + 4y = sin x dx d2y dy i ____   2  − 4 ___  + 5y = 25x2 − 7 dx dx d2y dy j ____2  − 2 ___  + 26y = ex dx dx 2 a Find a particular integral for the differential equation dy d  2y ____   2  − 5 ___  + 4y = x  2 − 3x + 2 dx  dx b Hence find the general solution. M06_IAL_FP2_44655_U06_105-124.indd 114 (6 marks) (3 marks) 25/04/2019 08:55 SECOND-ORDER DIFFERENTIAL EQUATIONS E/P CHAPTER 6 115 3 y satisfies the differential equation dy d   2y  ____2  − 6 ___  = 2x  2 − x + 1 dx  dx a Find the complementary function for this differential equation. b Hence find a suitable particular integral and write down the general solution to the differential equation. (7 marks) E/P E/P (3 marks) Hint Try a particular integral of the form λx + μx2 + νx3. 4 Find the general solution to the differential equation dy d  2y ____   2  + 4 ___  = 24x2 dx  dx (10 marks) 5 a Explain why λxe  xis not a suitable form for the particular integral for the differential equation d2y dy ____ (2 marks)   2  − 2 ___  + y = ex dx dx b Find the value of λ for which λx2ex is a particular integral for the differential equation. (5 marks) c Hence find the general solution. E/P (3 marks) d   2y dy ____ 6   2  + 4 ___  + 3y = kt + 5, where k is a constant and t . 0. dt dt  a Find the general solution to the differential equation in terms of k. (7 marks) For large values of t, this general solution may be approximated by a linear function. b Given that k = 6, find the equation of this linear function. (2 marks) Challenge Find the general solution of the differential equation d  2y ____   2  + y = 5xe  2x dx  6.3 Using boundary conditions You can use given boundary conditions to find a particular solution to a second-order differential equation. Since there are two arbitrary constants, you will need two boundary conditions to determine the complete particular solution. Example 11 d2y dy Find y in terms of x, given that ____   2  − y = 2ex, and that ___     = 0 and y = 0 at x = 0. dx dx d2y First consider the equation ____   2  − y = 0. dx m2 − 1 = 0 ⇒ m = ±1 So the complementary function is y = Aex + Be−x. M06_IAL_FP2_44655_U06_105-124.indd 115 Solve the auxiliary equation to find the values of m. 25/04/2019 08:55 116 CHAPTER 6 SECOND-ORDER DIFFERENTIAL EQUATIONS The particular integral cannot be λex, as this is part of the complementary function, so let y = λxex. dy d2y Then  ___  = λxex + λex and ____   2  = λxex + λex + λex dx dx 2y d Substitute into ____   2  − y = 2ex: dx x x x λxe + λe + λe − λxex = 2ex ⇒ λ=1 So a particular integral is xex. The general solution is y = Aex + Be−x + xex Since y = 0 at x = 0, 0 = A + B ⇒ A+B=0 Differentiating y = Aex + Be−x + xex with respect to x gives dy ___     = Aex − Be−x + ex + xex dx dy Since ___     = 0 at x = 0, 0 = A − B + 1 dx ⇒ A − B = −1 Solving the simultaneous equations gives __1 1 A = −__ 2 , and B =   2 __1 So y = −2  ex __1 +   2  e −x d2y As λex satisfies ____   2  − y = 0, it cannot also dx d2y ____ satisfy   2  − y = 2ex dx Substitute the boundary condition, y = 0 at x = 0, into the general solution to obtain an equation relating A and B. Substitute the second boundary condition, y d ___     = 0 at x = 0, into the derivative of the dx general solution, to obtain a second equation relating A and B. Solve the two equations to find values for A and B. + xex is the required solution. Example 12 Given that a particular integral is of the form λ sin 2t, find the solution to the differential d2x x  = 1 when t = 0. equation ____   2  + x = 3 sin 2t, for which x = 0 and d  ___ dt dt d2x First consider the equation ____   2  + x = 0. dt m2 + 1 = 0 ⇒ m = ±i So, the complementary function is x = A cos t + B sin t. The particular integral is λ sin 2t, so let x = λ sin 2t. dx d2x Then ___     = 2λ cos 2t and  ____   = −4λ sin 2t dt dt 2 d2x Substitute into ____   2  + x = 3 sin 2t: dt −4λ sin 2t + λ sin 2t = 3 sin 2t ⇒ λ = −1 M06_IAL_FP2_44655_U06_105-124.indd 116 Solve the auxiliary equation to find the values of m. Watch out Normally you would need to try a particular integral of the form λ sin 2t + μ cos 2t for this equation. However, in this case you are told that there is a particular integral in the form λ sin 2t. 25/04/2019 08:55 SECOND-ORDER DIFFERENTIAL EQUATIONS So a particular integral is −sin 2t. The general solution is x = A cos t + B sin t − sin 2t Since x = 0 at t = 0, A = 0. Differentiating x = B sin t − sin 2t with respect to t gives dx ___     = B cos t − 2 cos 2t dt dx Since  ___  = 1 at t = 0, 1 = B − 2 dt ⇒ B=3 CHAPTER 6 117 Use general solution = complementary function + particular integral. Substitute the initial condition, x = 0 at t = 0, into the general solution to obtain A = 0. dx Substitute the second initial condition, ___     = 1 at dt t = 0, into the derivative of the general solution, to obtain a second equation leading to B = 3. And so x = 3 sin t − sin 2t is the required solution. Exercise E E E 6C 1 a Find the general solution to the differential equation dy d  2y ____ (5 marks)   2  + 5 ___  + 6y = 12e  x dx  dx dy b Hence find the particular solution that satisfies y = 1 and ___     = 0when x = 0. (4 marks) dx 2 a Find the general solution to the differential equation dy d  2y ____ (5 marks)   2  + 2 ___  = 12e  2x dx  dx dy b Hence find the particular solution that satisfies y = 2 and ___     = 6when x = 0. (5 marks) dx dy 1 3 Given that y = 0 and ___     = _  6 when x = 0, find the particular solution to the differential equation dx dy ____ d  2y ___ (10 marks)   2  −     − 42y = 14 dx  dx 4 a Find the general solution to the differential equation d  2y ____ (6 marks)   2  + 9y = 16 sin x dx  dy b Hence find the particular solution that satisfies y = 1 and ___     = 8when x = 0. (6 marks) dx E 5 a Find the general solution to the differential equation dy d  2y (6 marks) 4____   2  + 4 ___  + 5y = sin x + 4 cos x dx  dx dy (6 marks) b Hence find the particular solution that satisfies y = 0 and ___     = 0when x = 0. dx E/P 6 a Find the general solution to the differential equation dx d  2x ____ (6 marks)   2  − 3 ___  + 2x = 2t − 3 dt dt  b Given that x = 1 when t = 0, and x = 2 when t = 1, find a particular solution of this differential equation. (6 marks) E M06_IAL_FP2_44655_U06_105-124.indd 117 25/04/2019 08:55 118 CHAPTER 6 E E/P E E E/P SECOND-ORDER DIFFERENTIAL EQUATIONS 7 Find the particular solution to the differential equation d  2x  ____2  − 9x = 10 sin t dt  dx that satisfies x = 2 and ___     = −1 when t = 0. dt 8 a i Find the value of λ for which y = λt3e2t is a particular solution to the differential equation d  2x dx ____   2  − 4 ___  + 4x = 3te  2t dt dt  ii Hence find the general solution to the differential equation. (6 marks) dx b Find the particular solution that satisfies x = 0 and ___     = 1when t = 0. (6 marks) dt 9 Find the particular solution to the differential equation d  2x 25 ____2  + 36x = 18 dt  dx that satisfies x = 1 and ___     = 0.6when t = 0. dt 10 a Find the general solution to the differential equation dx d  2x ____   2  − 2 ___  + 2x = 2t  2 dt dt  dx b Hence find the particular solution that satisfies x = 1 and ___     = 3when t = 0. dt 11 a Find the general solution to the differential equation dy d2y  ____2  – 3 ___  + 2y = 3e2x dx dx dy b Hence find the particular solution that satisfies y = 0, ___     = 0 when x = 0. dx E/P (10 marks) (12 marks) (6 marks) (6 marks) (7 marks) (6 marks) 12 Solve the differential equation d2y ____   2  + 9y = sin 3x dx dy subject to the boundary conditions y = 0,  ___  = 0 when x = 0. dx 2 d   x dx ____ E/P 13   2  + 5 ___  + 6x = 2e  −t dt dt  dx Given that x = 0 and  ___  = 2 at t = 0, dt a find x in terms of t. __ 2 √3  ____ b Show that the maximum value of x is    and justify that this is a maximum. 9 (14 marks) (8 marks) (7 marks) 6.4 Reducible second-order differential equations You can use a given substitution to reduce second-order differential equations into differential d  2y dy equations of the form a ____   2  + b ___     + cy = f(x) dx dx  M06_IAL_FP2_44655_U06_105-124.indd 118 25/04/2019 08:55 SECOND-ORDER DIFFERENTIAL EQUATIONS CHAPTER 6 119 Example 13 Given that x = eu, show that: d2y d2y ___ dy dy dy   2  −     a x  ___  = ___     b x2  ____2  = ___ dx du du dx du c Hence find the general solution to the differential equation d2y dy   2  + x  ___  + y = 0 x2 ____ dx dx dx a As x = eu ,  ___ = eu = x du From the chain rule, dy dy ___ dy dy dx ___   = ___   ×   = eu  ___ = x  ___, as required du dx du dx dx d2y d dy   ( ___   b ____2  = ___ du du ) du dy d = ___   (eu   ___) du dx dy d2y dx ___ = eu   + eu   ____2   ___ dx dx du 2y d dy dx = ___   + x2   ____2 , as  ___ = eu = x du du dx 2y 2y d dy d ____ ____ ___ So x2   2  =   2  −   as required. du dx du c Substitute the results from parts a and b into the differential equation d2y dy   x2 ____   + x   ___ + y = 0 2 dx dx dy dy d2y to obtain ____   2  −  ___ +  ___ + y = 0 du du du d  2y   ____2  + y = 0 du  m2 + 1 = 0 m = i or m = −i So the general solution in terms of u is y = A cos u + B sin u where A and B are arbitrary constants. x = eu ⇒ u = ln x and the general solution to d2y dy   the differential equation x2 ____   + x   ___ + y = 0 2 dx dx is y = A cos (ln x) + B sin (ln x) M06_IAL_FP2_44655_U06_105-124.indd 119 dy dy Use the chain rule to express  ___ in terms of  ___ dx du Differentiate this product using the product rule. dy Use the chain rule to differentiate ___   with dx respect to u, by differentiating with respect to x, d2y dx giving  ____2 , and then multiplying by ___   du dx d  2y dy This is in the form a  ____2  + b  ___  + cy = 0 with du du  a = 1, b = 0 and c = 1. Find the general solution by considering the roots of the auxiliary equation. The roots are complex, so the general solution will be in the form y = e  pu(A cos qu + B sin qu), with p = 0 and q = 1. Use u = ln x to give y in terms of x. 25/04/2019 08:55 120 CHAPTER 6 Exercise SECOND-ORDER DIFFERENTIAL EQUATIONS 6D 1 Find the general solution to each differential equation using the substitution x = eu, where u is a function of x. d2y dy d2y dy d2y dy ____ ___ ____ ___ ____ 2 2 2 a x   2  + 6x     + 4y = 0     + 6y = 0 b x   2  + 5x     + 4y = 0 c x   2  + 6x ___ dx dx dx dx dx dx 2 2 2 dy dy dy dy dy dy   2  + 4x ___     − 28y = 0   2  − 4x ___     − 14y = 0   2  + 3x ___     + 2y = 0 d x2 ____ e x2 ____ f x2 ____ dx dx dx dx dx dx E z 2 a Show that the transformation y =  __ x  transforms the differential equation d2y dy   2  + (2 – 4x) ___     − 4y = 0 (1) x2 ____ dx dx into the differential equation d2z dz ____   2  – 4 ___  = 0 (2) dx dx b Find the general solution to differential equation (2), giving z as a function of x. c Hence obtain the general solution to differential equation (1). E (4 marks) (1 mark) z 3 a Show that the substitution y = __   2  transforms the differential equation x d2y dy ____ ___ 2 (1) x   2  + 2x(x + 2)     + 2(x + 1)2 y = e–x Hint Use a particular integral of the form dx dx λe–x . ← Further Pure 2 Section 6.2 into the differential equation dz d2z (2) (6 marks)  ____2  + 2 ___  + 2z = e–x dx dx b Find the general solution to differential equation (2), giving z as a function of x. c Hence obtain the general solution to differential equation (1). E (6 marks) (7 marks) (1 mark) 4 a Use the substitution z = sin x to transform the differential equation dy d2y     − 2y cos3 x = 2 cos5 x  cos x  ____2  + sin x ___ dx dx into the equation d2y ___   2  − 2y = 2(1 − z2) dz d2y dy b Hence solve the equation cos x ____   2  + sin x ___     − 2y cos3 x = 2 cos5 x, dx dx giving y in terms of x. M06_IAL_FP2_44655_U06_105-124.indd 120 (6 marks) (8 marks) 25/04/2019 08:55 SECOND-ORDER DIFFERENTIAL EQUATIONS E CHAPTER 6 121 5 a Show that the transformation x = uttransforms the differential equation dx d  2x t  2 ____   2  − 2t  ___  = − 2(1 − 2t  2)x (1) dt dt  into the differential equation   2u d ____   2  − 4u = 0 (2) (6 marks) dt  b By solving differential equation (2), find the general solution to differential equation (1) in the form x = f(t). (8 marks) dx Given that x = 2 and ___     = 1at t = 1, dt c find the particular solution to differential equation (1). (5 marks) Challenge dy Use the substitution u = ___     to find the general solution to the differential equation dx dy d   2y ___ ____ x   2  +     = 12x dx  dx Chapter review 6 E 1 a Find the general solution to the differential equation dy π π      ___  + y tan x = e  x cos x, −  __  , x , __ 2 2 dx giving your answer in the form y = f(x). b Find the particular solution for which y = 1 at x = π. E (3 marks) dy 2 ___     − 3y = sin x dx Given that y = 0 when x = 0, find y in terms of x. E (6 marks) (7 marks) dy 3 ___     = x (4 − y  2) dx Given that y = 1 when x = 0, find y in terms of x. (7 marks) E d2y dy 4 Find the general solution to the differential equation  ____2  +   ___ + y = 0 dx dx (6 marks) E d2y dy 5 Find the general solution to the differential equation  ____2  − 12  ___ + 36y = 0 dx dx (6 marks) E d2y dy 6 Find the general solution to the differential equation  ____2  − 4   ___ = 0 dx dx (6 marks) E/P d2y 7 Find y in terms of k and x, given that   ____2  + k2y = 0 where k is a constant, and y = 1 dx dy ___ and     = 1 at x = 0. dx M06_IAL_FP2_44655_U06_105-124.indd 121 (8 marks) 25/04/2019 08:55 122 CHAPTER 6 E E E/P E/P SECOND-ORDER DIFFERENTIAL EQUATIONS d2y dy 8 Find the solution to the differential equation  ____2  − 2  ___ + 10y = 0 for which y = 0 dx dx dy ___ and     = 3 at x = 0. dx (8 marks) 9 a Find the value of k for which y = ke  2xis a particular integral of the differential equation dy d  2y (4 marks)  ____2  − 4 ___  + 13y = e  2x dx  dx b Using your answer to part a, find the general solution to the differential equation. (5 marks) 10 Find the general solution of the differential equation d  2y  ____2  − y = 4e  x dx  (7 marks) dy d2y 11 The differential equation  ____2  − 4   ___ + 4y = 4e2x is to be solved. dx dx a Find the complementary function. (3 marks) b Explain why neither λe2x nor λxe2x can be a particular integral for this equation. (2 marks) A particular integral has the form k x2e2x. c Determine the value of the constant k and find the general solution of the equation. E/P E 12 Find the particular solution of the differential equation d  2y  ____2  + 4y = 5 cos 3t dt  dy which satisfies the initial conditions that when t = 0, y = 1 and ___     = 2. dt (6 marks) (12 marks) 13 a Find the values of λ, μ and k such that y = λ + μx + kxe  2xis a particular integral of the differential equation dy d  2y ____ (5 marks)   2  − 3 ___  + 2y = 4x + e  2x dx  dx b Using your answer to part a, find the general solution of the differential equation. (5 marks) E/P E/P d2y dy 14 a Find the solution of the differential equation 16  ____2  + 8  ___  + 5y = 5x + 23 for which y = 3 dx dx dy ___ (8 marks) and     = 3 at x = 0. dx b Show that y ≈ x + 3 for large values of x. (2 marks) d2y dy 15 Find the solution of the differential equation  ____2  −   ___  − 6y = 3 sin 3x − 2 cos 3x for which dx dx y = 1 at x = 0 and for which y remains finite as x → ∞. (8 marks) M06_IAL_FP2_44655_U06_105-124.indd 122 25/04/2019 08:55 SECOND-ORDER DIFFERENTIAL EQUATIONS E/P CHAPTER 6 16 x satisfies the differential equation dx d  2x  ____2  + 8 ___  + 16x = cos 4t, t > 0 dt dt  a Find the general solution of the differential equation. b Find the particular solution of this differential equation for which, at t = 0, x = dx and ___     = 0. dt c Describe the behaviour of the function for large values of t. E/P 123 17 a Find the general solution to the differential equation d2y dy x2  ____2  + 4x  ___  + 2y = ln x, x . 0 dx dx using the substitution x = eu. (8 marks) 1  _2  (5 marks) (2 marks) (10 marks) b Find the equation of the solution curve passing through the point (1, 1) with gradient 1. E/P (3 marks) dy d2y 18 Solve the equation  ____2  + tan x  ___  + y cos2 x = cos2 x esin x, using the substitution z = sin x. dx dx dy (13 marks) Find the solution for which y = 1 and ___     = 3 at x = 0. dx Challenge 1 Use the substitution z = y2 to transform the differential equation dy 1 2(1 + x2) ___   + 2xy =  __ y dx into a differential equation in z and x. By first solving the transformed equation, a find the general solution of the original equation, giving y in terms of x. b Find the particular solution for which y = 2 when x = 0. 2 a Find the general solution of the differential equation d2y dy x2  ____2  + 4x  ___ + 2y = ln x, x . 0, dx dx using the substitution x = eu , where u is a function of x. b Find the equation of the solution curve passing through the point (1, 1) with gradient 1. 3 By means of a suitable substitution, show that the general solution to the differential equation d  2y dy 2 ____   2  = (    )  ___ dx  dx is given by y = A – ln(x + B), where A and B are arbitrary constants. M06_IAL_FP2_44655_U06_105-124.indd 123 25/04/2019 08:55 124 CHAPTER 6 SECOND-ORDER DIFFERENTIAL EQUATIONS Summary of key points 1 The natures of the roots α and β of the auxiliary equation determine the general solution dy d2y     + c = 0 to the second-order differential equation a  ____2  + b ___ dx dx You need to consider three different cases: • Case 1: b  2 > 4ac The auxiliary equation has two real roots α and β (α ≠ β). The general solution will be of the form y = Ae  αx + Be  βx where A and B are arbitrary constants. • Case 2: b  2 = 4ac The auxiliary equation has one repeated root α. The general solution will be of the form y = (A + Bx)e  αx where A and B are arbitrary constants. • Case 3: b  2 < 4ac The auxiliary equation has two complex conjugate roots α and β equal to p ± qi. The general solution will be of the form y = e  px(A cos qx + B sin qx) where A and B are arbitrary constants. dy d   2y     + cy = f(x) 2 To find the general solution to the differential equation a ____   2  + b ___ dx  dx d  2y ____ dy • Solve the corresponding homogeneous equation a  2  + b ___  + cy = 0to find the dx  dx complementary function, C.F. • C hoose an appropriate form for the particular integral, P.I., and substitute into the original equation to find the values of any coefficients. • The general solution is y = C.F. + P.I. 3 You can use a substitution to reduce a first-order differential equation into a form that you know how to solve, either by separating the variables, or by using an integrating factor. 4 You can use a given substitution to reduce second-order differential equations into differential equations of the form dy d  2y a  ____2  + b  ___  + cy = f(x) dx dx  M06_IAL_FP2_44655_U06_105-124.indd 124 25/04/2019 08:55 MACLAURIN AND TAYLOR SERIES CHAPTER 7 125 7 MACLAURIN AND TAYLOR SERIES 6.1 6.2 6.3 6.4 Learning objectives A�er completing this chapter you should be able to: ● Find and use higher derivatives of functions → pages 126–127 ● Derive and use Maclaurin series for simple functions → pages 128–132 ● Use series expansions of compound functions → pages 132–136 ● Derive and use Taylor series for simple functions → pages 136–139 ● Use the Taylor series method to find a series solution to a differential equation → pages 140–144 Prior knowledge check 1 Differentiate: a cos (1 + x 3 ) 1 b _______ ex sin x 2 ← Pure 3 Section 6.3 Find the general solution to the d2y dy differential equation ____2 + 2___ + 2y = 0 dx dx ← Further Pure 2 Section 6.1 M07_IAL_FP2_44655_U07_125-148.indd 125 Taylor series can be used to approximate functions by polynomials. Mathematicians and engineers use them to approximate and model solutions to complex differential equations such as those that describe the flow of air over an aircra� wing. In this chapter you will use Taylor series to find approximate solutions to differential equations that can’t be solved easily by other methods. 25/04/2019 08:54 126 CHAPTER 7 MACLAURIN AND TAYLOR SERIES 7.1 Higher derivatives You need to be able to find third, and higher, derivatives of given functions. You already know how to find first and second derivatives. dy If y = f(x), the first derivative of f(x) is given by  ___  = f9(x), and the second derivative of f(x) is given dx d  2y d dy by  ____2  =  ___ ( ___ ) = f 0(x). dx  dx dx d  3y d d  2y Similarly, the third derivative is given by  ____3  =  ___ ( ____2 ) = f -(x), and so on. dx  dx dx  You can find the nth derivative of f(x) by differentiating n times with respect to x. Example 1 d  3y 1     Given that y = ln(1 − x), find the value of ____   3  when x = __ 2 dx  Use the chain rule. dy ___   d 1     1   ___   1   × (−1) = −  _____   = _____     (1 − x) = _____ 1 − x dx 1−x 1−x dx ( ) dy ___ 2 d       2  = ___ dx dx dy d 1   × (−1) = −  _______ 1       = ___     (−(1 − x)−1) =  _______ ___ dx dx (1 − x)2 (1 − x)2 dy ___ d2y d 2 2   2   = ___     (−(1 − x)−2) =  _______   × (−1) = −  _______   ___ dx dx (1 − x)3 (1 − x)3 3 d   3  = ___     dx dx ( ) d3y −2 1  , ____   = −16 So when x =  __     3  = _______ 2 dx ( 1 − _ 21    )3 Example ← Pure 3 Section 6.3 Substitute x = _  12  2 f(x) = ex 2 a Show that f 9(x) = 2xf(x) b By differentiating the result in part a twice more with respect to x, show that: i f 0(x) = 2f(x) + 2xf 9(x) ii f -(x) = 2xf 0(x) + 4f 9(x) c Deduce the values of f 9(0), f 0(0) and f -(0). 2 2 d a f9(x) = ex ___    (x2) = 2xex dx = 2xf(x) b i f0(x) = 2f(x) + 2xf9(x) ii f-(x) = 2f9(x) + (2xf0(x) + 2f9(x)) = 2xf0(x) + 4f9(x) M07_IAL_FP2_44655_U07_125-148.indd 126 If f(x) = eu, then f9(x) = eu ___   du  dx f(x) = ex 2 Use the product rule. ← Pure 3 Section 6.4 Differentiate again. 25/04/2019 08:54 MACLAURIN AND TAYLOR SERIES CHAPTER 7 c f(0) = e0 = 1 f9(0) = 2 × 0 × e0 = 0 f0(0) = 2f(0) + 2 × 0 × f9(0) = 2f(0) = 2 f-(0) = 2 × 0 × f0(0) + 4f9(0) = 4f9(0) = 0 Exercise 127 Substitute x = 0 into f0(x). Substitute x = 0 into f-(x). 7A 1 For each function, f(x), find f9(x), f 0(x), f -(x) and f (n)(x). a e2x P b (1 + x)n c xex d ln(1 + x) dny 2 a Given that y = e2 + 3xfind an expression, in terms of y, for ____   n  dx 6 dy 1 b Hence evaluate ___   6  when x = ln (_  9 ) dx 3 Given that y = sin2 3x, dy a show that  ___  = 3 sin 6x dx d 4y π c Hence evaluate  ____4  when x = __     6 dx d2y d3y d4y b find expressions for  ____2  ,  ____3  and ____   4  dx dx dx 4 f(x) = x2e−x a Show that f -(x) = (6x − 6 − x2)e−x b Show that f -9(2) = 0 5 Given that y = sec x d2y a show that  ____2  = 2 sec3 x − sec x dx __ d3y π b show that the value of ____   3  when x = __     is 11√2  4 dx P 6 Given that y is a function of x, d2y d2 a show that  ____2  ( y2) = 2y  ____2  + 2 dx dx ( ) dy 2  ___   dx dy b Find an expression, in terms of y, ___    , dx dy ____ 2 d3y d3   2  and ___   3  , for ____   3  ( y2) dx dx dx ______ 7 Given that f(x) = ln (x + √ 1 + x2  ), show that: ______ a√1 + x2  f 9(x) = 1 b (1 + x2) f 0(x) + xf 9(x) = 0 c (1 + x2) f -(x) + 3xf 0(x) + f 9(x) = 0 d Deduce the values of f 9(0), f 0(0) and f -(0). M07_IAL_FP2_44655_U07_125-148.indd 127 25/04/2019 08:54 128 CHAPTER 7 MACLAURIN AND TAYLOR SERIES 7.2 Maclaurin series Many functions can be written as an infinite sum of terms of the form axn. You may have already encountered series expansions like these: 1  _____  = 1 + x + x  2 + x  3 + …, |x| , 1 1−x ____ x x  2 x   3 √ 1 + x  = 1 + __     −  __  +  ___  − …, |x| , 1 2 8 16 x  2 x   3 x  4 e  x = 1 + x +  __  +  __  +  ___  + …, x ∈ ℝ 2 6 24 Example Links The first two series expansions shown here are examples of the binomial expansion. ← Pure 4 Chapter 4.3 3 Given that f(x)can be differentiated infinitely many times and that it has a valid series expansion of the form f(x) = a 0 + a 1x + a 2x  2 + a 3x  3 + … + ar  x  r + …, where the ai are all real constants, show that the series expansion must be f 0(0)x  2 f  (r) (0)x  r     + … +   _______   + … f(x) = f(0) + f 9(0)x + ______ 2! r! Write f(x) = a 0 + a1x + a2x2 + a3x3 + … + arxr + … f(0) = a 0 Differentiating f(x) gives: f9(x) = a 1 + 2a 2x + 3a 3x  2 + … + rar  x  r−1 + … f0(x) = 2 × 1a 2 + 3 × 2a 3x + … + r (r − 1) ar  x  r−2 + … f-(x) = 3 × 2 × 1a 3 + … + r (r − 1)(r − 2) ar  x  r−3 + … The coefficient of a 0 can be found by setting x = 0. Successively differentiate with respect to x to obtain f9(x), f 0(x) and f -(x). Continuing in this way by differentiating r times: f  (r)( x) = r!ar  + terms in powers of x Evaluate each term at x = 0: f9(0) = a 1 ⇒ a 1 = f9(0) f0(0) f0(0) = 2!a 2 ⇒ a 2 = ____     2! f-(0)     f-(0) = 3!a 3 ⇒ a 3 = _____ 3! f  (r)(0)     f  (r)( 0) = r!ar  ⇒ ar  = ____ r! f0(0) f-(0) 3 f  (r)(0)     x  2 + _____       xr + … Therefore f(x) = f(0) + f9(0)x + _____   x  + … + _____ r! 2! 3! M07_IAL_FP2_44655_U07_125-148.indd 128 Find the coefficients a1, a2, a3,…, ar,… by substituting x = 0 into each result and rearranging. Substitute a  1 = f9(0),  f0(0) f-(0) a 2 = ____      , a 3 = _____     , … , 2! 3! ____ f  (r)( 0) ar  =     , … r! 25/04/2019 08:54 MACLAURIN AND TAYLOR SERIES CHAPTER 7 129 In this process, outlined in the worked example above, a polynomial in powers of x is being formed step by step. The process focuses on x = 0; substituting x = 0 into successive derivatives increases the power of the polynomial. For example, if you stop the process after finding f9(0) the polynomial is f0(0) 2   x  , after f-(0) it is cubic, linear, f(0) + f9(0)x, after f0(0) it is quadratic, f(0) + f9(0)x +   ____ 2! f0(0) 2 f-(0) f(0) + f9(0)x +   ____   x  +   ____   x  3and so on. 2! 3! Watch out Not all functions satisfy the The above argument assumes that the function can condition that f(0), f9(0), f0(0), … , f  (r)(0) all be written in the given form. This is only true if the have finite values. given series converges. The above reasoning also only 1  For example, when f(x) = ln x, f9(x) = __  x holds (i.e. remains true) if the function can be so f9(0)is undefined and therefore does not differentiated an infinite number of times, and have a finite value. if f(r)(0) is always finite. ■ The Maclaurin series expansion of a function f(x) is given by f(r)(0) f 0(0)     xr + … f(x) = f(0) + f9(0)x +_____     x2 + … +_____ 2! r! The series is valid provided that f(0), f9(0), f 0(0), … , f(r)(0), … all have finite values. The polynomial f(0) + f9(0)xis a Maclaurin polynomial of degree 1. f0(0) The polynomial f(0) + f9(0)x +   ____  x  2is a Maclaurin polynomial of degree 2. 2! (0) f0 f  (r)( 0) r   x  is a Maclaurin polynomial of degree r. The polynomial f(0) + f9(0)x +   ____  x  2 + … +   ____ 2! r! Even when f(r)(0) exists and is finite for all r, a Maclaurin series expansion is only valid for 1 values of x that give rise to a convergent series. For example, the Maclaurin series of _____     1−x 2 3 is 1 + x + x  + x   + … . 1 But when x = 2, the series gives 1 + 2 + 4 + 8 + … which does not converge to  _____  = −1. 1−2 Notation Example The range of validity for some individual Maclaurin series is given in the formulae booklet. If no range of validity is given in this chapter, you may assume that the expansion is valid for all x ∈ ℝ. 4 a Express ln(1 + x) as an infinite series in ascending powers of x. b Using only the first three terms of the series in part a, find estimates for: i ln 1.05 ii ln 1.25 iii ln 1.8 Comment on the accuracy of the estimates. a f(x) = ln (1 + x) ⇒ f(0) = ln 1 = 0 f9(x) = _____   1   = (1 + x)−1 ⇒ 1+x f0(x) = −(1 + x)−2 ⇒ f0(0) = −1 ⇒ f-(0) = 2! f-(x) = (−1)(−2)(1 + x)−3 f9(0) = 1 f(r)(x) = (−1)(−2)(−3)…(−(r − 1))(1 + x)−r ⇒ f(r)(0) = (−1)r − 1(r − 1)! M07_IAL_FP2_44655_U07_125-148.indd 129 Problem-solving The term ( −1)  rcan be used in the general term of alternating sequences, in which the terms are alternately positive and negative. 25/04/2019 08:54 130 CHAPTER 7 MACLAURIN AND TAYLOR SERIES (2!) So ln (1 + x) = 0 + 1x + ____   −1  x2 + ____     x3 + … 2! 3! (−1)r − 1(r − 1)! r + ____________        x + … r! x2 x3 xr ln(1 + x) = x − ___     +  ___  + … + (−1)r − 1  __ r  + … 2 3 b i 0.052 ______ 0.053 ln 1.05 = 0.05 − ______     +     − … 2 3 ≈ 0.0487916… This is correct to 5 d.p. ii 0.252 ______ 0.253 ln 1.25 = 0.25 − ______     +     − … 2 3 Substitute the values for f(0), f9(0), f 0(0), etc. into the Maclaurin series for f(x). Online This expansion is valid for −1 , x < 1. If you use a computer to generate the graphs of the successive Maclaurin polynomials you will see that they converge to the graph of ln (1 + x) between x = −1 and x = 1, but outside that interval they diverge rapidly. Explore this using GeoGebra. ≈ 0.223958… This is correct to 2 d.p. 0.82 _____ iii ln 1.8 = 0.8 −   0.83 _____   +   2   − … 3 ≈ 0.6506666… This is not correct to 1 d.p. Example The further away a value is from x = 0, the less accurate the approximation will be and the more terms of the series you need to take to maintain a required degree of accuracy. 5 a Find the first four terms in the Maclaurin series of sin x. b Using the first two terms of the series find an approximation for sin 10°. a f(x) = sin x f9(x) = cos x f0(x) = −sin x f-(x) = −cos x f-9(x) = sin x ⇒ ⇒ ⇒ ⇒ ⇒ f(0) = sin 0 = 0 f9(0) = cos 0 = 1 f0(0) = −sin 0 = 0 f-(0) = −cos 0 = −1 f-9(0) = sin 0 = 0 (−1)r 2r + 1 So sin x = x + ____  −1 x3 + __   1  x5 + ____  −1 x7 + … + _______   +…   x 5! 7! (2r + 1)! 3! = x − __   1   x3 + __   1   x5 − __   1   x7 + … 5! 7! 3! π π π 3 b sin 10° = sin  ___  ≈ ___     − __   1   ___     18 18 6 ( 18) ≈ 0.174532925 − 0.000886096 ≈ 0.173646829 M07_IAL_FP2_44655_U07_125-148.indd 130 f(n) = 0, if n is even, and the cycle of values 0, 1, 0, −1 repeats itself. This expansion is valid for all values of x. Watch out x must be in radians in expansions of trigonometric functions. This estimate is correct to 5 decimal places; even using sin x ≈ x, the approximation is correct to 2 d.p. 25/04/2019 08:54 MACLAURIN AND TAYLOR SERIES Exercise CHAPTER 7 131 7B 1 Use the formula for the Maclaurin series and differentiation to show that: Hint The binomial expansions of (1 + x)n , where n is fractional or negative and |x| , 1, are the Maclaurin series of the function. a (1 − x)−1 = 1 + x + x2 + … + xr + … _____ x x2 ___ x3     − __     +     − … b √1 + x  = 1 + __ 2 8 16 2 Use Maclaurin series and differentiation to show that the first three terms in the series x2     expansion of esin x are 1 + x + __ 2 P x2 x4 x2r     + __     + … + (−1)r ____     + … 3 a Show that the Maclaurin series of cos x is 1 − __ 2! 4! (2r)! b Using the first three terms of the series, show that it gives a value for cos 30° correct to 3 decimal places. Hint This expansion is valid for all values of x. 4 Using the series expansions for ex and ln(1 + x) respectively (i.e. in the same order as what has already been mentioned), find, correct to 3 decimal places, the values of: 6    ) a e b ln (__ 5 5 Use Maclaurin series and differentiation to expand, in ascending powers of x up to and including the term in x 4, b ln (1 + 2x) a e3x P c sin2 x 6 Using the addition formula for cos (A − B) and the series expansions of sin x and cos x, show that x2 x3 ___ x4 π 1    ) = ___   __  (1 + x − __     − __     +     + …) cos (x − __ 4 2 24 6 √  2 E 7 Given that f(x) = (1 − x)2 ln (1 − x), a show that f 0(x) = 3 + 2ln (1 − x) (2 marks) b find the values of f(0), f 9(0), f 0(0), and f -(0) c express (1 − x ln (1 − x) in ascending powers of x up to and including the term in x3. )2 E/P (1 mark) (3 marks) 8 a Using the series expansions of sin x and cos x, show that 3 sin x − 4x cos x + x = _  2  x3 − ___   120  x5 + … 3 17 3 sin x − 4x cos x + x        b Hence, find the limit, as x → 0, of ____________________ x3 M07_IAL_FP2_44655_U07_125-148.indd 131 (5 marks) (1 mark) 25/04/2019 08:54 132 CHAPTER 7 E MACLAURIN AND TAYLOR SERIES 9 Given that f(x) = ln cos x, a show that f 9(x) = −tan x (2 marks) b find the values of f 9(0), f 0(0), f -(0) and f -9(0) (1 mark) c express ln cos x as a series in ascending powers of x up to and including the term (3 marks) in x 4 π __ d show that using the first two terms of the Maclaurin series for ln cos x with x =     gives a 4 π2 π2     (1 + ___ value for ln 2 of ___    ) (2 marks) 16 96 E/P 1 2 10 Show that the Maclaurin series for tan x, as far as the term in x5, is x + __    x3 + ___    x5 3 15 Challenge (5 marks) Problem-solving The ratio test is a sufficient condition for the convergence of an infinite ∞ a r+1     ar converges if r→∞  lim ____   a    , 1and diverges series. It says that a series ∑ r r=1 a r+1 if  r→∞ lim ____   a    . 1 r | | | | a  r+1 If r→∞  lim ____   a    = 1or does not | | r exist then the ratio test is inconclusive. Use the ratio test to show that a the Maclaurin series expansion of ex converges for all x ∈ ℝ b the Maclaurin series expansion of ln (1 + x) converges for −1 , x , 1, and diverges for x . 1. 