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Cambridge International AS & A Level Mathematics: Pure Mathematics 1 Coursebook Copyright Material - Review Only - Not for Redistribution Copyright Material - Review Only - Not for Redistribution op y ve rs ity ni C U ev ie w ge -R am br id C op y Cambridge International AS & A Level Mathematics: w ge U R ni ev ie w C ve rs ity op y Pr es s -C Sue Pemberton Series Editor: Julian Gilbey s -C -R am br ev id ie Pure Mathematics 1 y op y op -R s es -C am br ev ie id g w e C U R ev ie w ni ve rs C ity Pr op y es s -C -R am br ev id ie w ge C U R ni ev ve ie w rs C ity op Pr y es Coursebook Copyright Material - Review Only - Not for Redistribution op y ve rs ity ni C U ev ie w ge am br id University Printing House, Cambridge CB2 8BS, United Kingdom -R One Liberty Plaza, 20th Floor, New York, NY 10006, USA 477 Williamstown Road, Port Melbourne, VIC 3207, Australia Pr es s -C 314–321, 3rd Floor, Plot 3, Splendor Forum, Jasola District Centre, New Delhi – 110025, India y 79 Anson Road, #06–04/06, Singapore 079906 op Cambridge University Press is part of the University of Cambridge. C op © Cambridge University Press 2018 y www.cambridge.org Information on this title: www.cambridge.org/9781108407144 ni R ev ie w C ve rs ity It furthers the University’s mission by disseminating knowledge in the pursuit of education, learning and research at the highest international levels of excellence. ie ev id br First published 2018 w ge U This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press. -R am 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 Printed in the United Kingdom by Latimer Trend s -C A catalogue record for this publication is available from the British Library es ISBN 978-1-108-40714-4 Paperback op y ve ® IGCSE is a registered trademark id ie w ge U Past exam paper questions throughout are reproduced by permission of Cambridge Assessment International Education. Cambridge Assessment International Education bears no responsibility for the example answers to questions taken from its past question papers which are contained in this publication. C ni R ev ie w rs C ity op Pr y Cambridge University Press has no responsibility for the persistence or accuracy of URLs for external or third-party internet websites referred to in this publication, and does not guarantee that any content on such websites is, or will remain, accurate or appropriate. Information regarding prices, travel timetables, and other factual information given in this work is correct at the time of first printing but Cambridge University Press does not guarantee the accuracy of such information thereafter. -C -R am br ev The questions, example answers, marks awarded and/or comments that appear in this book were written by the author(s). In examination, the way marks would be awarded to answers like these may be different. y op -R s es -C am br ev ie id g w e C U R ev ie w ni ve rs C ity Pr op y es s NOTICE TO TEACHERS IN THE UK It is illegal to reproduce any part of this work in material form (including photocopying and electronic storage) except under the following circumstances: (i) where you are abiding by a licence granted to your school or institution by the Copyright Licensing Agency; (ii) where no such licence exists, or where you wish to exceed the terms of a licence, and you have gained the written permission of Cambridge University Press; (iii) where you are allowed to reproduce without permission under the provisions of Chapter 3 of the Copyright, Designs and Patents Act 1988, which covers, for example, the reproduction of short passages within certain types of educational anthology and reproduction for the purposes of setting examination questions. Copyright Material - Review Only - Not for Redistribution ve rs ity am br id ev ie w ge 1.2 Completing the square 6 1.3 The quadratic formula C op 3 ni 1.1 Solving quadratic equations by factorisation U 10 w ge 1.4 Solving simultaneous equations (one linear and one quadratic) ev id ie 1.5 Solving more complex quadratic equations -C 1.8 The number of roots of a quadratic equation y rs ve 34 op 39 43 C U 2.3 Inverse functions w ge 2.4 The graph of a function and its inverse 2.7 Stretches es 67 70 ity C 72 3.2 Parallel and perpendicular lines 75 3.3 Equations of straight lines 78 U op ni ve rs 3.1 Length of a line segment and midpoint 82 e C 3.4 The equation of a circle w ie id g 3.5 Problems involving intersections of lines and circles 92 es s -R br am 88 ev End-of-chapter review exercise 3 -C w ie ev R 59 Pr op y End-of-chapter review exercise 2 51 57 s -C 2.8 Combined transformations 48 55 -R am br ev id ie 2.5 Transformations of functions 3 Coordinate geometry iii 33 ni ev 31 ity op C w ie 2.1 Definition of a function 2.2 Composite functions R 27 Pr End-of-chapter review exercise 1 2.6 Reflections 24 es 1.9 Intersection of a line and a quadratic curve 15 21 -R 1.7 Solving quadratic inequalities 11 17 s am br 1.6 Maximum and minimum values of a quadratic function 2 Functions y 1 y C 1 Quadratics w x ve rs ity op y Acknowledgements ev ie viii Pr es s -C How to use this book R vi -R Series introduction y Contents C U ni op y Contents Copyright Material - Review Only - Not for Redistribution ve rs ity C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 95 am br id ev ie w ge Cross-topic review exercise 1 118 121 C op 5.2 The general definition of an angle 123 U 5.3 Trigonometric ratios of general angles 127 w ge 5.4 Graphs of trigonometric functions id ie 5.5 Inverse trigonometric functions br ev 5.6 Trigonometric equations -R am 5.7 Trigonometric identities Pr y op iv ity 160 op 166 ni 6.3 Arithmetic progressions 6.4 Geometric progressions 156 ve 6.2 Binomial coefficients 155 171 C U R ev ie w 6.1 Binomial expansion of ( a + b ) n rs C 6 Series 175 w ge 6.5 Infinite geometric series br ev id ie 6.6 Further arithmetic and geometric series -R am End-of-chapter review exercise 6 es 190 191 C ity 7.1 Derivatives and gradient functions 198 7.3 Tangents and normals 7.4 Second derivatives 201 205 op w ni ve rs 7.2 The chain rule 209 -R s es am br ev ie id g w e C U End-of-chapter review exercise 7 -C ie 183 186 Pr op y 7 Differentiation ev 180 s -C Cross-topic review exercise 2 R 145 153 es End-of-chapter review exercise 5 140 149 s -C 5.8 Further trigonometric equations 136 y 5.1 Angles between 0° and 90° 116 ni ev ie w C 5 Trigonometry R 112 ve rs ity op y End-of-chapter review exercise 4 107 y 4.3 Area of a sector 101 104 Pr es s -C 4.2 Length of an arc 99 y 4.1 Radians -R 4 Circular measure Copyright Material - Review Only - Not for Redistribution ve rs ity C U ni op y Contents w ge 8 Further differentiation am br id ev ie 8.1 Increasing and decreasing functions 8.3 Practical maximum and minimum problems 221 227 Pr es s C w ev ie 235 ve rs ity op End-of-chapter review exercise 8 230 9 Integration 238 9.1 Integration as the reverse of differentiation 239 U R 9.3 Integration of expressions of the form ( ax + b ) C op 244 ni 9.2 Finding the constant of integration n 247 w ge 9.4 Further indefinite integration br ev id ie 9.5 Definite integration 9.8 Improper integrals 264 es s -C 260 268 Pr y 276 ity op ve 286 w ie 317 ev id br 319 y op -R s es -C am br ev ie id g w e C U R ev ie w ni ve rs C ity Pr op y es s -C -R am Glossary Index C U ni op 284 ge Answers v 280 rs C w ie Cross-topic review exercise 3 Practice exam-style paper ev 250 9.7 Area bounded by a curve and a line or by two curves End-of-chapter review exercise 9 R 249 253 -R am 9.6 Area under a curve 9.9 Volumes of revolution y y 8.5 Practical applications of connected rates of change y -C 8.4 Rates of change 213 216 -R 8.2 Stationary points 211 Copyright Material - Review Only - Not for Redistribution ve rs ity C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 w y ni C op C ve rs ity op y Pr es s -C -R Cambridge International AS & A Level Mathematics can be a life-changing course. On the one hand, it is a facilitating subject: there are many university courses that either require an A Level or equivalent qualification in mathematics or prefer applicants who have it. On the other hand, it will help you to learn to think more precisely and logically, while also encouraging creativity. Doing mathematics can be like doing art: just as an artist needs to master her tools (use of the paintbrush, for example) and understand theoretical ideas (perspective, colour wheels and so on), so does a mathematician (using tools such as algebra and calculus, which you will learn about in this course). But this is only the technical side: the joy in art comes through creativity, when the artist uses her tools to express ideas in novel ways. Mathematics is very similar: the tools are needed, but the deep joy in the subject comes through solving problems. w ev ie You might wonder what a mathematical ‘problem’ is. This is a very good question, and many people have offered different answers. You might like to write down your own thoughts on this question, and reflect on how they change as you progress through this course. One possible idea is that a mathematical problem is a mathematical question that you do not immediately know how to answer. (If you do know how to answer it immediately, then we might call it an ‘exercise’ instead.) Such a problem will take time to answer: you may have to try different approaches, using different tools or ideas, on your own or with others, until you finally discover a way into it. This may take minutes, hours, days or weeks to achieve, and your sense of achievement may well grow with the effort it has taken. Pr y es s -C -R am br ev id ie w ge U R ev ie am br id ge Series introduction y ve ni op This series of Cambridge International AS & A Level Mathematics coursebooks, written for the Cambridge Assessment International Education syllabus for examination from 2020, will support you both to learn the mathematics required for these examinations and to develop your mathematical problem-solving skills. The new examinations may well include more unfamiliar questions than in the past, and having these skills will allow you to approach such questions with curiosity and confidence. -R am br ev id ie w ge C U R ev ie w rs C ity op In addition to the mathematical tools that you will learn in this course, the problem-solving skills that you will develop will also help you throughout life, whatever you end up doing. It is very common to be faced with problems, be it in science, engineering, mathematics, accountancy, law or beyond, and having the confidence to systematically work your way through them will be very useful. vi Pr op y es s -C In addition to problem solving, there are two other key concepts that Cambridge Assessment International Education have introduced in this syllabus: namely communication and mathematical modelling. These appear in various forms throughout the coursebooks. y op C U w e ev ie id g es s -R br am -C R ev ie w ni ve rs C ity Communication in speech, writing and drawing lies at the heart of what it is to be human, and this is no less true in mathematics. While there is a temptation to think of mathematics as only existing in a dry, written form in textbooks, nothing could be further from the truth: mathematical communication comes in many forms, and discussing mathematical ideas with colleagues is a major part of every mathematician’s working life. As you study this course, you will work on many problems. Exploring them or struggling with them together with a classmate will help you both to develop your understanding and thinking, as well as improving your (mathematical) communication skills. And being able to convince someone that your reasoning is correct, initially verbally and then in writing, forms the heart of the mathematical skill of ‘proof’. Copyright Material - Review Only - Not for Redistribution ve rs ity C U ni op y Series introduction C ve rs ity op y Pr es s -C -R am br id ev ie w ge Mathematical modelling is where mathematics meets the ‘real world’. There are many situations where people need to make predictions or to understand what is happening in the world, and mathematics frequently provides tools to assist with this. Mathematicians will look at the real world situation and attempt to capture the key aspects of it in the form of equations, thereby building a model of reality. They will use this model to make predictions, and where possible test these against reality. If necessary, they will then attempt to improve the model in order to make better predictions. Examples include weather prediction and climate change modelling, forensic science (to understand what happened at an accident or crime scene), modelling population change in the human, animal and plant kingdoms, modelling aircraft and ship behaviour, modelling financial markets and many others. In this course, we will be developing tools which are vital for modelling many of these situations. ev ie w To support you in your learning, these coursebooks have a variety of new features, for example: ve ie w rs C ity op Pr y es s -C -R am br ev id ie w ge U R ni C op y ■ Explore activities: These activities are designed to offer problems for classroom use. They require thought and deliberation: some introduce a new idea, others will extend your thinking, while others can support consolidation. The activities are often best approached by working in small groups and then sharing your ideas with each other and the class, as they are not generally routine in nature. This is one of the ways in which you can develop problemsolving skills and confidence in handling unfamiliar questions. ■ Questions labelled as P , M or PS : These are questions with a particular emphasis on ‘Proof’, ‘Modelling’ or ‘Problem solving’. They are designed to support you in preparing for the new style of examination. They may or may not be harder than other questions in the exercise. ■ The language of the explanatory sections makes much more use of the words ‘we’, ‘us’ and ‘our’ than in previous coursebooks. This language invites and encourages you to be an active participant rather than an observer, simply following instructions (‘you do this, then you do that’). It is also the way that professional mathematicians usually write about mathematics. The new examinations may well present you with unfamiliar questions, and if you are used to being active in your mathematics, you will stand a better chance of being able to successfully handle such challenges. y op -R am We wish you every success as you embark on this course. y op U ev ie w ni ve rs C ity Pr op y es s -C Julian Gilbey London, 2018 br ev ie id g w e C Past exam paper questions throughout are reproduced by permission of Cambridge Assessment International Education. Cambridge Assessment International Education bears no responsibility for the example answers to questions taken from its past question papers which are contained in this publication. -R s es am The questions, example answers, marks awarded and/or comments that appear in this book were written by the author(s). In examination, the way marks would be awarded to answers like these may be different. -C R ev br id ie w ge C U R ni ev At various points in the books, there are also web links to relevant Underground Mathematics resources, which can be found on the free undergroundmathematics.org website. Underground Mathematics has the aim of producing engaging, rich materials for all students of Cambridge International AS & A Level Mathematics and similar qualifications. These high-quality resources have the potential to simultaneously develop your mathematical thinking skills and your fluency in techniques, so we do encourage you to make good use of them. Copyright Material - Review Only - Not for Redistribution vii op y ve rs ity ni am br id ev ie w ge C U How to use this book y ni C op Learning objectives indicate the important concepts within each chapter and help you to navigate through the coursebook. s es Pr ity Key point boxes contain a summary of the most important methods, facts and formulae. y op ni ev ve ie w C viii Prerequisite knowledge exercises identify prior learning that you need to have covered before starting the chapter. Try the questions to identify any areas that you need to review before continuing with the chapter. rs op y -C -R am br ev id ie w ge U R ev ie w C ve rs ity op y Pr es s -C -R Throughout this book you will notice particular features that are designed to help your learning. This section provides a brief overview of these features. ge C U R completing the square y It is important to remember to show appropriate calculations in coordinate geometry questions. Answers from scale drawings are not accepted. U op w ie ev -R s es am br ev ie id g w e C Explore boxes contain enrichment activities for extension work. These activities promote group work and peerto-peer discussion, and are intended to deepen your understanding of a concept. (Answers to the Explore questions are provided in the Teacher’s Resource.) -C R Worked examples provide step-by-step approaches to answering questions. The left side shows a fully worked solution, while the right side contains a commentary explaining each step in the working. ni ve rs C ity Pr op y es s -C -R am br ev id ie w Key terms are important terms in the topic that you are learning. They are highlighted in orange bold. The glossary contains clear definitions of these key terms. Copyright Material - Review Only - Not for Redistribution Tip boxes contain helpful guidance about calculating or checking your answers. ve rs ity In the Pure Mathematics 2 and 3 Coursebook, Chapter 7, you will learn how to expand these expressions for any real value of n. y op C w ni C op y ve rs ity C w U ie w ge id These questions focus on problem-solving. P M These questions focus on modelling. These questions focus on proofs. es s -C -R am br PS ev R Throughout each chapter there are multiple exercises containing practice questions. The questions are coded: Pr y ity op These questions are taken from past examination papers. rs C w y op ge w ie ev id -R es s -C am br The checklist contains a summary of the concepts that were covered in the chapter. You can use this to quickly check that you have covered the main topics. y op ie w ni ve rs C ity Pr op y The End-of-chapter review contains exam-style questions covering all topics in the chapter. You can use this to check your understanding of the topics you have covered. The number of marks gives an indication of how long you should be spending on the question. You should spend more time on questions with higher mark allocations; questions with only one or two marks should not need you to spend time doing complicated calculations or writing long explanations. C U R ni ev ve ie You should not use a calculator for these questions. You can use a calculator for these questions. At the end of each chapter there is a Checklist of learning and understanding. -R s es am br ev ie id g w e C U Cross-topic review exercises appear after several chapters, and cover topics from across the preceding chapters. -C ev Try the Sequences and Counting and Binomial resources on the Underground Mathematics website. Web link boxes contain links to useful resources on the internet. Rewind and Fast forward boxes direct you to related learning. Rewind boxes refer to earlier learning, in case you need to revise a topic. Fast forward boxes refer to topics that you will cover at a later stage, in case you would like to extend your study. Did you know? boxes contain interesting facts showing how Mathematics relates to the wider world. R WEB LINK Extension material goes beyond the syllabus. It is highlighted by a red line to the left of the text. Pr es s -C In Section 2.5 we learnt about the inverse of a function. Here we will look at the particular case of the inverse of a trigonometric function. ev ie E FAST FORWARD ev ie am br id REWIND -R ge U ni op y How to use this book Copyright Material - Review Only - Not for Redistribution ix ve rs ity C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 am br id ev ie w ge Acknowledgements op y Pr es s -C -R The authors and publishers acknowledge the following sources of copyright material and are grateful for the permissions granted. While every effort has been made, it has not always been possible to identify the sources of all the material used, or to trace all copyright holders. If any omissions are brought to our notice, we will be happy to include the appropriate acknowledgements on reprinting. ni C op y The following questions are used by permission of the Underground Mathematics website: Exercise 1F Question 9, Exercise 3C Question 16, Exercise 3E Questions 6 and 7, Exercise 4B Question 10. U R ev ie w C ve rs ity Past examination questions throughout are reproduced by permission of Cambridge Assessment International Education. w ge Thanks to the following for permission to reproduce images: br ev id ie Cover image iStock/Getty Images y op y op -R s es -C am br ev ie id g w e C U R ev ie w ni ve rs C ity Pr op y es s -C -R am br ev id ie w ge C U R ni ev ve ie w rs C x ity op Pr y es s -C -R am Inside (in order of appearance) English Heritage/Heritage Images/Getty Images, Sean Russell/Getty Images, Gopinath Duraisamy/EyeEm/Getty Images, Frank Fell/robertharding/Getty Images, Fred Icke/EyeEm/Getty Images, Ralph Grunewald/Getty Images, Gustavo Miranda Holley/Getty Images, shannonstent/Getty Images, wragg/Getty Images, Dimitrios Pikros/EyeEm/Getty Images Copyright Material - Review Only - Not for Redistribution op y ve rs ity ni C U ev ie w ge -R am br id Pr es s -C y ni C op y ve rs ity op C w ev ie rs w ge C U R ni op y ve ie w C ity op Pr y es s -C -R am br ev id ie w ge U R ev Chapter 1 Quadratics 1 id es s -C -R am br ev carry out the process of completing the square for a quadratic polynomial ax 2 + bx + c and use a completed square form find the discriminant of a quadratic polynomial ax 2 + bx + c and use the discriminant solve quadratic equations, and quadratic inequalities, in one unknown solve by substitution a pair of simultaneous equations of which one is linear and one is quadratic recognise and solve equations in x that are quadratic in some function of x understand the relationship between a graph of a quadratic function and its associated algebraic equation, and use the relationship between points of intersection of graphs and solutions of equations. Pr ity op y ni ve rs -R s es -C am br ev ie id g w e C U R ev ie w C op y ■ ■ ■ ■ ■ ■ ie In this chapter you will learn how to: Copyright Material - Review Only - Not for Redistribution ve rs ity ge C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 am br id ev ie w PREREQUISITE KNOWLEDGE What you should be able to do IGCSE® / O Level Mathematics Solve quadratic equations by factorising. b x 2 − 6x + 9 = 0 ve rs ity c 3x 2 − 17 x − 6 = 0 2 Solve: y b 3 − 2x ø 7 C op ni a 5x − 8 . 2 3 Solve: Solve simultaneous linear equations. a 2 x + 3 y = 13 w ge U R -R am br ev id ie 7 x − 5 y = −1 b 2 x − 7 y = 31 3x + 5 y = −31 4 Simplify: a ity op c rs C w 20 b ( 5 )2 Pr y es s -C Carry out simple manipulation of surds. 8 2 ni op y ve ie ev a x 2 + x − 12 = 0 Solve linear inequalities. IGCSE / O Level Additional Mathematics Why do we study quadratics? C U R 1 Solve: Pr es s -C y op C ev ie w IGCSE / O Level Mathematics IGCSE / O Level Mathematics 2 Check your skills -R Where it comes from -C -R am br ev id ie w ge At IGCSE / O Level, you will have learnt about straight-line graphs and their properties. They arise in the world around you. For example, a cell phone contract might involve a fixed monthly charge and then a certain cost per minute for calls: the monthly cost, y, is then given as y = mx + c, where c is the fixed monthly charge, m is the cost per minute and x is the number of minutes used. y You will have plotted graphs of quadratics such as y = 10 − x 2 before starting your -R s es am br ev ie id g w e C U op A Level course. These are most familiar as the shape of the path of a ball as it travels through the air (called its trajectory). Discovering that the trajectory is a quadratic was one of Galileo’s major successes in the early 17th century. He also discovered that the vertical motion of a ball thrown straight upwards can be modelled by a quadratic, as you will learn if you go on to study the Mechanics component. -C R ev ie w ni ve rs C ity Pr op y es s Quadratic functions are of the form y = ax 2 + bx + c (where a ≠ 0) and they have interesting properties that make them behave very differently from linear functions. A quadratic function has a maximum or a minimum value, and its graph has interesting symmetry. Studying quadratics offers a route into thinking about more complicated functions such as y = 7 x5 − 4x 4 + x 2 + x + 3. Copyright Material - Review Only - Not for Redistribution WEB LINK Try the Quadratics resource on the Underground Mathematics website (www.underground mathematics.org). ve rs ity C U ni op y Chapter 1: Quadratics w ge 1.1 Solving quadratic equations by factorisation am br id ev ie You already know the factorisation method and the quadratic formula method to solve quadratic equations algebraically. Pr es s -C -R This section consolidates and builds on your previous work on solving quadratic equations by factorisation. 2 x 2 + 3x − 5 = ( x − 1)( x − 2) ve rs ity w C op y EXPLORE 1.1 U ge 2x + 5 = x − 2 x = −7 y ev id Rearrange: w R Divide both sides by ( x − 1): C op ( x − 1)(2x + 5) = ( x − 1)( x − 2) ni Factorise the left-hand side: ie ev ie This is Rosa’s solution to the previous equation: -R am br Discuss her solution with your classmates and explain why her solution is not fully correct. es s -C Now solve the equation correctly. Solve: rs y ge Factorise. Use the fact that if pq = 0, then p = 0 or q = 0. x= w ie -R s 1 3 es or Divide both sides by the common factor of 3. 3x 2 − 13x − 10 = 0 Factorise. ity 9x 2 − 39x − 30 = 0 y ni ve rs ie op C U Solve. id g ev ie 2 or x = 5 3 -R s es -C am br x=− w e 3x + 2 = 0 or x − 5 = 0 TIP Divide by a common factor first, if possible. (3x + 2)( x − 5) = 0 ev R 5 2 Solve. Pr x= w C op y -C 2 x − 5 = 0 or 3x − 1 = 0 ev id am (2 x − 5)(3x − 1) = 0 b op 6x 2 − 17 x + 5 = 0 a C Write in the form ax 2 + bx + c = 0. U 6x 2 + 5 = 17 x br ni Answer ev b 9x 2 − 39x − 30 = 0 ve ie w a 6x 2 + 5 = 17 x R 3 ity C op Pr y WORKED EXAMPLE 1.1 Copyright Material - Review Only - Not for Redistribution ve rs ity ev ie w ge am br id Pr es s 21 2 − =1 2x x + 3 op ev ie 2 x 2 − 11x − 63 = 0 Factorise. U R ni (2 x + 7)( x − 9) = 0 Expand brackets and rearrange. ve rs ity w C 21( x + 3) − 4x = 2 x( x + 3) Multiply both sides by 2 x( x + 3). y y -C Answer -R 21 2 − = 1. 2x x + 3 Solve Solve. ie id br ev 7 or x = 9 2 -C -R am x=− w ge 2 x + 7 = 0 or x − 9 = 0 Pr 3x 2 + 26 x + 35 = 0. x2 + 8 ve rs Answer ni op y Multiply both sides by x 2 + 8. U R ev 3x 2 + 26 x + 35 =0 x2 + 8 Factorise. w ge 3x 2 + 26 x + 35 = 0 C ie w C Solve ity op y es s WORKED EXAMPLE 1.3 4 br ev id ie (3x + 5)( x + 7) = 0 -C -R am 3x + 5 = 0 or x + 7 = 0 Solve. es s 5 or x = −7 3 y op -R s es -C am br ev ie id g w e C U R ev ie w ni ve rs C ity Pr op y x=− C op WORKED EXAMPLE 1.2 C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 Copyright Material - Review Only - Not for Redistribution ve rs ity ge C U ni op y Chapter 1: Quadratics ev ie am br id w WORKED EXAMPLE 1.4 A rectangle has sides of length x cm and (6x − 7) cm. Pr es s -C Find the lengths of the sides of the rectangle. y Answer C w 6x 2 − 7 x − 90 = 0 Factorise. ni C op y (2 x − 9)(3x + 10) = 0 Solve. w 10 3 When x = 4 21 , 6x − 7 = 20. op Pr y es -C The rectangle has sides of length 4 21 cm and 20 cm. C 2 4(2 x + x − 6) = 1 ( x 2 − 3x + 1)6 = 1 C 2 ( x 2 − 3x + 1)(2 x + x − 6) = 1 ev ve ie B rs w 5 ity EXPLORE 1.2 A -R am ev ie Length is a positive quantity, so x = 4 21 . s x=− or id 9 2 br x= ge U R 2 x − 9 = 0 or 3x + 10 = 0 w Remember to check each of your answers. ev br -R am 3 State how many values of x satisfy: b equation B c equation C s -C a equation A TIP c equation C ie b equation B id a equation A C U 2 Solve: ge R ni op 1 Discuss with your classmates how you would solve each of these equations. y ev ie 6x – 7 Rearrange. ve rs ity op Area = x(6x − 7) = 6x 2 − 7 x = 90 x -R The area of the rectangle is 90 cm 2. y op -R s es -C am br ev ie id g w e C U R ev ie w ni ve rs C ity Pr op y es 4 Discuss your results. Copyright Material - Review Only - Not for Redistribution ve rs ity ev ie am br id b x 2 − 7 x + 12 = 0 d 5x 2 + 19x + 12 = 0 e 20 − 7 x = 6x 2 b 2 3 + =1 x x+2 d 5 3x + =2 x+3 x+4 f 3 1 1 + = x + 2 x − 1 ( x + 1)( x + 2) x2 + x − 6 =0 x2 + 5 c e 6x 2 + x − 2 =0 x2 + 7x + 4 ev 2 x 2 + 9x − 5 =0 x4 + 1 es b 4(2 x − 11x + 15) = 1 c 2( x − 4 x + 6) = 8 e ( x 2 + 2 x − 14)5 = 1 f ( x 2 − 7 x + 11)8 = 1 2 1 9 5 The diagram shows a right-angled triangle with sides 2 x cm, (2 x + 1) cm and 29 cm. rs w y op ni ev a Show that 2 x 2 + x − 210 = 0. U R b Find the lengths of the sides of the triangle. w ge C 2x + 1 ev id br -R am s x = 1. Pr es -C 7 Solve ( x − 11x + 29) op y PS x+3 x–1 (6 x 2 + x − 2) C ity 1.2 Completing the square w ev ie id g es s -R br am C U e ( x + d )2 = x 2 + 2 dx + d 2 and ( x − d )2 = x 2 − 2 dx + d 2 op y The method of completing the square aims to rewrite a quadratic expression using only one occurrence of the variable, making it an easier expression to work with. -C R ev ie w ni ve rs Another method we can use for solving quadratic equations is completing the square. If we expand the expressions ( x + d )2 and ( x − d )2 , we obtain the results: Check that your answers satisfy the original equation. WEB LINK ie 6 The area of the trapezium is 35.75 cm 2. Find the value of x. 2 TIP 29 2x ve ie 2 ity y op C 2 s -C 2 -R f 4 Find the real solutions of the following equations. a 8( x + 2 x − 15) = 1 x2 − 9 =0 7 x + 10 w b Pr ge id am br x2 − 2x − 8 =0 x 2 + 7 x + 10 d 3(2 x + 9 x + 2) = x(10 x − 13) = 3 y 3 1 + =2 x + 1 x( x + 1) 3 Solve: 3x 2 + x − 10 =0 a x2 − 7x + 6 6 f C op e ni 5x + 1 2 x − 1 − = x2 4 2 ve rs ity 6 =0 x−5 c d x 2 − 6x − 16 = 0 U R ev ie w C op y 2 Solve: a x− c Pr es s -C a x 2 + 3x − 10 = 0 -R 1 Solve by factorisation. ie EXERCISE 1A w ge C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 Copyright Material - Review Only - Not for Redistribution Try the Factorisable quadratics resource on the Underground Mathematics website. ve rs ity C U ni op y Chapter 1: Quadratics ev ie am br id w ge Rearranging these gives the following important results: Pr es s -C x 2 + 2 dx = ( x + d )2 − d 2 and x 2 − 2 dx = ( x − d )2 − d 2 -R KEY POINT 1.1 To complete the square for x 2 + 10 x , we can use the first of the previous results as follows: ve rs ity ev ie w C op y 10 ÷ 2 = 5 ւց x 2 + 10 x = ( x + 5)2 − 52 x 2 + 10 x = ( x + 5)2 − 25 U R ni C op y To complete the square for x 2 + 8x − 7, we again use the first result applied to the x 2 + 8x part, as follows: id ie w ge 8÷2 = 4 ւց x 2 + 8x − 7 = ( x + 4)2 − 42 − 7 br ev x 2 + 8x − 7 = ( x + 4)2 − 23 -R am To complete the square for 2 x 2 − 12 x + 5, we must first take a factor of 2 out of the first s -C two terms, so: w es Pr ity 7 2 x 2 − 12 x + 5 = 2 [( x − 3)2 − 9] + 5 = 2( x − 3)2 − 13 ve ie 6÷2 = 3 ւց x 2 − 6x = ( x − 3)2 − 32 , giving rs C op y 2 x 2 − 12 x + 5 = 2( x 2 − 6x ) + 5 y op id ie w ge WORKED EXAMPLE 1.5 C U R ni ev We can also use an algebraic method for completing the square, as shown in Worked example 1.5. -R -C s 2 x 2 − 12 x + 3 = p( x − q )2 + r Pr ity 2 x 2 − 12 x + 3 = px 2 − 2 pqx + pq 2 + r −12 = −2 pq (1) 3 = pq 2 + r (2) w e Substituting p = 2 and q = 3 in equation (3) therefore gives r = −15 es s -R br ev ie id g 2 x 2 − 12 x + 3 = 2( x − 3)2 − 15 am C U op Substituting p = 2 in equation (2) gives q = 3 (3) y 2= p ni ve rs Comparing coefficients of x 2 , coefficients of x and the constant gives -C R ev ie w C op y Expanding the brackets and simplifying gives: es Answer am br ev Express 2 x 2 − 12 x + 3 in the form p( x − q )2 + r, where p, q and r are constants to be found. Copyright Material - Review Only - Not for Redistribution ve rs ity ev ie w ge am br id WORKED EXAMPLE 1.6 C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 4x 2 + 20 x + 5 = ( ax + b )2 + c op y Expanding the brackets and simplifying gives: ve rs ity 4x 2 + 20 x + 5 = a 2 x 2 + 2 abx + b2 + c Pr es s -C Answer -R Express 4x 2 + 20 x + 5 in the form ( ax + b )2 + c, where a, b and c are constants to be found. (1) 20 = 2ab (2) (3) ni Equation (1) gives a = ± 2. 5 = b2 + c y ev ie w 4 = a2 C op C Comparing coefficients of x 2, coefficients of x and the constant gives U R Substituting a = 2 into equation (2) gives b = 5. w ge Substituting b = 5 into equation (3) gives c = −20. rs WORKED EXAMPLE 1.7 ni 5 3 + = 1. x+2 x−5 y Use completing the square to solve the equation op ev ve ie w C 8 ity op Pr y 4x 2 + 20 x + 5 = ( −2 x − 5)2 − 20 = (2 x + 5)2 − 20 es -C Substituting b = −5 into equation (3) gives c = −20. s Substituting a = −2 into equation (2) gives b = −5. -R br am Alternatively: ev id ie 4x 2 + 20 x + 5 = (2 x + 5)2 − 20 ie id Answer w ge C U R Leave your answers in surd form. 5 3 + =1 x+2 x−5 -R am br ev Multiply both sides by ( x + 2)( x − 5). -C 5( x − 5) + 3( x + 2) = ( x + 2)( x − 5) s 2 Pr ity y op C U 11 85 ± 2 2 ie ev -R 1 (11 ± 85 ) 2 s -C am x = es br id g x = 85 4 e R 11 =± 2 ni ve rs 2 x − 11 = 85 2 4 ev ie w C x − 11 − 11 + 9 = 0 2 2 x− Complete the square. w 2 es op y x 2 − 11x + 9 = 0 Expand brackets and collect terms. Copyright Material - Review Only - Not for Redistribution ve rs ity ev ie am br id EXERCISE 1B w ge C U ni op y Chapter 1: Quadratics 1 Express each of the following in the form ( x + a )2 + b. f c x 2 − 4x − 8 x 2 − 3x g x2 + 7x + 1 d x 2 + 15x h x 2 − 3x + 4 Pr es s x 2 + 4x + 8 -C e b x 2 + 8x -R a x 2 − 6x b 3x 2 − 12 x − 1 2 x 2 + 5x − 1 d 2x2 + 7x + 5 b 8x − x 2 c 4 − 3x − x 2 d 9 + 5x − x 2 y a 4x − x 2 ev ie c ve rs ity a 2 x 2 − 12 x + 19 3 Express each of the following in the form a − ( x + b )2 . w C op y 2 Express each of the following in the form a ( x + b )2 + c. U c 13 + 4x − 2 x 2 4x 2 + 20 x + 30 c 25x 2 + 40 x − 4 ev br b -R 9x 2 − 6x − 3 am a id 5 Express each of the following in the form ( ax + b )2 + c. d 2 + 5x − 3x 2 w b 3 − 12 x − 2 x 2 ge a 7 − 8x − 2 x 2 ie R ni C op 4 Express each of the following in the form p − q ( x + r )2 . b x 2 + 4x − 12 = 0 c x 2 − 2 x − 35 = 0 d x 2 − 9x + 14 = 0 e x 2 + 3x − 18 = 0 f x 2 + 9x − 10 = 0 op Pr y s -C a x 2 + 8x − 9 = 0 es 6 Solve by completing the square. d 9x 2 − 42 x + 61 9 b x 2 − 10 x + 2 = 0 c x 2 + 8x − 1 = 0 d 2 x 2 − 4x − 5 = 0 rs ity a x 2 + 4x − 7 = 0 f 2 x 2 − 8x − 3 = 0 y ve op 5 3 + = 2. Leave your answers in surd form. x+2 x−4 w ge 9 The diagram shows a right-angled triangle with sides x m, (2 x + 5) m and 10 m. br ev x -R Find the value of x. Leave your answer in surd form. am 10 ie id PS C U R 8 Solve e 2 x 2 + 6x + 3 = 0 ni ev ie w C 7 Solve by completing the square. Leave your answers in surd form. 2x + 5 PS 10 Find the real solutions of the equation (3x + 5x − 7) = 1. PS 49x 2 11 The path of a projectile is given by the equation y = ( 3 )x − , where x and y 9000 are measured in metres. es s 4 Pr op y -C 2 ity op y (x, y) Range br ev a Find the range of this projectile. am x ie id g O w e C U R ev ie w ni ve rs C y -R s es -C b Find the maximum height reached by this projectile. Copyright Material - Review Only - Not for Redistribution TIP You will learn how to derive formulae such as this if you go on to study Further Mathematics . ve rs ity C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 am br id -R If ax 2 + bx + c = 0, where a, b and c are constants and a ≠ 0, then Pr es s -C KEY POINT 1.2 −b ± b 2 − 4ac 2a C ve rs ity op y x= w We can solve quadratic equations using the quadratic formula. ev ie ge 1.3 The quadratic formula 2 U b c x+ =0 a a 2 w x+ b − b + c = 0 2a 2a a br ev id ie Rearrange the equation. 2 Write the right-hand side as a single fraction. -R am 2 x+ b = b − c 2a 4a 2 a 2 s b2 − 4ac 2a rs Write the right-hand side as a single fraction. y op w ie ev b from both sides. 2a −b ± b2 − 4ac 2a ie w ge id WORKED EXAMPLE 1.8 C U R x= es b ± 2a Pr x=− Subtract ity b2 − 4ac 2a C b =± 2a ve op x+ Find the square root of both sides. ni y -C 2 x + b = b − 4ac 2a 4a 2 10 y Complete the square. ge R x2 + Divide both sides by a. ni ax 2 + bx + c = 0 C op ev ie w The quadratic formula can be proved by completing the square for the equation ax 2 + bx + c = 0: -R am br ev Solve the equation 6x 2 − 3x − 2 = 0. -C Write your answers correct to 3 significant figures. es s Answer 3 + 57 3 − 57 or x = 12 12 y op -R s es am br ev ie id g w e C U or x = −0.379 (to 3 significant figures) -C w ie ev R x = 0.879 Pr x= ni ve rs −( −3) ± ( −3)2 − 4 × 6 × ( −2) 2×6 C x= ity op y Using a = 6, b = −3 and c = −2 in the quadratic formula gives: Copyright Material - Review Only - Not for Redistribution ve rs ity ev ie am br id EXERCISE 1C w ge C U ni op y Chapter 1: Quadratics 1 Solve using the quadratic formula. Give your answer correct to 2 decimal places. b x 2 + 6x + 4 = 0 d 2 x 2 + 5x − 6 = 0 e 4x 2 + 7 x + 2 = 0 c x 2 + 3x − 5 = 0 f 5x 2 + 7 x − 2 = 0 Pr es s -C -R a x 2 − 10 x − 3 = 0 The area of the rectangle is 63 cm 2. ve rs ity w C op y 2 A rectangle has sides of length x cm and (3x − 2) cm. Find the value of x, correct to 3 significant figures. y ev ie 3 Rectangle A has sides of length x cm and (2 x − 4) cm. U R ni C op Rectangle B has sides of length ( x + 1) cm and (5 − x ) cm. ge Rectangle A and rectangle B have the same area. id ie w Find the value of x, correct to 3 significant figures. br ev 5 2 + = 1. x −3 x +1 Give your answers correct to 3 significant figures. -C -R am 4 Solve the equation WEB LINK 5 Solve the quadratic equation ax − bx + c = 0, giving your answers in terms of a, b and c. es Pr y rs ity op C How do the solutions of this equation relate to the solutions of the equation ax 2 + bx + c = 0? w ve ie s 2 y op ni ev 1.4 Solving simultaneous equations (one linear and one quadratic) C ie y = x2 – 4 y op w ni ve rs (–1, –3) x ity C O Pr op y es s -C -R am ev y = 2x – 1 br id y (3, 5) ie C U The diagram shows the graphs of y = x 2 − 4 and y = 2 x − 1. ie id g w e The coordinates of the points of intersection of the two graphs are ( −1, −3) and (3, 5). -R s es am br ev It follows that x = −1, y = −3 and x = 3, y = 5 are the solutions of the simultaneous equations y = x 2 − 4 and y = 2 x − 1. -C ev R w ge U R In this section, we shall learn how to solve simultaneous equations where one equation is linear and the second equation is quadratic. Copyright Material - Review Only - Not for Redistribution Try the Quadratic solving sorter resource on the Underground Mathematics website. 11 ve rs ity C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 w ev ie (1) am br id (2) -R y = x2 − 4 y = 2x − 1 ge The solutions can also be found algebraically: Substitute for y from equation (2) into equation (1): -C 2x − 1 = x2 − 4 x2 − 2x − 3 = 0 ( x + 1)( x − 3) = 0 x = −1 or x = 3 C ve rs ity op y Pr es s Rearrange. Factorise. ni The solutions are: x = −1, y = −3 and x = 3, y = 5. id ie w ge In general, an equation in x and y is called quadratic if it has the form ax 2 + bxy + cy2 + dx + ey + f = 0, where at least one of a, b and c is non-zero. C op y Substituting x = 3 into equation (2) gives y = 6 − 1 = 5. U R ev ie w Substituting x = −1 into equation (2) gives y = −2 − 1 = −3. -C -R am br ev Our technique for solving one linear and one quadratic equation will work for these more general quadratics, too. (The graph of a general quadratic function such as this is called a conic.) y es s WORKED EXAMPLE 1.9 op Pr Solve the simultaneous equations. 12 ity R x − 4y = 8 y (1) 2 ni 2 op Answer 2x + 2 y = 7 ve rs x 2 − 4 y2 = 8 (2) U ev ie w C 2x + 2 y = 7 br 2 ev id ie w ge C 7 − 2y . 2 Substitute for x in equation (2): From equation (1), x = -C -R am 7 − 2 y − 4 y2 = 8 2 s 49 − 28 y + 4 y2 − 4 y2 = 8 4 es ity Rearrange. Factorise. id g ev ie 17 19 in equation (1) gives x = . 6 3 -R s es -C am br Substituting y = − w e C U 1 17 or y = 2 6 op (6 y + 17)(2 y − 1) = 0 y ni ve rs 12 y2 + 28 y − 17 = 0 y=− Multiply both sides by 4. Pr op y R ev ie w C 49 − 28 y + 4 y2 − 16 y2 = 32 Expand brackets. Copyright Material - Review Only - Not for Redistribution ve rs ity Pr es s From equation (1), 2 y = 7 − 2 x . C op y Rearrange. ni Factorise. U 3x 2 − 28x + 57 = 0 ie id 19 17 1 ,y=− and x = 3, y = . 3 6 2 op Pr y es s -C The solutions are: x = ity rs y = 6−x id 8x 2 − 2 xy = 4 y op x 2 + 3xy = 10 m 2 x + 3 y + 19 = 0 s x 2 + y2 + 4xy = 24 2 x − y = 14 o x − 12 y = 30 Pr n x + 2y = 5 2 y2 − xy = 20 x 2 + y2 = 10 ity 2x2 + 3 y = 5 l x + 2y = 6 y 2 = 8x + 4 es xy = 12 ev i k x − 4y = 2 2 4x − 3 y = 5 ni ve rs op y 2 The sum of two numbers is 26. The product of the two numbers is 153. op y a What are the two numbers? id g w e C U b If instead the product is 150 (and the sum is still 26), what would the two numbers now be? -R s es am br ev ie 3 The perimeter of a rectangle is 15.8 cm and its area is 13.5 cm 2. Find the lengths of the sides of the rectangle. -C C x 2 − 4xy = 20 -R br -C 5x − 2 y = 23 x − 5xy + y = 1 w f 2 x 2 − 3 y2 = 15 am xy = 8 ie x − 2y = 6 h 2y − x = 5 g 2x + y = 8 2 x 2 + y2 = 100 w ge e x 2 + 2 xy = 8 ie ni U R y = x2 j c 3 y = x + 10 b x + 4y = 6 ve ev a d y = 3x − 1 ev 13 1 Solve the simultaneous equations. ie w C EXERCISE 1D R -R br ev 19 or x = 3 3 am x= w ge (3x − 19)( x − 3) = 0 C ev ie x 2 − 49 + 28x − 4x 2 = 8 R Expand brackets. ve rs ity w C op y Substitute for 2 y in equation (2): x 2 − (7 − 2 x )2 = 8 ev ie am br id -C Alternative method: -R 1 into equation (1) gives x = 3. 2 19 17 1 The solutions are: x = , y = − and x = 3, y = . 3 6 2 Substituting y = w ge C U ni op y Chapter 1: Quadratics Copyright Material - Review Only - Not for Redistribution ve rs ity C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 am br id ev ie w ge 4 The sum of the perimeters of two squares is 50 cm and the sum of the areas is 93.25 cm 2. -R Find the side length of each square. Pr es s -C 5 The sum of the circumferences of two circles is 36 π cm and the sum of the areas is 170 π cm 2. 6 A cuboid has sides of length 5 cm, x cm and y cm. Given that x + y = 20.5 and the volume of the cuboid is 360 cm 3, find the value of x and the value of y. ve rs ity w C op y Find the radius of each circle. y C op U R ni ev ie 7 The diagram shows a solid formed by joining a hemisphere, of radius r cm, to a cylinder, of radius r cm and height h cm. h ie w ge The total height of the solid is 18 cm and the surface area is 205 π cm 2. 18 cm r The surface area, A, of a sphere with radius r is A = 4 π r 2. br ev id Find the value of r and the value of h. TIP s -C a Find the coordinates of the points A and B. -R am 8 The line y = 2 − x cuts the curve 5x 2 − y2 = 20 at the points A and B. Pr y es b Find the length of the line AB. rs C a Find the coordinates of the points A and B. ity op 9 The line 2 x + 5 y = 1 meets the curve x 2 + 5xy − 4 y2 + 10 = 0 at the points A and B. 14 ve ie w b Find the midpoint of the line AB. y op C U R ni ev 10 The line 7 x + 2 y = −20 intersects the curve x 2 + y2 + 4x + 6 y − 40 = 0 at the points A and B. Find the length of the line AB. id ie w ge 11 The line 7 y − x = 25 cuts the curve x 2 + y2 = 25 at the points A and B. br ev Find the equation of the perpendicular bisector of the line AB. -R am 12 The straight line y = x + 1 intersects the curve x 2 − y = 5 at the points A and B. es s -C Given that A lies below the x-axis and the point P lies on AB such that AP : PB = 4 : 1, find the coordinates of P. Pr ity ni ve rs WEB LINK op y 14 a Split 10 into two parts so that the difference between the squares of the parts is 60. es s -R br ev ie id g w e C U b Split N into two parts so that the difference between the squares of the parts is D. am R ev ie PS Find the equation of the perpendicular bisector of the line AB. -C w C op y 13 The line x − 2 y = 1 intersects the curve x + y2 = 9 at two points, A and B. Copyright Material - Review Only - Not for Redistribution Try the Elliptical crossings resource on the Underground Mathematics website. ve rs ity C U ni op y Chapter 1: Quadratics w ge 1.5 Solving more complex quadratic equations am br id ev ie You may be asked to solve an equation that is quadratic in some function of x. -R WORKED EXAMPLE 1.10 Pr es s -C Solve the equation 4x 4 − 37 x 2 + 9 = 0. op y Answer ve rs ity y U w ge 1 or y = 9 4 id es s -C -R am br ev 1 or x 2 = 9 4 1 or x = ± 3 x=± 2 Method 2: Factorise directly x2 = op y ve rs ity 15 ge C U R ni ev ie w C op Pr y 4x 4 − 37 x 2 + 9 = 0 (4x 2 − 1)( x 2 − 9) = 0 1 or x 2 = 9 x2 = 4 1 x=± or x = ± 3 2 ev id ie w WORKED EXAMPLE 1.11 -R -C Answer am br Solve the equation x − 4 x − 12 = 0. es Pr x. ity y2 − 4 y − 12 = 0 y = −2 Substitute U x = −2 has no solutions as x is never negative. ev -R s es am br ∴ x = 36 ie id g w e x = 6 or x = −2 x for y. C y = 6 or ni ve rs ( y − 6)( y + 2) = 0 -C R ev ie w C op y Let y = s x − 4 x − 12 = 0 y R (4 y − 1)( y − 9) = 0 y= Substitute x 2 for y. ni 4 y2 − 37 y + 9 = 0 ie ev ie Let y = x 2 . C op 4x 4 − 37 x 2 + 9 = 0 op w C Method 1: Substitution method Copyright Material - Review Only - Not for Redistribution ve rs ity ev ie w ge am br id WORKED EXAMPLE 1.12 C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 y ni w -R es s -C EXERCISE 1E am br ev id ie R x = −1 or x = 2 1 = 3−1 and 9 = 32. 3 U 1 or 3x = 9 3 Substitute 3x for y. ge 3x = ev ie 1 or y = 9 3 ve rs ity (3 y − 1)( y − 9) = 0 w C op y 3 y2 − 28 y + 9 = 0 y= Let y = 3x. Pr es s -C 3(3x )2 − 28(3x ) + 9 = 0 C op Answer -R Solve the equation 3(9x ) − 28(3x ) + 9 = 0. Pr b x6 − 7x3 − 8 = 0 c x 4 − 6x 2 + 5 = 0 e 3x 4 + x 2 − 4 = 0 f 8x 6 − 9x 3 + 1 = 0 h x 4 + 9x 2 + 14 = 0 i x8 − 15x 4 − 16 = 0 l 8 7 + =1 x6 x3 c 6x − 17 x + 5 = 0 f 3 x+ 32 x10 − 31x5 − 1 = 0 k 9 5 + =4 x 4 x2 b x ( x + 1) = 6 C ge 2 Solve: op j br ev id a 2 x − 9 x + 10 = 0 w x 4 + 2 x 2 − 15 = 0 ie g ve rs d 2 x 4 − 11x 2 + 5 = 0 y ity a x 4 − 13x 2 + 36 = 0 ni R ev ie w C 16 U op y 1 Find the real values of x that satisfy the following equations. d 10 x + x − 2 = 0 -R am e 8x + 5 = 14 x 5 = 16 x s -C 3 The curve y = 2 x and the line 3 y = x + 8 intersect at the points A and B. es op y a Write down an equation satisfied by the x-coordinates of A and B. Pr ity y 7 O 1 es s -R br ev ie id g w e C U op ni ve rs 4 The graph shows y = ax + b x + c for x ù 0. The graph crosses the x-axis 49 at the points ( 1, 0 ) and , 0 and it meets the y-axis at the point (0, 7). 4 Find the value of a, the value of b and the value of c. y Find the length of the line AB. am R ev ie w PS c -C C b Solve your equation in part a and, hence, find the coordinates of A and B. Copyright Material - Review Only - Not for Redistribution 49 4 x ve rs ity y ev ie w ge 5 The graph shows y = a (22 x ) + b(2 x ) + c. The graph crosses the axes at the points (2, 0), (4, 0) and (0, 90). Find the value of a, the value of b and the value of c. 90 2 O 4 x op y Pr es s -C -R am br id PS C U ni op y Chapter 1: Quadratics C op y The general form of a quadratic function is f( x ) = ax 2 + bx + c, where a, b and c are constants and a ≠ 0. ni ev ie w C ve rs ity 1.6 Maximum and minimum values of a quadratic function br A point where the gradient is zero is called a stationary point or a turning point. -R am es s -C Pr y ity op 17 If a , 0, the curve has a maximum point that occurs at the highest point of the curve. ge C U R ni op y ve rs C ie w If a . 0, the curve has a minimum point that occurs at the lowest point of the curve. ev TIP ev id ie w ge U R The shape of the graph of the function f( x ) = ax 2 + bx + c is called a parabola. The orientation of the parabola depends on the value of a, the coefficient of x 2. id ie w In the case of a parabola, we also call this point the vertex of the parabola. br ev Every parabola has a line of symmetry that passes through the vertex. -R am One important skill that we will develop during this course is ‘graph sketching’. es s -C A sketch graph needs to show the key features and behaviour of a function. C ● Pr WEB LINK op y Depending on the context we should show some or all of these. -R s es am br ev ie id g w e C U The skills you developed earlier in this chapter should enable you to draw a clear sketch graph for any quadratic function. -C R ev ie w ● the general shape of the graph the axis intercepts the coordinates of the vertex. ni ve rs ● ity op y When we sketch the graph of a quadratic function, the key features are: Copyright Material - Review Only - Not for Redistribution Try the Quadratic symmetry resource on the Underground Mathematics website for a further explanation of this. ve rs ity ev ie w ge am br id DID YOU KNOW? y U ni C op ev ie w C ve rs ity op y Pr es s -C -R If we rotate a parabola about its axis of symmetry, we obtain a three-dimensional shape called a paraboloid. Satellite dishes are paraboloid shapes. They have the special property that light rays are reflected to meet at a single point, if they are parallel to the axis of symmetry of the dish. This single point is called the focus of the satellite dish. A receiver at the focus of the paraboloid then picks up all the information entering the dish. WORKED EXAMPLE 1.13 R C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 w ge For the function f( x ) = x 2 − 3x − 4 : ie id a Find the axes crossing points for the graph of y = f( x ). -R Answer am br ev b Sketch the graph of y = f( x ) and find the coordinates of the vertex. es s -C a y = x 2 − 3x − 4 ( x + 1)( x − 4) = 0 x = −1 or x = 4 w Pr ity When y = 0, x 2 − 3x − 4 = 0 C ev ve ie Axes crossing points are: (0, −4), ( −1, 0) and (4, 0). 3 2 w ge y = x 2 – 3x – 4 ev –1 O x -R am br id ie x= y C U R ni op b The line of symmetry cuts the x-axis midway between the axis intercepts of −1 and 4. s es ni ve rs 2 3 3 3 , y = −3 − 4 2 2 2 25 y=− 4 Since a . 0, the curve is U-shaped. 3 25 Minimum point = , − 2 4 -R s es am br ev ie id g w e C U When x = y 3 . 2 op Hence, the line of symmetry is x = -C ev ie w ( 32 , –254 ity Pr –4 C op y -C 4 ( R y 18 rs op y When x = 0, y = −4 Copyright Material - Review Only - Not for Redistribution TIP Write your answer in fraction form. ve rs ity C U ni op y Chapter 1: Quadratics ev ie am br id w ge Completing the square is an alternative method that can be used to help sketch the graph of a quadratic function. 2 -R Completing the square for x 2 − 3x − 4 gives: 2 3 25 =x− − 2 4 2 C This part of the expression is a square so it will be at least zero. The smallest value it can be is 0. This 3 occurs when x = . 2 y 25 3 3 25 and this minimum occurs when x = . is − The minimum value of x − − 4 2 2 4 25 3 2 So the function f( x ) = x − 3x − 4 has a minimum point at . ,− 2 4 3 The line of symmetry is x = . 2 Pr 19 ity op WORKED EXAMPLE 1.14 w rs C es s -C if a , 0, there is a maximum point at ( h, k ). y ● -R am br ev If f( x ) = ax 2 + bx + c is written in the form f( x ) = a ( x − h )2 + k, then: b ● the line of symmetry is x = h = − 2a ● if a . 0, there is a minimum point at ( h, k ) y C U Answer ge id ie w This part of the expression is a square so ( x − 2)2 ù 0. The smallest value it can be is 0. This occurs when x = 2. br ev The maximum value of 9 − 4( x − 2)2 is 9 and this maximum occurs when x = 2. -R am Since this is being subtracted from 9, the whole expression is greatest when x = 2. s -C So the function f( x ) = 16x − 7 − 4x 2 has a maximum point at (2, 9). y es y O 31 2 1 2 C w ev s -R x=2 es am -C x ie –7 br id g e U R 3 x−2 = ± 2 1 x = 3 21 or x = 2 y = 16x – 7 – 4x2 ity ni ve rs ie w C When x = 0, y = −7 When y = 0, 9 − 4( x − 2)2 = 0 9 ( x − 2)2 = 4 (2, 9) Pr op y The line of symmetry is x = 2. op R op ni ev ve ie Sketch the graph of y = 16x − 7 − 4x 2 . Completing the square gives: 16x − 7 − 4x 2 = 9 − 4( x − 2)2 ev ie id KEY POINT 1.3 w ge U R ni C op w ev ie Pr es s 2 ve rs ity op y -C 3 3 x 2 − 3x − 4 = x − − − 4 2 2 Copyright Material - Review Only - Not for Redistribution ve rs ity ev ie am br id EXERCISE 1F w ge C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 1 Use the symmetry of each quadratic function to find the maximum or minimum points. y = x 2 − 6x + 8 b y = x 2 + 5x − 14 y = 2 x 2 + 7 x − 15 c d y = 12 + x − x 2 Pr es s -C a -R Sketch each graph, showing all axes crossing points. b Write down the equation of the line of symmetry for the graph of y = 2 x 2 − 8x + 1. ve rs ity w C op y 2 a Express 2 x 2 − 8x + 5 in the form a ( x + b )2 + c, where a, b and c are integers. 3 a Express 7 + 5x − x 2 in the form a − ( x + b )2, where a, and b are constants. y C op U R ni ev ie b Find the coordinates of the turning point of the curve y = 7 + 5x − x 2, stating whether it is a maximum or a minimum point. ge 4 a Express 2 x 2 + 9x + 4 in the form a ( x + b )2 + c, where a, b and c are constants. br ev id ie w b Write down the coordinates of the vertex of the curve y = 2 x 2 + 9x + 4, and state whether this is a maximum or a minimum point. op 20 es Pr y b Sketch the graph of y = 1 + x − 2 x 2 . -R -C 6 a Write 1 + x − 2 x 2 in the form p − 2( x − q )2. s am 5 Find the minimum value of x 2 − 7 x + 8 and the corresponding value of x. ity 8 Find the equations of parabolas A, B and C. ve ie w PS rs C 7 Prove that the graph of y = 4x 2 + 2 x + 5 does not intersect the x-axis. y ev y ev O –2 8 x 6 C es –4 Pr –6 op y 4 s 2 -R 2 –2 A ie id 4 br am 8 6 -C –4 10 C ge B w U R ni op 12 y op -R s es -C am br ev ie id g w e C U R ev ie w ni ve rs C ity –8 Copyright Material - Review Only - Not for Redistribution ve rs ity C 9 The diagram shows eight parabolas. w ge PS U ni op y Chapter 1: Quadratics am br id ev ie The equations of two of the parabolas are y = x 2 − 6x + 13 and y = − x 2 − 6x − 5. -R a Identify these two parabolas and find the equation of each of the other parabolas. Pr es s y E F A ev ie w C ve rs ity op y -C b Use graphing software to create your own parabola pattern. B x id w ie D C ev H -R am br G ge U R ni C op y O w s es rs 11 A parabola passes through the points ( −2, −3), (2, 9) and (6, 5). Find the equation of the parabola. y ni op 12 Prove that any quadratic that has its vertex at ( p, q ) has an equation of the form y = ax 2 − 2 apx + ap2 + q for some non-zero real number a. id ie w ge C U R P ve ie ev 21 ity op C PS Find the equation of the parabola. Pr 10 A parabola passes through the points (0, −24), ( −2, 0) and (4, 0). y PS -C [This question is an adaptation of Which parabola? on the Underground Mathematics website and was developed from an original idea from NRICH.] br ev 1.7 Solving quadratic inequalities -R am We already know how to solve linear inequalities. es s -C The following text shows two examples. Pr Expand brackets. 2 x + 14 , − 4 Subtract 14 from both sides. ity Divide both sides by 2. op C -R s es am w Divide both sides by −2. ie U Subtract 11 from both sides. br xø3 id g −2 x ù − 6 e Solve 11 − 2 x ù 5. -C R ev ie x , −9 y ni ve rs w 2 x , −18 ev C op y Solve 2( x + 7) , − 4 . Copyright Material - Review Only - Not for Redistribution ve rs ity C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 am br id ev ie w ge The second of the previous examples uses the important rule that: KEY POINT 1.4 Pr es s -C -R If we multiply or divide both sides of an inequality by a negative number, then the inequality sign must be reversed. y ni y U y = x2 – 5x – 14 ge Sketch the graph of y = x 2 − 5x − 14. id + -C –2 C es Pr br ev id ie w ge C U R -R s -C es Sketch the graph of y = 2 x 2 + 3x − 27. op U w id g e C For 2 x 2 + 3x − 27 ø 0 we need to find the range of values of x for which the curve is either zero or negative (below the x-axis). – es s -R br ev ie The solution is − 4 21 < x < 3. am O 2 So the x-axis intercepts are −4 21 and 3. -C ev + –4 1 ni ve rs C x = −4 21 or x = 3 ie w (2 x + 9)( x − 3) = 0 + ity When y = 0, 2 x 2 + 3x − 27 = 0 y = 2x2 + 3x – 27 y Pr op y Rearranging: 2 x 2 + 3x − 27 ø 0 y Answer am Solve 2 x 2 + 3x ø 27. R y – ni ev ve rs w ie The solution is x , −2 or x . 7. x 7 ity op y For x 2 − 5x − 14 . 0 we need to find the range of values of x for which the curve is positive (above the x-axis). For the sketch graph, you only need to identify which way up the graph is and where the x-intercepts are: you do not need to find the vertex or the y-intercept. s So the x-axis crossing points are −2 and 7. O -R br am ( x + 2)( x − 7) = 0 x = −2 or x = 7 WORKED EXAMPLE 1.16 + ev When y = 0, x 2 − 5x − 14 = 0 22 TIP op Answer w R Solve x 2 − 5x − 14 . 0. C op WORKED EXAMPLE 1.15 ie ev ie w C ve rs ity op y Quadratic inequalities can be solved by sketching a graph and considering when the graph is above or below the x-axis. Copyright Material - Review Only - Not for Redistribution 3 x ve rs ity ev ie Ivan is asked to solve the inequality 2x − 4 ù 7. x -R am br id EXPLORE 1.3 w ge C U ni op y Chapter 1: Quadratics Pr es s -C This is his solution: 2x − 4 ù 7x Multiply both sides by x: x ø− ve rs ity w C op y −4 ù 5x Subtract 2x from both sides: Divide both sides by 5: 4 5 y ni U R She writes: C op ev ie Anika checks to see if x = −1 satisfies the original inequality. w ge When x = −1: (2(−1) − 4) ÷ (−1) = 6 id ie Hence, x = −1is a value of x that does not satisfy the original inequality. -R am br ev So Ivan’s solution must be incorrect! -C Discuss Ivan’s solution with your classmates and explain Ivan’s error. op Pr y es s How could Ivan have approached this problem to obtain a correct solution? C rs ve e 6x 2 − 23x + 20 , 0 ev id br ( x − 6)( x − 4) ø 0 f (1 − 3x )(2 x + 1) , 0 c x 2 + 6x − 7 . 0 f 4 − 7x − 2x2 , 0 -R am -C b 15x , x 2 + 56 c x( x + 10) ø 12 − x d x + 4x , 3( x + 2) e ( x + 3)(1 − x ) , x − 1 f (4x + 3)(3x − 1) , 2 x( x + 3) h ( x − 2)2 . 14 − x i 6x( x + 1) , 5(7 − x ) es s a x 2 , 36 − 5x Pr 2 g ( x + 4)2 ù 25 ity op y c C d 14x 2 + 17 x − 6 ø 0 ie b x 2 + 7 x + 10 ø 0 C ni ve rs 4 Find the range of values of x for which 5 , 0. 2 x 2 + x − 15 op a x 2 − 3x ù 10 y 5 Find the set of values of x for which: C U and ( x − 5)2 , 4 w ie and x 2 − 2 x − 3 ù 0 s -R 2 es am 6 Find the range of values of x for which 2 x − 3x − 40 . 1. ev x2 + x − 2 . 0 br c id g e b x 2 + 4x − 21 ø 0 and x 2 − 9x + 8 . 0 -C w ie w ge a x 2 − 25 ù 0 3 Solve: ev y e (5 − x )( x + 6) ù 0 U R d (2 x + 3)( x − 2) , 0 b ( x − 3)( x + 2) . 0 ni ev a x( x − 3) ø 0 op ie w 1 Solve: 2 Solve: R 23 ity EXERCISE 1G Copyright Material - Review Only - Not for Redistribution ve rs ity x( x − 1) .x x +1 e x 2 + 4x − 5 ø0 x2 − 4 ev ie w b c x2 − 9 ù4 x −1 f x−3 x+2 ù x+4 x−5 Pr es s -R am br id x 2 − 2 x − 15 ù0 x−2 -C d ge 7 Solve: x ù3 a x −1 C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 1.8 The number of roots of a quadratic equation op y If f( x ) is a function, then we call the solutions to the equation f( x ) = 0 the roots of f( x ). −6 ± 62 − 4 × 1 × 9 2 ×1 −6 ± 0 x= 2 x = −3 or x = −3 −2 ± 2 2 − 4 × 1 × 6 2 ×1 −2 ± −20 x= 2 no real solution two equal real roots no real roots w ie ev -R am two distinct real roots C op U x= ge id 2 ×1 −2 ± 36 x= 2 x = 2 or x = −4 x= y x 2 + 2x + 6 = 0 −2 ± 2 2 − 4 × 1 × ( −8 ) br x= x 2 + 6x + 9 = 0 ni x 2 + 2x − 8 = 0 R ev ie w C ve rs ity Consider solving the following three quadratic equations of the form ax 2 + bx + c = 0 −b ± b2 − 4ac using the formula x = . 2a es s -C The part of the quadratic formula underneath the square root sign is called the discriminant. =0 ,0 w ge Nature of roots two distinct real roots C b 2 − 4 ac .0 U ni op y ve The sign (positive, zero or negative) of the discriminant tells us how many roots there are for a particular quadratic equation. id ie two equal real roots (or 1 repeated real root) no real roots br ev R ev ie w rs C ity The discriminant of ax 2 + bx + c = 0 is b 2 − 4ac. Pr op y KEY POINT 1.5 24 es ni ve rs x x C or a,0 x w e ev ie x a.0 or x s -R br am -C a,0 The curve is entirely above or entirely below the x-axis. id g no real roots or y op a.0 U R ,0 x The curve touches the x-axis at one point. es w ie Shape of curve y = = ax 2 + bx + c The curve cuts the x-axis at two distinct points. a.0 two equal real roots (or 1 repeated real root) ev =0 Pr two distinct real roots C .0 Nature of roots of ax 2 + bx + +c = 0 ity op y b 2 − 4 ac s -C -R am There is a connection between the roots of the quadratic equation ax 2 + bx + c = 0 and the corresponding curve y = ax 2 + bx + c. Copyright Material - Review Only - Not for Redistribution a,0 ve rs ity ge C U ni op y Chapter 1: Quadratics am br id ev ie w WORKED EXAMPLE 1.17 b2 − 4ac = 0 For two equal roots: U R ni WORKED EXAMPLE 1.18 y ve rs ity ev ie w C op k 2 = 16 k = −4 or k = 4 C op y k2 − 4 × 4 × 1 = 0 Pr es s -C Answer -R Find the values of k for which the equation 4x 2 + kx + 1 = 0 has two equal roots. ie id Answer w ge Find the values of k for which x 2 − 5x + 9 = k (5 − x ) has two equal roots. br ev x 2 − 5x + 9 = k (5 − x ) am Rearrange the equation into the form ax 2 + bx + c = 0 . -R x 2 − 5x + 9 − 5 k + kx = 0 s -C x 2 + ( k − 5)x + 9 − 5 k = 0 y es For two equal roots: b2 − 4ac = 0 Pr 2 y ev ve ie ( k + 11)( k − 1) = 0 rs k 2 + 10 k − 11 = 0 25 ity k − 10 k + 25 − 36 + 20 k = 0 w C op ( k − 5)2 − 4 × 1 × (9 − 5 k ) = 0 w ge C U R ni op k = −11 or k = 1 br ev id ie WORKED EXAMPLE 1.19 -R s -C es Pr ity b2 − 4ac . 0 2 4 k − 32 k . 0 4 k ( k − 8) . 0 + 0 k C ie id g w e Note that the critical values are where 4 k ( k − 8) = 0 . es s -R br ev Hence, k , 0 or k . 8. am 8 – U Critical values are 0 and 8. + ni ve rs ( −2 k )2 − 4 × k × 8 . 0 -C R ev ie w C For two distinct roots: y op y kx 2 − 2 kx + 8 = 0 op Answer am Find the values of k for which kx 2 − 2 kx + 8 = 0 has two distinct roots. Copyright Material - Review Only - Not for Redistribution ve rs ity ev ie am br id EXERCISE 1H w ge C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 b x 2 + 5x − 36 = 0 d 4x 2 − 4x + 1 = 0 2x2 − 7x + 8 = 0 Pr es s x 2 − 12 x + 36 = 0 -C a -R 1 Find the discriminant for each equation and, hence, decide if the equation has two distinct roots, two equal roots or no real roots. e c x 2 + 9x + 2 = 0 f 3x 2 + 10 x − 2 = 0 4 . x 3 The equation x 2 + bx + c = 0 has roots −5 and 7. ve rs ity C op y 2 Use the discriminant to determine the nature of the roots of 2 − 5x = ev ie w Find the value of b and the value of c. y C op ni b 4x 2 + 4( k − 2)x + k = 0 c ( k + 2)x 2 + 4 k = (4 k + 2)x d x 2 − 2 x + 1 = 2 k ( k − 2) e ( k + 1)x 2 + kx − 2 k = 0 f 4x 2 − ( k − 2)x + 9 = 0 ev ie ge id w x 2 + kx + 4 = 0 U a br R 4 Find the values of k for which the following equations have two equal roots. -R am 5 Find the values of k for which the following equations have two distinct roots. x 2 + 8x + 3 = k b 2 x 2 − 5x = 4 − k c kx 2 − 4x + 2 = 0 d kx 2 + 2( k − 1)x + k = 0 e 2 x 2 = 2( x − 1) + k 26 s es kx 2 + (2 k − 5)x = 1 − k Pr f op y -C a ity kx 2 + 2 kx = 4x − 6 d 2 x 2 + k = 3( x − 2) f y e b 3x 2 + 5x + k + 1 = 0 kx 2 + kx = 3x − 2 ge C 7 The equation kx 2 + px + 5 = 0 has repeated real roots. op 2 x 2 + 8x − 5 = kx 2 rs c ve kx 2 − 4x + 8 = 0 ni a U R ev ie w C 6 Find the values of k for which the following equations have no real roots. id ie w Find k in terms of p. -R am br ev 8 Find the range of values of k for which the equation kx 2 − 5x + 2 = 0 has real roots. 9 Prove that the roots of the equation 2 kx 2 + 5x − k = 0 are real and distinct for all real values of k. P 10 Prove that the roots of the equation x 2 + ( k − 2)x − 2 k = 0 are real and distinct for all real values of k. Pr ity 11 Prove that x 2 + kx + 2 = 0 has real roots if k ù 2 2. op -R s es -C am br ev ie id g w e C U R y For which other values of k does the equation have real roots? ev ie w P ni ve rs C op y es s -C P Copyright Material - Review Only - Not for Redistribution WEB LINK Try the Discriminating resource on the Underground Mathematics website. ve rs ity C U ni op y Chapter 1: Quadratics w ge 1.9 Intersection of a line and a quadratic curve ev ie -R Situation 2 Situation 3 one point of intersection no points of intersection The line cuts the curve at two distinct points. The line touches the curve at one point. This means that the line is a tangent to the curve. The line does not intersect the curve. C op y two points of intersection U R ni ev ie w C ve rs ity op y Pr es s -C Situation 1 am br id When considering the intersection of a straight line and a parabola, there are three possible situations. id ie w ge We have already learnt that to find the points of intersection of a straight line and a quadratic curve, we solve their equations simultaneously. .0 two distinct real roots =0 two equal real roots (repeated roots) one point of intersection (line is a tangent) ,0 no real roots Pr two distinct points of intersection no points of intersection op ni ev WORKED EXAMPLE 1.20 y ve ie w rs ity op C s Line and curve y Nature of roots es b 2 − 4 ac -R -C am br ev The discriminant of the resulting equation then enables us to say how many points of intersection there are. The three possible situations are shown in the following table. ie id Answer w ge C U R Find the value of k for which y = x + k is a tangent to the curve y = x 2 + 5x + 2. br ev x 2 + 5x + 2 = x + k -R am x 2 + 4x + (2 − k ) = 0 -C Since the line is a tangent to the curve, the discriminant of the quadratic must be zero, so: es s b2 − 4ac = 0 Pr ity op y ni ve rs -R s es -C am br ev ie id g w e C U R ev ie w C op y 42 − 4 × 1 × (2 − k ) = 0 16 − 8 + 4 k = 0 4 k = −8 k = −2 Copyright Material - Review Only - Not for Redistribution 27 ve rs ity ev ie w ge am br id WORKED EXAMPLE 1.21 C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 -C Answer -R Find the set of values of k for which y = kx − 1 intersects the curve y = x 2 − 2 x at two distinct points. y Pr es s x 2 − 2 x = kx − 1 x − ( k + 2)x + 1 = 0 2 ve rs ity + –4 U Critical values are −4 and 0. R + y k 2 + 4k . 0 k ( k + 4) . 0 ni ev ie w C b2 − 4ac . 0 ( k + 2) − 4 × 1 × 1 . 0 2 C op op Since the line intersects the curve at two distinct points, we must have discriminant . 0. – id ie w ge Hence, k , −4 or k . 0. k 0 es s -C -R am br ev This next example involves a more general quadratic equation. Our techniques for finding the conditions for intersection of a straight line and a quadratic equation will work for this more general quadratic equation too. op Pr y WORKED EXAMPLE 1.22 28 rs Answer ve ie w C ity Find the set of values of k for which the line 2 x + y = k does not intersect the curve xy = 8 . y w ge 2 x 2 − kx + 8 = 0 C U x( k − 2 x ) = 8 R op ni ev Substituting y = k − 2 x into xy = 8 gives: id ie Since the line and curve do not intersect, we must have discriminant , 0. br ev b2 − 4ac , 0 ( − k )2 − 4 × 2 × 8 , 0 + -R am + k 2 − 64 , 0 ( k + 8)( k − 8) , 0 8 s -C –8 op y es Critical values are −8 and 8. y op -R s es -C am br ev ie id g w e C U R ev ie w ni ve rs C ity Pr Hence, −8 , k , 8 . – Copyright Material - Review Only - Not for Redistribution k ve rs ity ev ie am br id EXERCISE 1I w ge C U ni op y Chapter 1: Quadratics -R 1 Find the values of k for which the line y = kx + 1 is a tangent to the curve y = x 2 − 7 x + 2. 2 Find the values of k for which the x-axis is a tangent to the curve y = x 2 − ( k + 3)x + (3k + 4). -C 5 . x−2 Can you explain graphically why there is only one such value of k? (You may want to use graph-drawing software to help with this.) Pr es s ve rs ity 4 The line y = k − 3x is a tangent to the curve x 2 + 2 xy − 20 = 0. w C op y 3 Find the value of k for which the line x + ky = 12 is a tangent to the curve y = y ev ie a Find the possible values of k. U R ni C op b For each of these values of k, find the coordinates of the point of contact of the tangent with the curve. w ge 5 Find the values of m for which the line y = mx + 6 is a tangent to the curve y = x 2 − 4x + 7. id ie For each of these values of m, find the coordinates of the point where the line touches the curve. -R am br ev 6 Find the set of values of k for which the line y = 2 x − 1 intersects the curve y = x 2 + kx + 3 at two distinct points. es s -C 7 Find the set of values of k for which the line x + 2 y = k intersects the curve xy = 6 at two distinct points. 29 9 Find the set of values of m for which the line y = mx + 5 does not meet the curve y = x 2 − x + 6. ity C op Pr y 8 Find the set of values of k for which the line y = k − x cuts the curve y = 5 − 3x − x 2 at two distinct points. y ve ie w rs 10 Find the set of values of k for which the line y = 2 x − 10 does not meet the curve y = x 2 − 6x + k. op ie id br ev 13 The line y = mx + c is a tangent to the curve ax 2 + by2 = c, where a, b, c and m are constants. abc − a . Prove that m2 = b y op -R s es -C am br ev ie id g w e C U R ev ie w ni ve rs C ity Pr op y es s -C -R am P w Prove that m2 + 8 m + 4c = 0. C 12 The line y = mx + c is a tangent to the curve y = x 2 − 4x + 4. ge P U R ni ev 11 Find the value of k for which the line y = kx + 6 is a tangent to the curve x 2 + y2 − 10 x + 8 y = 84. Copyright Material - Review Only - Not for Redistribution ve rs ity C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 ● completing the square ● using the quadratic formula x = −b ± b 2 − 4ac . 2a Pr es s -C factorisation -R Quadratic equations can be solved by: ● w ev ie am br id ge Checklist of learning and understanding op y Solving simultaneous equations where one is linear and one is quadratic Substitute this for x or y in the quadratic equation and then solve. Maximum and minimum points and lines of symmetry y ev ie w ● Rearrange the linear equation to make either x or y the subject. ve rs ity C ● ● if a , 0, there is a maximum point at ( h, k ). ie if a . 0, there is a minimum point at ( h, k ) br ev id ● w ge U R ni C op For a quadratic function f( x ) = ax 2 + bx + c that is written in the form f( x ) = a( x − h )2 + k : b ● the line of symmetry is x = h = − 2a -R am Quadratic equation ax 2 + bx + c = 0 and corresponding curve y = ax 2 + bx + c Discriminant = b 2 − 4ac. ● If b 2 − 4ac . 0, then the equation ax 2 + bx + c = 0 has two distinct real roots. ● If b 2 − 4ac = 0, then the equation ax 2 + bx + c = 0 has two equal real roots. op Pr y es s -C ● rs The condition for a quadratic equation to have real roots is b 2 − 4ac ù 0. y ve Intersection of a line and a general quadratic curve If a line and a general quadratic curve intersect at one point, then the line is a tangent to the curve at that point. ● Solving simultaneously the equations for the line and the curve gives an equation of the form ax 2 + bx + c = 0. ● b 2 − 4ac gives information about the intersection of the line and the curve. C w ie ge U ni op ● Nature of roots Line and parabola two distinct real roots two distinct points of intersection =0 two equal real roots one point of intersection (line is a tangent) ,0 no real roots no points of intersection -R y op -R s es -C am br ev ie id g w e C U R ev ie w ni ve rs C ity Pr op y es -C am .0 s b 2 − 4 ac ev id R ev ie w ● If b 2 − 4ac , 0, then the equation ax 2 + bx + c = 0 has no real roots. ity C ● br 30 Copyright Material - Review Only - Not for Redistribution ve rs ity ge C U ni op y Chapter 1: Quadratics ev ie am br id 1 w END-OF-CHAPTER REVIEW EXERCISE 1 A curve has equation y = 2 xy + 5 and a line has equation 2 x + 5 y = 1. [2] b Find the set of values of x that satisfy the inequality 9x 2 − 15x , 6. [2] ve rs ity 36 25 + 4 = 2. x4 x 3 Find the real roots of the equation 4 Find the set of values of k for which the line y = kx − 3 intersects the curve y = x 2 − 9x at two distinct points. C op y [4] U R Find the set of values of the constant k for which the line y = 2 x + k meets the curve y = 1 + 2 kx − x 2 at two distinct points. ie id [3] br ev b Find the values of the constant k for which the line y = kx + 3 is a tangent to the curve y = 4x 2 − 12 x + 7. -R s Pr y op For the case where the line is a tangent to the curve at a point C , find the value of k and the coordinates of C. [4] C U ie w ge Show that the curve lies above the x-axis. b Find the coordinates of the points of intersection of the line and the curve. [3] -R am br ev id a Write down the set of values of x that satisfy the inequality x 2 − 5x + 7 , 2 x − 3. [3] [1] s -C es A curve has equation y = 10 x − x 2 . Express 10 x − x 2 in the form a − ( x + b )2 . [3] b Write down the coordinates of the vertex of the curve. [2] c Find the set of values of x for which y ø 9. [3] ity Pr a ni ve rs op y op For the case where k = 2, the line and the curve intersect at points A and B. C U i y 10 A line has equation y = kx + 6 and a curve has equation y = x 2 + 3x + 2 k, where k is a constant. id g ev [5] [4] s -R Cambridge International AS & A Level Mathematics 9709 Paper 11 Q9 November 2011 es am ie Find the two values of k for which the line is a tangent to the curve. br ii w e Find the distance AB and the coordinates of the mid-point of AB. -C C w ie ev R [3] A curve has equation y = x 2 − 5x + 7 and a line has equation y = 2 x − 3. c 9 [1] ity rs For one value of k, the line intersects the curve at two distinct points, A and B, where the coordinates of A are ( −2, 13). Find the coordinates of B. ve w ie R ev c Show that the x-coordinates of the points of intersection of the curve and the line are given by the equation x 2 − 4x + (5 − k ) = 0. ni a es A curve has equation y = 5 − 2 x + x 2 and a line has equation y = 2 x + k, where k is a constant. b 8 [5] [4] -C C op y 7 [4] a Find the coordinates of the vertex of the parabola y = 4x 2 − 12 x + 7. am 6 w ge 5 [4] a Express 9x 2 − 15x in the form (3x − a )2 − b. ni ev ie w C op y 2 Pr es s -C -R The curve and the line intersect at the points A and B. Find the coordinates of the midpoint of the line AB. Copyright Material - Review Only - Not for Redistribution 31 ve rs ity am br id ev ie w ge C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 11 A curve has equation y = x 2 − 4x + 4 and a line has the equation y = mx, where m is a constant. For the case where m = 1, the curve and the line intersect at the points A and B. Find the non-zero value of m for which the line is a tangent to the curve, and find the coordinates of the point where the tangent touches the curve. op y ii ve rs ity C Express 2 x 2 − 4x + 1 in the form a ( x + b )2 + c and hence state the coordinates of the minimum point, A, on the curve y = 2 x 2 − 4x + 1. [4] y w [5] Cambridge International AS & A Level Mathematics 9709 Paper 11 Q7 June 2013 12 i ev ie [4] Pr es s -C Find the coordinates of the mid-point of AB. -R i U R ni C op The line x − y + 4 = 0 intersects the curve y = 2 x 2 − 4x + 1 at the points P and Q. w ge It is given that the coordinates of P are (3, 7). [3] id ie ii Find the coordinates of Q. iii Find the equation of the line joining Q to the mid-point of AP. ev br ity op Pr y es s -C -R am Cambridge International AS & A Level Mathematics 9709 Paper 11 Q10 June 2011 y op y op -R s es -C am br ev ie id g w e C U R ev ie w ni ve rs C ity Pr op y es s -C -R am br ev id ie w ge C U R ni ev ve ie w rs C 32 [3] Copyright Material - Review Only - Not for Redistribution op y ve rs ity ni C U ev ie w ge -R am br id Pr es s -C y ni C op y ve rs ity op C w ev ie rs op y ve w ge C U ni ie w C ity op Pr y es s -C -R am br ev id ie w ge U R ev R Chapter 2 Functions 33 id es s -C -R am br ev understand the terms function, domain, range, one-one function, inverse function and composition of functions identify the range of a given function in simple cases, and find the composition of two given functions determine whether or not a given function is one-one, and find the inverse of a one-one function in simple cases illustrate in graphical terms the relation between a one-one function and its inverse understand and use the transformations of the graph y = f( x ) given by y = f( x ) + a, y = f( x + a ), y = a f( x ), y = f( ax ) and simple combinations of these. Pr ity op y ni ve rs -R s es -C am br ev ie id g w e C U R ev ie w C op y ■ ■ ■ ■ ■ ie In this chapter you will learn how to: Copyright Material - Review Only - Not for Redistribution ve rs ity ge C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 am br id ev ie w PREREQUISITE KNOWLEDGE What you should be able to do IGCSE / O Level Mathematics Find an output for a given function. Pr es s -C op 2 If f( x ) = 2 x + 1 and g( x ) = 1 − x, find fg( x ). Find the inverse of a simple function. 3 If f( x ) = 5x + 4, find f −1( x ). Complete the square. 4 Express 2 x 2 − 12 x + 5 in the form a ( x + b )2 + c. ni C op y ve rs ity C w IGCSE / O Level Mathematics Chapter 1 ev ie 1 If f( x ) = 3x − 2 , find f(4). Find a composite function. y IGCSE / O Level Mathematics w ge U R Check your skills -R Where it comes from id ie Why do we study functions? -R am br ev At IGCSE / O Level, you learnt how to interpret expressions as functions with inputs and outputs and find simple composite functions and simple inverse functions. es Pr y w ve ni op y Modelling these situations using appropriate functions enables us to make predictions about real-life situations, such as: How long will it take for the number of bacteria to exceed 5 billion? w ge C U R ev ie ● WEB LINK ity ● rs C ● the temperature of a hot drink as it cools over time the height of a valve on a bicycle tyre as the bicycle travels along a horizontal road the depth of water in a conical container as it is filled from a tap the number of bacteria present after the start of an experiment. op ● 34 s -C There are many situations in the real world that can be modelled as functions. Some examples are: br ev id ie In this chapter we will develop a deeper understanding of functions and their special properties. Try the Thinking about functions and Combining functions resources on the Underground Mathematics website. -R am 2.1 Definition of a function es An alternative name for a function is a mapping. Pr op y s -C A function is a relation that uniquely associates members of one set with members of another set. C ity A function can be either a one-one function or a many-one function. ie w ni ve rs The function x ֏ x + 2, where x ∈ ℝ is an example of a one-one function. TIP y op C y = x+2 -R x s O es -C am br ev ie id g w e U R ev f(x) Copyright Material - Review Only - Not for Redistribution x ∈ ℝ means that x belongs to the set of real numbers. ve rs ity C U ni op y Chapter 2: Functions am br id ev ie w ge A one-one function has one output value for each input value. Equally important is the fact that for each output value appearing there is only one input value resulting in this output value. -R We can write this function as f : x ֏ x + 2 for x ∈ ℝ or f( x ) = x + 2 for x ∈ ℝ. y Pr es s -C f : x ֏ x + 2 is read as ‘the function f is such that x is mapped to x + 2’ or ‘f maps x to x + 2’. The function x ֏ x 2 , where x ∈ ℝ is a many-one function. ev ie w C ve rs ity op f( x ) is the output value of the function f when the input value is x. For example, when f( x ) = x + 2, f(5) = 5 + 2 = 7. y C op y = x2 -R am br ev id ie w ge U R ni f (x) O s -C x op Pr y es A many-one function has one output value for each input value but each output value can have more than one input value. rs f : x ֏ x 2 is read as ‘the function f is such that x is mapped to x 2 ’ or ‘f maps x to x 2 ’. ve ie w C ity We can write this function as f : x ֏ x 2 for x ∈ ℝ or f( x ) = x 2 for x ∈ ℝ. y op U y2 = x x Pr op y es s -C -R ev O am br id ie w ge C R y ni ev If we now consider the graph of y2 = x : ni ve rs y The set of input values for a function is called the domain of the function. U op When defining a function, it is important to also specify its domain. -R s es am br ev ie id g w e C The set of output values for a function is called the range (or codomain) of the function. -C R ev ie w C ity We can see that the input value shown has two output values. This means that this relation is not a function. Copyright Material - Review Only - Not for Redistribution 35 ve rs ity ev ie w ge am br id WORKED EXAMPLE 2.1 C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 -R f( x ) = 5 − 2 x for x ∈ ℝ, −4 < x < 5. b Sketch the graph of the function f . op y c Write down the range of the function f . a The domain is −4 < x < 5 . ve rs ity w C Answer Pr es s -C a Write down the domain of the function f . y ev ie b The graph of y = 5 − 2 x is a straight line with gradient −2 and y-intercept 5. ni C op When x = −4, y = 5 − 2( − 4) = 13 f(x) Pr y op range ity y id br ev Sketch the graph of the function. s -C -R am Find the range of f . Answer w The function f is defined by f( x ) = ( x − 3)2 + 8 for −1 < x < 9. ie ge C U WORKED EXAMPLE 2.2 R op ni ev ve ie rs c The range is −5 < f( x ) < 13. w (5, –5) domain C 36 x es O s -C -R am br ev id ie (–4, 13) w ge U R When x = 5, y = 5 − 2(5) = −5 . The circled part of the expression is a square so it will always be > 0 . The smallest value it can be is 0. This occurs when x = 3. ni ve rs ( x − 3)2 + 8 U op y The minimum value of the expression is 0 + 8 = 8 and this minimum occurs when x = 3. ie id g w e C So the function f( x ) = ( x − 3)2 + 8 will have a minimum point at the point (3, 8). s es am When x = 9, y = (9 − 3)2 + 8 = 44 -R br ev When x = −1, y = ( −1 − 3)2 + 8 = 24 -C R ev ie w C ity will be of the form Pr op y es f( x ) = ( x − 3)2 + 8 is a positive quadratic function so the graph Copyright Material - Review Only - Not for Redistribution ve rs ity y op y Pr es s -C range -R am br id ev ie (9, 44) (–1, 24) ve rs ity (3, 8) y domain U R ni The range is 8 < f( x ) < 44. x C op C O w ev ie w ge C U ni op y Chapter 2: Functions id ie w ge EXERCISE 2A 1 Which of these graphs represent functions? If the graph represents a function, state whether it is a one-one function or a many-one function. ev for x ∈ ℝ d y = 2x for x ∈ ℝ s for x ∈ ℝ, x > 0 x y = x2 − 3 es y Pr y= ity g C op e y = 2 x 3 − 1 for x ∈ ℝ 10 y= for x ∈ ℝ, x > 0 x b -R br am -C a y = 2 x − 3 for x ∈ ℝ c TIP x ∈ ℝ , x > 0 is sometimes shortened to just x > 0. f y = 3x 2 + 4 for x ∈ ℝ, x > 0 h y 2 = 4x 37 for x ∈ ℝ w rs 2 a Represent on a graph the function: y op U R ni ev ve ie 2 9 − x for x ∈ ℝ, −3 < x < 2 x֏ 2 x + 1 for x ∈ ℝ, 2 < x < 4 w ge C b State the nature of the function. id ie 3 a Represent on a graph the relation: -R am br ev 2 x + 1 for 0 < x < 2 y= 2 x − 3 for 2 < x < 4 es s -C b Explain why this relation is not a function. Pr y (1, 8) b y (–2, 20) ity a C op y 4 State the domain and range for the functions represented by these two graphs. U y (–3, 9) x O -R (1, –7) s es am -C (2, 4) x w ie ev (5, –8) br id g e R O op ev (–1, 4) y = 2x3 + 3x2 – 12x C ie w ni ve rs y = 7 + 2x – x2 Copyright Material - Review Only - Not for Redistribution ve rs ity C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 w ge 5 Find the range for each of these functions. b f( x ) = 2 x − 7 for −3 < x < 2 a f( x ) = x + 4 c f( x ) = 7 − 2 x for −1 < x < 4 e f( x ) = 2 x am br id ev ie for x > 8 d f : x ֏ 2x2 12 f f( x ) = x b f( x ) = 3 − 2 x 2 for x < 2 f : x ֏ x 2 + 3 for −2 < x < 5 d f( x ) = 7 − 3x 2 for −1 < x < 2 ve rs ity y op c f : x ֏ 8 − ( x − 5)2 for 4 < x < 10 1 2 for x > 4 b f( x ) = (2 x − 1)2 − 7 for x > for x > 2 d f( x ) = 1 + x − 4 2 y f( x ) = ( x − 2)2 + 5 C op a ni C f( x ) = x 2 − 2 for x ∈ ℝ 7 Find the range for each of these functions. w ev ie c Pr es s -C 6 Find the range for each of these functions. a for 1 < x < 8 -R for −5 < x < 4 for 1 < x < 4 w b f( x ) = 3x 2 − 10 x + 2 for x ∈ ℝ ev id ie f( x ) = x 2 + 6x − 11 for x ∈ ℝ br a ge U R 8 Express each function in the form a ( x + b ) + c , where a, b and c are constants and, hence, state the range of each function. -R f( x ) = 7 − 8x − x 2 for x ∈ ℝ b f( x ) = 2 − 6x − 3x 2 for x ∈ ℝ es s -C a am 9 Express each function in the form a − b( x + c )2 , where a, b and c are constants and, hence, state the range of each function. C 2 for 3 − x f( x ) = 3x − 7 for 0<x<2 Pr 38 2<x<4 ve ie w rs b Find the range of the function. ity op y 10 a Represent, on a graph, the function: y op ni ev 11 The function f : x ֏ x 2 + 6x + k , where k is a constant, is defined for x ∈ ℝ. C U R Find the range of f in terms of k. w ge 12 The function g : x ֏ 5 − ax − 2 x 2 , where a is a constant, is defined for x ∈ ℝ. br ev id ie Find the range of g in terms of a. -R am 13 f( x ) = x 2 − 2 x − 3 for x ∈ ℝ, − a < x < a s 14 f( x ) = x 2 + x − 4 for x ∈ ℝ , a < x < a + 3 es -C If the range of the function f is −4 < f( x ) < 5, find the value of a. Pr ity 15 f( x ) = 2 x 2 − 8x + 5 for x ∈ ℝ, 0 < x < k ni ve rs a Express f( x ) in the form a ( x + b )2 + c . op For your value of k from part b, find the range of f . C U c y b State the value of k for which the graph of y = f( x ) has a line of symmetry. w f( x ) = 2 x ie ev c f f( x ) = x−3−2 -R br am b f( x ) = x 2 + 2 1 e f( x ) = x−2 es f( x ) = 3x − 1 1 d f( x ) = x a s id g e 16 Find the largest possible domain for each function and state the corresponding range. -C R ev ie w C op y If the range of the function f is −2 < f( x ) < 16, find the possible values of a. Copyright Material - Review Only - Not for Redistribution ve rs ity C U ni op y Chapter 2: Functions w ge 2.2 Composite functions am br id ev ie Most functions that we meet can be described as combinations of two or more functions. (the function ‘multiply by 3’) f :x֏x−7 (the function ‘subtract 7’) y Pr es s -C g : x ֏ 3x -R For example, the function x ֏ 3x − 7 is the function ‘multiply by 3 and then subtract 7’. It is a combination of the two functions g and f, where: x f fg(x) U R ni g(x) y ev ie w C ve rs ity g C op op So, x ֏ 3x − 7 can be described as the function ‘first do g, then do f ’. w ge fg -R am br ev id ie When one function is followed by another function, the resulting function is called a composite function. s -C KEY POINT 2.1 op Pr y es fg( x ) means the function g acts on x first, then f acts on the result. 39 C ity There are three important points to remember about composite functions: rs ve ie w KEY POINT 2.2 y C U In general, fg( x ) ≠ gf( x ). R op ni ev fg only exists if the range of g is contained within the domain of f . br -R am EXPLORE 2.1 ev id ie w ge ff( x ) means you apply the function f twice. g( x ) = 3x − 1 for x ∈ ℝ es s -C f( x ) = 2 x − 5 for x ∈ ℝ ity Student B Student C gf( x ) = 2(3x − 1) − 5 = 6x − 7 gf( x ) = 3(2x − 5) − 1 = 6x − 16 gf( x ) = (3x − 1)(2x − 5) U = 6x 2 − 17x + 5 op ni ve rs Student A e C Discuss these solutions with your classmates. -R s es am br ev ie id g w Which student is correct? What error has each of the other students made? -C R ev ie w C Here are their solutions. y Pr op y Three students are asked to find the composite function gf( x ). Copyright Material - Review Only - Not for Redistribution ve rs ity y Answer ve rs ity w f is the function ‘subtract 4, square and then subtract 1’. ie br ev id WORKED EXAMPLE 2.4 w ge U R ni = 1 14 g( x ) = x 2 − 1 for x ∈ ℝ b c gf( x ) es y Answer = 2( x − 1) + 3 ve = (2 x + 3) − 1 f acts on x first and f( x ) = 2 x + 3 . ni g is the function ‘square and subtract 1’. U R 2 y gf( x ) = g(2 x + 3) ev b f is the function ‘double and add 3’. rs = 2x2 + 1 g acts on x first and g( x ) = x 2 − 1. op C 2 w ie Pr fg( x ) = f( x 2 − 1) ity op a 40 ff( x ) s -C fg( x ) -R a am f( x ) = 2 x + 3 for x ∈ ℝ Find: ge C = 4x 2 + 12 x + 9 − 1 id ie w = 4x 2 + 12 x + 8 ff( x ) = f(2 x + 3) ev f is the function ‘double and add 3’. br c es s -C -R am = 2(2 x + 3) + 3 = 4x + 9 Pr ity y ff( x ) U op b fg( x ) Answer g acts on x first and g( x ) = 3 − x 2 . ie id g w e fg( x ) = f(3 − x 2 ) 5 = (3 − x 2 ) − 2 5 = 1 − x2 ev f is the function ‘subtract 2 and then divide into 5’. es s -R br am -C a g( x ) = 3 − x 2 for x ∈ ℝ C w ie ev R a Find: 5 for x ∈ ℝ, x ≠ 2 x−2 ni ve rs C op y WORKED EXAMPLE 2.5 f :x֏ 2(4) + 3 11 = . 4−2 2 y 2 11 = − 4 − 1 2 g acts on 4 first and g(4) = C op C op 11 fg(4) = f 2 ev ie 2x + 3 for x ∈ ℝ, x > 2 x−2 Pr es s -C Find fg(4). g( x ) = -R f( x ) = ( x − 4)2 − 1 for x ∈ ℝ ev ie am br id WORKED EXAMPLE 2.3 w ge C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 Copyright Material - Review Only - Not for Redistribution ve rs ity ge ev ie w 5 ff ( x ) = f x−2 am br id -R = 5x − 10 9 − 2x g( x ) = 3x − 1 for x ∈ ℝ ge f( x ) = x 2 + 4x for x ∈ ℝ U R ni y ev ie WORKED EXAMPLE 2.6 C op 5( x − 2) 5 − 2( x − 2) Pr es s op Multiply numerator and denominator by ( x − 2). w C 5 5 −2 x−2 = y -C = ve rs ity b C U ni op y Chapter 2: Functions br Answer ev id ie w Find the values of k for which the equation fg( x ) = k has real solutions. am fg( x ) = (3x − 1)2 + 4(3x − 1) -R Expand brackets and simplify. = 9x + 6x − 3 When fg( x ) = k, s es ity 41 b2 − 4ac > 0 For real solutions: w rs C 9x 2 + 6x + ( −3 − k ) = 0 y op k > −4 a b fg(6) gf(4) ff( −3) c x ֏ x + 10 x for x ∈ ℝ, x . 0 k:x֏ es c x֏ x +5 b x+5 x֏ ni ve rs 3 f( x ) = ax + b for x ∈ ℝ ity a Pr Express each of the following in terms of h and/or k. op y Given that f(5) = 3 and f(3) = −3: C U a find the value of a and the value of b w -R s Solve the equation gf( x ) = 2 . es am a Find gf( x ). b 12 for x ∈ ℝ, x ≠ 1 1− x ie g:x֏ br 4 f : x ֏ 2 x + 3 for x ∈ ℝ ev id g e b solve the equation ff( x ) = 4. -C ie w C op y 2 h : x ֏ x + 5 for x ∈ ℝ, x . 0 ev x + 3 − 2 for x ∈ ℝ, x > −3 s -C Find: g( x ) = -R am 1 f( x ) = x 2 + 6 for x ∈ ℝ ev br id ie w ge C U R ni ev ve ie 62 − 4 × 9 × ( −3 − k ) > 0 144 + 36 k > 0 EXERCISE 2B R Rearrange and simplify. Pr 9x 2 + 6x − 3 = k op y -C 2 Copyright Material - Review Only - Not for Redistribution ve rs ity C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 5 g( x ) = x 2 − 2 for x ∈ ℝ ev ie w ge am br id a Find gh( x ). h( x ) = 2 x + 5 for x ∈ ℝ g( x ) = f( x ) = x 2 + 1 for x ∈ ℝ -C Pr es s Solve the equation fg( x ) = 5 . y 2 for x ∈ ℝ, x ≠ −1 x +1 Solve the equation hg( x ) = 11. h( x ) = ( x + 2)2 − 5 for x ∈ ℝ x +1 for x ∈ ℝ 2 Solve the equation gf( x ) = 1. 2x + 3 for x ∈ ℝ, x ≠ 1 x −1 ni ev ie g:x֏ y ve rs ity w C op 7 g( x ) = 8 f:x֏ 3 for x ∈ ℝ, x ≠ 2 x−2 C op 6 -R b Solve the equation gh( x ) = 14 . R x +1 for x ∈ ℝ, x . 0 2x + 5 Find an expression for ff( x ) , giving your answer as a single fraction in its simplest form. id ie w ge U 9 f( x ) = 10 f : x ֏ x 2 for x ∈ ℝ br ev g : x ֏ x + 1 for x ∈ ℝ b x ֏ x2 + 1 d x ֏ x4 e x ֏ x2 + 2x + 2 s es x ֏ x 4 + 2x2 + 1 ity rs U 13 f( x ) = x 2 − 3x for x ∈ ℝ w ge C g( x ) = 2 x − 5 for x ∈ ℝ op ni ve g( x ) = y C w ie f g( x ) = 2 x + 5 for x ∈ ℝ 2 for x ∈ ℝ, x ≠ 0 x Find the values of k for which the equation fg( x ) = x has two equal roots. ev x֏x+2 Show that the equation gf( x ) = 0 has no real solutions. 12 f( x ) = k − 2 x for x ∈ ℝ R c Pr y op 11 f( x ) = x 2 − 3x for x ∈ ℝ 42 -R x ֏ ( x + 1)2 -C a am Express each of the following as a composite function, using only f and/or g. id ie Find the values of k for which the equation gf( x ) = k has real solutions. -C -R am br ev x+5 1 for x ∈ ℝ, x ≠ 2x − 1 2 Show that ff( x ) = x. 14 f( x ) = es s 15 f( x ) = 2 x 2 + 4x − 8 for x ∈ ℝ, x > k Pr op y a Express 2 x 2 + 4x − 8 in the form a ( x + b )2 + c. ity ni ve rs 16 f( x ) = x 2 − 2 x + 4 for x ∈ ℝ op y a Find the set of values of x for which f( x ) > 7 . C Write down the range of f . w ie g( x ) = 2 x + 3 for x ∈ ℝ br 17 f( x ) = x 2 − 5x for x ∈ ℝ ev id g e c U b Express x 2 − 2 x + 4 in the form ( x − a )2 + b. -R s b Find the range of the function fg( x ). es am a Find fg( x ). -C R ev ie w C b Find the least value of k for which the function is one-one. Copyright Material - Review Only - Not for Redistribution ve rs ity am br id Pr es s 1 for x ∈ ℝ, x ≠ 0 x S( x ) = R( x ) = C w P( x ) = x 2 − 1 for x ∈ ℝ w Q( x ) = x + 2 for x ∈ ℝ ve rs ity y 19 op PS Find the values of x for which f( x ) = ff( x ). -C c -R b Show that if f( x ) = ff( x ) then x 2 + x − 2 = 0. ev ie ge 2 for x ∈ ℝ , x ≠ −1 x +1 a Find ff( x ) and state the domain of this function. 18 f( x ) = C U ni op y Chapter 2: Functions x + 1 − 1 for x ∈ ℝ, x > −1 y ev ie Functions P, Q, R and S are composed in some way to make a new function, f( x ). -R am br ev id ie w ge U R ni C op For each of the following, write f( x ) in terms of the functions P, Q, R and/or S, and state the domain and range for each composite function. 1 b f( x ) = x 2 + 1 c f( x ) = x d f( x ) = 2 + 1 a f( x ) = x 2 + 4x + 3 x 1 f f( x ) = x − 2 x + 1 + 1 g f( x ) = x − 1 e f( x ) = x+4 s -C 2.3 Inverse functions The inverse of a function f( x ) is the function that undoes what f( x ) has done. es ity 43 x rs KEY POINT 2.3 ff −1( x ) = f −1f( x ) = x ve ie w C op Pr y We write the inverse of the function f( x ) as f −1( x ) . y The range of f −1( x ) is the domain of f( x ). C U w f –1(x) br ev ie ge id It is important to remember that not every function has an inverse. -R am KEY POINT 2.4 es s -C An inverse function f −1( x ) exists if, and only if, the function f( x ) is a one-one mapping. We want to find the function f −1( x ), so if we write y = f −1( x ), then f( y ) = f(f −1( x )) = x, because f and f −1 are inverse functions. So if we write x = f( y ) and then rearrange it to get y = … , then the right-hand side will be f −1( x ). -R s es -C am br ev ie id g w e C U op y ni ve rs ev ie w C ity Pr op y You should already know how to find the inverse function of some simple one-one mappings. R y op ni ev The domain of f −1( x ) is the range of f( x ). R f(x) Copyright Material - Review Only - Not for Redistribution ve rs ity x = 3y − 1 Step 3: Rearrange to make y the subject. y= Pr es s -C x +1 3 x +1 . 3 If f and f −1 are the same function, then f is called a self-inverse function. 1 1 for x ≠ 0 , then f −1( x ) = for x ≠ 0 . For example, if f( x ) = x x 1 for x ≠ 0 is a self-inverse function. So f( x ) = x y C op R ni w C ve rs ity op y Hence, if f( x ) = 3x − 1, then f −1( x ) = ev ie w Step 2: Interchange the x and y variables. -R y = 3x − 1 am br id Step 1: Write the function as y = ev ie ge We find the inverse of the function f( x ) = 3x − 1 by following these steps: C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 ie f(x) ev f(x) = (x – 2)2 + 1 -R am br id The diagram shows the function f( x ) = ( x − 2)2 + 1 for x ∈ ℝ . Discuss the following questions with your classmates. w ge U EXPLORE 2.2 -C 1 What type of mapping is this function? es s 2 What are the coordinates of the vertex of the parabola? ve y op br ev id ie w ge C U R ni x + 2 − 7 for x ∈ ℝ, x > −2 -R am f( x ) = b Solve the equation f −1( x ) = f(62) . s -C a Find an expression for f −1( x ) . es f( x ) = x+2 −7 Step 1: Write the function as y = x+2 −7 Step 2: Interchange the x and y variables. x= y+2 −7 Step 3: Rearrange to make y the subject. x+7 = U ( x + 7) = y + 2 e 2 id g w y = ( x + 7)2 − 2 -R s es am br ev ie f −1( x ) = ( x + 7)2 − 2 -C C R y+2 op ity y= ni ve rs ie w C a Pr op y Answer y ev 6 If f has an inverse, what is it? If not, then how could you change the domain of f so that the function does have an inverse? WORKED EXAMPLE 2.7 ev x rs C w ie 5 Does this function have an inverse? O ity 4 What is the range of the function? Pr op y 3 What is the domain of the function? 