7.3 Series expansions of compound functions You can find the series expansions of compound functions using known Maclaurin series. In the last exercise you found the Maclaurin series of simple compound functions, such as e  3x and ln (1 + 2x). However, the resulting series could also be found by replacing x by 3xor x by 2xin the known expansions of e  x and ln (1 + x)respectively. When successive derivatives of a compound function are more difficult, or when there are products of functions involved, it is often possible to use one of the standard results. ■ The following Maclaurin series expansions are given in the formulae booklet: x   + … for all x • ex = 1 + x + __   x   + … +  __ 2! r! 2 3 x x xr + … __ __ • ln (1 + x) = x −     +    − … + (−1)r + 1  __ r 2 3 3 5 2r + 1   x   + __   x   − … + (−1)r ________ x   + … • sin x = x − __ 3! 5! (2r + 1)! 2 4 2r • cos x = 1 − __   x   + __   x   − … + (−1)r ____ x   + … 2! 4! (2r)! 3 5 2r + 1 • arctan x = x − ___   x   + ___   x   − … + (−1)r ______ x   + … 3 5 2r + 1 2 M07_IAL_FP2_44655_U07_125-148.indd 132 r −1 , x < 1 for all x for all x −1 < x < 1 25/04/2019 08:54 MACLAURIN AND TAYLOR SERIES Example CHAPTER 7 133 6 Write down the first four non-zero terms in the series expansion (in ascending powers of x) of cos (2x2). (2x2)2 ______ (2x2)4 ______ (2x2)6 cos (2x2) = 1 − _____     +     −     + … 2! 4! 6! Substitute 2x2 for x in the above series for cos x. 4 2 = 1 − 2x4 + __    x8 − ___    x12 + … 45 3 Example Watch out Make sure you simplify the coefficients as much as possible. 7 ______ √1 + 2x  Find the first three non-zero terms in the series expansion of ln (  _______ ) and state the values of 1 − 3x x for which the expansion is valid. ln(   _______ () a   = ln a − ln b Using ln  __ b   = ln √1 + 2x  − ln (1 − 3x) 1 − 3x ) _______ √ (1 + 2x)  _________ = __   21 ln (1 + 2x) − ln (1 − 3x) ( _1 Using ln a 2  = __  12  ln a ) (2x)2 _____ (2x)3 __1 _____ 1  __ 2 ln (1 + 2x) =   2 2x −   2  +   3  − … , −1 , 2x < 1 = x − x2 + __   43 x3 − … (−3x)2 − __   21 , x < __   21 (−3x)3 ln (1 − 3x) = (−3x) − _______       + _______   − …, −1 , −3x < 1 2 3 = −3x − __   92 x2 − 9x3 − … __1 1 −  __ 3 < x ,   3 ______ Substitute 2x for x in the expansion of ln(1 + x) Problem-solving You are substituting 2x into the series expansion of ln (1 + x), so the series is now only valid for −1 , 2x < 1, or − _  12  , x < _  12  √ 1 + 2x  ________ So ln     43 x3 − …)   = (x − x2 + __ 1 − 3x __ 9 − (−3x −   2 Example − __1 __1 9x3 − …), −   3 < x ,   3 3 = 4x + __   72 x2 + __   31 3 x + … , __1 1 −  __ 3 < x ,   3 x2 Substitute −3x for x in the expansion of ln(1 + x)0 You need both intervals to be satisfied. This is the case for −  _13 < x , _  13 8 Given that terms in xn with n . 4 may be neglected (i.e. deliberately ignored), use the series expansions for ex and sin x to show that x2 x4     − __     esin x ≈ 1 + x + __ 2 8 M07_IAL_FP2_44655_U07_125-148.indd 133 25/04/2019 08:54 134 CHAPTER 7 MACLAURIN AND TAYLOR SERIES Only two terms are used as the next term is kx  5. x  3 sin x = x − __     + … 3! So e  sin x = e  (x −   3!  + …) __ x3 __ x3 = e  x × e  −   6  × … Use e  a−b = e  a × e  −b x  2 x  3 ___ x  3 x  4 = ( 1 + x + __     + __     +     + …)(1 + ( −  __ ) + …) … 2 6 24 6 x  3 Substitute −  __  for xin the 6 expansion of e  x. x  2 x  3 x  4 __ x  3 x  4 = 1 + x + __     +  __  + ___     −     − __     + … 2 6 24 6 6 x  2 x  4 ≈ 1 + x +  __  −  __  2 8 7C Exercise P 1 Use the series expansions of ex, ln (1 + x) and sin x to expand the following functions as far as the fourth non-zero term. In each case state the values of x for which the expansion is valid. e2x × e3x b ________ ex  1 a __ ex  c e1 + x Hint For part f, write 2 + 3x as 2(1 + ___   3x ) 2 d ln (1 − x) x e sin  __   2 ( ) E/P Simplify as much as possible. f ln (2 + 3x) 2 a Using the Maclaurin series of ln (1 + x), show that 1+x x3 x5     + __     + …), −1 , x , 1 ln  _____   = 2(x + __ 5 3 1−x ( ) (4 marks) _____ √ 1+x b Deduce the series expansion for ln _____     , −1 , x , 1 1−x (2 marks) c By choosing a suitable value of x, and using only the first three terms of the series from 2 part a, find an approximation for ln ( _3   ), giving your answer to 4 decimal places. (2 marks) d Show that the first three terms of your series from part b, with x = _ 5 , give an approximation for ln 2, which is correct to 2 decimal places. (2 marks) 3 E/P 3 Show that, for small values of x, e2x − e−x ≈ 3x + _ 2 x2 (4 marks) E/P   2 x2 − __   8 x 4 − … 4 a Show that 3x sin 2x − cos 3x = −1 + __ (5 marks) 3 21 ( 59 ) 3x sin 2x − cos 3x + 1 _____________________    b Hence find the lim       x2 x→0 P (1 mark) 5 Find the series expansions, up to and including the term in x 4, of: a ln (1 + x − 2x2) b ln (9 + 6x + x2) Notation Factorise the quadratic first. and in each case give the range of values of x for which the expansion is valid. M07_IAL_FP2_44655_U07_125-148.indd 134 25/04/2019 08:54 MACLAURIN AND TAYLOR SERIES E/P CHAPTER 7 135 6 a Write down the series expansion of cos 2x in ascending powers of x, up to and including the term in x8. (3 marks) b Hence, or otherwise, find the first four non-zero terms in the series expansion for sin2 x. (3 marks) E/P 7 Show that the first two non-zero terms of the series expansion, in ascending powers of x, of ln (1 + x) + (x − 1)(ex − 1) are px3 and qx 4, where p and q are constants to be found. (6 marks) E/P 8 a By considering the product of the series expansions of sin x and (1 − x)−2, sin x expand  _______2  in ascending powers of x as far as the term in x 4. (1 − x) b Deduce the gradient of the tangent, at the origin, to the curve with sin x equation y = _______     (1 − x)2 P (6 marks) (3 marks) 9 Use the Maclaurin series, together with a suitable substitution, to show that: a (1 − 3x) ln (1 + 2x) = 2x − 8x2 + __   3 x3 − 12x 4 + … 26 b e2x sin x = x + 2x2 + __   6 x3 + x 4 + … 11 ______ c √1 + x2  e−x = 1 − x + x2 − _ 3 x3 + _  6 x 4 + … E/P 2 1 x2 − __  10 a Write down the first five non-zero terms in the series expansions of e 2 ∫ 1 (3 marks) x − __  2 b Using your result from part a, find an approximate value for   e 2 dx, giving your answer to 3 decimal places. E/P −1 3(p2 − 3) 3 11 a Show that e px sin 3x = 3x + 3px2 + _________   x + … where p is a constant.   2 b Given that the first non-zero term in the expansion, in ascending powers of x, of e px sin 3x + ln (1 + qx) − x is kx3, where k is a constant, find the values of p, q and k. E/P (3 marks) (5 marks) (4 marks) 12 f(x) = ex − ln x sin x, x . 0 a Show that, if x is sufficiently small and x 4 and higher powers of x are neglected, x2 f(x) ≈ 1 + x + __     (5 marks) 3 b Show that using x = 0.1 in the result from part a gives an approximation for f(0.1) which is correct to 6 significant figures. (2 marks) E/P 13 y = sin 2x − cos 2x d4y a Show that ___   4  = 16y dx b Find the first five terms of the Maclaurin series for y, giving each coefficient in its simplest form. M07_IAL_FP2_44655_U07_125-148.indd 135 (4 marks) (4 marks) 25/04/2019 08:54 136 CHAPTER 7 MACLAURIN AND TAYLOR SERIES Notation Challenge The Lorentz factor of a moving object, γ, is given by the formula 1   γ = _______   ______ √1 − β 2  A light year is the distance light travels in one year. where β = _ cv  is the ratio of v, the speed of the object, to c, the speed of light (3 × 108 m s−1). 1   in ascending ______ a Find the Maclaurin series expansion of γ =  _______ 1 − β 2  √ powers of β up to the term in β4. The theory of special relativity predicts that a period of time observed as T within a stationary frame of reference will be observed as a period of time __   T in a moving frame of reference. γ A spaceship travels from Earth to a planet 4.2 light years away. To an observer on Earth, the journey appears to take 20 years. b Use your answer to part a to estimate the observed journey time for a person on the spaceship. c Calculate the percentage error in your estimate. d Comment on whether your approximation would be more or less accurate if the spaceship was travelling at three times the speed. 7.4 Taylor series Earlier in this chapter you used Maclaurin series expansions to write a function of x as an infinite series in ascending powers of x. However, the conditions of the Maclaurin series expansion mean that some functions, such as ln x, cannot be expanded in this way. Links The Maclaurin series expansion requires that f(n)(0) exists and is finite for all n P ℕ. 1 If f(x) = ln x then f9(x) =  __ x  so f9(0) is undefined. ← Further Pure 2 Section 7.2 The construction of the Maclaurin series expansion focuses on x = 0 and, for a value of x very close to 0, a few terms of the series may well give a good approximation of the function. For values of x further away from 0, even if they are in the interval of validity, more and more terms of the series are required to give a good degree of accuracy. Notation An extreme example of this is in using x = 1 in the series for ln(1 + x) to find ln 2; thousands of terms of the series are required to reach 4 significant figure accuracy. To overcome these problems, a series expansion focusing on x = a can be derived. This series expansion, called a Taylor series, is a more general form of the Maclaurin series. Consider the functions f and g, where f(x + a) ; g(x). Then f(r)(x + a) = g(r)(x), r = 1, 2, 3, … , a ≠ 0 Notation For example, f(x) = ln x, g(x) = ln(x + 1) In particular, f(r)(a) = g(r)(0), r = 1, 2, 3, … M07_IAL_FP2_44655_U07_125-148.indd 136 25/04/2019 08:54 MACLAURIN AND TAYLOR SERIES CHAPTER 7 137 So the Maclaurin series expansion for g, g0(0) 2 g-(0) g(r)(0) r g(x) = g(0) + g9(0)x +   _____  x +   _____  x3 + … +   _____  x + … 2! 3! r! becomes Notation The Taylor 0(a) 2 f_____ -(a) 3 f_____ f(r)(a) r _____ series allows you to ■ f(x + a) = f(a) + f9(a)x +    x +    x + … +    x + … (A) 2! 3! r! approximate the value of f(x) close to x = a. Replacing x by x − a, gives a second useful form: r ( ) f 0(a) f -(a) f (a)    (x − a)r + … (B) ■ f(x) = f(a) + f9(a)(x − a) +   _____  (x − a)2 +   _____  (x − a)3 + … + _____ 2! 3! r! The expansions (A) and (B) given above are known as Taylor series expansions of f(x) at (or about) the point x = a. The Taylor series expansion is valid only if f(n)(a) exists and is finite for all n ∈ ℕ, and for values of x for which the infinite series converges. Example 9 Find the Taylor series expansion of e−x in powers of (x + 4) up to and including the term in (x + 4)3. Let f(x) = e−x and a = −4. f0(−4) f-(−4) f(x) = f(−4) + f9(−4)(x + 4) + _____       (x + 4)3 + …   (x + 4)2 + ______ 2! 3! Use the Taylor series expansion (B). f(x) = e−x ⇒ f(−4) = e4 f9(x) = −e−x ⇒ f9(−4) = −e4 f0(x) = f-(x) = e−x −e−x ⇒ f0(−4) = e4 You need to find f(−4), f9(−4), f0(−4) and f-(−4). ⇒ f-(−4) = −e4 Substituting the values in the series expansion gives e4 e4 e−x = e4 − e4(x + 4) + ___     (x + 4)2 − ___     (x + 4)3 + … 2! 3! e−x = e4 (1 − (x + 4) + __   21   (x + 4)2 − __   61 (x + 4)3 + …) Take a factor of e4 out of each term on the right-hand side. Example 10 π Express tan(x + __    )as a series in ascending powers of x up to and including the term x3. 4 π π Letf(x) = tan x, then tan (x + __    ) = f(x + __    ). 4 4 π f(x) = tan x ⇒ f(__    ) = 1 4 π f9(x) = sec2 x ⇒ f9( __ ) = 2 4 M07_IAL_FP2_44655_U07_125-148.indd 137 You need to use the Taylor series expansion (A) with π f(x) = tan x and a = __     4 25/04/2019 08:54 138 CHAPTER 7 MACLAURIN AND TAYLOR SERIES f0(x) = 2 × sec x × (sec x tan x) π = 2 × sec2 x × tan x ⇒ f0(__ ) = 2 × 2 × 1 = 4 4 f-(x) = 2 × sec2 x × sec2 x + 2 × tan x (2 × sec2 x tan x) π ⇒ f-(__ ) = 2 × 2 × 2 + 2 × 4 = 16 4 f0(a) f-(a) 3 Using f(x + a) = f(a) + f9(a)x +  ____  x2 +  ____   x + … 2! 3! 4 16 π tan (x + __    ) = 1 + 2x + __     x2 + ___     x3 +… 4 2! 3! = 1 + 2x + 2x2 + __   83  x3 + … Online Explore the Taylor series expansion of f(x) = tan x using GeoGebra. Watch out Make sure you simplify your coefficients as much as possible. Example 11 π π a Show that the Taylor series about __     of sin x in ascending powers of (x − __    )up to and including __ 6 6 √ 3  π 2 π π 2 1 ___ 1 __ __ __ __ __ the term (x −    ) is sin x =     +    (x −    ) −    (x −    ) 2 2 4 6 6 6 b Using the series in part a find, in terms of π, an approximation for sin 40°. π Find f(a), f9(a) and f0(a) where a = __     6 a f(x) = sin x, f9(x) = cos x,__f 0(x) = −sin x, √ 3  π π π 1 1 so f(__    ) = __    , f9(__ ) = ___    , f 0( __ ) = −  __  6 2 6 2 6 2 __ π π 2 1    ) − ______      ) − … so sin x =     +    (x − __   ( x − __ 2 2 6 2 × 2! 6 1 __ √ 3  ___ Substitute into Taylor series π expansion (B) with a = __     6 __ π π 2 1 =     +    (x − __    ) − __    (x − __    ) − … 2 2 6 4 6 1 __ √ 3  ___ 2π 2π b sin 40° = sin (___    ), so substituting x = ___     9 9 in to the series from part a gives __ √ 3  ___ π 1 π 2 1 ___ __    (___     sin 40° ≈     +    (   ) − __ 2 2 __18 4 18 ) π2 1 π√3  _____ ≈ __     + ____       −   2 36 1296 Exercise The percentage error in this approximation is about 0.1%. 7D __ 1 a Find the Taylor series expansion of √x in ascending powers of (x − 1) as far as the term in (x − 1)4. ___ b Use your answer in a to obtain an estimate for √ 1.2 , giving your answer to 3 decimal places. 2 Use a Taylor series expansion to express each function as a series in ascending powers of (x − a) as far as the term in (x − a)k, for the given values of a and k. π    , k = 3) c cos x (a = 1, k = 4) a ln x (a = e, k = 2) b tan x (a = __ 3 M07_IAL_FP2_44655_U07_125-148.indd 138 25/04/2019 08:54 MACLAURIN AND TAYLOR SERIES CHAPTER 7 139 3 a Use a Taylor series expansion to express each function as a series in ascending powers of x as far as the term in x4. π π i cos (x + __    ) ii ln (x + 5) iii sin (x − __    ) 4 3 b Use your result in ii to find an approximation for ln 5.2, giving your answer to 4 significant figures. E 4 Given that y = xe x dny a show that ____   n = (n + x)ex dx (3 marks) b find the Taylor series expansion of xex in ascending powers of (x + 1) up to and including (3 marks) the term in (x + 1)4 E 5 a Find the Taylor series for x3 ln x in ascending powers of (x − 1) up to and including (4 marks) the term in (x − 1)4 b Using your series from a, find an approximation for ln 1.5, giving your answer to 4 decimal places. E E E E/P (2 marks) 3 6 Find the Taylor series expansion of tan (x − α) about 0, where α = arctan(__    ), in ascending 4 (4 marks) powers of x up to and including the term in x2. π π 7 Find the Taylor series expansion of sin 2x about __    in ascending powers of ( x − __    ) up to 6 6 π 4 __ and including the term in ( x −    )  (4 marks) 6 1   ______  8 Given that y = ________ √ (1 + x)  dy d2y a find the values of  ___  and ____   2  when x = 3 (3 marks) dx dx 1 b find the Taylor series of ________   ______ , in ascending powers of (x − 3) up to and √(1 + x)  including the term in (x − 3)2 (4 marks) 9 Show that the Taylor series of ln x in powers of (x − 2) is ∞ (x − 2)  n ln 2 + ∑(−1)  n − 1  _______   n 2  n n=1 (6 marks) Challenge a Find the Taylor series expansion of ln (cos 2x) about π in ascending powers of (x − π) up to and including the term in (x − π)4 __ b (2 ) √3  Hence obtain an estimate for ln ___     M07_IAL_FP2_44655_U07_125-148.indd 139 25/04/2019 08:54 140 CHAPTER 7 MACLAURIN AND TAYLOR SERIES 7.5 Series solutions of differential equations You can use Taylor series to find series solutions of differential equations that can’t be solved using other techniques. This can allow you to find useful approximate solutions, and to find solutions that cannot be expressed using elementary functions. Links Suppose you have a first-order differential dy equation of the form ___   = f(x, y) and know dx the initial condition that at x = x0, y = y0, then dy you can calculate ___     by substituting x0 and y0 dx x0 into the original differential equation. Watch out | By successive differentiation of the original d2y d3y       differential equation, the values of  ____2    , ____ dx x0 dx3 x 0 and so on can be found by substituting previous results into the derived equations. | | You can use integrating factors or auxiliary equations to solve some first and second-order differential equations directly. ← Further Pure 2 Sections 5.2, 6.1 f(x, y) denotes a function of both x and y, such as x2y + 1, or exy. Such functions cannot always be written as a product of functions g(x)h(y). Notation | dy dy ___     is used to denote the value of ___     when dx x0 x = x0 dx dy ■ The series solution to the differential equation ___   = f(x, y) is found using the Taylor series dx expansion in the form dy d2y d3y (x − x0)2 ____ (x − x0)3 ____ y = y0 + (x − x0) ___     + ________     (C)     2    + ________     3    + … 2! 3! dx x0 dx x0 dx x0 | | | ■ In the situation where x0 = 0, this reduces to the Maclaurin series | | | dy x2 d2y x3 d3y y = y0 + x  ___  + ___     ____     ____   2    + ___       + … dx 0 2! dx 0 3! dx3 0 (D) Second-order and higher differential equations can be solved in the same manner. Example 12 Use the Taylor series method to find a series solution, in ascending powers of x up to and including the term in x3, of d2y ____   2  = y − sin x dx dy given that when x = 0, y = 1 and ___     = 2. dx M07_IAL_FP2_44655_U07_125-148.indd 140 25/04/2019 08:54 MACLAURIN AND TAYLOR SERIES CHAPTER 7 141 | dy The given conditions are x0 = 0, y0 = 1, ___     = 2 dx 0 d2y Substituting x0 = 0 and y0 = 1, into ____   2  = y − sin x dx 2y d gives ____       = 1 − sin 0 = 1 dx2 0 | d3y dy ____   3  = ___   − cos x (1) dx dx | d2y First find ____   2    dx 0 Differentiate the given differential equation with respect to x. | dy     = 2 into (1) Substituting x0 = 0 and ___ dx 0 | | d3y Find ____   3    dx 0 d3y gives ____   3    = 2 − cos 0 = 1. dx 0 Substituting the results into | | | 2 3 dy x2 d y x3 d y y = y0 + x ___     + ___      ____2    +  ___   ____3    + … dx 0 2! dx 0 3! dx 0 Then use the Taylor series expansion (D). x2 x3 y = 1 + x × 2 + ___     × 1 +  ___  × 1 + … 2! 3! gives x2 x3     +  ___  + … = 1 + 2x + ___ 2 6 Example 13 d2y dy dy Given that ____   2  + 2 ___  = xy and that y = 1 and ___     = 2 at x = 1, express y as a series in ascending dx dx dx powers of (x – 1) up to and including the term in (x – 1)4 You need to find d2y d3y d4y ____   2    , ____   3    and ____   4    dx 1 dx 1 dx 1 | dy The given conditions are x0 = 1, y0 = 1, ___       = 2 dx 1 | | | dy Substituting x0 = 1, y0 = 1 and ___       = 2 into dx 1 d y ____   d y ___   3   dx3 2   dx2 dy + 2   ___ dx = xy d2y dy + 2  ___2  = y + x___ dx dx | | d2y gives  ____2    = –3 dx 1 (1) | | dy d2y       = 2 and ____   2    = –3 into (1) Substituting x0 = 1, y0 = 1, ___ dx 1 dx 1 3 y d gives ____   3    = 9 dx 1 d4y d3y dy d2y dy ____   4  + 2  ____3  = ___   dx + x  ____2  + ___   dx (2) dx dx dx Differentiate the given equation with respect to x. | M07_IAL_FP2_44655_U07_125-148.indd 141 25/04/2019 08:54 142 CHAPTER 7 MACLAURIN AND TAYLOR SERIES Watch out | | | The initial conditions are given at x 0 = 1so you need to make sure you expand about this point in order to use the series solution. dy d2y d3y Substituting x0 = 1, ___       = 2, ____   2    = –3 and ____   3    = 9 dx 1 dx 1 dx 1 4 d y into (2) gives ____   4    = –17 dx 1 | Substituting all the values into | | | dy d2y d3y (x – x0)2 ____ (x – x0)3 ____ y = y0 + (x – x0) ___     + ________         2    + ________     3    + … 2! 3! dx 1 dx 1 dx 1 –17 –3 9 ____ ___ ____ gives y = 1 + 2(x – 1) +     (x – 1)2 +     (x – 1)3 +     (x – 1)4 + … 2! 4! 3! Then use the Taylor series expansion (C). 17 y = 1 + 2(x – 1) – __   32 (x – 1)2 + __   32 (x – 1)3 – ___   24  (x – 1)4 + … Example 14 dy Given that y satisfies the differential equation  ___  = y2 − x and that y = 1 at x = 0, find a series dx solution for y in ascending powers of x up to and including the term in x3. The given conditions are x0 = 0, y0 = 1. dy Substituting x0 = 0 and y0 = 1 into ___   = y2 − x dx dy gives ___     = 12 − 0 = 1 dx 0 You need to find d2y d3y d4y ____   2    , ____   3    and ____   4    dx 1 dx 1 dx 1 dy ___ Differentiate the given equation with respect to x. | 2 dy   2  = 2y___ −1 dx dx (1) | | | | dy Substituting y0 = 1 and ___     = 1 into (1) dx 0 | | d2y dy gives ____   2    = 2y0  ___   − 1 = 2 × 1 × 1 − 1 = 1 dx 0 dx 0 ( ) 3 d2y dy 2   3  = 2y___ 2  + 2 ___   dx dx dx dy ___ (2) Differentiate (1). | | dy d2y Substituting y0 = 1, ___     = 1 and ____   2    = 1 into (2) dx 0 dx 0 | | d3y d2y dy 2 gives ____   3    = 2y0____   2    + 2  ___     = 2 × 1 × 1 + 2 × 12 = 4 dx 0 dx 0 dx 0 | Substituting all of the values into | | | 2 3 dy x2 d y x3 d y y = y0 + x  ___   + __     ____   2    +  __  ____   3    + … dx 0 2! dx 0 3! dx 0 Use Taylor series expansion (D). gives y = 1 + x + __   21 x2 + __   23 x3 + … M07_IAL_FP2_44655_U07_125-148.indd 142 25/04/2019 08:54 MACLAURIN AND TAYLOR SERIES Exercise E E E E E CHAPTER 7 7E 1 Find a series solution, in ascending powers of x up to and including the term in x4, for the dy 1 d2y     differential equation  ____2  = x + 2y, given that at x = 0, y = 1 and  ___  = __ (8 marks) dx 2 dx dy d2y dy 2 The variable y satisfies (1 + x2)   ____2  + x ___  = 0 and at x = 0, y = 0 and  ___  = 1 dx dx dx Use the Taylor series method to find a series expansion for y in powers of x up to and including (8 marks) the term in x3. dy 3 Given that y satisfies the differential equation  ___  + y − ex = 0, and that y = 2 at x = 0, find dx a series solution for y in ascending powers of x up to and including the term in x3. (6 marks) dy d2y 4 Use the Taylor series method to find a series solution for  ____2  + x ___  + y = 0, given dx dx dy that at x = 0, y = 1 and  ___  = 2, giving your answer in ascending powers of x up to and dx 4 including the term in x . (8 marks) d2y dy 5 The variable y satisfies the differential equation  ____2  + 2   ___  = 3xy, and y = 1 and dx dx dy  ___  = −1 at x = 1. dx Express y as a series in powers of (x − 1) up to and including the term in (x − 1)3 E E/P 143 (8 marks) 6 Find a series solution, in ascending powers of x up to and including the term x4, to the dy dy d2y (8 marks) differential equation  ____2  + 2y ___  + y3 = 1 + x, given that at x = 0, y = 1 and  ___  = 1 dx dx dx dy 7 (1 + 2x)  ___  = x + 2y2 dx dy    (  ___ dx ) d 3y d2y a Show that (1 + 2x)   ____3  + 4(1 − y)   ____2  = 4 dx dx 2 (4 marks) dy b Given that y = 1 at x = 0, find a series solution of (1 + 2x)  ___  = x + 2y2, in ascending dx powers of x up to and including the term in x3. (4 marks) E/P π    )up to and including the term 8 Find the series solution in ascending powers of (x − __ 4 2 __ dy π π      for the differential equation sin x  ___  + y cos x = y2 given that y = √ 2  at x = __     in x − __ 4 4 dx (6 marks) ( M07_IAL_FP2_44655_U07_125-148.indd 143 ) 25/04/2019 08:54 144 CHAPTER 7 E/P MACLAURIN AND TAYLOR SERIES dy 9 The variable y satisfies the differential equation  ___  − x2 − y2 = 0 dx a Show that: dy d2y i  ____2  − 2y ___  − 2x = 0 dx dx (2 marks) dy    = 2 (  ___ dx ) d2y d3y ii  ____3  − 2y   ____2  − 2 dx dx 2 d4y d3y d2y dy b Derive a similar equation involving  ____4  ,   ____3  ,   ____2  , ___     and y. dx dx dx dx c Given also that y = 1 at x = 0, express y as a series in ascending powers of x in powers of x up to and including the term in x 4 E/P E/P dy 10 Given that cos x  ___  + y sin x + 2y3 = 0, and that y = 1 at x = 0, use the Taylor series dx 56 11 method to show that, close to x = 0, y ≈ 1 − 2x + ___     x2 − ___     x3 2 3 dy d2y 11 ____2  = 4x  ___  − 2y (1) dx dx a Show that   4y d d  3y d   5y  ____5  = px  ____4  + q  ____3  , dx  dx  dx  where p and q are integers to be determined. (2 marks) (3 marks) (4 marks) (6 marks) (4 marks) b Hence find a series solution, in ascending powers of (x − 1) up to the term in x5 of dy differential equation (1), given that y = ___     = 2 when x = 1. (5 marks) dx Chapter review 7 dny 1 a Given that y = e  1−2x, find an expression, in terms of y, for ___   n  dx 8y d e     b Hence show that ___   8  at x = ln 32 is __ dx 4 2 a For the function f(x) = ln (1 + ex), find the values of f9(0) and f 0(0). b Show that f -(0) = 0. c Find the series expansion of ln (1 + ex), in ascending powers of x up to and including the term in x2. E/P 3 a Write down the Maclaurin series of cos 4x in ascending powers of x, up to and including the (3 marks) term in x6. b Hence, or otherwise, show that the first three non-zero terms in the series expansion of 128 16 sin2 2x are 4x2 − ___     x4 + ____     x6 (3 marks) 3 45 E/P 4 Given that terms in x5 and higher powers may be neglected, use the Maclaurin series for ex and x2 x4 cos x to show that ecos x ≈ e(1 − __     + __    ) (5 marks) 2 6 M07_IAL_FP2_44655_U07_125-148.indd 144 25/04/2019 08:54 MACLAURIN AND TAYLOR SERIES CHAPTER 7 145 π 5 Using Taylor series, show that the first three terms in the series expansion of x − __      cot x, in 4 π π π 3 π 2 __ __ __ __ powers of x −      , are x −      − 2(x −      ) + 2 x −      4 4 4 4 ( E/P E/P P E E/P ) ( ) ( ) ( 6 Given that |2x| , 1, find the first two non-zero terms in the series expansion of ln ((1 + x)2(1 − 2x)) in ascending powers of x. ) (5 marks) 7 Use differentiation and Maclaurin series to express ln (sec x + tan x) as a series in ascending (5 marks) powers of x up to and including the term in x3. 8 Show that the results of differentiating the standard series expansions of e x,  sin x and cos x agree with d d d a ___   (ex) = ex b ___  (sin x) = cos x c ___   (cos x) = −sin x dx dx dx x2 x4 x2 5 4 (4 marks) 9 a Given that cos x = 1 − __     + __     − …, show that sec x = 1 + __     + ___     x + … 2 24 2! 4! x3 x5     + __     − …, find the first three b Using the result found in part a, and given that sin x = x − __ 3! 5! non-zero terms in the series expansion, in ascending powers of x, for tan x. (4 marks) 10 By using the series expansions of ex and cos x, or otherwise, find the expansion of ex cos 3x in ascending powers of x up to and including the term in x3. (5 marks) E/P 11 Find the first three derivatives of (1 + x)2 ln (1 + x). Hence, or otherwise, find the expansion of (5 marks) (1 + x)2 ln (1 + x) in ascending powers of x up to and including the term in x3. E/P 12 a Expand ln (1 + sin x) in ascending powers of x up to and including the term in x4. (4 marks) ∫ _π 6    b Hence find an approximation for   ln (1 + sin x) dx giving your answer to 0 3 decimal places. E/P P x3 13 a Using the first two terms, x + __     , in the expansion of tan x, show that 3 x2 x3 etan x = 1 + x + __     + __     + … 2 2 b Deduce the first four terms in the series expansion of e−tan x, in ascending powers of x. (3 marks) (3 marks) (3 marks) x2 x4 14 a Using Maclaurin series and differentiation, show that ln cos x = − __     − ___     + … 2 12 x b Using cos x = 2 cos2  __   − 1 and the result in part a, show that 2 x2 x4 ln (1 + cos x) = ln 2 − __     − ___     + … 4 96 ( ) E/P 15 y = e  3x − e  −3x d4y a Show that  ____4  = 81y (4 marks) dx b Find the first three non-zero terms of the Maclaurin series for y, giving each coefficient in its simplest form. (3 marks) c Find an expression for the nth non-zero term of the Maclaurin series for y. M07_IAL_FP2_44655_U07_125-148.indd 145 (2 marks) 25/04/2019 08:54 146 CHAPTER 7 MACLAURIN AND TAYLOR SERIES 16 a For the function f(x) = ln (1 + ex), find the values of f9(0) and f 0(0). b Show that f-(0) = 0. c Find the Taylor series expansion of ln (1 + ex), in ascending powers of x up to and including the term in x2. 17 a Write down the Taylor series for cos 4x in ascending powers of x, up to and including the term in x 6. b Hence, or otherwise, show that the first three non-zero terms in the series expansion of 128 16 sin2 2x are 4x2 − ___     x4 + ____     x6. 3 45 P E 18 Given that terms in x5 and higher powers may be neglected, use the Taylor series for ex and x2 x4 cos x, to show that ecos x ≈ e 1 − __     + __      2 6 ( ) dy 19 ___  = 2 + x + sin y dx a Given that y = 0, when x = 0, use the Taylor series method to obtain y as a series in (5 marks) ascending powers of x up to and including the term in x3. b Hence obtain an approximate value for y at x = 0.1 E 20 Given that |2x| , 1, find the first two non-zero terms in the Taylor series expansion of ln((1 + x)2(1 − 2x)) in ascending powers of x. (1 mark) (5 marks) E 21 Find the series solution, in ascending powers of x up to and including the term in x3, of the d2y dy dy differential equation  ____2  − (x + 2)  ___  + 3y = 0, given that at x = 0, y = 2 and  ___  = 4 (5 marks) dx dx dx E/P 22 a Use differentiation and Maclaurin series expansion to express ln (sec x + tan x) as a (4 marks) series in ascending powers of x up to and including the term in x3. sin x − ln (sec x + tan x)       b Hence find lim   ____________________   (4 marks) x→0 x(cos x − 1) 23 Show that the results of differentiating the series expansions xr x2 x3 ex = 1 + x + __     + __     + … + __     + … 2! 3! r! 3 5 7 (−1)rx2r + 1 x x x     + __     − __     + … + _________     + … sin x = x − __ 3! 5! 7! (2r + 1)! x2r x2 x4 __ x6     + __     −     + … + (−1)r ____     +… cos x = 1 − __ 2! 4! 6! (2r)! agree with the results d d d a ___  (ex) = ex b ___  (sin x) = cos x c ___  (cos x) = −sin x dx dx dx E dy d2y dy 24  ____2  + y  ___  = x, and at x = 1, y = 0 and  ___  = 2 dx dx dx Find a series solution of the differential equation, in ascending powers of (x − 1) up to (8 marks) and including the term in (x − 1)3 M07_IAL_FP2_44655_U07_125-148.indd 146 25/04/2019 08:54 MACLAURIN AND TAYLOR SERIES E E/P E E CHAPTER 7 147 x2 5 4 x2 x4 25 a Given that cos x = 1 − __     + __     − …, show that sec x = 1 + __     + ___     x + … (4 marks) 2! 4! 2 24 x3 x5     + __     − …, find the first b Using the result found in part a, and given that sin x = x − __ 3! 5! three non-zero terms in the series expansion, in ascending powers of x, for tan x. (4 marks) 26 By using the Taylor series expansions of ex and cos x, or otherwise, find the expansion (4 marks) of ex cos 3x in ascending powers of x up to and including the term in x3. d2y dy dy 27  ____2  + x2  ___  + y = 0 with y = 2 at x = 0 and  ___  = 1 at x = 0. dx dx dx a Use the Taylor series method to express y as a polynomial in x up to and including (4 marks) the term in x3. d4y ____ b Show that at x = 0,   4  = 0 (3 marks) dx 28 a Find the first three derivatives of (1 + x)2 ln (1 + x) (4 marks) b Hence, or otherwise, find the Taylor series expansion of (1 + x)2 ln (1 + x) in ascending (4 marks) powers of x up to and including the term in x3. E 29 a Expand ln (1 + sin x) in ascending powers of x up to and including the term in x 4. (6 marks) π __ ∫     6 b Hence find an approximation for   ln (1 + sin x) dx giving your answer to 0 3 decimal places. E/P E/P (3 marks) x3 30 a Using the first two terms, x + __     , in the Taylor series of tan x, show that 3 x2 x3 etan x = 1 + x + __     + __     + … (4 marks) 2 2 b Deduce the first four terms in the Taylor series of e−tan x, in ascending powers of x. (2 marks) dy ____ 2 ( ) dy ___ 2 31 y   2  +      + y = 0 dx dx d3y (5 marks)   3  a Find an expression for ____ dx dy Given that y = 1 and  ___  = 1 at x = 0, dx b find the series solution for y, in ascending powers of x, up to an including the term in x3. (5 marks) c Comment on whether it would be sensible to use your series solution from part b to give estimates for y at x = 0.2 and at x = 50. (2 marks) P x2 x4     + … 32 a Using Maclaurin series, and differentiation, show that ln cos x = −  __  − ___ 2 12 x b Using cos x = 2 cos2  __   − 1, and the result in part a, show that 2 x2 x4 ln (1 + cos x) = ln 2 − __     − ___     + … 4 96 ( ) P 33 a By writing 3x = ex ln 3, find the first four terms in the Taylor series of 3x. __ b Using your answer from part a, with a suitable value of x, find an approximation for √ 3 , giving your answer to 3 significant figures. M07_IAL_FP2_44655_U07_125-148.indd 147 25/04/2019 08:54 148 CHAPTER 7 E/P MACLAURIN AND TAYLOR SERIES πx 34 a Given that f(x) = ln(1 + 2 cos (___    )), find f9 and f 0. 2 b Hence, using Taylor series, show that the first two non-zero terms, in ascending πx π2 powers of (x − 1), of ln(1 + 2 cos (___    )) are − π(x − 1) −  __  (x − 1)  2 2 2 (4 marks) (2 marks) Challenge a Use induction to prove that the nth derivative of ln x is given by n (n − 1)! ____   d  n   ln x = (−1)  n+1  _______ x  n  dx  b Hence write down the Taylor series expansion about x = a of ln (x), where a . 0. Summary of key points 1 The Maclaurin series of a function f(x) is given by f(r)(0) r f 0(0) f(x) = f(0) + f9(0)x + _____     x2 + … +  _____   x + … 2! r! The series is valid provided that f(0), f9(0), f0(0), … , f(r)(0), … all have finite values. 