44 Copyright Material - Review Only - Not for Redistribution ve rs ity ge w 62 + 2 − 7 = 1 ev ie f(62) = am br id b C U ni op y Chapter 2: Functions ( x + 7)2 − 2 = 1 -R ( x + 7)2 = 3 -C x+7 = ± 3 ni C op y Hence, the only solution of f −1( x ) = f(62) is x = −7 + 3 . w ev ie The range of f is f( x ) > −7 so the domain of f −1 is x > −7. ve rs ity C op y x = −7 − 3 or x = −7 + 3 ge U WORKED EXAMPLE 2.8 R Pr es s x = −7 ± 3 ie id w f( x ) = 5 − ( x − 2)2 for x ∈ ℝ, k < x < 6 br ev a State the smallest value of k for which f has an inverse. -C -R am b For this value of k find an expression for f −1( x ) , and state the domain and range of f −1 . y (2, 5) es s Answer Pr When x = 6, y = 5 − 42 = −11 y = 5 – (x – 2)2 ity C op y a The vertex of the graph of y = 5 − ( x − 2)2 is at the point (2, 5). For the function f to have an inverse it must be a one-one function. rs w ie op y ve ni f( x ) = 5 − ( x − 2)2 C U ev R b y = 5 − ( x − 2)2 ie w ge id am br ev Step 2: Interchange the x and y variables. -R s -C y−2 = 5−x Pr es y =2+ 5−x Hence, f −1( x ) = 2 + 5 − x . ity op y ni ve rs The domain of f −1 is the same as the range of f . y Hence, the domain of f −1 is −11 < x < 5 . U op The range of f −1 is the same as the domain of f . -R s es am br ev ie id g w e C Hence, the range of f −1 is 2 < f −1 ( x ) < 6. -C C w ie x = 5 − ( y − 2)2 ( y − 2)2 = 5 − x Step 3: Rearrange to make y the subject. ev x Hence, the smallest value of k is 2. Step 1: Write the function as y = R O Copyright Material - Review Only - Not for Redistribution (6, –11) 45 ve rs ity ev ie am br id EXERCISE 2C w ge C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 c f( x ) = ( x − 5)2 + 3 for x ∈ ℝ , x > 5 e f( x ) = y f( x ) = ( x − 2)3 − 1 for x ∈ ℝ , x > 2 f y C op w ge R ni 5 for x ∈ ℝ, x > 2 2x + 1 a Find an expression for f −1( x ). f:x֏ U ev ie b Find an expression for f −1( x ) . 3 br ev id ie b Find the domain of f −1. f : x ֏ ( x + 1)3 − 4 for x ∈ ℝ, x > 0 -R am 4 s -C a Find an expression for f −1( x ) . C g : x ֏ 2 x 2 − 8x + 10 for x ∈ ℝ, x > 3 w rs a Explain why g has an inverse. Pr 5 ity op y es b Find the domain of f −1. 46 d ve rs ity op f : x ֏ x 2 + 4x for x ∈ ℝ, x > −2 f( x ) = x 2 + 3 for x ∈ ℝ , x > 0 8 f( x ) = for x ∈ ℝ , x ≠ 3 x−3 b a State the domain and range of f −1. w C 2 x+7 for x ∈ ℝ , x ≠ −2 x+2 Pr es s f( x ) = 5x − 8 for x ∈ ℝ -C a -R 1 Find an expression for f −1( x ) for each of the following functions. y op f : x ֏ 2 x 2 + 12 x − 14 for x ∈ ℝ, x > k C U R 6 ni ev ve ie b Find an expression for g −1( x ). ge a Find the least value of k for which f is one-one. br f : x ֏ x 2 − 6x for x ∈ ℝ -R am 7 ev id ie w b Find an expression for f −1( x ). -C a Find the range of f. Pr op y f( x ) = 9 − ( x − 3)2 for x ∈ ℝ, k < x < 7 find an expression for f −1( x ) ii state the domain and range of f −1. -R s es -C am br ev ie id g w e C U op i y ni ve rs b For this value of k: ity a State the smallest value of k for which f has an inverse. R ev ie w C 8 es s b State, with a reason, whether f has an inverse. Copyright Material - Review Only - Not for Redistribution ve rs ity y ge Pr es s -C ve rs ity y op C 5x − 1 for x ∈ ℝ, 0 , x < 3. x b State the domain of f. g( x ) = b − 5x for x ∈ ℝ y w The diagram shows the graph of y = f −1( x ) , where f −1( x ) = 10 f( x ) = 3x + a for x ∈ ℝ ev ie x O a Find an expression for f( x ). 5x – 1 x ev ie am br id w y= -R 9 C U ni op y Chapter 2: Functions U R ni C op Given that gf( −1) = 2 and g −1(7) = 1, find the value of a and the value of b. 11 f( x ) = 3x − 1 for x ∈ ℝ 3 for x ∈ ℝ, x ≠ 2 2x − 4 w ge g( x ) = id ie a Find expressions for f −1( x ) and g −1( x ). -R am br ev b Show that the equation f −1( x ) = g −1( x ) has two real roots. 12 f : x ֏ (2 x − 1)3 − 3 for x ∈ ℝ, 1 < x < 3 s es 47 ity C 13 f : x ֏ x 2 − 10 x for x ∈ ℝ, x > 5 Pr b Find the domain of f −1. op y -C a Find an expression for f −1( x ) . w rs a Express f( x ) in the form ( x − a )2 − b. y ve ie b Find an expression for f −1( x ) and state the domain of f −1. ni w ge C U R op ev 1 for x ∈ ℝ , x ≠ 1 x −1 a Find an expression for f −1( x ). 14 f( x ) = es s -C 15 Determine which of the following functions are self-inverse functions. 1 2x + 1 a f( x ) = for x ∈ ℝ , x ≠ 3 b f( x ) = for x ∈ ℝ , x ≠ 2 x−2 3−x 3x + 5 3 for x ∈ ℝ , x ≠ c f( x ) = 4x − 3 4 g : x ֏ 4 − 2 x for x ∈ ℝ ni ve rs 16 f : x ֏ 3x − 5 for x ∈ ℝ ity Pr op y op y a Find an expression for (fg) −1( x ). g −1 f −1( x ) . C ii w Comment on your results in part b. ie c f −1 g −1( x ) e i U R b Find expressions for: id g -R s -C am br ev Investigate if this is true for other functions. es C w ie ie -R am Give your answer in surd form. ev ev Find the values of x for which f( x ) = f −1( x ) . br c id b Show that if f( x ) = f −1( x ) , then x 2 − x − 1 = 0 . Copyright Material - Review Only - Not for Redistribution ve rs ity ge 2.4 The graph of a function and its inverse am br id ev ie Consider the function defined by f( x ) = 2 x + 1 for x ∈ ℝ , −4 < x < 2 . w C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 -R f( −4) = −7 and f(2) = 5. y op y Pr es s -C The domain of f is −4 < x < 2 and the range is −7 < f( x ) < 5. x −1 . The inverse of this function is f −1( x ) = 2 The domain of f −1 is the same as the range of f. O x C f–1 (–7, –4) f C op y Hence, the range of f −1 is −4 < f −1( x ) < 2. ni ev ie w The range of f −1 is the same as the domain of f. The representation of f and f −1 on the same graph can be seen in the diagram opposite. U R y=x (5, 2) ve rs ity Hence, the domain of f −1 is −7 < x < 5. (2, 5) (–4, –7) br ev id ie w ge It is important to note that the graphs of f and f −1 are reflections of each other in the line y = x. This is true for each one-one function and its inverse functions. -R am KEY POINT 2.5 -C The graphs of f and f −1 are reflections of each other in the line y = x. es s This is because ff −1( x ) = x = f −1 f( x ) C ity op Pr y When a function f is self-inverse, the graph of f will be symmetrical about the line y = x. 48 ve ie w rs WORKED EXAMPLE 2.9 y op ni ev f( x ) = ( x − 1)2 − 2 for x ∈ ℝ, 1 < x < 4 ie ev es s -C The function is one-one, so the inverse function exists. Pr f f Reflect f in y = x 2 C 8 x ie 6 ev 4 O –2 s es am –2 –2 -R br 2 w e id g O -C –2 f –1 4 U 2 6 y 4 y=x op w ni ve rs 6 ie y 8 ity C op y y 8 ev -R am When x = 4, y = 7. R The circled part of the expression is a square so it will always be > 0. The smallest value it can be is 0. This occurs when x = 1. The vertex is at the point (1, −2) . br y = ( x − 1)2 − 2 id Answer w ge C U R On the same axes, draw the graph of f and the graph of f −1. Copyright Material - Review Only - Not for Redistribution 2 4 6 8 x ve rs ity ge C U ni op y Chapter 2: Functions am br id ev ie w WORKED EXAMPLE 2.10 -C -R 2x + 7 for x ∈ ℝ, x ≠ 2 x−2 a Find an expression for f −1( x ) . f:x֏ f :x֏ 2x + 7 x−2 U R ni ev ie Step 1: Write the function as y = ie -R s Pr es 2x + 7 . x−2 49 ity rs y C w y id g–1 y=x -R x x ity O Pr op y Ali states that: -R s es am br ev ie id g w e C U Explain your answer. op Is Ali correct? y ni ve rs The diagrams show the functions f and g, together with their inverse functions f −1 and g−1. -C C w s f–1 O ie g es -C am br ev y=x ie ge U EXPLORE 2.3 R op ni ev ve ie w The graph of y = f( x ) is symmetrical about the line y = x. f ev y( x − 2) = 2 x + 7 2x + 7 y= x−2 f −1( x ) = f( x ) , so the function f is self-inverse. y R 2y + 7 y−2 ev id br am -C y op C b x= xy − 2 x = 2 y + 7 Step 3: Rearrange to make y the subject. Hence f −1( x ) = 2x + 7 x−2 w ge Step 2: Interchange the x and y variables. y= y w C a ve rs ity op Answer C op y Pr es s b State what your answer to part a tells you about the symmetry of the graph of y = f( x ). Copyright Material - Review Only - Not for Redistribution ve rs ity ev ie am br id EXERCISE 2D w ge C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 1 On a copy of each grid, draw the graph of f −1( x ) if it exists. ev ie w –2 U ev id 5 -R 1 1 2 3 4 6 O x 6 x w ie ev -R s y Pr es 4 for x ∈ ℝ, x > 0. x+2 ity State the domain and range of f −1. f O y ni ve rs d On a copy of the diagram, sketch the graph of y = f −1( x ) , making clear the relationship between the graphs. ie id g 3 3x − 1 for x ∈ ℝ, x ≠ 2x − 3 2 ev d f( x ) = 4 4x + 5 for x ∈ ℝ, x ≠ 3x − 4 3 es s -R br f( x ) = am c w e C U op 4 For each of the following functions, find an expression for f −1( x ) and, hence, decide if the graph of y = f( x ) is symmetrical about the line y = x. x+5 2x − 3 1 b f( x ) = for x ∈ ℝ, x ≠ for x ∈ ℝ, x ≠ 5 a f( x ) = 2x − 1 x−5 2 -C w C op y b Find an expression for f −1( x ) . ie 5 C ge id br am -C a State the range of f . ev 4 Sketch, on the same diagram, the graphs of y = f( x ) and y = f −1( x ) , making clear the relationship between the graphs. 3 The diagram shows the graph of y = f( x ) , where f( x ) = c 3 y ni U a Find an expression for f −1( x ). c 2 op f : x ֏ 2 x − 1 for x ∈ ℝ, −1 < x < 3 b State the domain and range of f −1. R 1 ve 5 rs w 2 ity 1 f 3 s 2 ie ev R 2 4 es f O y 6 C 50 d Pr op y 3 6 x 4 –6 br -C 4 2 –4 am 5 O –2 ie y 6 –4 –2 ge R –6 c –6 ni –4 6 x 4 y 2 C op O –2 4 2 ve rs ity –4 f Pr es s f 2 y 6 b w -C 4 y op C –6 -R y 6 a Copyright Material - Review Only - Not for Redistribution x ve rs ity w ev ie f( x ) = ge x+a 1 for x ∈ ℝ, x ≠ , where a and b are constants. bx − 1 b Prove that this function is self-inverse. ax + b d for x ∈ ℝ, x ≠ − , where a, b, c and d are constants. b g( x ) = cx + d c Find the condition for this function to be self-inverse. 5 a Pr es s -C -R am br id P C U ni op y Chapter 2: Functions op y 2.5 Transformations of functions y w ge U EXPLORE 2.4 R C op ni ev ie w C ve rs ity At IGCSE / O Level you met various transformations that can be applied to two-dimensional shapes. These included translations, reflections, rotations and enlargements. In this section you will learn how translations, reflections and stretches (and combinations of these) can be used to transform the graph of a function. id ie 1 a Use graphing software to draw the graphs of y = x 2, y = x 2 + 2 and y = x 2 − 3. -R 12 12 12 ,y= + 5 and y = − 4. x x x es y Pr d Can you generalise your results? op x − 2. s x, y = c Repeat part a using the graphs y = ity 2 a Use graphing software to draw the graphs of y = x 2, y = ( x + 2)2 and y = ( x − 5)2 . Discuss your observations with your classmates and explain how the second and third graphs could be obtained from the first graph. op ni ev b Repeat part a using the graphs y = x 3, y = ( x + 1)3 and y = ( x − 4)3 . c Can you generalise your results? C U R y ve ie w rs C x + 1 and y = b Repeat part a using the graphs y = -C am br ev Discuss your observations with your classmates and explain how the second and third graphs could be obtained from the first graph. w ge 3 a Use graphing software to draw the graphs of y = x 2 and y = − x 2 . -R am b Repeat part a using the graphs y = x 3 and y = − x 3. ev br id ie Discuss your observations with your classmates and explain how the second graph could be obtained from the first graph. -C c Repeat part a using the graphs y = 2 x and y = −2 x . es s d Can you generalise your results? Pr ity Discuss your observations with your classmates and explain how the second graph could be obtained from the first graph. ni ve rs b Repeat part a using the graphs y = 2 + x and y = 2 − x. c Can you generalise your results? op y 5 a Use graphing software to draw the graphs of y = x 2 and y = 2 x 2 and y = (2 x )2. C x , y = 2 x and y = w id g b Repeat part a using the graphs y = 2x . ie e U Discuss your observations with your classmates and explain how the second graph could be obtained from the first graph. c Repeat part a using the graphs y = 3 , y = 2 × 3 and y = 32 x . x ev x es s -R br am d Can you generalise your results? -C R ev ie w C op y 4 a Use graphing software to draw the graphs of y = 5 + x and y = 5 − x. Copyright Material - Review Only - Not for Redistribution 51 ve rs ity y w ge Translations C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 12 am br id ev ie The diagram shows the graphs of two functions that differ only by a constant. y = x2 – 2x + 4 10 -R y = x − 2x + 1 2 -C y = x2 − 2x + 4 8 op y Pr es s When the x-coordinates on the two graphs are the same ( x = x ) the y-coordinates differ by 3 ( y = y + 3). y = x2 – 2x + 1 6 4 Hence, the graph of y = x 2 − 2 x + 4 is a translation of the graph of y = x 2 − 2 x + 1 w ie s -C -R am br ev 0 The graph of y = f( x ) + a is a translation of the graph y = f( x ) by the vector . a es Now consider the two functions: op Pr y y = x2 − 2x + 1 52 C ity y = ( x − 3)2 − 2( x − 3) + 1 ve ni op y The graphs of these two functions are: ev 4 -R am br 6 es s -C 3 Pr 4 x U op y The curves have exactly the same shape but this time they are separated by 3 units in the positive x-direction. -R s es am br ev ie id g w e C You may be surprised that the curve has moved in the positive x-direction. Note, however, that a way of obtaining y = y is to have x = x − 3 or equivalently x = x + 3. This means that the two curves are at the same height when the red curve is 3 units to the right of the blue curve. -C ie ev R 6 ni ve rs 2 ity O w C op y 2 –2 ie y = (x – 3)2 – 2 (x – 3) + 1 id y = x2 – 2x + 1 w ge 8 C U R ev ie w rs We obtain the second function by replacing x by x − 3 in the first function. y 3 y –2 id KEY POINT 2.6 2 C op U ni 0 by the vector . 3 ge R ev ie w C ve rs ity This means that the two curves have exactly the same shape but that they are separated by 3 units in the positive y direction. Copyright Material - Review Only - Not for Redistribution O 2 4 x ve rs ity C U ni op y Chapter 2: Functions w ge Hence, the graph of y = ( x − 3)2 − 2( x − 3) + 1 is a translation of the graph of -R am br id ev ie 3 y = x 2 − 2 x + 1 by the vector . 0 Pr es s -C KEY POINT 2.7 y C op R ni KEY POINT 2.8 ve rs ity Combining these two results gives: ev ie w C op y a The graph of y = f( x − a ) is a translation of the graph y = f( x ) by the vector . 0 br ev id ie w ge U a The graph of y = f( x − a ) + b is a translation of the graph y = f( x ) by the vector . b -R am WORKED EXAMPLE 2.11 Pr y es s -C The graph of y = x 2 + 5x is translated 2 units to the right. Find the equation of the resulting graph. Give your answer in the form y = ax 2 + bx + c. y = x 2 + 5x Expand and simplify. C U R ni op y = x2 + x − 6 y ve y = ( x − 2) + 5( x − 2) ie Replace all occurrences of x by x − 2. rs w 2 ev 53 ity C op Answer id ie w ge WORKED EXAMPLE 2.12 br ev −5 2 x is translated by the vector . Find the equation of the resulting graph. 3 -R y= 2 x + 10 + 3 es 2( x + 5) + 3 Pr y= Replace x by x + 5 , and add 3 to the resulting function. -R s es -C am br ev ie id g w e C U op ie ev R y ni ve rs ity 2x op y y= w C s -C Answer am The graph of y = Copyright Material - Review Only - Not for Redistribution ve rs ity ev ie am br id EXERCISE 2E w ge C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 y = 2x2 -C 0 after translation by −2 c y = 7x2 − 2x 0 after translation by 1 d y = x2 − 1 0 after translation by 2 e y= 2 x −5 after translation by 0 f y= x x +1 3 after translation by 0 g y = x2 + x C op w ie ev Pr es -C 2 after translation by 3 y = 3x 2 − 2 s −1 after translation by 0 y h y ve rs ity ni U ge id br am ev ie w C op y Pr es s y=5 x b R 0 after translation by 4 -R a -R 1 Find the equation of each graph after the given transformation. to the graph y = x 2 + 5x + 2 b y = x3 + 2x2 + 1 to the graph y = x 3 + 2 x 2 − 4 c y = x 2 − 3x 6 y=x+ x y = 2x + 5 5 y = 2 − 3x x to the graph y = ( x + 1)2 − 3( x + 1) 6 to the graph y = x − 2 + x−2 to the graph y = 2 x + 3 5 to the graph y = − 3x + 10 ( x − 2)2 rs op y ve ni C w ie -R am 3 The diagram shows the graph of y = f( x ). b y = f ( x − 2) c y = f ( x + 1) − 5 y = f(x) 1 2 3 4 x –4 –3 –2 –1 O –1 –2 –3 –4 b 0 y = 2 x can be transformed to y = 2 x + 2 by a translation of . a C w e id g ie Find the value of a. es s -R br ev b y = 2 x can be transformed to y = 2 x + 2 by a translation of . 0 Find the value of b. am c op On the same diagram, sketch the graphs of y = 2 x and y = 2 x + 2. U 4 a -C R ev ie w ni ve rs ity Pr es y = f( x ) − 4 C a op y s -C Sketch the graphs of each of the following functions. y 4 3 2 1 y br f ev e U R d ity y = x 2 + 5x − 2 ge C w ie ev a id op 2 Find the translation that transforms the graph. 54 Copyright Material - Review Only - Not for Redistribution ve rs ity w ge 5 A cubic graph has equation y = ( x + 3 )( x − 2 )( x − 5 ). C U ni op y Chapter 2: Functions am br id ev ie 2 Write, in a similar form, the equation of the graph after a translation of . 0 -C -R 1 6 The graph of y = x 2 − 4x + 1 is translated by the vector . 2 WEB LINK Try the Between the lines resource on the Underground Mathematics website. y 2 7 The graph of y = ax 2 + bx + c is translated by the vector . −5 The resulting graph is y = 2 x 2 − 11x + 10 . Find the value of a, the value of b and the value of c. w ge The diagram shows the graphs of the two functions: br ev y = −( x 2 − 2 x + 1) 2 s -C -R am When the x-coordinates on the two graphs are the same ( x = x ), the y-coordinates are negative of each other ( y = − y ). rs Hence, the graph of y = −( x 2 − 2 x + 1) is a reflection of the graph of y = x 2 − 2 x + 1 in the x-axis. y C U R op ni ev ve ie w C ity op Pr y es This means that, when the x-coordinates are the same, the red curve is the same vertical distance from the x-axis as the blue curve but it is on the opposite side of the x-axis. KEY POINT 2.9 id ie w ge The graph of y = −f( x ) is a reflection of the graph y = f( x ) in the x-axis. br ev Now consider the two functions: -R am y = x2 − 2x + 1 -C y = ( − x )2 − 2( − x ) + 1 es s We obtain the second function by replacing x by −x in the first function. Pr op y The graphs of these two functions are demonstrated in the diagram. w 2 4 ie O C U id g –2 x -R s es -C am br –4 e 2 1 op y = x2 – 2x + 1 3 ev ie 4 y = (–x)2 – 2(–x) + 1 y ni ve rs w 6 R ev ity C y 5 y = x2 – 2x + 1 4 ie id y = x2 − 2x + 1 y 6 C op U 2.6 Reflections R y ve rs ity ni ev ie w C op PS Pr es s Find, in the form y = ax 2 + bx + c, the equation of the resulting graph. Copyright Material - Review Only - Not for Redistribution –2 O 2 4 x –2 –4 –6 y = –(x2 – 2x + 1) 55 ve rs ity C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 w ge The curves are at the same height ( y = y ) when x = − x or equivalently x = − x. -R am br id ev ie This means that the heights of the two graphs are the same when the red graph has the same horizontal displacement from the y-axis as the blue graph but is on the opposite side of the y-axis. op y Pr es s -C Hence, the graph of y = ( − x )2 − 2( − x ) + 1 is a reflection of the graph of y = x 2 − 2 x + 1 in the y-axis. U R ni WORKED EXAMPLE 2.13 y The graph of y = f( − x ) is a reflection of the graph y = f( x ) in the y-axis. C op ev ie w C ve rs ity KEY POINT 2.10 ie -R y es s y = − f( x ) is a reflection of y = f( x ) in the x-axis. Pr b The turning point is (5, 7). It is a maximum point. y = f( − x ) is a reflection of y = f( x ) in the y-axis. ity op C 56 y = f( − x ) ev id br -C Answer a b am y = − f( x ) a w ge The quadratic graph y = f( x ) has a minimum at the point (5, −7). Find the coordinates of the vertex and state whether it is a maximum or minimum of the graph for each of the following graphs. y ge C U EXERCISE 2F R op ni ev ve ie w rs The turning point is ( −5, −7) . It is a minimum point. y w 1 The diagram shows the graph of y = g( x ) . am b y = g( − x ) s es ie 3 2 y = g(x) 1 –4 –3 –2 –1 O –1 1 2 3 4 x –2 Pr op y –3 ity –4 ni ve rs C 2 Find the equation of each graph after the given transformation. y = 5x 2 after reflection in the x-axis. b y = 2 x 4 after reflection in the y-axis. c y = 2 x 2 − 3x + 1 after reflection in the y-axis. d y = 5 + 2 x − 3x 2 after reflection in the x-axis. -R s es am br ev ie id g w e C U op y a -C w ie ev R -R y = −g( x ) -C a br Sketch the graphs of each of the following functions. ev id 4 Copyright Material - Review Only - Not for Redistribution ve rs ity C U ni op y Chapter 2: Functions c w ev ie y = 2 x − 5x 2 onto the graph y = 5x 2 − 2 x y = x 3 + 2 x 2 − 3x + 1 onto the graph y = − x 3 − 2 x 2 + 3x − 1. y Pr es s -C d y = x 2 − 3x + 4 onto the graph y = x 2 + 3x + 4 -R b y = x 2 + 7 x − 3 onto the graph y = − x 2 − 7 x + 3 am br id a ge 3 Describe the transformation that maps the graph: op 2.7 Stretches y 10 ni y = 2( x 2 − 2 x − 3) ×2 y ev ie w y = x2 − 2x − 3 When the x-coordinates on the two graphs are the same ( x = x ), the y-coordinate on the red graph is double the y-coordinate on the blue graph ( y = 2 y ). y = x 2 – 2x – 3 w 5 ev id ie ge U R C op C ve rs ity The diagram shows the graphs of the two functions: -R am br This means that, when the x-coordinates are the same, the red curve is twice the distance of the blue graph from the x-axis. –2 O es ity rs op y ve ni w ie id KEY POINT 2.11 C U ● R –5 a stretch with scale factor 2 with the line y = 0 invariant a stretch with stretch factor 2 with the x-axis invariant a stretch with stretch factor 2 relative to the x-axis a vertical stretch with stretch factor 2. ge ev ● 6 x Pr y op C w ie ● 4 57 Note: there are alternative ways of expressing this transformation: ● 2 s -C –4 Hence, the graph of y = 2( x 2 − 2 x − 3) is a stretch of the graph of y = x 2 − 2 x − 3 from the x-axis. We say that it has been stretched with stretch factor 2 parallel to the y-axis. -R am br ev The graph of y = a f( x ) is a stretch of the graph y = f( x ) with stretch factor a parallel to the y-axis. y op -R s es -C am br ev ie id g w e C U R ev ie w ni ve rs C ity Pr op y es s -C Note: if a , 0, then y = a f( x ) can be considered to be a stretch of y = f( x ) with a negative scale factor or as a stretch with positive scale factor followed by a reflection in the x-axis. Copyright Material - Review Only - Not for Redistribution ×2 y = 2(x 2 – 2x – 3) ve rs ity C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 y = (2 x ) − 2(2 x ) − 3 2 ×1 2 ev ie am br id y = x2 − 2x − 3 w ge Now consider the two functions: y = (2x)2 – 2(2x) – 3 y 10 -R We obtain the second function by replacing x by 2x in the first function. ni y O 2 4 6 x ie ev –5 -R s es 1 parallel to the a op ni C U ie w ge ev A stretch parallel to the y-axis, factor 4, gives the function 4f( x ). Pr op y es The equation of the resulting graph is y = 20 − 2 x 2. s -C -R am br 1 2 x 2 4f( x ) = 20 − 2 x 2 Let f( x ) = 5 − id Answer y rs ve 1 2 x is stretched with stretch factor 4 parallel to the y-axis. 2 Find the equation of the resulting graph. ni ve rs C ity WORKED EXAMPLE 2.15 op C Express 4x 2 − 6x − 5 in terms of f( x ). e Let f( x ) = x 2 − 3x − 5 U Answer y Describe the single transformation that maps the graph of y = x 2 − 3x − 5 to the graph of y = 4x 2 − 6x − 5. ie id g w 4x 2 − 6x − 5 = (2 x )2 − 3(2 x ) − 5 -R s es am br ev = f(2 x ) 1 The transformation is a stretch parallel to the x-axis with stretch factor . 2 -C w –2 ity op C w ev ie WORKED EXAMPLE 2.14 The graph of y = 5 − R –4 Pr y x-axis. 58 ie y = x2 – 2x – 3 w ge id br am -C The graph of y = f( ax ) is a stretch of the graph y = f( x ) with stretch factor ev 5 C op y = x 2 − 2 x − 3 from the y-axis. We say that it has been stretched with stretch 1 factor parallel to the x-axis. 2 KEY POINT 2.12 R ×1 2 Hence, the graph of y = (2 x )2 − 2(2 x ) − 3 is a stretch of the graph of U R ev ie w C ve rs ity op y Pr es s -C The two curves are at the same height ( y = y ) when x = 2 x or equivalently 1 x = x. 2 This means that the heights of the two graphs are the same when the red graph has half the horizontal displacement from the y-axis as the blue graph. Copyright Material - Review Only - Not for Redistribution ve rs ity 1 The diagram shows the graph of y = f( x ). y –6 –4 4 6 x y = f(x) 2 –4 ni y –6 U 2 Find the equation of each graph after the given transformation. y = 3x 2 after a stretch parallel to the y-axis with stretch factor 2. b y = x 3 − 1 after a stretch parallel to the y-axis with stretch factor 3. 1 y = 2 x + 4 after a stretch parallel to the y-axis with stretch factor . 2 y = 2 x 2 − 8x + 10 after a stretch parallel to the x-axis with stretch factor 2. 1 y = 6x 3 − 36x after a stretch parallel to the x-axis with stretch factor . 3 ie ev id es e s -C -R am c d w ge a br R –2 O –2 ve rs ity op C w ev ie 2 Pr es s y = f(2 x ) b 4 C op y = 3f( x ) -C a y 6 -R Sketch the graphs of each of the following functions. ev ie am br id EXERCISE 2G w ge C U ni op y Chapter 2: Functions Pr y = x 2 + 2 x − 5 onto the graph y = 4x 2 + 4x − 5 b y = x 2 − 3x + 2 onto the graph y = 3x 2 − 9x + 6 c y = 2 x + 1 onto the graph y = 2 x + 1 + 2 d y= 59 ni y x − 6 onto the graph y = 3x − 6 ge C U op ve rs ity a R ev ie w C op y 3 Describe the single transformation that maps the graph: id ie w 2.8 Combined transformations br ev In this section you will learn how to apply simple combinations of transformations. -R am The transformations of the graph of y = f( x ) that you have studied so far can each be s -C categorised as either vertical or horizontal transformations. Vertical transformations es 0 translation a y = −f( x ) reflection in the x-axis y = a f( x ) vertical stretch, factor a y = f( x + a ) −a translation 0 y = f( − x ) reflection in the y-axis y = f( ax ) horizontal stretch, factor y 1 a op ni ve rs ity Pr y = f( x ) + a op y C -R s es am br ev ie id g w e C U When combining transformations care must be taken with the order in which the transformations are applied. -C w ie ev R Horizontal transformations Copyright Material - Review Only - Not for Redistribution ve rs ity Apply the transformations in the given order to triangle T Pr es s -C -R and for each question comment on whether the final images are the same or different. y 1 Combining two vertical transformations 1 T O 1 y 0 1 , then translate . 2 3 ni C op b Investigate for other pairs of vertical transformations. U R 3 x 2 ve rs ity w C op 0 1 a i Translate , then stretch vertically with factor . 2 3 ii Stretch vertically with factor ev ie y 2 ev ie am br id EXPLORE 2.5 w ge C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 ge 2 Combining one vertical and one horizontal transformation -R am br ev id ie w −2 a i Reflect in the x-axis, then translate . 0 −2 ii Translate , then reflect in the x-axis. 0 op 3 Combining two horizontal transformations ity 2 a i Stretch horizontally with factor 2, then translate . 0 2 ii Translate , then stretch horizontally with factor 2. 0 y op ni ev ve ie w rs C 60 Pr y es s -C b Investigate for other pairs of transformations where one is vertical and the other is horizontal. w ge C U R b Investigate for other pairs of horizontal transformations. br -R am KEY POINT 2.13 ev id ie From the Explore activity, you should have found that: es s -C • When two vertical transformations or two horizontal transformations are combined, the order in which they are applied may affect the outcome. C ity Pr op y • When one horizontal and one vertical transformation are combined, the order in which they are applied does not affect the outcome. op y We will now consider how the graph of y = f( x ) is transformed to the graph y = a f( x ) + k. w ie 0 translate k → a f( x ) + k add k to the function -R s es am br multiply function by a → a f( x ) → ev U e id g stretch vertically, factor a -C f( x ) → C This can be shown in a flow diagram as: R ev ie w ni ve rs Combining two vertical transformations Copyright Material - Review Only - Not for Redistribution ve rs ity C U ni op y Chapter 2: Functions am br id ev ie w ge This leads to the important result: KEY POINT 2.14 Pr es s -C -R Vertical transformations follow the ‘normal’ order of operations, as used in arithmetic. Combining two horizontal transformations ve rs ity replace x with bx id ie w ge U R This leads to the important result: KEY POINT 2.15 1 b → f( bx + c ) y → f( x + c ) → stretch horizontally, factor C op f( x ) → −c translate 0 replace x with x + c ni ev ie w C op y Now consider how the graph of y = f( x ) is transformed to the graph y = f( bx + c ). -C -R am br ev Horizontal transformations follow the opposite order to the ‘normal’ order of operations, as used in arithmetic. es ity Sketch the graph of y = 2f( x ) − 3. y 6 w y = f(x) 2 ni op y ve ie ev U R –4 –2 ie ev id br 4 6 x -R –6 y 6 es s -C y = 2f(x) 4 Pr op y 2 –4 y = 2f( x ) − 3 is a combination of two vertical transformations of y = f( x ) , hence the transformations follow the ‘normal’ order of operations. Step 1: Sketch the graph y = 2f( x ): O –2 w ge C –6 am Answer 61 4 rs C op The diagram shows the graph of y = f( x ) . Pr y s WORKED EXAMPLE 2.16 Stretch y = f( x ) vertically with stretch factor 2 . op C w e ev ie id g es s -R br am -C –4 y –6 U R ev ie w ni ve rs C ity 2 Copyright Material - Review Only - Not for Redistribution –2 O –2 –4 –6 2 4 6 x ve rs ity C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 –6 –4 –2 O –2 ve rs ity –4 ni C op y –6 w ge U WORKED EXAMPLE 2.17 ev es s -C y = g(x) 4 y = x2 Pr op y 3 ity 2 rs C 62 -R am br id ie The diagram shows the graph of y = x 2 and its image, y = g( x ), after a combination of transformations. y y 2 op 1 x 3 C O –1 ni R –2 U ev ve ie w 1 w ge a Find two different ways of describing the combination of transformations. br -R am Answer ev id ie b Write down the equation of the graph y = g( x ). es y = g(x) ity U R ev 2 y ni ve rs ie w 3 w 4 5 6 ie 3 x ev 2 -R 1 s am -C O es –1 br –2 id g e 1 C C y = x2 Pr y op op y s -C 4 1 a Translation of followed by a horizontal stretch, stretch factor . 2 0 4 y = 2f(x) – 3 2 Pr es s -C y op C w ev ie 4 -R am br id ev ie Translate y = 2f ( x ) by the vector 0 . −3 R y 6 w ge Step 2: Sketch the graph y = 2f ( x ) − 3: Copyright Material - Review Only - Not for Redistribution 2 4 6 x ve rs ity w ev ie 2 1 , followed by a translation of . 2 0 y y = g(x) Pr es s 4 ev ie w 2 –2 O –1 1 2 3 x id ie w ge –3 U R ni 1 y ve rs ity C op 3 C op y = x2 y -C Horizontal stretch, factor -R am br id ge OR C U ni op y Chapter 2: Functions s ity y ge C U EXERCISE 2H R op ni ev ve ie rs C 1 means ‘replace x by 2x’. 2 y = ( x − 4)2 becomes y = (2 x − 4)2 Hence, g( x ) = (2 x − 4)2. w y = ( x − 4)2 Pr Horizontal stretch, factor op y -C becomes es y = x2 The same answer will be obtained when using the second combination of transformations. You may wish to check this yourself. -R br am 4 Translation of means ‘replace x by x − 4 ’. 0 TIP ev b Using the first combination of transformations: 1 The diagram shows the graph of y = g( x ) . id ie w y 2 y = 2g( x ) + 1 y = 2 − g( x ) d y = 2g( − x ) + 1 f y = g(2 x ) + 3 g y = g(2 x − 6) h y = g( − x + 1) –3 –2 –1 O –1 1 2 3 x –2 es s y = −2g( x ) − 1 ity Pr –3 y op -R s es -C am br ev ie id g w e C U R ev ie w ni ve rs C op y e y = g(x) 1 -R br b -C c y = g( x + 2) + 3 am a ev Sketch the graph of each of the following. Copyright Material - Review Only - Not for Redistribution 63 ve rs ity 2 The diagram shows the graph of y = f( x ). 3 2 1 1 2 4 x 3 –3 –4 –4 –3 –2 –1–1O id ie ge y –2 1 C op –4 –3 –2 –1–1O U –2 –3 y 4 c 2 4 x 3 x w 2 3 –2 Pr es s -R 1 2 3 ni ev ie w 1 1 y = f(x) y 4 ve rs ity C 2 R ev ie am br id -C y op b 3 –4 1 –4 –3 –2 –1 O –1 y 4 –4 –3 –2 –1 O –1 w ge y 2 Write down, in terms of f( x ), the equation of the graph of each of the following diagrams. a C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 -R am br ev 3 Given that y = x 2 , find the image of the curve y = x 2 after each of the following combinations of transformations. es s -C 1 a a stretch in the y-direction with factor 3 followed by a translation by the vector 0 1 b a translation by the vector followed by a stretch in the y-direction with 0 factor 3 w rs ity Pr y op C 64 y op U R ni ev ve ie 4 Find the equation of the image of the curve y = x 2 after each of the following combinations of transformations and, in each case, sketch the graph of the resulting curve. -C -R am br ev id ie w ge C a a stretch in the x-direction with factor 2 followed by a translation by the 5 vector 0 5 b a translation by the vector followed by a stretch in the x-direction with 0 factor 2 s On a graph show the curve y = x 2 and each of your answers to parts a and b. es c Pr ni ve rs ity 0 a translation followed by a stretch parallel to the y-axis with stretch factor 2 −5 -R s es -C am br ev ie id g w e C U op y 2 b translation followed by a reflection in the x-axis 0 R ev ie w C op y 5 Given that f( x ) = x 2 + 1, find the image of y = f( x ) after each of the following combinations of transformations. Copyright Material - Review Only - Not for Redistribution –2 –3 –4 1 2 3 4 x ve rs ity C U ni op y Chapter 2: Functions The graph of y = g( x ) is reflected in the y-axis and then stretched with stretch factor 2 parallel to the y-axis. Write down the equation of the resulting graph. 2 b The graph of y = f( x ) is translated by the vector and then reflected in −3 -R am br id ev ie w ge 6 a Pr es s -C the x-axis. Write down the equation of the resulting graph. ve rs ity w C op y 7 Determine the sequence of transformations that maps y = f( x ) to each of the following functions. 1 b y = −f( x ) + 2 c y = f(2 x − 6) d y = 2f( x ) − 8 a y = f (x) + 3 2 9 Given that f( x ) = am y C op w x , write down the equation of the image of f( x ) after: -R 0 a reflection in the x-axis, followed by translation , followed by translation 3 1 0 , followed by a stretch parallel to the x-axis with stretch factor 2 Pr es s -C 65 0 b translation , followed by a stretch parallel to the x-axis with stretch 3 1 factor 2, followed by a reflection in the x-axis, followed by translation . 0 y ev ve ie w rs ity y op C ie ev id the curve y = 3 x onto the curve y = −2 3 x − 3 + 4 br c ge U R ni ev ie 8 Determine the sequence of transformations that maps: 1 a the curve y = x 3 onto the curve y = ( x + 5)3 2 1 b the curve y = x 3 onto the curve y = − ( x + 1)3 − 2 2 U R ni op 10 Given that g( x ) = x 2 , write down the equation of the image of g( x ) after: -R am br ev id ie w ge C −4 a translation , followed by a reflection in the y-axis, followed by translation 0 0 2 , followed by a stretch parallel to the y-axis with stretch factor 3 op y es s -C b a stretch parallel to the y-axis with stretch factor 3, followed by translation −4 0 y-axis, followed by translation , followed by reflection in the 0 . 2 Pr ity ni ve rs -R s -C am br ev ie id g w e C U op y 12 Find two different ways of describing the sequence of transformations that maps the graph of y = f( x ) onto the graph of y = f(2 x + 10) . es PS 11 Find two different ways of describing the combination of transformations that maps the graph of f( x ) = x onto the graph g( x ) = − x − 2 and sketch the graphs of y = f( x ) and y = g( x ). R ev ie w C PS Copyright Material - Review Only - Not for Redistribution WEB LINK Try the Transformers resource on the Underground Mathematics website. ve rs ity C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 am br id Functions ● A function can be either one-one or many-one. ● The set of input values for a function is called the domain of the function. ● The set of output values for a function is called the range (or image set) of the function. Pr es s ve rs ity ● fg only exists if the range of g is contained within the domain of f. ● In general, fg( x ) ≠ gf( x ). ni U ie w ge The inverse of a function f( x ) is the function that undoes what f( x ) has done. f f −1( x ) = f −1 f( x ) = x or if y = f( x ) then x = f −1( y ) id ● y fg( x ) means the function g acts on x first, then f acts on the result. The inverse of the function f( x ) is written as f −1 ( x ). ● The steps for finding the inverse function are: -R am br ev ● es Pr y The domain of f −1 ( x ) is the range of f( x ). op ity The range of f −1 ( x ) is the domain of f( x ). An inverse function f −1 ( x ) can exist if, and only if, the function f( x ) is one-one. ● The graphs of f and f −1 are reflections of each other in the line y = x. ● If f( x ) = f −1 ( x ), then the function f is called a self-inverse function. ● If f is self-inverse then ff( x ) = x. ● The graph of a self-inverse function has y = x as a line of symmetry. C U ni op y ve rs ● w ge R ev ie w C ● s -C Step 1: Write the function as y = Step 2: Interchange the x and y variables. Step 3: Rearrange to make y the subject. ● 0 The graph of y = f( x ) + a is a translation of y = f( x ) by the vector . a ● The graph of y = f( x + a ) is a translation of y = f( x ) by the vector ● The graph of y = − f( x ) is a reflection of the graph y = f( x ) in the x-axis. ● The graph of y = f( − x ) is a reflection of the graph y = f( x ) in the y-axis. es s -C Pr op y The graph of y = a f( x ) is a stretch of y = f( x ), stretch factor a, parallel to the y-axis. 1 ● The graph of y = f( ax ) is a stretch of y = f( x ), stretch factor , parallel to the x-axis. a ni ve rs Combining transformations When two vertical transformations or two horizontal transformations are combined, the order in which they are applied may affect the outcome. ● When one horizontal and one vertical transformation are combined, the order in which they are applied does not affect the outcome. ● Vertical transformations follow the ‘normal’ order of operations, as used in arithmetic ● Horizontal transformations follow the opposite order to the ‘normal’ order of operations, as used in arithmetic. -R s es am br ev ie id g w e C U op y ● -C ie w C ity ● ev −a . 0 -R am br ev id ● ie Transformations of functions R C op y op C w ev ie ● Inverse functions R -R A function is a rule that maps each x value to just one y value for a defined set of input values. -C ● Composite functions 66 w ev ie ge Checklist of learning and understanding Copyright Material - Review Only - Not for Redistribution ve rs ity ge C U ni op y Chapter 2: Functions Pr es s -C g : x ֏ 5x − x 2 ev ie Functions f and g are defined for x ∈ ℝ by: f : x ֏ 3x − 1 -R am br id 1 w END-OF-CHAPTER REVIEW EXERCISE 2 y Express gf( x ) in the form a − b( x − c )2, where a, b and c are constants. 2 [5] O x C op 3 U R y 2 ni ev ie w C ve rs ity op y es a Sketch the curve, showing the coordinates of any axes crossing points. Pr ity op ni ie [3] ev id Express − x 2 + 6x − 5 in the form a ( x + b )2 + c, where a, b and c are constants. br i [3] w ge b On the same diagram, sketch the graphs of f and f . [4] [3] C U R −1 67 y The function f : x ֏ x 2 − 2 is defined for the domain x > 0 . a Find f −1( x ) and state the domain of f −1. 5 [2] [2] 2 b The curve is translated by the vector , then stretched vertically with stretch factor 3. 0 Find the equation of the resulting curve, giving your answer in the form y = ax 2 + bx . ve w ie ev 4 [3] s A curve has equation y = x 2 + 6x + 8. rs C op y 3 -R -C am br ev id ie w ge The diagram shows a sketch of the curve with equation y = f( x ). 1 a Sketch the graph of y = − f x . 2 b Describe fully a sequence of two transformations that maps the graph of y = f( x ) onto the graph of y = f(3 − x ) . -R am The function f : x ֏ − x 2 + 6x − 5 is defined for x > m, where m is a constant. s -C ii State the smallest possible value of m for which f is one-one. [1] −1 −1 es Pr ity The function f : x ֏ x 2 − 4x + k is defined for the domain x > p , where k and p are constants. i Express f( x ) in the form ( x + a )2 + b + k, where a and b are constants. [2] ii State the range of f in terms of k. [1] op y ni ve rs 6 [4] Cambridge International AS & A Level Mathematics 9709 Paper 11 Q9 November 2015 w ie [4] s -R ev Cambridge International AS & A Level Mathematics 9709 Paper 11 Q8 June 2012 es br id g e iv For the value of p found in part iii, find an expression for f −1( x ) and state the domain of f −1, giving your answer in terms of k. am [1] C U iii State the smallest value of p for which f is one-one. -C R ev ie w C op y iii For the case where m = 5, find an expression for f ( x ) and state the domain of f . Copyright Material - Review Only - Not for Redistribution ve rs ity -R y y O ve rs ity op C x Pr es s -C 7 ev ie am br id w ge C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 y for − 1 < x < 1, C op for 1 < x < 4. U R ev ie 3x − 2 f( x ) = 4 5 − x ni w The diagram shows the function f defined for −1 < x < 4 , where w ge i State the range of f . [2] iii Obtain expressions to define the function f −1, giving also the set of values for which each expression is valid. [6] -R am br ev id ie ii Copy the diagram and on your copy sketch the graph of y = f −1( x ) . s -C Cambridge International AS & A Level Mathematics 9709 Paper 11 Q10 June 2014 es The function f is defined by f( x ) = 4x 2 − 24x + 11 , for x ∈ ℝ . C Pr i Express f( x ) in the form a ( x − b )2 + c and hence state the coordinates of the vertex of the graph of y = f( x ) . rs y ni op w ie ev ii State the range of g. -R am 2 [1] es [1] Pr op y iii Find the range of f. iv Find the expression for f −1( x ) and state the domain of f −1. [5] ity Cambridge International AS & A Level Mathematics 9709 Paper 11 Q8 June 2013 ni ve rs C Express x 2 − 2 x − 15 in the form ( x + a )2 + b. y [2] op U R The function f is defined for p < x < q, where p and q are positive constants, by id g w e C f : x ֏ x 2 − 2 x − 15 . ie The range of f is given by c < f( x ) < d, where c and d are constants. -R s -C am br ev ii State the smallest possible value of c. es w [3] s -C The value of k is now given to be 7. ie [4] C w ev id ie Express 2 x 2 − 12 x + 13 in the form a ( x + b )2 + c , where a, b and c are constants. ii The function f is defined by f( x ) = 2 x − 12 x + 13 , for x > k , where k is a constant. It is given that f is a one-one function. State the smallest possible value of k. 10 i [2] Cambridge International AS & A Level Mathematics 9709 Paper 11 Q10 November 2012 br i ge U R iii Find an expression for g −1( x ) and state the domain of g −1. 9 ev [4] The function g is defined by g( x ) = 4x 2 − 24x + 11 , for x < 1. ve 68 ity op y 8 [1] Copyright Material - Review Only - Not for Redistribution [1] ve rs ity am br id ev ie w ge C U ni op y Chapter 2: Functions For the case where c = 9 and d = 65, iv find an expression for f −1( x ). [3] Pr es s -C [4] -R iii find p and q, y Cambridge International AS & A Level Mathematics 9709 Paper 11 Q10 November 2014 w i Express f( x ) in the form a ( x − b )2 − c. ev ie ii State the range of f . ve rs ity C op 11 The function f is defined by f : x ֏ 2 x 2 − 12 x + 7 for x ∈ ℝ . [3] [1] ni C op y iii Find the set of values of x for which f( x ) < 21. U R The function g is defined by g : x ֏ 2 x + k for x ∈ ℝ . w ge iv Find the value of the constant k for which the equation gf( x ) = 0 has two equal roots. ie ev id br -R am es s g : x ֏ x 2 − 2. [2] ii Hence find the value of a for which fg( a ) = gf( a ). [3] iii Find the value of b ( b ≠ a ) for which g( b ) = b. [2] iv Find and simplify an expression for f −1g( x ) . [2] ity Pr Find and simplify expressions for fg( x ) and gf( x ). y ev ve ie w rs C op y -C f : x ֏ 2 x + 1, C U R ni op The function h is defined by w ge h : x ֏ x 2 − 2, for x < 0 . -R am br ev Cambridge International AS & A Level Mathematics 9709 Paper 11 Q11 June 2011 s es f : x ֏ 2 x 2 − 8x + 10 for 0 < x < 2, for 0 < x < 10. Pr g:x֏x ity Express f( x ) in the form a ( x + b )2 + c, where a, b and c are constants. iii State the domain of f −1. [1] C U op y iv Sketch on the same diagram the graphs of y = f( x ), y = g( x ) and y = f −1( x ), making clear the relationship between the graphs. w [4] [3] ev ie Cambridge International AS & A Level Mathematics 9709 Paper 11 Q11 November 2011 es s -R br id g e v Find an expression for f −1( x ). am [3] [1] ni ve rs ii State the range of f. -C ev ie w C op y -C 13 Functions f and g are defined by i [2] ie id v Find an expression for h −1( x ) . R [4] Cambridge International AS & A Level Mathematics 9709 Paper 11 Q9 June 2010 12 Functions f and g are defined for x ∈ ℝ by i [3] Copyright Material - Review Only - Not for Redistribution 69 op y ve rs ity ni C U ev ie w ge -R am br id Pr es s -C y ni C op y ve rs ity op C w ev ie C ity op Pr y es s -C -R am br ev id ie w ge U R 70 op y ve ni w ge C U R ev ie w rs Chapter 3 Coordinate geometry id es s -C -R am br ev find the equation of a straight line when given sufficient information interpret and use any of the forms y = mx + c, y − y1 = m( x − x1 ), ax + by + c = 0 in solving problems understand that the equation ( x − a )2 + ( y − b )2 = r 2 represents the circle with centre ( a, b ) and radius r use algebraic methods to solve problems involving lines and circles understand the relationship between a graph and its associated algebraic equation, and use the relationship between points of intersection of graphs and solutions of equations. y op -R s es -C am br ev ie id g w e C U R ev ie w ni ve rs C ity Pr op y ■ ■ ■ ■ ■ ie In this chapter you will learn how to: Copyright Material - Review Only - Not for Redistribution ve rs ity ge C U ni op y Chapter 3: Coordinate geometry What you should be able to do IGCSE / O Level Mathematics Find the midpoint and length of a line segment. Find the gradient of a line and state the gradient of a line that is perpendicular to the line. Pr es s -C w ie 3 The equation of a line is 2 y = x − 5. Write down: 3 a the gradient of the line ev id br c the x-intercept. -R am es s 4 a Complete the square for x 2 − 8x − 5. b Solve x 2 − 8x − 5 = 0.. Pr y 71 ity op C Why do we study coordinate geometry? rs w b State the gradient of the line that is perpendicular to the line AB. b the y-intercept Complete the square and solve quadratic equations. -C Chapter 1 ie 2 a Find the gradient of the line joining A( −1, 3 ) and B ( 5, 2 ). y Interpret and use equations of lines of the form y = mx + c. ge U IGCSE / O Level Mathematics 1 Find the midpoint and length of the line segment joining ( − 7, 4) and ( − 2, −8). C op ni ev ie w C ve rs ity op IGCSE / O Level Mathematics R Check your skills -R Where it comes from y am br id ev ie w PREREQUISITE KNOWLEDGE y op ge C U R ni ev ve This chapter builds on the coordinate geometry work that you learnt at IGCSE / O Level. You shall also learn about the Cartesian equation of a circle. Circles are one of a collection of mathematical shapes called conics or conic sections. -R am br ev id ie w A conic section is a curve obtained from the intersection of a plane with a cone. The three types of conic section are the ellipse, the parabola and the hyperbola. The circle is a special case of the ellipse. Conic sections provide a rich source of fascinating and beautiful results that mathematicians have been studying for thousands of years. Pr ity ni ve rs WEB LINK -R s es -C am br ev ie id g w e C U op y The Geometry of equations and Circles stations on the Underground Mathematics website have many useful resources for studying this topic. R ev ie w C op y es s -C Conic sections are very important in the study of astronomy. We also use their reflective properties in the design of satellite dishes, searchlights, and optical and radio telescopes. Copyright Material - Review Only - Not for Redistribution circle ellipse parabola hyperbola ve rs ity C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 w ge 3.1 Length of a line segment and midpoint TIP Pr es s -C -R am br id ev ie At IGCSE / O Level you learnt how to find the midpoint, M, of a line segment joining the points P ( x1, y1 ) and Q( x2 , y2 ) and the length of the line segment, PQ, using the two formulae in Key point 3.1. You need to know how to apply these formulae to solve problems. op y KEY POINT 3.1 Q (x2, y2) ( x2 − x1 )2 + ( y2 − y1 )2 ni P (x1, y1) w ie am br ev id WORKED EXAMPLE 3.1 ge U R y M C op To find the length of PQ: PQ = ev ie w C ve rs ity x + x2 y1 + y2 , To find the midpoint, M, of the line segment PQ: M = 1 2 2 It is important to remember to show appropriate calculations in coordinate geometry questions. Answers from scale drawings are not accepted. -C -R 3 The point M , −11 is the midpoint of the line segment joining the points P( −7, 4) and Q( a, b ). 2 y es s Find the value of a and the value of b. op Pr Answer y ve rs Decide which values to use for x1, y1, x2 , y2. ni C U x1 + x2 y1 + y2 3 , and midpoint = , −11 2 2 2 −7 + a , 4 + b = 3 , −11 2 2 2 w ev s es ni ve rs C w -R s -C am br ev ie id g w e C U op y b = −26. es ie ev and R Hence, a = 10 Pr 4+b = −11 2 4 + b = −22 b = −26 Equating the y-coordinates: -R −7 + a 3 = 2 2 −7 + a = 3 a = 10 op y -C Equating the x-coordinates: ity am br id R Using ( a, b ) ↑ ↑ ( x2 , y2 ) op ( −7, 4) ↑ ↑ ( x1, y1 ) ge ev ie w C ity Method 1: Using algebra ie 72 Copyright Material - Review Only - Not for Redistribution ve rs ity w ge C U ni op y Chapter 3: Coordinate geometry am br id ev ie Method 2: Using vectors P (−7, 4) 8 PM = 1 2 -R −15 1 2 b = −11 + ( −15) ∴ a = 10 and b = −26. Q (a, b) y C w ev ie ni U R WORKED EXAMPLE 3.2 w ge Three of the vertices of a parallelogram, ABCD, are A( −5, −1), B( −1, −4) and C(6, −2). br ev id ie a Find the midpoint of AC. -R ity Since ABCD is a parallelogram, the midpoint of BD is the same as the midpoint of AC. −1 + m −4 + n 1 3 = , ,− 2 2 2 2 rs y ni op ie Midpoint of BD = ev 73 ve w C op b Let the coordinates of D be ( m, n ). es s −5 + 6 −1 + −2 1 3 = , ,− 2 2 2 2 y -C a Midpoint of AC = Pr Answer am b Find the coordinates of D. −1 + m 1 = 2 2 −1 + m = 1 m=2 −4 + n 3 =− Equating the y-coordinates: 2 2 −4 + n = −3 n=1 D is the point (2, 1). Equating the x-coordinates: w ge O D (2, 1) br x C (6, −2) ev id ie A (−5, −1) -R am s B (−1, −4) Pr es -C op y ni ve rs ity WORKED EXAMPLE 3.3 y The distance between two points P ( −2, a ) and Q( a − 2, −7 ) is 17. Answer ( a − 2, −7 ) ↑ ↑ ( x1, y1 ) ↑ ↑ ( x2 , y2 ) e ( −2, a ) C U op Find the two possible values of a. ev ie id g w Decide which values to use for x1, y1, x2 , y2. es s -R br am -C C w ie y C U R ev R ( 32 , −11) ve rs ity and op ∴ a = 23 + 8 21 Pr es s y −15 M C op -C 8 ∴ MQ = Copyright Material - Review Only - Not for Redistribution ve rs ity ( x2 − x1 )2 + ( y2 − y1 )2 = 17 am br id ev ie w ge Using PQ = C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 ( a − 2 + 2)2 + ( −7 − a )2 = 17 Square both sides. 2 a 2 + 14a − 240 = 0 op -R Collect terms on one side. Divide both sides by 2. Factorise. ve rs ity C a 2 + 7 a − 120 = 0 Expand brackets. Pr es s a 2 + 49 + 14a + a 2 = 289 y -C a 2 + ( −7 − a )2 = 289 or y a=8 a + 15 = 0 a = −15 C op or Solve. ni a−8= 0 ie -C Tamar says that this triangle is right angled. 2√7 es C ve rs w EXERCISE 3A ie 4√3 ity op Pr y Explain your reasoning. 74 5√3 s Discuss whether he is correct. -R am The triangle has sides of length 2 7 cm, 4 3 cm and 5 3 cm. ev br id EXPLORE 3.1 w ge U R ev ie w ( a − 8)( a + 15) = 0 y op ni ev 1 Calculate the lengths of the sides of the triangle PQR. P( −4, 6), Q(6, 1), R(2, 9) w ge a C U R Use your answers to determine whether or not the triangle is right angled. br P(1, 6), Q( −2, 1) and R(3, −2). -R am 2 ev id ie b P( −5, 2), Q(9, 3), R( −2, 8) s -C Show that triangle PQR is a right-angled isosceles triangle and calculate the area of the triangle. es op y 3 The distance between two points, P ( a, −1) and Q( −5, a ), is 4 5. Pr ni ve rs Find the two possible values of b. ity 4 The distance between two points, P( −3, −2) and Q( b, 2b ), is 10. op y 5 The point ( −2, −3) is the midpoint of the line segment joining P( −6, −5) and Q( a, b ). C U Find the value of a and the value of b. ie id g w e 6 Three of the vertices of a parallelogram, ABCD, are A( −7, 3), B( −3, −11) and C(3, −5). br ev a Find the midpoint of AC. -R s Find the length of the diagonals AC and BD. es c am b Find the coordinates of D. -C R ev ie w C Find the two possible values of a. Copyright Material - Review Only - Not for Redistribution ve rs ity am br id w ev ie ge 7 The point P ( k, 2 k ) is equidistant from A(8, 11) and B(1, 12). Find the value of k. C U ni op y Chapter 3: Coordinate geometry -R 8 Triangle ABC has vertices at A( −6, 3), B(3, 5) and C(1, −4). Show that triangle ABC is isosceles and find the area of this triangle. y Pr es s -C 9 Triangle ABC has vertices at A( −7, 8), B (3, k ) and C(8, 5). Given that AB = 2 BC , find the value of k. ve rs ity w C op 5 10 The line x + y = 4 meets the curve y = 8 − at the points A and B. x Find the coordinates of the midpoint of AB. ev ie 11 The line y = x − 3 meets the curve y2 = 4x at the points A and B. ni C op y a Find the coordinates of the midpoint of AB. ie w 12 In triangle ABC , the midpoints of the sides AB, BC and AC are (1, 4), (2, 0) and ( −4, 1), respectively. Find the coordinates of points A, B and C . am br ev id PS ge U R b Find the length of the line segment AB. -R 3.2 Parallel and perpendicular lines op Pr y es s -C At IGCSE / O Level you learnt how to find the gradient of the line joining the points P ( x1, y1 ) and Q( x2 , y2 ) using the formula in Key point 3.2. ity Q(x2, y2) rs y2 − y1 x2 − x1 y op w ge C U R ni ev ie w Gradient of PQ = 75 ve C KEY POINT 3.2 ie -R am br ev id P(x1, y1) s -C You also learnt the following rules about parallel and perpendicular lines. Parallel lines gradient = m y op U R ev ie w ni ve rs C ity Pr op y es Perpendicular lines w ie If a line has gradient m, then every line 1 perpendicular to it has gradient − . m -R s es -C am br If two lines are parallel, then their gradients are equal. 1 m ev id g e C gradient = − Copyright Material - Review Only - Not for Redistribution ve rs ity C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 am br id ev ie w ge We can also write the rule for perpendicular lines as: KEY POINT 3.3 Pr es s -C -R If the gradients of two perpendicular lines are m1 and m2 , then m1 × m2 = −1. C ve rs ity op y You need to know how to apply the rules for gradients to solve problems involving parallel and perpendicular lines. C op ni U R Find the two possible values of k if A, B and C are collinear. w ge id ie If A, B and C are collinear, then they lie on the same line. -R am br ev gradient of AB = gradient of BC s es k + 15 k = 15 − k 2 k = 10 rs C or U ∴k = 3 Solve. ev id ie w ge k − 10 = 0 Factorise. -R am br R ( k − 3)( k − 10) = 0 or Collect terms on one side. ni ev k 2 − 13k + 30 = 0 k−3=0 Expand brackets. ve ie w 2 k + 30 = 15 k − k 2 Cross-multiply. ity 2( k + 15) = k (15 − k ) C 76 Simplify. Pr op y -C −k − k k − ( −15) = 10 − ( k − 5) 6 − 10 op Answer y The coordinates of three points are A( k − 5, −15), B (10, k ) and C (6, − k ). y ev ie w WORKED EXAMPLE 3.4 s -C WORKED EXAMPLE 3.5 op y es The vertices of triangle ABC are A(11, 3), B (2 k, k ) and C( −1, −11). Pr ni ve rs ity b Draw diagrams to show the two possible triangles. Answer ie id g w e Simplify the second fraction. ev + -R s es am br − y C k−3 −11 − k × = −1 2 k − 11 −1 − 2 k op U a Since angle ABC is 90°, gradient of AB × gradient of BC = −1. -C R ev ie w C a Find the two possible values of k if angle ABC is 90°. Copyright Material - Review Only - Not for Redistribution ve rs ity w ge C U ni op y Chapter 3: Coordinate geometry k−3 k + 11 × = −1 2 k − 11 2k + 1 ev ie am br id Multiply both sides by (2k − 11)(2k + 1). ( k − 3)( k + 11) = −(2 k − 11)(2 k + 1) 5k 2 − 12 k − 44 = 0 (5k − 22)( k + 2) = 0 y or k+2 = 0 ∴ k = 4.4 or k = −2 -R Factorise. ve rs ity op C w Collect terms on one side. Pr es s -C k 2 + 8 k − 33 = −4 k 2 + 20 k + 11 5 k − 22 = 0 Expand brackets. y U ge The two possible triangles are: ie x es s -C Pr C (−1, −11) ity y rs EXERCISE 3B ev ve ie w C op C (−1, −11) U R ni 1 The coordinates of three points are A( −6, 4), B(4, 6) and C(10, 7). ge C a Find the gradient of AB and the gradient of BC. id ie w b Use your answer to part a to decide whether or not the points A, B and C are collinear. br ev 2 The midpoint of the line segment joining P( −4, 5) and Q(6, 1) is M. -R am The point R has coordinates ( −3, −7). s -C Show that RM is perpendicular to PQ. es Pr Find the gradient of CD and the gradient of BC . ni ve rs ity 4 The coordinates of three of the vertices of a trapezium, ABCD, are A(3, 5), B( −5, 4) and C(1, −5). AD is parallel to BC and angle ADC is 90° . ie id g w Find the value of k if A, B and C are collinear. C e U 5 The coordinates of three points are A(5, 8), B ( k, 5) and C ( − k, 4). op y Find the coordinates of D. -R s es am Find the two possible values of k if angle ABC is 90°. ev br 6 The vertices of triangle ABC are A( −9, 2 k − 8), B (6, k ) and C ( k, 12). -C ev ie w C op y 3 Two vertices of a rectangle, ABCD, are A( −6, −4) and B(4, −8). R x -R am B (−4, −2) A (11, 3) O ev id br O B (8.8, 4.4) w A (11, 3) y y y R ni C op If k = −2, then B is the point ( −4, 2). op ev ie b If k = 4.4, then B is the point (8.8, 4.4). Copyright Material - Review Only - Not for Redistribution 77 ve rs ity 7 C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 w ge A is the point (0, 8) and B is the point (8, 6). am br id ev ie Find the point C on the y-axis such that angle ABC is 90°. -R 8 Three points have coordinates A(7, 4), B(19, 8) and C ( k, 2 k ). -C Find the value of the constant k for which: Pr es s a C lies on the line that passes through the points A and B x y − = 1, where a and b are positive constants, meets the x-axis at P and the y-axis at Q. a b 2 The gradient of the line PQ is and the length of the line PQ is 2 29. 5 Find the value of a and the value of b. ve rs ity ev ie w C 9 The line y op y b angle CAB is 90° . U R ni C op 10 P is the point ( a, a − 2) and Q is the point (4 − 3a, − a ). w ge a Find the gradient of the line PQ. id br ev Given that the distance PQ is 10 5 , find the two possible values of a. am c ie b Find the gradient of a line perpendicular to PQ. C es Pr b Find the value of a, the value of b and the value of c. c Find the perimeter of the rhombus. A (a, 1) B (8, 2) O w rs d Find the area of the rhombus. op ni C U 3.3 Equations of straight lines br ev id ie w ge At IGCSE / O Level you learnt the equation of a straight line is: KEY POINT 3.4 x y ve ie ev C (b, c) M ity op y a Find the coordinates of M. R D (4, 10) s -C M is the midpoint of BD. 78 y -R 11 The diagram shows a rhombus ABCD. es s -C -R am y = mx + c, where m is the gradient and c is the y-intercept, when the line is non-vertical. x = b when the line is vertical, where b is the x-intercept. y P (x, y) C ity Pr op y There is an alternative formula that we can use when we know the gradient of a straight line and a point on the line. A (x1, y1) y op -R s es -C am br ev ie id g w e C U R ev ie w ni ve rs Consider a line, with gradient m, that passes through the known point A( x1, y1 ) and whose general point is P ( x, y ). Copyright Material - Review Only - Not for Redistribution O x ve rs ity C U ni op y Chapter 3: Coordinate geometry y − y1 =m x − x1 y − y1 = m( x − x1 ) ge ev ie w Multiply both sides by ( x − x1 ). -R am br id Gradient of AP = m, hence Pr es s -C KEY POINT 3.5 C ve rs ity op y y − y1 = m( x − x1 ) ev ie w WORKED EXAMPLE 3.6 y The equation of a straight line, with gradient m, that passes through the point ( x1, y1 ) is: Answer w ge U R ni C op Find the equation of the straight line with gradient −2 that passes through the point (4, 1). y es s -C -R am br y − 1 = −2( x − 4) y − 1 = −2 x + 8 2x + y = 9 ity Find the equation of the straight line passing through the points ( −4, 3) and (6, −2). y ve op Decide which values to use for x1, y1, x2 , y2. ni (6, −2) ↑ ↑ ( x2 , y2 ) ie y op -R s es -C am br ev ie id g w e C U R ev ie w ni ve rs C ity Pr op y es s -C -R am br ev id y2 − y1 ( −2) − 3 1 = =− x2 − x1 6 − ( −4) 2 1 Using y − y1 = m( x − x1 ) with m = − , x1 = −4 and y1 = 3: 2 1 y − 3 = − ( x + 4) 2 2 y − 6 = −x − 4 x + 2y = 2 Gradient = m = w C U ( −4, 3) ↑ ↑ ( x1, y1 ) ge ie Answer ev 79 rs w C op Pr WORKED EXAMPLE 3.7 R ie ev id Using y − y1 = m( x − x1 ) with m = −2, x1 = 4 and y1 = 1: Copyright Material - Review Only - Not for Redistribution ve rs ity ev ie w ge am br id WORKED EXAMPLE 3.8 C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 -C Gradient of AB = 1 −2 − 1 −3 =− = 7 − ( −5) 12 4 op y Gradient of the perpendicular = 4 C op ni w ev Expand brackets and simplify. -R es s Multiply both sides by 2. Pr 1 Find the equation of the line with: ity op C ie ge id br am -C y EXERCISE 3C 80 ve ie w rs a gradient 2 passing through the point (4, 9) y op w ge 2 Find the equation of the line passing through each pair of points. br ev id ie ( 1, 0 ) and (5, 6) -R am b (3, −5) and ( −2, 4) (3, −1) and ( −3, −5) s -C c C U R ni ev b gradient −3 passing through the point (1, −4) 2 c gradient − passing through the point ( −4, 3). 3 a es 3 Find the equation of the line: Pr ity b parallel to the line x + 2 y = 6, passing through the point (4, −6) ni ve rs perpendicular to the line y = 2 x − 3, passing through the point (6, 1) op y d perpendicular to the line 2 x − 3 y = 12, passing through the point (8, −3). e C U 4 Find the equation of the perpendicular bisector of the line segment joining the points: ie id g w a (5, 2) and ( −3, 6) es s -R br ( −2, −7) and (5, −4). am c ev b ( −2, −5) and (8, 1) -C R ev ie w C op y a parallel to the line y = 3x − 5, passing through the point (1, 7) c x1 + x2 y1 + y2 , . 2 2 y ve rs ity Use midpoint = U R 2 y = 8x − 9 y2 − y1 . x2 − x1 Use m1 × m2 = −1. −5 + 7 1 + ( −2 ) 1 Midpoint of AB = , = 1, − 2 2 2 ∴ The perpendicular bisector is the line with gradient 4 passing 1 through the point 1, − . 2 1 Using y − y1 = m( x − x1 ) with x1 = 1, y1 = − and m = 4: 2 1 y + = 4( x − 1) 2 y = 4x − 4 21 C w ev ie Use gradient = Pr es s Answer -R Find the equation of the perpendicular bisector of the line segment joining A( −5, 1) and B(7, −2). Copyright Material - Review Only - Not for Redistribution ve rs ity C U ni op y Chapter 3: Coordinate geometry ev ie P is the point ( −4, 2) and Q is the point (5, −4). -R am br id 6 w ge 5 The line l1 passes through the points P( −10, 1) and Q(2, 10). The line l2 is parallel to l1 and passes through the point (4, −1). The point R lies on l2 , such that QR is perpendicular to l2 . Find the coordinates of R. Pr es s b Find the coordinates of the point R. c w ev ie a Find the equation of the line l. ve rs ity C op y -C A line, l, is drawn through P and perpendicular to PQ to meet the y-axis at the point R. Find the area of triangle PQR. U R ni C op y 7 The line l1 has equation 3x − 2 y = 12 and the line l2 has equation y = 15 − 2 x . The lines l1 and l2 intersect at the point A. w ge a Find the coordinates of A. id ie b Find the equation of the line through A that is perpendicular to the line l1. -R am br ev 8 The perpendicular bisector of the line joining A( −10, 5) and B( −2, −1) intersects the x-axis at P and the y-axis at Q. s -C a Find the equation of the line PQ. es Find the length of PQ. Pr c 81 9 The line l1 has equation 2 x + 5 y = 10. ity C op y b Find the coordinates of P and Q. ve ie w rs The line l2 passes through the point A( −9, −6) and is perpendicular to the line l1. y op ni ev a Find the equation of the line l2 . w ge C U R b Given that the lines l1 and l2 intersect at the point B, find the area of triangle ABO, where O is the origin. -R G x w ni ve rs C ity O ie op y H (5, −7) id g w e C U 11 The coordinates of three points are A( −4, −1), B(8, −9) and C ( k, 7). M is the midpoint of AB and MC is perpendicular to AB. Find the value of k. br ev ie 12 The point P is the reflection of the point ( −2, 10) in the line 4x − 3 y = 12. -R s es am Find the coordinates of P. -C ev R Pr op y F s E -C y es am br ev id ie 10 The diagram shows the points E , F and G lying on the line x + 2 y = 16. The point G lies on the x-axis and EF = FG. The line FH is perpendicular to EG. Find the coordinates of E and F . Copyright Material - Review Only - Not for Redistribution ve rs ity am br id ev ie w ge 13 The coordinates of triangle ABC are A( −7, 3), B(3, −7) and C(8, 8). P is the foot of the perpendicular from B to AC. a Find the equation of the line BP . -R b Find the coordinates of P. Find the lengths of AC and BP . Pr es s -C c C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 y d Use your answers to part c to find the area of triangle ABC . ni 15 The equations of two of the sides of triangle ABC are x + 2 y = 8 and 2 x + y = 1. Given that A is the point (2, −3) and that angle ABC = 90°, find: w ge PS C op b Find the coordinates of the point that is equidistant from P, Q and R. ie id -R am br ev b the coordinates of the point B. 16 Find two straight lines whose x-intercepts differ by 7, whose y-intercepts differ by 5 and whose gradients differ by 2. Pr ity [This question is based upon Straight line pairs on the Underground Mathematics website.] y ve 3.4 The equation of a circle op In this section you will learn about the equation of a circle. A circle is defined as the locus of all the points in a plane that are a fixed distance (the radius) from a given point (the centre). -R am br EXPLORE 3.2 ev id ie w ge C U ni ev ie w C 82 rs op y Is your solution unique? Investigate further. es s -C PS y ve rs ity ii PR PQ U ev ie R i a the equation of the third side R WEB LINK a Find the equation of the perpendicular bisectors of: w C op 14 The coordinates of triangle PQR are P(1, 1), Q(1, 8) and R(6, 6). s es Centre b ( x − 2)2 + ( y − 1)2 = 9 c ( x + 3)2 + ( y + 5)2 = 16 d ( x − 8)2 + ( y + 6)2 = 49 e x 2 + ( y + 4)2 = 4 f ( x + 6)2 + y 2 = 64 ity ni ve rs y x + y = 25 Radius 2 ie id g w e C U op a 2 Pr Equation of circle -R s es am br ev 2 Discuss your results with your classmates and explain how you can find the coordinates of the centre of a circle and the radius of a circle just by looking at the equation of the circle. -C R ev ie w C op y -C 1 Use graphing software to draw each of the following circles. From your graphs find the coordinates of the centre and the radius of each circle, and copy and complete the following table. Copyright Material - Review Only - Not for Redistribution Try the following resources on the Underground Mathematics website: • Lots of lines! • Straight lines • Simultaneous squares • Straight line pairs. ve rs ity C U ni op y Chapter 3: Coordinate geometry ev ie am br id w ge To find the equation of a circle, we let P ( x, y ) be any point on the circumference of a circle with centre C ( a, b ) and radius r. y -R P (x, y) Pr es s -C r C (a, b) C ve rs ity op y Q ev ie w O x ni C op y Using Pythagoras’ theorem on triangle CQP gives CQ2 + PQ2 = r 2 . U R Substituting CQ = x − a and PQ = y − b into CQ2 + PQ2 = r 2 gives: ie ev id -R am br KEY POINT 3.6 w ge ( x − a )2 + ( y − b )2 = r 2 s -C The equation of a circle with centre ( a, b ) and radius r can be written in completed square form as: ity 83 EXPLORE 3.3 ve ie w rs C op Pr y es ( x − a )2 + ( y − b )2 = r 2 y op b decreasing the value of a c increasing the value of b d decreasing the value of b. -R am br ev id ie w ge a increasing the value of a C U R ni ev The completed square form for the equation of a circle with centre ( a, b ) and radius r is ( x − a )2 + ( y − b )2 = r 2 . Use graphing software to investigate the effects of: DID YOU KNOW? s -C WORKED EXAMPLE 3.9 es Write down the coordinates of the centre and the radius of each of these circles. Pr b Centre = (2, 4), radius = 100 = 10 c Centre = ( −1, 8), radius = 12 = 2 3 op Centre = (0, 0), radius = 4 =2 -R s es am br ev ie id g w e U a C Answer y ni ve rs c ( x + 1)2 + ( y − 8)2 = 12 ity b ( x − 2)2 + ( y − 4)2 = 100 -C R ev ie w C op y a x 2 + y2 = 4 Copyright Material - Review Only - Not for Redistribution In the 17th century, the French philosopher and mathematician René Descartes developed the idea of using equations to represent geometrical shapes. The Cartesian coordinate system is named after this famous mathematician. ve rs ity -C Answer -R Find the equation of the circle with centre ( −4, 3) and radius 6. ev ie w ge am br id WORKED EXAMPLE 3.10 C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 ( x − ( −4))2 + ( y − 3)2 = 62 C op y ve rs ity WORKED EXAMPLE 3.11 ni ev ie w C op ( x + 4)2 + ( y − 3)2 = 36 Pr es s y Equation of circle is ( x − a )2 + ( y − b )2 = r 2 , where a = −4, b = 3 and r = 6. U R A is the point (3, 0) and B is the point (7, −4). ie x r -C C rs The centre of the circle, C , is the midpoint of AB. ve ie w ity B (7, −4) C 84 Pr op y es s O -R am A (3, 0) br y ev id Answer w ge Find the equation of the circle that has AB as a diameter. y op C U R ni ev 3 + 7 0 + ( −4) C = , = (5, −2) 2 2 w ie (5 − 3)2 + ( −2 − 0)2 = 8 ev r= br id ge Radius of circle, r, is equal to AC. es Pr ity op y x 2 − 2 ax + a 2 + y2 − 2by + b2 = r 2 x 2 + y2 − 2 ax − 2by + ( a 2 + b2 − r 2 ) = 0 U id g ie ev es s -R br am ● the coefficients of x 2 and y2 are equal there is no xy term. -C ● w e C When we write the equation of a circle in this form, we can note some important characteristics of the equation of a circle. For example: op y ni ve rs C w ie ( x − a )2 + ( y − b )2 = r 2 gives: Rearranging gives: ev R ( 8 )2 ( x − 5)2 + ( y + 2)2 = 8 Expanding the equation -R -C ( x − 5)2 + ( y + 2)2 = s am Equation of circle is ( x − a )2 + ( y − b )2 = r 2 , where a = 5, b = −2 and r = 8. Copyright Material - Review Only - Not for Redistribution ve rs ity C U ni op y Chapter 3: Coordinate geometry w x 2 + y 2 + 2 gx + 2 fy + c = 0 Pr es s -C g 2 + f 2 − c is the radius. You should not try to memorise the formulae for the centre and radius of a circle in this form, but rather work them out if needed, as shown in Worked example 3.12. -R KEY POINT 3.7 where ( − g, − f ) is the centre and TIP ev ie am br id ge We often write the expanded form of a circle as: ve rs ity WORKED EXAMPLE 3.12 ev ie w C op y This is the equation of a circle in expanded general form. C op ni U R ge Answer id ie w We answer this question by first completing the square. x 2 + 10 x + y2 − 8 y − 40 = 0 br ev Complete the square. -R ity Centre = ( −5, 4) and radius = 9. 85 rs C r 2 = 81 w ve ie es b=4 op a = −5 Pr y ( x + 5)2 + ( y − 4)2 = 81 y op ie P es s -C Pr op y The tangent to a circle at a point is perpendicular to the radius at that point. The perpendicular from the centre of a circle to a chord bisects the chord. w ni ve rs C ity The angle in a semicircle is a right angle. op y From these statements we can conclude that: If triangle ABC is right angled at B, then the points A, B and C lie on the circumference of a circle with AC as diameter. ● The perpendicular bisector of a chord passes through the centre of the circle. ● If a radius and a line at a point, P, on the circumference are at right angles, then the line must be a tangent to the curve. -R s es am br ev ie id g w e C U ● -C ie O -R am O O C ev A br id w ge C U R ni ev It is useful to remember the three following right angle facts for circles. B ev Collect constant terms together. Compare with ( x − a )2 + ( y − b )2 = r 2. s -C am ( x + 5)2 − 52 + ( y − 4)2 − 42 − 40 = 0 R y Find the centre and the radius of the circle x 2 + y2 + 10 x − 8 y − 40 = 0. Copyright Material - Review Only - Not for Redistribution ve rs ity A circle passes through the points P ( −1, 4), Q(1, 6) and R(5, 4). Q (1, 6) ve rs ity P (−1, 4) C op ge U R ni ev ie C y R (5, 4) w C op y y Pr es s -C Answer -R Find the equation of the circle. ev ie w ge am br id WORKED EXAMPLE 3.13 C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 ie w x id O br ev The centre of the circle lies on the perpendicular bisector of PQ and on the perpendicular bisector of QR. −1 + 1 4 + 6 , = (0, 5) 2 2 6−4 Gradient of PQ = =1 1 − ( −1) op s es Gradient of perpendicular bisector of PQ = −1 86 Pr y -C -R am Midpoint of PQ = C ity Equation of perpendicular bisector of PQ is: rs (1) y ve ev ie w ( y − 5) = −1( x − 0) y = −x + 5 ni C U id ie w ge R op 1+ 5 6 + 4 , = (3, 5) 2 2 4−6 1 Gradient of QR = =− 5−1 2 Midpoint of QR = br ev Gradient of perpendicular bisector of QR = 2 es s -C -R am Equation of perpendicular bisector of QR is: ( y − 5) = 2( x − 3) (2) y = 2x − 1 Pr ity x = 2, y = 3 C ni ve rs Radius = CR = (5 − 2)2 + (4 − 3)2 = 10 -R s es -C am br ev ie id g w e C U op Hence, the equation of the circle is ( x − 2)2 + ( y − 3)2 = 10 . R ev ie w Centre of circle = (2, 3) y op y Solving equations (1) and (2) gives: Copyright Material - Review Only - Not for Redistribution ve rs ity ev ie w ge am br id Alternative method: C U ni op y Chapter 3: Coordinate geometry The equation of the circle is ( x − a )2 + ( y − b )2 = r 2. -R The points ( −1, 4), (1, 6) and (5, 4) lie on the circle, so substituting gives: a 2 + 2 a + b2 − 8b + 17 = r 2 Pr es s -C ( −1 − a )2 + (4 − b )2 = r 2 (1) ve rs ity Then subtracting (1) − (3) and (2) − (3) gives two simultaneous equations for a and b, which can then be solved. U R EXERCISE 3D 2x 2 + 2 y2 = 9 c x 2 + ( y − 2)2 = 25 d ( x − 5)2 + ( y + 3)2 = 4 ( x + 7)2 + y2 = 18 f -R w id b -C e x 2 + y2 − 8x + 20 y + 110 = 0 ev 2( x − 3)2 + 2( y + 4)2 = 45 2 x 2 + 2 y2 − 14x − 10 y − 163 = 0 s h es g ie x 2 + y2 = 16 am a br ge 1 Find the centre and the radius of each of the following circles. C op y Finally, substituting into (1) gives r 2 . ni ev ie w C op y and similar for the other two points, giving equations (2) and (3). Pr centre ( −1, 3), radius 7 1 3 5 d centre , − , radius 2 2 2 y ev ve ie c b centre (5, −2), radius 4 ity a centre (0, 0), radius 8 rs w C op y 2 Find the equation of each of the following circles. w ge 4 A diameter of a circle has its end points at A( −6, 8) and B(2, −4). C U R ni op 3 Find the equation of the circle with centre (2, 5) passing through the point (6, 8). br ev id ie Find the equation of the circle. -R am 5 Sketch the circle ( x − 3)2 + ( y + 2)2 = 9. s -C 6 Find the equation of the circle that touches the x-axis and whose centre is (6, −5). es ity Pr Show that the centre of the circle lies on the line 4x + 2 y = 15. ni ve rs 8 A circle passes through the points (3, 2) and (7, 2) and has radius 2 2. y Find the two possible equations for this circle. op U 9 A circle passes through the points O(0, 0), A(8, 4) and B(6, 6). w e C Show that OA is a diameter of the circle and find the equation of this circle. -R s es am br ev ie id g 10 Show that x 2 + y2 − 6x + 2 y = 6 can be written in the form ( x − a )2 + ( y − b )2 = r 2 , where a, b and r are constants to be found. Hence, write down the coordinates of the centre of the circle and also the radius of the circle. -C R ev ie w C op y 7 The points P(1, −2) and Q(7, 1) lie on the circumference of a circle. Copyright Material - Review Only - Not for Redistribution 87 ve rs ity C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 am br id ev ie w ge 11 The equation of a circle is ( x − 3)2 + ( y + 2)2 = 25. Show that the point A(6, −6) lies on the circle and find the equation of the tangent to the circle at the point A. Pr es s -C -R 12 The line 2 x + 5 y = 20 cuts the x-axis at A and the y-axis at B. The point C is the midpoint of the line AB. Find the equation of the circle that has centre C and that passes through the points A and B. Show that this circle also passes through the point O(0, 0). b Find the equation of the circle that passes through the points P, Q and R. w ev ie a Show that angle PQR is a right angle. ve rs ity C op y 13 The points P ( −5, 6), Q( −3, 8) and R(3, 2) are joined to form a triangle. U R ni C op y 14 Find the equation of the circle that passes through the points (7, 3) and (11, −1) and has its centre lying on the line 2 x + y = 7. w ge 15 A circle passes through the points O(0, 0), P (3, 9) and Q(11, 11). br ev id ie Find the equation of the circle. i -R s Pr The radius of each green circle is 1 unit. Find the radius of the orange circle. ity op C w rs ii Use graphing software to draw the design. ii Use graphing software to draw this extended design. op The radius of each green circle is 1 unit. Find the radius of the blue circle. y ve i -C -R am br ev id ie w ge C U R ev ie b The design in part a is extended, as shown. ni 88 Try the following resources on the Underground Mathematics website: • Olympic rings • Teddy bear. es -C 17 a The design shown is made from four green circles and one orange circle. y PS am 16 A circle has radius 10 units and passes through the point (5, −16). The x-axis is a tangent to the circle. Find the possible equations of the circle. WEB LINK es s 3.5 Problems involving intersections of lines and circles TIP Line and parabola y Nature of roots ni ve rs b 2 − 4 ac two distinct points of intersection =0 two equal real roots one point of intersection (line is a tangent) ,0 no real roots C ie w e id g no points of intersection op two distinct real roots U .0 -R s es am br ev In this section you will solve problems involving the intersection of lines and circles. -C R ev ie w C ity Pr op y In Chapter 1 you learnt that the points of intersection of a line and a curve can be found by solving their equations simultaneously. You also learnt that if the resulting equation is of the form ax 2 + bx + c = 0, then b2 − 4ac gives information about the line and the curve. Copyright Material - Review Only - Not for Redistribution We can also describe an equation that has ‘two equal real roots’ as having ‘one repeated (real) root’. ve rs ity ge C U ni op y Chapter 3: Coordinate geometry am br id ev ie w WORKED EXAMPLE 3.14 -R The line x = 3 y + 10 intersects the circle x 2 + y2 = 20 at the points A and B. -C a Find the coordinates of the points A and B. Pr es s b Find the equation of the perpendicular bisector of AB and show that it passes through the centre of the circle. Factorise. ge y2 + 6 y + 8 = 0 Expand and simplify. U R (3 y + 10)2 + y2 = 20 Substitute 3 y + 10 for x. y x 2 + y2 = 20 C op Answer a ve rs ity Find the exact coordinates of P and Q. ni ev ie w C op y c The perpendicular bisector of AB intersects the circle at the points P and Q. ie y = −4 ev id or am When y = −2, x = 4 and when y = −4, x = −2. A and B are the points ( −2, −4) and (4, −2). −2 − ( −4) 1 = 4 − ( −2) 3 So the gradient of the perpendicular bisector = –3. -C Pr es s Gradient of AB = 89 Use m = −3, x1 = 1 and y1 = −3. ve ie w y − y1 = m( x − x1 ) ity −2 + 4 −4 + ( −2) Midpoint of AB = , = (1, −3) 2 2 rs C op y b -R br y = −2 w ( y + 2)( y + 4) = 0 y ev y − ( −3) = −3( x − 1) C U R ni op Perpendicular bisector is y = −3x. br x 2 + y2 = 20 s 2, y = −3 2 . Pr When x = − 2, y = 3 2 and when x = es x=± 2 ( ) ity P and Q are the points − 2, 3 2 and ( 2, −3 2 ), respectively. y op -R s es -C am br ev ie id g w e C U R ev ie w ni ve rs op y -C 10 x 2 = 20 C Substitute −3x for y. -R am c ev id ie Hence, the perpendicular bisector of AB passes through the point (0, 0), the centre of the circle x 2 + y2 = 20. w ge When x = 0, y = −3(0) = 0. Copyright Material - Review Only - Not for Redistribution ve rs ity ev ie w ge am br id WORKED EXAMPLE 3.15 C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 Answer -R Show that the line y = x − 13 is a tangent to the circle x 2 + y2 − 8x + 6 y + 7 = 0. x + ( x − 13) − 8x + 6( x − 13) + 7 = 0 2 y 2 x − 14x + 49 = 0 op 2 Expand and simplify. Factorise. C ve rs ity ( x − 7)( x − 7) = 0 x = 7 or x = 7 w ev ie Substitute x − 13 for y. Pr es s -C x 2 + y 2 − 8x + 6 y + 7 = 0 U w ge EXERCISE 3E ie R ni C op y The equation has one repeated root, hence y = x − 13 is a tangent. br ev id 1 Find the points of intersection of the line y = x − 3 and the circle ( x − 3)2 + ( y + 2)2 = 20. -R am 2 The line 2 x − y + 3 = 0 intersects the circle x 2 + y2 − 4x + 6 y − 12 = 0 at two points, D and E. s -C Find the length of DE . Pr y es 3 Show that the line 3x + y = 6 is a tangent to the circle x 2 + y2 + 4x + 16 y + 28 = 0. 4 Find the set of values of m for which the line y = mx + 1 intersects the circle ( x − 7)2 + ( y − 5)2 = 20 at two distinct points. rs ity op C 90 ve ie w 5 The line 2 y − x = 12 intersects the circle x 2 + y2 − 10 x − 12 y + 36 = 0 at the points A and B. y op ni ev a Find the coordinates of the points A and B. The perpendicular bisector of AB intersects the circle at the points P and Q. w ge c C U R b Find the equation of the perpendicular bisector of AB. br ev id ie Find the exact coordinates of P and Q. -R 6 Show that the circles x 2 + y2 = 25 and x 2 + y2 − 24x − 18 y + 125 = 0 touch each other. -C PS am d Find the exact area of quadrilateral APBQ. es s Find the coordinates of the point where they touch. Pr ity ni ve rs 7 Two circles have the following properties: the x-axis is a common tangent to the circles ● the point (8, 2) lies on both circles ● the centre of each circle lies on the line x + 2 y = 22. ie id g w a Find the equation of each circle. C e U op y ● br ev b Prove that the line 4x + 3 y = 88 is a common tangent to these circles. es s -R [Inspired by Can we find the two circles that satisfy these three conditions? on the Underground Mathematics website.] am R ev ie w PS -C C op y [This question is taken from Can we show that these two circles touch? on the Underground Mathematics website.] Copyright Material - Review Only - Not for Redistribution ve rs ity Midpoint, gradient and length of line segment Gradient of PQ is ● Length of segment PQ is -R ● Pr es s y2 − y1 . x2 − x1 ( x2 − x1 )2 + ( y2 − y1 )2 If the gradients of two parallel lines are m1 and m2 , then m1 = m2. ● If the gradients of two perpendicular lines are m1 and m2 , then m1 × m2 = −1. ni C op ● ge The equation of a straight line is: ie w y − y1 = m( x − x1 ), where m is the gradient and ( x1, y1 ) is a point on the line. id ● y Parallel and perpendicular lines x + x2 y1 + y2 , Midpoint, M, of PQ is 1 . 2 2 U R ev ie w C op P (x1, y1) ● ve rs ity M y -C Q (x2, y2) w ev ie am br id ge Checklist of learning and understanding C U ni op y Chapter 3: Coordinate geometry br ev The equation of a circle is: ( x − a )2 + ( y − b )2 = r 2 , where ( a, b ) is the centre and r is the radius. ● x 2 + y 2 + 2 gx + 2 fy + c = 0 , where ( − g, − f ) is the centre and g 2 + f 2 − c is the radius. 91 y op y op -R s es -C am br ev ie id g w e C U R ev ie w ni ve rs C ity Pr op y es s -C -R am br ev id ie w ge C U R ni ev ve ie w rs C ity op Pr y es s -C -R am ● Copyright Material - Review Only - Not for Redistribution ve rs ity ge C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 ev ie am br id 18 , where a is a constant. x−3 Find the set of values of a for which the line does not intersect the curve. Pr es s -C -R A line has equation 2 x + y = 20 and a curve has equation y = a + 1 y op y 2 w END-OF-CHAPTER REVIEW EXERCISE 3 [4] y = 6x + k y = 7√x w C ve rs ity B y C op O x ge U R ni ev ie A br For the case where k = 2, find the x-coordinates of A and B. [4] ii Find the value of k for which y = 6x + k is a tangent to the curve y = 7 x . [2] s -C -R am i es Cambridge International AS & A Level Mathematics 9709 Paper 11 Q5 June 2012 ity A is the point ( a, 3) and B is the point (4, b ). 1 The length of the line segment AB is 4 5 units and the gradientis − . 2 w rs C Pr y op 3 92 ie The curve and the line intersect at the points A and B. ev id w The diagram shows the curve y = 7 x and the line y = 6x + k , where k is a constant. [6] y op ni The curve y = 3 x − 2 and the line 3x − 4 y + 3 = 0 intersect at the points P and Q. C R 4 U ev ve ie Find the possible values of a and b. ie id ev The line ax − 2 y = 30 passes through the points A(10, 10) and B ( b, 10b ), where a and b are constants. am a Find the values of a and b. [1] s -C b Find the coordinates of the midpoint of AB. [3] -R br 5 [6] w ge Find the length of PQ. Find the area of triangle AOB in terms of t. [3] ity C i Pr The line with gradient −2 passing through the point P (3t, 2t ) intersects the x-axis at A and the y-axis at B. ni ve rs 6 [3] es op y c Find the equation of the perpendicular bisector of the line AB. op Cambridge International AS & A Level Mathematics 9709 Paper 11 Q6 June 2015 C U [4] -R s am br ev Find the coordinates of P. You must show all your working. ie id g w e The point P is the reflection of the point ( −7, 5) in the line 5x − 3 y = 18. -C 7 y Show that the mid-point of PC lies on the line y = x. R ev ii es ie w The line through P perpendicular to AB intersects the x-axis at C . Copyright Material - Review Only - Not for Redistribution [7] ve rs ity ev ie am br id 4 and the line x − 2 y + 6 = 0 intersect at the points A and B. x a Find the coordinates of these two points. -R The curve y = x + 2 − -C 8 w ge C U ni op y Chapter 3: Coordinate geometry b Find the perpendicular bisector of the line AB. Pr es s [4] The line y = mx + 1 intersects the circle x + y − 19x − 51 = 0 at the point P(5, 11). 2 2 a Find the coordinates of the point Q where the line meets the curve again. ve rs ity C op y 9 b Find the equation of the perpendicular bisector of the line PQ. w [4] [4] [3] y ev ie c Find the x-coordinates of the points where this perpendicular bisector intersects the circle. U R ni C op Give your answers in exact form. y ge 10 [4] s -C -R am br ev id ie w B (15, 22) op es Pr y C 93 x A (3, −2) w rs C ity O ve ie The diagram shows a triangle ABC in which A is (3, −2) and B is (15, 22). The gradients of AB, y op ni Find the gradient of AB and deduce the value of m. ii Find the coordinates of C . C U i br iii Find the coordinates of D. [4] -R am ie ev id The perpendicular bisector of AB meets BC at D. [2] [4] w ge R ev AC and BC are 2 m , −2 m and m respectively, where m is a positive constant. s -C Cambridge International AS & A Level Mathematics 9709 Paper 11 Q8 June 2010 es Find the equation of the perpendicular bisector of AB, giving your answer in the form y = mx + c. ii A point C on the perpendicular bisector has coordinates ( p, q ). The distance OC is 2 units, where O is the origin. Write down two equations involving p and q and hence find the coordinates of the possible positions of C . [5] [4] ni ve rs ity Pr i ie y op -R s -C am br ev ie id g w e C U R ev Cambridge International AS & A Level Mathematics 9709 Paper 11 Q7 November 2013 es w C op y 11 The point A has coordinates ( −1, 6) and the point B has coordinates (7, 2). Copyright Material - Review Only - Not for Redistribution ve rs ity am br id ev ie w ge C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 12 The coordinates of A are ( −3, 2) and the coordinates of C are (5, 6). -R The mid-point of AC is M and the perpendicular bisector of AC cuts the x-axis at B. Find the equation of MB and the coordinates of B. ii Show that AB is perpendicular to BC . y C op ni U R b Use your answer to part a to find the value of p. y [4] [1] [4] ie id w ge c Find the equation of the circle. 14 y ev ie 13 The points A(1, −2) and B(5, 4) lie on a circle with centre C (6, p ). a Find the equation of the perpendicular bisector of the line segment AB. D op Pr y es s -C -R am br ev A (13, 17) C (13, 4) B (3, 2) ity O x rs C 94 [2] Cambridge International AS & A Level Mathematics 9709 Paper 11 Q9 June 2012 ve rs ity op [2] iii Given that ABCD is a square, find the coordinates of D and the length of AD. C w [5] Pr es s -C i ve ie w ABCD is a trapezium with AB parallel to DC and angle BAD = 90°. y op ni ev a Calculate the coordinates of D. [2] ge C U R b Calculate the area of trapezium ABCD. [7] ie w 15 The equation of a curve is xy = 12 and the equation of a line is 3x + y = k , where k is a constant. br ev id a In the case where k = 20, the line intersects the curve at the points A and B. [4] -R am Find the midpoint of the line AB. [3] b Show that the perpendicular bisector of the line AB is 3x − 4 y = 17. [3] c A circle passes through A and B and has its centre on the line x = 15. Find the equation of this circle. [4] ni ve rs ity Pr a Find the equation of the line through A and B. op y 17 The equation of a circle is x 2 + y2 − 8x + 4 y + 4 = 0. [4] C U a Find the radius of the circle and the coordinates of its centre. [4] c Show that the point A(6, 2 3 − 2) lies on the circle. [2] ie ev br id g w e b Find the x-coordinates of the points where the circle crosses the x-axis, giving your answers in exact form. -R 3x + 3 y = 12 3 − 6. s es am d Show that the equation of the tangent to the circle at A is -C R ev ie w C op y es 16 A is the point ( −3, 6) and B is the point (9, −10). [4] s -C b Find the set of values of k for which the line 3x + y = k intersects the curve at two distinct points. Copyright Material - Review Only - Not for Redistribution [4] ve rs ity ge C U ni op y Cross-topic review exercise 1 w 4 17 + 18 = 2 . x4 x [4] y ve rs ity ni 4 x w ge U C op op C w ev ie R O –4 y Pr es s -C A (–2, 21) y 2 ev ie Solve the equation 1 -R am br id CROSS-TOPIC REVIEW EXERCISE 1 -R am br ev The diagram shows the graph of y = f( x ) for −4 < x < 4. ie id B (2, –11) Sketch on separate diagrams, showing the coordinates of any turning points, the graphs of: es Pr The graph of f( x ) = ax + b is reflected in the y-axis and then translated by the vector 0 . 3 The resulting function is g( x ) = 1 − 5x. Find the value of a and the value of b. op y The graph of y = ( x + 1)2 is transformed by the composition of two transformations to the graph of y = 2( x − 4)2. Find these two transformations. w ie [4] -R s es 1 –2 O Pr 1 3 2 –2 y = f(x) –3 es s -R br ev ie id g w e Sketch the graph of y = 2 − f( x ). C U The diagram shows the graph of y = f( x ) for −3 < x < 3. am x –1 y ity ni ve rs C w –1 op op y –3 -C [4] 2 ie ev R [4] ev br am y -C 6 95 C U The graph of y = x 2 + 1 is transformed by applying a reflection in the x-axis followed by a 3 translation of . Find the equation of the resulting graph in the form y = ax 2 + bx + c. 2 id 5 ge R ni ev 4 ve ie w rs C [2] ity op y b y = −2f( x ) 3 [2] s -C a y = f( x ) + 5 Copyright Material - Review Only - Not for Redistribution [4] ve rs ity am br id ev ie w ge C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 a Find the set of values of x for which f( x ) < x. -R The function f is such that f( x ) = x 2 − 5x + 5 for x ∈ ℝ. 7 [3] C 8 The line x + ky + k 2 = 0, where k is a constant, is a tangent to the curve y2 = 4x at the point P. Find, in terms of k, the coordinates of P. w [6] y A is the point (4, −6) and B is the point (12, 10). The perpendicular bisector of AB intersects the x-axis at C and the y-axis at D. Find the length of CD. U w ge a Given that AB = BC, show that a possible value of k is 4 and find the other possible value of k. [3] b For the case where k = 4, find the equation of the line that bisects angle ABC. [4] ev id br -R am A curve has equation xy = 12 + x and a line has equation y = kx − 9, where k is a constant. -C 11 es s a In the case where k = 2, find the coordinates of the points of intersection of the curve and the line. Pr [4] The function f is such that f( x ) = 2 x − 3 for x > k, where k is a constant. The function g is such that g( x ) = x 2 − 4 for x > −4. ve ie w rs C 12 y op ni ev a Find the smallest value of k for which the composite function gf can be formed. -R am br ev id ie 4 − 2 for x . 0 , x 4 g( x ) = for x > 0. 5x + 2 [3] es s -C i Find and simplify an expression for fg( x ) and state the range of fg. Pr [5] ity Cambridge International AS & A Level Mathematics 9709 Paper 11 Q8 November 2016 ni ve rs The equation x 2 + bx + c = 0 has roots −2 and 7. a Find the value of b and the value of c. y [2] U op b Using these values of b and c, find: w -R s es am br ev ie id g ii the set of values of x for which x 2 + bx + c , 10. C e i the coordinates of the vertex of the curve y = x 2 + bx + c -C ev ie w C op y ii Find an expression for g −1( x ) and find the domain of g −1. 14 [4] w ge The functions f and g are defined by f( x ) = [3] C U R b Solve the inequality gf( x ) . 45. 13 R [3] ity op y b Find the set of values of k for which the line does not intersect the curve. 96 [6] The points A, B and C have coordinates A(2, 8), B (9, 7) and C ( k, k − 2). 10 ie R ni 9 C op ev ie [3] ve rs ity op y Find the two possible values of m. Pr es s -C b The line y = mx − 11 is a tangent to the curve y = f( x ). Copyright Material - Review Only - Not for Redistribution [3] [3] ve rs ity ev ie am br id The line L1 passes through the points A( −6, 10) and B (6, 2). The line L2 is perpendicular to L1 and passes through the point C ( −7, 2). -R 15 w ge C U ni op y Cross-topic review exercise 1 [4] Pr es s -C a Find the equation of the line L2 . ev ie w C op 16 A curve has equation y = 12 x − x 2. [3] b State the maximum value of 12 x − x 2. [1] ni C op y The function g is defined as g: x ֏ 12 x − x 2, for x ù 6. −1 U R c State the domain and range of g −1. ie br ev id a Express 3x 2 + 12 x − 1 in the form a ( x + b )2 + c, where a, b and c are constants. [4] -R am c Find the set of values of k for which 3x 2 + 12 x − 1 = kx − 4 has no real solutions. s es The function f is such that f( x ) = 2 x + 1 for x ∈ ℝ . Pr The function g is such that g( x ) = 8 − ax − bx 2 for x > k, where a, b and k are constants. The function fg is such that fg( x ) = 17 − 24x − 4x for x > k. ity [3] b Find the least possible value of k for which g has an inverse. [4] ni op y ve rs w ie ev 97 2 U ie w ge A circle has centre (8, 3) and passes through the point P (13, 5) . [2] C c For the value of k found in part b, find g −1( x ). R [3] [2] 2 a Find the value of a and the value of b. 19 [3] b Write down the coordinates of the vertex of the curve y = 3x + 12 x − 1. -C C op y 18 [2] w ge d Find g ( x ). 17 [4] a Express 12 x − x 2 in the form a − ( x + b )2, where a and b are constants to be determined. ve rs ity y b Find the coordinates of the point of intersection of lines L1 and L2 . [4] br ev id a Find the equation of the circle. [3] b Find f −1( x ) and g −1( x ). [3] ni ve rs ity Pr es The function f is such that f( x ) = 3x − 7 for x ∈ ℝ . 18 for x ∈ ℝ , x ≠ 5. The function g is such that g( x ) = 5−x a Find the value of x for which fg( x ) = 5. [3] op A curve has equation y = 2 − 3x − x 2. C U 21 y c Show that the equation f −1( x ) = g −1( x ) has no real roots. id g w e a Express 2 − 3x − x 2 in the form a − ( x + b )2, where a and b are constants. [2] [1] [3] [3] -R c Find the two values of m for which the line y = mx + 3 is a tangent to the curve y = 2 − 3x − x 2. am br ev ie b Write down the coordinates of the maximum point on the curve. es s d For each value of m in part c, find the coordinates of the point where the line touches the curve. -C R ev ie w C op y 20 [5] s -C Give your answer in the form ax + by = c. -R am b Find the equation of the tangent to the circle at the point P. Copyright Material - Review Only - Not for Redistribution ve rs ity ev ie am br id w ge C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 a Find the coordinates of the centre of the circle. [2] Pr es s -C b Find the radius of the circle. [2] -R A circle, C , has equation x 2 + y2 − 16x − 36 = 0. 22 [3] y The function f is such that f( x ) = 3x − 2 for x > 0. 