2 The following Maclaurin series are given in the formulae booklet: x2 xr ex = 1 + x + __     + … +  __  + … for all x 2! r! xr x2 x3   r  + …     − … + (−1)r + 1 __ −1 < x < 1 ln (1 + x) = x − __     + __ 2 3 x2r + 1 x3 x5     + … for all x sin x = x − __     +  __  − … + (−1)r _______ 3! 5! (2r + 1)! x2r x3 x5     + … for all x cos x = 1 − __     +  __  − … + (−1)r _______ 3! 5! (2r + 1)! x2r + 1 x3 x5       − … + (−1)r _______   + … −1 < x < 1 arctan x = x − __     + __ 5 3 2r + 1 -(a) f(r)(a) f 0(a) 2 f _____     xr + … (A) 3 f(x + a) = f(a) + f9(a)x +   _____   x +     x3 + … + _____ 2! 3! r! f _____ 0(a) f f(r)(a) (a)     (x − a)r + … (B)   (x − a)3 + … + _____ f(x) = f(a) + f9(a)(x − a) +     (x − a)2 +   _____ 2! 3! r! The expansions (A) and (B) given above are known as Taylor series expansions of f(x) at (or about) the point x = a. The Taylor series expansion is valid only if f(n)(a) exists and is finite for all n ∈ ℕ, and for values of x for which the infinite series converges. dy 4 •The series solution to the differential equation ___     = f(x, y) is found using the Taylor series dx expansion in the form dy d2y d3y (x − x0)2 ____ (x − x0)3 ____   + _________         2  + _________     3  + … (C) y = y0 + (x − x0) ___ 2! 3! dx x0 dx x0 dx x0 • In the situation where x0 = 0, this reduces to the Maclaurin series 2 3 dy x2 d y x3 d y y = y0 + x  ___  + __      ____2  + __      ____3  + … (D) dx 0 2! dx 0 3! dx 0 | | M07_IAL_FP2_44655_U07_125-148.indd 148 | | | | 25/04/2019 08:54 POLAR COORDINATES CHAPTER 8 149 8 POLAR COORDINATES 7.1 7.2 Learning objectives After completing this chapter you should be able to: ● Convert between polar and Cartesian coordinates → pages 150–153 ● Sketch curves with r given as a function of θ → pages 153–158 ● Find the area enclosed by a polar curve → pages 158–161 ● Find tangents parallel to, or at right angles to, the initial line → pages 162–165 Prior knowledge check 1 Find the exact value of π ∫0 sin2θ dθ ← Pure 3 Section 7.3 2 y = cos x + sin x cos x Find, in the interval 0 , x , π, the values dy of x for which ___ = 0 dx ← Pure 3 Section 6.4 3 a On an Argand diagram, show the locus of points given by values of z that satisfy |z – 3i| = 3 b Find the area of the region defined by the set of points, R, where π R = {z : |z − 3i| < 3} ∩ {z : 0 < arg z < __} 2 ← Further Pure 2 Section 4.3 M08A_IAL_FP2_44655_U08_149-167.indd 149 Polar coordinates describe positions in terms of angles and distances. GPS navigation systems use polar coordinates to triangulate the position of a ship or an aircraft. 25/04/2019 08:54 150 CHAPTER 8 POLAR COORDINATES 8.1 Polar coordinates and equations Polar coordinates are an alternative way of describing the position of a point P in two-dimensional space. You need two measurements: firstly, the distance the point is from the pole (usually the origin O ), r, and secondly, the angle measured anticlockwise from the initial line (usually the positive x-axis), θ. Polar coordinates are written as (r, θ). Notation When working in polar coordinates y the axes might also be labelled like this: θ= x P(x, y) or (r, θ) r y θ O π 2 O x Initial line The coordinates of P can be written in either Cartesian form as (x, y) or in polar form as (r, θ). You can convert between Cartesian coordinates and polar coordinates using right-angled triangle trigonometry. From the diagram above you can see that: ■ r cos θ = x r sin θ = y Watch out ■ r 2 = x2 + y2 Always draw a sketch diagram to check in which quadrant the point lies, and always measure the polar angle from the positive x-axis. θ = arctan ( __ ) Example y x 1 Find polar coordinates of the points with the Cartesian coordinates: a (3, 4) b (5, −12) __ c (−√3 , −1) a y Draw a sketch. r O 4 θ 3 ________ √ 32 + 42  x r= =5 4 θ = arctan  __ = 0.927… 3 So the polar coordinates are (5, 0.927) M08A_IAL_FP2_44655_U08_149-167.indd 150 Use Pythagoras’ theorem to find r. Use trigonometry to find θ. Give your answer in radians. 25/04/2019 08:54 POLAR COORDINATES b CHAPTER 8 y Draw a sketch. 5 O α x 12 r Use Pythagoras’ theorem to find r. ___________ 2 + (−12)2  = 13 r = √ 5    12 α = arctan  __ 5 = 1.176… So θ = −1.176 So the polar coordinates are (13, −1.176) y c – 3 r 151 α r The sketch shows that the point is in the 3rd quadrant. O x –1 ___________ __ (− = √ 3) 2 + (−1)2  = 2 √ π 1 __  = __ α = arctan  ___     √ 3  6 π 7π So θ = π +  __  = ___     6 6 7π    ) So the polar coordinates are ( 2, ___ 6 Example Use trigonometry to find θ, taking care to ensure it is in the correct quadrant. You could also write this point as (13, 5.107) since −1.176 + 2π = 5.107 Use Pythagoras’ theorem to find r. 7π The point is in the third quadrant so use θ = ___     6 7π 5π You could also use θ =  ___  − 2π = −  ___  6 6 2 Convert the polar coordinates into Cartesian form. The angles are measured in radians. 4π 2π a(10, ___    ) b(8, ___    ) 3 3 4π a x = r cos θ = 10 cos  ___  = −5 3 __ 4π ___ y = r sin θ = 10 sin     = −5√3  3 __ So the Cartesian coordinates are (−5, −5√3 ) 2π b x = r cos θ = 8 cos  ___  = − 4 3 __ 2π ___ y = r sin θ = 8 sin     = 4√3  3 __ So the Cartesian coordinates are (−4, 4√3 ) Polar equations of curves are usually given in the form r = f(θ). For example, r = 2 cos θ r = 1 + 2θ r=3 In this example r is constant. You can convert between polar equations of curves and their Cartesian forms. M08A_IAL_FP2_44655_U08_149-167.indd 151 25/04/2019 08:54 152 CHAPTER 8 Example POLAR COORDINATES 3 Find Cartesian equations of the curves: b r = 2 + cos 2θ a r=5 π c r 2 = sin 2θ, 0 , θ < __     2 a r=5 You need to replace r with an equation in x and y. Use r 2 = x2 + y2. So the equation r = 5 represents a circle with centre O and radius 5. Square both sides to get r 2 = 25 So a Cartesian equation is x2 + y2 = 25 b r = 2 + cos 2θ You need an equation in x and y, so use x = r cos θ. This means first writing cos 2θ in terms of cos θ. r = 1 + (1 + cos 2θ) r = 1 + 2 cos2 θ Now use x = r cos θ and r 2 = x2 + y2 Multiply by r2: r3 = r2 + 2r2 cos2 θ __ 3 (x2 + y2)   2 = x2 + y2 + 2x2 Or Watch out Polar coordinates often give rise to complicated Cartesian equations, which cannot be written easily in the form y = … __ 3 (x2 + y2)   2 = 3x2 + y2 π c r2 = sin 2θ, 0 , θ <  __ 2 r2 = 2 sin θ cos θ Problem-solving Multiply by r2: You need to use the substitutions x = r cos θ and y = r sin θ. Use sin 2θ ≡ 2 sin θ cos θ and then multiply by r2. r4 = 2 × r sin θ × r cos θ (x2 + y2)2 = 2xy Example 4 Find polar equations for: a y2 = 4x b x2 − y2 = 5 a y2 = 4x r 2 sin2 θ = 4r cos θ r sin2 θ = 4 cos θ 4 cos θ r = _______     = 4 cot θ cosec θ sin2 θ So a polar equation is r = 4 cot θ cosec θ __ c y√3  = x + 4 Substitute x = r cos θ and y = r sin θ Divide by r and simplify. b x2 − y2 = 5 r2 cos2 θ − r2 sin2 θ = 5 Substitute x = r cos θ and y = r sin θ r2 cos 2θ = 5 Use cos 2θ ≡ cos2 θ − sin2 θ r2 (cos2 θ − sin2 θ) = 5 So a polar equation is r2 = 5 sec 2θ M08A_IAL_FP2_44655_U08_149-167.indd 152 25/04/2019 08:54 POLAR COORDINATES CHAPTER 8 153 __ c y√3  = x + 4 __ r√3  sin θ = r cos θ + 4 __ r(√3  sin θ − cos θ) = 4 __ √3  ___ r(    sin θ − __   42   21 cos θ) = __ 2 π r sin (θ − __    ) = 2 6 π So a polar equation is r = 2 cosec (θ − __     ) 6 Exercise Substitute x = r cos θ and y = r sin θ and then try to simplify the trigonometric expression. Use the sin(A − B) formula. 8A 1 Find polar coordinates of the points with the Cartesian coordinates: a (5,12) b (−5, 12) d (2, −3) e (√3 , −1) __ 2 Convert the polar coordinates into Cartesian form. π π    ) b(6, − __ ) a(6, __ 6 6 5π ___ e (2, π) d(10,    ) 4 c (−5, −12) 3π c(6, ___    ) 4 3 Find Cartesian equations for the curves, where a is a positive constant. a r=2 d r = 4a tan θ sec θ g r = 4(1 − cos 2θ) b r = 3 sec θ e r = 2a cos θ h r= 2 cos2 θ c r = 5 cosec θ f r = 3a sin θ i r2 = 1 + tan2 θ 4 Find polar equations for the curves. a x2 + y2 = 16 b xy = 4 c (x2 + y2)2 = 2xy d x2 + y2 − 2x = 0 e (x + y)2 = 4 f x−y=3 g y = 2x h y = −√3 x + a i y = x(x − a) __ Challenge Show that the distance, d, between the two points (r1, θ1) and (r2, θ2) in polar coordinates is ______________________ d = √ r     2  + r2  2  − 2r1r2 cos (θ1 − θ2)  1 8.2 Sketching curves You can sketch curves given in polar form by learning the shapes of some standard curves. ■ r = a is a circle with centre O and radius a. ■ θ = α is a half-line through O and making an angle α with the initial line. ■ r = aθ is a spiral starting at O. M08A_IAL_FP2_44655_U08_149-167.indd 153 25/04/2019 08:54 154 CHAPTER 8 Example POLAR COORDINATES 5 Sketch the curves 3π b θ =  ___  4 where a is a positive constant. a r=5 c r = aθ θ=π a 2 5 r =5 θ =0 5 O 5 This is a standard curve: a circle with centre O and radius 5. Initial line 5 b θ = 3π 4 θ= π 2 O This is another standard graph: a half-line. Notice it is only ‘half’ of the line y = −x. The other half of π 7π the line would have equation θ = −__     or θ = ___     4 4 θ=0 Initial line θ=π c 2 r = aθ θ=0 O Initial line You can also sketch curves by drawing up a table of values of r for particular values of θ. It is common to choose only values of θ that give positive values of r. M08A_IAL_FP2_44655_U08_149-167.indd 154 This is another standard curve: a spiral. It crosses the horizontal axis at −aπ, 0 and 2aπ and aπ 3aπ the vertical axis at ___     and – ____    . The curve here 2 2 drawn for values of θ in the range 0 < θ < 2π. Watch out Some graph-drawing programs and graphical calculators will sketch polar curves for negative values of r so take care when using these tools to help you. 25/04/2019 08:54 POLAR COORDINATES CHAPTER 8 Example 155 6 Sketch the curves. a r = a(1 + cos θ) b r = a sin 3θ a r = a(1 + cos θ) θ 0 r 2a π __ 3π ___ π     2 a     2 a 0 c r2 = a2 cos 2θ 2π 2a θ= π 2 r = a (1 + cos θ) a When sketching polar curves it is useful to plot points for key values of θ. Make a table of values π for θ at multiples of __    to determine the points 2 at which the curve meets or intersects the coordinate axes. θ=0 2a O Problem-solving Initial line a b r = a sin 3θ Need to consider 4π 5π π 2π 0 < θ < __    , ___     < θ < π and ___     < θ < ___     3 3 3 3 π π __ __         0 θ 6 3 0 0 r a θ= π 2 r = a sin 3θ θ=0 O Initial line This curve is ‘heart’ shaped and is known as a cardioid. Since we only draw the curve when r > 0 you need to determine the values of θ required. Choose values of θ which give exact values of r. The values shown here define the first loop of the curve. The values of r will be the same in the other two loops. Problem-solving The curve given by r = a sin 3θ is typical of the patterns that arise in polar curves for equations of the form r = a cos nθ or r = a sin nθ. They will have n loops symmetrically arranged around O. a c r2 = a2 cos 2θ You need values of θ in the ranges 5π 3π π π − __  < θ < __     and ___     < θ < ___     4 4 4 4 π π 3π 5π __ __     ___     π ___     θ −  4  0 4 4 4 0 0 0 0 r a a Establish the values of θ for which the curve exists. Draw up a table of values and sketch the curve. θ= π 2 r2 = a2 cos 2θ a M08A_IAL_FP2_44655_U08_149-167.indd 155 O a θ=0 Initial line Online Explore curves given in polar form using GeoGebra. 25/04/2019 08:54 156 CHAPTER 8 POLAR COORDINATES Curves with equations of the form r = a(p + q cos θ) are defined for all values of θ if p > q. An example of this, when p = q, was the cardioid seen in Example 6a. These curves fall into two types, those that are ‘egg’ shaped (i.e. a convex curve) and those with a ‘dimple’ (i.e. the curve is concave at θ = π). The conditions for each type are given below: Links You can prove these conditions by considering the number of tangents to the curve that are perpendicular to the initial line. ‘egg’ shape when p > 2q Example ‘dimple’ shape when q < p , 2q 7 Sketch the curves. a r = a(5 + 2 cos θ) b r = a(3 + 2 cos θ) a r = a(5 + 2 cos θ) θ 0 r 7a π __     2 5a 3π ___ π     2 5a 3a Draw up a suitable table of values. θ= π 2 5a r = a (5 + 2cos θ) Since 5 . 2 × 2 there is no ‘dimple’. θ=0 3a 7a O Initial line 5a b r = a(3 + 2 cos θ) θ 0 r 5a π __     2 3a 3π ___ π     2 3a a Draw up a suitable table of values. θ=π 2 3a r = a (3 + 2cos θ) Since 3 , 2 × 2 there will be a ‘dimple’ for θ close to π. θ=0 a O 5a Initial line 3a M08A_IAL_FP2_44655_U08_149-167.indd 156 25/04/2019 08:54 POLAR COORDINATES CHAPTER 8 You may also need to find a polar curve to represent a locus of points on an Argand diagram. Example 157 Links If the pole is taken as the origin, and the initial line is taken as the positive real axis, then the point (r, θ) will represent the complex number re  iθ ← Further Pure 2 Section 8.1 8 a Show on an Argand diagram the locus of points given by the values of z satisfying |z − 3 − 4i| = 5 b Show that this locus of points can be represented by the polar curve r = 6 cos θ + 8 sin θ a Im |z – 3 – 4i| = 5 8 This locus is a circle with centre 3 + 4i and radius 5. 3 + 4i O 6 Re b In Cartesian form, (x − 3)2 + (y − 4)2 = 25 (r cos θ − 3)2 + (r sin θ − 4)2 = 25 r  2 cos  2 θ − 6r cos θ + 9 + r  2 sin  2 θ − 8r sin θ + 16 = 25 2 2 r  (cos  θ + sin  2 θ) − 6r cos θ − 8r sin θ = 0 r  2 = 6r cos θ + 8r sin θ r = 6 cos θ + 8 sin θ Exercise 8B 1 Sketch the curves. g r = a sin θ h r = a(1 − cos θ) π c θ = − __  4 π    ) f r = 2 sec (θ − __ 3 i r = a cos 3θ m r = a(2 + sin θ) n r = a(6 + sin θ) o r = a (4 + 3 sin θ) a r=6 d r = 2 sec θ j r = a(2 + cos θ) p r = 2θ E Substitute for x and y in polar form. 5π b θ = ___     4 e r = 3 cosec θ k r = a(6 + cos θ) l r = a (4 + 3 cos θ) q r2 = a2 sin θ r r 2 = a2 sin 2θ 2 Sketch the graph with polar equation π     – θ) r = k sec (__ 4 where k is a positive constant, giving the coordinates of any points of intersection with the coordinate axes in terms of k. (4 marks) M08A_IAL_FP2_44655_U08_149-167.indd 157 25/04/2019 08:54 158 CHAPTER 8 E POLAR COORDINATES 3 a Show on an Argand diagram the locus of points given by the values of z satisfying |z − 12 − 5i| = 13 b Show that this locus of points can be represented by the polar curve r = 24 cos θ + 10 sin θ E 4 a Show on an Argand diagram the locus of points given by the values of z satisfying |z + 4 + 3i| = 5 b Show that this locus of points can be represented by the polar curve r = − 8 cos θ − 6 sin θ (2 marks) (4 marks) (2 marks) (4 marks) 8.3 Area enclosed by a polar curve You can find areas enclosed by a polar curve using integration. θ= π 2 ■ The area of a sector bounded by a polar curve and the half-lines θ = α and θ = β, where θ is in radians, is given by the formula Area β Area = _  1  ∫     r   2 dθ 2α Example β 9 O α Initial line Find the area enclosed by the cardioid with equation r = a(1 + cos θ) Problem-solving θ= π 2 r = a (1 + cos θ) a θ=0 2a O Start by sketching the curve. You can simplify your calculation by using the fact that the curve is symmetric about the initial line. Hence you can integrate from 0 to π and then double your answer. Initial line a The curve is symmetric about the initial line and so finding the area above this line and doubling it gives: a2 π Area = 2 × ___     ∫     (1 + cos θ)2 dθ 2 0 π 2 = a ∫     (1 + 2 cos θ + cos2 θ) dθ Use the formula for area. Remember to square the expression for r. You can use trigonometric identities for cos 2θ to integrate terms in cos2 θ or sin2 θ: 1 + cos 2θ cos2 θ ≡ _________   ← Pure 3 Section 4.3   2 0 π   32  + 2 cos θ + __   21   cos 2θ)dθ = a2∫     (__ 0 = π [ 2 + 2 sin θ +   4  sin 2θ]  0  a2 __   3 θ __1 = a2 ((__   32 π + 0 + 0) − 0) 3a2π =  _____   2 M08A_IAL_FP2_44655_U08_149-167.indd 158 Watch out Unlike Cartesian integration, areas in the third and fourth quadrants do not produce negative integrals. You could obtain the same result by integrating between 0 and 2π: 2π 3a  2π _   12  ∫    a2 (1 + cos θ )  2 dθ = _____     2 0 25/04/2019 08:54 POLAR COORDINATES CHAPTER 8 159 Example 10 Find the area of one loop of the curve with polar equation r = a sin 4θ Find the values of θ which will give the beginning and end of a loop by solving r = 0. r = a sin 4θ will have one loop for π 0 < θ < __     4 __ π a2 __  π  1   4 2     r dθ = ___    ∫ 4    sin2 4θ dθ Area =  __ 2 ∫ 20 0 a2 __ 4π ___ =    ∫     (1 − cos 8θ) dθ 40 __   π sin 8θ 4 a2 ___ ______ =     [θ −    ]     4 8 0 2 π sin 2π a =  ___  (__     − ______ − 0     4 4 8 ) 2 a π = ____     16 Watch out Use the area formula. Use the trigonometric identity for cos 2θ. In this 1 − cos 8θ case, sin2 4θ ≡ _________     2 Remember sin 2π = 0. Online Explore the area enclosed by a loop of the polar curve with the form r = a sin θ using GeoGebra. r = sin nθ has n loops and so a simple way of finding the area of one loop would appear to be 2π aπ to find  _1 ∫    r2 dθ and divide by n. This would give ____     2 2 8 0 The reason why this is not the correct answer is because when you take r 2 in the integral you are also including the n loops given by r , 0. You need to choose your limits carefully so that r > 0 for all values within the range of the integral. Example 11 a On the same diagram, sketch the curves with equations r = 2 + cos θ and r = 5 cos θ b Find the polar coordinates of the points of intersection of these two curves. c Find the exact area of the region which lies within both curves. θ= π a A table of values would consider π 3π θ = 0, __    , π, ___     2 2 2 r = 5 cos θ r = 2 + cos θ θ=0 O Initial line b The points of intersection are given by 2 + cos θ = 5 cos θ So 4 cos θ = 2 π θ = ± __     3 π   52 , ± __    ) So the polar coordinates are ( __ 3 M08A_IAL_FP2_44655_U08_149-167.indd 159 This is the region required in part c. Form a suitable equation to find the points of intersection. Solve for θ and then substitute in r = 5 cos θ to find the value of r. 25/04/2019 08:54 160 CHAPTER 8 c POLAR COORDINATES θ= π 2 Remember that the area formula gives areas of sectors. So you need the sector formed by the red curve and the sector formed by the blue curve. Again you can use symmetry about the initial line. θ=0 O Initial line __ __   π   π Area = 2 × __   21 ∫ 3    (2 + cos θ)2 dθ + 2 × __   21 ∫__π  2  (5 cos θ)2 dθ __ π   3   __ π 3 0   = ∫     (4 + 4 cos θ + cos2 θ) dθ + ∫__π  2  25 cos2 θ dθ   3 0 Square and use the trigonometric identity for cos 2θ. __ __   π 9 cos 2θ   π 25 (1 + cos 2θ) dθ   2 + 4 cos θ + ______     dθ + ∫ __π  2  ___ = ∫ 3    (__ ) 2 __   3 2 0 __ π     π sin 2θ 3 ___ sin 2θ 2 25 __ ______ ______ 9 = [  2 θ + 4 sin θ +       +   2 [ θ +           4 ]0 2 ]__ 3π __ __ = (    + +     ) − (0) + (     + 0) − (     +     2 4 6 8 8 ) 43π √__ = ____     − 3  12 3π ___ Exercise E E/P __ 2√3  √ 3  ___ 25π ____ 25π ____ 25√3  ______ 2π Use the exact value of sin  ___  3 8C 1 Find the area of the finite region bounded by the curve with the given polar equation and the half-lines θ = α and θ = β. π π π π π a r = a cos θ, α = 0, β = __     b r = a(1 + sin θ), α = − __ , β = __     c r = a sin 3θ, α = __    , β = __     2 2 2 4 6 π π __ __ 2 2 2 2 e r = a tan θ, α = 0, β =     f r = 2aθ, α = 0, β = π d r = a cos 2θ, α = 0, β =     4 4 π g r = a(3 + 2 cos θ), α = 0, β = __     2 2p2 + q2 2 Show that the area enclosed by the curve with polar equation r = a(p + q cos θ) is _______     πa2 2 3 Find the area of a single loop of the curve with equation r = a cos 3θ 187π 4 A curve has equation r = a + 5 sin θ, a . 5. The area enclosed by the curve is _____     2 Find the value of a. (5 marks) 5 The diagram shows the curves with equations r = a sin 4θ π and r = a sin 2θ for 0 < θ < __     2 The finite region R is contained within both curves. θ= π 2 r = a sin 2θ Find the area of R, giving your answer in terms of a. (8 marks) R O M08A_IAL_FP2_44655_U08_149-167.indd 160 r = a sin 4θ Initial line 25/04/2019 08:54 POLAR COORDINATES E/P CHAPTER 8 161 π 2 θ= 6 The diagram shows the curves with equations r = 1 + sin θ and r = 3 sin θ. The finite region R is contained within both curves. Find the area of R. (8 marks) r = 3 sin θ R r = 1 + sin θ Initial line O E/P 7 The set of points, A, is defined by π     < arg z < 0} ∩ { z : |z − 4 + 3i | < 5} A = {z : − __ 4 a Sketch on an Argand diagram the set of points, A. (4 marks) Given that the locus of points given by the values of z satisfying |z − 4 + 3i| = 5 can be expressed in polar form using the equation r = 8 cos θ − 6 sin θ b find, correct to 3 significant figures, the area of the region defined by A. (8 marks) E/P 8 The set of points, A, is defined by π A = {z :  __  < arg z < π} ∩ { z : |z + 12 − 5i | < 13} 2 a Sketch on an Argand diagram the set of points, A. b Find, correct to 3 significant figures, the area of the region defined by A. E/P E/P 9 The diagram shows the curve C with polar equation π r = 1 + cos 3θ, 0 < θ <  __  __ 3 2 + √ 2  _____ At points A and B, the value of r is     2 Point A lies on C and point B lies on the initial line. Find, correct to 3 significant figures, the finite area bounded by the curve, the line segment AB and the initial line, shown shaded in the diagram. (9 marks) 10 The diagram shows the curves r = 1 + sin θ and r = 3 sin θ. Find the shaded area, giving your answer correct to 2 decimal places. (8 marks) θ= (4 marks) (8 marks) π 2 C O A B θ= Initial line π 2 r = 3 sin θ r = 1 + sin θ O Challenge The cross-section of a shell is modelled using the curve with polar equation r = k θ, 0 < θ < 4π, where k is a positive constant. The horizontal diameter of the shell, as shown in the diagram, is 3 cm. a Find the exact value of k. b Hence find the total shaded area of the crosssection. M08A_IAL_FP2_44655_U08_149-167.indd 161 Initial line 3 cm π 2 θ= Initial line r = kθ 25/04/2019 08:54 162 CHAPTER 8 POLAR COORDINATES 8.4 Tangents to polar curves If you are given a curve r = f(θ) in polar form, you can write it as a parametric curve in Cartesian form, using θ as the parameter: x = r cos θ = f(θ) cos θ y = r sin θ = f(θ) sin θ By differentiating parametrically, you can find the gradient of the curve at any point: dy __     dy dθ ___     = __     dx ___ dx     dθ dy When __     = 0, a tangent to the curve will be horizontal. dθ dx When ___     = 0, a tangent to the curve will be vertical. dθ You need to be able to find tangents to a polar curve that are parallel or perpendicular to the initial line. dy ■ To find a tangent parallel to the initial line set ___     = 0 dθ dx ■ To find a tangent perpendicular to the initial line set ___   = 0 dθ Example 12 Find the coordinates of the points on r = a (1 + cos θ) where the tangents are parallel to the initial line θ = 0. y = r sin θ = a(sin θ + sin θ cos θ) dy ___   = a(cos θ + cos2 θ − sin2 θ) dθ So 0 = 2 cos2 θ + cos θ − 1 0 = (2 cos θ − 1)(cos θ + 1) π either cos θ = __   21 ⇒ θ = ± __     3 3a     so r = a (1 + __   21 ) = ___ 2 or cos θ = −1 ⇒ θ = π, and so r = 0 So the tangents parallel to the initial line 3a π    , ± __ are at ( ___    ) and (0, π). 2 3 dy Find an expression for y and then solve ___     = 0 dθ Solve the equations to find θ and then substitute back to find r. Problem-solving You can see these tangents on a sketch of y = a(1 + cos θ) ( 3a2 , π3 ) (O, π) ( 3a2 , – π3 ) M08A_IAL_FP2_44655_U08_149-167.indd 162 25/04/2019 08:54 POLAR COORDINATES CHAPTER 8 163 Example 13 π Find the equations and the points of contact of the tangents to the curve r = a sin 2θ, 0 < θ <  __  2 that are: a parallel to the initial line b perpendicular to the initial line. Give answers to 3 significant figures where appropriate. a y = r sin θ = a sin θ sin 2θ dy ___   = a(cos θ sin 2θ + 2 cos 2θ sin θ) dθ = 2a sin θ(cos2 θ + cos2 θ − sin2 θ) dy  ___ = 0 ⇒ sin θ = 0 ⇒ θ = 0 dθ __ or 2 cos2 θ = sin2 θ ⇒ tan θ = ± √ 2  ⇒ θ = 0.955 So θ = 0 or 0.955 __ √ 2  __ 1 ___ and r = 0 or r = 2a ×     × ___   __  √ √ 3  3 __ 2a√2     , 0.955) So the points are (0, 0) and (_____ 3 The equation of the initial line is θ = 0 and that is the tangent through (0, 0). The equation of the tangent through __ √ 2  2a _____ (   3 , 0.955 ) __ __ __ √ 2  2a√2  2a√2  ___ __ 4a _____ _____ is y =     × sin θ =     ×     = ____   __  3 3 √ 3  3√3  So the equation of the tangent is 4a r = ____   __  cosec θ 3√3  b x = r cos θ = a cos θ sin 2θ dx  ___ = −a sin θ sin 2θ + 2a cos θ cos 2θ dθ = 2a cos θ (−sin2 θ + cos2 θ − sin2 θ) dx π ___   = 0 ⇒ cos θ = 0 ⇒ θ =  __  2 dθ So the y-axis is a tangent. 1   __  Or cos2 θ − 2 sin2 θ = 0 ⇒ tan θ = ± ___ √ 2  So θ = 0.615__ __ √ 2  ___ __ 2a√2  1 _____ ___ __ and r = 2a ×     ×     =     √ 3  √ 3  __ 3 2a√2     , 0.615) The tangent is at (_____ 3 __ __ __ √ 2  2a√2  2a√2  ___ __ 4a _____ _____ x =     × cos α =     ×     = ____   __  3 3 √ 3  3√3  So the equation of the tangent is: 4a r = ____   __  sec θ 3√3  M08A_IAL_FP2_44655_U08_149-167.indd 163 Form an expression for y and differentiate using the product rule. Use sin 2θ ≡ 2 sin θ cos θ and then take out the common factor. Then use cos 2θ ≡ cos2 θ – sin2 θ Choose values of θ within the range given in the question. __ If tan a =__√ 2 then drawing a triangle shows that √2  1   __  sin α = ___   __  and cos α = ___ √ √ 3  3  Use sin 2A ≡ 2 sin A cos A to find r. Use y = r sin θ to find the equation of the tangent and write it in polar form using r = y cosec θ. Form an expression for x and differentiate using the product rule. Use sin 2θ = 2 sin θ cos θ and then take out the common factor. Then use a formula for cos 2θ. 1 If tan α = ___   __ then drawing a triangle shows that √ __ 2  √2  1 ___ cos α =   __  and sin α = ___   __  √ √ 3  3  Use sin 2A ≡ 2 sin A cos A to find r. Use x = r cos θ to find the equation of the tangent in the form r = x sec θ 25/04/2019 08:54 164 CHAPTER 8 POLAR COORDINATES Example 14 The curve C has equation r = (p + q cos θ), where p and q are positive constants and p . q. Prove that the curve is convex for p > 2q, and has a dimple for p , 2q. θ= Problem-solving π 2 If the curve is not convex then there will be more than two tangents to the curve that are perpendicular to the initial line. a θ=0 2a O Initial line –a x = r cos θ = p cos θ + q cos2 θ dx ___   = 0 ⇒ 0 = −p sin θ − 2q cos θ sin θ dθ ⇒ 0 = −sin θ(p + 2q cos θ) This has solutions sin θ = 0 when θ = 0 or π p and cos θ = − ___ 2q If p , 2q then there will be two solutions to this equation in the second and third quadrants (the green tangents). In this case the curve is not convex and has a dimple. Find an expression for x and differentiate. Solve the equation and consider all possible cases. The two tangents at the two points represented by these solutions have the same equation. If p = 2q then the solution is θ = π and so there are only two tangents (the blue ones). In this case the curve is convex. If p . 2q then there is no solution to this equation and only the two blue tangents are possible. In this case the curve is convex. Hence the curve is convex for p > 2q, and has a dimple for p , 2q. Exercise 8D 1 Find the points on the cardioid r = a(1 + cos θ) where the tangents are perpendicular to the initial line. 2 Find the points on the spiral r = e2θ, 0 < θ < π, where the tangents are: a perpendicular to the initial line b parallel to the initial line. Give your answers to 3 significant figures. M08A_IAL_FP2_44655_U08_149-167.indd 164 25/04/2019 08:54 POLAR COORDINATES CHAPTER 8 165 π π 3 a Find the points on the curve r = a cos 2θ, − __     < θ < __    where the tangents are parallel to 4 4 the initial line, giving your answers to 3 significant figures where appropriate. b Find the equations of these tangents. E E E E/P 4 Find the points on the curve with equation r = a(7 + 2 cos θ) where the tangents are parallel to the initial line. (6 marks) 5 Find the equations of the tangents to r = 2 + cos θ that are perpendicular to the initial line. (6 marks) π    , where the tangent is 6 Find the point on the curve with equation r = a(1 + tan θ), 0 < θ , __ 2 perpendicular to the initial line. (6 marks) 7 The curve C has polar equation π r = 1 + 3 cos θ, 0 < θ < __     2 The tangent to C at a point A on the curve is parallel to the initial line. Point O is the pole. (7 marks) Find the exact length of the line OA. E/P 8 The diagram shows a cardioid with polar equation r = 2(1 + cos θ) θ= π 2 The shaded area is enclosed by the curve and the vertical line segment which is tangent to the curve and perpendicular to the initial line. Find the shaded area, correct to 3 significant figures. (8 marks) r = 2 (1 + cos θ) O Initial line Chapter review 8 E 1 Determine the area enclosed by the curve with equation r = a(1 +  _2 sin θ), 1 a . 0, 0 < θ , 2π, giving your answer in terms of a and π. (6 marks) E/P 2 a Sketch the curve with equation r = a(1 + cos θ) for 0 < θ < 2π, where a . 0. (2 marks) π π __ __ b Sketch also the line with equation r = 2a sec θ for −     , θ ,    , on the same diagram. (2 marks) 2 2 π c The half-line with equation θ = α, 0 , α , __    , meets the curve at A and the line with equation 2 r = 2a sec θ at B. If O is the pole, find the value of cos α for which OB = 2OA. (5 marks) E/P 3 Sketch, in the same diagram, the curves with equations r = 3 cos θ and r = 1 + cos θ and find the area of the region lying inside both curves. (9 marks) M08A_IAL_FP2_44655_U08_149-167.indd 165 25/04/2019 08:54 166 CHAPTER 8 E E/P E E/P E POLAR COORDINATES 4 Find the polar coordinates of the points on r2 = a2 sin 2θ where the tangent is perpendicular (7 marks) to the initial line. 5 a Shade the region R for which the polar coordinates r, θ satisfy π π     r < 4 cos 2θ for −  __  < θ < __ 4 4 b Find the area of R. (2 marks) (5 marks) 6 Sketch the curve with polar equation r = a(1 − cos θ), where a . 0, stating the polar coordinates of the point on the curve at which r has its maximum value. 7 a On the same diagram, sketch the curve C1 with polar equation π π r = 2 cos 2θ, −  __  , θ < __     4 4 π     and the curve C2 with polar equation θ = ___ 12 b Find the area of the smaller region bounded by C1 and C2. (3 marks) (6 marks) 8 a Sketch on the same diagram the circle with polar equation r = 4 cos θ and the line with (4 marks) polar equation r = 2 sec θ b State polar coordinates for their points of intersection. E/P (5 marks) 9 The diagram shows a sketch of the curves with polar equations (4 marks) θ=π 2 r = a(1 + cos θ) and r = 3a cos θ, a . 0 P a Find the polar coordinates of the point of (4 marks) intersection P of the two curves. b Find the area, shaded in the figure, bounded by the two curves and by the initial line θ = 0, giving your answer in terms of a and π. (7 marks) E/P E/P O Initial line 10 Obtain a Cartesian equation for the curve with polar equation a r 2 = sec 2θ (4 marks) b (4 marks) r2 = cosec 2θ 11 a Show on an Argand diagram the locus of points given by the values of z satisfying __ |z − 1 − i | = √ 2  (2 marks) b Show that this locus of points can be represented by the polar curve r = 2 cos θ + 2 sin θ (4 marks) The set of points, A, is defined by __ π π A = {z :  __  < arg z < __    } ∩ { z : |z − 1 − i | < √ 2  } 2 6 c Show, by sketching on your Argand diagram, the set of points, A. (2 marks) d Find, correct to 3 significant figures, the area of the region defined by A. (5 marks) M08A_IAL_FP2_44655_U08_149-167.indd 166 25/04/2019 08:54 POLAR COORDINATES E CHAPTER 8 12 The diagram shows the curve C with polar equation π r = 4 cos 2θ, 0 < θ <  __  4 At point A the value of r is 2. Point A lies on C and point B lies on the initial line vertically below A. Find, correct to 3 significant figures, the area of the finite region bounded by the curve, the line segment AB and the initial line, shown shaded in the diagram. (9 marks) E/P 13 The diagram shows the curve with polar equation π r = 4 sin 2θ, 0 < θ <  __  2 The shaded region is bounded by the curve, the initial line and the tangent to the curve which is perpendicular to the initial line. Find, correct to 2 decimal places, the area of the shaded region. (8 marks) 167 θ=π 2 A O Initial line B θ=π 2 r = 4 sin 2θ O Initial line Challenge __ The curve C has polar equation r = √ 2  θ Show that an equation for the tangent to the curve at the point where π θ = __    is 2(π − 4)y + 2(π + 4)x = π2 4 Summary of key points 1 For a point P with polar coordinates (r, θ) and Cartesian coordinates (x, y), r cos θ = x and r sin θ = y y r2 = x2 + y2, θ = arctan (__   x ) Care must be taken to ensure that θ is in the correct quadrant. 2 • r = a is a circle with centre O and radius a. • θ = α is a half-line through O and making an angle α with the initial line. • r = aθ is a spiral starting at O. 3 The area of a sector bounded by a polar curve and the half-lines θ = α and θ = β, where θ is in radians, is given by the formula Area = _  12 ∫     r2 dθ β α dy 4 • To find a tangent parallel to the initial line set ___    = 0 dθ dx • To find a tangent perpendicular to the initial line set ___    = 0 dθ M08A_IAL_FP2_44655_U08_149-167.indd 167 25/04/2019 08:54 168 2 REVIEW EXERCISE Review exercise E 1 Find, in the form y = f(x), the general solution to the differential equation dy 4 (5)     y = 6x − 5, x . 0  ___  + __ dx x E ← Further Pure 2 Section 5.2 E 2 Solve the differential equation dy y  ___  − __     = x2, x . 0 dx x ← Further Pure 2 Section 5.2 3 Find the general solution to the differential equation dy 1  x , x . 0 (x + 1) ___  + 2y = __ dx giving your answer in the form y = f(x). ← Further Pure 2 Section 5.2 E (5) ← Further Pure 2 Section 5.2 E 4 Obtain the solution to dy π      ___  + y tan x = e2x cos x, 0 < x , __ 2 dx for which y = 2 at x = 0, giving your answer in the form y = f(x). (6) 5 Find the general solution to the differential equation dy π      ___  + 2y cot 2x = sin x, 0 , x , __ 2 dx giving your answer in the form y = f(x). c draw a sketch of the particular solution curve. 6 Solve the differential equation dy (1 + x)  ___  − xy = xe−x dx given that y = 1 at x = 0. E/P (5) (6) 9 Given that θ satisfies the differential equation dθ d2θ ____ 2 + 4___ + 5θ = 0 dt dt dθ and that, when t = 0, θ = 3 and  ___  = −6, dt express θ in terms of t. (8) 10 Given that 3x sin 2x is a particular integral of the differential equation d2y ____2 + 4y = k cos 2x dx where k is a constant, a calculate the value of k M08B_IAL_FP2_44655_RE2_168-177.indd 168 (2) ← Further Pure 2 Sections 6.1, 6.3 E/P ← Further Pure 2 Section 5.2 (5) ← Further Pure 2 Section 5.2 ← Further Pure 2 Section 5.2 E 8 a Find the general solution to the differential equation dy  ___  + 2y = x dx Given that y = 1 at x = 0, b find the exact values of the coordinates of the minimum point of the particular solution curve, (3) ← Further Pure 2 Section 5.2 E 7 a Find the general solution to the differential equation dy (5) cos x  ___  + (sin x)y = cos3 x dx b Show that, for 0 < x < 2π, there are two points on the x-axis through which all the solution curves for this differential equation pass. (4) c Sketch the graph, 0 < x < 2π, of the particular solution for which y = 0 at x = 0. (3) giving your answer for y in terms of x. (5) E 2 (3) 25/04/2019 08:53 REVIEW EXERCISE 2 169 b find the particular solution of the differential equation for which at x = 0, π π     (8) y = 2, and for which at x = __    , y = __ 2 4 c Sketch the curve with equation x = f(t), 0 < t < π, showing the coordinates, as multiples of π, of the points where the curve cuts the t-axis. (2) ← Further Pure 2 Sections 6.2, 6.3 ← Further Pure 2 Sections 6.2, 6.3 E/P 11 Given that a + bx is a particular integral of the differential equation d2y dy ____2 − 4 ___ + 4y = 16 + 4x dx dx a find the values of the constants a and b (3) E/P b find the particular solution to this differential equation for which y = 8 dy and ___     = 9 at x = 0. (8) dx ← Further Pure 2 Sections 6.2, 6.3 d2y dy ____   + 4  ___  + 5y = 65 sin 2x, x . 0 E/P 12  dx dx 2 a Find the general solution to the differential equation. ← Further Pure 2 Sections 6.2, 6.3 E/P (8) b Show that for large values of x this general solution may be approximated by a sine function and find this sine function. (2) 13 a Find the general solution to the differential equation dy d 2y (8)   ___2  + 2  ___  + 2y = 2e−t dt dt b Find the particular solution to this differential equation for which y = 1 dy and ___     = 1 at t = 0. (2) dt ←Further Pure 2 Sections 6.2, 6.3 E/P 14 a Find the general solution to the differential equation dx d 2x   ___2  + 2  ___  + 5x = 0 (8) dt dt dx b Given that x = 1 and ___     = 1 at t = 0, dt find the particular solution to the differential equation, giving your answer in the form x = f(t). (2) M08B_IAL_FP2_44655_RE2_168-177.indd 169 16 a Find the value of λ for which λx cos 3x is a particular integral of the differential equation d 2y (3)   ___2  + 9y = −12 sin 3x dx b Hence find the general solution to this differential equation. (6) The particular solution of the differential dy     = 2 at equation for which y = 1 and ___ dx x = 0, is y = g(x). ← Further Pure 2 Section 6.2 E/P 15 a Find the general solution to the differential equation dy d2y (8) 2   ___2  + 7  ___  + 3y = 3t2 + 11t dt dt b Find the particular solution to this differential equation for which y = 1 dy and ___     = 1 when t = 0. (2) dt c For this particular solution, calculate the value of y when t = 1. (2) c Find g(x). (2) d Sketch the graph of y = g(x), 0 < x < π. (2) ← Further Pure 2 Sections 6.2, 6.3 E d2y dy 17  ___2  − 6   ___  + 9y = 4e3t, t > 0 dt dt a Show that Kt 2 e3t is a particular integral of the differential equation, where K is a constant to be found. (3) b Find the general solution to the differential equation. (6) Given that a particular solution satisfies dy y = 3 and ___     = 1 when t = 0, dt 25/04/2019 08:53 170 2 REVIEW EXERCISE c find this solution.(2) Another particular solution which satisfies dy y = 1 and ___     = 0 when t = 0, has equation dt y = (1 − 3t + 2t 2)e3t d For this particular solution, draw a sketch graph of y against t, showing where the graph crosses the t-axis. Determine also the coordinates of the minimum point on the sketch graph. (4) d Find the particular solution to the differential equation for which y = 0 at π x = 0 and at x = __     (2) 2 e Show that a local minimum value of the solution in part d is 3 ______ 2 3 arccos(__  π   )− __    √π 2 − 4  (4) 2 ← Further Pure 2 Sections 6.2, 6.3 E ← Further Pure 2 Sections 6.2, 6.3 E/P 18 a Find the general solution to the differential equation dx d 2x 2   ___2  + 5  ___  + 2x = 2t + 9 (8) dt dt b Find the particular solution of this differential equation for which x = 3 ___  = −1 when t = 0. (2) and   dx dt E 19 Given that x = At 2 e−t satisfies the differential equation dx d2x   ___2  + 2  ___  + x = e−t, dt dt (3) a find the value of A. b Hence find the solution to the differential equation for which x = 1 dx (7) and  ___  = 0 at t = 0. dt c Use your solution to prove that for (2) t > 0, x < 1. M08B_IAL_FP2_44655_RE2_168-177.indd 170 22 a Use the substitution y = vx to transform the equation dy (4x + y)(x + y)      , x . 0 (1)  ___ = _____________ dx x2 into the equation dv x ___ = (2 + v)2 (2) (4) dx b Solve differential equation (2) to find (4) v in terms of x. c Hence show that x y = −2x − _______     , where c is an ln x + c arbitrary constant, is the general solution to differential equation (1). (3) ← Further Pure 2 Section 5.3 ← Further Pure 2 Sections 6.2, 6.3 20 Given that y = kx is a particular solution of the differential equation d2y   ___2  + y = 3x, dx a find the value of the constant k. (3) b Find the most general solution to this differential equation for which y = 0 at x = 0. (6) c Prove that all curves given by this solution pass through the point (π, 3π) and that they all have equal gradients π (3) when x = __     2 1 ← Further Pure 2 Section 5.3 ← Further Pure 2 Sections 6.2, 6.3 E 21 a By using the substitution y = _  2 (u − x), or otherwise, find the general solution of the differential equation dy (4)  ___ = x + 2y dx Given that y = 2 at x = 0, (3) b express y in terms of x. E 23 a Show that the substitution y = vx transforms the differential equation dy 3x − 4y   (1)  ___ = _______   dx 4x + 3y into the differential equation dv 3v2 + 8v − 3 x ___    = −   __________   (2) (4) 3v + 4 dx b Find the general solution of differential equation (2). (4) c Given that y = 7 at x = 1, show that the particular solution to differential equation (1) can be written as (3) (3y − x) (y + 3x) = 200 ← Further Pure 2 Section 5.3 25/04/2019 08:53 REVIEW EXERCISE E 2 24 a Use the substitution μ = y−2 to transform the differential equation dy (1)  ___ + 2xy = xe− x y3 dx into the differential equation dμ  ___ − 4xμ = −2xe− x (2) (4) dx b Find the general solution to differential equation (2). (4) 171 E 2 2 c Hence obtain the solution to differential equation (1) for which y = 1 at x = 0. (3) ← Further Pure 2 Section 5.3 E 25 a Show that the transformation y = xv transforms the equation d2y dy x2   ____2  − 2x  ___ + (2 + 9x2) y = x5 (1) dx dx into the equation d2v   ____2  + 9v = x2 (2) dx (6) b Solve differential equation (2) to find (4) v as a function of x. c Hence state the general solution to differential equation (1). (2) ← Further Pure 2 Section 6.4 E 27 Given that x = ln t, t . 0, and that y is a function of x, dy dy (2) a find ___    in terms of  ___ and t dx dt d2y d2y dy b show that   ____2  = t 2   ___2  + t  ___  (4) dt dx dt c Show that the substitution x = ln t transforms the differential equation dy d2y   ____2  − (1 − 6e x)  ___ + 10y e2x dx dx x x 2 = 5e sin 2e (1) into the differential equation d2y dy   ___2  + 6  ___ + 10y = 5 sin 2t (2) (6) dt dt d Hence find the general solution to (1), giving your answer in the form y = f(x). (6) ← Further Pure 2 Section 6.4 E/P 28 Given that x is so small that terms in x3 and higher powers of x may be neglected, show that 11 sin x − 6 cos x + 5 = A + Bx + Cx2 stating the values of the constants A, B and C. (6) ← Further Pure 2 Sections 7.2, 7.3 _1 26 Given that x = t    2 , x . 0, t . 0, and that y is a function of x, E/P 29 Show that for x . 1, dy dy ___ ___ a find    in terms of    and t. (2) ln(x2 − x + 1) + ln(x + 1) − 3 ln x dx dt n−1 (−1) 1 1 ___ __ ______ d2y d2y dy   −     + … +   =     + … (6) ____ ___ ___ Assuming that   2  = 4t   2  + 2     x3 2x6 nx3n dt dx dt ← Further Pure 2 Sections 7.2, 7.3 _1 b show that the substitution x = t    2  transforms the differential equation E/P 30 Given that x is so small that terms in x4 2y d y d 1 and higher powers of x may be neglected,   x )  ___ − 16x2y = 4x2 e2 x   ____2  + (6x − __ dx dx find the values of the constants A, B, C (1) and D for which into the differential equation e−2x cos 5x = A + Bx + Cx2 + Dx3 (6) dy d 2y ___ ___ t 2 (2) (6)   2  + 3    − 4y = e ← Further Pure 2 Sections 7.2, 7.3 dt dt c Hence find the general solution to E/P 31 a Find the first four terms of the (6) (1) giving y in terms of x. expansion, in ascending powers of x, of ← Further Pure 2 Section 6.4 2 (2x + 3)−1, |x| ,   __ (3) 3 2 M08B_IAL_FP2_44655_RE2_168-177.indd 171 25/04/2019 08:53 172 2 REVIEW EXERCISE b Hence, or otherwise, find the first four non-zero terms of the expansion, in ascending powers of x, of 2 sin 2x ______ (5)    , |x| ,   __ 2x + 3 3 E ← Further Pure 2 Sections 7.2, 7.3 E/P 32 a By using the Maclaurin series for cos x and ln(1 + x), find the series expansion for ln(cos x) in ascending powers of x up to and including the term in x4. (6) b Hence, or otherwise, obtain the first two non-zero terms in the series expansion for ln(sec x) in ascending powers of x. (4) ← Further Pure 2 Section 7.4 E 33 Given that π f(x) = ln(1 + cos 2x), 0 < x , __ 2 Show that: E a f9(x) = −2 tan x (2) b f 00(x) = −(f 09(x) f9(x) − (f 0(x)2) (5) c Find the Maclaurin series expansion of f(x), in ascending powers of x, up (4) to and including the term in x4. 34 a Find the Taylor series of cos 2x in π ascending powers of (x − __    ) up to 4 π 5 and including the term in ( x − __    )  (4) 4 b Use your answer to part a to obtain an estimate of cos 2, giving your answer to 6 decimal places. (2) E 35 a Find the Taylor series of ln (sin x) in π ascending powers of ( x − __    ) up to 6 π 3    )  (4) and including the term in ( x − __ 6 b Use your answer to part a to obtain an estimate of ln(sin 0.5), giving your answer to 6 decimal places. (2) ← Further Pure 2 Section 7.4 M08B_IAL_FP2_44655_RE2_168-177.indd 172 d2y dy 38 (1 − x2)   ____2  − x  ___ + 2y = 0 dx dx dy At x = 0, y = 2 and ___    = −1 dx d3y a Find the value of  ____3  at x = 0. dx b Express y as a series in ascending powers of x, up to and including the term in x3. (4) (4) dy 39 (1 + 2x) ___    = x + 4y2 dx a Show that d2y dy (1 + 2x)   ____2  = 1 + 2 (4y − 1)  ___  (1) dx dx (4) b Differentiate equation (1) with respect to x to obtain an equation involving d3y d2y dy    , x and y. (4)   ____3 ,   ____2 , ___ dx dx dx 1 Given that y = _  2  at x = 0, ← Further Pure 2 Section 7.4 E (3) ← Further Pure 2 Section 7.5 ← Further Pure 2 Sections 7.1, 7.2, 7.3 E 37 Find the Taylor series of ln x about x = 1. ← Further Pure 2 Section 7.4 ← Further Pure 2 Sections 7.2, 7.3 E 36 Given that y = tan x, d3y dy d2y (3) a find  ___ ,   ____2  and   ____3  dx dx dx b Find the Taylor series of tan x in π ascending powers of ( x − __    ) up to 4 π 3    )  (4) and including the term in ( x − __ 4 c Hence show that 3π π π2 π3 tan  ___  ≈ 1 + ___     +   ____  +   _____  (3) 10 10 200 3000 c find a series solution for y, in ascending powers of x, up to and including the term in x3. (4) ← Further Pure 2 Sections 6.1, 7.5 E/P dy 40 ___ = y2 + xy + x, y = 1 at x = 0 dx a Use the Taylor series method to find y as a series in ascending powers of x, up to and including the term in x3. (6) 25/04/2019 08:53 REVIEW EXERCISE 2 173 b Use your series to find y at x = 0.1, giving your answer to 2 decimal places. (4) a a circle, centre O and radius 2 (1) b a line perpendicular to the initial line and passing through the point with polar coordinates (3, 0) (2) c a straight line through the points with π      . (2) polar coordinates (4, 0) and 4, __ 3 ← Further Pure 2 Section 7.5 41 dy x + 3 y   ___ = _____     dx y + 1 Given that y = 1.5 at x = 0, a use the Taylor series method to find the series solution for y, in ascending powers of x, up to and including the (6) term in x3. b Use your result to part a to estimate, to 3 decimal places, the value of y at x = 0.1. (4) ( E E/P 42 y   E/P   +       + y = 0 dx2 ( dx ) 2 dy ___ 2 d3y (4) a Find an expression for  ____3  dx dy    = 1 at x = 0, Given that y = 1 and ___ dx b find the series solution for y, in ascending powers of x, up to and (4) including the term in x3. c Comment on whether it would be sensible to use your series solution to give estimates for y at x = 0.2 and (2) at x = 50. d2y dy 43  ____2  − 4 ___    + 3y2 = 6, dx dx dy with y = 1 and ___    = 0 at x = 0. dx a Use the Taylor series method to obtain y as a series of ascending powers of x, up to and including the term in x4. (6) b Hence find the approximate value (3) of y when x = 0.2. ← Further Pure 2 Section 7.5 E 44 Relative to the origin O as pole and initial line θ = 0, find an equation in polar coordinate form for: M08B_IAL_FP2_44655_RE2_168-177.indd 173 46 a Sketch the curve with polar equation π π r = 3 cos 2θ, − __  < θ ,  __  (2) 4 4 b Find the area of the smaller finite region enclosed between the curve and π (6) the half-line θ = __     6 c Find the exact distance between the two tangents which are parallel to the initial line. (6) ← Further Pure 2 Sections 8.2, 8.3, 8.4 E/P ← Further Pure 2 Section 7.5 E/P 45 a Sketch the curve with polar equation (2) r = a cos 3θ, 0 < θ , 2π b Find the area enclosed by one loop of this curve. (6) ← Further Pure 2 Sections 8.2, 8.3 ← Further Pure 2 Section 7.5 dy ____ ) ← Further Pure 2 Section 8.2 47 a Sketch, on the same diagram, the curves defined by the polar equations r = a and r = a(1 + cos θ), where a is a positive constant and −π , θ < π. (4) b By considering the stationary values of r sin θ, or otherwise, find equations of the tangents to the curve r = a(1 + cos θ) which are parallel to the initial line. (6) c Show that the area of the region for which a , r , a(1 + cos θ) is (π + 8)a2   ________   (6) 4 ← Further Pure 2 Sections 8.2, 8.3, 8.4 E/P 48 The curve C has polar equation π π r = 3a cos θ, − __  < θ , __    . The curve D has 2 2 polar equation r = a(1 + cos θ), −π < θ , π. Given that a is positive, a sketch, on the same diagram, the graphs of C and D, indicating where each curve cuts the initial line. (4) 25/04/2019 08:53 174 2 REVIEW EXERCISE The graphs of C and D intersect at the pole O and at the points P and Q. b Find the polar coordinates of P and Q. At the point A, where A is distinct from O, on this curve, the tangent to the curve π is parallel to θ = __     2 b Determine the polar coordinates of the point A, giving your answer to 3 significant figures. (6) (3) c Use integration to find the exact value of the area enclosed by the curve D π and the lines θ = 0 and θ = __     (6) 3 The region R contains all points which lie outside D and inside C. ← Further Pure 2 Sections 8.2, 8.4 E Given that the value of the smaller area enclosed by the curve C and the line π θ = __     is 3 __ 3a2   ___  (2π − 3√3 ) 16 d show that the area of R is πa2. (6) ← Further Pure 2 Sections 8.2, 8.3, 8.4 E b Sketch on the same diagram the graphs of C and D, indicating where each cuts the initial line. (4) 49 a Show on an Argand diagram the locus of points given by the values of z satisfying |z − 3 + 4i | = 5 The graphs of C and D intersect at the points P and Q. (2) c Find the polar coordinates of P and Q. b Show that this locus of points can be represented by the polar curve r = 6 cos θ − 8 sin θ.(4) The set of points A is defined by π A = {z : −  __  < arg z < 0} ∩ {z:|z − 3 + 4i| < 5} 2 c Find, correct to 3 significant figures, the area of the region defined by A. (4) 52 The curve C has polar equation π π r = 6 cos θ, − __  < θ , __     2 2 and the line D has polar equation π 5π π r = 3 sec(__     − θ), − __  < θ , ___     6 6 3 a Find a Cartesian equation of C and a Cartesian equation of D. (4) (3) ← Further Pure 2 Sections 8.1, 8.2 E 53 θ = π 2 C ← Further Pure 2 Sections 8.2, 8.3 E/P 50 a Sketch the curve with polar equation π π r = cos 2θ, − __  < θ < __     (2) 4 4 At the distinct points A and B on this curve, the tangents to the curve are parallel to the initial line, θ = 0. b Determine the polar coordinates of A and B, giving your answers to 3 significant figures. (6) ← Further Pure 2 Sections 8.2, 8.3 E/P O Initial line The figure shows a sketch of the curve C with polar equation π r2 = a2 sin 2θ, 0 < θ < __    , 2 where a is a constant. Find the area of the shaded region enclosed by C. (6) ← Further Pure 2 Section 8.2 51 a Sketch the curve with polar equation π r = sin 2θ, 0 < θ < __     (2) 2 M08B_IAL_FP2_44655_RE2_168-177.indd 174 25/04/2019 08:53 REVIEW EXERCISE E/P 2 54 θ = π 2 175 π The figure show the half-lines θ = 0, θ = __     2 and the curves with polar equations π 1 r =  _2, 0 < θ < __    , and 2 π     r = sin 2θ, 0 < θ < __ 2 a Find the exact values of θ at the two points where the curves cross. (4) m C O Initial line The figure shows a curve C with polar π equation r = 4a cos 2θ, 0 < θ < __    , and a 4 π __ line m with polar equation θ =    . The 8 shaded region, shown in the figure, is bounded by C and m. Use calculus to show that the area of the shaded region 1 (6) is  _2 a2(π − 2) b Find by integration the area of the shaded region, shown in the figure, which is bounded by both curves. (6) ← Further Pure 2 Sections 8.2, 8.3 E/P 57 θ= π 2 P ← Further Pure 2 Section 8.3 E/P 55 θ= π 2 O a – 0.5a C Initial line 1 r = a(1 + cosθ) 2 1.5a Initial line O Q The curve C, shown in the figure, has polar equation __ r = a(3 + √5  cos θ), −π < θ , π –a The curve shown in the figure has polar equation 1 r = a (1 + __     cos θ), a . 0, 0 , θ < 2π. 2 Determine the area enclosed by the curve, giving your answer in terms of a and π. (6) ← Further Pure 2 Section 8.3 E/P 56 θ = π 2 a Find the polar coordinates of the points P and Q where the tangents to C are parallel to the initial line. (6) The curve C represents the perimeter of the surface of a swimming pool. The direct distance from P to Q is 20 m. b Calculate the value of a. (2) c Find the area of the surface of the pool. (6) ← Further Pure 2 Sections 8.2, 8.3, 8.4 r = sin 2θ r=1 2 O M08B_IAL_FP2_44655_RE2_168-177.indd 175 Initial line 25/04/2019 08:53 176 2 E/P REVIEW EXERCISE 58 θ= W π 2 C1 : r = 3a(1 − cos θ), −π < θ , π and C2 : r = a(1 + cos θ), −π < θ , π X The curves meet at the pole O and at the points A and B. a Find, in terms of a, the polar coordinates of the points A and B. (2) b Show __ that the length of the line AB 3√3  ____ (3) is     a. 2 The region inside C2 and outside C1 is shaded in the figure. c Find, in terms of a, the area of this region. (6) A badge is designed which has the shape of the shaded region. Given that the length of the line AB is 4.5 cm, d calculate the area of this badge, giving your answer to 3 significant figures. (3) A O Initial line B Z Y The figure shows a sketch of the cardioid C with equation r = a(1 + cos θ), −π , θ < π. Also shown are the tangents to C that are parallel and perpendicular to the initial line. These tangents form a rectangle WXYZ. a Find the area of the finite region, shaded in the figure, bounded by the curve C. (6) b Find the polar coordinates of the points A and B where WZ touches the curve C. (6) c Hence find the length of WX. __ (2) 3√3 a _____ Given that the length of WZ is    , 2 d find the area of the rectangle WXYZ.(2) A heart-shape is modelled by the cardioid C, where a = 10 cm. The heart shape is cut from the rectangular card WXYZ, shown in the figure. e Find a numerical value for the area of card wasted in making this heart shape. (3) ← Further Pure 2 Sections 8.3, 8.4 E/P 59 C1 θ=π 2 B C2 Initial line O A The figure is a sketch of two curves C1 and C2 with polar equations M08B_IAL_FP2_44655_RE2_168-177.indd 176 ← Further Pure 2 Sections 8.2, 8.3 E/P 60 r = a(5 – 2 cos θ) θ=π 2 D O r = a(3 + 2 cos θ) B A Initial line C A logo is designed which consists of two overlapping closed curves. The polar equations of these curves are r = a(3 + 2 cos θ), 0 < θ , 2π and r = a(5 − 2 cos θ), 0 < θ , 2π The figure is a sketch (not to scale) of these two curves. a Write down the polar coordinates of the points A and B where the curves meet the initial line. (2) b Find the polar coordinates of the points C and D where the two curves meet. (4) 25/04/2019 08:53 REVIEW EXERCISE 2 177 c Show that the area of the overlapping region, which is shaded in the figure, is __ 2   a__  (49π − 48√3 ) (6) 3 ← Further Pure 2 Sections 8.2, 8.3 Challenge The diagram shows the curve C with polar equation r = f(θ). The line l is a tangent to the curve at the point P(r, θ), and α is the acute angle between l and the radial line at P. θ= π 2 l C α P (r, θ) O θ Initial line r Show that tan α = _____     dr ___ (  ) dθ ← Further Pure 2 Section 8.4 M08B_IAL_FP2_44655_RE2_168-177.indd 177 25/04/2019 08:53 178 EXAM PRACTICE Exam practice Mathematics International Advanced Subsidiary/ Advanced Level Further Pure 2 Time: 1 hour 30 minutes You must have: Mathematical Formulae and Statistical Tables, Calculator Answer ALL questions 1 Use algebra to find the set of values of x for which x 1 _____     , _____     x+1 x+3 2 (6) Prove that n n(pn + q) 1 ∑     ____________     = ______________       (r + 2)(r + 4) 24(n + 3)(n + 4) r=1 (5) where p and q are constants to be found. 3 A complex number z has argument θ and modulus 1. 1   n  = 2i sin nθ, n ∈ ℤ  + a Show that z  n− __ z  b Hence, show that 8 sin  4 θ = cos 4θ − 4 cos 2θ + 3 4 (2) (5) a Show that the locus of points given by the values of z satisfying |z + 12 + 5i| = 13 can be represented by the polar curve with equation r = −2(12 cos θ + 5 sin θ) (4) b Show on an Argand diagram the set of points A defined by A = { z : |z + 12 + 5i| < 13} ∩ { z : −π < arg z < −  __   4} c Find, correct to 3 significant figures, the area of the region defined by A. (4) a Find the general solution to the differential equation dy cos x  ___  + y sin x = cos3 x dx b Find the particular solution which satisfies the condition that y = 3 when x = 0. (9) 3π 5 Z01_IAL_FP2_44655_EXP_178-179.indd 178 (5) (2) 25/04/2019 08:52 EXAM PRACTICE 6 179 a Show that the substitution x = et transforms the differential equation d2y dy     + 12y = 0, x2  ____2  + 8x ___ x>0 dx dx into the differential equation dy d2y ___   2  + 7  ___  + 12y = 0(8) dt dt b Hence find the general solution of the original differential equation.(5) dy d   2 y    )  + 2y = 0 ___ 7 ____2  + ( d x  dx 2 dy Given that when x = 0, y = ___     = 1, find a series solution for y in ascending powers of x, dx up to and including the term in x3. 8 (9) The point P represents a complex number z in an Argand diagram. Given that _ √  2 |z − i| = |z − 4| a find a Cartesian equation for the locus of P, simplifying your answer (3) b sketch the locus of P. (2) c On your sketch from part b, shade the region for which _ π √  2 |z − i| , |z − 4| and     |arg (z + 1)| , __ 2 d Find the complex numbers for which _ π √  2 |z − i| = |z − 4| and     |arg (z + 1)| = __ 2 Z01_IAL_FP2_44655_EXP_178-179.indd 179 (2) (4) TOTAL FOR PAPER: 75 MARKS 25/04/2019 08:52 180 GLOSSARY GLOSSARY acute (angle) an angle less than 90° algebraic representing mathematical information with symbols (i.e. using letters and numbers) alternating sequences sequences in which successive terms change repeatedly between being positive and negative approximate not exact, but close enough to be used approximation a number that is not exact arbitrary something based on a random choice rather than a reason arbitrary constant a constant to which various values may be assigned; it is unaffected by the changes in the values of the variables of the equation arc a smooth curve joining two points Argand diagram a diagram using Cartesian axes on which complex numbers are represented geometrically argument the specific input of a function argument of a complex number gives the angle between the positive real axis and the line joining the point to the origin ascending increasing asymptote a line that a curve approaches but never quite reaches auxiliary equation an equation of the form am2 + bm + c = 0 which is derived from a linear differential equation axis (plural axes) either of the two lines by which the positions of points are measured in a graph binomial an algebraic expression of the sum or difference of two terms. For example, (a + b)n is the general form of a binomial expression binomial expansion the algebraic expansion of powers of a binomial bound forming the edge of an area boundary the line (real or imaginary) that marks the edge of an area boundary conditions restrictions used to find a particular solution to a second-order differential equation cardioid a curve that is heart-shaped Cartesian coordinates a unique point in a plane specified by a pair of numerical coordinates chain rule a formula used to differentiate composite functions, or functions of another function circumference the boundary of a curved geometric shape coefficient in 4x3, the coefficient of x3 is 4 Z02_IAL_FP2_44655_GLS_180-182.indd 180 complementary function (C.F.) the general solution of a homogeneous differential equation complex conjugate each of a pair of complex numbers having their real parts identical and their imaginary parts of equal magnitude but opposite sign. If z = a + bi, then z* = a – bi, where a, b ∈ ℝ complex number a number that can be expressed in the form a + bi (where a and b are real numbers and i is a solution of the equation x2 = −1); called an imaginary number because there is no real number that satisfies this equation consecutive following one after another, without being interrupted constant a term that does not include a variable. In the expression x2 + 3x − 6, the constant term is −6 convention the way something is usually done converge (of a series) approaching a limiting value as the number of terms increases converse opposite convert change coordinates a set of values that show an exact position. In a two-dimensional grid, the first number represents a point on the x-axis and the second number represents a point on the y-axis corresponding an equivalent; connected with what you have just mentioned critical value a value that is important in the context of solving a problem cubic a polynomial of degree 3 deduce to conclude from a known or assumed fact degree the degree of a polynomial is equal to the highest power in that polynomial de Moivre’s theorem (r(cos θ + i sin θ))n = rn(cos nθ + i sin nθ) denominator the bottom part of a fraction derivative a way to represent the rate of change, in dy d2y other words,  __  is the first derivative, and ___   2  is the dx dx second derivative differentiate the process of finding the instantaneous rate of change of a function with respect to one of its variables differentiation the instantaneous rate of change of a function with respect to one of its variables distinct not equal diverge to move away from a limit as the argument (input) of the function increases or decreases or as the number of terms of the series increases 25/04/2019 08:52 GLOSSARY enclosed surrounded on all sides enlargement a transformation of a shape that involves increasing or decreasing the length of each side by a scale factor equate to make equal equation a statement where values of two mathematical expressions are equal. Solving an equation consists of determining the value(s) of the variable Euler’s relation eiθ = cos θ + i sin θ exponential form involving exponents expression any group of algebraic terms. For example, 2x + 6y + 3z is an algebraic expression factorise to rewrite an expression using brackets. We factorise x2 + 3x + 2 to get (x + 1)(x + 2) finite having a fixed size; not infinite finite series the sum of the values of a finite sequence first-order differential equation an equation in which f(x, y) is a function of two variables defined on a region in the xy-plane. For example, the terms in the 1 1 _ 1 1 __ 1 1 sequence 1, _    , _    ,    , __    ,    , ..., (    )n − 1 converge _ 2 4 8 16 32 2 toward a value of zero as n tends toward infinity fractional involving fractions function the relationship between a set of inputs and a set of outputs, where each input is related to exactly one output gradient the slope of a line half-line a straight line extending infinitely in a single direction from a point homogeneous differential equation a differential d2y dy equation of the form a ___   2  + b  __  + c = 0 dx dx identity an equality that holds true without being affected by the values chosen for its variables imaginary number a number that is expressed in terms of the square root of a negative number a, where __ a ∈ ℝ, and i = √ −1 , so i2 = −1 infinite without limit; not finite infinite series the sum of the values in an infinite sequence initial line a line, usually the x-axis, that the angle θ is measured from when using in polar coordinates integer a whole number. The symbol for integers is ℤ integrating factor a function that is used as a multiplier for another function in order to allow that function to be solved intersection the point where two lines meet or cross over intersects meets or crosses at a point interval the range of a set of numbers. For example, 3, 4, 5, 6, 7 are the members of the set of numbers satisfied by the interval 2 < x < 8, where x is an integer Z02_IAL_FP2_44655_GLS_180-182.indd 181 181 invariant (point or line) a fixed point or line that does not move under a transformation inverse operations that reverse each_other. For example, the inverse of y = x2 is x = √ y  limit a value toward which an expression converges as one or more variables approach certain values locus (of a set of points) the set of points that satisfies given conditions or a rule loop a closed curve Maclaurin series an expansion series of a function, where the approximate value of the function is determined as a sum of the derivatives of that function major arc an arc of a circle having measure greater than or equal to 180° (π radians) mapping a relationship such that each element of a given set (the domain of the function) is associated with an element of another set (the range of the function) method of differences a way to calculate a polynomial using its values at several consecutive points midpoint (of a line segment) a point on a line that divides it into two equal parts minor arc an arc of a circle having measure less than 180° (π radians) modulus (of a complex number) the distance of a point from the _____ origin. For any complex number z = a + bi, |z| =√ a2 + b2  modulus−argument form the modulus−argument form of a complex number is z = r(cos θ + i sin θ), where r = |z| and θ = arg z non-homogeneous differential equation a d2y dy differential equation of the form a ____   2  + b __  + c = f(x) dx dx numerator the top part of a fraction numerical relating to numbers origin the point where the y-axis and x-axis intersect on a flat coordinate plane parallel two lines side-by-side, the same distance apart at every point parametric equation a set of equations that express a set of quantities as functions of a number of independent variables, known as parameters. c For example, x = ct, y = _    , t ∈ ℝ are the parametric t equations of a hyperbola particular integral (P.I.) is a function that satisfies a differential equation particular solution the exact solution to a differential equation perpendicular one line meeting another at 90° perpendicular bisector a perpendicular line that divides a line segment into two equal parts 25/04/2019 08:52 182 GLOSSARY plane a flat two-dimensional surface extending into infinity polar coordinate (system) a two-dimensional coordinate system in which each point on a plane is determined by a distance from a reference point and an angle from a reference direction pole the point, usually the origin, from which the distance of a point is measured polynomial an expression of two or more algebraic terms with positive whole number indices. For example, 2x + 6x2 + 7x6 is a polynomial with positive with integer exponents power another name for an index number product 2 × 3 = 6, so 6 is the product of 2 and 3 product rule a method for differentiating problems where one function is multiplied by another proof by mathematical induction a special form of deductive reasoning (i.e. using the information you have to form an opinion). It is used to prove a fact about all the elements in an infinite set by performing a finite number of steps quadrant the area when y two-dimensional axes Second First are divided into four quadrant quadrant quadrants Third quadrant O Fourth quadrant x quadratic expressions such as x2 + 3x are quadratic, where the highest power of any variable is 2 radian describes an angle subtended by a circular arc as the length of the arc divided by the radius of the arc. One radian is the angle subtended at the centre of a circle by an arc that is equal in length to the radius of the circle rational number a number that can be expressed as an integer or fraction real number a value that can be represented along a number line and includes all rational and irrational numbers. The symbol for real numbers is ℝ roots the set of all possible solutions. A quadratic equation has up to 2 roots rotation a transformation of an object about its centre by a specified angle scale factor a number which multiplies some quantity. The ratio of any two corresponding lengths in two similar geometric figures is called a scale factor second-order differential equations differential d2y equations contains second derivatives e.g. ___   2  dx segment (i) part of a line connecting two points (ii) the area of a circle cut off by a chord Z02_IAL_FP2_44655_GLS_180-182.indd 182 separating the variables a method used to solve certain types of differential equations sequence a series of numbers following a set rule. 4, 9, 14, 19, ... is an example of an arithmetic sequence series the sum of terms in a sequence series expansions a method for calculating a function that cannot be expressed more easily series solution a solution to certain types of differential equations subtended an angle subtended by an arc, line segment, or any other section of a curve is one whose two rays pass through the endpoints of the arc successive coming or following on after the other sum the addition of two or more numbers. For example, 2 + 3 = 5, so 5 is the sum of 2 and 3 surd a number that cannot_be simplified to remove a square root. For example, √ 2 _is a surd because it is an irrational number, whereas √ 4 = ±2 which is a rational number and not a surd symmetrical when a shape looks the same following a transformation such as reflection or rotation tangent (i) a trigonometric function that is equal to the ratio of the side opposite an acute angle (in a rightangled triangle) to the adjacent side (ii) a line that touches a curve at a point without crossing over and matches the gradient of the curve at that point Taylor series an expansion series of a function, where the approximate value of the function is determined as a sum of the derivatives of that function transform to map linearly transformation a linear mapping that is either a reflection, rotation or stretch translation a function that moves an object a certain distance nth root of unity the solutions to zn = 1 valid true variable a quantity that is able to be changed, i.e. not constant vector an object that has both a magnitude and a direction. Geometrically, a vector is a directed line segment, whose length is the magnitude of the vector and with an arrow indicating the direction verify show vertex (plural vertices) where two lines meet at an angle, especially in a polygon w-plane a geometric representation of the complex numbers, w = u + vi where u is represented by the horizontal axis and vi is represented by the vertical axis z-plane a geometric representation of the complex numbers, z = x + yi where x is represented by the horizontal axis and yi is represented by the vertical axis 25/04/2019 08:52 ANSWERS ANSWERS CHAPTER 1 Prior knowledge check 1 a x , −  __13  or x . 