23 ev ie [2] d The point P lies on the circle and the line L is a tangent to C at the point P. Given that the line L has 4 gradient , find the equation of the perpendicular to the line L at the point P. 3 ve rs ity w C op y c Find the coordinates of the points where the circle meets the x-axis. U R ni C op The function g is such that g( x ) = 2 x 2 − 8 for x < k, where k is a constant. ge a Find the greatest value of k for which the composite function fg can be formed. id ie w b For the case where k = −3: [2] br ev i find the range of fg [4] -R am ii find (fg)−1( x ) and state the domain and range of (fg)−1. A curve has equation xy = 20 and a line has equation x + 2 y = k, where k is a constant. s -C 24 [3] es Find: i the coordinates of the points A and B ity [3] ii the equation of the perpendicular bisector of the line AB. [4] ve ie w rs C 98 Pr op y a In the case where k = 14, the line intersects the curve at the points A and B. [4] y op y op -R s es -C am br ev ie id g w e C U R ev ie w ni ve rs C ity Pr op y es s -C -R am br ev id ie w ge C U R ni ev b Find the values of k for which the line is a tangent to the curve. Copyright Material - Review Only - Not for Redistribution op y ve rs ity ni C U ev ie w ge -R am br id Pr es s -C y ni C op y ve rs ity op C w ev ie Chapter 4 Circular measure y op w ge C U R ni ev ve rs w C ity op Pr y es s -C -R am br ev id ie w ge U R ie 99 id -R am br ev understand the definition of a radian, and use the relationship between radians and degrees 1 use the formulae s = rθ and A = r 2θ to solve problems concerning the arc length and sector 2 area of a circle. y op -R s es -C am br ev ie id g w e C U R ev ie w ni ve rs C ity Pr op y es s -C ■ ■ ie In this section you will learn how to: Copyright Material - Review Only - Not for Redistribution ve rs ity ge C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 What you should be able to do IGCSE / O Level Mathematics Find the perimeter and area of sectors. Pr es s -C y 5 cm br 40° w -R am x cm 6 cm ev id 3 8 cm ie ge U ni y Find the value of x and the value of y. Solve problems involving the sine and cosine rules for any triangle and the formula: Find the value of x and the area of the triangle. es s -C y cm x° 12 cm 1 Area of triangle = ab sin C 2 Pr y rs C ity op At IGCSE / O Level, you will have always worked with angles that were measured in degrees. Have you ever wondered why there are 360° in one complete revolution? The original reason for choosing the degree as a unit of angular measure is unknown but there are a number of different theories. y ve w ● The ancient Babylonians divided the circle into 6 equilateral triangles and then subdivided each angle at O into 60 further parts, resulting in 360 divisions in one complete revolution. op Ancient astronomers claimed that the Sun advanced in its path by one degree each day and that a solar year consisted of 360 days. O 360 has many factors that make division of the circle so much easier. -R ● am br ev id ie w ge C U R ● ni ie 2 ve rs ity C w ev ie R IGCSE / O Level Mathematics Another measure for angles ev 1 Find the perimeter and area of a sector of a circle with radius 6 cm and sector angle 30°. Use Pythagoras’ theorem and trigonometry on right-angled triangles. op IGCSE / O Level Mathematics 100 Check your skills -R Where it comes from C op am br id ev ie w PREREQUISITE KNOWLEDGE op y es s -C Degrees are not the only way in which we can measure angles. In this chapter you will learn how to use radian measure. This is sometimes referred to as the natural unit of y op -R s es -C am br ev ie id g w e C U R ev ie w ni ve rs C ity Pr angular measure and we use it extensively in mathematics because it can simplify many formulae and calculations. Copyright Material - Review Only - Not for Redistribution ve rs ity 1 rad 1 radian is sometimes written as 1 rad, but often no symbol at all is used for angles measured in radians. -R O r B Pr es s -C r r ev ie In the diagram, the magnitude of angle AOB is 1 radian. am br id A w ge 4.1 Radians C U ni op y Chapter 4: Circular measure y KEY POINT 4.1 C op ni U br ev id ie w ge R KEY POINT 4.2 2 π radians = 360° π radians = 180° y It follows that the circumference (an arc of length 2 πr) subtends an angle of 2 π radians at the centre, therefore: ev ie w C ve rs ity op An arc equal in length to the radius of a circle subtends an angle of 1 radian at the centre. -R am When an angle is written in terms of π, we usually omit the word radian (or rad). s -C Hence, π = 180°. es Converting from degrees to radians π π , 45° = etc. 2 4 We can convert angles that are not simple fractions of 180° using the following rule. y π . 180 C w ge U R To change from degrees to radians, multiply by op ni ve KEY POINT 4.3 ie ev rs w C ity op Pr y Since 180° = π, then 90° = id ie Converting from radians to degrees ni ve rs C y op -R s -C am br ev ie id g w e C U R 180 ≈ 57°.) π es w (It is useful to remember that 1 radian = 1 × ev ie 180 . π ity To change from radians to degrees, multiply by Pr op y KEY POINT 4.4 es s -C -R am br ev π π = 30°, = 18° etc. 6 10 We can convert angles that are not simple fractions of π using the following rule. Since π = 180°, Copyright Material - Review Only - Not for Redistribution 101 ve rs ity Pr es s -C Method 2: y Method 2: w ge Method 1: ni R b π radians 6 5π 5 π 180 ° × radians = 9 π 9 br ev id ie π radians = 180° 5π radians = 100° 9 -R am π radians = 20° 9 op Pr y es s -C 5π radians = 100° 9 102 π radians 6 30° = U ev ie w 180 ° = π radians 6 6 30° = π radians 30° = 30 × 180 ve rs ity op C 180° = π radians C op Method 1: y a -R a Change 30° to radians, giving your answer in terms of π. 5π radians to degrees. b Change 9 Answer ev ie w ge am br id WORKED EXAMPLE 4.1 C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 π radians. 6 There are other angles, which you should learn, that can be written as simple multiples of π. 45° 60° 90° 180° 270° Radians 0 π 6 π 4 π 3 π 2 π 3π 2 360° ie w 2π br ev id C 30° op y ve 0° ge Degrees U ni There are: R ev ie w rs C ity In Worked example 4.1, we found that 30° = -R es s -C EXERCISE 4A am We can quickly find other angles, such as 120°, using these known angles. b 40° c 150° g 135° ity Pr a 20° i m 9° n 35° j 300° op y o 600° π 12 9π 20 7π 12 i l 4π 15 m 5π 4 n w h 7π 3 -R 3π 10 e 4π 3 j 9π 2 o 9π 8 C d s 7π 5 π 6 g id g k br 4π 9 am f c e U 2 Change these angles to degrees. π π b a 2 3 225° ie 540° e 5° ev l d 50° es k 65° 25° h 210° ni ve rs f -C R ev ie w C op y 1 Change these angles to radians, giving your answers in terms of π. Copyright Material - Review Only - Not for Redistribution ve rs ity C U ni op y Chapter 4: Circular measure 47° c d 200° w b 32° am br id a 28° ev ie ge 3 Write each of these angles in radians, correct to 3 significant figures. e 320° d 1.52 rad e 0.79 rad b 0.8 rad c 1.34 rad -C a 1.2 rad -R 4 Write each of these angles in degrees, correct to 1 decimal place. Degrees 0 Radians 0 30 90 135 180 225 270 315 360 π 2π 60 90 120 150 180 210 240 270 300 330 360 π 2π y 0 45 U cos(0.9) π tan 5 w π sin 3 f ie e id c br π d cos 2 b tan(1.5) ge a sin(0.7) ev R ni 6 Use your calculator to find: C op C ev ie w b 0 Radians Pr es s Degrees ve rs ity a op y 5 Copy and complete the tables, giving your answers in terms of π. -R am 7 Calculate the length of QR. es Pr y 103 ity op ve op Q 5 cm ie Robert is told the size of angle BAC in degrees and he is then asked to calculate the length of the line BC. He uses his calculator but forgets that his calculator is in radian mode. Luckily he still manages to obtain the correct answer. Given that angle BAC is between 10° and 15°, use graphing software to help you find the size of angle BAC, correct to 2 decimal places. -R s es -C am br ev ie id g w e C U op ie ev R y ni ve rs ity Pr es s -C op y ev am B -R br id 6 cm A w C w C ge 8 C U R ni 1 rad y rs C w ie ev P PS You do not need to change the angle to degrees. You should set the angle mode on your calculator to radians. s -C R TIP Copyright Material - Review Only - Not for Redistribution ve rs ity ev ie am br id EXPLORE 4.1 w ge C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 chord arc sector Pr es s -C Explain what is meant by: minor arc and major arc minor sector and major sector minor segment and major segment. ● A r r O C op y Given that the radius of a circle is r cm and that the angle subtended at the centre of the circle by the chord AB is θ °, discuss and write down an expression, in terms of r and θ , for finding each of the following: ● w ie ● length of chord AB area of minor sector AOB area of minor segment AOB. ev id ● -R am ● length of minor arc AB perimeter of minor sector AOB perimeter of minor segment AOB br ● ge U R ● B θ° ni ev ie w C ● segment ve rs ity op y ● -R Discuss and explain, with the aid of diagrams, the meaning of each of these words. es s -C What would the answers be if the angle θ was measured in radians instead? WEB LINK op Pr y DID YOU KNOW? ity A geographical coordinate system is used to describe the location of any point on the Earth’s surface. The coordinates used are longitude and latitude. ‘Horizontal’ circles and ‘vertical’ circles form the ‘grid’. The horizontal circles are perpendicular to the axis of rotation of the Earth and are known as lines of latitude. The vertical circles pass through the North and South poles and are known as lines of longitude. y op w ge -R am br ev From the definition of a radian, an arc that subtends an angle of 1 radian at the centre of the circle is of length r. Hence, if an arc subtends an angle of θ radians at the centre, the length of the arc is rθ . es s -C KEY POINT 4.5 ie ev -R s es am br π 3 = 5 π cm = 15 × id g Arc length = rθ w e Answer C U op π radians at the centre of a circle with radius 15 cm. 3 Find the length of the arc in terms of π. An arc subtends an angle of y ni ve rs WORKED EXAMPLE 4.2 -C R ev ie w C ity Pr op y Arc length = rθ rθ B ie id 4.2 Length of an arc Try the Where are you? resource on the Underground Mathematics website. C U R ni ev ve ie w rs C 104 Copyright Material - Review Only - Not for Redistribution r θ O r A ve rs ity ge C U ni op y Chapter 4: Circular measure am br id ev ie w WORKED EXAMPLE 4.3 A sector has an angle of 1.5 radians and an arc length of 12 cm. Pr es s -C ni WORKED EXAMPLE 4.4 y ev ie w C ve rs ity op y Arc length = rθ 12 = r × 1.5 r = 8 cm ev br -R am a the length of arc CD b the length of AD es Pr b AB = 2 × 8 cos 0.9 = 9.9457… AD = AB − DB = 9.9457… − 8 c = 17.1cm (to 3 significant figures) br b radius 7 cm and angle es s -R am -C Pr b radius 3.5 cm and angle 0.65 radians. ity a radius 10 cm and angle 1.3 radians ni ve rs 3 Find, in radians, the angle of a sector of: U op 4 The High Roller Ferris wheel in the USA has a diameter of 158.5 metres. y b radius 12 cm and arc length 9.6 cm. a radius 10 cm and arc length 5 cm ev ie id g es s -R br am π radians. 16 w e C Calculate the distance travelled by a capsule as the wheel rotates through -C C op y 2 Find the arc length of a sector of: 3π 7 7π d radius 24 cm and angle . 6 ev id ie w ge C U ni op y ve ie 1 Find, in terms of π, the arc length of a sector of: π a radius 8 cm and angle 4 3π c radius 16 cm and angle 8 w Perimeter = DC + CA + AD = 7.2 + 8 + 1.945… = 1.95 cm (to 3 significant figures) EXERCISE 4B ie B D ity C w Arc length = rθ = 8 × 0.9 = 7.2 cm ev R 8 cm 0.9 rad A rs op y Answer ev 8 cm s -C c the perimeter of the shaded region. a C ie ge id Find: w U R Triangle ABC is isosceles with AC = CB = 8 cm. CD is an arc of a circle, centre B, and angle ABC = 0.9 radians. R C op Answer -R Find the radius of the sector. Copyright Material - Review Only - Not for Redistribution 105 ve rs ity C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 w 2.1 rad -R Pr es s 6 cm ve rs ity y w ie 7 cm O es Pr ABCD is a rectangle with AB = 5 cm and BC = 24 cm. rs y ve O is the midpoint of BC . E C U ge R a the length of AO w B ev -R am es s -C A θ 10 cm B O ity Pr op y C e U op x− y . Show that P = 4 xy + π( x + y ) + 2( x − y ) sin −1 x + y x y ni ve rs 10 The diagram shows the cross-section of two cylindrical metal rods of radii x cm and y cm. A thin band, of length P cm, holds the two rods tightly together. -R s es am br ev ie id g w [This question is based upon Belt on the Underground Mathematics website.] -C C w ie C O C 9 The diagram shows a semicircle with radius 10 cm and centre O. Angle BOC = θ radians. The perimeter of sector AOC is twice the perimeter of sector BOC. π−2 . a Show that θ = 3 b Find the perimeter of triangle ABC . ev D ie id the perimeter of the shaded region. br c A op ni ev OAED is a sector of a circle, centre O. Find: b angle AOD, in radians R 7 cm ity op y the perimeter of the shaded segment. B 2 rad s b the length of chord AB C A -R -C a the length of arc AB PS O ev id br am AB is a chord and angle AOB = 2 radians. Find: 8 6 cm C op the perimeter of the shaded area. 7 The circle has radius 7 cm and centre O. c P R U c 8 cm Q ge R b the length of QR w 8 cm 5 cm ni a angle POQ, in radians ie 4.3 rad 6 The circle has radius 6 cm and centre O. PQ is a tangent to the circle at the point P. QRO is a straight line. Find: ev ie w C op y -C 1.2 rad 106 c ev ie b am br id a ge 5 Find the perimeter of each of these sectors. Copyright Material - Review Only - Not for Redistribution y ve rs ity w ge ev ie am br id B r θ O A r op y Pr es s -C -R 4.3 Area of a sector C U ni op y Chapter 4: Circular measure ve rs ity y area of sector angle in the sector = area of circle complete angle at the centre ni U w ge θ × πr 2 2π -R am br ev id ie R area of sector θ = 2π πr 2 area of sector = C op When θ is measured in radians, the ratio becomes: s 1 2 rθ 2 ity 107 WORKED EXAMPLE 4.5 ni ev Find the area of a sector of a circle with radius 9 cm and angle w ge Answer C U R Give your answer in terms of π. -R s es Pr ni ve rs ity WORKED EXAMPLE 4.6 op y The circle has radius 6 cm and centre O. AB is a chord and angle AOB = 1.2 radians. Find: id g w e C U a the area of sector AOB b the area of triangle AOB -R s es am br ev ie c the area of the shaded segment. -C ev ie w C op y -C am br ev id ie 1 2 rθ 2 1 π = × 92 × 2 6 27 π cm 2 = 4 Area of sector = R π radians. 6 y ve ie w rs C op Pr y Area of sector = es -C KEY POINT 4.6 op ev ie w C To find the formula for the area of a sector, we use the ratio: Copyright Material - Review Only - Not for Redistribution A B 1.2 rad 6 cm 6 cm O ve rs ity ev ie w ge am br id Answer 1 2 rθ 2 1 = × 62 × 1.2 2 = 21.6 cm 2 Pr es s -C 1 ab sin C 2 1 = × 6 × 6 × sin1.2 2 = 16.7767… ve rs ity y Area of triangle AOB = w C op b -R Area of sector AOB = a ev ie C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 C op ni Area of shaded segment = area of sector AOB − area of triangle AOB = 21.6 − 16.7767… w ie = 4.82 cm 2 (to 3 significant figures) -R am br ev id ge U c R y = 16.8 cm 2 (to 3 significant figures) es s -C WORKED EXAMPLE 4.7 op Pr y The diagram shows a circle inscribed inside a square of side length 10 cm. 108 C ity A quarter circle, of radius 10 cm, is drawn with the vertex of the square as centre. y op ie br C -R β ( 10 + 10 ) = 5 2 cm 2 5 es s 2 10 C U Shaded area = area of segment PQR − area of segment PQS ie id g w e 1 1 1 1 = × 52 × β − × 52 × sin β − × 102 × 2θ − × 102 × sin 2θ 2 2 2 2 s es am -R br ev = 21.968 − 7.3296 = 14.6 cm 2 (to 3 significant figures) 5√2 θ θ O y sin θ sin α = 5 10 θ = 0.4867 rad α α Q ni ve rs Sine rule: -C R ev ie w Hence, β = 2 π − 2α = 2.4189 rad S Pr op y -C 1 1 Pythagoras: (diagonal of square) = 2 2 52 + (5 2 )2 − 102 Cosine rule: cos α = 2×5×5 2 α = 1.932 rad ity am Radius of inscribed circle = 5 cm P R op OQ = 10 cm ev id Answer w ge C U R ni ev ve ie w rs Find the shaded area. Copyright Material - Review Only - Not for Redistribution ve rs ity ev ie 2π radians 5 4π d radius 9 cm and angle radians. 3 b radius 10 cm and angle b radius 2.6 cm and angle 0.9 radians. ve rs ity a radius 34 cm and angle 1.5 radian C op y 2 Find the area of a sector of: Pr es s -C 1 Find, in terms of π, the area of a sector of: π a radius 12 cm and angle radians 6 2π c radius 4.5 cm and angle radians 9 -R am br id EXERCISE 4C w ge C U ni op y Chapter 4: Circular measure ev ie w 3 Find, in radians, the angle of a sector of: C op ni AOB is a sector of a circle, centre O, with radius 8 cm. U R 4 b radius 6 cm and area 27 cm 2 . y a radius 4 cm and area 9 cm 2 w ge The length of arc AB is 10 cm. Find: b the area of the sector AOB. br ev id ie a angle AOB, in radians Q s es O b Find the length of PX . 109 Find the area of the shaded region. ni op ve ie ev R 8 cm C U O ie id P π 3 w ge Find the exact area of the shaded region. Q y w rs P 6 The diagram shows a sector, POR, of a circle, centre O, with radius 8 cm π and sector angle radians. The lines OR and QR are perpendicular 3 and OPQ is a straight line. R X Pr a Find angle POQ, in radians. c 4 cm ity C op y -C -R am 5 The diagram shows a sector, POQ, of a circle, centre O, with radius 4 cm. The length of arc PQ is 7 cm. The lines PX and QX are tangents to the circle at P and Q, respectively. es s -C P Pr π 3 b Find the exact area of the shaded region. O A ity 5 cm ni ve rs 8 The diagram shows three touching circles with radii 6 cm, 4 cm and 2 cm. 2 op -R s es -C am br ev ie id g w e C U R y Find the area of the shaded region. ev ie w C op y a Find the exact length of AP. PS B -R am br ev 7 The diagram shows a sector, AOB, of a circle, centre O, with radius π 5 cm and sector angle radians. The lines AP and BP are tangents 3 to the circle at A and B, respectively. Copyright Material - Review Only - Not for Redistribution 2 6 4 6 4 ve rs ity C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 am br id w FH is the arc of a circle, centre E. Find the area of: F ev ie ge 9 The diagram shows a semicircle, centre O, with radius 8 cm. b sector FOG c d the shaded region. 2 rad -R a triangle EOF E Pr es s -C sector FEH a The perimeter of the shaded region is P cm. r Show that P = (3 + 3 3 + π ). 3 b The area of the shaded region is A cm 2. r2 (3 3 − π ). Show that A = 6 E ve rs ity F C op y G U R H G O r cm ni ev ie w C op y 10 The diagram shows a sector, EOG, of a circle, centre O, with radius r cm. The line GF is a tangent to the circle at G, and E is the midpoint of OF . O w ge 11 The diagram shows two circles with radius r cm. ie es ity A quarter circle, of radius 10 cm, is drawn from each vertex of the square. Find the exact area of the shaded region. y op 13 The diagram shows a circle with radius 1cm, centre O. w ge PS C U R ni ev ve ie w rs C Pr y 12 The diagram shows a square of side length 10 cm. op PS 110 s -C -R am br Find, in terms of r, the exact area of the shaded region. ev id The centre of each circle lies on the circumference of the other circle. O br ev id ie Triangle AOB is right angled and its hypotenuse AB is a tangent to the circle at P. -R am Angle BAO = x radians. -C a Find an expression for the length of AB in terms of tan x. x es P B B Pr ity op area of inner circle 2 = . area of sector 3 -R s es am br ev ie id g w e C U b Show that y ni ve rs a Show that R = 3r. -C C w ie ev R A 14 The diagram shows a sector, AOB, of a circle, centre O, with radius R cm and π sector angle radians. 3 An inner circle of radius r cm touches the three sides of the sector. op y P s b Find the value of x for which the two shaded areas are equal. Copyright Material - Review Only - Not for Redistribution π 3 O A ve rs ity am br id r Pr es s -C r -R Radians and degrees 1 rad r w C ve rs ity op y O One radian is the size of the angle subtended at the centre of a circle, radius r, by an arc of length r. ● π radians = 180° ● To change from degrees to radians, multiply by ● To change from radians to degrees, multiply by w π . 180 ie 180 . π ev br id ge U ni C op y ● R ev ie w ev ie ge Checklist of learning and understanding C U ni op y Chapter 4: Circular measure -R am Arc length and area of a sector θ r A 111 y op ni ev ve ie w rs C ity O Pr op y r es s -C B When θ is measured in radians, the length of arc AB is rθ . 1 2 ● When θ is measured in radians, the area of sector AOB is r θ. 2 y op -R s es -C am br ev ie id g w e C U R ev ie w ni ve rs C ity Pr op y es s -C -R am br ev id ie w ge C U R ● Copyright Material - Review Only - Not for Redistribution ve rs ity ge C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 R -R Y M C ve rs ity op y Pr es s -C 1 ev ie am br id w END-OF-CHAPTER REVIEW EXERCISE 4 w P X y C op ni R the total perimeter of the shaded region U ev ie The diagram shows an equilateral triangle, PQR, with side length 5 cm. M is the midpoint of the line QR. An arc of a circle, centre P, touches QR at M and meets PQ at X and PR at Y . Find in terms of π and 3: a br 2 α rad O A 8 cm es s -C Pr y C ity op C rs op y ve Find α in terms of π. ni w ie ev i ii Find the perimeter of the complete figure in terms of π. [3] [2] C -R am br ev id ie w Cambridge International AS & A Level Mathematics 9709 Paper 11 Q3 June 2013 α rad Pr op y es s -C D 2 cm B E 4 cm ity A ni ve rs C op the area of the shaded region, C U i y The diagram shows triangle ABC in which AB is perpendicular to BC . The length of AB is 4 cm and angle CAB is α radians. The arc DE with centre A and radius 2 cm meets AC at D and AB at E . Find, in terms of α , w ie ev -R s es am [3] [3] Cambridge International AS & A Level Mathematics 9709 Paper 11 Q6 June 2014 br id g e ii the perimeter of the shaded region. -C w ie C ge U R ev [3] In the diagram, OAB is a sector of a circle with centre O and radius 8 cm . Angle BOA is α radians. OAC is a semicircle with diameter OA. The area of the semicircle OAC is twice the area of the sector OAB. 3 R [5] B -R am 8 cm ev id ie w ge b the total area of the shaded region. 112 Q Copyright Material - Review Only - Not for Redistribution ve rs ity Pr es s y ve rs ity op C B θ rad -R C -C 4 ev ie am br id w ge C U ni op y Chapter 4: Circular measure r θ rad r A y ev ie w O ge U R ni C op The diagram represents a metal plate OABC , consisting of a sector OAB of a circle with centre O and radius r, together with a triangle OCB which is right-angled at C . Angle AOB = θ radians and OC is perpendicular to OA. B 113 r D y C A ge C U R O op θ rad ni ev ve ie w rs C ity op Pr y es s -C 5 -R am br ev id ie w [3] Find an expression in terms of r and θ for the perimeter of the plate. 1 [3] ii For the case where r = 10 and θ = π, find the area of the plate. 5 Cambridge International AS & A Level Mathematics 9709 Paper 11 Q5 November 2011 i br Find AC in terms of r and θ . am i ev id ie w The diagram shows a sector OAB of a circle with centre O and radius r. Angle AOB is θ radians. The point C on OA is such that BC is perpendicular to OA. The point D is on BC and the circular arc AD has centre C . [1] s es Pr A piece of wire of length 24 cm is bent to form the perimeter of a sector of a circle of radius r cm. Show that the area of the sector, A cm 2 , is given by A = 12 r − r 2. ni ve rs i [3] ii Express A in the form a − ( r − b )2, where a and b are constants. y [2] C w ev ie id g es s -R br am -C [2] Cambridge International AS & A Level Mathematics 9709 Paper 11 Q5 June 2015 e U op iii Given that r can vary, state the greatest value of A and find the corresponding angle of the sector. R ev ie w C 6 [6] Cambridge International AS & A Level Mathematics 9709 Paper 11 Q6 November 2012 ity op y -C -R 1 ii Find the perimeter of the shaded region ABD when θ = π and r = 4, giving your answer as an 3 exact value. Copyright Material - Review Only - Not for Redistribution ve rs ity am br id ev ie w ge C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 -R C y C E D ve rs ity op y B w ev ie r A Pr es s -C 7 U Show that the radius of the larger circle is r 2 . w i ge R ni C op The diagram shows a circle with centre A and radius r. Diameters CAD and BAE are perpendicular to each other. A larger circle has centre B and passes through C and D. es s B D r ity op Pr y C θ A y op C U R O r ni ev ve ie w rs C 114 ie -R am br Cambridge International AS & A Level Mathematics 9709 Paper 11 Q7 November 2015 -C 8 [6] ev id ii Find the area of the shaded region in terms of r. [1] ie id Express the perimeter of the shaded region in terms of r, θ and π. [4] ev br i w ge In the diagram, AOB is a quarter circle with centre O and radius r. The point C lies on the arc AB and the point D lies on OB. The line CD is parallel to AO and angle AOC = θ radians. es s -C Pr A C D id g C w e U 4 cm B op α rad y ity ni ve rs C w ie ev R O -R s es am br ev ie In the diagram, AB is an arc of a circle with centre O and radius 4 cm. Angle AOB is α radians. The point D on OB is such that AD is perpendicular to OB. The arc DC, with centre O, meets OA at C . -C [3] Cambridge International AS & A Level Mathematics 9709 Paper 11 Q7 June 2016 op y 9 -R am ii For the case where r = 5 cm and θ = 0.6, find the area of the shaded region. Copyright Material - Review Only - Not for Redistribution ve rs ity ev ie am br id i w ge C U ni op y Chapter 4: Circular measure Find an expression in terms of α for the perimeter of the shaded region ABDC . [4] -R 1 π, find the area of the shaded region ABDC , giving your answer in the form 6 kπ , where k is a constant to be determined. [4] Pr es s -C ii For the case where α = A B r r y O α rad E U R ni ev ie w C ve rs ity 10 C op op y Cambridge International AS & A Level Mathematics 9709 Paper 11 Q8 November 2014 D id ie w ge C ity It is now given that the shaded and unshaded pieces are equal in area. iii Find α in terms of π. w [3] Pr ii the area of the metal plate. [3] es the perimeter of the metal plate, [2] rs C op y i s -C -R am br ev The diagram shows a metal plate made by fixing together two pieces, OABCD (shaded) and OAED (unshaded). The piece OABCD is a minor sector of a circle with centre O and radius 2r. The piece OAED is a major sector of a circle with centre O and radius r. Angle AOD is α radians. Simplifying your answers where possible, find, in terms of α , π and r, y op y op -R s es -C am br ev ie id g w e C U R ev ie w ni ve rs C ity Pr op y es s -C -R am br ev id ie w ge C U R ni ev ve ie Cambridge International AS & A Level Mathematics 9709 Paper 11 Q6 November 2013 Copyright Material - Review Only - Not for Redistribution 115 op y ve rs ity ni C U ev ie w ge -R am br id Pr es s -C y ni C op y ve rs ity op C w ev ie op y ve ni w ge C U R ev ie w Chapter 5 Trigonometry rs C ity op Pr y es s -C -R am br ev id ie w ge U R 116 id es s -C -R am br ev sketch and use graphs of the sine, cosine and tangent functions (for angles of any size, and using either degrees or radians) use the exact values of the sine, cosine and tangent of 30°, 45°, 60°, and related angles use the notations sin −1 x, cos −1 x, tan −1 x to denote the principal values of the inverse trigonometric relations sin θ = tan θ and sin2 θ + cos2 θ = 1 use the identities cos θ find all the solutions of simple trigonometrical equations lying in a specified interval. y op -R s es -C am br ev ie id g w e C U R ev ie w ni ve rs C ity Pr op y ■ ■ ■ ■ ■ ie In this section you will learn how to: Copyright Material - Review Only - Not for Redistribution ve rs ity ge C U ni op y Chapter 5: Trigonometry am br id ev ie w PREREQUISITE KNOWLEDGE What you should be able to do IGCSE / O Level Mathematics Use Pythagoras’ theorem and trigonometry on right-angled triangles. Check your skills 1 y Find each of the following in terms of r. y b sin θ ni C op c cos θ U ge 2 a Convert to radians. es ity rs ii 720° op y ve ni 45° b Convert to degrees. π i 6 7π ii 2 13π iii 12 Pr y op C w ie ev i iii 150° s -C -R am br ev id ie w Convert between degrees and radians. Solve quadratic equations. b Solve 2 x 2 + 7 x − 15 = 0. ev id ie w ge C U 3 a Solve x 2 − 5x = 0. br -R am FAST FORWARD Pr op y es s -C You should already know how to calculate lengths and angles using the sine, cosine and tangent ratios. In this chapter you shall learn about some of the special rules connecting these trigonometric functions together with the special properties of their graphs. The graphs of y = sin x and y = cos x are sometimes referred to as waves. ity Oscillations and waves occur in many situations in real life. A few examples of these are musical sound waves, light waves, water waves, electricity, vibrations of an aircraft wing and microwaves. Scientists and engineers represent these oscillations/waves using trigonometric functions. w e C U op y ni ve rs C -R s es am br ev ie id g Try the Trigonometry: Triangles to functions resource on the Underground Mathematics website. -C w ie ev R B a BC Why do we study trigonometry? WEB LINK r cm d tan θ IGCSE / O Level Mathematics R θ° A ve rs ity op C w ev ie R Chapter 4 C 1 cm Pr es s -C -R Where it comes from Copyright Material - Review Only - Not for Redistribution In the Pure Mathematics 2 & 3 Coursebook, Chapter 3 you shall learn about the secant, cosecant and cotangent functions, which are closely connected to the sine, cosine and tangent functions. You shall also learn many more rules involving these six trigonometric functions. 117 ve rs ity C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 am br id You should already know the following trigonometric ratios. cos θ = y ni y sin θ iii tan θ br ev id ie w ge 1 − tan2 θ 3 5 −6 . b Show that = cos θ + sin θ 5 Answer -R Pr ity rs w ve ie 3 C w ev s es y op C w ity Pr ) =5−2 5 +2 5 −4 =1 es s 3 5 −6 = 5 ( 5 + 2 )( 5 − 2 ) ie 3 Multiply the numerator and denominator by 5 − 2. ev e ( 5 − 2) 5 ( 5 + 2 )( 5 − 2 ) ( id g am br = ) Multiply the numerator and denominator by 15. -R ( 3 5 +2 U 5 Simplify. ni ve rs -C = R -R 2 2 1− 5 1 − tan2 θ = cos θ + sin θ 5 2 + 3 3 1 5 = 5 + 2 3 -C √5 2 . 5 br am w C op y b θ° ie id 2 3 x y ni 32 − ( 5 )2 = 2 U R x= ge ev Using Pythagoras’ theorem: iii From the triangle, tan θ = ie cos2 θ means (cos θ )2 5 5 × 3 3 5 = 9 ii A right-angled triangle with angle θ is shown in this diagram. ∴ sin θ = ev TIP s 2 = y op C 118 5 = 3 es -C am a i cos2 θ = cos θ × cos θ op R ii U cos2 θ i ve rs ity 5 , where 0° < θ < 90°. 3 a Find the exact values of: cos θ = ev ie w C op WORKED EXAMPLE 5.1 C op x opposite adjacent y tan θ = x tan θ = -R -C θ° adjacent hypotenuse x cos θ = r opposite hypotenuse y sin θ = r sin θ = y Pr es s r ev ie w ge 5.1 Angles between 0° and 90° Copyright Material - Review Only - Not for Redistribution ve rs ity C U ni op y Chapter 5: Trigonometry ev ie -R am br id y Pr es s -C Consider a right-angled isosceles triangle whose two equal sides are of length 1 unit. √2 45° 2 1 ve rs ity ev ie w C op We find the third side using Pythagoras’ theorem: 12 + 12 = Triangle 2 30° U R ni C op y Consider an equilateral triangle whose sides are of length 2 units. br ev id ie 2 60° The value 1 can be 2 1 2 1 2 op ity 1 3 written as 2 . 2 3 2 1 2 3 The value w C 1 ev -C -R am WORKED EXAMPLE 5.2 ev es Rationalise the denominator. -R s es -C am br ev ie id g w e C U R y 1 1 × 2 2 1 = 2 2 1× 2 = 2 2 × 2 2 = 4 sin 30° cos 45° = ni ve rs a op Answer c 3 Pr sin π π sin 4 6 2 π 2 π + sin cos 3 3 2 cos 2 π ity op y C b sin 30° cos 45° ie w s Find the exact value of: a 1 can be 3 3 . written as 3 ie ve rs 3 2 1 2 U π 3 TIP ge θ = 60° = tan θ id π 4 cos θ ni sin θ 119 br R ev ie w C op These two triangles give the important results: θ = 45° = -R s y es 3 π 6 1 Pr -C am We can find the height of the triangle using Pythagoras’ theorem: θ = 30° = √3 w ge The perpendicular bisector to the base splits the equilateral triangle into two congruent right-angled triangles. 22 − 12 = 1 y Triangle 1 w ge We can obtain exact values of the sine, cosine and tangent of 30°, 45° and 60° or π , π and π from the following two triangles. 6 4 3 Copyright Material - Review Only - Not for Redistribution ve rs ity ev ie -R Pr es s The denominator simplifies to 1 3 + = 1. 4 4 y Rationalise the denominator. C op ve rs ity ev ie 2 = 2 ni 1 2 = U s e c sin2 θ + cos2 θ f 5 1 + cos θ c 1 − sin2 θ f 5− rs op e 2 sin θ + 1 1 and that θ is acute, find the exact value of: 4 b tan θ ie d sin θ cos θ tan θ am 1 1 + tan θ sin θ -R cos θ br a ev id 3 Given that sin θ = -C e es sin 45° + cos 30° e sin2 45° 2 + tan 60° f sin2 30° + cos2 30° 2 sin 45° cos 45° c 1 − 2 sin2 ni ve rs π 4 y cos π 3 cos f -R s es am -C C U op − w tan 1 br e 1 ev d π π − tan 6 3 π sin 4 e sin id g C w ie c 5 Find the exact value of each of the following. π π π a sin cos b cos2 4 4 3 R ev sin2 45° Pr sin 60° sin 30° b ity d op y sin 30° cos 60° tan θ sin θ s 4 Find the exact value of each of the following. a y ve ni cos θ sin θ 3 − sin θ 3 + cos θ ge d U w ie ev R sin θ f 2 and that θ is acute, find the exact value of: 5 b cos θ 2 Given that tan θ = a 1 − sin2 θ cos θ C 5 tan θ 2 sin θ cos θ w d c Pr sin θ ie C 120 4 and that θ is acute, find the exact value of: 5 b tan θ ity a op y 1 Given that cos θ = es -C -R am br ev id ie w ge R 2 2 2 = 2 EXERCISE 5A 2 π π means sin . sin 3 3 am br id -C y w C op c w ge π π π = sin × sin 3 3 3 3 3 = × 2 2 3 = 4 1 1 π π 2 × × 2 cos sin 2 2 4 6 = 2 2 2 π 2 π + sin cos 1 + 3 3 3 2 2 sin2 b C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 Copyright Material - Review Only - Not for Redistribution π 6 π π + tan 3 6 π sin 3 ve rs ity C w π 1 1 and the missing function is from the list sin θ , tan θ , and . cos θ 2 tan θ ev ie 6 In the table, 0 ø θ ø am br id PS ge U ni op y Chapter 5: Trigonometry …… 1 cos θ 1 2 -R ve rs ity θ =… 3 1 3 1 2 …… …… 2 y …… ge U R ni C op op C w 1 sin θ ev ie θ =… Pr es s θ =… y -C Without using a calculator, copy and complete the table. w 5.2 The general definition of an angle id ie second quadrant br P -R am To do this we need a general definition for an angle: θ es op Pr y third quadrant rs C w y ve ni op ie WORKED EXAMPLE 5.3 ev fourth quadrant ity The Cartesian plane is divided into four quadrants, and the angle θ is said to be in the quadrant where OP lies. In the previous diagram, θ is in the first quadrant. C -R br am Answer -C a 120° is an anticlockwise rotation. b y x P 430° 70° op y ni ve rs w ie Acute angle made with x-axis = 60° ev ie id g w e C U Acute angle made with x-axis = 70° es s -R br am x O ity O y Pr 60° -C − 430° is an anticlockwise rotation. es 120° op y d s P C 3π 4 w c 430° ev id b ie ge U R Draw a diagram showing the quadrant in which the rotating line OP lies for each of the following angles. In each case, find the acute angle that the line OP makes with the x-axis. a 120° ev x O s -C An angle is a measure of the rotation of a line segment OP about a fixed point O. The angle is measured from the positive x-direction. An anticlockwise rotation is taken as positive and a clockwise rotation is taken as negative. R first quadrant ev We need to be able to use the three basic trigonometric functions for any angle. y Copyright Material - Review Only - Not for Redistribution 2π 3 121 ve rs ity am br id d − w 3π is an anticlockwise rotation. 4 c 2 π is a clockwise rotation. 3 ev ie ge C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 y 3π 4 x ni 2π 3 y P π 4 Acute angle made with x-axis = π 3 w ge U x O C op C w ev ie π 3 – ve rs ity op O Acute angle made with x-axis = R Pr es s π 4 y -C -R y P -R s -C EXERCISE 5B am br ev id ie The acute angle made with the x-axis is sometimes called the basic angle or the reference angle. Pr y w y op ni U d ie ev id br x θ = –500° es s -C x O -R am O y w ge y x O ve ie ev R θ = –320° x O θ = 200° y rs θ = 110° c b C a C 122 ity op y es 1 For each of the following diagrams, find the basic angle of θ . Pr am -C f 2π 3 h − C op y −150° w 5π 3 17 π − 8 ie j ev 13π 9 d -R i −100° s 7π 6 es g ity 400° ni ve rs e b U 310° e c id g 100° br a R ev ie w C op y 2 Draw a diagram showing the quadrant in which the rotating line OP lies for each of the following angles. On each diagram, indicate clearly the direction of rotation and state the acute angle that the line OP makes with the x-axis. Copyright Material - Review Only - Not for Redistribution ve rs ity C U ni op y Chapter 5: Trigonometry am br id ev ie w ge 3 In each part of this question you are given the basic angle, b, the quadrant in which θ lies and the range in which θ lies. Find the value of θ . a b = 55°, second quadrant, 0° , θ , 360° b = 32°, fourth quadrant, 360° , θ , 720° Pr es s π , third quadrant, 0 , θ , 2 π 4 π e b = , second quadrant, 2 π , θ , 4 π 3 π f b = , fourth quadrant, −4 π , θ , − 2 π 6 d b= y C op U R ni ev ie w C ve rs ity op y -C c -R b b = 20°, third quadrant, −180° , θ , 0° 5.3 Trigonometric ratios of general angles ie y P(x, y) ev id r -R y x y , cos θ = , tan θ = , when x ≠ 0 r r x y θ O x x y es s -C sin θ = am br KEY POINT 5.1 w ge In general, trigonometric ratios of any angle θ in any quadrant are defined as: x 2 + y2 . op Pr Where x and y are the coordinates of the point P and r is the length of OP, where r = 123 rs cos θ = op y ve ni x r tan θ = y x w y r U sin θ = ge R ev ie EXPLORE 5.1 C w C ity You need to know the signs of the three trigonometric ratios in each of the four quadrants. -R cos θ y + = =+ r + x + = =+ r + y + = =+ x + x − = =− r + y = x x = r y − = =+ x − x = r y = x s C U y w e On a copy of the diagram, record which ratios are positive in each quadrant. ie id g The first quadrant has been completed for you. -R s es am br ev (All three ratios are positive in the first quadrant.) sin cos tan y 4th quadrant y − = =− r + op y = r es 3rd quadrant tan θ Pr y = r ni ve rs 2nd quadrant -C R ev ie w C op y 1st quadrant sin θ ity -C am br ev id ie By considering whether x and y are positive or negative ( + or − ) in each of the four quadrants, copy and complete the table. (r is positive in all four quadrants.) Copyright Material - Review Only - Not for Redistribution O x ve rs ity C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 90° am br id ev ie w ge The diagram shows which trigonometric functions are positive in each quadrant. Sin All -R 180° y C ve rs ity op WORKED EXAMPLE 5.4 0°, 360° O Tan Pr es s -C We can memorise this diagram using a mnemonic such as ‘All Students Trust Cambridge’. Cos 270° y cos( −130°) U Answer y S br ev id ie In the second quadrant, sin is positive. es Pr A ity ve y −130° ni ev T ie w ge id ev -R br am es s -C 3 and that 180° ø θ ø 270°, find the value of sin θ and the value of tan θ . 5 Pr op y Answer θ is in the third quadrant. sin is negative and tan is positive in this quadrant. y ity S y2 = 25 − 9 = 16 −3 x C w e ev ie id g es s -R br am 5 T −4 4 4 −4 = . ∴ sin θ = = − and tan θ = −3 3 5 5 -C O y U Since y < 0, y = −4. A θ ni ve rs y2 + ( −3)2 = 52 C C w ie C C U R Given that cos θ = − x 50° O op C w C S WORKED EXAMPLE 5.5 ev x y In the third quadrant only tan is positive, so cos is negative. cos( −130°) = − cos 50° ie 40° O T rs op y b The acute angle made with the x-axis is 50°. R A 140° s -C -R am sin 140° = sin 40° 124 w ge a The acute angle made with the x-axis is 40°. y R b C op sin 140° op a ni ev ie w Express in terms of trigonometric ratios of acute angles: Copyright Material - Review Only - Not for Redistribution ve rs ity ge C U ni op y Chapter 5: Trigonometry ev ie am br id w WORKED EXAMPLE 5.6 b sin 120° -C a -R Without using a calculator, find the exact values of: Pr es s Answer ∴ sin 120° is positive. y S ve rs ity C op y a 120° lies in the second quadrant. 7π 6 cos A T w ge U R 7π lies in the third quadrant. 6 7π ∴ cos is negative. 6 7π π −π= Basic acute angle = 6 6 3 7π π ∴ cos = − cos = − 6 6 2 ie -R s es C Pr T ity 125 rs y ve ev b c cos 50° op w -R am 230° lies in the third quadrant. S A 230° es Basic acute angle = 230° − 180° = 50° x O 50° Pr ∴ sin 230° = − sin 50° = −b y s -C ∴ sin 230° is negative. C ity T y op -R s es -C am br ev ie id g w e C U R ev ie w ni ve rs C op y d ie br ev id Answer tan 40° C ni U sin 230° ge R x O π 6 Given that sin 50° = b, express each of the following in terms of b. a A 7π 6 ev id am -C y op C w ie WORKED EXAMPLE 5.7 a C y S br b x O y 3 2 ni ev ie ∴ sin 120° = sin 60° = 120° 60° C op w Basic acute angle = 180° − 120° = 60° Copyright Material - Review Only - Not for Redistribution tan 140° ve rs ity am br id 1 − b2 = 1 − b2 1 -R 1 50° b 50° √1 – b2 C op U R ni ev ie 40° 1 d 140° lies in the second quadrant. -R 1 − b2 b es ity tan 125° ni 4π 5 f 9π 8 g 7π cos − 10 sin d cos( −245°) h tan sin 225° d tan( −300°) π tan − 6 h cos d cos 245° y c op cos cos 305° C e b ve sin 190° R a rs 1 Express the following as trigonometric ratios of acute angles. U C w C Pr y op EXERCISE 5C ie x O T s -C ∴ tan 140° = − tan 40° = − 40° ev id br am Basic acute angle = 180° − 140° = 40° ev A 140° ie S w ge y ∴ tan 140° is negative. 126 y 1 − b2 b ve rs ity ∴ tan 40° = w C op y c Show 40° on the triangle: b √ 1 – b2 Pr es s -C ∴ cos 50° = w b Draw the right-angled triangle showing the angle 50°: ev ie ge C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 11π 9 am br 4π 3 w c f cos 7π 3 ie sin tan 330° ev e b g -R cos 120° id a ge 2 Without using a calculator, find the exact values of each of the following. 10 π 3 s y ni ve rs U sin θ 5 and that 180° ø θ ø 360°, find the value of: 12 b cos θ e a w ie c ev sin 25° cos 65° s -R b es br tan 205° am a id g 7 Given that tan 25° = a, express each of the following in terms of a. -C C w ie ev R 6 Given that tan θ = − op sin θ 1 and that 180° ø θ ø 270°, find the value of: 3 b tan θ 5 Given that cos θ = − C a Pr cos θ a 2 and that θ is obtuse, find the value of: 5 b tan θ ity op y 4 Given that sin θ = es -C 3 Given that sin θ < 0 and tan θ < 0, name the quadrant in which angle θ lies. Copyright Material - Review Only - Not for Redistribution ve rs ity C U ni op y Chapter 5: Trigonometry cos A -C a y w ev ie ve rs ity sin A −1 sin θ …… 1 2 1 cos θ −2 − 2 ie ev 1 3 − 1 2 s …… es id br am op Pr y -C θ = 210° w …… ge θ = …… -R U R ni C op Without using a calculator, copy and complete the table. θ = 120° 1 1 and . sin θ tan θ y ev ie w PS 11 In the table, 0° ø θ ø 360° and the missing function is from the list cos θ , tan θ , …… 127 rs y ve ni ev ie EXPLORE 5.2 op w C ity 5.4 Graphs of trigonometric functions -C C w -R am br ev id ie ge U R Consider taking a ride on a Ferris wheel with radius 50 metres that rotates at a constant speed. You enter the ride from a platform that is level with the centre of the wheel and the wheel turns in an anticlockwise direction through one complete turn. es s 1 Sketch the following two graphs and discuss their properties. Pr ni ve rs ity b The graph of your horizontal displacement from the centre of the wheel plotted against the angle turned through. -R s es am br ev ie id g w e C U op y 2 Discuss with your classmates what the two graphs would be like if you turned through two complete turns. -C ev ie w C op y a The graph of your vertical displacement from the centre of the wheel plotted against the angle turned through. R cos 347° 2 3 and cos B = , where A and B are in the same quadrant, find the value of: 3 4 b cos A c sin B d tan B C op d sin 257° 5 4 and cos B = − , where A and B are in the same quadrant, find the value of: 13 5 b tan A c sin B d tan B 10 Given that tan A = − a c tan13° -R 9 Given that sin A = b Pr es s sin 77° am br id a ge 8 Given that cos 77° = b, express each of the following in terms of b. Copyright Material - Review Only - Not for Redistribution ve rs ity The graphs of y = sin x and y = cos x w ge y am br id ev ie Suppose that OP makes an angle of x with the positive horizontal axis and that P moves around the unit circle, through one complete revolution. P(cos x, sin x) 1 -R x The coordinates of P will be (cos x, sin x ). x O op y Pr es s -C C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 C ve rs ity The height of P above the horizontal axis changes from 0 → 1 → 0 → −1 → 0. 360 x 270 -R –1 -C am br ev y = sin x w 180 ie 90 id O ge U R ni 1 y ev ie y C op w The graph of sin x against x for 0° < x < 360° is therefore: es s The displacement of P from the vertical axis changes from 1 → 0 → −1 → 0 → 1. Pr op y The graph of cos x against x for 0° ø x ø 360° is therefore: y = cos x ity y C 128 O y 360 x 270 180 ie w ge br ev id –1 C U R 90 op ni ev ve ie w rs 1 s es Pr O 90 –1 270 360 x –360 –270 –90 O C w e ev ie id g es s -R br am -C –180 –1 U R ev ie 180 y –90 y = cos x op –180 ity –270 y 1 ni ve rs op y -C y 1 y = sin x w C –360 -R am The graphs of y = sin x and y = cos x can be continued beyond 0° ø x ø 360° : Copyright Material - Review Only - Not for Redistribution 90 180 270 360 x ve rs ity C U ni op y Chapter 5: Trigonometry am br id ev ie w ge The sine and cosine functions are called periodic functions because they repeat themselves over and over again. -R The period of a periodic function is defined as the length of one repetition or cycle. -C The sine and cosine functions repeat every 360° . Pr es s We say they have a period of 360° (or 2π radians). C ve rs ity op y The amplitude of a periodic function is defined as the distance between a maximum (or minimum) point and the principal axis. C op ni w -R am br ● ie ● ev ● sin( − x ) = − sin x sin(180° − x ) = sin x sin(180° + x ) = − sin x sin(360° − x ) = − sin x sin(360° + x ) = sin x U R ● ge ● y The symmetry of the curve y = sin x shows these important relationships: id ev ie w The functions y = sin x and y = cos x both have amplitude 1. s -C EXPLORE 5.3 2 cos(180° − x ) = 4 cos(360° − x ) = 5 cos(360° + x ) = 3 rs y op ni C U -R s –90 90 O es –180 180 270 360 450 540 x w ie U op The tangent function behaves very differently to the sine and cosine functions. y y = tan x ni ve rs C ity Pr op y -C am br ev id ie w ge y w e C The tangent function repeats its cycle every 180° so its period is 180° (or π radians). s es am The tangent function does not have an amplitude. -R br ev ie id g The red dashed lines at x = ± 90°, x = 270° and x = 450° are called asymptotes. The branches of the graph get closer and closer to the asymptotes without ever reaching them. -C ev R 129 cos(180° + x ) = ve ev The graph of y = tan x R Pr cos( − x ) = ity 1 ie w C op y es By considering the shape of the cosine curve, complete the following statements, giving your answers in terms of cos x. Copyright Material - Review Only - Not for Redistribution ve rs ity ev ie am br id EXPLORE 5.4 w ge C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 2 tan(180° − x ) = 4 tan(360° − x ) = 5 tan(360° + x ) = 3 tan(180° + x ) = op y Pr es s tan( − x ) = -C 1 -R By considering the shape of the tangent curve, complete the following statements, giving your answers in terms of tan x. Transformations of trigonometric functions ve rs ity C ev br am x 360 s 270 es -C 180 y = sin x op Pr y 90 -R y = 2 sin x 1 rs C ity –2 130 y ie w ge id y 2 –1 U The graph of y = a sin x O C op ni w ev ie R REWIND These rules for the transformations of the graph y = f( x ) can be used to transform graphs of trigonometric functions. These transformations include y = a f( x ), y = f( ax ), y = f(x ) + a and y = f( x + a ) and simple combinations of these. ve ni op y It is a stretch, stretch factor 2, parallel to the y-axis. The amplitude of y = 2 sin x is 2 and the period is 360°. ge C U R ev ie w The graph of y = 2 sin x is a stretch of the graph of y = sin x. ie ev id br y am s 180 360 270 es -C y = sin 2x x Pr op y 90 -R y = sin x 1 O w The graph of y = sin ax C ity –1 y op C U w e ev ie id g es s -R br am -C R ev ie w ni ve rs The graph of y = sin 2 x is a stretch of the graph of y = sin x. 1 It is a stretch, stretch factor , parallel to the x-axis. 2 The amplitude of y = sin 2 x is 1 and the period is 180°. Copyright Material - Review Only - Not for Redistribution In Section 2.6, you learnt some rules for the transformation of the graph y = f( x ). Here we will look at how these rules can be used to transform graphs of trigonometric functions. ve rs ity C U ni op y Chapter 5: Trigonometry ev ie -R y = 1 + sin x -C 1 y O 270 180 y = sin x op 90 C y U ni C op w ev ie The graph of y = 1 + sin x is a translation of the graph of y = sin x. 0 It is a translation of . 1 R x ve rs ity –1 360 Pr es s am br id y 2 w ge The graph of y = = a + sin x w ge The amplitude of y = 1 + sin x is 1 and the period is 360°. am y y = sin(x + 90) O C y = sin x rs w –1 ve ie 131 360 x 270 180 ity 90 Pr op y es s -C 1 -R br ev id ie The graph of y = sin(x + a ) y w ge C U −90 . It is a translation of 0 R op ni ev The graph of y = sin( x + 90) is a translation of the graph of y = sin x. br ie ev id The amplitude of y = sin( x + 90) is 1 and the period is 360°. -C -R am WORKED EXAMPLE 5.8 es s On the same grid, sketch the graphs of y = sin x and y = sin( x − 90) for 0° < x < 360°. ity Pr 90 y = sin( x − 90) is a translation of the graph y = sin x by the vector . 