1 2 a x . 2 or x , −  __43  __ 183 y b __ b − 2 − √6  , x , − 2 + √6  b __32  , x , __   52  y = x3 + 2x2 – 3x Exercise 1A 1 a −1 , x , 6 b x< −3 or x > 2 __ __ c −1 , x , 1 d − √3  , x , −1 or 1 , x , √3  e 0 < x , 1 or x >  __23  f x , −1 or 0 , x , 2 g x , −2 or −1 , x , 1 or x . 2 h −1 , x , 0 or 0 , x , 2 i x , 4 or x . __   14   3 j −2 , x , 5 or x . __   17   2 1 _ 2 a {x : x .   3  } ∪ {x : − 5 , x , 0} b {x : x , 0} ∪ {x : 2 , x , 5} c {x : x , − 2} ∪ {x : 0 , x , 1} d {x : x , − 3} ∪ {x : − 1 , x , 1}   21  } e {x : −  __13  , x , 0} ∪ {x : 0 , x , __ f {x : −1 , x , −  __13  } ∪ {x : x . __   12  } __ __ 3 − 5 , x , − 4 and −___ 1 − √7  , x , − 1 + √7  ___ 5 − √29  5 + √29  4 {x : −  __21  , x ,________     } ∪ {x : 3 , x ,________     } 2 2 5 a The student did not square the denominators before cross-multiplying. Multiplying by negative values does not preserve the inequality. b −  __43  , x , − 1 or 0 , x , 4 O –3 x 1 y c y= 1 –1 O 1 x+1 x 6 {x : − 2 , x , − 1} ∪ {x : − __  21  , x , 0} Challenge x , ln  __12  or x . ln 1 1 a y d Exercise 1B y y = x2 – 5x + 6 6 y= 4x 1 – 2x O 1 2 x –2 O Z03_IAL_FP2_44655_ANS_183-229.indd 183 2 3 x 25/04/2019 08:52 184 ANSWERS y 2a y d 4 y = x2 – 2x + 1 4 y= x 4x – 2 O y = 4 – 4x2 1 –1 1 4 O 1 4 2 3 x x 1 y = 4 – 3x y b 1 3 a (7,  __4 ) b (4, 2) and (−1, −3) c (−2, 0), (0, −4), (4, 12) y=x y 4 a 1 –1 O –1 x 1 y=x–1 y= 1 x c O x 1 4 y= x–1 y y = 2x – 1 –4 b (3, 2) and (−1, −2) c −1 , x , 1 or x . 3 5 O y= 3 x–2 –1 – 32 1 2 2 a y x y = 32 x 2 3 O x 3 y= b (−3,  _13  ) and (  _32  ,  _43  ) 2 3–x c −3 , x , _   32  or x . 3 Online Worked solutions are available in SolutionBank. Z03_IAL_FP2_44655_ANS_183-229.indd 184 25/04/2019 08:52 ANSWERS Challenge y a y 6 a 185 (x – 2)2 + (y – 4)2 = 10 4x (x – 1)2 y= 4 O y= 3x 2–x O x 12 –1 –3 2 x 5 b (−1, 3), (1, 1), (3, 7), (5, 5), y c (0, 0), (  _53  , 15) and (−1, −1) b   53 or 2 , x c x < −1 or 0 < x , 1 or 1 , x < _ y 7 a (x – 2)2 + (y – 4)2 = 10 4 y=x–2 –2 O –2 y = 4x – 5 x–2 x 2 3 y= 6(2 – x) (x + 2)(x – 3) 2 1 a x , __   67  y ___ ___ b __12  (−√13 − 1) , t , __   12  (√13 − 1) __ x x+2 x 5 Exercise 1C c x , −2 or 0 < x < 1 or 2 < x , 3 y= –1 d − 1 , x , 1 and 3 , x , 5 b (0, −2), (1, −1) and (2, 0) 8 a 5 2O __ c −7 , x , −2 − √ 7 or −2 + √ 7   , x , 3 y= 1 x d x > 1 or x < − 2 e x . 1 or x , − 3 f x . 1 or x , −  __13  2 a y 1 –2 O x y = |3x – 2| b (−1, −1) and (2,  _12  ) c −2 , x , −1 or 0 , x , 2 4 2 y = 2x + 4 –2 O 2 3 x b {x : −  __25  < x < 6} Z03_IAL_FP2_44655_ANS_183-229.indd 185 25/04/2019 08:52 186 ANSWERS b f(x) = (x + 1)(x + 5)(x − 3) y 3 a y y = |x2 – 4| y= 4 x2 – 1 O –2 –1 1 x 3 –5 –1 O 3 –15 y = f(x) –4 __ x __ b − √  5 < x , − 1 or 1 , x < √ 5   4 {x : − 1 , x ,  __31  } __ __ c __ 5 {x : x , −1 − √ 3   } ∪ {x : −√2   , x , √ 3 − 1} 6 a __ __ __ __ x = –5, 1 − √ 5   < x < 1 − √ 3 , 1 + √ 3   < x < 1 + √ 5   Chapter review 1 y 1 0 < x < 2 or x > 4 __ __ 2 − 2 , x , 1 − √   6 or x . 1 + √ 6   3 0 , x , 2 or x .  _72  4 {x : 0 , x ,  _32  } ∪ {x : 3 , x , 4} 4a 5 {x : x , − 1} ∪ {x : 1 , x , 11} y 6 a y= y = 2x + 4 x–2 1 x–a 2 –1 y = 4|x – a| O a x O –2 x 2 y = 2x + 2 __ __ b 1 − √ 5   , x , 2 or x . 1 + √ 5   y 7 a b x , a or x . a + __  12  y = 2x2 – 4 x –2 7 −2 , x , 0 or x . 2 8 a The student hasn’t checked which critical values actually correspond to intersections of the graphs. 2 b 1,x,5 O 2 x Challenge a f(−1) = (−1)3 + 3(−1)2 − 13(−1) − 15 = −1 + 3 + 13 − 15 = 0 So by the factor theorem (x + 1) is a factor. y = 2 – 4x __ __ b − √2   , x , − 1 or 0 , x , √   2 or x . _   32  Online Worked solutions are available in SolutionBank. Z03_IAL_FP2_44655_ANS_183-229.indd 186 25/04/2019 08:52 ANSWERS y 8 a y= x–2 3x – 1 y b 2 1 3 y = |2x2 + x – 6| 1 O –2 y= 187 1 3 x 2 6 2 x+2 2 –2 O ___ x 3 2 y = 6 – 3x ___ b − 2 , x , 3 − √ 11  and __   31  , x , 3 + √ 11  y 9 a __ __ c x , − 1 − √ 7 or 0 , x , 1 or x . − 1 + √  7 16 a y = 2x – 1 x+4 y y = |2x – 1| y= x+1 x–2 2 1 –4 O 4 y= x 2 |x2 – 4| 1 –2 __ O 1 2 x 2 __ b x = −1 − √ 6  , − 1, − 1 + √ 6  , 3 __ __ c x , −1 − √ 6 , or −1 , x , −1 + √ 6   or x . 3 __ __ b x , − 4, 5 − 3√3   , x , 2, 5 + 3√3   , x 10 1 , x , 5 11 − 3 , x , 3 12 x ,  _27  __ Challenge Solving x  2− 5x + 2 = x − 3 and x  2−__5x + 2 = __3 − x we find that the critical values are x = 2 − √5 , 1, 2 + √5 , 5 Sketching the graphs we have y __ y = |x – 3| 13 x , √ 3 − 1 or x . √ 3 + 1 14 a y = |x2 – 5x + 2| y 3 2 3 y = |2x – 3| O 2– 51 __ 2+ 5 5 x __ {x : x , 2 − √5 } ∪ {x : 1 , x , 2 + √5 } ∪ {x : x . 5} O –1 1 5 x 3 2 CHAPTER 2 Prior knowledge check y = 5x – 1 b x._   47  __ __ 15 a x = −1 − √ 7 , 0, 1, −1 + √   7 Z03_IAL_FP2_44655_ANS_183-229.indd 187 1 a 1098 b 10 761 619.5 2 a Use the following: n n(n + 1)(2n + 1) r 2 = _______________   ∑        6 r=1 n n(n + 1) r 2 = ________       ∑  2 r=1 and simplify to get the answer. b 10 073 25/04/2019 08:52 188 ANSWERS Exercise 2A 1 a __   12  (r(r + 1) − r(r − 1)) = __  21  (r  2 + r − r  2 + r) = __  12  (2r) = r n n n r=1 r=1 r=1 b ∑  r =  __12  ∑  r(r + 1) −  __12  ∑  r(r − 1) r = 1:  __21 × 1 × 2 − __  12 × 1 × 0 r= r= ⋮ 2:  __21 × 3:  __21 × 2×3− 3×4− __  12 × __  12 × 3 a ( r + 1)  3 − (r − 1)  3 = (r  3+ 3r  2+ 3r + 1) − (r  3− 3r  2+ 3r − 1) = 6r  2+ 2 n n n b   ∑ (6r2 + 2) = 6  ∑ r  2+   ∑ 2 = 2n  3+ 3n  2+ 3n r=1 2×1 3×2 n r=1 n r=1 r=1 n r=1 So 6  ∑ r 2 = 2n  3+ 3n  2 + n = n(2n  2+ 3n + 1) r=1 r = n − 1:  __12 × (n − 1)(n) − __  12  (n − 1)(n − 2) r = n:  __12  n(n + 1) − __  12  n(n − 1) When you add, all terms cancel except __  12  n(n + 1) n Hence ∑  r =  __12  n(n + 1) r=1 n(n + 3) 2 ______________         4(n + 1)(n + 2) n(3n + 5) 1 1            b ______________ 3 a  ___  − ________ 2r 2(r + 2) 4(n + 1)(n + 2) 1 1 1     = _____   − _____       4 a ____________ (r + 2)(r + 3) r + 2 r + 3 n   b ________   3(n + 3) (r + 1)! − r! ___________ r!(r + 1 − 1) _______ 1 r 1        ≡ ___________     ≡        ≡     5 a __  − _______ r! (r + 1)! r!(r + 1)! r!(r + 1)! (r + 1)! 1   b 1 − _______   (n + 1)! n(n + 2)   6  ________ (n + 1)2 1 1    , which 7 a Method of differences yields ___     − _________ 10 2(2n + 5) n _________  , so a = 10 and b = 25. simplifies to   10n + 25 1 1   = ________    . Assume true for n = k. b For n = 1, _____   5 × 7 10 + 25 Let n = k + 1, then k+1 1 k 1       = _________   ∑ ______________     + _______________       10k + 25 (2k + 5)(2k + 7) r = 1 (2r + 3)(2r +5) k+1 2k2 + 7k + 5        = _____________       =   _______________ (2k + 1)(2k + 7) 10(k + 1) +25 Therefore true for all values of n. 1 4 1   8 Method of differences yields  __ (______   − ______     , 3 3r − 2 3r + 4 ) n(15n + 17)    which simplifies to _______________     , so a = 15 and (3n + 1)(3n + 4) b = 17. 9 Method of differences yields ( n + 1)  2+ n  2 − 1  2 − 0  2= 2n  2+ 2n = 2n(n + 1) so a = 2. Chapter review 2 r=1 n So 6  ∑ r  2+   ∑ 2 = 6 ∑ r  2+ 2n = 2n  3+ 3n  2+ 3n 2 1 1 1 a ___________       = _____     − _____     (r + 2)( r + 4) r + 2 r + 4 n n 1 1 2         =  ∑( _____  − _____ b  ∑ ____________ r + 4) r = 1 (r + 2)(r + 4) r=1 r + 2 1 1 _____ 1 1 = __     + __     −     − _____     3 4 n+3 n+4 2 + 25n 7n        =   _______________ 12(n + 3)(n + 4) 1 1 4     = ______     − ______     2 a  ______________ (4r − 1)(4r + 3) 4r − 1 4r + 3 n 4 4n + 3 − 3 1 1 _______     =  __   −     = __________     b ∑  ______________ 3 4n + 3 3(4n + 3) r=1 (4r − 1)( 4r + 3) 4n   = _________   3(4n + 3) c 0.00126 n = n(n + 1)(2n + 1) So   ∑ r 2 = _ 61 n(n + 1)(2n + 1) r=1 2 2 4     − _____    ) 4   ∑  ____________     =   ∑ (_____ n n r = 1 (r + 1)(r + 3) r=1 r+1 r+3 n(5n + 13) 2 2 _____ 2 2 = __     + __     −        − _____     = _____________       2 3 n+2 n+3 n   ∑ ((r + 1)3 − (r − 1))3 = n(2n2 + 3n + 3) 5 3(n + 2)(n + 3) r=1 6 7 8 9 Calculate (2n)(2(2n)2 + 3(2n) + 3) − (n − 1)(2(n − 1)2 + 3(n − 1) + 3) which gives that a = 14, b = 15, c = 3, d = 2 3n a Method of differences yields  _________ , 12n + 16 so a = 3, b = 12 and c = 16. 6n 3n − 3   b Simplify: _________   − ________     24n + 16 12n + 4 1 1 The general term __     +  _____ is a um, not a difference, r r+1 so the terms will not cancel out, and the method of differences cannot be used in this case. n 1 Recognise this is  ∑   _______ and apply method of r − 1 r(r + 1) 1 1 1 1       − ________   + ________     to differences. Simplify __     + __ 2 4 ( 2(n + 1) 2(n + 2) ) 2n + 3 3      , stating a = 2 and b = 3.     − ______________ obtain __ 4 2(n + 1)(n + 2) 1 1     a ______  − ______ 2r + 1 2r + 5 b 0.0218 Challenge a k = 11 b a = 11, b = 48, c = 49 CHAPTER 3 Prior knowledge check π b __     3 e 4 1 a 8 π d  __  2 2 c 16 π f __  6 Im 2 1 –2 –1 O 1 2 Re –1 –2 3 4032 Online Worked solutions are available in SolutionBank. Z03_IAL_FP2_44655_ANS_183-229.indd 188 25/04/2019 08:52 ANSWERS Exercise 3A b 6e  2 ___ 5πi  −  ____ 6 c 4e___ e √29  e−1.19i 3.02i d √ 65  __ e____  3πi  f 2√6 e 4 g 2√2 e 4 h 8e 6 __ i 9e  iθ= cos θ + i sin θ, e  −iθ= cos θ − i sin θ So e  iθe   −iθ= (cos θ + i sin)(cos θ − i sin θ) LHS = e  i(θ−θ) = e  0= 1 RHS = cos  2 θ − i  2 sin  2 θ = cos  2 θ + sin  2 θ Hence cos  2 θ + sin  2 θ ≡ 1 πi __     1 a 3eπi πi  −  __ __  πi  πi  −  __ 5 2e __ √   3 1  +  ___ 2 a  __   i 2 2 c 3 + 3i b −4 Challenge a n = 1; LHS = (re  iθ)  1 = re  iθ RHS = r  1e   iθ = re  iθ As LHS = RHS, the equation holds for n = 1. Assume the equation holds for n = k, k ∈ ℤ  +. i.e. (re  iθ)  k= r  ke   ikθ With n = k + 1, the equation becomes (re  iθ)  k+1 = (re  iθ)  k × re  iθ= r  ke   ikθ × re  iθ= r  k+1e   i(kθ+θ)= r  k+1e   i(k+1)θ Therefore, the equation holds when n = k + 1. If the equation holds for n = k, then it has been shown to be true for n = k + 1. As the equation holds for n = 1, it is now also true for all n ∈ ℤ  +by mathematical induction. 1 1 b Given n ∈ ℤ  +, we have: (re  iθ)  −n = _______    = r  −ne   iθ n  = ______   −inθ (re  )  r ne  inθ __ d 4√3   + 4i __ √ 3   __ ___ f −     +  1  i 2 2 h −3 − 3i e −3i g −1 __ i −4 + 4i√3   10π 10π ____ 3 a cos(−     + i sin(−  ____ ) 13 ) 13 3π 3π b 4 cos(−  ___ ) + i sin (−  ___ ) ( 5 5 ) 7π 7π ___ ___ c 5 cos(   ) + i sin(   ) ( 8 8 ) 4 eiθ = cos θ + i sin θ e−iθ = cos (−θ) + i sin (−θ) = cos θ − i sin θ (1) − (2): eiθ − e−iθ = 2i sin θ 1  __ (eiθ − e−iθ) = sin θ 2i 1 ⇒ sin θ =  __ (eiθ − e−iθ) (as required) 2i (1) (2) Exercise 3C Exercise 3B _ 7π 7π 1 a cos  ___  + i sin  ___  12 12 _   c 3i√ 2 _ _ √   3 3 √   5π 5π b  ___  cos  ___  + i  ___  sin  ___  7 7 4 4 c −i 3 a e  5iθ b e  πi 4 a e  3iθ _ _ b 3 √ 5  cos 4θ + 3i √ 5  sin 4θ 2 a −  __14 5 a 6√ 3 e   6 5πi ___ πi __ _ c 6e   3 _ 1 a cos__6θ + i sin 6θ √ 3   1 c −  ___  + __     i 2 2 e 1 2 a e  iθ b e   2iθ −iθ d e  e e   11iθ b cos 12θ__+ i sin 12θ   1 √ 3 d −  __  +  ___ i 2 2 f i c e  −6iθ fe  5iθ 3 a 1 b −1 4 a (1 + i)5 = −4 − 4i c (1 − i)6__= 8i __ e ( __   23  − __   21 √  i 3 )9 = 81i√ 3   c 1 8 = 4096 b (−2 + 2i) __ 6 √ (1 − i  3 ) = 64 d __ πi __ d 3√ 2 e   4 πi πi __ __ b 2√ 2 e   4 c __   34 e −  2 _ ___ 7πi 7πi πi ___ __ − √ b 3 e   12 c 18e   12 d 6e   4 (cos 9θ + i sin 9θ)(cos 4θ + i sin 4θ) ______   4iθ e  9iθe            =   7iθ  = e  9iθ+4iθ−7iθ 6  _______________________________ cos 7θ + i sin 7θ e  = e  6iθ= cos 6θ + i sin 6θ π 5πi ___ 7 2e 6 i, 2e −  6 πi __   2 8 a 2e b n = 1: LHS = (1 + i) 1= 1 + i _ _ πi 1 __ __ π π 1_ 1 RHS = 2   2 e       4 = √ 2   cos  __  + i sin  __  = √ 2   ___   _  + i  ___ ( ( √   4 4) 2 √ 2   ) =1+i As LHS = RHS, the equation holds for n =1. Assume the equation holds for n = k, k ∈ ℤ  +. k __ kπi ___ i.e. (1 + i)  k = 2  2e     4  With n = k + 1, the equation becomes k __ kπi ___ (1 + i)  k+1 = (1 + i)  k× (1 + i) = 2  2e     4 × (1 + i) k __ kπi πi 1 __ ___ __   4  = 2  2e   × 2   2 e     4  (k + 1)πi k kπi πi k+1 1 __ __ ___ __ ____ _____ = 2   2 +  2 e    4   +   4  = 2    2 e     4  Therefore, the equation holds when n = k + 1. If the equation holds for n = k, then it has been shown to be true for n = k + 1. As the equation holds for n = 1, it is now also true for all n ∈ ℤ  +by mathematical induction. c 256 Z03_IAL_FP2_44655_ANS_183-229.indd 189 189 5 6 7 8 9 __ f (−2√ 3 − 2i)5 = 512√ 3 − 512i __ __ (3 + √ 3 i)5__= −432 + 144i√ 3  −8 + 8i √ 3   −27i 2π __ a e    3 i b 3 Write a + bi and a − bi as r(cos θ + i sin θ) and r(cos θ − i sin θ) respectively. Then by de Moivre’s theorem, (a + bi)n + (a − bi)n = rn(cos nθ + i sin nθ) + rn(cos nθ − i sin nθ) = 2rncos nθ which is always real. Challenge Given n ∈ ℤ  +, we have: 1 1 (r(cos θ + i sin))  −n = _______________     n  = _________________       (r(cos θ + i sin))  r  n(cos nθ + i sin nθ) by de Moivre’s theorem for positive integer exponents. cos nθ − i sin nθ 1     × ______________    =  _________________       r  n(cos nθ + i sin nθ) cos nθ − i sin nθ cos nθ − i sin nθ cos nθ − i sin nθ   = __________________             = ____________________      r  n(cos  2 nθ − i  2 sin  2 nθ) r  n(cos  2 nθ + sin  2 nθ) = r  −n(cos nθ − i sin nθ) = r  −n(cos (−nθ) + i sin (−nθ)) 25/04/2019 08:52 190 ANSWERS = cos  5 θ − 10 cos  3 θ(1 − cos  2 θ) + 5 cos θ(1 − 2 cos  2 θ + cos  4 θ) = 16 cos  5 θ − 20 cos  3 θ + 5 cos θ b 0.475, 1.57, 2.67 (3 s.f.) 1 3 a Let z = cos θ + i sin θ, then 2 cos θ = z + __     z 6 1 6 __ 6 ( z +  z )  = (2 cos θ) = 64 cos  θ 1 1 1 3 ___ = z  6 + 6 z  5(  ___ )+ 15 z  4 ___ (  z  2 )+ 20 z  (  z  3 ) z  1 1 1 ___ ___ + 15 z  2 ___ (  z  4 )+ 6z(  z  5 )+ (   z  6 ) 1 1 1   6 )+ 6 z  4 + ___   4  + 15 z  2 + ___   2  + 20 = (z  6 + ___ ( ( z  z  ) z  ) Exercise 3D 1 a ( cos θ + i sin θ)  3= cos 3θ + i sin 3θ = cos  3 θ + 3i cos  2 θ sin θ + 3i  2 cos θ sin  2 θ + i  3 sin  3 θ = cos  3 θ + 3i cos  2 θ sin θ − 3 cos θ sin  2 θ − i sin  3 θ ⇒ cos 3θ + i sin 3θ = cos  3 θ + 3i cos  2 θ sin θ − 3 cos θ sin  2 θ − i sin  3 θ Equating the imaginary parts: sin 3θ = 3 cos  2 θ sin θ − sin  3 θ = 3 sin θ(1 − sin  2 θ) − sin  3 θ = 3 sin θ − 4 sin  3 θ b (cos θ + i sin θ)  5= cos 5θ + i sin 5θ = cos  5 θ + 5i cos  4 θ sin θ + 10 i  2 cos  3 θ sin  2 θ + 10 i  3 cos  2 θ sin  3 θ + 5 i  4 cos θ sin  4 θ + i  5 sin  5 θ ⇒ cos 5θ + i sin 5θ = cos  5 θ + 5i cos  4 θ sin θ − 10 cos  3 θ sin  2 θ − 10i cos  2 θ sin  3 θ + 5 cos θ sin  4 θ + i sin  5 θ Equating the imaginary parts: sin 5θ = 5 cos  4 θ sin θ − 10 cos  2 sin  3 θ + sin  5 θ = 5(1 − sin  2 θ )  2 sinθ − 10(1 − s in 2 θ)sin  3 θ + sin  5 θ = 16 sin  5 θ − 20 sin  3 θ + 5 sin θ c (cos θ + i sin θ)  7= cos 7θ + i sin 7θ = cos  7 θ + 7i cos  6 θ sin θ + 21 i  2 cos  5 θ sin  2 θ + 35 i  3 cos  4 θ sin  3 θ + 35 i  4 cos  3 θ sin  4 θ + 21 i  5 cos  2 θ sin  5 θ + 7 i  6 cos θ sin  6 θ + i  7 sin  7 θ ⇒ cos 7θ + i sin 7θ = cos  7 θ + 7i cos  6 θ sin θ − 21 cos  5 θ sin  2 θ − 35i cos  4 θ sin  3 θ + 35 cos  3 θ sin  4 θ + 21i cos  2 θ sin  5 θ − 7 cos θ sin  6 θ − i sin  7 θ Equating the real parts: cos 7θ = cos  7 θ − 21 cos  5 θ sin  2 θ + 35 cos  3 θ sin  4 θ − 7 cos θ sin  6 θ = cos  7 θ − 21 cos  5 θ (1 − cos  2 θ) + 35 cos  3 θ (1 − cos  2 θ)  2− 7 cos θ (1 − cos  2 θ)  3 = 64 cos  7 θ − 112 cos  5 θ + 56 cos  3 θ − 7 cos θ d Let z = cos θ + i sin θ 1 4 ( z + _ z )  = ( 2 cos θ)  4 = 16 cos  4 θ = z  4+ 4z  3( _   1z )+ 6z  2( _   z1  2 )+ 4z(_   z1  3 ) + _   z1  4  4 = 2 cos 6θ + 6(2 cos 4θ) + 15(2 cos 2θ) + 20 64 cos  6 θ = 2 cos 6θ + 6(2 cos 4θ) + 15(2 cos 2θ) + 20 32 cos  6 θ = cos 6θ + 6 cos 4θ + 15 cos 2θ + 10 __   6π 9 __ 5π b ∫   cos6 θ dθ = ___     + ___       √ 3 0 96 64 1 a If z = cos θ + i sin θ, then 2 cos θ = z + __     and z 1 2i sin θ = z − __     z 2 4 1 1 So, 2  2 cos  2 θ × (2i)  4 sin  4 θ = ( z + __    )  ( z − __    )  z z 2 2 2 2 1 1 1 1 1    )(z − __    ))  (z − __    )  = z  2 − ___   2    (z − __    )  = ( (z + __ ( ) z z z z z  1 1   4  z  2− 2 + ___   2  = z  4− 2 + ___ ( z  )( z  ) 1 ___ 2 1 ___ 6 4 2     = z  − 2z  − z  + 4 −   2  −   4  + ___ z  z  z  6 1 1 1   6  − 2 z  4 + ___   4  − z  2 + ___   2  + 4 = z  6 + ___ ( ( z  ) ( z  ) z  ) b 5 a 6 a = (z  4 + _   z1  4 )+ 4(z  2 + _   z1  2 )+ 6 = 2 cos 4θ + 4(2 cos 2θ) + 6 16 cos  4 θ = 2 cos 4θ + 4(2 cos 2θ) + 6 = 2(cos 4θ + 4 cos 2θ + 3) ⇒ cos  4 θ = __  81  (cos 4θ + 4 cos 2θ + 3) e Let z = cos θ + i sin θ 1 ( z − _ z )  = ( 2i sin θ)  5 = 32i  5 sin  5 θ = 32i sin  5 θ 3 2 = z  5+ 5z  4(−  _1z )+ 10z  3 (−  _1z )  + 10z  2 (−  _1z )  5 + 5z (−  _1z )  + ( −  _1z )  4 5 5 = z  5− 5z  3+ 10z − _  10   +  _   − _   z1  5  z  3 z = (z  5 − _   z1  5 )− 5(z  3 − _   z1  3 )+ 10(z − _  1z ) = 2i sin 5θ − 5(2i sin 3θ) + 10(2i sin θ) 32i sin  θ = 2i sin 5θ − 10i sin 3θ + 20i sin θ 1 ⇒ sin  5 θ = __  16   (sin5θ − 5 sin 3θ + 10 sin θ) 5 2 a(cos θ + i sin θ)  5= cos 5θ + i sin 5θ = cos  5 θ + 5C1 cos  4 θ(i sin θ) + 5C2 cos  3 θ (i sin θ)  2 + 5C3 cos  2 θ (i sin θ)  3 + 5C4 cos θ (i sin θ)  4 + (i sin θ)  5 = cos  5 θ + 5i cos  4 θ sin θ − 10 cos  3 θ sin  2 θ −10i cos  2 θ sin  3 θ + 5 cos θ sin  4 θ + i sin  5 θ Equating the real parts gives cos 5θ = cos  5 θ − 10 cos  3 θ sin  2 θ + 5 cos θ sin  4 θ = cos  5 θ − 10 cos  3 θ(1 − cos  2 θ) + 5 cos θ (1 − cos  2 θ)  2 b = 2 cos 6θ − 2(2 cos 4θ) − 2 cos 2θ + 4 So, 64 cos  2 θ sin  4 θ = 2 cos 6θ − 4 cos 4θ − 2 cos 2θ + 4 ⇒ 32 cos  2 θ sin  4 θ = cos 6θ − 2 cos 4θ − cos 2θ + 2 π ___     48 π 5π 1 67 ____ b  ___  + ___     c   6144  ___  32 64 48 (cos θ + i sin θ)  6= cos 6θ + i sin 6θ = cos  6 θ + 6C1 cos  5 θ(i sin θ) + 6C2 cos  4 θ (i sin θ)  2 + 6C3 cos  3 θ (i sin θ)  3 + 6C4 cos  2 θ (i sin θ)  4 + 6C5 cos θ (i sin θ)  5 + (i sin θ)  6 = cos  6 θ + 6i cos  5 θ sin θ − 15 cos  4 θ sin  2 θ − 20i cos  3 θ sin  3 θ + 15 cos  2 θ sin  4 θ + 6i cos θ sin  5 θ − sin  6 θ Equating the real parts gives cos 6θ = cos  6 θ − 15 cos 4 θ sin  2 θ + 15 cos 2 θ sin  4 θ − sin  6 θ = cos  6 θ − 15 cos  4 θ(1 − cos  2 θ) + 15 cos  2 θ (1 − cos  2 θ)  2 − (1 − cos  2 θ)  3 = cos  6 θ − 15 cos  4 θ(1 − cos  2 θ) + 15 cos  2 θ(1 − 2 cos  2 θ + cos  4 θ) − (1 − 3 cos 2 θ + 3 cos  4 θ − cos  6 θ) = 32 cos  6 θ − 48 cos  4 θ + 18 cos  2 θ − 1 π 5π 7π cos  ___ ≈ 0.985, cos  ___ ≈ 0.643, cos  ___ ≈ 0.342, 18 18 18 13π 17π 11π cos  ____ ≈ −0.342, cos  ____ ≈ −0.643, cos  ____ ≈ −0.985 18 18 18 7 a cos 4θ + i sin 4θ = ( cos θ + i sin θ)4 4 = cos  θ + 4i cos  3 θ sin θ + 6 i  2 cos  2 θ sin  2 θ + 4 i  3 cos θ sin  3 θ + i  4 sin  4 θ = cos  4 θ + 4i cos  3 θ sin θ − 6 cos  2 θ sin  2 θ − 4i cos θ sin  3 θ + sin  4 θ Equating the imaginary parts: sin 4θ = 4 cos  3 θ sin θ − 4 cos θ sin  3 θ Online Worked solutions are available in SolutionBank. Z03_IAL_FP2_44655_ANS_183-229.indd 190 25/04/2019 08:52 ANSWERS b Equating the real parts: cos 4θ = cos  4 θ − 6 cos  2 θ sin  2 θ + sin  4 θ sin 4θ __________________________ 4 cos3 θ sin θ − 4 cos θ sin3 θ   tan 4θ = ______     =       cos 4θ cos4 θ − 6 cos2 θ sin2 θ + sin4 θ 1 ______ 3 3 (   cos4 θ )(4 cos θ sin θ − 4 cos θ sin θ)       = ___________________________________       1 4 2 2 4 ______ (  cos4 θ )(cos θ − 6 cos θ sin θ + sin θ) 4 tan θ − 4 tan3 θ         =   __________________ 1 − 6 tan2 θ + tan4 θ c x = 0.20, 1.50, −5.03, −0.67 (2 d.p.) 191 b – –2 O Re 3 1 – i 2 2 centre (−1, 0) radius 1 2π 2π 4π 4π z = 1, cos ( ___ ) + i sin (___    ), cos (___    ) + i sin (___    ), 5 5 5 5 2π 2π 4π 4π cos (−  ___ ) + i sin (−  ___ ), cos (−  ___ ) + i sin (−  ___ ) 5 5 5 5 z  1 + z  2 + z  3 + z  4 + z  5 = 0 2π 2π 2π 4π 4π 1 + cos  ___  + i sin  ___  + cos  ___  + i sin  ___ + cos(−  ___ ) 5 5 5 5 5 2π 4π 4π + i sin(−  ___ )+ cos(−  ___ )+ i sin(−  ___ )= 0 5 5 5 2π 4π 4π 2π ⇒ 1 + cos  ___  + i sin  ___  + cos  ___  + i sin  ___  5 5 5 5 2π 2π 4π 4π    )− i sin(___    )+ cos(___    )− i sin(___    )= 0 + cos(___ 5 5 5 5 4π 2π ___ ___ ⇒ 1 + 2 cos     + 2 cos    = 0 5 5 4π 2π cos  ___  + cos  ___  = −  __12 5 5 2π r = 4, θ = −  ___  3__ __ __ −  __ __  ____ __ −  ____ πi  5πi  2πi   πi  z=√ 2 e 6 , √ 2 e 3 , √ 2 e 6 , √ 2 e 3 Im – Exercise 3E 1 a z = 1,__i, −1, −i __ √ √     3 3 b z =  ___  + _ 12 i, −  ___  + _ 12 i, −i 2 __2 __   3 3√3 3√    i, −  _32 −  ____   i c z = 3, −  _32 +  ____ 2 2 d z = 2 + 2i, −2 + 2i, 2 − 2i, −2 − 2i e z = 1 __+ i, −1 + i, 1__ − i, −1 − i f z=√   3 − i, 2i, −√3   − i 2π  + i sin  ___ 2π , 2 a z = cos 0 + i sin 0, cos  ___ 7 7 4π , cos  ___ 6π  6π  + i sin  ___ 4π  + i sin  ___ cos  ___ 7 7 7 7 2π   + i sin −  ___ 2π   , cos −  ___ 4π   + i sin −  ___ 4π   , cos −  ___ 7 7 7 7 6π   + i sin −  ___ 6π   cos −  ___ 7 7 π   + i sin −  __ π     , b z = 2 cos −  __ 8 8 3π   + i sin   ___ 3π     , 2 cos    ___ 7π   + i sin   ___ 7π     , 2 cos    ___ 8 8 8 8 5π   + i sin −  ___ 5π     2 cos −  ___ 8 8 π  + i sin  __ π   , 2 cos  ___ 3π  + i sin  ___ 3π   , c z = 2 cos  __ 5 5 5 5 π   + i sin −  __ π     , 2(cos π + i sin π), 2 cos −  __ 5 5 3π   + i sin −  ___ 3π     2 cos −  ___ 5 5 __ __ π π 3π  + i sin  ___ 3π   , ___ ___ √ 2 cos     + i sin      , √ 2   cos  ___ d z =   12 12 4 4 __ − 7π     7π   + i sin   ____   cos −  ___ √2 12 12 __ π π      , ___ √ 2 cos −      + i sin −  ___ e z =   12 12 __ __ 5π   + i sin    ___ 5π     , √ 11π   + i sin    ____ 11π     ,   cos   ___   2 cos   ____ √2 12 12 12 12 __ 7π   + i sin −  ___ 7π       cos −  ___ √2 12 12 5π   + i sin −  ___ 5π     , f z = 4 cos −  ___ 18 18 7π   + i sin    ___ 7π     , 4 cos −  ____ 17π   + i sin −  ____ 17π     4 cos    ___ 18 18 18 18 ( ( ) ) ( ( ( ( ) ( ( ( ( ) ( ( ) ) ( ( ( ( )) ( ( ) ( )) ) ( ( ( ( ) ) ( )) ( ( ) ( ( ) )) ( ) ( ) ) )) ( 1   __ 4 0.23i 1   __ 4 1.80i 1   __ 4 −1.34i c 5 a b ) ( )) 6 a b ) πi 2e 3 ( )) ( 5πi 2e 6 ) )) ( ( ) ( )) ( ( ) ( )) ( ( ) ( ( ) ( )) ( )) ( ( ) ( ( ) ( ( – 2e πi 3 Re )) )) – 2e __ 7πi ____ __ __ πi ___ 2πi 3 5πi ____ __ 11πi _____ −     −             7 √ 2 e 12 , √ 2  e 12, √ 2  e 12 , √ 2  e 12 Im 5πi 2e 12 1   __ 4 −2.91i , 5 e __ , 5 e __ , 5 e 3 a z = 5 __e −0.29i, √ 1.80i, √ −2.39i b z=√  e 3 3  e  e 3 __ __ __ __ c z=√  e 2 0.57i__, z = √ 2  e2.14i, __ z=√ 2  e−1.00i, z = √ 2  e−2.57i √ √     3 3 4 a z = −   __12 +   ___  i, −2, −  __12 −   ___  i 2 2 11πi π 2 2e 12 π 2 – 2e Z03_IAL_FP2_44655_ANS_183-229.indd 191 Im 3 1 + i 2 2 π 2 π 2 Re – 2e πi 12 7πi 12 25/04/2019 08:52 192 ANSWERS 1   2  = 2 cos 6θ + 6 cos 2θ b z2 + __ ( z ) c a = _ 14 , b = _ 34  __ π 8 a r=√ 8  , θ = __     6 2πi 8πi  4πi   ____   ____ −  ____ b w = 4e 9 , w = 4e 9 , w = 4e 9 __  πi  ____  3πi  ____  5πi  3  __6π ∫ b Expressing as a product_of the linear factors: _ _ _ √ √ √   ___     √ 2   2 2 2 ___ (z + 1)(z − i)(z + i)(z −     −     i)(z −  ___  +  ___  i) 2 _2 2 2 _ _ _ √ √   √ 2     √ 2   2 2 z +  ___  +  ___  i)(z +  ___  −  ___  i) ( 2 2 2 2 _ _ 2 z + 1)(z  2+ √ 2   z + 1) = (z + 1)(z  2+ 1)(z  2− √   2 4 = (z + 1)(z  + 1)(z  + 1) Therefore (z2 + 1) and (z4 + 1) are factors. Challenge πi   __ πi  −  __ 2πi   ____ 2πi  −  ____ 3 a 1, e 3 , e 3 , e 3 , e __   π d   cos32 θ dθ =   (_  4 cos 6θ + _ 4cos 2 θ)dθ ____  7πi  9 ae 4 , i, e 4 , −1, e 4 , −i, e 4 , e  πi 0 6 6 1 3 0 _   π 3 √ 1 sin 6θ +  _38 sin 2θ]    = __   16   3  = [  __ 24 6 __ 0 1 6 a If z = cos θ + i sin θ,then 2 cos θ = z + __     z 5 1 So 2  5 cos  5 θ = ( z + __    )  z 1 1 1 1 1 ___ ___ 2 ___ 5 5    ) + 5C2z  3 ___ = z  5 + 5C1z  4(__ (  z  2 ) + C3z  (  z  3 ) + C4z(  z  4 ) +   z  5  z 10 5 ___ 1     + ___   3  +   5  = z  5+ 5z  3+ 10z + ___ z  z  z 1 1 1   5  + 5 z  3 + ___   3  + 10(z + __    ) = z  5 + ___ ( ( z z  ) z  ) 1 b Rewrite the equation as ( 1 + __    )  = 1. z kπi ___     1 Then 1 + __     = e 3 for some k ∈ ℤ, by a. z ___  kπi  1 So, __     = e 3 − 1 z ∫ = 2 cos 5θ + 5(2 cos 3θ) + 10(2 cos θ) So 32 cos  5 θ = 2 cos 5θ + 10 cos 3θ + 20 cos θ 1 cos  5 θ = __  16 (cos 5θ + 5 cos 3θ + 10 cos θ) __ b   16 15 Chapter review 3 1 a e   iθ= cos θ + i sin θ, e  −iθ= cos θ − i sin θ e  iθ + e  −iθ= 2 cos θ, so cos θ = __  12 (e  iθ + e  −iθ) b cos A cos B = __  12 (e   iA + e  −iA) × __   12 (e   iB + e  −iB) =  __14 (e   iA + e  −iA)(e  iB + e  −iB) = __   14 (e   i(A+B) + e  i(A−B) + e  i(B−A) + e  −i(A+B)) = __   14 ((e  i(A+B) + e  −i(A+B)) + (e  i(A−B) + e  −i(A−B))) = __   14 (2 cos(A + B) + 2 cos(A − B)) cos(A + B) + cos(A − B)          =   _______________________ 2 2 n = 1; LHS = r(cos θ + i sin θ) RHS = r  1(cos θ + i sin θ) = r(cos θ + i sin θ) As LHS = RHS, the equation holds for n = 1. Assume the equation holds for n = k, k ∈ ℤ  +. i.e. z  k = r  k (cos kθ + i sin kθ) With n = k + 1, the equation becomes: z  k+1 = z  k × z = rk(cos kθ + i sin kθ) × r(cos θ + i sin θ) = rk+1((cos kθ cos θ − sin kθ sin θ) + i(sin kθ cos θ + cos kθ sin θ)) = rk+1(cos(k + 1) θ + i sin (k + 1) θ) by the addition formulae. Therefore, the equation holds when n = k + 1. 1 − 3 tan2 θ 1 − 3 cot−2 θ b cot 3θ =  ______________               =  _______________   3 3 tan θ − tan θ 3 cot−1 θ − cot−3 θ 3 cot θ − 3 cot θ         =   _____________ 3 cot2 θ − 1 3 cos 7x + i sin 7x 4 a 16 b 256 5 a Let z = cos θ + i sin θ z  n = (cos θ + i sin θ)  n= cos nθ + i sin nθ 1 ___   n  = z  −n = (cos θ + i sin θ)  −n= cos (−nθ) + i sin (−nθ) z  = cos nθ − i sin nθ 1 ⇒ z  n + ___   n  = cosnθ + i sinnθ + cosnθ − i sinnθ = 2 cosnθ z  1 7 a If z = cos θ + i sin θ, then 2i sin θ = z − __     z 6 1 So (2i)  6 sin  6 θ = ( z − __    )  z 1 1 1 6 5 __ 3 ___ 6 6 = z  − C1z  (   ) + 6C2z  4 ___ (  z  2 ) − C3z  (  z  3 ) z 1 1 1 ___ ___ 6 + 6C4z  2 ___ (  z  4 ) − C5z(  z  5 ) +   z  6  15 6 1   2  −  ___4  + ___     = z  6− 6z  4+ 15z  2− 20 + ___ z  z  z  6 1 1 1   6  − 6 z  4 + ___   4  + 15 z  2 + ___   2  − 20 = z  6 + ___ ( ( ( z  ) z  ) z  ) = 2 cos 6θ − 6(2 cos 4θ) + 15(2 cos 2θ) − 20 So, − 64 sin  6 θ = 2 cos 6θ − 12 cos 4θ + 30 cos 2θ − 20 1 b c os  6 θ ≡ __   32   (cos 6θ + 6 cos 4θ + 15 cos 2θ + 10) π __ c     4 8(cos θ + i sin θ)  6= cos 6θ + i sin 6θ = cos  6 θ + 6C1 cos  5 θ(i sin θ) + 6C2 cos  4 θ (i sin θ)  2 + 6C3 cos  3 θ (i sin θ)  3 + 6C4cos  2 θ (i sin θ)  4 + 6C5cos θ (i sin θ)  5 + (i sin θ)  6 = cos  6 θ + 6i cos  5 θ sin θ − 15 cos  4 θ sin  2 θ − 20i cos  3 θ sin  3 θ + 15 cos  2 θ sin  4 θ + 6i cos θ sin  5 θ − sin  6 θ Equating imaginary parts gives sin 6θ = 6 cos  5 θ sin θ − 20 cos  3 θ sin  3 θ + 6 cos θ sin  5 θ = 2 sin θ cos θ(3 cos  4 θ − 10 cos  2 θ sin  2 θ + 3 sin  4 θ) = sin 2θ(3 cos  4 θ − 10 cos  2 θ(1 − cos  2 θ) + 3 (1 − cos  2 θ)  2) = sin 2θ(3 cos4 θ − 10 cos2 θ(1 − cos2 θ) + 3(1 − 2 cos2 θ + cos4 θ)) = sin 2θ(16 cos  4 θ − 16 cos  2 θ + 3) 9 a(cos θ + i sin θ)  5= cos 5θ + i sin 5θ = cos  5 θ + 5C1cos  4 θ(i sin θ) + 5C2cos  3 θ (i sin θ)  2 + 5C3cos  2 θ (i sin θ)  3 + 5C4cos θ (i sin θ)  4 + (i sin θ)  5 = cos  5 θ + 5i cos  4 θ sin θ − 10 cos  3 θ sin  2 θ − 10i cos  2 θ sin  3 θ + 5 cos θ sin  4 θ + i sin  5 θ Equating real parts gives cos 5θ = cos  5 θ − 10 cos  3 θ sin  2 θ + 5 cos θ sin  4 θ = cos  5 θ − 10 cos  3 θ(1 − cos  2 θ) + 5 cos θ (1 − cos  2 θ)  2 = cos  5 θ − 10 cos  3 θ(1 − cos  2 θ) + 5 cos θ(1 − 2 cos  2 θ + cos  4 θ) = 16 cos  5 θ − 20 cos  3 θ + 5 cos θ _ _ b –1, __   14 (1 + √ 5 ) ≈ 0.809,  __14 (1 − √ 5 ) ≈ − 0.309 Online Worked solutions are available in SolutionBank. Z03_IAL_FP2_44655_ANS_183-229.indd 192 25/04/2019 08:52 ANSWERS 10 a Let z = cos θ + i sin θ 1 5 __ 5 5 5 5 ( z −  z ) = (2i sin θ)  = 32 i  sin  θ = 32i sin  θ 1 1 2 = z  5+ 5 z  4(−  __ )+ 10 z  3 (−  __ ) z z 5 1 3 1 4 1 __ __ 2 + 10 z  (−    ) + 5z (−    ) + ( −  __ )  z z z 10 5 ___ 1     + ___   3  −   5  = z  5− 5 z  3+ 10z − ___ z z  z  1 1 1   5 )− 5(z  3 − ___   3 )+ 10(z − __    ) = ( z  5 − ___ z z  z  = 2i sin 5θ − 5(2i sin 3θ) + 10(2i sin θ) So 32i sin  5 θ = 2i sin 5θ − 10i sin 3θ + 20i sin θ 1 ⇒ sin  5 θ = __  16 (sin 5θ − 5 sin 3θ + 10 sin θ) π ___ 5π __ b 0,    ,     6 6 11 a (cos θ + i sin θ)  5= cos 5θ + i sin 5θ = cos  5 θ + 5i cos  4 θ sin θ + 10 i  2 cos  3 θ sin  2 θ + 10 i  3 cos  2 θ sin  3 θ + 5 i  4 cos θ sin  4 θ + i  5 sin  5 θ ⇒ cos 5θ + i sin 5θ = cos  5 θ + 5i cos  4 θ sin θ − 10 cos  3 θ sin  2 θ − 10i cos  2 θ sin  3 θ + 5 cos θ sin  4 θ − i sin  5 θ Equating the real parts: cos 5θ = cos  5 θ − 10 cos  3 θ sin  2 θ + 5 cos θ sin  4 θ = cos θ(cos  4 θ − 10cos  2 θ(1 − cos  2 θ) + 5(1 − cos  2 θ)  2) = cos θ(16 cos  4 θ − 20 cos  2 θ + 5) b If cos 5θ = 0, then cos θ(16 cos  4 θ − 20 cos  2 θ + 5) = 0 If x = cos θ, then x(16x  4− 20x  2_ + 5) = 0 which has _   20 ± √ 80  _______ 5 ± √ 5 _________ 2 solutions x = 0 and x  =     =    , by the 32 8 quadratic formula. π π Since θ = ___    is a solution to cos 5θ = 0, x = cos  ___  10 10 must be a solution to x(16x  4− 20_ x  2+ 5) = 0. 5 5 ± √   π Since x ≠ 0, cos  2 ___     = x  2 =  _______  , for some choice ( 10 ) 8 of sign. 3π To find which, note that θ = ___    gives another 10 3π π solution and cos  ___  . cos  ___ by looking at the graph. 10 10 π Hence θ = ___    corresponds to the larger of the two 10 _ 5 5 + √   π     =  _______   solutions and cos  2 ___ ( 10 ) 8 __ __     5 − √ 5 5 − √ 5 3π 7π    ) =  _______    ) =  _______  , cos2(___  , c cos2(___ 10 8 __ 10 8   5 + √ 5 9π   cos2( ___ ) =  _______ 10 8 3 tan θ − tan3 θ         12 a tan3θ ≡   ______________ 1 − 3 tan2 θ __ π π   cos −  __  + i sin −  __  13 a 4√ 2 ( 4 )) ( ( 4) __ b z = √ 2 e c πi   −  ___ 20 __ ____  7πi  __ ____  3πi  __ , √ 2 e 20 , √ 2 e 4 , √ 2 e 9πi  −  ____ 20 __ , √ 2 e 17πi  −  _____ 20 193 _ iπ   −  ___ 12 14 a √ 2 e b _ ____  7iπ  __ , √ 2 e 12 , √ 2 e Im Im B M 2π 3 2π 3 2π 3 A Re C _ √   2 π c r =  ___ , θ = __     4 2 i d −  __  8 Challenge 6 1 Rewrite the equation as (1 + __    )  = 1. z ___  kπi  1     = e 3 for some k ∈ ℤ, since it is a sixth root of Then 1 + __ z unity. ___  kπi  1 So __     = e 3 − 1 z kπi  −  ___ ie 6   1 1 1   = _____________   = _____________        = −  _______        ___ z = _______ kπi kπi kπi kπi kπi ___ ___ ___ ___         kπ          −    kπ  2 sin  ___ e 3 − 1 e 6 (e 6 − e 6 ) e 6 (2i sin  ___ 6 6) kπ kπ kπ kπ ___ ___ ___ ___ i(cos     − i sin    ) sin     + i cos     kπ 6 6 6 6 1  − __             = −  _________________  1  i cot  ___    = −  _______________   = −  __ 2 2 6 kπ kπ ___ ___ 2 sin     2 sin     6 6 1  + it for t ∈ ℝ. So the points lie on the straight line z = −  __ 2 CHAPTER 4 Prior knowledge check Im 1 z1 4 3 2 z2 1 –4 –3 –2 –1 O –1 –2 –3 9iπ  −  ____ 12 1 2 3 4 Re z3 –4 z2 z3 O z1 Re z5 z4 Z03_IAL_FP2_44655_ANS_183-229.indd 193 25/04/2019 08:52 194 ANSWERS g 2 Im h Im 6 4 Im –7 –3 O 1 3 Re 2 (1, 1) 5+i 1 (–3, –4) O 2 1 –1 4 3 6 5 –4 Re 6 Re –8 –4 –2 i 5|i –3 Im 5 O –4 10 Re –1 3 Im 4 –6 3 –1 + i 3 2 –11 1 2 a –4 (5, –6) –3 –2 –1 O –1 –1 – i 3 –2 2 1 Im 4 Re 3 8 (5, 4) –3 O –4 Re ___ Exercise 4A 1 a Im b Im 10 6 O 6 O c Im O d 1 3 3 O 5 Re Re Im b X (0, –3) Im (0, 9) 3 1 O –2 (0, 4) (–1, 0) O –1 –5 O (0, –1) 5 Re 8 C(4, 3) f Im (5, –7) b ( x − 5)  2 + ( y + 7)  2= 25 π c 2 arctan(__   75 ) − __    = −0.330 rad (3 s.f.) 2 4 a (x – 4)2 + (y – 3)2 = 82 Im –2 e 5 Re –3 2 O Re 10 Re ___ √39  )i (4 + √ 39  b i __ )i and (4 − __ ii 5 + 4 √3 and 5 − 4 √3   3 a Im Re 4 Re Y c |z|min = 3, |z|max = 13 Online Worked solutions are available in SolutionBank. Z03_IAL_FP2_44655_ANS_183-229.indd 194 25/04/2019 08:52 ANSWERS 5 a 195 j Im Im 2 3+2 (0, 11) (10, 5) 2 3 (–2, 2 3) (–6, 1) 2 3–2 O –4 (11 , 0) 4 O Re Re π b __  2 6 a Im y = –4x + 11 c 2.51 rad b 7 a Im __ 5 9 √  b ____  10 Im (0, 6) O 2 6 Re O –8 Re 4 ( 32 , 3) (0, 94 ) (3, 0) (– 92 , 0) O x = –2 x=4 c (0, 11 ) 2 8 O Im 8 a Im d Im Re Re y = 2.5 y = –3 –6 O (3, 1) Re O –3 (11 , 3 0) Re ___ e f Im (2, 2) h Im (7, 5) –1 O –3 Im –4 O 2 –2 Re 3 6 Re (3, –2) (8, –2) 7 Re –5 (0, –6) 10 a Im b Im y = 3x – 6 Im 4 Re 1 x=2 i (0, 1) c Im (–4, 2) O O 5 Re (2, 0) O –3 (0, 9) (0, 5) –7 2 –6 y = –x (–3, 5) Re y= (–2, –2) g –1 Re O Im 3 9 a –4 O 11 √13  c _______   13 Im b b y = −  __32 x + __  11   2 Im 3 (0, 6) π 4 π 3 O (0, 94 ) Re (–3, 0) O Re ( 92 , 0) (–3, 0) O Z03_IAL_FP2_44655_ANS_183-229.indd 195 y = – 12 x + 94 Re 25/04/2019 08:52 196 ANSWERS c Im d O (–2, –2) π 4 π 2 O __ Im Re b z = (2 + √ 3 ) + 3i 15 a,b Im (0, 8) y=5 π 4 (0, 2) 1 f Im Im 3π 4 2 |z – 2i| = |z – 8i| –4 Re e arg(z – 2 – i) = π 4 O O Re π 2 c z = 6 + 5i Im 16 a,b Re |z – 3 + 2i| = 4 (1, 1) O g Re h Im 2π 3 Re (3, –2) Re (3, –4) π 2 (1, –3) i __ 17 a z = 4 + 4i√3  O 18 a Im 2π b arg(z − 8) =___     3 Re π dmin 3 O (–4, 0) Im O __ __ 3π 2 11 a arg(z – 1) = – π 4 c a = 3 + 2√2 , b = −2 − 2√2   Im (0, 4) –4 Re Im O O π 4 O 1 Re Re __ b Hence the minimum value of |z| is |z|min = 2√3   19 a Im –2 2 b Use the cosine__rule to find + 2|z| − 5 = 0, solve to get |z| = − 1 + √ 6   __ __ 12 a |zmax| = 6 √2 + 4 and |zmin| = 6 √2 − 4 b (− 2.38, π) 13 a Im (–8, 4) |z|2 O Re b Im 2 –4 O 5 Re O (7, 2) θ Re Maximum value of arg(z + 15 − 2i) = θ θ 2 2 sin(__   ___     ) = ________   _______  = ____ 2 √ 2  2 + 7  2  √ 53  2   ___  ⇒ θ = 2 arcsin ____ (√ 53 ) b z = −4 + 3i 14 a Im 4 __ __ __ __ c (− 8 + √ 2  , 4 − √ 2  )and (− 8 − √ 2  , 4 + √ 2  ) 2 O Challenge 0.37 < θ < 2.77 (2, 2) 2 4 Re Online Worked solutions are available in SolutionBank. Z03_IAL_FP2_44655_ANS_183-229.indd 196 25/04/2019 08:52 ANSWERS Exercise 4B f 1 a Im 197 Im (x – 83) + y = 169 2 (x – 6)2 + y2 = 9 3 3 O 9 Re 6 O 4 3 ( ) 2 12 3 8 3 Re –3 b + y = 256 (x + 19 15) 225 2 2 c Im π 4 3 2 – 3 15 –7 3 z 2 a arg z + 3 = π 4 Im O – 19 15 Re O –3 2 –3 ( zz +– 3i4 ) = 6π b arg Im O Re Re Im π 6 –1 3 3 –5 3 O –4 Re c Im ( –3 x2 + y + 5 3 ) 2 2 = 16 9 O Re d Im π 3 O ( z –z2 ) = 3π arg Re d (14, –5) Im 3 5 O (x – 14)2 + (y + 5)2 = 100 e Im Re π 4 (x2 – 4)2 + (y – 6)2 = 20 ( zz –– 3i5 ) = 4π arg e Im (4, 6) O π 3 O Z03_IAL_FP2_44655_ANS_183-229.indd 197 Re arg z – arg(z – 2 + 3i) = π 3 2 + 3i Re 25/04/2019 08:53 198 ANSWERS f ( zz –+ 4i4 ) = 2π arg Im 4 –2 + 2i O Re –4 3 a Substituting x + iy for z and squaring gives (x + 1)2 + (y + 1)2 = 4((x + 4)2 + (y − 2)2) which can be rearranged to (x + 5)2 + (y − 3)2 = 8, which is the equation of a circle with centre (−5, 3). 8 (x + 1)2 + (y + 2)2 = 8, y . 0 z+2 π 9 a arg(_____         = __ z + 5) 4 z−i π     b arg( ______ ) = __ z − 4i 6 z−6−i 2π         = ___ c arg(_________ z − 1 − 2i ) 3 10 a Substituting x + iy for z and squaring gives (x + 3)2 + y2 = 9((x − 5)2 + y2) which can be rearranged to x2 + y2 − 12x + 27 = 0. b Im |z + 3| = 3|z – 5| _ b 2√ 2 4 Im a argz – arg(z + 4) = π 4 3 O π 4 b c d e 5 a O –4 Re (−2, 2) _ 2√ 2 (x + 2)2 + (y − 2)2 = 8 6π + 4 Substituting x + iy for z and squaring gives x2 + y2 = 4((x + 4)2 + y2) which can be rearranged to (x + _  16 )2 + y2 = _  64 , 3 9 16 which is the equation of a circle with centre ( −  _ , 0) 3 8 _ and radius  3 b Im |z| = 2|z + 4| _ π π c 3 √ 3 cos  __ + i sin  __  ( 6 6) 11 a z 1= 6i, z 2 = 3, k = 2 b Substituting x + iy for z and squaring gives x2 + (y − 6)2 = 4((x − 3)2 + y2) which can be rearranged to x2 + y2 − 8x + 4y = 0. _ _ 3π c α = −  ___  d (4 − √ 10 , − 2 − √ 10 ) 4 Challenge The locus is an ellipse with foci at a and at −a, and major axis of length b. Exercise 4C 1 a |z| < 3 O O ( ) 2 Re –2 O Re 2 Re –2 c c −  _83 < Im(z) < _   83 π 3π 6  __  < arg z < ___     4 4 7 a arg w – 8i = π 2 w+6 Im 5 b Im 8 3 (– 163, 0) Re 6 Im 4 2 Im –7 –6 –5 –4 –3 –2 –1O –2 8 1 2 Re –4 –6 (–3, 4) Im d –6 –6 O b ( x + 3)  2 + ( y − 4)  2= 25, x , 0, y . 0 π c a = __     , b = π 2 d − 8 , Re(z) , 0 O –2 4 Re Re –2 – 8i Online Worked solutions are available in SolutionBank. Z03_IAL_FP2_44655_ANS_183-229.indd 198 25/04/2019 08:53 ANSWERS Challenge Im 3 e 199 Im 2 O 1 –3 –2 –1 O –1 1 Re 3 Re 2 –2 (–5, –8) –3 Im f g –4 –2 O –2 2 2 Im –2 O –2 4 Re –4 –4 –6 –6 –8 –8 –10 –10 2 4 6 8 Re Exercise 4D 1 a b Im Im 2 R R 1 Im O 4 c O –1 Re d Im Re Im R O O Re O 1 Re –1 π 3 π 4 R Re Im 2 Im 3 1 2 3 |z| = |z – 6i| 3 –3 O Re O Re R 4 arg(z – 4 – 2i) = π 2 Im |z – 2| = |z – 6 – 8i| R O 5 __ |z| = 5 3 Im 3π 4 arg(z – 4 – 2i) = 0 Re R √ 2 x − 2√ 2  ii (x + 1)2 + y2 = 9 a i y = −2 __ __ b z = − √ 2  + 2i √ 2  or z = −2i Im c Im 3 O O –1 4 R 1 __ Re –3 π 4 O Re arg(z + 3) = π 4 3 Re –3 Z03_IAL_FP2_44655_ANS_183-229.indd 199 05/09/19 11:56 PM 200 ANSWERS 5 a Exercise 4E Im –1 O 1 a i Re w3 –1 (–1, –2) π 4 4 3 w2 2 w1 –2 1 R b Im v 1 u –3 –2 –1 O −3 ii Translation ( ) 2 b i v Re(z) = 7 2 O 5 R 7 6 a 4 9 Re 5π 7π b ___     , arg z , ___     6 6 Im 3 w3 4 3 2 w1 1 w2 |z + 6| < 3 –9 O –3 O Re –6 –3 b 16 7 a Im w3 π 4 8 8 a 2 3 4 5 u 4 w2 3 2 w1 Re(z) = lm(z) O 1 ii Enlargement by scale factor 2 with centre O c i v 1 1 u –5 –4 –3 –2 –1 O π __ ii Rotation    anticlockwise about O followed by 2 −2 translation ( ) 1 d i v Re Im π 3 5 2 w3 4 3 –1 O Re 1 –2 _ 16π b ____    − 4 √ 3 3 2 w1 w2 O 1 2 3 4 5 6 7 u ii Enlargement by scale factor 3 with centre O 0 followed by translation (   ) −2 2 w = 4z − 8 + 12i c 4 Challenge Im 4 2 O R 3 6 Re Online Worked solutions are available in SolutionBank. Z03_IAL_FP2_44655_ANS_183-229.indd 200 05/09/19 11:56 PM ANSWERS 201 | | 2w − 1 2w − 1 9 Rearrange to z = _______     , then _______ = 2 w w 1 ⇒ |2w − 1|= 2|w| ⇒ |w − _  2 | = |w| 3 w = 4iz 4 (u + 1)2 + (v − 3)2 = 64 5 a v This is the perpendicular bisector of (0, 0) and ( _12 , 0), so is a line in the w-plane. v l 5 |w – 2i| = 3 (0, 2) O O –3 1 2 u 3 u –1 u= 1 4 v b iw − i 2iw 10 a Rearrange to get z = −  ______  ⇒ z − i = −  ______  w−1 w−1 2w     = 1 ⇒ 2|w| = |w − 1| So |z − i| = ______ w−1 Substituting u + iv for w and squaring gives 4(u2 + v2) = (u − 1)2 + v2, which can be rearranged to give (u + _  13 )2 + v2 = _  49 , which is the equation of a 1 _ circle with centre (−  3 , 0) and radius _  23 b v arg (w – i) = π 4 (0, 1) u O c | π 4 v v = 2u + 4 | (u + 13) + v = 49 4 2 2 3 –2 (– 13 , 0) O u O 6 a Circle with centre (0, 0) and radius _  12 π b Half-line from (0, 0) at an angle of −  __  4 _ √ 5 c Circle with centre (−1, − _12 ) and radius  ___  2 7 a The circle in the z-plane is |z| = 3, so the corresponding locus in the w-plane will be such that |w| = |z|2 = 9, i.e. a circle with centre (0, 0) and radius 9. arg w = arg (z2) = 2arg z Thus, if z moves around the circle |z| = 3 once, w will move around the circle |w| = 9 twice. b The non-negative real axis: v = 0, u > 0 c The non-positive real axis: v = 0, u < 0 8 a i Substituting u + iv for w and squaring gives u2 + (v − 2)2 = 4(u2 + v2) which can be rearranged to u2 + (v − _  23 )2 = _  16 , which is the equation of a circle. 9 ii Centre (0,  _23 ) , radius _   43 b 2w − 3 11 Rearrange to get z = _______   . Substitute x + iy for z and w 2u  2− 3v + 2 v  2   u + iv for w and rearrange to get x = ______________         u  2+ v  2 3v _______ and y =   2   u  + v  2 Then the equation of line 2y = x gives 6v = 2u2 − 3v + 2v2, which can be rearranged to (u − _  34 )2 + (v − _32 )2 = _  45 , which is the equation of a circle 16 _ 5 3 √ 3 _ 3 _ ____ with centre (   4 ,  2 ) and radius     4 12 a v = 0 v b R O R O u v=0 v (0, 23 ) 2 u c |z| = 2 ⇒ 2|w + i| = |w − i|, then substituting u + iv for w and squaring gives 4(u2 + (v+ 1)2) = u2 + (v − 1)2, which can be rearranged to give u2 + (v + _   53 )2 = _  16 , 9 4 3 u which is the equation of a circle with centre (0, −  _35 ) and radius _  43 w − 3i 13 Rearrange to get z = ______    . |z| = 3 ⇒ 3|w − 4| = |w− 3i|, w−4 then substituting u + iv for w and squaring gives 9((u − 4)2 + v2) = u2 + (v − 3)2, which can be rearranged to give (u − _   92 )2 + (v + _   38 )2 = _  225 , which is the equation of 64 a circle with centre ( _   29 , −  _83 ) and radius _  15 8 Z03_IAL_FP2_44655_ANS_183-229.indd 201 25/04/2019 08:53 202 ANSWERS 1 − iw 14 a Rearrange to get z = ______   then substituting w u + iv for w and rearranging gives u u   2+ v  2 + v z =  _______  − i  ___________  , so the real axis, y = 0,    u  2+ v  2 ( u  2+ v  2 ) u  2+ v  2 + v  = 0 ⇒ u2 + v2 + v = 0, which    becomes ___________   u  2+ v  2 θ 3 3 1   ___  = ___   __  sin(__    ) = ________   _______  = ____ 2   √3  2 + 6  2  √45  √5 π π 1    + 2 arcsin ___   __  ⇒ __     + θ = __ (√ 2 2 5   ) 4 a Im  12 )2 = _  41 , which is a can be rearranged to u2 + (v + _ 4  12 ) and radius _ circle with centre (0, _  12 u  = 4 ⇒ u = 4u2 + 4v2 b The line x = 4 becomes  _______ u  2+ v  2 1  64 , which which can be rearranged to (u − _  18 )2 + v2 = _ is a circle with centre ( _   18 , 0)and radius _  18 4z*   2  = z + z* = 2Re(z) 15 w = z +  _4z = z + ____ |z| O b (3.96, 3.86) and (1.14, −1.03) c −π , θ , −0.41, 0.41 , θ , π Im 5 a Since |z| = 2, −2 < 2Re(z) < 2, so w ∈ [−4, 4]. k = 4. Re (5, 0) (–5, 5) ( 12 , 4) 1 − 3w 16 Rearrange to get z =  _______, then substituting u + iv w (6, 3) ( u  2+ v  2) v u − 3u  − 3v    , so the line       − i _______ for w gives z = _____________   2 2 u  2+ v  2 2x − 2y + 7 = 0 becomes 2(u − 3u2 − 3v2) + 2v + 7(u2 + v2) = 0. This can be rearranged to (u + 1)2 + (v + 1)2 = 2, which is the _ equation of a circle with centre (−1, −1) and radius √ . 2 5 (– 22 , 0) O Re __ √ 5  c ___  10 5 11 b y = ___    x + __     2 4 6 a y = __  12 x + 3 Challenge w = iz + 2i b 6 + 6i c Chapter review 4 Im 1 a (x + 1)2 + ( y − 1)2 = 1 Im b 2 (–1, 1) –2 |z + 1 – i| = 1 1 O –1 –6 Re d |z − 1|min = √ 5  − 1 __ |z|max = √ 2  + 1 __ |z − 1|max = √ 5  + 1 arg (z – 2 + 4i) = π 4 O Re 7 a i y=x−2 ii (x − 2)2 + y2 = 8 b −2i, 4 + 2i c |z – 2| = 2 2 Im |z – 3 + i| = |z – 1 – i| Re π 4 __ O __ c |z|min = √ 2  − 1 Im 3 π 4 __ 2 a (0, 54 ) O (2, –4) (2, 0) Re b 3√2   3 Im 3 8 a i x=2 ii line; perpendicular bisector of (0, 0) and (4, 0) 6 –3 π θ 2 O b i(x − _  16   + y  2 = _   64 3) 9 2 Re 16 ii Circle with centre ( _ , 0) and radius _  38 3 π Max value = __     + θ 2 Online Worked solutions are available in SolutionBank. Z03_IAL_FP2_44655_ANS_183-229.indd 202 25/04/2019 08:53 ANSWERS 9 a Im 13 a A circular arc anticlockwise from 5 + 2i to 1 + 6i. π Since θ = __     , it is a semicircle. Centre is (3, 4) and 2 _ radius_is 2 √ . 2 b 5 + 2√ 2 14 a Im |z – 2 + i| = 3 O Re (2, –1) 203 3 3 b m = √ 3 π c arg(z + i) = __     _3 √ 3 − 1) d a =  _12 + i( _ 2 10 a (x −  _43 )2 + y2 = _  25 9 _ (2, 3) O _ 4 + √ 34 4 + √ 34  + i( ________   b  ________ ) 6 6 c Im Re R O |z + 2| = |2z – 1| 11 a Im c yes v u 1   = _______ 15 a z = __  − i _______   , so the image of x = _    12   w u  2+ v  2 ( u  2+ v  2) u   = _   1 , which can be rearranged to is _______   2 u  + v  2 2 (u − 1)2 + v2 = 1, which is the equation of a circle with centre (1, 0) and radius 1. b v (u – 1)2 + v2 = 1 R ( 43 , 0) Re b 5π argz = π 4 O R 2 _ ( ) arg z – 4 – 2i = π 2 z – 6i 6 u 16 a Im 2 –2 O Re –2 (2, 4) (0, –4) (4, 2) |z + 4i| = 2 –6 O Re _ b 2√ 2 12 a Substituting x + iy for z and squaring gives 4((x + 3)2 + y2) = (x − 3)2 + y2 which can be rearranged to x2 + y2 + 10x + 9 = 0. b Im 2|z + 3| = |z – 3| 4 5 c tan θ = ±  _34 Z03_IAL_FP2_44655_ANS_183-229.indd 203 O Re b 6 c i ii iii iv Circle with centre (0, −8) and radius 4 Circle with centre (4, 0) and radius 2 Circle with centre (−4, 0) and radius 2 Circle with centre (0, 4) and radius 2 2 − iw 17 a z = ______     w−1 − u(u − 1) − v(2 + v) (2 + v) (u − 1) − uv  + i ___________________                =  __________________ ( (u − 1)  2+ v  2 ) (u − 1)  2+ v  2 (2 + v) (u − 1) − uv  = 0,    So the line x = 0 has image __________________      (u − 1)  2+ v  2 and this can be rearranged to v = 2u − 2, which is a line in the w-plane. b y = x has image with equation (2 + v)(u − 1) − uv = −u2 + u − 2v − v2 which can be rearranged to (u + _  12 )2 + (v + _  12 )2 = _  52 , which is the _equation of_a circle with centre ( _   12 ,  _12 ) _ 10 and radius      10   =  _ √ _52   = √ _ 4 2 √ 25/04/2019 08:53 204 ANSWERS 4 − iw 18 z = ______    , so the image of the circle |z| = 1 is such that w+1 4 − iw    = 1 ⇒ |w + 4i| = |w + 1|, and then substituting ______ w+1 u + iv for w gives u2 + (v + 4)2 = (u + 1)2 + v2, which can be rearranged to 2u − 8v − 15 = 0, which is the equation of line l. w−6 19 Rearrange to get z =  ______ . |z| = 2 ⇒ 2|w + 3i| = |w − 6|, w + 3i then substituting u + iv for w and squaring gives | 6 a y= 2– x y | y=– x= 1 2 2 x–1 O 4(u2 + (v + 3)2) = (u − 6)2 + v2, which can be rearranged to give (u + 2)2 + (v + 4)2 = 20, which is the of _ _equation a circle with centre (−2, −4) and radius √ 20 = 2 √ . 5  52 , b = 0, c = −  _52 20 a a = _ b ω=5 x 2 _ 13 13  17 21 a a = _ , b = − _ , c = − _ 5 5 5 b 3 ±  _45 √ 10 22 a v = u − 1 b x + y + 1 = 0 has image v + (u − 1) + ((u − 1)2 +v2) = 0 ⇒ u2 + v2 − u + v = 0 This can be written as (u −  _12 )2 + (v + _  12 )2 = _  12 , which 1 _ is the equation of a circle with centre ( 2 , −  _12 ) and c x , 0, 1 , x , 3 b (0, 2) and (3, −1) y 7 a _ 2x (x + 1)2 y= √ 2 radius  ___  2 c v –1 O 2 l x 1 O y= u 4x 2–x ( 12 , – 12 ) –1 C 20 b (0, 0) and ( −  __52 , −  __ 9) Challenge 1 a Im __ b 3 √2  − 3 c{x : x < − __52 } ∪ {x : x . 2} ∪ {x : x = 0} 8 a O z = –3 – 3i y y = | 3x – 2| Re 5 y = | x – 5| 2 f(z) = −iz* + 1 + i 2 Review exercise 1 O 1 x , −4, −1 , x , 2 x 5 2 3 b(−  __32 , __  13 , __   7 , __  13 2 ) (4 4 ) 2 {x : x , 0} ∪ {x : 2 , x , 4} 3 {x : −3 , x , 0} ∪ {x : x . 4} c x , −  __32 , x . __   74 4 {x : −  __12 , x , 0} ∪ {x : x . 3} 5 {x : x , −4k} ∪ {x : −2k , x , 0} ∪ {x : x . 2k} 9 a b x.2 y 2 y = | x + 2| –2 O x Online Worked solutions are available in SolutionBank. Z03_IAL_FP2_44655_ANS_183-229.indd 204 25/04/2019 08:53 ANSWERS 18 Using partial fractions, 2 2 2  ____________     = _____   − _____       (r + 1)(r + 2) r + 1 r + 2 y 10 a y = | x – 2a| Using the method of differences, n n 2 2 2 2 2     − _____     = __       =  ∑(_____     − _____    ∑____________   r + 2) 2 n + 2 r=1 r + 1 r=1 (r + 1)(r + 2) n + 2 − 2 _____ n 2   = _________     =     = 1 − _____   n+2 n+2 n+2 2a O b x , __   13 a x 2a 19 Using partial fractions, 4 2 2     = _____  ____________   − _____       (r + 1)(r + 3) r + 1 r + 3 Using the method of differences, n n 2 2 4     − _____          =  ∑(_____  ∑____________ r + 3) r=1 (r + 1)(r + 3) r=1 r + 1 5 2 2 _____ 2 2 2 2 =  __  + __     −       − _____     = __     − _____   − _____     2 3 n+2 n+3 3 n+2 n+3 5(n + 2)(n + 3) − 3(n + 3) − 6(n + 2)            = __________________________________   3(n + 2)(n + 3) 5n  2+ 25n + 30 − 6n − 18 − 6n − 12            = __________________________________   3(n + 2)(n + 3) 2 n(5n + 13) 5 n  + 13n                   = ______________ = ______________ 3(n + 2)(n + 3) 3(n + 2)(n + 3) __ 11 {x : x , 6 − 2√3 } ∪ { x : 4 , x , 6} 12 a y y=x y = | 2x – 1| 1 O x 1 2 1 __ 1 __ b (  3 ,  3 )and (1, 1) c {x : x .  __13 } ∪ { x : x . 1} 13 {x : −5 , x ,  __13 } 14 −  __1 a < x < −  __1 a 3 15 a 7 y y = |x2 – 6x + 8| O 2 3 x 23 a Using partial fractions, 4 2 2  _______  = __         − _____ r(r + 2) r r + 2 Using the method of differences, n n 2 2 4 2 __ 2 2 2     − _____     =  __   +     − _____    ∑   _______  =   ∑ (__   − _____     r + 2) 1 2 n + 1 n + 2 r=1 r r = 1 r(r + 2) 2 2   − _____     = 3 − _____   n+1 n+2 3(n + 1)(n + 2) − 2(n + 2) − 2(n + 1)       = __________________________________       (n + 1)(n + 2) 3n  2+ 9n + 6 − 2n − 4 − 2n − 2           = _____________________________   (n + 1)(n + 2) 2 n(3n + 5) 3n  + 5n             = _____________       = _____________ (n + 1)(n + 2) (n + 1)(n + 2) The curve meets the x-axis at (2, 0) and (4, 0). The line meets the x-axis at (3, 0). 7 __ 3 __ b (  2 ,  4 ), (5, 3) x ,  __72 , x . 5 y 16 a y = |(x – 2)(x – 4)| 8 6 O 2 3 x 4 y = 6 – 2x __ __ b 2 − √ 2 and 4 − √ 2   17 a x = −  __52 , x = −  __74 or x = 1 __ __ c 2 − √ 2  , x , 4 − √ 2  b {x : x , −  __52 } ∪ {x : −  __74 , x , 1} Hence, a = 5, b = 13, c = 3 (r + 1)( r + 1) − r(r + 2) r+1 r 20 a _____  −   _____  = ____________________            r+2 r+1 (r + 1)(r + 2) 1 r  2+ 2r + 1 − r  2− 2r ____________         =       = ___________________   (r + 1)(r + 2) (r + 1)(r + 2) n   b  ________ 2(n + 2) 1   −   _____ 2   +   _____ 1   21 a f(x) =   _____ x+1 x+2 x+3 1   +   _____ 1   __   −   _____ b  1 6 n+2 n+3 2r − 1   22 a   ________ r2(r − 1)2 b Using the method of differences, n n 1 1 2r − 1 1 1 1   =  ∑ _______   − ___   2  = ___     2     − ___    = 1 − ___   2   ∑ _________ 2 r  ) 1  2 n  2 r=2 r  (r − 1)  r=2 ( (r − 1)  2 n  2y = 3x – 9 4 205 Hence a = 3, b = 5 b 0.0398 24 a Using partial fractions, 2 1 1  _______   = ______     − ______     4r  2− 1 2r − 1 2r + 1 Using the method of differences, n n 2 1 1 1   =   ∑( ______       ∑  _______   − ______     = 1 − _______   2 r = 1 2r − 1 2r + 1 ) 2n + 1 r = 1 4r  − 1 20 b ___ 861 Z03_IAL_FP2_44655_ANS_183-229.indd 205 25/04/2019 08:53 206 ANSWERS 25 a A = 24, B = 2 b Using the identity from part a, n n r=1 r=1  ∑ (24r  2+ 2) = ∑   ((2r + 1)  3 − (2r − 1)  3) n n n r=1 r=1 r=1   2 =  ∑((2r + 1)  3 − (2r − 1)  3) 24  ∑r  2 + ∑ Using the method of differences, n 24  ∑r  2 = 8n  3+ 12n  2+ 6n + 1 − 1 − 2n r=1 = 8n  3+ 12n  2+ 4n = 4n(n + 1)(2n − 1) 4n(n + 1) (2n − 1) __ 1      =     n(n + 1)(2n + 1) ∑   r  2 =  _________________ r=1 24 6 c 194 380 26 Using partial fractions, 1 1 1 1  _____________     = ___     + ________       − _____   r(r + 1)(r + 2) 2r r + 1 2(r + 2) n Using the method of differences, 2n 2n 1 1 1 1     − _____     + ________        = ∑  ___ ∑   _____________ r + 1 2(r + 2) ) r=1 ( 2r r=1 r(r + 1)(r + 2) 1 1 1 =  __  − _________     + ________     4 2(2n + 1) 4(n + 1) (n + 1)(2n + 1) − 2(n + 1) + (2n + 1)            = __________________________________   4(n + 1)(2n + 1) 2 2n  + 3n + 1 − 2n − 2 + 2n + 1           = _____________________________   4(n + 1)(2n + 1) n(2n + 3) 2n  2+ 3n             = _______________       = _______________ 4(n + 1)(2n + 1) 4(n + 1)(2n + 1) Hence a = 2, b = 3, c = 4 (r − 1) r(r + 1) + (r + 1) − r 1 1            = _________________________       − _____   27 aRHS = r − 1 + __ r(r + 1) r r+1 2 r( r  − 1) + 1 _________ r  3 − r + 1      =     = LHS =  ____________ r(r + 1) r(r + 1) n(n2 + 1) b _________  2(n + 1) 1   28 1 −   _________ 3n(n + 1) cos 2x + i sin 2x cos 2x + i sin 2x     = __________________         29 ______________ cos 9x − i sin 9x cos (−9x) + i sin (−9x) e  2xi   = e  11xi= cos 11x + i sin 11x =  ____ e  −9xi Hence n = 11 30 a cos 5θ + i sin 5θ = (cos θ + i sin θ)5 = cos5 θ + 5i cos4 θ sin θ – 10 cos3 θ sin2 θ – 10i cos2 θ sin3 θ + 5 cos θ sin4 θ + i sin5 θ Equating real parts: cos 5θ = cos5 θ – 10 cos3 θ sin2 θ + 5 cos θ sin4 θ = cos5 θ – 10 cos3 θ(1 – cos2 θ) + 5 cos θ(1 – cos2 θ)2 = 16 cos5 θ – 20 cos3 θ + 5 cos θ b x = 0.809, −0.309, −1 31 a cos 5θ + i sin 5θ = (cos θ + i sin θ)5 = cos5 θ + 5i cos4 θ sin θ – 10 cos3 θ sin2 θ – 10i cos2 θ sin3 θ + 5 cos θ sin 4θ + i sin 5θ Equating imaginary parts: sin 5θ = 5 cos4 θ sin θ – 10 cos2 θ sin3 θ + sin5 θ = sin θ(5 cos4 θ – 10 cos2 θ sin2 θ + sin4 θ) = sin θ(5 cos4 θ – 10 cos2 θ(1 – cos2 θ) + (1 – cos2 θ)2) = sin θ(16 cos4 θ – 12 cos2 θ + 1) π 3π b 0,  __ ,  ___ , 1.209 (3 d.p.) and 1.932 (3 d.p) 4 4 e   iθ − e  −iθ z − z  −1 32 asin θ = ________    , if z = e  iθ, then sin θ = _______     2i 2i 5 z − z  −1       sin  5 θ = ( _______ 2i ) 1 ( 5   z  − 5z  3+ 10z − 10 z  −1+ 5z  −3− z  −5) = ____   32i 5(z  3– z  −3) __________ 10(z − z  −1) 1 z  5− z  −5 __________   −     +     = ___    ( ________ ) 2i 2i 2i 16 1 = ___     (sin 5θ − 5 sin 3θ + 10 sin θ) 16 π π   __  2 __     2 ∫ ∫ 1   16   (sin 5θ − 5 sin 3θ + 10 sin θ)dθ b   sin   5 θ dθ = __ 0 0 π   __  2 1 −  __1 cos 5θ + __  53 cos 3θ − 10 cos θ]    =  __ 16 [ 5 0 8 1   16 = __ (0 − ( −  __15 + __   53 − 10)) = __   15 33 a z−n = cos (−nθ) + i sin (−nθ) = cos nθ – i sin nθ zn + z−n = cos nθ + i sin nθ + cos nθ – i sin nθ = 2 cos nθ 1 b ___ (cos 6θ + 6 cos 4θ + 15 cos 2θ + 10) 32 π __     2 ∫ c   cos6 θ dθ 0 π __     2 ∫ 1 = __   32   (cos 6θ + 6 cos 4θ + 15 cos 2θ + 10)dθ 0 π   __  2 1 __   1 sin 6θ + __  64 sin 4θ + __  15 sin 2θ + 10θ]    =  __ 32 [ 6 2 0 5π 1    (5π − 0) =   ___  = ___ 32 32 __ ___   πi   __ ____   9πi  __ _____ __ ____   17πi  −   7πi  34 a z = √ 2 e 20, √ 2 e 20 , √ 2 e 20 , √ 2 e Im b 9πi 2e 20 20 __ , √ 2 e 3πi  −  ____ 4 17πi 2e 20 πi 20 Re 2e – 3πi 4 π  i __ 2e 7π  i ___ 35 a z = 4e 9, 4e 9 , 4e 7πi 20 5π  −i ___ 9 b z  9 = ( 4e ) , ( 4e ) , (4e π  9 i __ 9 – 7π  9 i ___ 9 ) 5π  9 −i ___ 9 = 4  9 e  iπ, 4 9 e  7iπ, 4 9 e  −5iπ The value of all three of these expressions is −49 = −218 Hence the solutions satisfy z9 + 2k = 0, where k = 18. 36 z = cos θ + i sin θ, where π 9π 7π 3π π  ,  5π ___   =   __ θ =  ___   ,   ___  , − ___     , − ___ 10 10 ( 2 ) 10 10 10 ___   πi   ____  7πi  _____  13πi  37 a 2e15, 2e 15 , 2e 15 , 2e πi   − ___ 15 , 2e 11πi  −  _____ 15 b Vertices of a regular pentagon, inscribed in a circle radius 2, centred on the origin. Online Worked solutions are available in SolutionBank. Z03_IAL_FP2_44655_ANS_183-229.indd 206 25/04/2019 08:53 ANSWERS 38 a Im 207 Im 41 P O Re 1 2 C 3 O Re Q (–3, –1) 42 a __ b Maximum value of |z| is 3 + √__5   minimum value of |z| is 3 − √   5 39 a Im Im O (–3, –1) 2 2 43 44 O b 4 40 a Re (2, –1) c − __12  − __   12 i b __12  π __ π __ a Because     ,     b Centre (1, 0) 4 2 a −1 + 3i _ 11π 7 5√ 3 − 4 b θ = ____    , b = −  __  +  ________  i 12 2 2 Radius 2, centre (−1, 2) _ √   a b 2 y z Im 2 z – 2i z+2 3 O –2 3 O __ 45 46 Re Re x 47 a L and M are both circles, so are similar. _ √ 14  2 b _____   5 3π __  48 ____ 2√2  49 p = __   43 50 Im 2π π < arg(z – 1) < 3 4 __ 3√2  3√2  b z = −  ____     i   + ( 3 + ____ 2 2 ) 2π 3 O Z03_IAL_FP2_44655_ANS_183-229.indd 207 1 π 4 Re 25/04/2019 08:53 208 ANSWERS 51 Im 59 a a b w = eiθ + ae−iθ = A cos θ + iB sin θ So A + B = 2 and A − B = 2a ⇒ A = 1 + a, B = 1 − a Splitting w into real and imaginary parts gives u = (1 + a)cos θ and v = (1 − a)sin θ u 2 v 2 ⇒  _____    +  _____    = 1 (1 + a) ( 1 − a) ⇒ u2(1 − a)2 + v2(1 + a)2 = (1 − a2)2 c y |z – 3i| = 3 3π 4 3 (3, 3) – 1 __    e  iθ π 2 1 3 O O 52 Re Im x Challenge π 5π 1 0 , x , __    , ___     , x , π 6 6 2 a n will be of one of the forms 3k, 3k + 1, 3k − 1: n = 3k: 1  + ( e  3 )  + ( e  3 )  ______________ 1 + e  + e  ____________________            =        = 3k π 3 O Re 10 53 a k = 4 54 55 56 57 58 b 10π q−p2 By sketching region, deduce that ( _____     = 2x.   ___ 2 ) So (q − p)2 = 8x ⇒ q − p = √ 8x  (For non-zero area, q . p ⇒ q − p . 0) a 72 b Im(z) = 10 2w − 1 Rearrange to get z = _______    . |z| = 1 ⇒ |w − 2| = |2w − 1| w−2 Substituting u + iv for w and squaring gives (u − 2)2 + v2 = (2u − 1)2 + v2, which can be rearranged to give u2 + v2 = 1, which is the circle |w| = 1. −i a Rearrange to get z = ______    , and then subtitute u + iv w−1 1−u −v ____________   + i ____________     for w to get z =     (u − 1)  2+ v  2 (u − 1)  2+ v  2 1−u   = __   1 , and rearranging      So Im(z) = __  12 ⇒ ____________ (u − 1)  2+ v  2 2 gives u2 + v2 = 1, or |w| = 1. (5 − i)z − 2i b w = ___________        z 1 − iw a Rearrange to get z = ______    , and then subtitute w−1 u + iv for w to get (v + 1)(u − 1) − uv u(1 − u) − v(v + 1) z =  _________________   + i  _________________               (u − 1)  2+ v  2 (u − 1)  2+ v  2 π When argz =  __ , x = y, so 4 (v + 1)(u − 1) − uv = u(1 − u) − v(v + 1) and rearranging this gives u2 + v2 = 1, or |w| = 1. b v = −u y y c d P(0, 1) O 1 x O 3k 2πi ____     6k 2πki 3 1+1+1 _________  = 1   3 n = 3k + 1: 5π 6 5 2πi ____     ( 12, – 12) Q 4πki 3 + ( e    3 )  1 + e  2πki+  3  + e  4πki+  3  1  3k+1 + ( e    3 )  _________________________                  = ____________________     2πi ____ 3k+1 2πi ____ 6k+2 3 2πi ____ 4πi ____ 3 4πi 2πi ____     ____     1 + e  3 + e  3     = 0 = _____________   3 n = 3k − 1: + ( e    3 )  1 + e  2πki−  3  + e  4πki−  3  1  3k−1 + ( e    3 )  _________________________                 = ____________________     2πi ____ 3k−1 3 2πi ____ 6k−2 2πi ____ 2πi − ____  3 4πi ____ 4πi − ____  1 + e  3 + e  3     = 0 = _______________   3 b Consider jth term of f(x), ajx j. f(1) + f(ω) + f(ω2) The corresponding terms in  _______________      are: 3 aj(1) j aj(ω) j aj(ω2) j   _____   +   _____   +   ______   3 3 3 From part a, this expression is equal to aj if j is 0 or a multiple of 3, and 0 otherwise. f(1) + f(ω) + f(ω2)   _______________     is the sum of all such expressions 3 for all terms in f(x), so is equal to the sum of all aj where j is 0 or a multiple of 3, as required. 45  ) x  r. c (1 + x)45 = ∑(45 r=0 r So the sum of the coefficients of powers of x that 15 45 are 0 or multiples of 3 is ∑(  ) . r = 0 3r From part b, this is equal to (1 + 1)45 + (1 + ω)45 + (1 + ω2)45   ____________________________          3 45 2 45 45 − 2 2 + (−ω ) + (−ω)45 2      =   _______   =   ___________________ 3 3 x Online Worked solutions are available in SolutionBank. Z03_IAL_FP2_44655_ANS_183-229.indd 208 25/04/2019 08:53 ANSWERS CHAPTER 5 Prior knowledge check 1 y = (x − 1)ex + c 2 y2 = 4 − x2 3 a−  __32 ln | 50 − 2t | + c 209 x3 3 y = ___     + c where c is constant 3 y b−  __14 ln | cos (4x) | + c y 1 y= x 3 1 y 2 x3 3 y y Exercise 5A x2 3 x3 3 1 y x3 3 4 + c where c is constant y 2 3 y x2 3 2 y x2 2 1 y x2 1 3 yx 2 y x2 1 1 y x2 2 O x 2 2 1 2 2 y = Aex where A is constant y x3 3 y 3ex y 2ex y ex 3 2 2 3 4 4 y = ln Ax, where A is constant y 4 y ln (3x) 3 y ln (2x) y ln (x) 2 y ln ( 12 x) 1 1 y ln ( 3 x) 6 4 2 1O 2 3 4 5 1 2 x O 1 y = Ax2, where A is constant y 4 y 2x2 y x2 x O 2 y ln 3x y ln 2x y ln x y ln 12 x y ln 13 x 6 x y 1 1 2 x2 2 3 y ex y 3ex y 2ex x O y 12 x2 y x2 y 2x2 Z03_IAL_FP2_44655_ANS_183-229.indd 209 25/04/2019 08:53 210 ANSWERS 6 y2 − x2 = 2c, where c is constant 10 y = A sin x, where A is constant y2 x2 4 y2 x2 16 y 4 y2 x2 1 y 2 x2 0 y y 3 sin x 2 1 4 O 3 y2 x2 16 1 2 y 3 sin x 3 1 2 3 4 x 2 y sin x 4 y2 x2 4 y2 x2 1 x 6 4 1 7 y = ln  ________ , where c is constant (−x − c) 1 y ln x y ln y ln (11 x) y 4 x tan t 1 1 1x y ln 4 (21 x ) y 2 sin x π π 11 x = tan t + c for − __  < t < __    , where c is constant 2 2 3 y ln π x π 2 1 2 4 3 2 1 O 1 y 2 sin x y sin x 3 1 2x 2 x tan t O 2π π 2 2 3 x tan t 1 2 1 6 5 4 3 2 1 O 1 x tan t 2 t x tan t 2 4 6 1 Aet 12 x = _______    , where A is constant 1 + Aet 2 x x 1 2 3 4 x= 4 y x 4x 1 3 y 3x x1 2 y x 2x 1 1 y x x 1 x O 3et 1 + 3et 0.5 Ax 8 y = _____    , x > 0, where A is constant x+1 y 13 a x= x= et 1 + et et 1 + 12 et 1 2 t O y x O 9 y = sin x + c, where c is constant y2 x y y2 4x y2 9x y sin x 1 O x y sin x y sin x 1 y2 16x b y2 = 9x y sin x 2 Online Worked solutions are available in SolutionBank. Z03_IAL_FP2_44655_ANS_183-229.indd 210 25/04/2019 08:53 ANSWERS 1 e x − __  12  e x + __  12  6 y =  __ x x x 4x  1   + ___   7 y = −  ____ 3 3x2   2 c   8 a y =  _13 (x2 + 1)2 + _______ (x + 1)   22   b y =  _13 (x2 + 1)2 − ________ 3(x + 1) c ____________ 9 a y = 1 +     sec x + tan x 1 or y = 1 + ________   cos x b y = 1 + ____________     sec x + tan x 1 + sin x 10 a y = A cos x + sin x b y = − 3 cos x + sin x π c x =  __  ⇒ y = A × 0 + 1 = 1 2 3π x = ___     ⇒ y = A × 0 − 1 = −1 2 3π π    , 1 and ( ___    , 1)lie on all possible solution curves. so __ (2 ) 2 bx ___   11 y = e a + c 14 a y2 + 5x2 = 45 y b 7 6 5 4 3 2 1 3 x O 1 3 2 3 4 Exercise 5C 5 6 7 Exercise 5B c __ 1 a y = __  1 x sin x +  x  b y = xe2x − e2x + cex c y = 3x cosec x + c cosec x d y = xex_________ + cx e y = ln(__   1  + __ f y = ± __   1  x2 + ___   c     c2 ) 2 x 2x 6 x+c   → 0 b y 2 a y =  _____ ex 3 a y = 1 + __  1 + __   c   x x2 1 + ____ b 1 +  __   1   x 4x2 1 + __   1   1 +  __ x x2 1 + __   5   y = 1 +  __ x x2 y √ 1 x + 5 x2 y=1+ 1 x + 1 4x2 y=1+ 1 x + 1 x2 y=1 √ (–2, 34 ) √ O – 12 A(x + 1) ln  ________  (x + 2) 4 a y = __________   ln x 5 a y =  _13 ex + ce−2x c y = xecos x + cecos x 2 __  + c cos x e y =   x 2 g y = x ln (x + 2) + cx ( i j ) x (x + 1) ___   _______   ln   16 3 (x + 2)    b y =  ____________ ln x b y = −cot x + c cosec x d y = e2x + cex c 1 ln x + __ f y =  __  x x h y =  _14 x + cx− 3 y = (x + 2) ln (x + 2) + c (x + 2) 1   e x − __ y =  __  14  e x + __  c4  x x x3 Z03_IAL_FP2_44655_ANS_183-229.indd 211 1 a y2 = 2x2(ln x + c) b y3 = 3x3(ln x + c) −x c y =  _______  d y3 = x3(Ax − 1) ln x + c 1 __ dy −     = x3(Ax − 1) 2 a Given z = y−2, y = z   2 and ___ dx dy So ___     + (  __12 tan x) y = −(2 sec x) y3 dx − __3 −__1 1 −__3 dz ⇒ − __  z   2 ___     + ( __   12 tan x)z   2 = −2sec x z   2 2 dx dz ∴  ___  − z tan x = 4 sec x dx ______ cos x      b y = ______ 4x + c 1 __ dx dz 3 a Given that z = x   2 , x = z2 and ___     = 2z ___  dt1 dt __ dx So the equation ___     + t2x = t2x   2 becomes dt dz ___ 2z     + t2z2 = t2z dt dz   12 t 2 Divide through by 2z:  ___  +  __12 t 2z = __ dt 2 1 2 __ b x = ( 1 + ce  −  4 t )  dy dz     = −z2 ___     4 a Let z = y−1, then y = z−1 and ___ dx dx dy 1 (x + 1)3 2     y becomes     y = _______ So  ___  − __ x dx x (x + 1)3 dz 1 −1 ________     − __     z =   x  z−2 −z−2 ___ dx x (x + 1)3 dz 1 Multiply through by −z2: ___     + __   x  z = −  _______ x  dx 4x     b y =  ____________ 4c − (x + 1)4 ______ dz x+c      b y = ______ 5 a (1 + x2)  ___  + 2xz = 1 dx 1 + x2 ______ x + 4      c y = ______ 1 + x2 dy dy ___ dz dz 1   × ___         ×     = __________   6 ___  = ___ dx dz dx −(n − 1)y−n dx So differential equation becomes yn dz −  _____  × ___     + Py = Qyn n − 1 dx dz ⇒ ___     − (n − y)Py−(n − 1) = −Q(n − 1) dx dz and then ___     − (n − 1)Pz = −Q(n − 1) dx du 1     + u 7 a Differential equation becomes  ___  = __ dx 1 1 2 __ b This solves to give u +   2 u = x + c. √ 2 y=1+ 211 1 _ 2(y + 2x) + (y + 2x)2 −2x = k (k = 2c) ⇒ 4x2 + 4xy + y2 + 2y + 2x = k 25/04/2019 08:53 212 ANSWERS Challenge 1 dy 1 dv   v , ___ Substitute y = __     = −  __2   ___  dx v dx Differential equation becomes x __ 1 dv 1 __ x2 −  __2  ___ ( v   dx ) −   v  =   v2  dv 1 ⇒ x  ___  + v = −  __ x  dx Integrate both sides to get xv = −ln x + C −x 1 Substitute v = __     to get y = ________     y ln x + C Chapter review 5 1 y = 2 sin x + c cos x 1 _ 2 y = 5 + c (1 − x2) 2 x c __ __ 3 y = −    +  x 2 3 _ c 4 y =  _25 x 2 + __  x −x2 1 _ 5 y =  2 + ce ______ 6 y = 2x + c x√ 1 − x2  keλx kxn+1 7 a y =   _____  + ceax b y =   _____  eax + ceax λ–a n+1 8 y = sin x + A cosec x dy dz 9 a Given that z = y –1, then y = z –1 so ___     = –z –2  ___  dx dx dy The equation x  ___  + y = y2 ln x becomes dx dz –xz –2  ___  + z –1 = z –2 ln x dx dz __ z ln x   –     = –  ____   Dividing through by –xz –2 gives ___ x dx x 1    , where c is a constant. b y =  ____________ 1 + cx + ln x dy 1 –  __12___ 1 __ dz     = __   z     10 a Given that z = y2, y= z   2 and ___ dx 2 dx the differential equation becomes 1 – __1 dz – __1 __ cos x z   2  ___  – z   2 sin x + z   2= 0 dx – __1 dz Divide through by z   2: cos x  ___  – z sin x = –1 dx b z = c sec x – x sec x c y2 = c sec x – x sec x, where c is a constant dy y dz     = z + x  ___  11 a Given that z =  __ , y = zx so ___ x dx dx dy The equation (x2 – y2)  ___  – xy = 0 becomes dx dz (x2 – z2x2)(z + x  ___ )– xzx = 0 dx dz ⇒ (1 – z2)z + (1 – z2)x  ___  – z = 0 dx z dz   – z   ⇒ x  ___  = ______ dx 1 – z2 z3 dz ⇒ x  ___  =  ______2  dx 1 – z b 2y2 (ln y + c) + x2 = 0, where c is a constant y dy dz 12 a z = __     ⇒ y = xz and ___     = z + x  ___  x dx dx dy y(x + y) dz xz(x + xz)     becomes z + x  ___  = __________     So  ___  = ________ x(xz – x) dx x(y – x) dx dz z(1 + z)     ⇒ z + x  ___  = ________ (z – 1) dx 2z dz z(1 + z)   – z = _____       So x  ___  = ________ z–1 z–1 dx y __1 1 ___ __ b     –  2 ln y =  2 ln x + c, where c is a constant. 