0 ni ve rs y op y y = sin x -R s es am -C 360 x 270 C 180 ie –1 90 ev O y = sin(x – 90) br id g e U R ev ie 1 w w C op y Answer Copyright Material - Review Only - Not for Redistribution ve rs ity am br id y Step 2: Sketch the graph of y = cos( x + 90). op O y = cos(x + 90) ni C op O U ge w ie -R s es 90 180 270 360 x Pr ity –2 y y = 2 cos(x + 90) + 1 y ve 3 op ni C U 2 O 90 –1 -R am br ev id ie w ge 1 –2 es s -C ni ve rs ity Pr op y a Write down the period and amplitude of f. op y b Write down the coordinates of the maximum and minimum points on the curve y = f( x ). w -R s es am br ev ie id g e d Use your answer to part c to sketch the graph of y = 1 + 3 cos 2 x. C U c Sketch the graph of y = f( x ). -C w C f( x ) = 3 cos 2 x for 0° < x < 360°. ie y = 2 cos(x + 90) –1 rs w ie ev R 360 x y O 0 Translate y = 2 cos( x + 90) by the vector . 1 Period = 360° WORKED EXAMPLE 5.9 ev 270 1 Step 4: Sketch the graph of y = 2 cos( x + 90) + 1. Amplitude = 2. R 180 2 ev id br am -C y op C 132 90 –1 Stretch y = cos( x + 90) with stretch factor 2, parallel to the y-axis. Amplitude = 2 360 x 1 Step 3: Sketch the graph of y = 2 cos( x + 90). Period = 360° 270 y ve rs ity C w ev ie R Amplitude = 1 180 2 −90 Translate y = cos x by the vector 0 . Period = 360° 90 –1 Pr es s Amplitude = 1 y = cos x y -C Period = 360° y 2 1 -R Step 1: Start with a sketch of y = cos x. w for 0° < x < 360°, we can build up the transformation in steps. ev ie ge To sketch the graph of a trigonometric function, such as y = 2 cos( x + 90) + 1 C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 Copyright Material - Review Only - Not for Redistribution 180 270 360 x ve rs ity Answer w ge C U ni op y Chapter 5: Trigonometry am br id ev ie 360° = 180° 2 Amplitude = 3 Period = b y = cos x has its maximum and minimum points at: Pr es s -C -R a Hence, f( x ) = 3 cos 2 x has its maximum and minimum points at: y 4 3 2 1 270 180 y w ie ev s -R am Pr es y = 3 cos 2x + 1 ity 360 x 270 rs 180 op y ve 90 133 ge C U R ni y -C y 5 4 3 2 1 op C w ie x 360 0 y = 1 + 3 cos 2 x is a translation of the graph y = 3 cos 2 x by the vector . 1 O –1 –2 –3 –4 ev C op ni U 90 br O –1 –2 –3 –4 y = 3 cos 2x ge R ev ie w c d ve rs ity (0°, 3), (90°, −3), (180°, 3), (270°, −3) and (360°, 3) id C op y (0°, 1), (180°, −1), (360°, 1), (540°, −1) and ( 720°, 1 ) br ev id ie w WORKED EXAMPLE 5.10 -R am a On the same grid, sketch the graphs of y = sin 2 x and y = 1 + 3 cos 2 x for 0° < x < 360°. -C b State the number of solutions of the equation sin 2 x = 1 + 3 cos 2 x for 0° < x < 360°. es ity y = 1 + 3 cos 2x 360 y = sin 2x x y 270 op 180 w e C U 90 ni ve rs O –1 –2 –3 –4 Pr y 5 4 3 2 1 ie id g b The graphs of y = sin 2 x and y = 1 + 3 cos 2 x intersect each other at four points in the interval. -R s es am br ev Hence, the number of solutions of the equation sin 2 x = 1 + 3 cos 2 x is four. -C R ev ie w C op y a s Answer Copyright Material - Review Only - Not for Redistribution ve rs ity 1 Write down the period of each of these functions. ev ie am br id EXERCISE 5D w ge C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 f 1 x° 2 y = 5 cos(2 x + 45)° y = 5 cos 2 x° c y = 7 sin y = 4 sin(2 x + 60)° f 1 x° 2 y = 2 sin(3x + 10)° + 5 c y = tan 3x f y = 2 sin 3x − 1 i y = tan( x − 90) b y = sin 2 x° d y = 1 + 2 sin 3x° e y = tan( x − 30)° c Pr es s -R y = cos x° -C a y = 3tan a y = sin x° d y = 2 − 3 cos 4x° b ve rs ity w C op y 2 Write down the amplitude of each of these functions. e y C op ie y = 2 cos( x + 60) w h y = sin( x − 45) id g ge U R ni ev ie 3 Sketch the graph of each of these functions for 0° < x < 360°. 1 a y = 2 cos x b y = sin x 2 d y = 3 cos 2 x e y = 1 + 3 cos x ii ev am y = 2 sin x -C i π iii y = sin 2 x + 4 π y = cos x − 2 -R br 4 a Sketch the graph of each of these functions for 0 < x < 2 π . es s b Write down the coordinates of the turning points for your graph for part a iii. Pr 6 a On the same diagram, sketch the graphs of y = 2 sin x and y = 2 + cos 3x for 0 < x < 2 π. ve ie w rs C b State the number of solutions of the equation sin 2 x = 1 + cos 2 x for 0° < x < 360°. ity op y 5 a On the same diagram, sketch the graphs of y = sin 2 x and y = 1 + cos 2 x for 0° < x < 360°. 134 y op ni ev b Hence, state the number of solutions, in the interval 0 < x < 2 π, of the equation 2 sin x = 2 + cos 3x. C U R 7 a On the same diagram, sketch and label the graphs of y = 3sin x and y = cos 2 x for the interval 0 < x < 2 π. ie -R s -C es 6 Pr op y 5 O π – π 3π – 2π x 2 e U 2 y 1 op 2 ni ve rs ity 3 R ie -R s -C am br Find the value of a, the value of b and the value of c. ev id g w Part of the graph y = a sin bx + c is shown above. es ev ie w C 4 C 7 am 9 8 ev id y br 8 w ge b State the number of solutions of the equation 3sin x = cos 2 x in the interval 0 < x < 2 π. Copyright Material - Review Only - Not for Redistribution ve rs ity -C 1 w ev ie -R 2 O 120 180 240 300 360 x Part of the graph of y = a + b cos cx is shown above. ve rs ity Write down the value of a, the value of b and the value of c. ev ie 10 a Sketch the graph of y = 2 sin x for −π < x < π. ni U R w br y ev id ie State the coordinates of the other points where the line intersects the curve. -R s es P π 2 π 135 2π x y ev ve ie w rs 3π 2 ity O Pr 5 C op y -C am 11 ge b Find the value of k. Give your answer in terms of π. c w s es Pr ity 13 f( x ) = a − b cos x for 0° < x < 360°, where a and b are positive constants. ni ve rs The maximum value of f( x ) is 8 and the minimum value is −2. op y a Find the value of a and the value of b. C U b Sketch the graph of y = f( x ). id g w e 14 f( x ) = a + b sin cx for 0° < x < 360°, where a and b are positive constants. -R s es am Find the value of a, the value of b and the value of c. ev br ie The maximum value of f( x ) is 9, the minimum value of f( x ) is 1 and the period is 120°. -C ie w C op y a the value of a and the value of b ev -R br am -C 7π Given that f(0) = 3 and that f = 2, find: 6 b the range of f . ie ev id ge C U R ni op Part of the graph of y = a tan bx + c is shown above. π The graph passes through the point P , 8 . 4 Find the value of a, the value of b and the value of c. 12 f( x ) = a + b sin x for 0 < x < 2 π R C op The straight line y = kx intersects this curve at the maximum point. y w C op y 60 Pr es s 3 am br id 5 4 C y ge 9 U ni op y Chapter 5: Trigonometry Copyright Material - Review Only - Not for Redistribution ve rs ity C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 am br id -R a Write down the value of A and the value of B. -C b Write down the amplitude of f( x ). y 16 The graph of y = sin x is reflected in the line x = π and then in the line y = 1. 17 The graph of y = cos x is reflected in the line x = π and then in the line y = 3. 2 ni C op y Find the equation of the resulting function. U R ev ie w PS Find the equation of the resulting function. ve rs ity C op PS Sketch the graph of f( x ). Pr es s c w The maximum value of f( x ) is 7 and the period is 60°. ev ie ge 15 f( x ) = A + 5 cos Bx for 0° < x < 120° ge 5.5 Inverse trigonometric functions -R am br ev id ie w The functions y = sin x, y = cos x and y = tan x for x ∈ R are many-one functions. If, however, we suitably restrict the domain of each of these functions, it is possible to make the function one-one and hence we can define each inverse function. y 1 C 136 π – 2 y = sin –1x π – y op ni x O –1 1 -R y = sin x π π <x< 2 2 range: −1 < sin x < 1 s es π –– 2 y = sin −1 x op y domain: − 1 < x < 1 π π range: − < sin −1 x < 2 2 -R s es -C am br ev ie id g w e C U R ev ie w ni ve rs C ity Pr op y -C domain: − REWIND In Chapter 2 you learnt about functions and that only one-one functions can have an inverse function. You also learnt that if f and f −1 are inverse functions, then the graph of f −1 is a reflection of the graph of f in the line y = x . x ev id br am w C 2 ie R –1 U O π –– 2 ge ev ve ie w rs y = sin x In Section 2.5 you learnt about the inverse of a function. Here we will look at the particular case of the inverse of a trigonometric function. y ity op Pr y es s -C The graphs of the suitably restricted functions y = sin x, y = cos x and y = tan x and their inverse functions y = sin −1 x, y = cos −1 x and y = tan −1 x, together with their domains and ranges are: REWIND Copyright Material - Review Only - Not for Redistribution ve rs ity ge C U ni op y Chapter 5: Trigonometry ev ie w y π am br id π x y Pr es s π 2 w y = cos x ev ie domain: 0 < x < π –1 x w ge id ie range: 0 < cos −1 x < π ev y = tan x y y = tan –1 x π – 2 es s -C -R am br 1 domain: −1 < x < 1 π x 2 137 O op y x C U R ni ev ve ie w rs O ity –π 2 Pr y op C O y = cos −1 x U R y y = cos –1 x 2 ni range: −1 < cos x < 1 ve rs ity C op –1 π – C op -C O -R y = cos x y y 1 br ev id ie w ge π –– 2 am y = tan x domain: x ∈ R π π range: − < tan −1 x < 2 2 -R π π <x< 2 2 range: tan x ∈ R y = tan −1 x Pr op y es s -C domain: − C U op y ni ve rs x = sin−1 0.5 π Using a calculator gives x = . 6 The angle that the calculator gives is the one that lies in the range of the function sin −1. (This is sometimes called the principal angle.) -R s es am br ev ie id g w e The principal angle is the angle that lies in the range of the inverse trigonometric function. 5π , that satisfies sin x = 0.5 with 0 < x < π. We can There is a second angle, x = 6 find this second angle either by using skills learnt earlier in this chapter or by using the symmetry of the curve y = sin x. -C R ev ie w C ity When solving the equation sin x = 0.5 for 0 < x < π, we can find one solution using the inverse functions: Copyright Material - Review Only - Not for Redistribution ve rs ity ev ie w ge am br id WORKED EXAMPLE 5.11 C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 b 3 cos−1 2 ev ie ve rs ity a sin −1 0 means the angle whose sine is 0, where −90° < angle < 90°. Hence, sin −1 0 = 0°. w C op y Answer tan −1( −1) c Pr es s sin −1 0 -C a -R The output of the sin −1, cos −1 and tan −1 functions can be given in degrees if that is needed. Without using a calculator, write down, in degrees, the value of: U R ni C op y 3 3 b cos −1 , where 0° < angle < 180°. means the angle whose cosine is 2 2 id ie w ge 3 = 30°. Hence, cos −1 2 br ev c tan −1( −1) means the angle whose tangent is −1, where −90° < angle < 90°. es s -C -R am Hence, tan −1 ( −1) = −45°. op Pr y WORKED EXAMPLE 5.12 ity x The function f( x ) = 3sin − 1 is defined for the domain −π < x < π. 2 a Sketch the graph of y = f( x ) and explain why f has an inverse function. rs ie ev -R Pr ni ve rs w C -R s es am br ev ie id g w e x f( x ) = 3sin − 1 2 -C c U b Range is −4 < f( x ) < 2. op f has an inverse function because f is a one-one function with this domain. y ity C –2 –4 ie π x π 2 O s π 2 f es am -C – y 2 op y –π w ge id br a ev C U R c Find f −1( x ) and state its domain. Answer R op ni ev b Find the range of f . y ve ie w C 138 Copyright Material - Review Only - Not for Redistribution ve rs ity C U ni op y Chapter 5: Trigonometry ge x y = 3sin − 1 2 am br id ev ie w Step 1: Write the function as y = ve rs ity y = 2 sin −1 x + 1 for −4 < x < 2 . 3 U R ni C op The inverse function is f −1( x ) = 2 sin −1 id ie w ge EXERCISE 5E x + 1 3 y y op C w x+ +1 y = sin 3 2 y x + 1 = sin −1 2 3 Pr es s Step 3: Rearrange to make y the subject. ev ie y y x = 3sin − 1 2 -R -C Step 2: Interchange the x and y variables. -R am br ev 1 Without using a calculator, write down, in degrees, the value of: 1 a cos −1 1 b sin −1 2 tan −1 − 3 ( ) tan −1 3 f 1 cos −1 − 2 es e s sin −1 ( −1 ) -C d c Pr sin −1 0 d 1 tan −1 − 3 tan −1 1 c 1 cos −1 2 e 1 cos −1 − 2 f 3 sin −1 − 2 y op rs ni ve w ie ev b ity a C op y 2 Without using a calculator, write down, in terms of π, the value of: w π π <x< . 2 2 br ev 4 The function f( x ) = 3sin x − 4 is defined for the domain − ie id ge C U R 3 3 Given that θ = cos−1 , find the exact value of: 5 b tan2 θ a sin2 θ b Find f −1( x ) . -R am a Find the range of f . s es Pr a Find the range of f and sketch the graph of y = f( x ). b Explain why f has an inverse and find the equation of this inverse. 6 The function f( x ) = 5 − 2 sin x is defined for the domain π < x < p. 2 C U b For this value of p, find f −1( x ) and state the domain of f −1. op a Find the largest value of p for which f has an inverse. y ity Sketch the graph of y = f −1( x ) on your graph for part a. ni ve rs c w ie Find f −1( x ) and state its range. s -R b es am br a Find the range of f . ev id g e x 7 The function f( x ) = 4 cos − 5 is defined for the domain 0 < x < 2 π. 2 -C R ev ie w C op y -C 5 The function f( x ) = 4 − 2 cos x is defined for the domain 0 < x < π. Copyright Material - Review Only - Not for Redistribution 139 ve rs ity C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 One solution is given by x = sin −1(0.5) = π (or 30°). 6 w -R am br id Consider solving the equation sin x = 0.5 for −360° < x < 360°. ev ie ge 5.6 Trigonometric equations ve rs ity C w O −90 ni ev ie 90 30 360 x 270 180 150 U R −360 −270 −180 –330 –210 y = 0.5 y op y 1 C op y The graph of y = sin x for −360° < x < 360° is: Pr es s -C There are, however, many more values of x for which sin x = 0.5. y = sin x w id ie ge −1 -R am br ev The graph shows there are four values of x, between −360° and 360°, for which sin x = 0.5. es s -C We can use the calculator value of x = 30°, together with the symmetry of the curve to find the remaining answers. C ve ie w rs WORKED EXAMPLE 5.13 y op w ge Answer C U R ni ev Solve cos x = −0.7 for 0° < x < 360°. Use a calculator to find cos−1( −0.7), correct to 1 decimal place. id ie cos x = −0.7 ev br One solution is x = 134.4° -R am y es Pr op y 134.4 O 225.6 180 270 360 w x y = –0.7 ni ve rs C ity 90 y –1 ie U The sketch graph shows there are two values of x, between − 0° and 360°, for which cos x = − 0.7 . w e C Using the symmetry of the curve, the second value is (360° − 134.4°) = 225.6°. ie ev (correct to 1 decimal place) s -R x = 134.4° or 225.6° es am br id g Hence, the solution of cos x = −0.7 for 0° < x < 360° is: -C ev R y = cos x s -C 1 op 140 ity op Pr y Hence, the solution of sin x = 0.5 for − 360° < x < 360° is: x = −330°, −210°, 30° or 150° Copyright Material - Review Only - Not for Redistribution ve rs ity ge C U ni op y Chapter 5: Trigonometry am br id ev ie w WORKED EXAMPLE 5.14 Pr es s -C Answer -R Solve tan 2 A = −2.1 for 0° < A < 180°. tan x = −2.1 A solution is x = −64.54°. Let 2A = x. Use a calculator to find tan −1( −2.1). ve rs ity C op y tan 2 A = −2.1 115.46 90 -R Pr y ity 2 A = x: 2 A = −64.54° A = −32.3° 2 A = 115.46° A = 57.7° y ve ni op Hence, the solution of tan 2 A = −2.1 for 0° < A < 180° is: ev 141 2 A = 295.46° A = 147.7° rs op C w ie x = (115.46° + 180°) = 295.46° es x = ( −64.54° + 180°) = 115.46° s x = −64.54° id ie w ge C U R A = 57.7° or 147.7° (correct to 1 decimal place) -R am br ev WORKED EXAMPLE 5.15 es s -C π Solve sin 2 A + = 0.6 for 0 < A < π. 6 Pr y Use a calculator to find sin−1 ( 0.6 ). x = 0.6435 radians -R s es -C am br ev ie id g w e C U R π = x. 6 op sin x = 0.6 ev Let 2 A + ity π sin 2 A + = 0.6 6 ni ve rs w C op y Answer ie y = –2.1 ev br am -C Using the symmetry of the curve: Using 360 x 270 180 ie –90 295.46 w O id ge U R –64.54 C op y y = tan x ni ev ie w y Copyright Material - Review Only - Not for Redistribution ve rs ity 2.498 π 2π x 3π – 2 U R y = 2.498 C op π = x: 6 π 2 A + = 0.6435 6 1 π A = 0.6435 − 2 6 2A + x = π − 0.6435 ni ev ie w x = 0.6435 π = 2.498 6 1 π A = 2.498 − 2 6 br ev id ie ge 2A + w C Using the symmetry of the curve: ve rs ity op y y = sin x –1 Using -R y = 0.6 Pr es s 2 ev ie w ge π – 0.6435 -C O am br id y 1 C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 A = 0.0600 -R am A = 0.987 es s -C π Hence, the solution of sin 2 A + = .0.6 for 0 < A < π is: 6 ity op Pr y A = 0.0600 or 0.987 radians (correct to 3significant figures) es Pr ity ni ve rs C w x 2 + y2 = r 2 op C U x y and sin θ = . r r -R s es am br ev ie id g e Use cos θ = w 2 y Divide both sides by r 2. x + y =1 r r -C ie ev C w -R s -C op y sin θ for all θ with cos θ ≠ 0. cos θ Rule 2 2 ie y x and cos θ = . r r ev Use sin θ = am br id ge Divide numerator and denominator by r. KEY POINT 5.2 tan θ = y ve x U y tan θ = x y θ° ni Rule 1 y r = x r R r Two very important rules can be found using this triangle. R ev ie w rs Consider a right-angled triangle. op C 142 Copyright Material - Review Only - Not for Redistribution ve rs ity ev ie am br id KEY POINT 5.3 w ge C U ni op y Chapter 5: Trigonometry -R cos2 θ + sin2 θ = 1 for all θ . op y Pr es s -C If we use the unit circle definition of the trigonometric functions, we discover that these two important rules are true for all valid values of θ . We can use them to help solve more complicated trigonometric equations. w C ve rs ity WORKED EXAMPLE 5.16 y ni R Answer C op ev ie Solve 3 cos2 x − sin x cos x = 0 for 0° < x < 360°. U 3 cos2 x − sin x cos x = 0 Factorise. w ge cos x(3 cos x − sin x ) = 0 ev -R s es tan x = 3 x = 71.6 or 180 + 71.6 x = 71.6° or 251.6° y Pr op y -C am br id ie cos x = 0 or 3 cos x − sin x = 0 x = 90°, 270° sin x = 3 cos x 3 143 y y = tan x ge U R 360 x 270 op 251.6 C 180 s -C -R am br ev id The solution of 3 cos2 x − sin x cos x = 0 for 0° < x < 360° is: x = 71.6°, 90°, 251.6° or 270° w 90 ni ev 71.6 ie O ve ie w rs C ity y=3 Pr op y es WORKED EXAMPLE 5.17 ity ni ve rs Answer 2 sin2 x + 3 cos x − 3 = 0 op y Replace sin2 x with 1 − cos2 x. 2(1 − cos2 x ) + 3 cos x − 3 = 0 C U Expand brackets and collect terms. id g w e 2 cos2 x − 3 cos x + 1 = 0 ie Factorise. -R s es am br ev (2 cos x − 1)(cos x − 1) = 0 -C R ev ie w C Solve 2 sin2 x + 3 cos x − 3 = 0 for 0 < x < 2 π. Copyright Material - Review Only - Not for Redistribution ve rs ity w ge C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 Pr es s -C -R am br id ev ie 1 cos x = 1 or 2 π π π x= or 2 π − x x = 0 or 2 π 3 3 3 π 5π x= or 3 3 cos x = op y y 1 π 3 π 2 π 2π x C op y = cos x -C ie ev -R am br id The solution of 2 sin2 x + 3 cos x − 1 = 0 for 0 < x < 2 π is: π 5π x = 0, , or 2 π 3 3 w ge U –1 R 3π 5π 2 3 ni ev ie w O y C ve rs ity y = 0.5 y es s EXERCISE 5F Pr tan x = 1.5 b sin x = 0.4 c cos x = 0.7 d sin x = −0.3 e cos x = −0.6 f tan x = −2 g 2 cos x − 1 = 0 h 5 sin x + 3 = 0 d sin x = −0.7 h 5 tan x + 7 = 0 ity C w ve ie a rs op 1 Solve each of these equations for 0° < x < 360°. 144 y c tan x = 3 f cos x = −0.5 g 4 sin x = 3 cos 2 x = 0.6 b sin 3x = 0.8 e 3 cos 2 x = 2 f 5 sin 2 x = −4 w c tan 2 x = 4 d sin 2 x = −0.5 g 4 + 2 tan 2 x = 0 h 1 − 5 sin 2 x = 0 b π cos x + = −0.5 for 0 < x < 2 π 6 3sin(2 x − 4) = 2 x π 2 sin + = 1 for 0 < x < 4 π 3 4 s -C ev a -R br 3 Solve each of these equations for 0° < x < 180°. am op cos x = 0.5 C b ie tan x = −3 ni e U sin x = 0.3 ge a id R ev 2 Solve each of the these equations for 0 < x < 2 π. sin( x − 60°) = 0.5 c ity es a cos(2 x + 45°) = 0.8 for 0° < x < 180° d e x 2 tan + 3 = 0 for 0° < x < 540° 2 f -R s es -C am br ev ie id g w e C U op y ni ve rs Pr for 0° < x < 360° R ev ie w C op y 4 Solve each of these equations for the given domains. Copyright Material - Review Only - Not for Redistribution for 0 < x < π ve rs ity C U ni op y Chapter 5: Trigonometry w 2 sin x − 3 cos x = 0 d 3 cos 2 x − 4 sin 2 x = 0 b 5 sin2 x − 3sin x = 0 d sin2 x + 2 sin x cos x = 0 f sin x tan x = 4 sin x b 4 tan2 x = 9 ev ie b 4 sin x + 7 cos x = 0 -R c 2 sin x = cos x am br id a ge 5 Solve each of these equations for 0° < x < 360°. Pr es s -C 6 Solve 4 sin(2 x + 0.3) − 5 cos(2 x + 0.3) = 0 for 0 < x < π. a sin x cos( x − 60) = 0 c tan2 x = 5 tan x e 2 sin x cos x = sin x ve rs ity ev ie w C op y 7 Solve each of these equations for 0° < x < 360°. 3 cos2 x − 2 cos x − 1 = 0 br d 2 sin2 x − cos x − 1 = 0 f cos x + 5 = 6 sin2 x h 1 + tan x cos x = 2 cos2 x b 2 cos2 x + 5 sin x = 4 s 2 cos2 x − sin2 x − 2 sin x − 1 = 0 es g tan2 x + 2 tan x − 3 = 0 -R am 3 cos2 x − 3 = sin x -C e ie c b ev 2 sin2 x + sin x − 1 = 0 id a w ge 9 Solve each of these equations for 0° < x < 360°. C op ni 4 cos2 x = 1 U R a y 8 Solve each of these equations for 0° < x < 360°. Pr a 4 tan x = 3 cos x ity C op y 10 Solve each of these equations for 0 < x < 2 π. y ev ve ie w rs 11 Solve sin2 x + 3sin x cos x + 2 cos2 x = 0 for 0 < x < 2 π. U R ni op 5.7 Trigonometric identities C ge x + x = 2 x is called an identity because it is true for all values of x. br ev id ie w When writing an identity, we often replace the = symbol with a ≡ symbol to emphasise that it is an identity. Pr op y es s -C -R am Two commonly used trigonometric identities are: sin x sin2 x + cos2 x ≡ 1 and tan x ≡ cos x In this section you will learn how to use these two identities to simplify expressions and to prove other more complicated identities that involve sin x, cos x and tan x. op y ni ve rs -R s es -C am br ev ie id g w e C U R ev ie w C ity When proving an identity, it is usual to start with the more complicated side of the identity and prove that it simplifies to the less complicated side. Copyright Material - Review Only - Not for Redistribution 145 ve rs ity ev ie w ge am br id WORKED EXAMPLE 5.18 C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 ≡ 4 cos2 x − 3 + 3 cos2 x ve rs ity TIP y WORKED EXAMPLE 5.19 ni LHS ≡ 1 + sin x cos x + cos x 1 + sin x ≡ (1 + sin x )2 + cos2 x cos x(1 + sin x ) Expand the brackets in the numerator. ≡ 1 + 2 sin x + sin2 x + cos2 x cos x(1 + sin x ) Use sin2 x + cos2 x ≡ 1 . ≡ 2 + 2 sin x cos x(1 + sin x ) ≡ 2(1 + sin x ) cos x( 1 + sin x ) ≡ 2 cos x es s -C -R am br ev Add the two fractions. ity y C op Divide numerator and denominator by 1 + sin x. br ev id ie w ge U ni ve rs Factorise the numerator. -R am ≡ RHS Pr y op R ev ie w C 146 C ity op y Pr 2 1 + sin x 1 ≡ tan x + . Prove the identity 1 − sin x cos x Add the two fractions. w ie ev br -R s (1 + sin )2 2 es -C am sin x + 1 ≡ cos x sin x . cos x C 2 id g sin x 1 ≡ + cos x cos x Use tan x = U 2 e 1 RHS ≡ tan x + cos x op w ni ve rs Answer ie es s -C WORKED EXAMPLE 5.20 ev ie id w ge U R 1 + sin x cos x 2 Prove the identity + ≡ . cos x 1 + sin x cos x y ev ie w C op ≡ 7 cos2 x − 3 Answer R Replace sin2 x with 1 − cos2 x. Pr es s y -C 4 cos2 x − 3sin2 x ≡ 4 cos2 x − 3(1 − cos2 x ) C op Answer -R Express 4 cos2 x − 3sin2 x in terms of cos x. Copyright Material - Review Only - Not for Redistribution LHS means left-hand side and RHS means right-hand side. ve rs ity w (1 + sin x )2 1 − sin2 x Replace cos2 x with 1 − sin2 x in the denominator. ev ie ≡ ge (1 + sin x )2 cos2 x (1 + sin x )2 (1 + sin x )(1 − sin x ) ≡ 1 + sin x 1 − sin x Pr es s ≡ Divide numerator and denominator by 1 + sin x. y U w ge EXPLORE 5.5 id ie R ni C op ≡ LHS ev ie Use 1 − sin2 x = (1 + sin x )(1 − sin x ). ve rs ity w C op y -C -R am br id ≡ C U ni op y Chapter 5: Trigonometry -R s cos x es tan 2 x cos 3 x cos 3 x + cos x sin 2 x 147 ity op C sin 2 x tan 2 x cos x Pr y -C am 1 sin x tan x + cos x 1 – sin 2 x cos x sin 2 x sin x tan x cos 3 x sin 2 x cos x (1 – cos x)(1 + cos x) rs 1 – sin 2 x y op ni ev ve ie w ev br Equivalent trigonometric expressions: id ie Create trigonometric expressions of your own that simplify to sin x. w ge C U R Discuss why each of the trigonometric expressions in the coloured boxes simplifies to cos x. es s -C Compare your answers with those of your classmates. -R am br ev (Your expressions must contain at least two different trigonometric ratios.) Pr op y EXERCISE 5G ity 1 + sin x − sin2 x ≡ cos x + tan x cos x op y d w e id g ie ev cos 4 x + sin2 x cos2 x ≡ cos2 x -R s es am 1 − cos2 x ≡ tan x sin x cos x f br e b C cos2 x ≡ 1 + sin x 1 − sin x cos2 x − sin2 x + sin x ≡ cos x cos x + sin x -C R c cos x tan x ≡ sin x U a ni ve rs 2 Prove each of these identities. ev ie w C 1 Express 2 sin2 x − 7 cos2 x + 4 in terms of sin x. Copyright Material - Review Only - Not for Redistribution ve rs ity C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 c 2 − (sin x + cos x )2 ≡ (sin x − cos x )2 b cos2 x − sin2 x ≡ 1 − 2 sin2 x d cos 4 x + sin2 x ≡ sin 4 x + cos2 x b sin 4 x − cos 4 x ≡ 2 sin2 x − 1 ve rs ity cos 4 x − sin4 x ≡ 1 − tan2 x cos2 x e sin x − cos x tan x − 1 ≡ sin x + cos x tan x + 1 g tan x + 1 ≡ sin x + cos x sin x tan x + cos x d cos x 1 ≡ 1+ tan x(1 − sin x ) sin x f 1 1 − cos x 1 sin x − tan x ≡ 1 + cos x ie sin2 x (1 − cos2 x ) ≡ tan 4 x cos2 x (1 − sin2 x ) b tan x + -R s es 1 − cos x ≡ sin x tan x cos x e sin x sin x 2 tan x + ≡ 1 − sin x 1 + sin x cos x Pr c 1 1 ≡ tan x sin x cos x sin x 1 + cos x 2 + ≡ 1 + cos x sin x sin x ity d 1 + cos x 1 − cos x 4 − ≡ 1 − cos x 1 + cos x sin x tan x f y ve rs C w ie h ev id br am op y -C 6 Prove each of these identities. cos x 1 a − ≡ tan x cos x 1 + sin x 148 2 w ge U ni c y tan2 x − sin2 x ≡ tan2 x sin2 x op C w (cos2 x − 2)2 − 3sin2 x ≡ cos 4 x + sin2 x C op c 5 Prove each of these identities. cos2 x − sin2 x ≡ cos x + sin x a cos x − sin x ev ie d 2(1 + cos x ) − (1 + cos x )2 ≡ sin2 x -R cos2 x − sin2 x ≡ 2 cos2 x − 1 y a Pr es s -C 4 Prove each of these identities. R b w (sin x + cos x )2 ≡ 1 + 2 sin x cos x am br id a ev ie ge 3 Prove each of these identities. op U R ni ev 7 Show that (1 + cos x )2 + (1 − cos x )2 + 2 sin2 x has a constant value for all x and state this value. w ge C 8 a Express 7 sin2 x + 4 cos2 x in the form a + b sin2 x. -R am br 9 a Express 4 sin θ − cos2 θ in the form (sin θ + a )2 + b. ev id ie b State the range of the function f( x ) = 7 sin2 x + 4 cos2 x, for the domain 0 < x < 2 π. s 1 − sin θ 1 2(1 + sin θ ) , show that = . 2 cos θ a cos θ es 10 a Given that a = op y PS P -C b Hence, state the maximum and minimum values of 4 sin θ − cos2 θ , for the domain 0 < θ < 2 π. y op -R s es -C am br ev ie id g w e C U R ev ie w ni ve rs C ity Pr b Hence, find sin θ and cos θ in terms of a. Copyright Material - Review Only - Not for Redistribution ve rs ity C U ni op y Chapter 5: Trigonometry w ge 5.8 Further trigonometric equations -R am br id ev ie This section uses trigonometric identities to help solve some more complex trigonometric equations. Pr es s 1 − tan2 θ ≡ 2 cos2 θ − 1. 1 + tan2 θ ni U Use tan θ = sin θ 1− cos θ 2 sin θ 1+ cos θ 2 w ie ev -R am Simplify. rs 1 − tan2 θ = 5 cos θ − 3 1 + tan2 θ Use the result from part a. 2 cos2 θ − 1 = 5 cos θ − 3 Rearrange. br ev id ie w ge b C U R ni op ≡ RHS y ve w ie 149 ity C ≡ cos2 θ − (1 − cos2 θ ) ≡ 2 cos2 θ − 1 Replace sin2 θ with 1 − cos2 θ . Pr op y ≡ cos2 θ − sin2 θ ev -R am 2 cos2 θ − 5 cos θ + 2 = 0 s -C (2 cos θ − 1)(cos θ − 2) = 0 1 or cos θ = 2 2 1 θ = cos −1 2 Pr es cos θ = 2 has no solutions. ity θ = 60° Factorise. θ = 360° − 60° or ni ve rs y op -R s -C am br ev ie id g w e C U R ev Solution is θ = 60° or θ = 300°. es w C op y cos θ = ie Use sin2 θ + cos2 θ = 1. s cos2 θ − sin2 θ cos2 θ + sin2 θ es -C ≡ sin θ . cos θ Multiply numerator and denominator by cos2 θ . br ≡ 1 − tan2 θ 1 + tan2 θ ge R a LHS ≡ id ev ie Answer 1 − tan2 θ = 5 cos θ − 3 for 0° < θ < 360°. 1 + tan2 θ y b Hence, solve the equation C op w C op y a Prove the identity ve rs ity -C WORKED EXAMPLE 5.21 Copyright Material - Review Only - Not for Redistribution ve rs ity ev ie am br id EXERCISE 5H w ge C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 1 a Show that the equation cos θ + sin θ = 5 cos θ can be written in the form tan θ = k. -R b Hence, solve the equation cos θ + sin θ = 5 cos θ for 0° < θ < 360°. ve rs ity b Hence, solve the equation 3sin2 θ + 5 sin θ cos θ = 2 cos2 θ for 0° < θ < 180°. 3 a Show that the equation 8 sin2 θ + 2 cos2 θ − cos θ = 6 can be written in the form 6 cos2 θ + cos θ − 2 = 0. w C op y Pr es s -C 2 a Show that the equation 3sin2 θ + 5 sin θ cos θ = 2 cos2 θ can be written in the form 3tan2 θ + 5 tan θ − 2 = 0. y ev ie b Hence, solve the equation 8 sin2 θ + 2 cos2 θ − cos θ = 6 for 0° < θ < 360°. U R ni C op 4 a Show that the equation 4 sin 4 θ + 14 = 19 cos2 θ can be written in the form 4x 2 + 19x − 5 = 0, where x = sin2 θ . id ie w ge b Hence, solve the equation 4 sin 4 θ + 14 = 19 cos2 θ for 0° < θ < 360°. br ev 5 a Show that the equation sin θ tan θ = 3 can be written in the form cos2 θ + 3 cos θ − 1 = 0. -R am b Hence, solve the equation sin θ tan θ = 3 for 0° < θ < 360°. es s -C 6 a Show that the equation 5(2 sin θ − cos θ ) = 4(sin θ + 2 cos θ ) can be written in the form tan θ = op Pr y b Hence, solve the equation 5(2 sin θ − cos θ ) = 4(sin θ + 2 cos θ ) for 0° < θ < 360°. sin θ 1 + cos θ 2 + ≡ . 1 + cos θ sin θ sin θ ity 7 a Prove the identity sin θ 1 + cos θ + = 1 + 3 sin θ for 0° < θ < 360°. 1 + cos θ sin θ ni br ev id ie w ge C U R y cos θ 1 ≡ − 1. tan θ (1 + sin θ ) sin θ cos θ = 1 for 0° < θ < 360°. b Hence, solve the equation tan θ (1 + sin θ ) 8 a Prove the identity op ev ve ie b Hence, solve the equation rs w C 150 1 1 2 . + ≡ 1 + sin θ 1 − sin θ cos2 θ 1 1 b Hence, solve the equation cos θ + = 5 for 0° < θ < 360°. 1 + sin θ 1 − sin θ es 1 + cos θ . 1 − cos θ Pr ≡ 2 = 2 for 0° < θ < 360°. y ni ve rs 1 1 + b Hence, solve the equation sin θ tan θ op 2 ity 10 a Prove the identity 1 + 1 sin θ tan θ C 1 for 0° < θ < 360°. 2 ie ev -R s es am br id g e b Hence, solve the equation cos 4 θ − sin 4 θ = w U 11 a Prove the identity cos 4 θ − sin 4 θ ≡ 2 cos2 θ − 1. -C R ev ie w C op y s -C -R am 9 a Prove the identity Copyright Material - Review Only - Not for Redistribution 13 . 6 ve rs ity C U ni op y Chapter 5: Trigonometry Exact values of trigonometric functions 1 2 3 2 1 3 θ = 45° = π 4 1 2 1 2 1 θ = 60° = π 3 1 2 3 Pr es s ve rs ity y Angles measured anticlockwise from the positive x-direction are positive. ● Angles measured clockwise from the positive x -direction are negative. 180° ie -R es Cos ity op Pr y 0°, 360° O Tan 151 270° rs C Useful mnemonic: ‘A ll Students Trust Cambridge’. y ve w ie All s -C Sin ev id am br 90° w ge U ● ● C op ni y op C w ev ie 3 2 Diagram showing where sin, cos and tan are positive ev w π 6 Positive and negative angles tan θ -R cos θ -C sin θ θ = = 30° = R ev ie am br id ge Checklist of learning and understanding op 1 180 270 360 x 270 360 x s -C -R am –1 90 ie –90 O br –360 –270 –180 ev id w ge y = sin x es y Pr op y 1 y = cos x 90 180 op y –1 -R s es am br ev ie id g w e C U R ev ie w ni ve rs C ity –360 –270 –180 –90 O -C C y U R ni Graphs of trigonometric functions Copyright Material - Review Only - Not for Redistribution ve rs ity ev ie Pr es s –180 –90 O y ni U ge The graph of y = sin( ax ) is a stretch of y = sin x , stretch factor -R am br ev id ie w 1 , parallel to the x -axis. a 0 The graph of y = a + sin x is a translation of y = sin x by the vector . a −a . 0 es s -C The graph of y = sin( x + a ) is a translation of y = sin x by the vector ity y π – 2 y op 2 y = sin −1 x domain: −1 < x < 1 π π range: − < sin −1 x < 2 2 s 1 op y ni ve rs ity Pr y = tan −1 x domain: x ∈ R π π range: − < tan −1 x < 2 2 -R s es am br ev ie id g w e sin2 x + cos2 x ≡ 1 C U Trigonometric identities sin x ● tan x ≡ cos x -C π –– 2 x y = cos −1 x domain: −1 < x < 1 range: 0 < cos −1 x < π op y C w O es –1 -R br am -C π –– 2 ● O ev id 1 y = cos –1 x ie x y = tan –1 x C π – ge O w U R ni ev ve ie y= y π sin –1x rs 2 w Pr y op C y π – –1 ie 360 x C op C w ev ie R 270 ● Inverse trigonometric functions ev 180 The graph of y = a sin x is a stretch of y = sin x , stretch factor a, parallel to the y -axis. ● R 90 ve rs ity op –270 ● ● 152 y = tan x -R am br id -C w y –360 y C ge U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 Copyright Material - Review Only - Not for Redistribution x ve rs ity ge C U ni op y Chapter 5: Trigonometry -R y 3 Pr es s op y 1 –π w C ve rs ity 3π –– 2 ev ie y = a + b sin x 2 -C 1 ev ie am br id w END-OF-CHAPTER REVIEW EXERCISE 5 1π –– 2 O 1 π – 2 y C op ni ge U R State the values of the constants a and b. w ie id Find the value of x satisfying the equation sin−1( x − 1) = tan−1(3). ev br -R am Given that θ is an acute angle measured in radians and that cos θ = k , find, in terms of k, an expression for: s tan θ c cos( π − θ ) es b Pr [1] [1] Solve the equation sin 2 x = 5 cos 2 x, for 0° < x < 180°. 6 Solve the equation rs 5 ni op y ve [4] id br -R am [4] 1 for 0 < θ < 2 π. 2 ii Write down the number of roots of the equation 2 cos 2θ − 1 = 0 in the interval 0 < θ < 2 π. Sketch, on a single diagram, the graphs of y = cos 2θ and y = es s -C Pr iii Deduce the number of roots of the equation 2 cos 2θ − 1 = 0 in the interval 10π < θ < 20 π. [3] [1] [1] ity op y w ev Cambridge International AS & A Level Mathematics 9709 Paper 11 Q3 November 2014 C Cambridge International AS & A Level Mathematics 9709 Paper 11 Q3 November 2011 Show that the equation 2 tan2 θ sin2 θ = 1 can be written in the form 2 sin 4 θ + sin2 θ − 1 = 0. C w ev ie id g es s -R br am -C [2] [4] Cambridge International AS & A Level Mathematics 9709 Paper 11 Q5 June 2011 e U R ii Hence solve the equation 2 tan2 θ sin2 θ = 1 for 0° < θ < 360°. y i op 9 ni ve rs w ie [4] Solve the equation 2 cos2 x = 5 sin x − 1 for 0° < x < 360°. i [4] ie ge 13sin2 θ + cos θ = 2 for 0° < θ < 180°. 2 + cos θ C U R ev [1] ity y op C sin θ ie w a Solve the equation cos −1(8x 4 + 14x 2 − 16) = π. 8 [3] Cambridge International AS & A Level Mathematics 9709 Paper 11 Q2 November 2014 4 7 [2] Cambridge International AS & A Level Mathematics 9709 Paper 11 Q1 June 2014 -C 3 2π x –1 The diagram shows part of the graph of y = a + b sin x. 2 ev 3π – 2 π Copyright Material - Review Only - Not for Redistribution 153 ve rs ity ev ie am br id 10 i w ge C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 Solve the equation 4 sin2 x + 8 cos x − 7 = 0 for 0° < x < 360°. [4] -C -R 1 1 ii Hence find the solution of the equation 4 sin2 θ + 8 cos θ − 7 = 0 for 0° < θ < 360°. 2 2 Pr es s y C op 2 − sin x 3 for 0 < x < 2 π. = 1 + 2 sin x 4 U R 12 a Solve the equation y ve rs ity Prove the identity ni op C w ev ie Cambridge International AS & A Level Mathematics 9709 Paper 11 Q4 November 2013 sin x tan x 1 ≡ 1+ . [3] 1 − cos x cos x sin x tan x ii Hence solve the equation + 2 = 0, for 0° < x < 360°. [3] 1 − cos x Cambridge International AS & A Level Mathematics 9709 Paper 11 Q4 November 2010 11 i ev id br [1] s -C -R am [1] es [2] op 14 i ity Cambridge International AS & A Level Mathematics 9709 Paper 11 Q7 November 2010 Solve the equation 2 cos2 θ = 3sin θ , for 0° < θ < 360°. [4] rs C w [3] Pr y iv Obtain an expression, in terms of x, for f −1( x ). ni op y ve ie ii The smallest positive solution of the equation 2 cos2 ( nθ ) = 3sin( nθ ), where n is a positive integer, is 10°. State the value of n and hence find the largest solution of this equation in the interval 0° < θ < 360°. ev [4] ie 1 13 A function f is defined by f : x → 3 − 2 tan x for 0 < x < π. 2 i State the range of f. ii State the exact value of f 2 π . 3 iii Sketch the graph of y = f(x ). C [3] Cambridge International AS & A Level Mathematics 9709 Paper 11 Q7 November 2012 w ge U R [3] w ge b Solve the equation sin x − 2 cos x = 2 ( 2 sin x − 3 cos x ) for −π < x < π. 154 [2] s -C -R am br ev id Show that ie sin θ cos θ 1 + ≡ . [3] sin θ + cos θ sin θ − cos θ sin2 θ − cos2 θ sin θ cos θ + = 3, for 0° < θ < 360°. [4] ii Hence solve the equation sin θ + cos θ sin θ − cos θ Cambridge International AS & A Level Mathematics 9709 Paper 11 Q5 June 2013 15 i es 4 cos θ + 15 = 0 can be expressed as 4 sin2 θ − 15 sin θ − 4 = 0. tan θ 4 cos θ + 15 = 0 for 0° < θ < 360°. ii Hence solve the equation tan θ Show that the equation [3] ity Pr [3] ni ve rs Cambridge International AS & A Level Mathematics 9709 Paper 11 Q4 November 2015 C U op y 17 The function f : x → 5 + 3 cos 1 x is defined for 0 < x < 2 π. 2 i Solve the equation f( x ) = 7, giving your answer correct to 2 decimal places. id g w e ii Sketch the graph of y = f(x ). ie iii Explain why f has an inverse. ev [3] [2] [1] [3] s -R Cambridge International AS & A Level Mathematics 9709 Paper 11 Q8 June 2015 es am br iv Obtain an expression for f −1( x ). -C R ev ie w C op y 16 i Copyright Material - Review Only - Not for Redistribution op y ve rs ity ni C U ev ie w ge -R am br id Pr es s -C y ni C op y ve rs ity op C w ev ie rs op y ve w ge C U ni ie w C ity op Pr y es s -C -R am br ev id ie w ge U R ev R Chapter 6 Series 155 id es s -C -R am br ev use the expansion of ( a + b ) n, where n is a positive integer recognise arithmetic and geometric progressions use the formulae for the nth term and for the sum of the first n terms to solve problems involving arithmetic or geometric progressions use the condition for the convergence of a geometric progression, and the formula for the sum to infinity of a convergent geometric progression. y op -R s es -C am br ev ie id g w e C U R ev ie w ni ve rs C ity Pr op y ■ ■ ■ ■ ie In this chapter you will learn how to: Copyright Material - Review Only - Not for Redistribution ve rs ity ge C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 w IGCSE / O Level Mathematics Expand brackets. b (1 − 3x )(1 + 2 x − 3x 2 ) 2 Simplify: b ( −2 x 3 )5 3 Find the nth term of these linear sequences. ni y Find the nth term of a linear sequence. a U R (2 x + 3)2 C op IGCSE / O Level Mathematics a a (5x 2 )3 ve rs ity ev ie 1 Expand: Pr es s y op Simplify indices. w C IGCSE / O Level Mathematics Check your skills -R What you should be able to do -C Where it comes from ev ie am br id PREREQUISITE KNOWLEDGE 5, 7, 9, 11, 13, … br ev id ie w ge b 8, 5, 2, −1, −4, … -R am Why study series? y es s -C At IGCSE / O Level you learnt how to expand expressions such as (1 + x )2 . In this chapter you will learn how to expand expressions of the form (1 + x ) n, where n can be any positive integer. Expansions of this type are called binomial expansions. ve op ni Binomial means ‘two terms’. br The expansion of ( a + b )2 can be used to expand ( a + b )3: -R am C ev id ie You should already know that ( a + b )2 = a 2 + 2 ab + b2. ( a + b )3 = ( a + b )( a 2 + 2 ab + b2 ) s -C = a 3 + 2 a 2b + ab2 + a 2b + 2 ab2 + b3 Pr op y es = a 3 + 3a 2b + 3ab2 + b3 Similarly, it can be shown that ( a + b )4 = a 4 + 4a 3b + 6a 2b2 + 4ab3 + b 4. WEB LINK 1a + 1b 1a 2 + 2 ab + 1b2 ( a + b )3 = 3 1a 3 + 3a 2b + 3ab2 + 1b3 ( a + b )4 = 4 1a 4 + 4a 3b + 6a 2b2 + 4ab3 + 1b 4 op 2 -R s es am br ev ie id g w e C U ( a + b )2 = y ni ve rs ( a + b )1 = -C R ev ie w C ity Writing the expansions of ( a + b ) n in full in order: 1 FAST FORWARD Properties of binomial expansions are also used in probability theory, which you will learn about if you go on to study the Probability and Statistics 1 Coursebook, Chapter 7. w ge The word is used in algebra for expressions such as x + 3 and 5x − 2 y. ( a + b )0 = 1 In the Pure Mathematics 2 and 3 Coursebook, Chapter 7, you will learn how to expand these expressions for any real value of n. y 6.1 Binomial exapansion of ( a + b ) n U R ev ie w rs C ity op Pr This chapter also covers arithmetic and geometric progressions. Both the mathematical and the real world are full of number sequences that have particular special properties. You will learn how to find the sum of the numbers in these progressions. Some fractal patterns can generate these types of sequences. 156 FAST FORWARD Copyright Material - Review Only - Not for Redistribution Try the Sequences and Counting and binomials resources on the Underground Mathematics website. ve rs ity C U ni op y Chapter 6: Series ev ie -R The first term is a 4 and then the power of a decreases by 1 while the power of b increases by 1 in each successive term. All of the terms have a total index of 4 ( a 4, a 3b, a 2b2 , ab3 and b 4 ). Pr es s -C ● am br id ● w ge If you look at the expansion of ( a + b )4, you should notice that the powers of a and b form a pattern. There is a similar pattern in the other expansions. ve rs ity ց ւ 6 The next row is then: 10 10 1 5 id 1 br s es DID YOU KNOW? ni U 1 6 15 6 4 10 10 20 1 5 15 y ev R 5 1 op 4 1 1 1 3 1 6 1 ge There are many number patterns to be found in Pascal’s triangle. 4 s 1 10 20 Pr ni ve rs ity 1 What do you notice if you find the total of each row in Pascal’s triangle? Can you explain your findings? op y 2 Can you find the Fibonacci sequence (1, 1, 2, 3, 5, 8, 13, …) in Pascal’s triangle? You may want to add terms together. -R s es am br ev ie id g w e C U 3 Pascal’s triangle has many other number patterns. Which number patterns can you find? -C R ev ie w C op y These numbers are called tetrahedral numbers. es -C -R am br ev id For example, the numbers 1, 4, 10 and 20 have been highlighted. C 3 ve ie 2 rs w 1 1 1 ity C 1 Pascal’s triangle is named after the French mathematician Blaise Pascal (1623–1662). Pr y op 1 Each number is the sum of the two numbers in the row above it. -R -C am ( a + b )5 = 1a 5 + 5a 4b + 10 a 3b2 + 10 a 2b3 + 5ab 4 + 1b5 EXPLORE 6.1 Each row always starts and finishes with a 1. ev This row can then be used to write down the expansion of ( a + b )5: w n = 5: 1 5 4 w 4 ie 1 y 1 + 2 1 ց ւ 1 3 + 3 1 TIP C op 1 ni 1 U n = 1: n = 2: n = 3: n = 4: 1 ie n = 0: ge R ev ie w C op y The coefficients also form a pattern that is known as Pascal’s triangle. Copyright Material - Review Only - Not for Redistribution 157 ve rs ity ev ie w ge am br id WORKED EXAMPLE 6.1 C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 Use Pascal’s triangle to find the expansion of: (5 − 2 x )4 -C b (3x + 2)3 (3x + 2)3 = 1(3x )3 + 3(3x )2 (2) + 3(3x )(2)2 + 1(2)3 4 = 27 x 3 + 54x 2 + 36x + 8 y (5 − 2 x ) b ev ie The index is 3 so use the row for n = 3 in Pascal’s triangle (1, 3, 3, 1). ve rs ity w C op y a Pr es s Answer -R (3x + 2)3 a U R (5 − 2 x )4 = 1(5)4 + 4(5)3 ( −2 x ) + 6(5)2 ( −2 x )2 + 4(5)( −2 x )3 + 1( −2 x )4 br ev id ie w ge = 625 − 1000 x + 600 x 2 − 160 x 3 + 16x 4 C op ni The index is 4 so use the row for n = 4 in Pascal’s triangle (1, 4, 6, 4, 1). -R am WORKED EXAMPLE 6.2 s es -C a Use Pascal’s triangle to expand (1 − 2 x )5. op Pr y b Find the coefficient of x 3 in the expansion of (3 + 5x )(1 − 2 x )5. 158 ity a (1 − 2 x )5 rs w C Answer y ev ve ie The index is 5 so use the row for n = 5 in Pascal’s triangle (1, 5, 10, 10, 5, 1). U R = 1 − 10 x + 40 x 2 − 80 x 3 + 80 x 4 − 32 x5 w ge b (3 + 5x )(1 − 2 x )5 = (3 + 5x )(1 − 10 x + 40 x 2 − 80 x 3 + 80 x 4 − 32 x5 ) br ev id ie The term in x 3 comes from the products: -R am (3 + 5x )(1 − 10 x + 40 x 2 − 80 x 3 + 80 x 4 − 32 x5 ) -C 3 × ( −80 x 3 ) = −240 x 3 and 5x × 40 x 2 = 200 x 3 ni ve rs s b (1 − x )4 c ( x + y )3 d e ( x − y )4 f (2 x + 3 y )3 g (2 x − 3)4 C x2 + 3 2 x3 w e ev ie id g es s -R br am h (2 − x )3 y ( x + 2)3 U a op 1 Use Pascal’s triangle to find the expansions of: -C ie w C ity EXERCISE 6A ev es Pr op y Coefficient of x 3 = −240 + 200 = −40. R C ni op (1 − 2 x )5 = 1(1)5 + 5(1)4 ( −2 x ) + 10(1)3 ( −2 x )2 + 10(1)2 ( −2 x )3 + 5(1)( −2 x )4 + 1( −2 x )5 Copyright Material - Review Only - Not for Redistribution 3 ve rs ity C U ni op y Chapter 6: Series (2 x − 1)4 (3 + x )5 + (3 − x )5 ≡ A + Bx 2 + Cx 4 w f c (3 − x )5 d (4 + x )4 g (4x + 3)4 h 5− x 2 ev ie (1 + x )5 4 Pr es s -C 3 ( x − 2)5 b -R e ( x + 3)4 am br id a ge 2 Find the coefficient of x 3 in the expansions of: y Find the value of A, the value of B and the value of C . ve rs ity Find the possible values of the constant a. ev ie 5 a Expand (2 + x )4 . U R ni C op b Use your answer to part a to express (2 + 3 )4 in the form a + b 3 . w ge 6 a Expand (1 + x )3 . id br ev (1 + 5 )3 in the form a + b 5 (1 − 5 )3 in the form c + d 5 . -R am ii Use your answers to part b to simplify (1 + 5 )3 + (1 − 5 )3. es s -C c op Pr y 7 Expand (1 + x )(2 + 3x )4. 159 ity 8 a Expand ( x 2 − 1)4 . b Find the coefficient of x 6 in the expansion of (1 − 2 x 2 )( x 2 − 1)4. 4 ve ie w rs C ie b Use your answer to part a to express: i y op ni 4 w ge U 3 10 Find the term independent of x in the expansion of x 2 − 2 . x C ev 2 9 Find the coefficient of x 2 in the expansion of 3x − . x R y w C op 4 The coefficient of x 2 in the expansion of (3 + ax )4 is 216. id ie 11 a Find the first three terms, in ascending powers of y, in the expansion of (1 + y )4 . -R am br ev b By replacing y with 5x − 2 x 2, find the coefficient of x 2 in the expansion of (1 + 5x − 2 x 2 )4. -C 12 The coefficient of x 2 in the expansion of (1 + ax )4 is 30 times the coefficient of x in the expansion of 3 s es Pr 4 ity 1 13 Find the power of x that has the greatest coefficient in the expansion of 3x 4 + . x ni ve rs 14 a Write down the expansion of ( x + y )5 . 5 b Without using a calculator and using your result from part a, find the value of 10 , correct to 4 y 1 op U C the nearest hundred. ie id g w e 4 4 q 1 1 15 a Given that x 2 + − x 2 − = px5 + , find the value of p and the value of q. x x x 4 4 -R s es am br ev 1 1 b Hence, without using a calculator, find the exact value of 2 + − 2 − . 2 2 -C R ev ie w C op y 1 + ax . Find the value of a. 3 Copyright Material - Review Only - Not for Redistribution ve rs ity C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 1 x 1 a Express x 3 + 3 in terms of y. x 1 b Express x5 + 5 in terms of y. x Pr es s -C -R am br id ev ie w ge 16 y = x + PS y 6.2 Binomial coefficients y U R ni EXPLORE 6.2 C op ev ie w C ve rs ity op Pascal’s triangle can be used to expand ( a + b ) n for any positive integer n, but if n is large it can take a long time to write out all the rows in the triangle. Hence, we need a more efficient method to find the coefficients in the expansions. The coefficients in the binomial expansion of (1 + x ) n are known as binomial coefficients. w ge Consider the expansion: 10 10 5 1 op ie es 1 Use your calculator to find the values of: Pr y s -C Find the nCr function on your calculator. On some calculators this may be n C r n or . r ity 5 5 5 5 5 5 0 , 1 , 2 , 3 , 4 and 5 . op C w ie -R am br ev id ge U R ni ev 2 What do you notice about your answers to question 1? 3 Complete the following four statements. 5 The coefficient of x 2 in the expansion of (1 + x )5 is . … … The coefficient of x r in the expansion of (1 + x ) n is . … y ve ie w rs C 160 C ity Pr op y es s -C 5 5 The coefficient of the 4th term in the expansion of ( 1 + x ) is . … … n The coefficient of the ( r + 1)th term in the expansion of ( 1 + x ) is . … y op w ev ie id g es s -R br am C U e n n n n If n is a positive integer, then (1 + x ) n = + x + x 2 + … + x n . 2 0 1 n -C R ev ie w ni ve rs We write the binomial expansion of (1 + x ) n , where n is a positive integer as: KEY POINT 6.1 5 To find , key in 2 5 nCr 2 . -R am The coefficients are: 1 5 ev br id (1+ x )5 = 1+ 5x + 10x 2 + 10x 3 + 5x 4 + x 5 TIP Copyright Material - Review Only - Not for Redistribution ve rs ity C U ni op y Chapter 6: Series am br id ev ie w ge We can therefore write the expansion of (1 + x ) n using binomial coefficients; the result is known as the Binomial theorem. n -C -R b We can use the Binomial theorem to expand ( a + b ) n, too. We can write ( a + b ) n = a n 1 + a (assuming that a ≠ 0 ). Pr es s KEY POINT 6.2 y ni ev ie WORKED EXAMPLE 6.3 (2 − 3x )10 ie ev -R am 15 15 15 2 15 3 (1 + x )15 = x+ x + x +… + 0 1 2 3 = 1 + 15x + 105x 2 + 455x 3 + … es s -C ity Pr 10 7 10 10 10 9 10 8 (2 − 3x )10 = 2 + 2 ( −3x )1 + 2 ( −3x )2 + 2 ( −3x )3 + … 1 2 3 0 = 1024 − 15 360 x + 103680 x 2 − 414 720 x 3 + … ve ie w rs C op y b w b br Answer a U (1 + x )15 ge a id R Find, in ascending powers of x, the first four terms in the expansion of: y ev You should also know how to work out the binomial coefficients without using a calculator. w ge C U R ni op 5 5 From Pascal’s triangle, we know that = 1 and = 1. 0 5 br ev id ie In general, we can write this as: -R am KEY POINT 6.3 Pr y 5 5×4×3×2 =5 = 4 4× 3× 2 ×1 C 5 5×4×3 = 10 = 3 3× 2 ×1 op ni ve rs U 5 5×4 = 10 = 2 2 ×1 -R s es am br ev ie id g w e 5 5 = =5 1 1 -C ev ie w C 5 5 5 5 We write , , and as: 1 2 3 4 ity op y es s -C n n 0 = 1 and n = 1 R C op w C ve rs ity op y n n n n ( a + b ) n = a n + a n −1b1 + a n − 2 b 2 + … + b n 0 1 2 n Copyright Material - Review Only - Not for Redistribution 161 ve rs ity C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 am br id ev ie w ge In general, if r is a positive integer less than n, then: KEY POINT 6.4 y Pr es s -C -R n n × ( n − 1) × ( n − 2) × … × ( n − r + 1) r = r × ( r − 1) × ( r − 2) × … × 3 × 2 × 1 ve rs ity C op y 8 a Without using a calculator, find the value of . 