2x y dy dz 13 a Given that z = __    , y = zx and ___     = z + x  ___  x dx dx dy –3xy   becomes The equation ___     = ________   dx y2 – 3x2 dz –3x2z z + x  ___  =  _________   dx z2x2 – 3x2 dz –z3 –3z So x  ___  = ______   – z =  ______     dx z2 – 3 z2 – 3 2 3x b ln y +  ____2  = c, where c is a constant. 2y dy du 14 a Let u = x + y, then ___     = 1 +  ___  and so dx dx dy ___     = (x + y + 1)(x + y – 1) becomes dx du ___     – 1 = (u + 1)(u – 1) = u2 – 1 dx du ⇒  ___  = u2 dx –1 b y = _____     – x, where c is a constant x+c du dy     =  ___  – 1 15 a Given that u = y – x – 2, ___ dx dx dy du     + 1 = u2 So ___     = ( y – x – 2)2 becomes ___ dx dx du ⇒ ___  = u2 – 1 dx 1 + Ae2x  , where A is a positive constant. b y = x + 2 +  ________ 1 – Ae2x −__1 dv −__3 du     = −  __12 u   2  ___  16 a v = u   2, ___ dt dt −__3 du −__1 −__3    2 which Equation becomes −  __12 u   2  ___  × t + u   2= 2t  3u dt du 2u     −  ___ = − 4t  2. rearranges to ___ t dt 1 __ b Using integrating factor e  −2∫    t dt= e  −2lnt= t  −2, get d  ___  (u t  −2)= − 4 ⇒ u t  −2= − 4t + c, and u = −4t3 + ct2. dt Then the general solution for the original equation 1 _________  is v =  __________ √t  2 (c − 4t)  CHAPTER 6 Prior knowledge check __   23 x  3 + c b y = _______     x x __ x __ 1 b y =  __ x (x2 + 5) 2 a y = e   2 ( e   2 + 1) 2 1 a y = Ax  2+ 1 Exercise 6A 1 a c e g 2 a c e g 3 a c e g h y = Ae−3x + Be−2x b y = Ae−5x + Be3x d y = A + Be−5x f _1 y = Ae− 4x+ Be2x h y = (A + Bx)e−5x b y = (A + Bx)e−xx d −__ y = (A + Bx)e   45x f −__ y = (A + Bx)e   2 h y = A cos 5x + B sin 5x b y = A cos x + B sin x d y = e−4x(A cos x + B sin x) f y = e−10x(A cos 3x + B sin 3x) __ 3 y = e −2 x(A cos  _32 x + B sin  _32 x) y = Ae2x + Be6x y = Ae7x + Be−4x _1 y = Ae− 3x+ Be−2x 2x __ −_15 x y = Ae + Be   3 y = (A + Bx)e9x y = (A + Bx)e4x x __ y = (A + Bx)e   2 __ −√3 x y = (A + Bx)e y = A cos 9x + B sin 9x y = A cos  _43 x + B sin  _43 x y = e2x(A cos x + B sin x) __ √ Online Worked solutions are available in SolutionBank. Z03_IAL_FP2_44655_ANS_183-229.indd 212 25/04/2019 08:53 ANSWERS 4 a b c d e y = (A + Bx)e−7x y = Ae−4x + Be3x y = e−2x(A cos 3x + B sin 3x) 3x __ y = (A + Bx)e   4 1 _ y = e 3x(A cos  _23 x + B sin  _23 x) _2 −__x y = Ae 3x+ Be   2 f 5 a i x = Ae  (–k+√k – 9 )t______ + Be  (–k–√k – 9 )t ______ –t ii x = e (A sin((√9 – k2 )t) + B cos((√9 – k2 )t) iii x = (A + Bt)e–kt__ __ b i x = e–2t(A cos (√5 t)+ B sin (√5 t)) ii x → 0 6 From auxiliary equation: b α = –  ___  (using quadratic formula) 2a b2 = 4ac (setting discriminant = 0) y = (A + Bx)eax dy ___     = αeαx (A + Bx) + Beαx dx d2y   ____2  = Bαeαx + α2eαx (A + Bx) + αBeαx dx Substituting these 5 relationships into d2y dy a   ____2  + b  ___  + cy dx dx yields a result of 0, so (A + Bx)eαx is a solution ______ 2 ______ 2 7 Substitute y = Af(x) + Bg(x) into differential equation: a(Af″( x) + Bg″(x)) + b(Af′(x) + Bg′( x)) + c(Af(x) + Bg(x)) = A(af″(x) + bf′(x) + cf(x)) + B(ag″(x) + bg′(x) + cg(x)) = A(0) + B(0) =0 Challenge Aeαx + Beβx = Aepx eqi x + Bepx e–qi x = epx((A + B) cosqx + i(A − B) sinqx) Set B = A*, so that A = λ + μi and B = λ − μi, λ, μ ∈ ℝ. Then A + B = 2λ and i(A − B) = −2μ Hence setting λ =  __12 C and μ = − __12 D gives the required result. Exercise 6B 1 a b c d e f g y = Ae−x + Be−5x + 2 y = Ae6x + Be2x + 2 + 3x y = Ae−4x + Be3x − 2e2x y = Ae−5x + Be3x −  _13 y = (A + Bx)e4x + 1 +  _12 x y = (A + Bx)e−x + 4 sin 2x − 3 cos 2x y = A cos 9x + B sin 9x + _ 16 e3x h y = A cos 2x + B sin 2x + _ 13 sin x i y = e2x(A cos x + B sin x) + 3 + 8x + 5x2 1 x j y = ex(A cos 5x + B sin 5x) + __  25 e 7 1 2 1 __ __ __ 2 a  4 x  −   8 x +  32 7 b y = Ae  4x + Be  x + __   14 x  2 − __   18 x + __  32 3 a A e  6x + B 17 1 2 b y = Ae  6x + B − __  19 x  3 + __   36 x  − ___   108 x 4 y = A + Be−4x + 2x3 − __   32 x2 − __   34 x 5 a C.F. contains a term in xex. Results in setting up equation in the form of ex = 0. Not possible. b λ =  _12 c y = (A + Bx +  _12 x2) ex 4k kt 6 a y = Ae–t + Be–3t + __   53 – ___     + ___     9 3 b y = 2t –1 Z03_IAL_FP2_44655_ANS_183-229.indd 213 213 Challenge y = A cosx + B sinx + ( x − __  45 )e 2x Exercise 6C 1 a y = Ae−3x + Be−3x + ex b y = e−3x − e−2x + ex 3 2x 3 3 __ −2x 2 a y = A + Be +     e b y = 2 −  __  e−2x + __     e2x 2 2 2 1 1 1 3 y = __     e−6x + __     e7x − __     3 6 6 4 a y = A cos 3x + B sin 3x + 2 sin x b y = cos 3x + 2 sin 3x + 2 sin x 1 __ 5 a y = e −  2 x (A cos x + B sin x) + sin x 1 __ b y = sin x (1 − e −  2 x ) 6 a x = Ae2t + Be2t + t b x = et + e2t + t 7 a x = e3t + e−3t − sin t 1 1 8 a x = Ae2t + Bte2t + __    t3e2t b x = (t + __    t3)e2t 2 2 6 6 1 9 a x = __     (cos  __  t + sin  __ t + 1) 2 5 5 10 a x = et(A cos t + B sin t) + t2 + 2t + 1 b x = et sin t + 1 + 2t + t2 or x = et sin t + (1 + t)2 11 a y = Aex + Be2x + 3xe2x b y = 3ex − 3e2x + 3xe2x x 1 ___ __ 12 a y =     sin 3x −     cos 3x 18 6 13 a x = e−t − e−3t dx 1 b Setting ___     = 0 gives t = __     ln3, then substituting 2 dy __ 2√3      this into x = e−t − e−3t gives x = ____ 9 d2x Since ____   2 < 0, this is the maximum. dt Exercise 6D A B 1 1 a y = __   4  + __   b y = (A + B ln x) × __   2  x x x B A A   3  d y =  __7  + Bx4   2  + __ c y = __ x x x B 1 e y = Ax7 + __   2  f y = __   (A cos ln x + B sin ln x) x x dy dz z     2 a y = __     ⇒ xy = z and x ___  + y = ___ x dx dx dy d2z d2y dy ___     +     =  ____2  Also x  ____2  + ___ dx dx dx dx dy d2y So the equation x  ____2 + (2 – 4x) ___  – 4y = 0 dx dx dz d2z     – y) – 4y = 0 becomes ____2 – 4(___ dx dx dz d2z which rearranges to give  ____2  – 4 ___  = 0 dx dx A B 4x c y = __     + __     e  b z = A + Be4x x x z __ 3 a y =   2  ⇒ x2y = z x dy dz     + 2xy = ___     (1) So x2___ dx dx 2y 2 d dy dy dz (2) and x2  ____2 + 2x ___  + 2x ___  + 2y =   ____2  dx dx dx dx The differential equation becomes d2y dy dy (x2  ____2 + 4x ___  + 2y) + ( 2x2___     + 4xy)+ 2x2y = e–x dx dx dx Using results (1) and (2), d2z dz   ____2  + 2 ___  + 2z = e–x dx dx b z = e−x(A cos x + B sin x + 1) e–x c y =  ___   (A cos x + B sin x + 1) x2 25/04/2019 08:53 214 ANSWERS dz 4 a z = sin x ⇒ ___     = cos x dx dy dy ___     × cos x So     = ___ dx dz 2y d2y dy d     sin x and  ____2  =   ____2  cos2x – ___ dz dx dz The equation becomes d2y dy dy cos3 x   ____2 – cos x sin x  ___  + cos x sin x  ___  – dz dz dz 2y cos3 x = 2 cos5 x Dividing by cos3 x gives 2y d   ____2 – 2y = 2 cos2 x = 2(1 – z2) dz __ __ √ √ b y = Ae 2 sin x+ Be– 2  sin x+ sin2 x dx du du d  2 x d  2 u 5 a x = ut, ___     = u + t  ___ ,  ____2  = 2  ___  + t  ____   dt dt d t  dt d t  2 So differential equation becomes du du d  2u t  2 2  ___  + t  ____   − 2t(u + t  ___ )= − 2(1 − 2t  2)ut ( dt dt dt  2) d   2u which rearranges to give t  3  ____ ( dt  2  − 4u)= 0 2 d  u ⇒ ____ 2 − 4u = 0 dt  3 5 c x = t(____   2  e2t + ____   −2  e−2t) b x = t(Ae  2t + Be  −2t ) 4e 4e Challenge y = A ln x + B + 3x  2 Chapter review 6 1 a y = e  x cos x + A cos x b y = e  x cos x − (1 + e π) cos x 1 1 3x 2 y = −  __ (3 sin x + cos x) + __  10 e  10 2(3e2x  − 1)      3 y =   ___________ 3e2x  + 1 __ __ _1 √ 3 √3 4 y = e −2 x(A cos  ___  x + B sin  ___  x) 2 2 5 y = (A + Bx)e6x 6 y = A + Be4x 7 y = cos kx + __  1 sin kx k 8 y = e x sin 3x 1 __ 9 a k =   9 b y = e2x (A cos 3x + B sin 3x) +  _19e2x 10 y = Aex + Be−x + 2xex 11 a y = (A + Bx)e2x b They are part of the complementary function. c k = 2 and y = (A + Bx + 2x2)e2x 12 y = sin 2t + 2 cos 2t − cos 3t 13 a k = 1, μ = 2, λ = 3 b y = Aex + Be2x + xe2x + 2x + 3 1 __ 14 a y = 4e  –  4x sin  __12 x + x + 3 1 1 _ _ b As x →∞, e− 4 x→ 0, so 4e− 4 xsin _ 12 x → 0 and y ≈ x + 3. 5 −2x 1 _ _ 15 y =  6 e +  6 (cos 3x − sin 3x) 1 16 a x = Ae  −4t + Bt e  −4t + __   32 sin 4t 2 2 1 b x = __  12 e  −4t + __   15 t e  −4t + __   32 sin 4t 8 1 c Will oscillate as a sine wave with amplitude  __ and 32 π __ period     2 9 4 A B __1  34  b y =  __  – ____   1 ln x – __  34     +   ln x – __     + __ 17 a y = __     + __ x x2 2 x 4x2 2 5 1 1  2 sin (sin x) + __  2 esinx 18 y = __  2 cos (sin x) + __ Challenge ______ ______ √ x+c x+4 1 a y = ± ______      b y = ______      1 + x2 1 + x2 9 4 B A b y = __  x − ____   2  + _ 12 ln x − _ 34   2  + _ 12 ln x − _ 34 2 a y = __  x + __ x 4x dy du     = u2 3 a Let u = ___    , so equation becomes ___ dx dx 1 1 ⇒ ___   2  du = dx ⇒ –  __  = x + B u u dy 1 ___ _____   ⇒ y = A – ln(x + b) ⇒     = –   x+B dx √ ∫ ∫ CHAPTER 7 Prior knowledge check 1 a −3x  2 sin(1 + x3) − (sin x + cos x) b  ______________         e  x sin  2 x 2 Auxilliary equation λ  2+ 2λ + 2 = 0 has solution λ = −1 ± i, so general solution is y(x) = Ae  −x sin x + Be  −x cos x Exercise 7A 1 a f  ′(x) = 2e2x, f  ″(x) = 4e2x, f‴ (x) = 23e2x = 8e2x, f (n)(x) = 2ne2x b f  ′(x) = n(1 + x)n − 1, f  ″(x) = n(n − 1)(1 + x)n − 2, f‴ (x) = n(n − 1)(n − 2)(1 + x)n − 3, f (n)(x) = n! c f  ′(x) = ex + xex, f  ″(x) = 2ex + xex, f‴ (x) = 3ex + xex, f (n)(x) = nex + xex d f  ′(x) = (1 + x)−1, f  ″(x) = −(1 + x)−2, f‴ (x) = 2(1 + x)−3, f (n)(x) = (−1)n − 1(n − 1)!(1 + x)−n dny 2 a  ____n = 3ne2+3x = 3ny b e2 dx dy 3 a ___  = 3 × cos 3x × 2 sin 3x dx = 6 sin x cos x = 3 sin 6x d2y d3y d4y ____ b  2  = 18 cos 6x,  ____3  = −108 sin 6x,  ____4  = −648 cos 6x dx dx dx c 648 4 af′ (x) = 2xe  −x− x  2e   −  x −x f″(x) = (2e  − 2xe  −  x) − (2xe  −x− x  2e   −  x) = e  −  x(2 − 4x + x  2) f‴(x) = e  −  x(−4 + 2x) − e  −  x(2 − 4x + x  2) = e  −  x(−6 + 6x − x  2) bf″  ″ (x) = e  −  x(6 − 2x) − e  −  x(−6 + 6x − x  2) = e  −  x(12 − 8x + x  2) so f″  ″ (2) = e  −  2(12 − 16 + 4) = 0 dy 5 a Given that y = sec x,  ___  = sec x tan x dx d  2y ____   2 = sec x(sec  2 x) + (sec x tan x)tan x dx  = sec x(sec  2 x + tan  2 x) = 2 sec  3 x − sec x ____ d  3y b  3  = 6 sec  2 x(sec x tan x) − sec x tan x dx  = sec x tan x(6 sec  2 x − 1) 3 _ π d  y (√_) When x =  __ , ____   (1)(6(2) − 1) = 11√ 2     2 = 2 4 dx  2 2 dy d  y dy d d   2    (2y ___ )= 2y ____2  + 2(___    )  6 a ____2 (y  2) = ___ dx dx dx dx  dx  dy ____ d  3y d  2y ____ ___ b 2 (y  dx  3  + 3  dx  ×  dx  2 ) x 1 1 _ × 1 + _______      _  = _______   _  7 af′ (x) =  ___________ 2  2  ) ( √ √ √ x + 1 + x   1 + x   1 + x  2  _ 2 √ So 1 + x    f′ (x) = 1 Online Worked solutions are available in SolutionBank. Z03_IAL_FP2_44655_ANS_183-229.indd 214 25/04/2019 08:53 ANSWERS b Differentiating this equation w.r.t. x, _ x √ 1 + x  2  f″ (x) + _______   _ f′ (x) = 0 √ 1 + x  2  ⇒ (1 + x  2) f″ (x) + xf′ (x) = 0 c Differentiating this equation w.r.t. x ((1 + x  2) f‴ (x) + 2xf″ (x))+ (f′ (x) + xf″ (x))= 0 ⇒ (1 + x  2) f‴ (x) + 3xf″ (x) + f′ (x) = 0 d f9(0) = 1, f 0(0) = 0, f -(0) = −1 Exercise 7B 1 a f(x) = (1 − x)  −  1 ⇒ f(0) = 1 f′ (x) = −1(1 − x)  −  2(−1) = (1 − x)  −  2 ⇒ f′ (0) = 1 f″ (x) = −2(1 − x)  −  3(−1) = 2(1 − x)  −  3 ⇒ f″ (0) = 2 f‴ (x) = − (3 × 2)(1 − x)  −  4(−1) = (3 × 2)(1 − x)  −  4 ⇒ f‴ (0) = 3! General term: f  (r)(x) = r(r − 1)…2(1 − x)  −(r+1) = r!(1 − x)  −(r+1) ⇒ f  (r)(0) = r! f″ (0) f  (r)(0)    x  2+ … + ______ Using f(x) = f(0)+ f′ (0)x + _____    x  r+ … 2! r! r! 2    x  2+ … + __    x  r+ … (1 − x)  −1= 1 + x + ___ 2! r! = 1 + x + x  2+ … + x  r+ … _ 1 _ b f(x) = √ 1 + x = (1 + x)    2  1 _ f ′ (x) = _  12 (1 + x)  −  2  f″ (x) = _  12 (−  _12 )( 1 + x)  2 − _3  ⇒ f(0) = 1 ⇒ f′ (0) = _   21  ⇒ f″ (0) = −  _14  5 _ f‴ (x) = _  12 (−  _12 )(−  _32 )( 1 + x)  −  2  ⇒ f‴ (0) =  _38  Using Maclaurin’s expansion, 3 1 _ _ _ (  8 ) (−   4 ) √ 1 + x = 1 + _  12 x + ____    x  2 + ___    x  3− … 2! 3! x  2 ___ x  3 x ___ __ = 1 +     −     +    − … 2 8 16 sinx 2 f(x) = e  ⇒ f(0) = 1 f′ (x) = cos xe  sinx ⇒ f′ (0) = 1 f″ (x) = cos  2 xe  sinx− sin xe  sinx ⇒ f″ (0) = 1 Using Maclaurin’s expansion, 1 e  sinx = 1 + x + ___    x  2+ … = 1 + x + _  12 x  2… 2! 3 a f(x) = cos x ⇒ f(0) = 1 f′ (x) = − sin x ⇒ f′ (0) = 0 f″ (x) = − cos x ⇒ f″ (0) = −1 f‴ (x) = sin x ⇒ f‴ (0) = 0 f″  ″ (x) = cos x ⇒ f″  ″(0)   = 1 The process repeats itself every 4th derivative. Using Maclaurin’s expansion, (−1)r −1 1 cos x = 1 + ___     x  2 + ___     x  4+ … +  _____  x  2r+ … 2! 4! (2r)! (−1)r x  2 x  4    + … +  _____  x  2r+ … = 1 −  ___  + ___ 2! 4! (2r)! x  4 x  2 ___ π ___    , b Using cos x ≈ 1 −     +     with x = __ 2! 4! 6 π  4 π   2    = 0.86605… which is cos x ≈ 1 − ___     + _______ 72 31104 correct to 3 d.p. 4 a e = 2.718 (3 d.p.) b ln ( __   65 )= 0.182 (3 d.p.) 215 5 a 1 + 3x +  __92 x 2 + __   92 x 3 + __   27 x4 + … 8 b 2x − 2x2 + __   83 x 3 − 4x4 + … 4 x c x2 −  __ + … 3 π π π     = cos x cos  __ + sin x sin  __  6 cos x − __ ( 4) 4 4 1 = ___   _  (cos x + sin x) √ 2   x  2 x  4 x  3 x   5 1_ ___     + ___     + ___ =    ((1 − ___    − …)+ ( x − ___    − …)) 2! 4! 3! 5! √ 2   x  2 x  3 x  4 1_  (1 + x − ___     −  ___  +  ___ − …) =  ___ 2 24 √ 6 2   2 7 a f(x) = (1 − x) ln (1 − x) −1 f9(x) = (1 − x)  2 × _____    + 2(1 − x)(−1) ln(1 − x) 1−x = x − 1 − 2(1 − x) ln(1 − x) −1 f 0(x) = 1 − 2((1 − x) × _____    + (−1) ln(1 − x)) 1−x = 3 + 2 ln(1 − x) b f(0) = 0, f9(0) = −1, f 0(0) = 3, f -(0) = −2 c −x + __   32 x 2 − __  13 x3 x  3 x  5 1 8 a sin x = x − ___     +  ___ − … = x − __   3 + ___   120 x   5− …  16 x 3! 5! x  2 x  4 1 4   2 + __   24 x   − … cos x = 1 −  ___  +  ___ − … = 1 −  __12 x 2! 4! 3 sin x − 4x cos x + x 1 1 5 = 3(x − __  16 x   3 + ___   120 x   5− …)− 4(x − __  12 x   3 + __   24 x   − …) + x 17 5 = __   32 x   3 − ___   120 x   + … b __   32 1 9 a f9(x) =  _____ × (− sin x) = − tan x cos x b f9(0) = 0, f 0(0) = −1, f -(0) = 0, f -(0) = −2 −x2 ___ x4 c   ____   −     2 12 1 __ π d ln cos __  = ln(2  −  2 ) = −  __12 ln 2 ( 4) And by the Maclaurin series we have also π 2 π 4  __     __    (4) (4) π __ _____ _____   ln cos    ≈ −     −   ( 4) 2 12 2 2 π π ln 2 ≈  ___  1 +  ___  16 ( 96 ) 10 f(x) = tan x ⇒ f(0) = 0 f′ (x) = sec  2 x ⇒ f′ (0) = 1 f″ (x) = 2 sec  2 tan x ⇒ f″ (0) = 0 f‴ (x) = 4 sec2 x tan2 x + 2 sec4 x = 6 sec4 x − 4 sec2 x = 2(sec  4 x + 2 sec  2 x tan  2 x) ⇒ f‴ (0) = 2 f″  ″ (x) = 24 sec4 x tan x − 8 sec2 x tan x ⇒ f″  ″(0)   = 0 f‴  ″ (x) = 24 sec4 x tan2 x + 24 sec6 x − 8 sec2 x tan2 x − 8sec4 x ⇒ f‴  ″(0)   = 16 So the Maclaurin series is 0 0 16 2 0 + 1x +  ___  x  2 + ___     x  3 + ___     x  4 + ___     x  5+ … 2! 3! 4! 5! 2 1    x5 + … = x +  __ x3 + ___ 3 15 Challenge x   2 x   3 x   r ae  x= 1 + x + ___     +  ___ … +  ___ + … 2! 3! r! x  r x   r+1    ; ar+1 = _______     ar = ___ (r + 1)! r! |x| a x   r+1 r! r+1 ____ im  _______     = lr→∞   × ___     = lim  _____  , 1 lr→∞ im    ar (r + 1)! x  r r→∞ r + 1 | | Z03_IAL_FP2_44655_ANS_183-229.indd 215 | | 25/04/2019 08:53 216 ANSWERS b x  2 x  3 x  r (1 + x) = x − ___ ln     +  ___ − … + (−1)  r+1___    + … r 2 3 x  r x  r+1 r+1___ r+2_____ ( ) ( )   ar = −1      ; ar+1 = −1     r r+1 r+2 r+1 a ( ) −1   x   rx r+1 r im  __________       = lim  _____   ×  _________   lr→∞ im    ____  = lr→∞ ar ( −1)  r+1x  r r→∞ r + 1 r+1 x = lr→∞ im   ______  = | x| 1 1 + __     r So ln(1 + x) converges for −1 , x , 1 and diverges for x . 1. | | | | | | Exercise 7C | | __  −   x __ + … valid for all values of x 1 a 1 − x +  x 2 6 3 x  _____ b 1 + 4x + 8x2 +  32 + … valid for all values of x 3 2 3 __  +   x __ + … valid for all values of x c e(1 + x +  x ) 2 6 2 3 __  −   x __  −   x __ − … d −x −  x 2 3 4 2 3 4 −1 < x , 1 2 7 5 3840 645 120 of x x 2 +  9 x 3 + … − _2  , x < _ 2  x 3 ___ ___ f ln 2 + ___     −  9 3 3 2 8 8 x   3 ___ x  4 ___ x  5 x  2 ___ ___ 2 a ln (1 + x) = x −     +     −     +    − …, −1 , x < 1 2 3 4 5 x   3 ___ x  4 ___ x  5 x  2 ___ ___ ln (1 − x) = −x −     −     −     +    − …, −1 < x , 1 2 3 4 5 1+x x  3 x  5 _____     +  ___ + …)   = ln (1 + x) − ln (1 − x) = 2(x + ___ ln(  1 − x) 3 5 As x must be in both the intervals −1 , x < 1 and −1 < x , 1, x must be in the interval −1 , x , 1. ( 48 ) −1 , x , 1 3 5 1 _ c x = −  5 ; −0.0027% (4 d.p.) x x bx +  ___  +   ___ + … , 3 5 __   1  d 2 1 + __  35     = __     1 ln (4) = ln (2) ln _____ ( 1 − __ 3 ) 2 5 and the series from b gives 3 __ (   5 )      +  __35 + ____ 3 3 __ (   5 )  ____    + 0.69 … 5 3 5 Which is ln 2 correct to 2 d.p. (2x)  2 (2x)  3 4x  3     +  _____    + … 3 e  2x= 1 + 2x + _____  + … = 1 + 2x + 2x  2 + ____ 3 2! 3! (−x)2 _____ (−x)3 x  3 x  2 ___ _____ ___ −x   +    + … = 1 − x +     −    + … e  = 1 − x +   2 2! 3! 6 4  32 x   2if terms in x  3and above may be So e  2x − e  −x≈ 3x + __ neglected. 4x3 a 3x sin 2x = 3x(2x −  ____ + …) 3 9x2 _____ 27x4 ____ cos 3x = ( 1 −     +    − …) 2 8 So we get that 9x2 27x4 3x sin 2x − cos 3x = 6x2 – 4x3 – ( 1 –  ____  +   _____   + … 2 8 ) b 5 a 21  __ 2 x 2 – __   59 x 4 + … = –1 + __  21 2 8 5x 7x 17x x –  ____  +  ____  –  _____   12   + …,  __21  , x < __ 2 3 b e   2x sin x (2x)  2 ( 2x)  3 _____ (2x)  4 x   3 1 + 2x + _____ =(     + _____    + …)     +    + …)(x − ___ 2! 3! 4! 3! 4x  3 2x  4 x  3     + ____    + …)    + …)(x − ___ = (1 + 2x + 2x  2 + ____ 3 3 6 = x + 2x  2 + _   11   x  3+ x  4+ … 6 _ x x  −   ___ x   +   _____ x   −   ________   + … valid for all values e __ 3 2x4 4x6 2x8 6 a 1 – 2x2 +   ____  –   ____  +   ____  – … 3 45 315 x4 2x6 x8 b x2 –   __  +   ____  –   ____  + … 3 45 315 7 p =  __23 , q = –  __18 17x3 _____ 11x4 8 a x + 2x2 +   _____   +     + … 3 6 b 1 9 a (1 − 3x) ln (1 + 2x) 8x  3 = (1 − 3x)(2x − 2x  2 + ____    − 4x  4+ …) 3   26   x  3− 12x  4+ … = 2x − 8x  2 + _ 3 4 2 3 4 2x x2 2x3 x4 b 2 ln 3 +  ___  –   __  +   ____  –   ____ + …, –3 , x < 3 3 9 81 162 1 _ √ 1 + x  2  e  −x= (1 + x  2 )    2  e  −x c (x  2)  2 x   2 x   3 x   4 =( 1 +   _12  x  2+ ( _   12 )(−  _12 )  _____     −  ___  +  ___ + …)  + …)(1 − x + ___ 2! 2! 3! 4! x  2 x  4 x  2 x  3 x   4     −  ___  +  ___ + …) = ( 1 +  ___  −  ___ + …)(1 − x + ___ 2 8 2 24 6   23  x  3 + _   16  x  4+ … = 1 − x + x  2 − _ x8 x2 x4 x6 10 a 1 –   __  +   __  –   ___  +   ____  – … 2 8 48 384 b 1.711 (3 d.p.) 11 a e   px sin 3x (px)  2 (px)  3 ( 3x)  3 = (1 + px + _____     + _____    + …)    + …)(3x − _____ 2! 3! 3!   2 _____   3 p  2x _____ p  3x 9 x   3    + …) = (1 + px +     +    + …)(3x − ____ 2 2 6 3(p2 − 3)x3   + … = 3x + 3px2 +   __________ 2 13 2 __   b q = –2 p =  3  k = –  __ 2 e  x x−lnx x −lnx x 12 a e   = e  × e  = e  × e  lnx  −1 = ___     x   x sin x e ⇒ e  x−lnx sin x = _______     x x  2 x  3 x  3     +  ___ + …)( x − ___    + …) (1 + x + ___ 2 6 6 ________________________________      ,x.0 f(x) =       x ___ ___ x  2 ___ x  2 x  3 = (1 + x +     +    + …)(1 −    + …) 2 6 6 x  2 = 1 + x + ___    ignoring terms in x  4and above 3 e  0.1 sin 0.1  = 1.103329… b f(0.1) = __________   0.1 Using the approximation in part a, f(0.1) = 1 + 0.1 + 0.00333333 = 1.103333… . This result is correct to 6 s.f. d4y ____ 13 a   4  = 16(sin 2x − cos 2x) = 16y dx b y = −1 + 2x + 2x  2 − __   43 x   3 − __   23 x   4+ … Challenge a γ = 1 + __  12 β   2 + __   38 β   4 b 19.6 years (3 s.f.) c 0.0027% d As β is larger, the error in γ is larger, so the approximation would be less accurate. Online Worked solutions are available in SolutionBank. Z03_IAL_FP2_44655_ANS_183-229.indd 216 25/04/2019 08:53 ANSWERS Exercise 7D 5 1 1 a 1 + __  12  (x − 1) − __  18  (x − 1)2 + __   16   (x − 1)3 − ___  128   (x − 1)4 + … b 1.095 (3 d.p.) (x − e)2 x − e _______ 2 a 1 +  _____  + …   −   e 2e2 __ __ π π 2 40 π 3 √3  + 4 x − __    ) + 4√3  (x −   __ )  +   __ x − __    )  + … b 3 ( ( 3 3 3 cos 1   (x − 1)2 c cos 1 − sin 1 (x − 1) −   _____ 2 sin 1 cos 1   (x − 1)4 + …   (x − 1)3 +   _____ +   _____ 24 6__ √ 2  1 4 x − …) 3 a i ___  (1 − x −   __12 x2 +   __16 x3 +   __ 24 2 1 1 2 1 1 __ __ ___ 3  2500 x4 + … ii ln5 +   5 x −   50 x +   375 x − ____ __ __ __ √ √     3 3 1 3 + x +  ___  x2 − ___     x3 −  ___  x4 + …) iii __12 (−√  2! 3! 4! b 1.649 (4 s.f.) 4 b e−1(−1 + __  12 (x + 1)2 + __  13 (x + 1)3 + __  18 (x + 1)4 + …) 5 a (x − 1) + __  52 (x − 1)2 + __  11 (x − 1)3 + __  14 (x − 1)4 + … 6 b 0.4059 (4 d.p.) 6 − __34 + __   25 x − __  75 x2 + … 16 64 __ __ √ 3  π π 2 2 π 3    )− √ 3 (x − __    )  − __   3 (x − __    )  7 ___ + 1(x − __ 2 __ 6 6 6 √ 3  π 4    )  + … + ___    (x − __ 3 6 dy 1 8 a ___  = −  __ 16 dx 3 2 dy   3   ____2  = ___ dx 3 128 1 3 1   12 − __   16 (x − 3) + ___  256 (x − 3)2 + … b y = ________   _______  = __ √(1 + x)  2 1 1   3    2  , f ′′′(x) = ___ 9 f(x) = ln x, f ′(x) = __   , f ′′(x) = − ___ x x x (k − 1) ! (k − 1) !   ⇒ f  k (2) = (−1)  k−1  ________   f  k (x) = (−1)  k−1  ________ x  k 2  k Substituting into the Taylor series expansion gives ∞ (n − 1) !   (x − 2)  n   1   (−1)  n−1  ________ f(x) = ln 2 + ∑ _ n! 2  n n=1 ∞ ( x − 2)  n   = ln 2 + ∑ (−1)  n−1  ________ n 2  n n=1 | | Challenge 4   3  (x − π)  4− ... a ln (cos 2x) = − 2 (x − π)  2 − _ b – 0.1433 (4 d.p.) Exercise 7E x x3 x4 1 y = 1 + __     + x2 +  __  +   __ + … 2 3 6 x3 2 y = x −  __ + … 6 x3 3 y = 2 − x + x2 −  __  … 6 4 y = 1 + 2x −  __12 x2 − __  23 x3 + __  18 x4 + … 5 y = 1 − (x − 1) + __  52 (x − 1)2 − __  53 (x − 1)3 + … 6 y = 1 + x − x2 + __  12 x4 + … dy 7 a Differentiating (1 + 2x)  ___  = x + 2y2 with respect to x dx dy dy d2y (1) (1 + 2x)   ____2  + 2  ___  = 1 + 4y  ___  dx dx dx Z03_IAL_FP2_44655_ANS_183-229.indd 217 217 Differentiating (1) gives d2y d2y d2y dy 2 d3y    )  (1 + 2x)   ____3  + 2   ____2  + 2   ____2  = 4y   ____2  + 4(___ dx dx dx dx dx d2y dy 2 d3y ____ ____ ___ (2) ⇒ (1 + 2x)   3 + 4(1 − y)   2  = 4(   )  dx dx dx 5 8  3 x3 ... b y = 1 + 2x + __  2 x2 + __ __ __ __ 3√2  π π 2    ) + ____    (x − __    )  + ... 8 y = √ 2 + √ 2 (x − __ 4 2 4 dy 9 a i Differentiating  ___  − x2 − y2 = 0 with respect to x dx dy d2y (1) gives   ____2 − 2y  ___  − 2x = 0 dx dx iiDifferentiating (1) gives 2 d2y dy d3y    )  − 2 = 0   ____3 − 2y   ____2  − 2(___ dx dx dx d2y dy 2 d3y ____ ____    )  = 2 (2) So   3 − 2y   2  − 2(___ dx dx dx 4 3 2 dy dy dy d y b  ____4  − 2y   ____3 − 6(___    )   ____2 = 0 dx dx dx dx c y = 1 + x + x2 + __  43 x3 + __  76 x4 + ... dy (1) 10 Differentiating cos x  ___  + y sin x + 2y3 = 0 dx with respect to x gives d2y dy dy dy cos x  ____2 − sin x  ___  + y cos x + sin x  ___  + 6y2  ___  = 0 (2) dx dx dx dx Differentiating again d2y dy d2y d3y cos x   ____3  − sin x   ____2  − y sin x + cos x  ___  + 6y2   ____2  + dx dx dx dx 2 dy ___ (3) 12y(   )  = 0 dx dy ___ Substituting x0 = 0, y0 = 1 into (1) gives       + 2(1) = 0, dx 0 dy       = −2 so ___ dx 0 dy       = −2 into (2) gives Substituting x0 = 0, y0 = 1, ___ dx 0 d2y d2y   ____2    + 1 + 6(1)(−2) = 0, so  ____2    = 11 dx 0 dx 0 dy d2y       = −2,   ____2    = 11 Substituting x0 = 0, y0 = 1, ___ dx 0 dx 0 into (3) gives d3y   ____3    + (1)(−2) + 6(1)(11) + 12(1)(−2)2, dx 0 | | | | | | | | | d 3y so   ____2    = −112 dx 0 Substituting these values into the Taylor series, gives (−112) 3 11 y = 1 + (−2)x + ___     x2 + _______     x + ... 2! 3! 56 3 11 2 __ __ y = 1 − 2x +  2 x −  3 x + ... Ignoring terms in x4 and higher powers, y ≈ 1 − 2x + __  11 x2 − __  56 x3 2 3 11 a Repeated differentiation gives: d   3 y dy dy d  2 y dy d   2 y ____   3  = 4  ___+ 4x  ____2  − 2  ___ = 2  ___+ 4x  ____2  dx dx dx d x  d x  d x  d  y d   y d   y d  y d  y d   y ____   = 2  ____  + 4  ____ + 4x  ____  = 6  ____ + 4x  ____    4 2 2 3 2 3 d x  4 d x  2 d x  2 d x  3 d x  2 d x  3 d   3 y d   4 y d  4 y d  3 y d  5 y d  3 y  ____5  = 6  ____3  + 4  ____3 + 4x  ____4 = 4x  ____4  + 10  ____3  d x   d x   d x   d x   d x  d x  p = 4 and q = 10 25/04/2019 08:53 218 ANSWERS b y = 2 + 2(x – 1) + 2(x – 1)2 + __  10 (x – 1)3 + __  13 (x – 1)4 3 3 77 +  __ (x – 1)5 + … 15 10 f ′(x) = (1 + x)(1 + 2 ln (1 + x)) Chapter review 7 d  ny ____ 1 a   n  = ( −2)  ne   1−2x dx  d  8y   1−2ln32 = 256e  1+ln32  b  ____8  = ( −2)  8e dx  1   ln _____   256  e  1 = __   e   = 256(e  1)( e 1024) = _____ 1024 4 2 a f ′(0) = __  12 , f″(0) = __  14   x − e  x2(1 + e  x)e   x e  x(1 − e  x) (1 + e  x)  2e   = __________            b f ′′′(x) =  __________________________ (1 + e  x)  4 (1 + e  x)  3 f ′′′(0) = 0 x x2 c ln 2 + __     +   __  + … 2 8 3 a 1 – 8x2 + __   32 x 4 – ___   256 x 6 + … 3 45 −2 b cos 4x = 1 − 2 sin 2 2x so 2 sin  2 2x = 1 − cos 4x = 8 x  2 − __   32 x  4 + ___   256 x  6+ … 3 45 so sin  2 2x = 4x  2 − __   16 x  4 + ___   128 x  6+ … 3 45 x  2 x   3 x   4     +  ___  +  ___ + … and 4Using e  x= 1 + x + ___ 2 6 24 x  2 x4 cos x = 1 − ___     +  ___  − … 2 24 x   __  +   ___ (1 −  x 2 24) 2 e  = e cosx ( 4 ( 2) __  −  x 2 2 x     ___ 24 4 = e × e × e ( 2) )( ) 24 x  2 x x  2 x   4 x  4   4     +  ___  +  ___ + …)= e(1 − ___     +  ___ + …) = e(1 − ___ 2 8 24 2 6 –3x2 – 2x3 – … x3 x +   __  + … 6 d a  ___  (e  x) dx x  2 x   3 x  4 x x  r+1   r d     +  ___  +  ___ + … +  ___  + ________    + …     1 + x + ___ = ___ ) 2! 3! 4! r! (r + 1)! dx ( r 2 3 (r + 1)x  3x  ____ 4x  2x ____ ___ _________  + … = 0 + 1 +     +     +    + … +   2! 3! 4! (r + 1)! 2 3 r x  x  x  = 1 + x + ___     +  ___ + … +  ___  + … = e  x 2! 3! r! x   3 ___ x  2r+1 d x  5 d r ___ ___ ___    + … b    (sin x) =     x −     +     − … + (−1)  _________ ) 3! 5! (2r + 1)! dx dx ( 2r 2 4 (2r + 1) x   5x  3x  ____ r__________ ____  + … = 1 −     +     − … + (−1)    3! 5! (2r + 1)! 2 4 2r   x  x  x r    … = cos x = 1 − ___     +  ___  − … + (−1)  _____ 2! 4! (2r)! 2 4 x   x  2r d x   x   6 d     +  ___  −  ___ … + (−1)  r _____      1 − ___ c  ___ (cos x) = ___ 2! 4! 6! (2r)! dx dx ( x  2r+2 r+1    + … + (−1)  _________ ) (2r + 2)! 3 5 2r−1 2rx  4x  ____ 6x  2x ____ r_______ ___   = −     +     −    … + (−1)    2! 4! (2r)! 6! 2r+1 (2r + 2) x   r+1      + . + (−1)  ____________ (2r + 2)! ___  3 ___ x x  2r+1 x  5 r = − x −     +     − … + (−1)   _________  = − sin x ( 3! 5! (2r + 1)! ) x  2 x  4     −  ___ + …) a cos x = 1 − (___ 2 24 −1 x  2 x   4 1     −  ___ + …))    = 1 − ( ___ ⇒ sec x = _____   cos x ( 2 24 x  2 x  4     −  ___ )) sec x = 1 + (−1)(− (___ 2 24 2 (−1)(−2) x  2 x  4 ________     −  ___ ))  + …   (− (___ +   2 24 2! 5 4 = 1 +  __12 x   2 + __   24 x   + … 2 2 x  1 ___ x   x  = e 1 + −  ___  + __   2 −       + … 1 + ___    + … 5 6 7 8 2 x3 2 5 b x +   __  +  __ x + … 3 15 13 3 __ 2 9 1 + x – 4x –   3 x + … 4 f ′′(x) = 3 + 2 ln (1 + x) 2 f ′′′(x) =  _____  1+x (1 + x)2 ln (1 + x) = x + __   32 x 2 + __   13 x 3 + … 2 3 4 x x x __ __ ___ 11 a x –     +     –     + … 2 6 12 b 0.116 (3 d.p.) 3 __ + … x +  x 3 12 a f(x) = e  tanx = e __   x 3 = e  x… × e 3 2 x  x  3 x  3 = (1 + x + ___     +  ___ + …)(1 + ___    + …) 3 2! 3! x  3 x  2 ___ ___ = 1 + x +     +    + … 2 2 x2 x3 b 1 – x +   __  –   __  + … 2 2 13 a f(x) = ln cos x f(0) = 0 − sin x ______ f ′(x) =    = − tan x f ′(0) = 0 cos x f ′′(x) = − sec  2 x f ′′(0) = − 1 f ′′′(x) = − 2 sec  2 x tan x f ′′′(0) = 0 f ′′′′(x) = − 2 sec  4 x − 4 sec  2 x tan  2 x f ′′′′(0) = −2 Substituting into Maclaurin, x   2   4 x x  4   2 x ln cos x = (−1)  ___  + (−2)  ___ + … = −  ___  −  ___ + … 2 12 2! 4! x b Using 1 + cos x = 2 cos  2  __ , 2 x x ln (1 + cos x) = ln 2 cos  2  __ = ln 2 + 2 ln cos  __  2 2 x 2 1 __ x 4    )  − _   12  (   )  − …) so ln (1 + cos x) = ln 2 + 2(−  _12  (__ 2 2 x   4   2 x     −  ___ − … = ln 2 − ___ 4 96 d2y dy ___ 14 a     = 3(e3x + e−3x),  ____2 = 9(e3x + e−3x), dx dx d4y d3y ____ 3x −3x   3 = 27(e + e ),  ____4  = 81(e3x – e–3x) = 81y dx dx 5 b y = 6x + 9x  3 + __   81 x   + … 20 2(3)  2n−1x   2n−1 ___________ c       (2n − 1)! π π     ⇒ f(a) = 0 15 Let f(x) = ( x −  __ )c ot x and a = __ 4 4 π f ′(x) = ( x −  _  )(− cosec  2 x) + cot x ⇒ f ′(a) = 1 4 π f ′′(x) = ( x −  _   2 cot x cosec  2 x + (−2 cosec  2 x) ⇒ f ′′(a) = − 4 4) π ( f ′′′(x) = ( x −  _  ) − 2 cosec  4 x − 4 cot  2 x cosec  2 x) 4 + 6 cot x cosec  2 x ⇒ f ′′′(a) = 12 Substituting into the Taylor series expansion gives π π 2 12 π 3 −4 f(x) = 0 + 1 x −  __  + ___     x −  __    + ___     x −  __    + . . . ( 4) 4) 4) 2! ( 3! ( π π 2 π 3 = x −  __  − 2 x −  __    + 2 x −  __    + . . . as required ( ( ( 4) 4) 4)  14 16 a f ′(0) = __  12 , f ″(0) = __ 1(1 − 1) _________ ex(ex− 1)   ⇒ f ′′′(0) =  ________3  = 0 b f ′′′(x) =   x (e + 1)3 (1 + 1) x2 x __ __ c ln2 +     +    + … 2 8 x < 0 17 a 1 − 8x2 + __  32 x4 − ___  256 x6 + … 3 45 Online Worked solutions are available in SolutionBank. Z03_IAL_FP2_44655_ANS_183-229.indd 218 25/04/2019 08:53 ANSWERS 18 ecosx= e(ecosx−1) 2 x _ ( − 2 + . . .)  x2 ___ x4 ___ ____________   + …) = e(1 +  −      +     + …   +     2 ( 2 ) 24 2 4 x  ___ x  2 x x  ___ x     4 ___ = e(1 −      +      +     + …)≈ e(1 −  ___   +  ___ ) 2 24 8 2 6  12 x3 + … b 0.2155 a y = 2x + __  32 x2 + __ −3x2 − 2x3 − … y = 2 + 4x + x2 − __  23 x3 + … x3 2 __ a y = x +    + … b  _3  6 d d x2 x3 x4 xr xr + 1 a ___ (ex) = ___     1 + x +  ___  +   ___  +   ___ + ... +  __  +   _______ + ... ) 2! 3! 4! r! (r + 1)! dx dx ( (r + 1)xr 3x2 ____ 4x3 2x ____ ___ ________   + ... = 1 +     +     +     + ... +   2! 3! 4! (r + 1)! 2 3 r x x x = 1 + x +   ___  +   ___  + ... +  __  + ... 2! 3! r! x =e d x3 x5 x2r + 1 d b ___  (sin x) = ___     x −   ___  +   ___ − ... + (−1)r  ________ + ... ) 3! 5! (2r + 1)! dx dx ( 2r 2 4 (2r + 1)x 5x 3x = 1 −  ____  +  ____ − ... + (−1)r  __________ + ... 3! 5! (2r + 1)! 2 4 6 x2r x x x = 1 −  ___  +  ___  −  ___ + ... + (−1)r  _____ + ... = cos x 2! 4! 6! (2r)! x2 x4 x6 x2r d d     1−   ___  +   ___  −   ___ + ... + (−1)r  _____  + c ___ (cos x) = ___ 2! 4! 6! (2r)! dx dx ( x2r + 2 (−1)r + 1  ________ + ... ) (2r + 2)! 3 5 2rx2r − 1 4x 6x 2x   + = −  ___  +  ____  −  ____ + ... + (−1)r   _______ 2! 4! (2r)! 6! (2r + 2)x2r + 1     + ... (−1)r + 1   ____________ (2r + 2)! 3 5 x2r + 1 x x = −x +  ___  −  ___ + ... + (−1)r + 1   ________ + ... 3! 5! (2r + 1)! x3 x5 x2r + 1 = – x −   ___  +   ___ − ... + (−1)r  ________ + ... ( ) 3! 5! (2r + 1)! = −sin x y = 2(x − 1) + __  12 (x − 1)2 − __  12 (x − 1)3 + ... x2 x4 a You can write cos x = 1 − (   __  −   ___ + ...); it is not 2 24 necessary to have higher powers 1 1 sec x = _____       = ________________     4 cos x x2 ___ 1 − (   __ −   x + ...) 2 19 20 21 22 23 24 25 )) 24 Using the binomial expansion but only requring powers up to x4 x2 x4 sec x = 1 + (−1)(− (   __  −   ___ )) 2 24 2 (−1)(−2) x2 x4 ________   ( − (   __  −   ___ ))  + ... +   2 24 2! x2 ___ x4 x4 __ __ = 1 + (     −    ) +    + higher powers of x 2 24 4 x2 __ 5 4 __ = 1 +     +   24 x + ... 2 x3 2 5 b x +  __  +  __ x + ... 15 3 26 1 + x − 4x2 − __  13 x3 + ... 3 x3 27 a y = 2 + x − x2 −  __ + ... 6 b Differentiating with respect to x gives d 2y dy d3y d2y d2y d4y   ____4 + 2x   ____2  + 2  ___  + x2   ____3 + 2x   ____2  +   ____2 = 0 dx dx dx dx dx dx (1) ( (2 24 −1 x x = 1 −   __  −   ___ + ...   