4 n b Find an expression, in terms of n, for . 4 U R ni ev ie w C op WORKED EXAMPLE 6.4 w ge Answer 8 8×7×6×5 4 = 4 × 3 × 2 × 1 = 70 b n n × ( n − 1) × ( n − 2) × ( n − 3) n( n − 1)( n − 2)( n − 3) = 4 = 4×3×2 ×1 24 op Pr y es s -C -R am br ev id ie a 162 rs n x When 1 − is expanded in ascending powers of x, the coefficient of x 2 is 4. Given that n is the positive 3 integer, find the value of n. y op ie -R s Pr -R s es -C am br ev ie id g w e C U op ie ev R y ni ve rs As n is a positive integer, n = 9. ity op y ( n − 9)( n + 8) = 0 n = 9 or n = −8 w C n2 − n − 72 = 0 es -C am n × ( n − 1) =4 18 n( n − 1) = 72 ev br id 2 n x n × ( n − 1) x 2 n × ( n − 1) 2 x Term in x 2 = − = × = 3 2 ×1 9 18 2 w ge Answer C U R ni ev ve ie w C ity WORKED EXAMPLE 6.5 Copyright Material - Review Only - Not for Redistribution ve rs ity ge C U ni op y Chapter 6: Series am br id ev ie w WORKED EXAMPLE 6.6 8 Term in x 4 = (2)4 ( kx )4 = 1120 k 4 x 4 4 ve rs ity ev ie w C op y 8 Term in x5 = (2)3 ( kx )5 = 448 k 5 x5 5 Pr es s -C Answer -R When (2 + kx )8 is expanded, the coefficient of x5 is two times the coefficient of x 4 . Given that k > 0, find the value of k. ni C op y Coefficient of x5 = 2 × coefficient of x 4 U w id 448 k 4 ( k − 5) = 0 ge 448 k 5 − 2240 k 4 = 0 ie R 448 k 5 = 2 × 1120 k 4 br ev k = 0 or kk==50 or k = 5 s -C -R am As k is a positive integer, k = 5 . op Pr y es WORKED EXAMPLE 6.7 163 ity b Use your answer to part a to estimate the value of 1.99 × 1.029. s Let x = 0.01. es (2 − x )(1 + 2 x )9 = 2 + 35x + 270 x 2 + … ity ni ve rs 1.99 × 1.029 ≈ 2.377 Pr 1.99 × 1.029 ≈ 2 + 35(0.01) + 270(0.01)2 w w e C U op y n There is an alternative formula for calculating . To be able to understand and apply r the alternative formula, we need to first know about factorial notation. -R s es am br ev ie id g We write 6! to mean 6 × 5 × 4 × 3 × 2 × 1, and call it ‘6 factorial’. -C ie ev R op ev = 2 + (2 × 18 − 1)x + (2 × 144 − 18)x 2 + … = 2 + 35x + 270 x 2 + … -R am br id ie w = (2 − x )(1 + 18x + 144x 2 + …) = 2(1 + 18x + 144x 2 + …) − x(1 + 18x + 144x 2 + …) -C C op y b C ni 9 9 9 (2 − x )(1 + 2 x )9 = (2 − x ) + (2 x )1 + (2 x )2 + … 2 0 1 ge R a U ev Answer y ve ie w rs C a Obtain the first three terms in the expansion of (2 − x )(1 + 2 x )9. Copyright Material - Review Only - Not for Redistribution ve rs ity C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 am br id ev ie w ge In general, if n is a positive integer, then: KEY POINT 6.5 -C w ie am n n! to find the value of: Use the formula = r r! ( n − r )! 8 a 4 164 s es Pr op y 9 b 3 ity rs ve ie w C Answer 8 8! 8! = = 70 a = − 4!(8 4)! 4! 4! 4 -R br ev id WORKED EXAMPLE 6.8 ge U R ni n n! r = r! ( n − r )! y w KEY POINT 6.6 C op C ve rs ity op y n The formula for then becomes: r ev ie -R Pr es s -C n! = n × ( n − 1) × ( n − 2) × ( n − 3) × … × 3 × 2 × 1 y op id ie w ge C U R ni ev 9 9! 9! b = = = 84 3!(9 − 3)! 3! 6! 3 -R am br ev WORKED EXAMPLE 6.9 9 s -C 5 Find the term independent of x in the expansion of x + 2 . x op y es Answer op y ni ve rs The term that is independent of x is the term that when simplified does not involve x. 5 The x terms cancel each other out when the power of x is double the power of 2 . x Also, the sum of these powers must be 9. ev -R s es am br 9 6 5 3 125 6 3 x x 2 = 84 × x × x 6 = 10 500 ie id g w e C U 9 Hence, we are looking for powers of 6 and 3, respectively, and the corresponding binomial coefficient is . 3 The term independent of x is: -C R ev ie w C ity Pr 3 2 1 9 x + 5 = 9 x 9 + 9 x8 5 + 9 x 7 5 + 9 x 6 5 + … 3 x 2 0 2 x 2 1 x 2 x2 Copyright Material - Review Only - Not for Redistribution ve rs ity ev ie am br id EXERCISE 6B w ge C U ni op y Chapter 6: Series 1 Without using a calculator, find the value of each of the following. 9 6 -R b Pr es s 7 3 -C a c 12 4 c n 3 d 15 6 d 12 7 n 2 b n 1 ve rs ity a b 8 5 c 14 3 w C op ni 10 2 ge R a y n n! 3 Use the formula = to find the value of of each of the following. r r! ( n − r )! U ev ie w C op y 2 Express each of the following in terms of n. 7 f -R 3+ x 2 (2 − x )13 g (2 + x 2 )8 (1 + x 2 )12 h x2 2 + 2 d 3− x 3 9 s -C e am br ev id ie 4 Find, in ascending powers of x, the first three terms in each of the following expansions. 7 x d b (1 − 3x )10 c 1+ a (1 + 2 x )8 2 (1 + 3x )12 c 2+ x 4 12 C es b Pr (1 − x )9 7 10 ity a 6 Find the coefficient of x 4 in the expansion of (2 x + 1)12. w rs C op y 5 Find the coefficient of x 3 in each of the following expansions. y op ni ev ve ie 7 Find the term in x5 in the expansion of (5 − 2 x )8. U R 8 Find the coefficient of x8 y5 in the expansion of ( x − 2 y )13. id ie w ge 3 9 Find the term independent of x in the expansion of x − 2 . x ev b (1 + 2 x )(1 − 3x )10 x (1 + x ) 1 − 2 c -R (1 − x )(2 + x )7 am a br 10 Find, in ascending powers of x, the first three terms of each of the following expansions. 8 s -C 11 a Find, in ascending powers of x, the first three terms in the expansion of (2 + x )10. 8 ity Pr x 12 a Find, in ascending powers of x, the first three terms in the expansion of 1 − . 2 8 x b Hence, obtain the coefficient of x 2 in the expansion of (2 + 3x − x 2 ) 1 − . 2 ie w ni ve rs C op y es b By replacing x with 2 y − 3 y2, find the first three terms in the expansion of (2 + 2 y − 3 y2 )10. y op ev 13 Find the first three terms, in ascending powers of x, in the expansion of (2 − 3x )4 (1 + 2 x )10. id g n -R s es am br ev ie x 15 The first two terms, in ascending powers of x, in the expansion of (1 + x ) 2 − are p + qx 2. 4 Find the values of n , p and q. -C PS w e C U R 14 The first four terms, in ascending powers of x, in the expansion of (1 + ax + bx 2 )7 are 1 − 14x + 91x 2 + px 3. Find the values of a, b and p. Copyright Material - Review Only - Not for Redistribution 165 ve rs ity C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 w ge 6.3 Arithmetic progressions am br id ev ie At IGCSE / O Level you learnt that a number sequence is a list of numbers and that the numbers in the sequence are called the terms of the sequence. The notation used for arithmetic progressions is: d = common difference op y a = first term Pr es s -C -R A linear sequence such as 5, 8, 11, 14, 17, … is also called an arithmetic progression. Each term differs from the term before by a constant. This constant is called the common difference. l = last term y C op a + 3d term 4 U a + 2d term 3 a + 4d term 5 w ge a+d term 2 a term 1 ni The first five terms of an arithmetic progression whose first term is a and whose common difference is d are: R ev ie w C ve rs ity The common difference is also allowed to be zero or negative. For example, 10, 6, 2, −2, … and 5, 5, 5, 5, … are both arithmetic progressions. br -R am KEY POINT 6.7 ev id ie From this pattern, you can see that the formula for the nth term is given by: y es s -C nth term = a + ( n − 1)d C ity op Pr WORKED EXAMPLE 6.10 166 rs y ni op Use a = −3, d = 4 and nth term = 237. U R nth term = a + ( n − 1)d Solve. -R am br ev id ie w ge 237 = −3 + 4( n − 1) n − 1 = 60 n = 61 C Answer ve ev ie w Find the number of terms in the arithmetic progression −3, 1, 5, 9, 13, … , 237. es s -C WORKED EXAMPLE 6.11 ity Pr op y The fourth term of an arithmetic progression is 7 and the tenth term is 16. Find the first term and the common difference. ni ve rs 10th term = 16 ⇒ a + 9d = 16 (2) e d = 1.5 C U (2) − (1) gives 6d = 9 (1) y ⇒ a + 3d = 7 op 4th term = 7 R id g w Substituting into (1) gives a + 4.5 = 7 -R s -C am br ev ie a = 2.5 First term = 2.5, common difference = 1.5 es ev ie w C Answer Copyright Material - Review Only - Not for Redistribution ve rs ity ge C U ni op y Chapter 6: Series am br id ev ie w WORKED EXAMPLE 6.12 Pr es s -C Answer -R The nth term of an arithmetic progression is 5 − 6n. Find the first term and the common difference. y 1st term = 5 − 6(1) = −1 ve rs ity y ev ie Substitute n = 2 into nth term = 5 − 6 n. Common difference = 2nd term − 1st term = −6 w C op 2nd term = 5 − 6(2) = −7 Substitute n = 1 into nth term = 5 − 6 n. ni C op The sum of an arithmetic progression ie br ev id EXPLORE 6.3 w ge U R When the terms in a sequence are added together we call the resulting sum a series. -R am 1+ 2 + 3 + 4 + … + 97 + 98 + 99 + 100 = ? ity 1 Can you complete Gauss’s method to find the answer? w rs C op Pr y es s -C It is said that, at the age of seven or eight, the famous mathematician Carl Gauss was asked to find the sum of the numbers from 1 to 100. His teacher expected this task to keep him occupied for some time but Gauss surprised him by writing down the correct answer almost immediately. His method involved adding the numbers in pairs: 1 + 100 = 101 , 2 + 99 = 101, 3 + 98 = 101, … y b 5 + 10 + 15 + 20 + … + 185 + 190 + 195 + 200 c 6 + 9 + 12 + 15 + … + 93 + 96 + 99 + 102 op 2 + 4 + 6 + 8 + … + 494 + 496 + 498 + 500 ev id ie w ge C U ni a br R ev ve ie 2 Use Gauss’s method to find the sum of: es s -C -R am 3 Use Gauss’s method to find an expression, in terms of n, for the sum: 1 + 2 + 3 + 4 + … + ( n − 3) + ( n − 2) + ( n − 1) + n Pr op y The sum of an arithmetic progression, Sn, can be written as: -R s es -C am br ev ie id g w e C U R n [2 a + ( n − 1)d ] 2 y Sn = op or ity n (a + l ) 2 ni ve rs Sn = ev ie w C KEY POINT 6.8 Copyright Material - Review Only - Not for Redistribution 167 ve rs ity C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 w ev ie + ( a + d ) + ( a + 2d ) + … + (l − 2d ) + (l − d ) + + (l − d ) + (l − 2d ) + … + ( a + 2d ) + ( a + d ) + a l am br id Reversing: Sn = Sn = ge We can prove this result as follows, by writing out the series in full. l a ve rs ity op y Pr es s -C -R 2Sn = ( a + l ) + ( a + l ) + ( a + l ) + … + ( a + l ) + ( a + l ) + ( a + l ) 2Sn = n( a + l ), as there are n terms in the series n So Sn = ( a + l ). 2 n Using l = a + ( n − 1)d , this can be rewritten as Sn = [2a + ( n − 1)d ]. 2 Adding: y C op KEY POINT 6.9 ni ev ie w C It is useful to remember the following rule that applies for all sequences. w ie br ev id WORKED EXAMPLE 6.13 ge U R nth term = Sn − Sn − 1 s -C -R am In an arithmetic progression, the 1st term is −12, the 17th term is 12 and the last term is 45. Find the sum of all the terms in the progression. ity Use nth term = 12 when n = 17 and a = −12. Solve. op y ve rs 12 = −12 + 16d 3 d = 2 ni ev ie w C nth term = a + ( n − 1)d Pr op y We start by working out the common difference. 168 es Answer C U R We now determine the number of terms in the whole sequence. Use nth term = 45 when a = −12 and d = id ie w ge nth term = a + ( n − 1)d 3 45 = −12 + ( n − 1) 2 n − 1 = 38 n = 39 s es ity Pr Use a = −12 , l = 45 and n = 39. -R s es -C am br ev ie id g w e C U op ev R y ni ve rs = 643 21 ie w C op y -C Finally, we can work out the sum of all the terms. n Sn = ( a + l ) 2 39 ( −12 + 45) S39 = 2 -R am br ev Solve. Copyright Material - Review Only - Not for Redistribution 3 . 2 ve rs ity ge C U ni op y Chapter 6: Series am br id ev ie w WORKED EXAMPLE 6.14 The 10th term in an arithmetic progression is 14 and the sum of the first 7 terms is 42. -C -R Find the first term of the progression and the common difference. Pr es s Answer ni U br 4 into equation (1) gives a = 2. 3 4 First term = 2, common difference = 3 ev id ie w ge w (1) − (2) gives 6d = 8 4 d = 3 Use n = 7 and S7 = 42. y n [2 a + ( n − 1)d ] 2 7 42 = (2 a + 6d ) 2 (2) 6 = a + 3d Sn = ev ie R (1) C op 14 = a + 9d op -R s Pr y es -C am Substituting d = 169 C ity WORKED EXAMPLE 6.15 ve ie w rs The sum of the first n terms, Sn, of a particular arithmetic progression is given by Sn = 4 n2 + n . y op ni ev a Find the first term and the common difference. id ie w ge C U R b Find an expression for the nth term. Answer a S1 = 4(1)2 + 1 = 5 br ev First term = 5 -R am S2 = 4(2)2 + 2 = 18 C es Pr U op y Method 2: nth term = Sn − Sn − 1 = 4 n2 + n − [4( n − 1)2 + ( n − 1)] -R s es am br ev ie id g w e C = 4 n2 + n − (4 n2 − 8 n + 4 + n − 1) = 8n − 3 -C ev ie w ni ve rs = 5 + 8( n − 1) = 8n − 3 Use a = 5, d = 8. ity op y First term = 5, common difference = 8 b Method 1: nth term = a + ( n − 1)d First term + second term = 18 s -C Second term = 18 − 5 = 13 R Use nth term = 14 when n = 10. ve rs ity C op y nth term = a + ( n − 1)d Copyright Material - Review Only - Not for Redistribution ve rs ity ev ie am br id EXERCISE 6C w ge C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 1 The first term in an arithmetic progression is a and the common difference is d . -R Write down expressions, in terms of a and d , for the seventh term and the 19th term. 13 + 17 + 21 + … + 97 op y a Pr es s -C 2 Find the number of terms and the sum of each of these arithmetic series. b 152 + 149 + 146 + … + 50 b 4 + 1 + ( −2) + … (38 terms) c 1 1 2 … (20 terms) + + + 3 2 3 d − x − 5x − 9x − … (40 terms) C op y 5 + 12 + 19 + … (17 terms) R ni ev ie ve rs ity a w C 3 Find the sum of each of these arithmetic series. w ge U 4 The first term of an arithmetic progression is 15 and the sum of the first 20 terms is 1630. Find the common difference. br ev id ie 5 In an arithmetic progression, the first term is −27, the 16th term is 78 and the last term is 169. s -C b Find the sum of the terms in this progression. -R am a Find the common difference and the number of terms. op Pr y es 6 The first two terms in an arithmetic progression are 146 and 139. The last term is −43. Find the sum of all the terms in this progression. ity 7 The first two terms in an arithmetic progression are 2 and 9. The last term in the progression is the only number that is greater than 150. Find the sum of all the terms in the progression. ve ie w rs C 170 y op C U R ni ev 8 The first term of an arithmetic progression is 15 and the last term is 27. The sum of the first five terms is 79. Find the number of terms in this progression. w ge 9 Find the sum of all the integers between 100 and 300 that are multiples of 7. -R am br ev id ie 10 The first term of an arithmetic progression is 2 and the 11th term is 17. The sum of all the terms in the progression is 500. Find the number of terms in the progression. op y es s -C 11 Robert buys a car for $8000 in total (including interest). He pays for the car by making monthly payments that are in arithmetic progression. The first payment that he makes is $200 and the debt is fully repaid after 16 payments. Find the fifth payment. Pr ity a Find the first term and the common difference. ni ve rs b Given that the nth term of this progression is −59, find the value of n. C U op y 13 The sum of the first n terms, Sn, of a particular arithmetic progression is given by Sn = 4 n2 + 3n. Find the first term and the common difference. -R s es am br ev ie id g w e 14 The sum of the first n terms, Sn, of a particular arithmetic progression is given by Sn = 12 n − 2 n2 . Find the first term and the common difference. -C R ev ie w C 12 The sixth term of an arithmetic progression is −3 and the sum of the first ten terms is −10. Copyright Material - Review Only - Not for Redistribution ve rs ity C U ni op y Chapter 6: Series am br id ev ie w ge 15 The sum of the first n terms, Sn, of a particular arithmetic progression is given by Sn = Find an expression for the nth term. 1 (5 n2 − 17 n ). 4 Pr es s -C -R 16 A circle is divided into ten sectors. The sizes of the angles of the sectors are in arithmetic progression. The angle of the largest sector is seven times the angle of the smallest sector. Find the angle of the smallest sector. b Find the 65th term in terms of a. y 18 The tenth term in an arithmetic progression is three times the third term. Show that the sum of the first ten terms is eight times the sum of the first three terms. w 19 The first term of an arithmetic progression is sin2 x and the second term is 1. U R ni P C op ev ie a Find d in terms of a. ve rs ity C op y 17 An arithmetic sequence has first term a and common difference d. The sum of the first 20 terms is seven times the sum of the first five terms. ge a Write down an expression, in terms of sin x , for the fifth term of this progression. ie id br ev 20 The sum of the digits in the number 67 is 13 (as 6 + 7 = 13). -R am PS w b Show that the sum of the first ten terms of this progression is 10 + 35 cos2 x. a Show that the sum of the digits of the integers from 19 to 21 is 15. y es s -C b Find the sum of the digits of the integers from 1 to 99. op Pr 6.4 Geometric progressions 171 rs ni op r = common ratio U R a = first term y ve The notation used for a geometric progression is: ev ie w C ity The sequence 2, 6, 18, 54, … is called a geometric progression. Each term is three times the preceding term. The constant multiplier, 3, is called the common ratio. ar 2 term 3 ar 3 term 4 ar 4 term 5 -R am br ar term 2 ie a term 1 ev id w ge C The first five terms of a geometric progression whose first term is a and whose common ratio is r are: s -C This leads to the formula for the nth term of a geometric progression: op y es KEY POINT 6.10 op -R s es am C ie ev Use nth term = 1 when n = 5 and r = br nth term = ar n − 1 id g Answer w e U 1 The fifth term of a geometric progression is 1 and the common ratio is . 2 Find the eighth term and an expression for the nth term. y ni ve rs WORKED EXAMPLE 6.16 -C R ev ie w C ity Pr nth term = ar n − 1 Copyright Material - Review Only - Not for Redistribution 1 . 2 ve rs ity ev ie w ge 4 -R am br id 1 1= a 2 a = 16 C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 7 n −1 WORKED EXAMPLE 6.17 y ev ie w C ve rs ity op y 1 nth term = ar n − 1 = 16 2 Pr es s -C 1 1 8th term = 16 = 2 8 ie br 4 ar 4 40.5 = ar 12 27 r3 = 8 3 r= 2 3 Substituting r = into equation (1) gives a = 8. 2 n −1 3 3 First term = 8, common ratio = , nth term = 8 . 2 2 rs y op ie w ge ev br id WORKED EXAMPLE 6.18 C U R ni ev ve ie w C 172 ity op Pr y es s -C (2) ÷ (1) gives -R (2) am 40.5 = ar (1) ev 12 = ar id Answer w ge U R ni C op The second and fifth terms in a geometric progression are 12 and 40.5, respectively. Find the first term and the common ratio. Hence, write down an expression for the nth term. n -R -C am 2 The nth term of a geometric progression is 9 − . Find the first term and the common ratio. 3 es s Answer 1 Pr op y 2 1st term = 9 − = −6 3 2 ni ve rs y 2nd term 4 2 = =− 1st term −6 3 op Common ratio = e C U 2 This is also clear from the formula directly: each term is − times the previous one. 3 -R s es am br ev ie id g w 2 First term = −6, common ratio = − . 3 -C R ev ie w C ity 2 2nd term = 9 − = 4 3 Copyright Material - Review Only - Not for Redistribution ve rs ity ev ie am br id EXPLORE 6.4 w ge C U ni op y Chapter 6: Series -R In this Explore activity you are not allowed to use a calculator. Pr es s -C 1 Consider the sum of the first 10 terms, S10 , of a geometric progression with a = 1 and r = 3. a Multiply both sides of the previous equation by the common ratio, 3, and complete the following statement. ve rs ity C op y S10 = 1+ 3 + 32 + 33 + … + 37 + 38 + 39 w 3S10 = 3 + 32 + 3… + 3… + … + 3… + 3… + 3… y C op ni ev ie b How does this compare to the original expression? Can you use this to find a simpler way of expressing the sum S10? b a + ar + ar 2 + … ev br (10 terms) (10 terms) -R am a + ar + ar 2 + … (n terms) es s -C c ie 1+ r + r 2 + … id a w ge U R 2 Use the method from question 1 to find an alternative way of expressing each of the following. C ity op Pr y You will have discovered in Explore 6.4 that the sum of a geometric progression, Sn, can be written as: rs y a ( r n − 1) r −1 op Sn = C U ge ie id ev -R br ● Use the first formula when −1 < r < 1. Use the second formula when r > 1 or when r ø −1. am ● w Either formula can be used but it is usually easier to: rSn = ar + ar + … + ar n − 3 + ar n − 2 + ar n − 1 + ar n s n ni ve rs a ( r n − 1) r −1 w e C U op Multiplying the numerator and the denominator by −1 gives the alternative formula a (1 − r n ) Sn = . 1− r -R s es am br ev ie id g Can you see why this formula does not work when r = 1 ? -C ie ev Sn = (2) ity ( r − 1)Sn = a ( r n − 1) w C (2) − (1): rSn − Sn = ar − a R (1) es op y r × (1): 2 Pr -C This is the proof of the formulae in Key point 6.11. Sn = a + ar + ar 2 + … + ar n − 3 + ar n − 2 + ar n − 1 y or ve a (1 − r n ) 1− r R ev Sn = TIP ni ie w KEY POINT 6.11 173 Copyright Material - Review Only - Not for Redistribution These formulae are not defined when r = 1 . ve rs ity ev ie w ge am br id WORKED EXAMPLE 6.19 C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 Simplify. ni WORKED EXAMPLE 6.20 y C w ev ie Use a = 3, r = 2 and n = 12. ve rs ity op y Pr es s -C a ( r n − 1) r −1 3(212 − 1) S12 = 2 −1 = 12 285 Sn = C op Answer -R Find the sum of the first 12 terms of the geometric series 3 + 6 + 12 + 24 + … . ie id Answer w ge U R The third term of a geometric progression is nine times the first term. The sum of the first six terms is k times the sum of the first two terms. Find the value of k. br ev 3rd term = 9 × first term es ity rs ie br ev id EXERCISE 6D w ge C U R ni op y ve When r = 3 , k = 91 and when r = −3, k = 91. Hence, k = 91. ev Rearrange to make k the subject. Pr y a ( r 6 − 1) ka ( r 2 − 1) = r −1 r −1 6 r −1 k= 2 r −1 op ie w C 174 Divide both sides by a (which we assume is non-zero) and solve. s -C Use -R am ar 2 = 9a r = ±3 S6 = kS2 -C -R 2, 4, 8, 14, … b 7, 21, 63, 189, … d 1 2 4 7 , , , ,… 9 9 9 9 e 1, 0.4, 0.16, 0.64, … c 81, −27, 9, −3, … f 1, −1, 1, −1, … Pr es s a op y ni ve rs ity 2 The first term in a geometric progression is a and the common ratio is r. Write down expressions, in terms of a and r, for the sixth term and the 15th term. op y 3 The first term of a geometric progression is 270 and the fourth term is 80. Find the common ratio. C U 4 The first term of a geometric progression is 50 and the second term is −30. Find the fourth term. br ev ie id g w e 5 The second term of a geometric progression is 12 and the fourth term is 27. Given that all the terms are positive, find the common ratio and the first term. -R s es am 6 The sum of the second and third terms in a geometric progression is 84. The second term is 16 less than the first term. Given that all the terms in the progression are positive, find the first term. -C C w ie ev R am 1 Identify whether the following sequences are geometric. If they are geometric, write down the common ratio and the eighth term. Copyright Material - Review Only - Not for Redistribution ve rs ity C U ni op y Chapter 6: Series w ge 7 Three consecutive terms of a geometric progression are x, 4 and x + 6 . Find the possible values of x. -R 3 + 6 + 12 + 24 + … 1− 2 + 4 − 8 +… b 128 + 64 + 32 + 16 + … d 243 + 162 + 108 + 72 + … Pr es s -C c am br id a ev ie 8 Find the sum of the first eight terms of each of these geometric series. 10 A ball is thrown vertically upwards from the ground. The ball rises to a height of 8 m and then falls and 3 bounces. After each bounce it rises to of the height of the previous bounce. 4 a Write down an expression for the height that the ball rises after the nth impact with the ground. y ve rs ity ev ie w C op y 9 The first four terms of a geometric progression are 0.5, 1, 2 and 4. Find the smallest number of terms that will give a sum greater than 1000 000. U R ni C op b Find the total distance that the ball travels from the first throw to the fifth impact with the ground. w ge 11 The second term of a geometric progression is 24 and the third term is 12( x + 1). id ie a Find, in terms of x, the first term of the progression. -R am br ev b Given that the sum of the first three terms is 76, find the possible values of x. s -C 12 The third term of a geometric progression is nine times the first term. The sum of the first four terms is k times the first term. Find the possible values of k. Pr rs b Find the total value of the donations made during the years 2010 to 2016, inclusive. C U ni op S3 n − S2 n = r 2 n. Sn w ge R Show that y 14 A geometric progression has first term a, common ratio r and sum to n terms Sn. ev P ve w 1 1 1 1 , 9, , 27, , 81, ,…. 3 9 27 81 Show that the sum of the first 2n terms of the sequence is 1 (2 + 3n − 31 − n ). 2 15 Consider the sequence 1, 1, 3, P 16 Let Sn = 1 + 11 + 111 + 1111 + 11111 + … to n terms. id ie P s Pr ni ve rs 1 , so it begins 2 br es C w s -R These sums are getting closer and closer to 4. am ie ev id g e U 1 1 1 2, 1, , , , … . We can work out the sum of the first n terms of this: 2 4 8 1 3 7 S1 = 2, S2 = 3, S3 = 3 , S4 = 3 , S5 = 3 , and so on. 2 4 8 op Consider the infinite geometric progression where a = 2 and r = y An infinite sequence is a sequence whose terms continue forever. -C R ev ie w 6.5 Infinite geometric series es 10 n + 1 − 10 − 9 n . 81 ity Show that Sn = C op y -C -R am br ev ie 175 a Find the value of the donation in 2016. ity C op y es 13 A company makes a donation to charity each year. The value of the donation increases exponentially by 10% each year. The value of the donation in 2010 was $10 000. Copyright Material - Review Only - Not for Redistribution ve rs ity C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 am br id ev ie w ge The diagram of the 2 by 2 square is a visual representation of this series. If the pattern of rectangles inside the square is continued, the total area of the rectangles approximates the area of the whole square (which is 4) increasingly well as more rectangles are included. 1 1 1 … + + + is 4, 2 4 8 because the sum of the first n terms gets as close to 4 as we like as n gets larger. We write 1 1 1 2 + 1 + + + + … = 4. We also say that the sum to infinity of this series is 4, and that 2 4 8 the series converges to 4. A series that converges is also known as a convergent series. 2 -R Pr es s -C y op w ge U ni C op y ve rs ity C w ev ie R br ev id ie to be very useful, and gives us answers that work consistently when we try to do more mathematics with them. op y ve ni ie id DID YOU KNOW? w ge C U R ev ie w rs C 176 ity op Pr y es s -C -R am You are probably also familiar with a very important example of an infinite geometric series without realising it! What do we mean by the recurring decimal 0.3333…? 3 3 3 + + + … . If we work out the sum We can write this as a series: 0.3333… = 10 100 1000 3 33 of the first n terms of this geometric series, we find S1 = = 0.3, S2 = = 0.33, 10 100 333 1 S3 = = 0.333 and so on. These sums are getting as close as we like to , so we say that 1000 3 1 1 the sum of the infinite series is equal to , and we write = 0.3333… . This justifies what 3 3 you have been writing for many years. Using the formula we will be working out shortly, we can easily write any recurring decimal as an exact fraction. s -C -R am br ev The first person to introduce infinite decimal numbers was Simon Stevin in 1585. He was an influential mathematician who popularised the use of decimals more generally as well, through a publication called De Thiende (‘The tenth’). Pr op y es EXPLORE 6.5 y a=− 1 , r = −2 2 op d 1 5 C a = 3, r = − w ni ve rs b ie 2 3 U a = 6, r = e c 3 , r = −2 5 id g a a= br ev 2 Find other convergent geometric series of your own. In each case, find the sum to infinity. -R s es am 3 Can you find a condition for r for which a geometric series is convergent? -C R ev ie w C ity 1 Investigate whether these infinite geometric series converge or not. You could use a spreadsheet to help with the calculations. If they converge, state their sum to infinity. Copyright Material - Review Only - Not for Redistribution 1 4 2 We therefore say that the sum of the infinite geometric series 2 + 1 + You might be wondering why we can say this, as no matter how many terms we add up, the answer is always less than 4. The simplest answer is because it works. Mathematicians and philosophers have struggled with the idea of infinity for thousands of years, and 1 1 1 whether something like ‘2 + 1 + + + + …’ even makes sense. But over the past few 2 4 8 1 1 1 hundred years, we have worked out that writing ‘2 + 1 + + + + … = 4’ turns out 2 4 8 1 8 1 2 1 2 ve rs ity C U ni op y Chapter 6: Series -R am br id ev ie w ge Consider the geometric series a + ar + ar 2 + ar 3 + … + ar n − 1. a (1 − r n ) . The sum, Sn, is given by the formula Sn = 1− r n If −1 < r < 1 , then as n gets larger and larger, r gets closer and closer to 0. We say that as n tends to infinity, r n tends to zero, and we write ‘as n → ∞, r n → 0’. y Pr es s -C a (1 − r n ) a (1 − 0) a → = . 1− r 1− r 1− r Hence, as n → ∞, ve rs ity KEY POINT 6.12 y a provided that −1 < r < 1. 1− r U R ni S∞ = C op ev ie w C op This gives the result: br ev id ie w ge If r ù 1 or r ø −1, then r n does not converge, and so the series itself does not converge. So an infinite geometric series converges when and only when −1 < r < 1. -R am WORKED EXAMPLE 6.21 s -C The first four terms of a geometric progression are 5, 4, 3.2 and 2.56. y es a Write down the common ratio. rs y ni op Use a = 5 and r = C w -R am br ev id 4 1− 5 = 25 a (1 − r 3 ) 1− r 3 2 a 1− − 3 y 2 Use S3 = 63 and r = − . 3 C U S3 = ie id g ev Simplify. es s -R br 2 1− − 3 am 63 = w e a op Answer ni ve rs b Find the sum to infinity. ity Pr A geometric progression has a common ratio of − 2 and the sum of the first three terms is 63. 3 a Find the first term of the progression. -C ev ie w C op y es s -C WORKED EXAMPLE 6.22 R 4 . 5 ie = a 1− r 5 U ev R b S∞ = second term 4 = first term 5 ve Common ratio = ie w a ity Answer 177 ge C op Pr b Find the sum to infinity. Copyright Material - Review Only - Not for Redistribution ve rs ity ev ie w ge 35 27 am br id Solve. 5 3 -R 63 = a× C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 -C a = 81 a 1− r 81 = 2 1− − 3 Pr es s y C op ni ev ie w = 48 53 w ge U EXERCISE 6E R 2 Use a = 81 and r = − . 3 ve rs ity C op y b S∞ = br ev id ie 1 Find the sum to infinity of each of the following geometric series. 2 … 2 2 a 2+ + + + b 1 + 0.1 + 0.01 + 0.001 + … 3 9 27 -R am 40 − 20 + 10 − 5 + … -C c d −64 + 48 − 36 + 27 − … es s 2 The first four terms of a geometric progression are 1, 0.52, 0.54 and 0.56. Find the sum to infinity. op Pr y 3 The first term of a geometric progression is 8 and the second term is 6. Find the sum to infinity. ity 4 The first term of a geometric progression is 270 and the fourth term is 80. Find the common ratio and the sum to infinity. w rs C 178 y op C U R ni ev ve ie ɺ ɺ as the sum of a geometric progression. 5 a Write the recurring decimal 0.57 ɺ ɺ can be written as 19 . b Use your answer to part a to show that 0.57 33 ev id ie w ge 6 The first term of a geometric progression is 150 and the sum to infinity is 200. Find the common ratio and the sum of the first four terms. -R am br 7 The second term of a geometric progression is 4.5 and the sum to infinity is 18. Find the common ratio and the first term. es s -C 8 Write the recurring decimal 0.315151515… as a fraction. Pr ity e C U op 10 The third term of a geometric progression is 16 and the sixth term is − 1 . 4 a Find the common ratio and the first term. es s -R br ev ie id g w b Find the sum to infinity. am y b the sum to infinity. ni ve rs a the common ratio and the first term -C R ev ie w C op y 9 The second term of a geometric progression is 9 and the fourth term is 4. Given that the common ratio is positive, find: Copyright Material - Review Only - Not for Redistribution ve rs ity C U ni op y Chapter 6: Series am br id ev ie w ge 11 The first three terms of a geometric progression are 135, k and 60. Given that all the terms in the progression are positive, find: a the value of k -R b the sum to infinity. Pr es s a Find the value of k. b Find the sum to infinity. ve rs ity 13 The fourth term of a geometric progression is 48 and the sum to infinity is five times the first term. Find the first term. y ev ie w C op y -C 12 The first three terms of a geometric progression are k + 12 , k and k − 9, respectively. w ge U R ni C op 14 A geometric progression has first term a and common ratio r. The sum of the first three terms is 3.92 and the sum to infinity is 5. Find the value of a and the value of r. id ie 15 The first term of a geometric progression is 1 and the second term is 2 cos x, where 0 < x < 16 A circle of radius 1cm is drawn touching the three edges of an equilateral triangle. -R PS am br ev Find the set of values of x for which this progression is convergent. π . 2 s es Pr This process is then repeated an infinite number of times, as shown in the diagram. 179 a Find the sum of the circumferences of all the circles. op w pattern 2 pattern 3 pattern 4 br ev pattern 1 ie id ge C U R ni 17 y ve rs b Find the sum of the areas of all the circles. w ie P ev 1 cm ity C op y -C Three smaller circles are then drawn at each corner to touch the original circle and two edges of the triangle. -R am We can construct a Koch snowflake as follows. Starting with an equilateral triangle (pattern 1), we perform the following steps to produce pattern 2. es s -C Step 1: Divide each line segment into three equal segments. Pr Step 3: Remove the line segments that were used as the base of the equilateral triangles in step 2. ni ve rs ity These three steps are then repeated to produce the next pattern. -R s es am br ev ie id g w e C U op y a Let pn be the perimeter of pattern n. Show that the sequence p1, p2 , p3 , … tends to infinity. 8 b Let An be the area of pattern n. Show that the sequence A1, A2 , A3 , … tends to times the area of the original 5 triangle. 8 c The Koch snowflake is the limit of the patterns. It has infinite perimeter but an area of of the original 5 triangle, as you have shown. This snowflake pattern is an example of a fractal. Use the internet to find out about the Sierpinski triangle fractal. -C R ev ie w C op y Step 2: Draw an equilateral triangle, pointing outwards, that has the middle segment from step 1 as its base. Copyright Material - Review Only - Not for Redistribution ve rs ity C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 am br id ev ie w ge 6.6 Further arithmetic and geometric series -R EXPLORE 6.6 Pr es s -C a, b, c,… op y 1 Given that a, b and c are in arithmetic progression, find an equation connecting a, b and c. y U R ni WORKED EXAMPLE 6.23 C op ev ie w C ve rs ity 2 Given that a, b and c are in geometric progression, find an equation connecting a, b and c. id ie w ge The first, second and third terms of an arithmetic series are x, y and x 2 . The first, second and third terms of a geometric series are x, x 2 and y. Given that x < 0 , find: br ev a the value of x and the value of y -R am b the sum to infinity of the geometric series es s -C c the sum of the first 20 terms of the arithmetic series. op a Arithmetic series is: x + y + x 2 + … rs (1) Geometric series is: x + x + y + … y x2 2 = x x y = x3 (2) Use common ratios. (1) and (2) give 2 x 3 = x 2 + x Divide by x (since x ≠ 0) and rearrange. y ev id ie w ge C U R ni op 2 ve ie w 2 y = x2 + x ev Use common differences. ity y − x = x2 − y C 180 Pr y Answer 2x − x − 1 = 0 (2 x + 1)( x − 1) = 0 br 2 1 or x = 1 2 1 1 Hence, x = − and y = − . 2 8 a b S∞ = 1− r 1 − 1 2 =− S∞ = 3 1 1− − 2 -R am Factorise and solve. s Pr es x ≠ 1 since x < 0. C U op y ni ve rs ity Use a = − 1 and r = − 1 . 2 2 1 1 3 Use n = 20 , a = − 1 , d = − − − = . 8 2 8 2 n [2 a + ( n − 1)d ] 2 20 3 −1 + 19 S20 = 8 2 s es = 61.25 -R br ev ie id g w e Sn = am c -C R ev ie w C op y -C x=− Copyright Material - Review Only - Not for Redistribution ve rs ity ev ie am br id DID YOU KNOW? w ge C U ni op y Chapter 6: Series op y Pr es s -C -R Georg Cantor (1845–1918) was a German mathematician who is famous for his work on set theory and for formalising many ideas about infinity. He developed the theory that there are infinite sets of different sizes. He showed that the set of natural numbers (1, 2, 3, ...) and the set of rational numbers (all fractions) are actually the same size, whereas the set of real numbers is actually larger than either of them. ve rs ity ni b geometric U R a arithmetic y 1 The first term of a progression is 16 and the second term is 24. Find the sum of the first eight terms given that the progression is: ge 2 The first term of a progression is 20 and the second term is 16. C op ev ie w C EXERCISE 6F id ie w a Given that the progression is geometric, find the sum to infinity. -R am br ev b Given that the progression is arithmetic, find the number of terms in the progression if the sum of all the terms is −160. s es Pr a the value of r 181 b the sixth term of each progression. ity C op y -C 3 The first, second and third terms of a geometric progression are the first, fourth and tenth terms, respectively, of an arithmetic progression. Given that the first term in each progression is 12 and the common ratio of the geometric progression is r, where r ≠ 1 , find: y op w ge C U R ni ev ve ie w rs 1 4 A geometric progression has eight terms. The first term is 256 and the common ratio is . 2 1 An arithmetic progression has 51 terms and common difference . 2 The sum of all the terms in the geometric progression is equal to the sum of all the terms in the arithmetic progression. Find the first term and the last term in the arithmetic progression. -R am br ev id ie 5 The first, second and third terms of a geometric progression are the first, sixth and ninth terms, respectively, of an arithmetic progression. Given that the first term in each progression is 100 and the common ratio of the geometric progression is r, where r ≠ 1 , find: s -C a the value of r es op y b the fifth term of each progression. Pr ity a Find the common difference of this progression. y ni ve rs The first, third and nth terms of this arithmetic progression are the first, second and third terms, respectively, of a geometric progression. op w e 7 The first term of a progression is 2x and the second term is x 2 . C U b Find the common ratio of the geometric progression and the value of n. -R s es am br ev ie id g a For the case where the progression is arithmetic with a common difference of 15, find the two possible values of x and corresponding values of the third term. 1 b For the case where the progression is geometric with a third term of − , find the sum to infinity. 16 -C R ev ie w C 6 The first term of an arithmetic progression is 16 and the sum of the first 20 terms is 1080. Copyright Material - Review Only - Not for Redistribution ve rs ity C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 Binomial expansions w Pr es s -C n n n! n × ( n − 1) × ( n − 2) × … × ( n − r + 1) or = the formulae = . r × ( r − 1) × ( r − 2) × … × 3 × 2 × 1 r r! ( n − r )! r op y ● -R n Binomial coefficients, denoted by n C r or , can be found using: r ● Pascal’s triangle ev ie am br id ge Checklist of learning and understanding ni We can extend this rule to give: y n n n n n (1 + x ) n = + x + x 2 + … + x n , where the ( r + 1)th term = x r. r 2 0 1 n C op ev ie w C ve rs ity If n is a positive integer, the Binomial theorem states that: w ge U R n n−r r n n n n b. ( a + b ) n = a n + a n − 1b1 + a n − 2 b 2 + … + b n , where the ( r + 1)th term = a r 2 1 0 n s -C Arithmetic series -R br n( n − 1) 2 n( n − 1)( n − 2) 3 x + x + … + xn 2! 3! am (1 + x ) n = 1 + nx + ev id ie We can also write the expansion of (1 + x ) n as: y Pr the kth term is a + ( k − 1)d op ● es For an arithmetic progression with first term a, common difference d and n terms: the last term is l = a + ( n − 1)d n n ( a + l ) = [2 a + ( n − 1)d ]. ● the sum of the terms is Sn = 2 2 rs ni ev Geometric series y ve ie w C ity ● op 182 ● the last term is ar n − 1 ● sum of the terms is Sn = C U the kth term is ar k − 1 w ge ● -R a . 1− r y op -R s es -C am br ev ie id g w e C U R ev ie w ni ve rs C ity Pr op y es -C When an infinite geometric series converges, S∞ = s am The condition for an infinite geometric series to converge is −1 < r < 1. ev ie a (1 − r n ) a ( r n − 1) = . 1− r r −1 id br R For a geometric progression with first term a, common ratio r and n terms: Copyright Material - Review Only - Not for Redistribution ve rs ity ge C U ni op y Chapter 6: Series ev ie 5 3 Find the coefficient of x 2 in the expansion of 2 x + 2 . x -R [3] x In the expansion of 1 − (5 + x )6, the coefficient of x 2 is zero. a Find the value of a. [3] Pr es s 2 . Find the term independent of x in the expansion of 3x − 5x y U 1 . Find the term independent of x in the expansion of 3x 2 − 2 x3 [4] 8 a Find the first three terms in the expansion of ( x − 3x 2 )8, in descending powers of x. am [3] b Find the coefficient of x in the expansion of (1 − x )( x − 3x ) . [2] a Find the first three terms in the expansion of (1 + px )8, in ascending powers of x. w es s 2 8 Pr 2 [4] ii (3 − x ) y (1 + 2 x )5 op i ve Find the first three terms, in ascending powers of x, in the expansion of: ni C 5 w Find the coefficient of x 2 in the expansion of [(1 + 2 x )(3 − x )]5. [2] [2] [3] id ie ev R b 183 rs ity Given that the coefficient of x in the expansion of (1 − 2 x )(1 + px ) is 204, find the possible values of p. U 10 a [3] 8 ge y op b -R br 15 [4] 5 ev id ie ge 7 C w [4] 2 Find the coefficient of x5 in the expansion of x 3 + 2 . x 9 ie 5 6 -C R [3] In the expansion of (2 + ax )7 , where a is a constant, the coefficient of x is −2240. Find the coefficient of x 2 . ni 5 6 C op 4 ve rs ity ev ie w C op y 3 [3] In the expansion of ( a + 2 x )6, the coefficient of x is equal to the coefficient of x 2. Find the value of the constant a. -C 2 am br id 1 w END-OF-CHAPTER REVIEW EXERCISE 6 -R am br ev 11 The first term of an arithmetic progression is 1.75 and the second term is 1.5. The sum of the first n terms is −n . Find the value of n. [4] s b the first term c the sum to infinity. es the common ratio [3] Pr a [1] ni ve rs ity [2] Find d in terms of a. b Write down an expression, in terms of a, for the 50th term. [3] [2] id g w e U a C op y 13 An arithmetic progression has first term a and common difference d . The sum of the first 100 terms is 25 times the sum of the first 20 terms. ev br -R s Given that the nth term of the progression is −19, find the value of n. es b Find the first term of the progression and the common difference. am a ie 14 The tenth term of an arithmetic progression is 17 and the sum of the first five terms is 190. -C R ev ie w C op y -C 12 The second term of a geometric progression is −1458 and the fifth term is 432. Find: Copyright Material - Review Only - Not for Redistribution [4] [2] ve rs ity ev ie [3] The seventh term of an arithmetic progression is 19 and the sum of the first twelve terms is 224. Find the fourth term. [4] -R The fifth term of an arithmetic progression is 18 and the sum of the first eight terms is 186. Find the first term and the common difference. 1 The first term of a geometric progression is 32 and the fourth term is . Find the 2 sum to infinity of the progression. ve rs ity y C op [3] [3] An arithmetic progression has first term −4. The nth term is 8 and the (2 n )th term is 20.8. Find the value of n. [4] br ev id ie w A second geometric progression has first term 5a, common ratio 3r and sum to infinity 10S . Find the value of r. -R am b [4] A geometric progression has first term a, common ratio r and sum to infinity S. ge 17 a A geometric progression has first term 3 and common ratio r. A second geometric progression has 1 first term 2 and common ratio r. The two progressions have the same sum to infinity, S. Find the 5 value of r and the value of S. ni R ev ie w b U C op y 16 a Pr es s -C b am br id 15 a w ge C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 Pr Model 1: Increase the prize money by $1000 each day. Model 2: Increase the prize money by 10% each day. ve ie w rs C 184 ity op y es s -C 18 A television quiz show takes place every day. On day 1 the prize money is $1000. If this is not won the prize money is increased for day 2. The prize money is increased in a similar way every day until it is won. The television company considered the following two different models for increasing the prize money. if Model 1 is used, ii if Model 2 is used. br y op ev id ie [3] -R am Cambridge International AS & A Level Mathematics 9709 Paper 11 Q8 June 2011 es s -C 19 a The first two terms of an arithmetic progression are 1 and cos2 x respectively. Show that the sum of the first ten terms can be expressed in the form a − b sin2 x, where a and b are constants to be found. [3] b The first two terms of a geometric progression are 1 and 1 tan2 θ respectively, 3 1 where 0 < θ < π . 2 [2] i Find the set of values of θ for which the progression is convergent. 1 ii Find the exact value of the sum to infinity when θ = π . [2] 6 Cambridge International AS & A Level Mathematics 9709 Paper 11 Q7 June 2012 op y ni ve rs ity Pr op y C w ev ie id g -R For the case where the progression is geometric with a sum to infinity of 8, find the third term. [4] [4] s Cambridge International AS & A Level Mathematics 9709 Paper 11 Q8 November 2015 es am ii For the case where the progression is arithmetic with a common difference of 12, find the possible values of x and the corresponding values of the third term. br i e U 20 The first term of a progression is 4x and the second term is x 2. -C C w ie ev R [4] w ge i C U R ni ev On each day that the prize money is not won the television company makes a donation to charity. The amount donated is 5% of the value of the prize on that day. After 40 days the prize money has still not been won. Calculate the total amount donated to charity Copyright Material - Review Only - Not for Redistribution ve rs ity am br id ev ie w ge C U ni op y Chapter 6: Series -R 1 2 21 a The third and fourth terms of a geometric progression are and respectively. Find the 3 9 sum to infinity of the progression. Pr es s ni C op b A geometric progression has first term a, common ratio r and sum to infinity 6. A second geometric progression has first term 2a, common ratio r 2 and sum to infinity 7. Find the values of a and r. U w ge [5] 185 w rs C ity op Pr y es s -C -R am br ev id ie Cambridge International AS & A Level Mathematics 9709 Paper 11 Q9 November 2013 ni op y ve ie ev y -R s es am br ev ie id g w e C U op ie w ni ve rs C ity Pr op y es s -C -R am br ev id ie w ge C U R ev R -C [5] y 22 a In an arithmetic progression the sum of the first ten terms is 400 and the sum of the next ten terms is 1000. Find the common difference and the first term. R [4] Cambridge International AS & A Level Mathematics 9709 Paper 11 Q7 June 2015 ve rs ity ev ie w C op y -C b A circle is divided into 5 sectors in such a way that the angles of the sectors are in arithmetic progression. Given that the angle of the largest sector is 4 times the angle of the smallest sector, find the angle of the largest sector. [4] Copyright Material - Review Only - Not for Redistribution ve rs ity ge C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 ev ie am br id 1 w CROSS-TOPIC REVIEW EXERCISE 2 3 6 2 a Find the first three terms in the expansion of 3x − 2 , in descending powers of x. x y op [2] a Find the first three terms when (1 − 2 x )5 is expanded, in ascending powers of x. [3] C op ni U w id am br ev b Find the sum to infinity. -R [3] Pr op b Find the sum to infinity. [2] ity The sum of the 1st and 2nd terms of a geometric progression is 50 and the sum of the 2nd and 3rd terms is 30. Find the sum to infinity. ve ie w rs C s es -C y a Find the value of k. 7 [2] The first three terms of a geometric progression are 3k + 14, k + 14 and k, respectively. All the terms in the progression are positive. 