Z03_IAL_FP2_44655_ANS_183-229.indd 219 2 4 28 a b 29 a 30 a b 31 a b c 32 a b 33 a b 34 a | | dy d2y Substituting x = 0,  ___    = 1,   ____2    = −2 dx 0 dx 0 d3y ____ and   3    = −1 into (1) gives, dx 0 d4y d4y at x = 0,  ____4 + 2(1) + (−2) = 0, so  ____4 = 0 dx dx 1 f ′(x) = (1 + x)2 _____   + 2(1 + x)ln(1 + x) 1+x = (1 + x)(1 + 2ln(1 + x)) 2   f ″(x) = (1 + x)(_____ + (1 + 2ln(1 + x)) 1 + x) = 3 + 2ln(1 + x) 2 f ′″(x) =  _____ 1+x  31 x3 + … x + __  32 x2 + __ 2 x x3 x 4 x −  __  +   __  −   ___ + … b 0.116 (3 d.p.) 2 6 12 x x __ x +  __ + ... tan x x 3 = e  = e × e    3 f(x) = e (As only terms up to x3 are required, only first two terms of tan x are needed.) x2 x3 x3 = (1 + x +  ___  +   ___ + ...)(1 +  __ + ...) 3 2! 3! no other terms required. x3 x2 x3 = ( 1 +  __  + x +  ___  +   ___ + ...) 3 2! 3! x2 x3 = 1 + x +  __  +   __ + ... 2 2 x2 __ x3 __ 1 − x +     −    + … 2 2 dy ____ d2y d3y 1 ___ ____ __   3  = −   (  (3   2 + 1)) y dx dx dx 5x3 y = 1 + x − x2 +  ____ + … 6 The approximation is best for small values of x (close to 0). x = 0.2, therefore, would be acceptable, but not x = 50. f(x) = ln cos x f(0) = 0 −sin x f ′(x) =  ______ = −tan x f ′(0) = 0 cos x f ″(0) = −1 f ″(x) = −sec2 x f ″′(x) = −2 sec2 x tan x f ″′(0) = 0 f ″″(x) = −2 sec4 x − 4 sec2x tan2x f ″″(0) = −2 Substituting into Maclaurin: x2 x4 x2 x4 ln cos x = (−1)  __  + (−2)   ___ + ... = −  __  −   ___ − ... 2 12 2! 4! x    ), Using 1 + cos x ≡ 2 cos2(__ 2 x x ln(1 + cos x) = ln 2 cos2( __ ) = ln 2 + 2 ln cos( __ ) ( 2 ) 2 x 2 __ x 4 1 __ so ln(1 + cos x) = ln 2 + 2 − __12 (__ (   2 )  −   12 (  2 )  − ...) x2 x4 = ln 2 −  __  −   ___ − ... 4 96 Let y = 3x, then ln y = ln 3x = x ln 3 ⇒ y = ex ln 3 so 3x = ex ln 3 3(ln 3)3 x2(ln 3)2 x 1 + x ln 3 +  ________   +   ________  + … 2 6 1.73 (3 s.f.) f(x) = cosec x f ′(x) = −cosec x cot x i f ″(x) = −cosec x (−cosec2x) + cot x (cosec x cotx) = cosec x (cosec2x + cot2x) = cosec x (cosec2x + (cosec2x − 1)) = cosec x (2 cosec2x − 1) | 4 2 219 1 3 25/04/2019 08:53 220 ANSWERS ii f ″′(x) = cosec x (−4 cosec2x cot x) − cosec x cot x (2 cosec2x − 1) = −cosec x cot x (6 cosec2x −__1) __ __ __ √2  3 π π 2 11√2  π 3 __ ____ b √2 − √ 2 (x −    ) +     (x − __    )  − _____     (x − __    )  + … 4 2 4 4 6 πx   − π sin( __ 2)      35 a f ′(x) = ____________     πx __ 1 + 2 cos(  2 ) π  2 sin  2(  __     π  2 cos( __ 2) 2)     2         − _______________    f ′′(x) = −  _______________ πx __ πx __ 2(1 + 2 cos(  2 )) 1 + 2 cos(   )   ( 2 ) πx πx b f(1) = 0, f ′(1) = − π and f ′′(1) = − π  2, so π  2 f(x) = − π(x − 1) − ___     (x − 1)  2+ ... 2 4 a r=4 c r2 = sin 2θ 4 e r2 = _________     1 + sin 2θ g θ = arctan 2 i r = tan θ sec θ + a sec θ θ=π 2 r=6 6 θ=0 O b 1 0.5π π 5π 2 __ , ___     6 6 3 a Circle centre (0,3), radius 3 c Initial line θ=π 2 5π 4 O θ = 5π 4 3i d θ=0 Initial line θ=π 2 O O ) ) Exercise 8B CHAPTER 8 Prior knowledge check Im ( ( Challenge Consider the triangle formed by the two points and the origin and use the cosine rule to find d. 1 a Challenge d 1 a Base case: n = 1 we have  ___  ln x = __     x dx (n − 1) ! d  n  , then Suppose that ____   n  ln x = (−1)  n+1  ________ x  n dx  n+1 (n ! − 1) n! d d  _____   = (−1)  n+2  ____       (−1)  n+1  ________   n+1  ln x = ___ x  n dx dx  x  n+1 n ∞ (x − a)    b ln x = ln a +   ∑( −1)  n+1  _______ na  n n=1 b r2 = 8 cosec 2θ d r = 2 cos θ 3__   sec θ + __  π   f r =  ___ 4 √ 2   π   h r = __  a  cosec θ + __ 2 3 π 4 θ=0 Initial line π θ=– 4 θ=π 2 Re θ=0 9π ___ b     2 O 2 Initial line r = 2 sec θ Exercise 8A 1 a (13, 1.176) b (13, 1.966) _ c (13, –1.966) d (√ 13 , – 0.983) π e 2, −  __  ( __ 6 ) __ √3 , −3) 2 a (3√3 __ , 3) __ b (3 __ __ √2 , 3√2 ) √2 , −5√2 ) c (−3 d (−5 e (−2, 0) 3 a x2 + y2 = 4 b x=3 x2 c y=5 d x2 = 4ay or y =  ___ 4a 2 2 2 2 2 e x + y = 2ax or (x − a) + y = a 2 2 ____ f x2 + y2 = 3ay or x2 + y − ___  3a   =   9a   2 4 _3 _3 g (x2 + y2) 2= 8y2 h (x2 + y2) 2= 2x2 2 i x =1 ( e θ=π 2 3 r = 3 cosec θ θ=0 O Initial line ) Online Worked solutions are available in SolutionBank. Z03_IAL_FP2_44655_ANS_183-229.indd 220 25/04/2019 08:53 ANSWERS θ=π 2 f l 221 θ=π 2 r = a(4 + 3 cos θ) 4a r = 2 sec (θ – 4 3 O π ) 3 θ=0 4 Initial line θ=π 2 g 7a 3a r = a sin θ 2a Initial line θ=π 2 r = a(1 – cos θ) θ=π 2 7a θ=0 r = a(6 + sin θ) Initial line θ=0 a O 6a i r = a cos 3θ θ=π 2 θ=π 6 7a r = a(4 + 3 sin θ) Initial line 7π 6 θ =– π 6 θ=0 O a 4a 3π 2 4a π θ=0 aO 3a r = 2θ O 2π Initial line Initial line θ=π 2 p r = a(2 + cos θ) 2a Initial line θ=π 2 o θ=0 O θ=π 2 6a 5a θ = 5π 6 j Initial line a n O 2a O a 2a r = a(2 + sin θ) θ=0 θ=0 O Initial line θ=π 2 m a 2 h θ=0 a O θ=0 4π Initial line 2a k θ=π 2 6a 3π q r = a(6 + cos θ) θ=0 5a O 7a θ=π 2 a r2 = a2 sin θ Initial line θ=0 6a Z03_IAL_FP2_44655_ANS_183-229.indd 221 O Initial line 25/04/2019 08:53 222 ANSWERS θ=π 2 r b Cartesian equation is (x + 4)2 + (y + 3)2 = 25 Convert to polar coordinates: (r cos θ + 4)2 + (r sin θ + 3)2 = 25 Then rearrange to get r = −8 cos θ − 6 sin θ r2 = a2 sin 2θ θ=0 O 2 Exercise 8C 2 2 ____ _____ 1 a  πa   b  3πa   8 4 2 (π + 2)a2 __    d  a c   ________ 48 __ 4 2 ln √2  2 ln 2 2π3 _______ ______ ______   or  a   f  2a   e  a 2 4 3 2 __  (11π + 24) g  a 4 π 1 2 Area = 2 ×  __   a2(p + q cos θ)2 dθ 20 Initial line θ=π 2 ∫ π ∫ = a2   (p2 + 2pq cos θ + q2 cos2 θ) dθ 0 π a2q2 π = a2[ p2θ + 2pq sin θ]  0  +   ____   (cos 2θ + 1) dθ 2 0 π 2q2 a 1 ____ __ 2 2 = a p π +     [   sin 2θ + θ]    2 2 0 2 + q2 a2q2π 2p   =   ________   πa2 = a2p2π +   ______ 2 2 k 2 ∫ θ=0 O 3 a k 2 Initial line 2 ____  3  πa 12 4 a=9 __ 2 π 3√3  5 a__  (__     −   ____ ) 4 4 16 6 ___   5π  4 7 a Im Im |z – 12 – 5i| = 13 10 |z – 4 + 3i| = 5 12 + 5i O O 24 A Re Re 8 4 – 3i –6 b Cartesian equation is (x − 12)2 + (y − 5)2 = 169 Convert to polar coordinates: (r cos θ − 12)2 + (r sin θ − 5)2 = 169 Then rearrange this to get r = 24 cos θ + 10 sin θ 4 a Im arg z = π 4 b 35.1 8 a Im |z + 4 + 3i| = 5 O –8 Re A –4 – 3i 10 –12 + 5i –6 –26 O Re b 385 Online Worked solutions are available in SolutionBank. Z03_IAL_FP2_44655_ANS_183-229.indd 222 25/04/2019 08:53 ANSWERS 9 0.0966 10 0.79 5 a θ=π 2 θ=π 4 Challenge 3 a k = ___     7π 12π b ____     7 Exercise 8D ( ) ( r = 4 cos 2θ R ( 4 6 b 2π ) θ=π 2 r = a(1 – cos θ) a θ=0 O 2a Chapter review 8 Initial line a 2 _____ 1   9πa   8 Maximum value at (2a, π) 7 a θ=π 2 π θ=π 2 2 a, b Initial line θ = –π 4 ) 5 r cos θ = 3 r cos θ = −1 r = 3 sec θ r = −sec θ 6 2a, __  π   4_ 3 + √ 73    7 ________ 4 8 0.212 θ=0 O ) __  , ___ __  , ____ 1 (2a, 0),   a     2π   and   a  −2π 2 3 2 3 2 a (9.15, 1.11) b (212, 2.68) __ a√6  2a ___ 3 a   , ±0.421 b r = ±   ____   cosec θ 3 9 15 __ 4 (  2 a, ±1.32 ) ( 223 4 C2 r = 2a sec θ a θ=0 O Initial line C1 θ=0 2a Initial line O a –π 4 __ π √ 3  b __  −   ___  8 6 8 a θ=π 2 r = a (1 + cos θ) __ √ 5 − 1 _______ c cos α =   2 π θ= 2 3   2 2 r = 4 cos θ 2 θ=0 4 Initial line r = 3 cos θ θ=0 O Initial line 5π Area = ___     4 ___ ___ __ __ √ 3  __ ___ √3  7π π     , ___  π       and 0, __ 4 a     ,      , (a___ 2 6 2 6) 2 ) √ ( (2 ) ( ) 11 a ) __ 5π  a2 b ___ 8 1   b y =  ___ 2x Im |z – 1 – i| = 2 2 O Z03_IAL_FP2_44655_ANS_183-229.indd 223 r = 2 sec θ π  b 2√2 , __  π   , (2√2 , − __ 4 4) __   a, __  π   9 a  3 3 10 a y2 = x2 − 1 r = 1 + cos θ (√ __ 2 Re 25/04/2019 08:53 224 ANSWERS b Cartesian equation is (x − 1)2 − (y + 1)2 = 2 Convert to polar coordinates: (r cos θ − 1)2 + (r sin θ − 1)2 = 2 Then rearrange to get r = 2 cos θ + 2 sin θ c Im 2 1 1 A x O Re 2 9 θ = 3 e−2t cos t 10 a k = 12 b y = 2 cos 2x − __   π  sin 2x + 3x sin 2x 4 11 a a = 5, b = 1 b y = e2x (3 + 2x) + 5 + x 12 a y = e−2x (A cos x + B sin x) + sin 2x − 8cos 2x b As x → ∞, e  −kx → 0 ⇒ y → sin 2x − 8cos 2x Let sin 2x − 8cos 2x = R sin (2x − α) = R sin 2x cos α − R cos2x sin α Equating the coefficients of cos 2x and sin 2x _ ⇒ R = √ 65  , tan α = 8 Hence, for large x, y can be approximated by the _ sine function √ 65  sin (2x − α), where tan α = 8 (α ≈ 82.9°) 13 a y = e−t (A cos t + B sin t) + 2e−t b y = e−t (2 sin t − cos t) + 2e−t 14 a x = e−t (A cos 2t + B sin 2t) b x = e−t (cos 2t + sin 2t) c x d 3.59 12 2.09 13 1.52 Challenge _ _ x = r cos θ = √ 2   θ cos θ, y = r sin θ = √ 2   θ sin θ _ _ _ dy dx √_ ___ ___ √ √     =   2 cos θ − 2   θ sin θ,     = 2   sin θ + √ 2   θ cos θ dθ dθ dy sin θ + θ cos θ            = ____________ So ___ π dx cos θ − θ sin θ 1 +  __  4+π π 4 _____ __ _____   At θ =    the gradient of the tangent is   π  =   4 1 −  __  4 − π 4 4+π     x + c So the tangent is of the form y = (_____ 4 − π) π __ π π   2 __ _______   Substituting in the point     ,      , c =   (4 4) 2(π − 4) So the equation for the tangent is 4+π π2 y = (_______   +   ________  4 − πx ) (2π − 4) Rearranging, this is 2(π − 4)y + 2(π + 4)x = π  2 1 Review exercise 2 1 y= 1 ( 2 ln 5, 4 ln 5) arg z = π 6 O x2 y c x = f(t) c − x + __ x4 3 __  + cx 2 y =   x 2 c x + ln x +  3 y =   ___________    (x + 1)2 4 y =  _12 (e2x + 3) cos x 2 sin3 x c 5 y =   ________  +   ______ 3 sin 2x sin 2x xe−x   −   ________ 5ex   −   ________ e−x   6 y =   ________ 4(1 + x) 2(1 + x) 4(1 + x) 7 a y = sin x cos x + c cos x π 3π bcos x = 0, 0 < x < 2π ⇒ x = __     ,  ___  2 2 π 3π     , 0 and ( ___     , 0)lie on all of the The points __ (2 ) 2 solution curves for the differential equation. c y O 3π 8 7π 8 _1 2 t+ B e−3t + t2 − t + 1 15 a y = A e− _1 b y =  _45 (e− 2 t− e−3t )+ t2 − t + 1 c 16 a b c d 1.45 (3 s.f.) λ=2 y = A cos 3x + B sin 3x + 2x cos 3x y = (1 + 2x) cos 3x y 1 2 O t y = g(x) π 2 x __ 8 a y =   __  −   1   + ce−2x 2 4 π 3π 2 2π x 1 O π 6 π 2 5π 6 x b ( _12 ln 5,  _14 ln 5 ) Online Worked solutions are available in SolutionBank. Z03_IAL_FP2_44655_ANS_183-229.indd 224 25/04/2019 08:53 ANSWERS dy d  2y 17 a y = Kt  2 e  3t ,  ___ = 2Kt e  3t+ 3K t  2 e  3t ,  ____2 = 2K e  3t dt dt  + 12Kt e  3t+ 9K t  2 e  3t Substituting into the differential equation 2K e  3t+ 12Kt e  3t+ 9Kt  2 e  3t− 12Kt e  3t− 18Kt  2 e  3t + 9Kt  2 e  3t = 4 e  3t ⇒ 2K = 4 ⇒ K = 2 2t  2 e  3tis a particular integral of the differential equation b y = (A + Bt + 2t2)e3t c y = (3 − 8t + 2t2)e3t d y 5 1 5 ( 6 ,– 9 e 2 ) O 1 2 1 t _1 18 a x = A e− 2 t+ Be−2t + t + 2 −2t b x=e +t+2 19 a A =  _12 b x = (1 + t +  _12 t2 )e−t dx 1 1 c ___ = (1 + t) e  −t− ( 1 + t + __     t  2) e  −t = −  __  t  2 e  −t < 0, 2 2 dt for all real t When t = 0, x = 1 and x has a negative gradient for all positive t, x is a decreasing function of t. Hence, for t > 0, x < 1. 20 a k = 3 b y = A sin x + 3x c At x = π, y = A sin π + 3π = 3π This is independent of the value of A. Hence, all curves given by the solution in part a pass through (π, 3π). dy ___     = A cos x + 3 dx π π dy At x = __     ,  ___  = A cos  __ + 3 = 3 2 dx 2 This is independent of the value of A. Hence, all curves given by the solution in part a have an equal π gradient of 3 at x = __     2 3π  sin x d y = 3x −  ___ 2 dy 3π e For a minimum  ___ = 3 − ___     cos x = 0 2 dx 2 2   cos x = __  π  ⇒ x = arccos  __ π d   2 y ___ 3π ____   =     sin x   d x  2 2 2 π d  y In the interval 0 < x < __     ,  ____2  > 0 ⇒ minimum 2 dx  π   2− 4 4 2 2       2  = ______ sin  x = 1 − cos  x = 1 − ___ π  π  2 _ √ π  2− 4  π   In the interval 0 < x < __     , sin x =  ________ 2 π  2 _ 3√ 2 2 __ y = 3 arccos  __ π  −  2  π  − 4  Z03_IAL_FP2_44655_ANS_183-229.indd 225 225 9e2x − 2x − 1 Ce2x − 2x − 1 21 a y =  _____________         b y =  _____________   4 4 dy dv 22 a y = vx,  ___  = x  ___  + v dx dx (4x + vx)(x + vx) dv      = 4 + 5v + v2   x  ___  + v = ________________ dx x2 dv ⇒ x  ___  = 4 + 4v + v2 = (2 + v)2 dx 1 b v = −2 − _______   ln x + c dy dv 23 a y = vx, ___     = x  ___  + v dx dx 3 − 4v 3x − 4vx _______ dv   =       x  ___  + v = _________ 4x + 3vx 4 + 3v dx 2 3v + 8v − 3 dv 3 − 4v      − v = −   ____________     x  ___  = _______ 3v + 4 dx 4 + 3v C   2  b 3v2 + 8v − 3 = __ x 8y 3y2 ___ y C   +     − 3 = __   2  c y = xv ⇒ v = __     ⇒   ____ x x x2 x ⇒ 3y2 + 8yx − 3x2 = C y = 7 at x = 1 ⇒ C = 200 Factorising the LHS, (3y − x)( y + 3x) = 200 dy dy y   3dμ dμ 24 a ___  = −2y  −3  ___  ⇒ ___     = −  ___   ___  2 dx dx dx dx d μ So −  __12  ___  + 2xμ = xe  −x  dx dμ ___ ⇒     − 4xμ = −2xe  −x  dx 2 2 2 2 1 c  __2  = _  13 e−x + _  23 e2x b μ =  _13 e−x + C e2x y 2 dy dv dv d y d2v 25 a ___  = v + x  ___ ,  ____2  = 2  ___  + x   ____2  dx dx dx dx dx d 2v dv dv ____ ___ 2 So x x   2  + 2     − 2x(v + x  ___ ) + (2 + 9x2)vx = x5 ( dx dx ) dx d2v d2v ⇒ x3   ____2 + 9x3v = x5 ⇒   ____2 + 9v = x2 dx dx 2 b v = A cos 3x + B sin 3x + __  19 x2 − __  81 2 2 2 c y = Ax cos 3x + Bx sin 3x +  __19 x3 − __  81 x 1 dy _ 26 a 2t  2   ___ dt 2y 1 1 dy __ __ dy d 1   __1  2t  2  ___  − 16ty = 4te2t b 4t   ____2  + 2  ___  + 6t  2 − __ dt ( dt dt t  2) d2y dy ⇒ 4t   ____2 + 12t  ___  − 16ty = 4te2t dt dt d2y dy 2 2 2 + __   16 e2x ⇒   ____2  + 3  ___  − 4y = e2t c y = Ae x + Be−4x dt dt dy ___ 27 a t   dt d2y ___ dt d dy d dy ____     = t  ___  (t  ___ )     ___ b  2  =     × ___ dx dt ( dx ) dt dt dx dy d2y dy d2y     + t   ____2 ) = t2   ____2  + t  ___  = t (___ dt dt dt dt d2y dy dy ____ ___ ___ 2 2 2 c (t   dt2  + t   dt ) − (1 − 6t)t   dt  + 10yt = 5t sin 2t dy d2y ⇒   ____2  + 6  ___  + 10y = 5 sin 2t dt dt   16 sin(2 ex) − __  13 cos(2 ex) d y = e−3e (A cos(ex) + B sin(ex)) + __ x 25/04/2019 08:53 226 ANSWERS x  2 x x   2   4 28cos x = 1 − ___     +  ___ − … = 1 −  ___  ,neglecting terms in x3 2! 4! 2! and higher powers x  3 x  5 sin x = x − ___     +  ___ − … = x,neglecting terms in x3 and 3! 5! higher powers x  2 11 sin x − 6 cos x + 5 = 11x − 6(1 − ___    )+ 5 2 = −1 + 11x + 3x  2 A = −1, B = 11, C = 3 29 LHS = ln (x  2 − x + 1)+ ln (x + 1) − 3 ln x = ln((x  2 − x + 1)(x + 1))− ln x  3 x  3+ 1 1 = ln  ______ = ln 1 + ___   3  ( x  3  ( ) x  ) 1 Substituting  ___3  for x and n for r in the series x  (−1)  r + 1 x  r x  2 x   3 ln (1 + x) = x − ___     +  ___ + … +  ___________   +… r 2 3 n − 1 ( −1)  1 1    + …  + … + _________   LHS = ___   3  − ____ x  2 x  6 n x  3n 21 , D = __   71 30 A = 1, B = −2, C = −  __ 2 3   1  (x − 1)  3 + . . . 37 (x − 1) − __  1  (x − 1)  2 + __ 2 3 d3y b 2 − x − 2x2 + __  16 x3 + … 38 a  ____3 = 1 dx 39 a Differentiate the equation with respect to x: dy d2y dy 2  ___  + (1 + 2x)   ____2 = 1 + 8y  ___  dx dx dx dy dy dy d2y (1 + 2x)   ____2 = 1 + 8y  ___  − 2  ___  = 1 + 2(4y − 1)  ___  dx dx dx dx d3y dy 2 d2y d2y   )   + 2(4y − 1)  ____2  … b 2   ____2 + (1 + 2x)   ____3  = 8 (___ dx dx dx dx  32 x2 + __  43 x3 + … c __12 + x + __ 40 a 1 + x + 2x2 + 2x3 + … b 1.12 (2 d.p.) 41 a 1.5 + 0.8 x − 0.208 x2 + 0.131 982 x3 + … b 1.578 (3 d.p.) 2 1 dy d y 42 a −  __  ___ (3   ____2 + 1) b 1 + x − x2 + __  56 x3 + … y dx dx c The series expansion up to and including the term in x3 can be used to estimate y if x is small. So it would be sensible to use it at x = 0.2 but not at x = 50. 43 a 1 +  _32 x2 + 2x3 +  _54 x4 + … 8 3 4 2 31 a __13 − __   29 x + __   27 x − __   81 x + … 44 a r = 2 8 4 4 3 b __23 x − __   49 x2 − __   27 x + __   81 x + … x2 x 4 x2 x 4 b   __  +   ___  + … 32 a −   __  −   ___  − … 2 12 2 12 33 a L et u = 1 + cos 2x, then f(x) = ln u du ___     = −2 sin 2x dx du 1 du 1 f′ (x) = f′ (u)  ___  =  __   ___  =  __________  × (− 2 sin 2x) dx u dx 1 + cos 2x − 2 sin x − 4 sin x cos x ________ =       = − 2 tan x = _____________   cos x 2 cos  2 x __ b 1.08 (2 d.p.) b r = 3 sec θ ( ) p   c r = 2√3  sec θ −  __ 6 45 a r = a cos 3θ θ= O a Initial line b f″ (x) = − 2 sec  2 x θ=– f‴ (x) = − 4 sec  2 x tan x f″  ″ (x) = − 8 sec x sec x tan x tan x − 4 sec  2 x sec  2 x = − 8 sec  2 x t an  2 x − 4 sec  4 x c ln 2 − x2 − __   16 x4 + … π π 3 π5 4 34 a −2 x −   __  +   __43 x −   __    −   __ x −   __    + … 15 ( ( ( 4) 4) 4) b −0.416 147 (6 d.p.) 35 a p   a2 b ___ 12 46 a 2 θ=π 6 π 4 θ= O __ 3 3 θ=– b −0.735 166 (6 d.p.) dy 36 a  ___   = sec2 x dx 2y d____   2  = 2 sec2 x tan x dx d3y   ____3  = 4 sec2 x tan2 x + 2 sec4 x dx π π2 π3       +   __83 x − __       + … b 1 + 2 x −  __  + 2 x − __ ( ( ( 4) 4) 4) π π 3π     = ___     c Let x = ___     ⇒ x – __ 10 4 20 π π 2 8 ___ π 3 3π     + 2 ___     + __         tan  ___ = 1 + 2 ___ ( 20 ) ( 20 ) 10 3 ( 20 ) π π2 π3 = 1 + ___     +   ____  +   _____  10 200 3000 π 6 r = 3 cos 2θ 4√3  π π π −ln 2 + √ 3  x −   __  − 2 x −   __    + ____     x − __       + … ( ( 3 ( 6) 6) 6) __ π 6 π 4 __ __ 3 __ b  32 (2π − 3√3 ) θ=π 2 47 a Initial line 2√6  c  ____   3 r = a (1 + cos θ) r=a O a 2a Initial line Online Worked solutions are available in SolutionBank. Z03_IAL_FP2_44655_ANS_183-229.indd 226 25/04/2019 08:53 ANSWERS b I n Cartesian form: (x − 3)  2 + ( y + 4)  2= 25 ⇒ ( r cos θ − 3)  2 + ( r sin θ + 4)  2= 25 ⇒ r  2 cos  2 θ − 6r cos θ + 9 + r  2 sin  2 θ + 8r sin θ + 16 = 25 ⇒ r  2(cos  2 θ + sin  2 θ)− 6r cos θ + 8r sin θ = 0 ⇒ r  2( cos  2 θ + sin  2 θ) = r(6 cos θ − 8 sin θ) ⇒ r = (6 cos θ − 8 sin θ) c 63.3 π 50 a A θ= r = cos 2θ 4 __ 3√3 a b r =   _____   cosec θ 4 __ 3√3 a   cosec θ r = −  _____ 4 θ = π c The circle and the cardioid meet when π a = a(1 + cos θ) ⇒ cos θ = 0 ⇒ θ = ±  __  2 π __     1 2 A = 2 × __     ∫     r  2 dθ 20 π __     227 π __     ∫   2  r  2 dθ= ∫   2  a  2 (1 + cos θ)  2 dθ 0 0 π __     O = ∫     a  (1 + 2 cos θ + cos  θ) dθ 2 0 2 2 π B 4 b (0.667, 0.421) and (0.667, −0.421) 51 a θ = π 2 2 1 1     cos 2θ + __    ) dθ = ∫     a  2(1 + 2 cos θ + __ 2 2 0 π __     2 3 1     cos 2θ + __    ) dθ = a  2 ∫     (2 cos θ + __ 2 2 0 π  __  [ (4 4 2 ]0 The required area is A less half the circle = r = sin 2θ ) 3θ 2 3π 1 2 sin θ + __     sin 2θ + ___         = a  2 ___    + 2 A 3π π+8 2 1    + 2) − __     πa  2= ( _____   a    a  2(___ 2 4 4 ) 48 a θ=π 2 P O C D O 2a 3a Initial line θ=– π __     a  2 1 Initial line Initial line b (0.943, 0.615) 2 2 52 a (x − 3) __ + y = 9 x + √ 3 y = 6 b θ=π 2 D Q π   __   a, __ __   a, −  __ b P   3  π   , Q   3 2 3 2 3 __ a2   (4π + 9√3  ) c  ___ 16 d Let the smaller area enclosed by C and the half-line π θ = __     be A2: 3 2 3a    )  − 2A1  − 2A2  R = π (___ 2 _ _ 9 a  2 π ____ 6 a  2( 2 a  2(   ) − ____     )   −     4π + 9 √ 3   2π − 3 √ 3 = ______   4 16 16 _ _ 2 2 2 2 9 √ 3  a  2 9 a  π ____ a  π 9 √ 3  a  ______ 3 a  π _______   −     −  _______   −     +     = πa  2 =  ______ 4 2 8 4 8 49 a Im ( ) ( ) O 6 Re 2 3 –8 Z03_IAL_FP2_44655_ANS_183-229.indd 227 P C 6 O c P 3, __  π   , Q(6, 0) 3 53  _12 a2 ( Q Initial line ) π  __  π  __  1 4 1 4 54 A = __     ∫__  π     r  2 dθ= __     ∫__  π     16 a  2 cos  2 2θ dθ 2 8 2 8 π __     π __     4 4 1     sin 4θ]    = 4 a  2 ∫__  π     (1 + cos 4θ) dθ= 4 a  2 [θ + __ π __ 4     8 8 π 1 1     − __     = __     a  2(π − 2) = 4 a  2(__ 8 4) 2 55 _98 πa2 3 – 4i A __ 3  ___ 5π  π   −   √ 56 a θ = ___   π  ,  ___ b ___   12 12 12 16 57 a P(4a, 1.107), Q(4a, −1.107) __ 5√5  b  ____   m 4 ______   m2 c  2875π 32 25/04/2019 08:53 228 ANSWERS 58 a _32 πa2 2π   2π   , B:   1 __   a,  ___ __   a, − ___ b A:   1 2 3 2 3 c  _94 a __ 27√3  2 d  _____   a 8 e 113 cm2 (3 s.f.) π 3 __ π   , B:  __ __   a, −  __ 59 a A:   3 ( 2 a,   3 ) _ 2 3   π 3 √ 3 3   a b AB = 2 × __     a sin  __  =  _____ 2 3 2 __ √3  − 4π)a2 c (9 d 9.07 cm2 (3 s.f.) ( ) ( ( ) Exam Practice __ ) n ) ( ) π = a  2 ∫__  π     (9 + 12 cos θ + 4 cos  2 θ) dθ 3 π ∫ π     (11 + 12 cos θ + 2 cos 2θ) dθ = a  2 = a  2 [11θ __ 3 _ π   22π 13 √ 3 + 12 sin θ + sin 2θ]  __  π    = a  2(____     −  ______   3 3 2 ) π __     1 3 A 2 = 2 ×  __    a  2 (5 − 2 cos θ)  2 dθ 20 ∫ π __     3 ∫ = a  2   (25 − 20 cos θ + 4 cos  2 θ) dθ 0 π __     3 ∫ = a  2   (27 − 20 cos θ + 2 cos 2θ) dθ [ 0 n 1 1 1 ∑  ___________     = ∑    _______  − _______   2(r + 4) r=1 (r + 2)( r + 4) r=1 2(r + 2) n(7n + 25) 1 1 1 1 ________     −        −  ________  = ______________       = __     + __ 6 8 2(n + 3) 2(n + 4) 24(n + 3)(n + 4) 60 a A:(5a, 0), B:(3a, 0) 5π   , D: 4a, __ b C: 4a,  ___  π   3 3 π 1 π r   2 dθ __ cA 1= 2 ×     ∫__  π     = ∫__  π     a  2 (3 + 2 cos θ)  2 dθ 2 3 3 ( __ 1 x . √ 3 , − √3  , x , −1, x , −3 2 p = 7, q = 25 Using partial fractions, 1 1 1 ___________       = _______   − _______       (r + 2)( r + 4) 2(r + 2) 2(r + 4) Using the method of differences, ] π __     _ 3   27π 19 √3     −  _____   = a  2 27θ − 20 sin θ + sin 2θ     = a  2(____ 3 _ 2 ) 0 _   3 22π 13 √3 27π 19 √      −  _____     −  _____   + a  2(____   A 1   + A 2 = a  2(____ 3 2 ) 3 2 ) _ a  2   ) =  ___ (49π − 48 √3 3 Challenge dx dr 1 x = rcosθ ⇒ ___    = −rsinθ +  ___  cosθ dθ dθ dy dr y = rsinθ ⇒ ___     = rcosθ +  ___  sinθ dθ dθ dr r cos θ +  ___ sin θ dθ ________________ So l has gradient         = tan(α + θ) dr − r sin θ +  ___ cos θ dθ dr r cos θ +  ___ sin θ dθ tan α + tan θ _____________ ________________ Thus          =         1 − tan α tan θ dr − r sin θ +  ___ cos θ dθ Rearrange and cancel to get dr dr     cos2θ tanα = rcos2θ − ___     tanα sin2θ −rsin2θ + ___ dθ dθ dr r     ⇒  ___  tanα = r ⇒ tan α = ___ dθ dr ___     dθ 3 a z = cos θ + i sin θ z  n = ( cos θ + i sin θ)  n z  n= cos nθ + i sin nθ 1  ___n = cos nθ − i sin nθ z  1   n = 2 sin nθ z  n − ___ z  4 1 1 1     (2i sin θ)4 = __    (z − __    )  b 8 sin4θ = __ 2 2 z 4 1 1   2  + __   4  =  __  z4 − 4z2 + 6 − __ 2( z z ) 1 __ =    (2 cos 4θ − 8 cos 2θ + 6) 2 = cos 4θ − 4 cos 2θ + 3 4 a(x + 12)  2 + ( y + 5)  2= 169 (r cos θ + 12)  2 + ( r sin θ + 5)  2= 169 r  2 cos  2 θ + 24r cos θ + 144 + r  2 sin  2 θ + 10r sin θ + 25 = 169 r  2 = − 24r cos θ − 10r sin θ r = − 2(12 cos θ + 5 sin θ) b Im |z + 12 + 5i| = 13 –24 O Re –12 – 5i –10 arg z = – 3π 4 c 252 5 a y = cos x (sin x + A) b y = cos x (sin x + 3) Online Worked solutions are available in SolutionBank. Z03_IAL_FP2_44655_ANS_183-229.indd 228 25/04/2019 08:53 ANSWERS dy dy ___ dt 6 a ___  = ___     ×     dx dt dx 1 dy dy __ 1 dy      ___      ×     = __  ___  = ___ dx dt et ( x dt ) dy 1 d2y d2y −1 ___   2      +  ___2   ____    ____2  = ___ dx x dt x dt2 substituting into dy d2y ___ x2  ____2  + 8x     + 12y = 0 dx dx gives 1 d2y ___ 1 dy dy     −     + 8x ⋅  __   ___  + 12y = 0 x2 ⋅  ___2  (____ x dt x dt2 dt ) dy d2y ____ ___   2  + 7     + 12y = 0 dt dt B A   4  b y = ___   3  + ___ x x 3x2 2x2 7 y = 1 + x −   ____  +   ____  2 3 Z03_IAL_FP2_44655_ANS_183-229.indd 229 229 8 a (x + 4)2 + (y − 2)2 = 34 b, c Im arg(z + 1) = π 2 (x + 4)2 + (y – 2)2 = 34 34 (–4, 2) –1 O Re arg(z + 1) = – π 2 d −1 + 7i and −1 − 3i 26/04/2019 12:01 230 INDEX INDEX A alternating sequence 129 angles between vectors 38 circle properties 57–61 half-line 51–2, 57, 64, 65–7, 72 half-line in polar coordinates 153–4, 158 subtended at an arc 57–61 subtended at centre of circle 57, 60 see also argument; polar coordinates answers to questions 183–229 arbitrary constants 91, 95, 106–7, 115–17 arc of a circle 57–61 areas enclosed by polar curves 158–61 regions on Argand diagrams 63–9 Argand diagrams 46–82, 86–8 distances in constant ratio 55–6 geometric property of argument 51–2, 56–63 loci on 47–63, 157 modulus and argument 24, 31, 39, 40 multiplying complex numbers 27 polar curves 157 regions given by complex numbers 63–5 regions given by inequalities 65–9 roots of unity 37–8 transformations of complex plane 70–8 argument 8, 24, 31, 37–40 loci in terms of 51–2, 56–61 maximum value 48–9 principal argument 66 asymptotes 6 auxiliary equation 106–8, 119, 140 finding complementary functions 112–14, 115, 116 B binomial expansion 32–3, 34–5, 128, 131 boundary conditions 91, 106, 115–18 Z04_IAL_FP2_44655_IND_230-238.indd 230 C cancelling terms 3, 4 method of differences 15, 16, 17, 18 cardioids 155, 158, 162 concave/convex 156, 164 Cartesian coordinates 81, 150–3 Cartesian equations 47–53, 59–60, 162 converting to/from polar 151–3 locus in w-plane 72–3 chain rule 119, 126 circles angle subtended at an arc 57–61 angle subtended at centre 57, 60 on Argand diagram 47, 48–9, 52, 55–6, 157 Cartesian equation 47, 48–9, 52, 55 circle properties 49, 57–61 polar equation 153–4 regions on Argand diagrams 63–4, 66–8 transformations 70–1, 72, 74–5 complementary function 110, 112–14, 115–17 complex conjugate 75 complex conjugate roots 107, 108 complex numbers 22–45, 85–6 de Moivre’s theorem 29–42 exponential form 23–9, 30, 34, 40 modulus-argument form 23–5, 26, 29–30, 31, 37, 39 multiplying and dividing 26–9 nth roots 37–42 powers of 29–31 regions on Argand diagrams 63–5 trigonometric identities 32–6 see also Argand diagrams compound functions 132–6 compound transformations 71–2 concave polar curves 156, 158, 162, 164 constant of integration 92 constant ratio 55 convergent series 129, 130, 132, 137 convex polar curves 156, 159–60, 164 critical values 2–4, 5–6, 8–9, 10 D de Moivre’s theorem 29–32 deriving trigonometric identities 32–6 nth roots of a complex number 37–42 derivatives 106, 126–7 see also differential equations differences, method of 15–21 differential equations 90–124, 168–71 boundary conditions 115–18 complementary function 110, 112–14, 115–17 family of solution curves 91–4 first order 90–104 general solution see general solution integrating factor 95–7 particular integral 110–14, 116–17 particular solution 91, 93–4, 106, 115–18 reducible 98–102, 118–21 second order 105–24 second order homogeneous 106–10, 112 second order non-homogeneous 110–15 separating the variables 91–4, 99, 100 series solutions 140–4 solve by auxiliary equation 106–10, 140 substitution 98–102, 118–21 ‘dimple’ shaped curves 156, 158, 162, 164 distance between two points 47, 56 in constant ratio 55–6 from origin to a point 49, 50, 150 see also polar coordinates division, complex numbers 26, 27–9 E ‘egg’ shaped curves 156, 159–60 endpoints 57–61 enlargements 70–1, 72 equation of a circle 60, 74–5 Cartesian form 47, 48–9, 52, 55 polar form 153–4 equation of a half-line 51–2, 153–4 equation of a line 50, 163–4 25/04/2019 08:52 INDEX Euler’s identity 23 Euler’s relation 23, 29 exam practice 178–9 exponential form 23–9, 30, 34, 40 exponential function 23, 112 Maclaurin series 132, 133–4 Taylor series 137 F family of solution curves 91–4 finite series, sum of 15–21 first order differential equations see differential equations fractions in inequalities 2–8 sum of a series 16–18 functions compound functions 132–6 and differential equations 95–8, 110–15 higher derivatives 126–7 Maclaurin series expansion 128–36 polynomial approximations to 125 Taylor series expansions 136–9 see also exponential function; trigonometric functions G general solution 91–8, 106, 110, 112–15 family of curves 91–4 and roots of auxiliary equation 107–10 using substitutions 99–102, 119–21 geometrical reasoning 27 glossary 180–2 graphs see sketching graphs H half-line 51–2, 57, 59, 71–2 polar equation of 153–4 regions formed by 64, 65–7, 158 higher derivatives 126–7 homogeneous second order differential equations 106–10, 112 horizontal asymptotes 6 I imaginary axis 48, 81 imaginary numbers 23 imaginary roots 107, 108 inequalities 1–13, 83–4, 89 modulus inequalities 8–11 regions on Argand diagrams 65–9 Z04_IAL_FP2_44655_IND_230-238.indd 231 solving algebraically 2–5 solving graphically 5–10 infinite series 23, 129–30, 137 initial line 150, 157, 158 tangents parallel to 162–3 tangents perpendicular to 156, 162, 163–4 integrating factor 95–7, 99–100 integration constant of 92 finding areas 158, 159, 160 polar equations 158–61 intersection points 5–9, 159 intersections 64, 67–8, 159–60 intervals 2–4 J Julia set 46 L lines 50, 71–4, 75–6 locus of points on Argand diagrams 47–63, 157 from fixed points in constant ratio 55 represented by polar curves 157 transformations 70–8 logarithmic functions 129–30, 132, 133, 136 loops 155, 159 Lorentz factor 136 M Maclaurin polynomials 129–30 Maclaurin series 128–37, 144–8, 171–3 compound functions 132–6 limitations 136–7 simple functions 128–32 major arc 58–60 maximum value of argument 48–9 method of differences 15–21 minor arc 58–9 modulus 70–1, 73, 74, 75 minimum/maximum values 49–50 modulus inequalities 8–11 modulus-argument form 26, 29–30, 31, 37, 39 and exponential form 23–5 modulus-argument rules 26–7 multiplication, complex numbers 26–9 N non-homogeneous second order differential equations 110–15 nth roots 37–42 nth roots of unity 38 231 P parabolas 73, 94 parametric equations 73, 162 particular integral 110–14, 116–17 perpendicular bisector 47, 50, 64, 74 points distance between two 47, 55–6 distance from origin to 49, 50 of intersection 5–9, 159 loci on Argand diagram 47–63, 157 polar coordinates 149–67, 173–7 converting to/from Cartesian 150–3 intersections of polar curves 159 tangents to polar curves 162–5 polar curves 151–65, 173–7 area enclosed by 158–61 intersections 159–60 loci on Argand diagram 157 sketching 153–8, 159–61, 164 tangents to 162–5 polar equations 151–3, 158–61 polynomials 113, 125, 129–30, 136–8 population growth model 90 positive integer exponents 29–30 power series 23 see also Maclaurin series; Taylor series powers of complex numbers 29–31 principal argument 66 principle of superposition 110 product rule 95, 99, 119, 126 proof by induction 15, 19, 29–30 Pythagoras’ theorem 150–1 R range of validity 129, 130, 136 ratio test 132 real axis 48, 82 real roots 107 reducible differential equations 98–102, 118–21 repeated root 107, 108 review questions 83–9, 168–77 Argand diagrams 78–81, 86–8 complex numbers 42–4, 85–6 differential equations 102–4, 121–3, 168–71 inequalities 11–13, 83–4, 89 Maclaurin and Taylor series 144–8, 171–3 polar coordinates 165–7, 173–7 series 20–1, 84–5, 89, 144–8, 171–3 roots 10, 107–8, 113, 119 nth roots 37–42 rotation 65, 66, 72 25/04/2019 08:52 232 INDEX S second derivatives 106 second order differential equations 105–24 homogeneous 106–10, 112 non-homogeneous 110–15 semicircle 57, 58, 61 separating the variables 91–4, 99, 100 sequence, alternating 129 series 14–21, 84–5, 89 binomial expansion 32–3, 34–5, 128, 131 convergent 129, 130, 132, 137 infinite 23, 129–30, 137 see also Maclaurin series; Taylor series series expansions 23, 128–9 binomial 32–3, 34–5, 128, 131 see also Maclaurin series; Taylor series series solutions 140–4 set notation 3–4, 10, 64, 67 simultaneous equations 111, 112, 116 Z04_IAL_FP2_44655_IND_230-238.indd 232 sketching graphs identifying intervals 2–4 modulus graphs 8–11 polar coordinates 150–1, 162 polar curves 153–8, 159–61, 164 solution curves 91–4 to solve inequalities 5–8 see also Argand diagrams solution curves 91–4 solution sets 2–4, 6 spiral 153–4 substitution 98–102, 118–21 subtended angles 57–9, 60 sum of a finite series 15–21 symmetric curves 73, 155, 158 trigonometric functions complex numbers 23–5, 26, 27, 29–39 differential equations 95, 97, 111, 112, 116–17 polar coordinates 150–3, 155–7, 158–64 series expansions 23, 130–5, 137–9 trigonometric identities 32–6, 158, 159, 160 trigonometry 49, 150–1 T V tangents 49, 156, 162–5 Taylor series 125, 136–9, 145–8, 171–3 differential equations solutions 140–4 transformations of the complex plane 70–8 translations 70–2 U unity, nth roots of 38 variables multiplying inequalities by 2 separating 91–4, 99, 100 vectors 38, 70, 71, 72, 81 vertical asymptotes 6 vertices of regular n-gon 38 video games 1 25/04/2019 08:52 233 Z04_IAL_FP2_44655_IND_230-238.indd 233 25/04/2019 08:52 234 Z04_IAL_FP2_44655_IND_230-238.indd 234 25/04/2019 08:52 235 Z04_IAL_FP2_44655_IND_230-238.indd 235 25/04/2019 08:52 236 Z04_IAL_FP2_44655_IND_230-238.indd 236 25/04/2019 08:52 237 Z04_IAL_FP2_44655_IND_230-238.indd 237 25/04/2019 08:52 238 Z04_IAL_FP2_44655_IND_230-238.indd 238 25/04/2019 08:52 PEARSON EDEXCEL INTERNATIONAL A LEVEL STUDENT BOOK Pearson Edexcel International A Level Further Pure Mathematics 2 Student Book provides comprehensive coverage of the Further Pure Mathematics 2 unit. This book is designed to provide students with the best preparation possible for the examination: • • • • • • • Content is fully mapped to the specification to provide comprehensive coverage and easy reference Engaging and relevant international content in a real-world context Exam-style questions at the end of each chapter, and an exam practice paper at the end of the book, provide practice for exam writing technique Signposted transferable skills prepare for further education and employment Reviewed by a language specialist to ensure the book is written in a clear and accessible style Glossary of key Mathematics terminology, and full answers, included at the back of the book Interactive practice activities also included IAL FURTHER PURE MATHS 1 Student Book ISBN: 9781292244648 An Online Teacher Resource Pack (9781292244624) provides further planning, teaching and assessment support. This Student Book supports the following qualifications: Pearson Edexcel International Advanced Subsidiary in Further Mathematics (XFM01) Pearson Edexcel International Advanced Level in Further Mathematics (YFM01) For first teaching September 2018 IAL FURTHER PURE MATHS 3 Student Book ISBN: 9781292244662 www.pearsonglobalschools.com IAL_FPM2_Cover.indd 1-3 PEARSON EDEXCEL INTERNATIONAL A LEVEL FURTHER PURE MATHEMATICS 2 STUDENT BOOK FURTHER PURE MATHEMATICS 2 MATHEMATICS eBook included PEARSON EDEXCEL INTERNATIONAL A LEVEL FURTHER PURE MATHEMATICS 2 STUDENT BOOK 16/04/2019 14:53

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