186 [3] ie a Find the fourth term. 6 [2] The first term of a geometric progression is 50 and the second term is −40. ge R Find the value of a. y b In the expansion of (3 + ax )(1 − 2 x )5, the coefficient of x 2 is zero. 5 [3] 6 2 2 b Hence, find the coefficient of x 2 in the expansion of 1 + 3x − . x x ve rs ity C [3] Pr es s -C -R 1 Find the term independent of x in the expansion of 4x − 2 . x 4 [2] 6 2 ev ie w 4 Find the highest power of x in the expansion of (5x 4 + 3)8 + (1 − 3x 3 )5 (4x 2 − 5x5 )6 . [6] y op i Show that cos x ≡ 1 − 2 sin2 x + sin 4 x . U R 8 ni ev Cambridge International AS & A Level Mathematics 9709 Paper 11 Q5 November 2016 4 [5] w ge C ii Hence, or otherwise, solve the equation 8 sin 4 x + cos 4 x = 2 cos2 x for 0° < x < 360°. [1] ie a Show that the area, A cm 2, of the sector is given by A = 30 r − r 2. [2] b Express 30 r − r 2 in the form a − ( r − b )2, where a and b are constants. [2] -R [1] Pr c find the value of r at which A is a maximum es s -C op y Given that r can vary: ity [1] -R s es -C am br ev ie id g w e C U op ie ev R y ni ve rs d find this stationary value of A. w C ev br A sector of a circle, radius r cm, has a perimeter of 60 cm. am 9 id Cambridge International AS & A Level Mathematics 9709 Paper 11 Q6 November 2016 Copyright Material - Review Only - Not for Redistribution ve rs ity -C r cm ve rs ity The diagram shows a metal plate consisting of a rectangle with sides x cm and r cm and two identical sectors of a circle of radius r cm. The perimeter of the plate is 100 cm. [2] b Express 50 r − r 2 in the form a − ( r − b )2, where a and b are constants. [2] C op U w ge c find the value of r at which A is a maximum ev id br -R es s -C am rm lm Pr y The diagram shows a running track. The track has a perimeter of 400 m and consists of two straight sections of length l m and two semicircular sections of radius r m. ity op C [1] y C U [1] y -R s es -C am br ev ie id g w e C U op ie w ni ve rs C ity Pr op y es s -C -R am br ev id ie d find this stationary value of A. ev [3] [2] w ge c show that A has a maximum value when l = 0 R [2] op ni ev ve ie w rs a Show that the area, A m2, of the region enclosed by the track is given by A = 400 r − π r 2 . 2 a b b Express 400 r − πr 2 in the form − π r − , where a and b are constants. π π Given that l and r can vary: R [1] ie d find this stationary value of A. 11 y a Show that the area, A cm2, of the plate is given by A = 50 r − r 2. ni ev ie w C op y x cm Given that r can vary: R r cm Pr es s 10 -R am br id ev ie w ge C U ni op y Cross-topic review exercise 2 Copyright Material - Review Only - Not for Redistribution 187 ve rs ity C1 op y Pr es s -C ev ie w C ve rs ity P 8 cm T Q 2 cm w ge U R ni C op C2 y 12 -R am br id ev ie w ge C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 S ie id R -C -R am br ev The diagram shows two circles, C1 and C2 , touching at the point T . Circle C1 has centre P and radius 8 cm; circle C2 has centre Q and radius 2 cm. Points R and S lie on C1 and C2 respectively, and RS is a tangent to both circles. [2] es s i Show that RS = 8 cm. op C S y op U r T ge P ie id 2θ O br C ni ev R B w A ev 13 rs Cambridge International AS & A Level Mathematics 9709 Paper 11 Q9 November 2010 ve w [4] ity iii Find the area of the shaded region. ie [2] Pr y ii Find angle RPQ in radians correct to 4 significant figures. 188 s -C -R am In the diagram, OAB is an isosceles triangle with OA = OB and angle AOB = 2θ radians. Arc PST has centre O and radius r, and the line ASB is a tangent to the arc PST at S. Pr ity [5] ni ve rs y The function f is such that f( x ) = 2 sin2 x − 3 cos2 x for 0 < x < π. e C ii State the greatest and least values of f( x ). w ie [2] [2] [3] ev Cambridge International AS & A Level Mathematics 9709 Paper 11 Q5 June 2010 es s -R id g iii Solve the equation f( x ) + 1 = 0. br op U i Express f( x ) in the form a + b cos2 x , stating the values of a and b. am 14 [4] Cambridge International AS & A Level Mathematics 9709 Paper 11 Q9 June 2011 -C R ev ie w C op y es i Find the total area of the shaded regions in terms of r and θ . 1 ii In the case where θ = π and r = 6 , find the total perimeter of the shaded regions, leaving your 3 answer in terms of 3 and π. Copyright Material - Review Only - Not for Redistribution ve rs ity am br id ev ie w ge C U ni op y Cross-topic review exercise 2 sin θ 1 1 − ≡ . 1 − cos θ sin θ tan θ sin θ 1 − = 4 tan θ for 0° , θ , 180°. ii Hence solve the equation 1 − cos θ sin θ Pr es s ve rs ity The function f is defined by f : x ֏ 4 sin x − 1 for − i State the range of f . w [3] Cambridge International AS & A Level Mathematics 9709 Paper 11 Q9 June 2014 y op C 16 [4] -R i Prove the identity -C 15 1 1 π < x < π. 2 2 [2] y ev ie ii Find the coordinates of the points at which the curve y = f( x ) intersects the coordinate axes. R ni C op iii Sketch the graph of y = f( x ) . −1 −1 ge U iv Obtain an expression for f ( x ), stating both the domain and range of f . [4] w ie id ev a The first term of a geometric progression in which all the terms are positive is 50. The third term is 32. Find the sum to infinity of the progression. -R am [3] Pr y es s -C b The first three terms of an arithmetic progression are 2 sin x , 3 cos x and (sin x + 2 cos x ) respectively, where x is an acute angle. 4 [3] i Show that tan x = . 3 ii Find the sum of the first twenty terms of the progression. [3] ity Cambridge International AS & A Level Mathematics 9709 Paper 11 Q9 June 2016 y op y op -R s es -C am br ev ie id g w e C U R ev ie w ni ve rs C ity Pr op y es s -C -R am br ev id ie w ge C U R ni ev ve ie w rs C op [2] Cambridge International AS & A Level Mathematics 9709 Paper 11 Q11 June 2016 br 17 [3] Copyright Material - Review Only - Not for Redistribution 189 op y ve rs ity ni C U ev ie w ge -R am br id Pr es s -C y ni C op y ve rs ity op C w ev ie C op y ve ni w ge C U R ev ie w rs Chapter 7 Differentiation ity op Pr y es s -C -R am br ev id ie w ge U R 190 id es s -C -R am br ev understand that the gradient of a curve at a point is the limit of the gradients of a suitable sequence of chords dy d2 y and use the notations f ′( x ) , f ′′( x ) , for the first and second derivatives dx dx 2 use the derivative of x n (for any rational n), together with constant multiples, sums, differences of functions, and of composite functions using the chain rule apply differentiation to gradients, tangents and normals. y op -R s es -C am br ev ie id g w e C U R ev ie w ni ve rs C ity Pr op y ■ ■ ■ ■ ie In this chapter you will learn how to: Copyright Material - Review Only - Not for Redistribution ve rs ity ge C U ni op y Chapter 7: Differentiation am br id ev ie w PREREQUISITE KNOWLEDGE What you should be able to do IGCSE / O Level Mathematics Use the rules of indices to simplify expressions to the form ax n . a 3x x ni C op y ve rs ity b 5 3 x2 x c 2 x 1 d 2x 3 e x2 2 f − 2x 3 5 x ev id ie k in the form ( ax + b ) n w U ge Write -C -R am br k ( ax + b )− n . Find the gradient of a perpendicular line. es Pr y ity op Find the equation of a line with a given gradient and a given point on the line. y 4 Find the equation of the line with gradient 2 that passes through the point (2, 5). op ni ev ve ie w rs Chapter 3 2 Write in the form k ( ax + b )− n : 4 a ( x − 2)3 2 b (3x + 1)5 3 The gradient of a line is 2 . 3 Write down the gradient of a line that is perpendicular to it. s Chapter 3 C U R 1 Write in the form ax n : Pr es s -C y op C w ev ie R IGCSE / O Level Mathematics C Check your skills -R Where it comes from w ge Why do we study differentiation? -R am br ev id ie Calculus is the mathematical study of change. Calculus has two basic tools, differentiation and integration, and it has widespread uses in science, medicine, engineering and economics. A few examples where calculus is used are: designing effective aircraft wings ● the study of radioactive decay ● the study of population change ● modelling the financial world. In this chapter you will be studying the first of the two basic tools of calculus. You will learn the rules of differentiation and how to apply these to problems involving gradients, tangents and normals. In Chapter 8 you will then learn how to apply these rules of differentiation to more practical problems. y ni ve rs op 7.1 Derivatives and gradient functions br ev ie id g w e C U At IGCSE / O Level you learnt how to estimate the gradient of a curve at a point by drawing a suitable tangent and then calculating the gradient of the tangent. This method only gives an approximate answer (because of the inaccuracy of drawing the tangent) and it is also very time consuming. -R s es am In this chapter you will learn a method for finding the exact gradient of the graph of a function (which does not involve drawing the graph). This exact method is called differentiation. -C R ev ie w C ity Pr op y es s -C ● Copyright Material - Review Only - Not for Redistribution WEB LINK Try the Calculus resources on the Underground Mathematics website. 191 ve rs ity ev ie am br id EXPLORE 7.1 w ge C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 -R Consider the quadratic function y = x 2 and a point P ( x, x 2 ) on the curve. y = x2 y Pr es s -C 1 Let P be the point (2, 4). op y The points A(2.2, 4.84), B (2.1, 4.41) and C (2.01, 4.0401) also lie on the curve and are close to the point P (2, 4). ii the chord PB ev ie O ni iii the chord PC. x y the chord PA C op w i P (2, 4) ve rs ity C a Calculate the gradient of: U R b Discuss your results with those of your classmates and make suggestions as to what is happening. id ie w ge c Suggest a value for the gradient of the curve y = x 2 at the point (2, 4). br ev 2 Let P be the point (3, 9). -C -R am The points A(3.2, 10.24), B (3.1, 9.61) and C (3.01, 9.0601) also lie on the curve and are close to the point P (3, 9). es Pr ii the chord PB iii the chord PC. w rs C the chord PA ity op y i U R ni ev c Suggest a value for the gradient of the curve y = x 2 at the point (3, 9). y ve ie b Discuss your results with those of your classmates and make suggestions as to what is happening. op 192 s a Calculate the gradient of: w ge C 3 Use a spreadsheet to investigate the value of the gradient at other points on the curve y = x 2. -R am br ev id ie 4 Can you suggest a general formula for the gradient of the curve y = x 2 at the point ( a, a 2 )? What would be the gradient at ( x, x 2 )? es Pr e y = x2 op y δy δx x ie id g O A C P U R ev ie w ni ve rs C ity y w op y Take a point P ( x, y ) on the curve y = x 2 and a point A that is close to the point P. -R s es am br ev The coordinates of A are ( x + δx, y + δy ), where δx is a small increase in the value of x and δy is the corresponding small increase in the value of y. -C WEB LINK s -C The general formula for the gradient of the curve y = x 2 at the point (x, x 2 ) can be proved algebraically. Copyright Material - Review Only - Not for Redistribution There are other ways of thinking about the gradient of a curve. Try the following resources on the Underground Mathematics website Zooming in and Mapping a derivative. ve rs ity C U ni op y Chapter 7: Differentiation ( 2 y op x 2 + 2 x δx + ( δx ) − x 2 δx 2 ve rs ity C We use the Greek symbol delta, δ, to denote a very small change in a quantity. -R ( x + δx ) − x 2 x δx + ( δx ) = δx = 2 x + δx w ev ie ( x + δx )2 − x2 2 = TIP Pr es s -C = ). ev ie y2 − y1 x2 − x1 am br id Gradient of chord PA = w ge We can also write the coordinates of P and A as ( x, x 2 ) and x + δx, ( x + δx ) U R ni C op y As δx tends towards 0, A tends to P and the gradient of the chord PA tends to a value. We call this value the gradient of the curve at P. w ge In this case, therefore, the gradient of the curve at P is 2x. id ie This process of finding the gradient of a curve at any point is called differentiation. s -C -R am br ev Later in this chapter, you will learn some rules for differentiating functions without having to calculate the gradients of chords as we have done here. The process of calculating gradients using the limit of gradients of chords is sometimes called differentiation from first principles. es Notation rs y op -C -R am br ev id ie w ge C U R ni ev ve ie w C ity op Pr y There are three different notations that are used to describe the previous rule. dy = 2 x. 1. If y = x 2, then dx 2. If f( x ) = x 2, then f ′( x ) = 2 x. d ( x2 ) = 2x 3. dx dy is called the derivative of y with respect to x. If y is a function of x, then dx Likewise f ′( x ) is called the derivative of f(x). dy or f ′( x ) is sometimes also called the If y = f( x ) is the graph of a function, then dx gradient function of this curve. d ( x 2 ) = 2 x means ‘if we differentiate x 2 with respect to x, the result is 2 x’. dx Pr ni ve rs ity EXPLORE 7.2 y 1 Use a spreadsheet to investigate the gradient of the curve y = x 3. C U op 2 Can you suggest a general formula for the gradient of the curve y = x 3 at the point ( x, x 3 )? -R s es am br ev ie id g w e 3 Differentiate y = x 3 from first principles to confirm your answer to question 2. -C R ev ie w C op y es s You do not need to be able to differentiate from first principles but you are expected to understand that the gradient of a curve at a point is the limit of a suitable sequence of chords. Copyright Material - Review Only - Not for Redistribution DID YOU KNOW? Gottfried Wilhelm Leibniz and Isaac Newton are both credited with developing the modern calculus that we use today. Leibniz’s notation for derivatives dy was . Newton’s dx dy notation for dx was yɺ . The notation f ′( x ) is known as Lagrange’s notation. 193 ve rs ity C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 w ge Differentiation of power functions d ( x 6 ) = 6x 5 dx Pr es s d ( x 5 ) = 5x 4 dx -C d ( x 4 ) = 4x 3 dx -R am br id ev ie d d ( x 3 ) = 3x 2. ( x 2 ) = 2 x and that dx dx Investigating the gradient of the curves y = x 4 , y = x5 and y = x 6 would give the results: We now know that op y This leads to the general rule for differentiating power functions: d ( x n ) = nx n – 1 dx y ve rs ity U R ni C op This is true for any real power n, not only for positive integer values of n. id op Pr y es s -C -R am br ev So for the earlier example where y = x 2: dy = 2 × x 2 −1 dx = 2 x1 = 2x y op U w ie ev s -R br am -C id g 1 1 −2 x 2 1 = 2 x e 1 1 2 −1 x 2 1 Multiply by the power and then subtract 2 one from the power. C 1 = 1 Write x as x 2 . x f( x ) = x 2 f ′( x ) = ev es Pr ni ve rs f( x ) = Multiply by the power −2 and then subtract one from the power. −3 ity op y = −2 x 2 =− 3 x 1 as x −2. x2 s am Write -R br id ie w Multiply by the power 7 and then subtract one from the power. = −2 x −2 − 1 C w ie ev c y=2 d x C ge d ( x7 ) = 7x7 − 1 dx = 7x6 d 1 d = ( x −2 ) dx x 2 dx -C b f( x ) = U ni ev R Answer a R c y rs Find the derivative of each of the following. 1 b a x7 x2 ve ie w C ity WORKED EXAMPLE 7.1 es 194 w ‘Multiply by the power n and then subtract one from the power.’ ie ge You may find it easier to remember this rule as: op ev ie w C KEY POINT 7.1 Copyright Material - Review Only - Not for Redistribution ve rs ity ev ie Multiply by the power and then subtract one from the power. -R y = 2x 0 TIP w ge Write 2 as 2 x 0 . am br id y=2 d C U ni op y Chapter 7: Differentiation op y Pr es s -C dy = 0x 0 − 1 dx =0 It is worth remembering that when you differentiate a constant, the answer is always 0. Scalar multiple rule C op ge U KEY POINT 7.2 R y If k is a constant and f( x ) is a function then: ni ev ie w C ve rs ity You need to know and be able to use the following two rules. -R am br ev id ie w d d [ kf( x )] = k [f( x )] dx dx Addition/subtraction rule es s -C If f( x ) and g( x ) are functions then ity 195 y C U w ge 1 4 + + 5 with respect to x. x 2x2 br ev id ie R WORKED EXAMPLE 7.2 Differentiate 3x 4 − op ni ev ve rs d d d [f( x ) ± g( x )] = [f( x )] ± [g( x )] dx dx dx ie w C op Pr y KEY POINT 7.3 -R am Answer Pr 1 d 1 d d −2 d (x4 ) − ( x −2 ) + 4 x +5 (x0 ) dx 2 dx dx dx 1 −3 1 ( −2 x −3 ) + 4 − x 2 + 5(0 x −1 ) 2 2 ie w ni ve rs = 3(4x 3 ) − ity C op y =3 es s -C 1 − d 4 1 4 d 4 1 −2 3x − 3x − x + 4x 2 + 5x 0 + + 5 = 2 dx dx 2 x 2x − 3 y op 2 x3 C 1 − x3 -R s es -C am br ev ie id g e U R = 12 x 3 + w ev = 12 x 3 + x −3 − 2 x 2 Copyright Material - Review Only - Not for Redistribution ve rs ity ev ie w ge am br id WORKED EXAMPLE 7.3 C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 -C Answer -R Find the gradient of the tangent to the curve y = x(2 x − 1)( x + 3) at the point (1, 4). y = 2 x + 5x − 3x 2 y 3 y C op w ie br ev id WORKED EXAMPLE 7.4 ge U ni ev ie w C ve rs ity op dy = 6x 2 + 10 x − 3 dx dy = 6(1)2 + 10(1) − 3 When x = 1, dx dx = 13 Gradient of curve at (1, 4) is 13. R Expand brackets and simplify. Pr es s y = x(2 x − 1)( x + 3) -R am The curve y = ax 4 + bx 2 + x has gradient 3 when x = 1 and gradient −51 when x = −2. es s -C Find the value of a and the value of b. Pr y Answer op y = ax 4 + bx 2 + x ity dy = 4ax 3 + 2bx + 1 dx dy Since = 3 when x = 1: dx rs y ve op 4a (1)3 + 2b(1) + 1 = 3 4a + 2b = 2 2a + b = 1 -R s es am op y -C 4a ( −2)3 + 2b( −2) + 1 = −51 −32 a − 4b = −52 8a + b = 13 ev id ie dy = −51 when x = −2: dx br w ity ni ve rs C ∴a = 2 Substitute a = 2 into (1): 4 + b = 1 (2) Pr (2) − (1) gives 6a = 12 y op -R s -C am br ev ie id g w e C U R ev ie ∴ b = −3 es Since w ge (1) C U R ni ev ie w C 196 Copyright Material - Review Only - Not for Redistribution ve rs ity ev ie am br id EXERCISE 7A w ge C U ni op y Chapter 7: Differentiation -R 1 The points A(0, 0), B(0.5, 0.75), C(0.8, 1.44), D(0.95, 1.8525), E(0.99, 1.9701) and F(1, 2) lie on the curve y = f(x ). AF BF Gradient 2 2.5 CF DF b Use the values in the table to predict the value of w ev ie Pr es s Chord ve rs ity C op y -C a Copy and complete the table to show the gradients of the chords CF, DF and EF. y C op d y= c ev f 3 g es -C x2 b f( x ) = 3x5 e f( x ) = 5 3x 2 f f( x ) = −2 Pr c ve g 6 x 2 4x f( x ) = x f( x ) = a d y = ( x + 5)( x − 4) h w )2 y = 7 − 3x + 5x 2 ev f br 2x − 5 x2 y= 4x 2 + 3x − 2 x es s -C -R am y = 7 x2 − y= ie ( y = 2x2 − 3 c C U ge id e op C dy when x = 2. dx id g e 8 Given that xy = 12, find the value of w U 7 Find the gradient of the curve y = (2 x − 5)( x + 4) at the point (3, 7) . y ni ve rs ity Pr op y -R s es am br ev ie 9 Find the gradient of the curve y = 5x 2 − 8x + 3 at the point where the curve crosses the y-axis. -C C w 3 x 2x x f( x ) = 3x 3 f( x ) = d 5 1 3 2 + h y = 3x + − i x x2 x 2 x dy for each curve at the given point. 6 Find the value of dx a y = x 2 + x − 4 at the point (1, −2) 2 b y =5− at the point (2, 4) x 3x − 2 at the point ( −2, −2) c y= x2 g ie h 197 dy for each of the following. dx y = 5x 2 − x + 1 b y = 2 x 3 + 8x − 4 5 Find R ity f( x ) = 2 x 4 rs a x3 × x2 ni ev ie w C op 4 Find f ′( x ) for each of the following. 1 x x5 x2 d y 8 y e x −4 op x9 12 at (2, 6) x s am b x5 w ge id y = x 2 − 2 x + 3 at (0, 3) ie y = 2 x at (4, 4) b -R ni c U y = x 4 at (1, 1) br R a a ev dy when x = 1. dx 2 By considering the gradient of a suitable sequence of chords, find a value for the gradient of the curve at the given point. 3 Differentiate with respect to x: R EF Copyright Material - Review Only - Not for Redistribution ve rs ity C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 ev ie 11 Find the gradient of the curve y = the x-axis. 5 x − 10 at the point where the curve crosses x2 -R am br id w ge 10 Find the coordinates of the points on the curve y = x 3 − 3x − 8 where the gradient is 9. Pr es s -C 12 The curve y = x 2 − 4x − 5 and the line y = 1 − 3x meet at the points A and B. b Find the gradient of the curve at each of the points A and B. ve rs ity 13 The gradient of the curve y = ax 2 + bx at the point (3, −3) is 5. Find the value of a and the value of b. y ev ie w C op y a Find the coordinates of the points A and B. U R ni C op 14 The gradient of the curve y = x 3 + ax 2 + bx + 7 at the point (1, 5) is −5. Find the value of a and the value of b. ge b has gradient 16 when x = 1 and gradient −8 when x2 x = −1. Find the value of a and the value of b. br ev id ie w 15 The curve y = ax + -C -R am 16 Given that the gradient of the curve y = x 3 + ax 2 + bx + 3 is zero when x = 1 and when x = 6, find the value of a and the value of b. s dy , 0. dx dy 18 Given that y = 4x 3 + 3x 2 − 6x − 9, find the range of values of x for which > 0. dx 19 A curve has equation y = 3x 3 + 6x 2 + 4x − 5. Show that the gradient of the curve is never negative. Try the following resources on the Underground Mathematics website: • Slippery slopes • Gradient match. op y ve ni 7.2 The chain rule U R ev ie w rs C 198 ity op Pr y es 17 Given that y = 2 x 3 − 3x 2 − 36x + 5, find the range of values of x for which WEB LINK ev id ie w ge C To differentiate y = (3x − 2)7, we could expand the brackets and then differentiate each term separately. This would take a long time to do. There is a more efficient method available that allows us to find the derivative without expanding. -R am br Let u = 3x − 2, then y = (3x − 2)7 becomes y = u7. This means that y has changed from a function in terms of x to a function in terms of u. op y es s -C We can find the derivative of the composite function y = (3x − 2)7 using the chain rule: Try the Chain mapping resource on the Underground Mathematics website. ity C -R s es -C am br ev ie id g w e C U op ev R y ni ve rs dy dy du = × du dx dx ie w WEB LINK Pr KEY POINT 7.4 Copyright Material - Review Only - Not for Redistribution ve rs ity ge C U ni op y Chapter 7: Differentiation am br id ev ie w WORKED EXAMPLE 7.5 y = (3x − 2)7 Let u = 3x − 2 so y = u7 and dy = 7u6 du ve rs ity y y ni U R = 7(3x − 2)6 × 3 Use the chain rule. C op ev ie w C op du =3 dx dy dy du = × dx du dx = 7u6 × 3 Pr es s -C Answer -R Find the derivative of y = (3x − 2)7. id ie w ge = 21(3x − 2)6 br ev With practice you will be able to do this mentally. -R am Consider the ‘inside’ of (3x − 2)7 to be 3x − 2. y op Pr 3 Step 2: Differentiate the ‘inside’: s 7(3x − 2)6 Step 1: Differentiate the ‘outside’: es -C To differentiate (3x − 2)7: 199 rs op es y ni ve rs -R s -C am br ev ie id g w e C U R = −10(3x 2 + 1)−6 × 6x 60 x =− (3x 2 + 1)6 ity Pr Use the chain rule. op am C w ie C w dy = −10 u −6 du ev and op y -C du = 6x dx dy dy du = × dx du dx = −10 u −6 × 6x -R y = 2 u −5 s so Let u = 3x 2 + 1 ev y ve ni id 2 (3x 2 + 1)5 br y= ie Answer 2 . (3x + 1)5 2 ge R Find the derivative of y = U ev ie WORKED EXAMPLE 7.6 es w C ity Step 3: Multiply these two expressions: 21(3x − 2)6 Copyright Material - Review Only - Not for Redistribution ve rs ity w ge C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 6x ev ie Step 2: Differentiate the ‘inside’: Pr es s −10(3x 2 + 1)−6 60 x (3x 2 + 1)6 The curve y = ax + b passes through the point (12, 4) and has gradient 1 at this point. 4 ie (1) -C 1 y = ( ax + b ) 2 and dy 1 − 2 = u du 2 ity rs y op ni C U ie w ge Substitute x = 12 and ev (2) -R am br id ev Use the chain rule. ve 1 1 −2 u ×a 2 a dy = dx 2 ax + b a 1 = 4 2 12 a + b 2 a = 12 a + b = R es y = u2 dy dy du = × dx du dx s es Pr ity 4 = 24 + b 16 = 24 + b b = −8 ∴ a = 2, b = −8 y op -R s es -C am br ev ie id g w e C U R ni ve rs ie w C op y -C (1) and (2) give 2 a = 4 a=2 Substituting a = 2 into (1) gives: ev ax + b in the form ( ax + b ) 2 . Pr y op C w ie so 1 du =a dx 1 Write s 1 Let u = ax + b 200 -R am 4 = 12 a + b Substitute x = 12 and y = 4. ev ax + b br y= id Answer w ge U R Find the value of a and the value of b. y WORKED EXAMPLE 7.7 ni ev ie w C ve rs ity op y Step 3: Multiply the two expressions: −60 x(3x 2 + 1)−6 = − C op -C Step 1: Differentiate the ‘outside’: -R am br id Alternatively, to differentiate the expression mentally: 2 as 2(3x 2 + 1)−5. Write 2 (3x + 1)5 Copyright Material - Review Only - Not for Redistribution dy 1 = . dx 4 ve rs ity ev ie (2 x + 3)8 e (5x − 2)8 4 f 5(2 x − 1)5 i ( x 2 + 3)5 j (2 − x 2 )8 2 Differentiate with respect to x: 1 a b x+2 4 f e (3x + 1)6 3 Differentiate with respect to x: 3 x−5 3 2(3x + 1)5 ni ve rs ity -C y op C w ev ie (3 − 4x )5 d g 2(4 − 7 x )4 h k ( x 2 + 4 x )3 l c g 2x + 3 c f 2 3x + 1 8 3 − 2x 8 x2 + 2 x 2 x2 − 1 1 2x − 5 d h w br 5 − 2x b ie 3 e id x−5 a d g h ev ge U R c y b Pr es s ( x + 4)6 a -R 1 Differentiate with respect to x: C op am br id EXERCISE 7B w ge C U ni op y Chapter 7: Differentiation 1 x + 1 2 9 1 (3x − 1)7 5 5 x2 − 5 x 16 x2 + 2 7 (2 x 2 − 5 x )7 x 3 − 5x 6 3 2 − 3x -R am 4 Find the gradient of the curve y = (2 x − 3)5 at the point (2, 1). 6 at the point where the curve crosses the y-axis. ( x − 1)2 3 6 Find the gradient of the curve y = x − at the points where the curve crosses the x-axis. x+2 es Pr y op ity 7 Find the coordinates of the point on the curve y = y ve ie C U O w x ni ve rs ity Pr op y es normal C dy at the point A( x1, y1 ) is m, then the equation of the tangent dx at A is given by: ev ie id g br KEY POINT 7.5 w e C U If the value of es s -R y − y1 = m( x − x1 ) am y The line perpendicular to the tangent at the point A is called the normal at A. -C w ie -R s -C A(x1, y1) ie id br am tangent ev ge y = f(x) y op R ni ev op w a 3 passes through the point (2, 1) and has gradient − at this point. bx − 1 5 Find the value of a and the value of b. 8 The curve y = 7.3 Tangents and normals ev R ( x − 10 x + 26) where the gradient is 0. 2 rs C s -C 5 Find the gradient of the curve y = Copyright Material - Review Only - Not for Redistribution TIP We use the numerical form for m in this formula (not the derivative formula). 201 ve rs ity C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 am br id ev ie w ge The normal at the point ( x1, y1 ) is perpendicular to the tangent, so the gradient of the 1 normal is − and the equation of the normal is given by: m -R KEY POINT 7.6 Pr es s -C 1 ( x − x1 ) m op y y − y1 = − y ie id Answer 8 − 9 at the point where x = 2 . x2 w ge U R Find the equation of the tangent and the normal to the curve y = 2 x 2 + C op WORKED EXAMPLE 7.8 ni ev ie w C ve rs ity This formula only makes sense when m ≠ 0. If m = 0, it means that the tangent is horizontal and the normal is vertical, so it has equation x = x1 instead. s -C -R am br ev y = 2 x 2 + 8x −2 − 9 dy = 4x − 16x −3 dx When x = 2, y = 2(2)2 + 8(2)−2 − 9 = 1 es dy = 4(2) − 16(2)−3 = 6 dx Tangent: passes through the point (2, 1) and gradient = 6 Pr y ve ni C br ev id ie w ge R 1 y − 1 = − ( x − 2) 6 x + 6y = 8 1 6 U ev Normal: passes through the point (2, 1) and gradient = − op w ie rs y − 1 = 6( x − 2) y = 6x − 11 C 202 ity op y and -R am WORKED EXAMPLE 7.9 )3 ( s es -C A curve has equation y = 4 − x . Pr op y The normal at the point P (4, 8) and the normal at the point Q(9, 1) intersect at the point R. ni ve rs y ( ( )2 = −3 ( )2 3 4− 9 dy 1 =− =− dx 2 9 2 Copyright Material - Review Only - Not for Redistribution s -C When x = 9, es am br 3 4− 4 dy =− When x = 4, dx 2 4 ev ) -R id g e ( )2 ie 1 3 4− x 2 1 − dy = 3 4− x − x 2 = − dx 2 x 2 op U )3 C ( y = 4− x w Answer a ity b Find the area of triangle PQR. R ev ie w C a Find the coordinates of R. ve rs ity 1 3 am br id ev ie Normal at P: passes through the point (4, 8) and gradient = w ge C U ni op y Chapter 7: Differentiation 1 ( x − 4) 3 3 y = x + 20 -R y−8= (1) y = 2 x − 17 (2) Solving equations (1) and (2) gives: y ni C op w 3(2 x − 17) = x + 20 x = 14.2 ev ie Pr es s y − 1 = 2( x − 9) ve rs ity C op y -C Normal at Q: passes through the point (9, 1) and gradient = 2 U R When x = 14.2, y = 2(14.2) − 17 = 11.4 w 10.4 Q(9, 1) 5.2 5 203 x ity O Pr 7 y op es s -C P(4, 8) -R am 3.4 C ie R(14.2, 11.4) 10.2 ev y br b id ge Hence, R is the point (14.2, 11.4). ve ie w rs Area of triangle PQR = area of rectangle − sum of areas of outside triangles y op C ie w = 44.2 units2 br -C -R am EXERCISE 7C ev id ge U R ni ev 1 1 1 = 10.2 × 10.4 − × 5 × 7 + × 5.2 ×10.4 + × 10.2 × 3.4 2 2 2 = 106.08 − [ 17.5 + 27.04 + 17.34 ] s b y = (2 x − 5)4 at the point (2, 1) x3 − 5 y= at the point ( −1, 6) x y = 2 x − 5 at the point (9, 4) ity y op d ni ve rs c es y = x 2 − 3x + 2 at the point (3, 2) Pr a C w e ie ev -R s es d id g c br b y = 3x 3 + x 2 − 4x + 1 at the point (0, 1) 3 y= 3 at the point ( −2, −3) x +1 y = (5 − 2 x )3 at the point (3, −1) 20 y= 2 at the point (3, 2) x +1 am a U 2 Find the equation of the normal to each curve at the given point. -C R ev ie w C op y 1 Find the equation of the tangent to each curve at the given point. Copyright Material - Review Only - Not for Redistribution ve rs ity C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 am br id ev ie w ge 8 1 . 3 A curve passes through the point A 2, and has equation y = 2 ( x + 2)2 a Find the equation of the tangent to the curve at the point A. -R b Find the equation of the normal to the curve at the point A. Pr es s -C 4 The equation of a curve is y = 5 − 3x − 2 x 2 . C op ni ev ie 5 The normal to the curve y = x 3 − 5x + 3 at the point ( −1, 7) intersects the y-axis at the point P. Find the coordinates of P. U R y b Find the coordinates of the point at which the normal meets the curve again. ve rs ity w C op y a Show that the equation of the normal to the curve at the point ( −2, 3) is x + 5 y = 13. id ie w ge 6 The tangents to the curve y = 5 − 3x − x 2 at the points ( −1, 7) and ( −4, 1) meet at the point Q. -R am br ev Find the coordinates of Q. s -C 7 The normal to the curve y = 4 − 2 x at the point P (16, −4) meets the x-axis at the point Q. es b Find the coordinates of Q. ity 10 8 The equation of a curve is y = 2 x − 2 + 8. x dy . a Find dx 5 b Show that the normal to the curve at the point −4, − meets the y-axis at 8 the point (0, −3). y op C w 6 at the point (3, 6) meets the x-axis at P and x−2 id ge 9 The normal to the curve y = the y-axis at Q. ie U R ni ev ve ie w rs C 204 Pr op y a Find the equation of the normal PQ. -R am br ev Find the midpoint of PQ. es s -C 10 A curve has equation y = x5 − 8x 3 + 16x . The normal at the point P (1, 9) and the tangent at the point Q( −1, −9) intersect at the point R. )3 Pr ( ni ve rs a Find the coordinates of R. ity 11 A curve has equation y = 2 x − 1 + 2 . The normal at the point P (4, 4) and the normal at the point Q(9, 18) intersect at the point R. op y b Find the area of triangle PQR. U 12 and passes through the points A(2, 12) and x B (6, 20) . At each of the points C and D on the curve, the tangent is parallel to AB. ie id g w e C 12 A curve has equation y = 3x + br ev a Find the coordinates of the points C and D. Give your answer in exact form. -R s es am b Find the equation of the perpendicular bisector of CD. -C R ev ie w C op y Find the coordinates of R. Copyright Material - Review Only - Not for Redistribution ve rs ity C U ni op y Chapter 7: Differentiation am br id ev ie w ge 13 The curve y = x( x − 1)( x + 2) crosses the x-axis at the points O(0, 0), A(1, 0) and B ( −2, 0) . The normals to the curve at the points A and B meet at the point C. Find the coordinates of the point C . 5 and passes through the points P ( −1, 1) . Find 2 − 3x the equation of the tangent to the curve at P and find the angle that this tangent Pr es s -C -R 14 A curve has equation y = y makes with the x-axis. 12 − 4 intersects the x-axis at P. The tangent to the curve at 2x − 3 P intersects the y-axis at Q. Find the distance PQ. ve rs ity w C op 15 The curve y = y C op ni ev ie 16 The normal to the curve y = 2 x 2 + kx − 3 at the point (3, −6) is parallel to the line x + 5 y = 10. U R a Find the value of k. Try the Tangent or normal resource on the Underground Mathematics website. id ie w ge b Find the coordinates of the point where the normal meets the curve again. br ev 7.4 Second derivatives dy . dx dy is called the first derivative of y with respect to x. dx dy d dy with respect to x we obtain If we then differentiate , which is usually dx dx dx 2 d y written as . dx 2 d2 y is called the second derivative of y with respect to x. dx 2 f ′′( x ) = 6x + 10 -R s es Pr op y 2 5 , find d y . 3 (2 x − 3) dx 2 ity y = 5(2 x − 3)−3 op y Use the chain rule. C U Use the chain rule. -R s es am br ev ie id g w e d2 y = 120(2 x − 3)−5 × 2 dx 2 240 = (2 x − 3)5 ni ve rs dy = −15(2 x − 3)−4 × 2 dx = −30(2 x − 3)−4 -C ie w C Answer ev y ev br -C am WORKED EXAMPLE 7.10 Given that y = op or w f ′( x ) = 3x 2 + 10 x − 3 U or id ge R dy = 3x 2 + 10 x − 3 dx d2 y = 6x + 10 dx 2 f( x ) = x 3 + 5x 2 − 3x + 2 C ni or ie So for y = x 3 + 5x 2 − 3x + 2 ev -R rs ve ie w C ity op Pr y es s -C am If we differentiate y with respect to x we obtain R WEB LINK Copyright Material - Review Only - Not for Redistribution 205 ve rs ity ev ie w ge A curve has equation y = x 3 + 3x 2 − 9x + 2. y = x 3 + 3x 2 − 9x + 2 ni U 3x 2 + 6x − 9 , 0 x2 + 2x − 3 , 0 ( x + 3)( x − 1) , 0 −3 , x , 1 −3 -R s 2 y 1 op 0 rs −1 ve −2 ni w ie ev −3 U ∴ −3 , x , − 1 R es (2) ity op y x , −1 Combining (1) and (2) on a number line: −4 C 2 ie ev id 2 d 2 y dy . ≠ dx 2 dx -C -R am es s EXERCISE 7D d y = (2 x − 3)4 g y= Pr 4x − 9 f h y = 2 x 2 (5 − 3x + x 2 ) i e id g ie ev ( x − 3) -R f( x ) = x 2 2x − 3 x x2 15 f( x ) = 3 2x + 1 f( x ) = f s e c es br f( x ) = 1 − 3x am d op U 2x − 5 x2 6 x2 2 y= 3x + 1 5x − 4 y= x y=2− y ity y= ni ve rs e 2 Find f ′′( x ) for each of the following functions. 5 3 4x 2 − 3 b f( x ) = a f( x ) = 2 − 2x x 2 x5 -C C w ie ev R c C op y a d2 y for each of the following functions. dx 2 y = x 2 + 8x − 4 b y = 5x 3 − 7 x 2 + 5 1 Find w Hence, w ge d2 y dy = 6 x + 6 and = (3x 2 + 6 x − 9)2 . 2 dx dx br b 1 Pr -C 6x + 6 , 0 C 206 + − ev id (1) br am d2 y = 6x + 6 dx 2 d2 y , 0 when: dx 2 w + ge R dy , 0 when: dx ie ev ie w dy = 3x 2 + 6x − 9 dx y ve rs ity C op Answer a -R 2 d 2 y dy ≠ . dx 2 dx y b Show that dy d2 y and are negative. dx dx 2 Pr es s -C a Find the range of values of x for which C op am br id WORKED EXAMPLE 7.11 C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 Copyright Material - Review Only - Not for Redistribution x ve rs ity am br id 4 Given that f( x ) = x 3 + 2 x 2 − 3x − 1, find: b f(1) f ′(1) c -R a C dy d2 y and . dx dx 2 w 3 Given that y = 4x − (2 x − 1)4 , find ev ie ge U ni op y Chapter 7: Differentiation f ′′(1) 3 , find f ′′( x ) . (2 x − 1)8 2 , find the value of f ′′( −4). 6 Given that f( x ) = 1 − 2x y 3 4 7 w 6 -R Pr y es s -C 8 A curve has equation y = x 3 − 6x 2 − 15x − 7. Find the range of values of x for dy d2 y and which both are positive. dx dx 2 d2 y dy = 2 y. 9 Given that y = x 2 − 2 x + 5, show that 4 2 + ( x − 1) dx dx d2 y dy 10 Given that y = 4 x , show that 4 x 2 + 4x = y. 2 dx dx 207 w rs ity op C 5 ie id br am d2 y dx 2 2 ge dy dx 1 C op 0 ev R x ni ve rs ity 7 A curve has equation y = 2 x 3 − 21x 2 + 60 x + 5. Copy and complete the table dy d2 y and to show whether are positive ( + ), negative ( − ) or zero (0) for the dx dx 2 given values of x. U ev ie w C op y Pr es s -C 5 Given that f ′( x ) = y op C w y -R s es -C am br ev ie id g w e C U op ev ie w ni ve rs C ity Pr op y es s -C -R am br ev id dy ax + b . Given that = 0 and 12 A curve has equation y = 2 dx x d2 y 1 = when x = 2, find the value of a and the value of b. 2 dx 2 R WEB LINK Try the Gradients of gradients resource on the Underground Mathematics website. ie ge U R ni ev ve ie 11 A curve has equation y = x 3 + 2 x 2 − 4x + 6 . dy 2 = 0 when x = −2 and when x = . a Show that dx 3 2 d2 y b Find the value of when x = −2 and when x = . 2 3 dx Copyright Material - Review Only - Not for Redistribution ve rs ity C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 am br id Gradient of a curve d ( x n ) = nx n − 1 dx Power rule: op y ● -C The four rules of differentiation d d [ kf( x )] = k [f( x )] dx dx ● Addition/subtraction rule: d d d [f( x ) ± g( x )] = [f( x )] ± [g( x )] dx dx dx ● Chain rule: dy dy du = × dx du dx y C op s -C -R am br ev id ie dy at the point ( x1, y1 ) is m, then: dx ● the equation of the tangent at that point is given by y − y1 = m( x − x1 ) 1 ● the equation of the normal at that point is given by y − y1 = − ( x − x1 ) . m If the value of es Second derivatives Pr y d dy d2 y = dx dx dx 2 y op y op -R s es -C am br ev ie id g w e C U R ev ie w ni ve rs C ity Pr op y es s -C -R am br ev id ie w ge C U R ni ev ve ie w rs C 208 ity op ● w Tangents and normals ge U ni w ve rs ity Scalar multiple rule: C ● R ev ie -R dy represents the gradient of the curve y = f( x ). dx Pr es s ● w ev ie ge Checklist of learning and understanding Copyright Material - Review Only - Not for Redistribution ve rs ity ge C U ni op y Chapter 7: Differentiation ev ie 3x5 − 7 with respect to x. 4x 8 Find the gradient of the curve y = at the point where x = 2. 4x − 5 A curve has equation y = 3x 3 − 3x 2 + x − 7. Show that the gradient of the curve is never negative. 6 The normal to the curve y = 5 x at the point P (4, 10) meets the x-axis at the point Q. ve rs ity Find the gradient of the curve y = Find the coordinates of Q. y [4] C op ni U b br ev id ie 12 The equation of a curve is y = 5 x + 2 . x dy . a Find dx b Show that the normal to the curve at the point (2, 13) meets the x-axis at the point (28, 0). 12 at the point (9, 4) meets the x-axis at P and the y-axis at Q. The normal to the curve y = x -R [2] Pr es s [3] [6] The curve y = x( x − 3)( x − 5) crosses the x-axis at the points O(0, 0), A(3, 0) and B (5, 0) . The tangents to the curve at the points A and B meet at the point C . Find the coordinates of the point C . 2 . 10 A curve passes through the point A(4, 2) and has equation y = ( x − 3)2 a Find the equation of the tangent to the curve at the point A. [6] op Find the length of PQ, correct to 3 significant figures. ity 9 rs w ni op y ve ie ev w ge Find the equation of the normal to the curve at the point A. [5] [2] ie id b C U R [4] [1] w ge Find the equation of the normal PQ. -C y 8 [3] 15 at the point where x = 5. x2 − 2 x am 7 C dy d y and . dx dx 2 5 a [3] 2 The equation of a curve is y = (3 − 5x )3 − 2 x. Find C w [3] 4 R ev ie Pr es s op y 3 [3] -R Differentiate -C 2 am br id 1 w END-OF-CHAPTER REVIEW EXERCISE 7 -R am br ev 10 11 A curve passes through the point P (5, 1) and has equation y = 3 − . x a Show that the equation of the normal to the curve at the point P is 5x + 2 y = 27. [4] ii Find the midpoint of PQ. s Find the coordinates of Q. [3] es i [1] op C w ev ie id g es s -R br am -C [7] Cambridge International AS & A Level Mathematics 9709 Paper 11 Q8 June 2016 e U y ni ve rs ity 4 and passes through the points A(1, −1) and B (4, 11). x At each of the points C and D on the curve, the tangent is parallel to AB. Find the equation of the perpendicular bisector of CD. 12 A curve has equation y = 3x − R ev ie w C op y b Pr -C The normal meets the curve again at the point Q. Copyright Material - Review Only - Not for Redistribution 209 ve rs ity -R y C C y=2 − A ve rs ity op y Pr es s -C 13 ev ie am br id w ge C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 x y U R C op B ni ev ie w O 18 , which crosses the x-axis at A and the y-axis at B. 2x + 3 The normal to the curve at A crosses the y-axis at C . i Show that the equation of the line AC is 9x + 4 y = 27. ii Find the length of BC . -R am br ev id ie w ge The diagram shows part of the curve y = 2 − es s -C Pr y 14 The equation of a curve is y = 3 + 4x − x 2. i Show that the equation of the normal to the curve at the point (3, 6) is 2 y = x + 9. ii Given that the normal meets the coordinate axes at points A and B, find the coordinates of the mid-point of AB. iii Find the coordinates of the point at which the normal meets the curve again. ity rs w ve ie [2] [4] y op U ni ev y C R 1 br ev id ie w ge y = (6x + 2) 3 B Pr op y es s -C E -R am A(1, 2) C x 1 y ni ve rs C ity O w -R s Cambridge International AS & A Level Mathematics 9709 Paper 11 Q11 November 2012 es am br ev ie id g w e C U op The diagram shows the curve y = (6x + 2) 3 and the point A(1, 2) which lies on the curve. The tangent to the curve at A cuts the y-axis at B and the normal to the curve at A cuts the x-axis at C . [5] i Find the equation of the tangent AB and the equation of the normal AC. [3] ii Find the distance BC . iii Find the coordinates of the point of intersection, E , of OA and BC , and determine whether E is the mid-point of OA. [4] -C ie [4] Cambridge International AS & A Level Mathematics 9709 Paper 11 Q10 November 2010 15 ev [6] [2] Cambridge International AS & A Level Mathematics 9709 Paper 11 Q7 June 2010 op C 210 R 18 2x + 3 Copyright Material - Review Only - Not for Redistribution op y ve rs ity ni C U ev ie w ge -R am br id Pr es s -C y ni C op y ve rs ity op C w ev ie Chapter 8 Further differentiation y op w ge C U R ni ev ve rs w C ity op Pr y es s -C -R am br ev id ie w ge U R ie 211 id -R am br ev apply differentiation to increasing and decreasing functions and rates of change locate stationary points and determine their nature, and use information about stationary points when sketching graphs. y op -R s es -C am br ev ie id g w e C U R ev ie w ni ve rs C ity Pr op y es s -C ■ ■ ie In this chapter you will learn how to: Copyright Material - Review Only - Not for Redistribution ve rs ity ge C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 w Chapter 1 Solve quadratic inequalities. ev ie What you should be able to do Check your skills y Pr es s -C Where it comes from -R am br id PREREQUISITE KNOWLEDGE Find the first and second derivatives of x n. Chapter 7 Differentiate composite functions. 2 2 Find dy and d y2 for the following. dx dx a y = 3x 2 − x + 2 3 b y= 2x2 c y = 3x x y U ni C op C w ev ie a dy for the following. dx y = (2x − 1)5 b y= 3 Find 3 (1 − 3x )2 s -C -R am br ev id ie w ge R x2 − 2x − 3 . 0 a b 6 + x − x2 . 0 ve rs ity op Chapter 7 1 Solve: es Why do we study differentiation? FAST FORWARD C y ve ni R ev ie w rs In this chapter you will build on this knowledge and learn how to apply differentiation to problems that involve finding when a function is increasing (or decreasing) or when a function is at a maximum (or minimum) value. You will also learn how to solve practical problems involving rates of change. op 212 ity op Pr y In Chapter 7, you learnt how to differentiate functions and how to use differentiation to find gradients, tangents and normals. C w es s -C -R am br ev id ie manufacturers of canned food and drinks needing to minimise the cost of their manufacturing by minimising the amount of metal required to make a can for a given volume doctors calculating the time interval when the concentration of a drug in the bloodstream is increasing economists might use these tools to advise a company on its pricing strategy scientists calculating the rate at which the area of an oil slick is increasing. y op -R s es -C am br ev ie id g w e C U R ev ie w ni ve rs C ity Pr op y • • • • ge U There are many situations in real life where these skills are needed. Some examples are: In the Mechanics Coursebook, Chapter 6 you will learn to apply these skills to problems concerning displacement, velocity and time. Copyright Material - Review Only - Not for Redistribution WEB LINK Explore the Calculus meets functions station on the Underground Mathematics website. ve rs ity ev ie am br id EXPLORE 8.1 w ge C U ni op y Chapter 8: Further differentiation Section A: Increasing functions y y = f(x) -R Consider the graph of y = f( x ). Pr es s -C 1 Complete the following two statements about y = f( x ). y x O ‘As the value of x increases the value of y…’ 2 Sketch other graphs that satisfy these statements. ve rs ity w C op ‘The sign of the gradient at any point is always …’ These types of functions are called increasing functions. 1 Complete the following two statements about y = g( x ). U R y y ni Consider the graph of y = g( x ). C op ev ie Section B: Decreasing functions w ge ‘As the value of x increases the value of y …’ O ie br These types of functions are called decreasing functions. Pr y 8.1 Increasing and decreasing functions es s -R am 2 Sketch other graphs that satisfy these statements. -C Try the Choose your families resource on the Underground Mathematics website. ev id ‘The sign of the gradient at any point is always …’ WEB LINK x y = g(x) rs C w Likewise, a decreasing function f( x ) is one where the f( x ) values decrease whenever the x value increases, or f ( a ) > f ( b ) whenever a < b. ni op y ve ie ev 213 ity op As you probably worked out from Explore 8.1, an increasing function f( x ) is one where the f( x ) values increase whenever the x value increases. More precisely, this means that f ( a ) < f ( b ) whenever a < b. id ie w ge C U R Sometimes we talk about a function increasing at a point, meaning that the function values are increasing around that point. If the gradient of the function is positive at a point, then the function is increasing there. -C Now consider the function y = h( x ), shown on the graph. Pr ity (a, b) O y op C U w e ev ie id g es s -R br am -C dy >0 dx dy dx < 0 ni ve rs C ev ie w h( x ) is increasing when x . a, i.e. R y = h(x) es dy . 0 for x . a. dx dy h( x ) is decreasing when x , a, i.e. , 0 for x , a. dx op y • • y s We can divide the graph into two distinct sections: -R am br ev In the same way, we can talk about a function decreasing at a point. If the gradient of the function is negative at a point, then the function is decreasing there. Copyright Material - Review Only - Not for Redistribution x ve rs ity ev ie w ge am br id WORKED EXAMPLE 8.1 C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 -C y ni w ie am br ev id WORKED EXAMPLE 8.2 ge U R ev ie w C ve rs ity op y dy = −3 − 2 x dx dy , 0, y is decreasing. When dx −3 − 2 x , 0 2x . − 3 3 x . − 2 Pr es s y = 8 − 3x − x 2 C op Answer -R Find the set of values of x for which y = 8 − 3x − x 2 is decreasing. -R For the function f( x ) = 4x 3 − 15x 2 − 72 x − 8: es s -C a Find f ′( x ). Pr y b Find the range of values of x for which f( x ) = 4x 3 − 15x 2 − 72 x − 8 is increasing. C ity op c Find the range of values of x for which f( x ) = 4x 3 − 15x 2 − 72 x − 8 is decreasing. y op ev f ′( x ) = 12 x 2 − 30 x − 72 rs f( x ) = 4x 3 − 15x 2 − 72 x − 8 ve ie a ni w Answer U R b When f ′( x ) > 0, f(x) is increasing. + br -R -C 3 and 4. 2 Pr op y es 3 ∴ x < − and x > 4. 2 ity 3 <x<4 2 y op -R s es -C am br ev ie id g w e C U R ev ie ni ve rs C c When f ′( x ) < 0, f(x) is decreasing. w – s am (2 x + 3)( x − 4) > 0 ∴− 4 –3 2 ev id ie 2 x 2 − 5x − 12 > 0 Critical values are − + w ge 12 x 2 − 30 x − 72 > 0 C 214 Copyright Material - Review Only - Not for Redistribution x ve rs ity ge C U ni op y Chapter 8: Further differentiation am br id ev ie w WORKED EXAMPLE 8.3 5 3 for x . . Find an expression for f ′( x ) and determine whether f is an 2x − 3 2 increasing function, a decreasing function or neither. Pr es s -C -R A function f is defined as f( x ) = Answer y 5 2x − 3 = 5(2 x − 3)−1 C op y Differentiate using the chain rule. U ni ev ie f ′( x ) = −5(2 x − 3)−2 (2) 10 =− (2 x − 3)2 R Write in a form ready for differentiating. ve rs ity w C op f( x ) = id ie w ge 3 If x . , then (2 x − 3)2 . 0 for all values of x. 2 Hence, f ′( x ) , 0 for all values of x in the domain of f. -C -R am br ev ∴ f is a decreasing function. y es s EXERCISE 8A Pr f( x ) = x 2 − 8x + 2 c f( x ) = 5 − 7 x − 2 x 2 e f( x ) = 2 x 3 − 15x 2 + 24x + 6 f( x ) = x 3 − 12 x 2 + 2 f f( x ) = 16 + 16x − x 2 − x 3 op c f( x ) = 2 x 3 − 21x 2 + 60 x − 5 d e f( x ) = −40 x + 13x 2 − x 3 ev id ie w f( x ) = 10 + 9x − x 2 -C f( x ) = 11 + 24x − 3x 2 − x 3 -R br f f( x ) = x 3 − 3x 2 − 9x + 5 1 ( 5 − 2x )3 + 4x is increasing. 6 s 4 for x > 1. Find an expression for f ′( x ) and determine whether f is 1 − 2x an increasing function, a decreasing function or neither. 5 2 for x > 0. Find an expression for f ′( x ) and determine 5 A function f is defined as f( x ) = 2 − x+2 ( x + 2) whether f is an increasing function, a decreasing function or neither. es op y x2 − 4 is an increasing function. x U 6 Show that f( x ) = ni ve rs ity Pr op y id g w e C 7 A function f is defined as f( x ) = (2 x + 5)2 − 3 for x ù 0. Find an expression for f ′( x ) and explain why f is an increasing function. ie 2 − x 2 for x . 0. Show that f is a decreasing function. x4 9 A manufacturing company produces x articles per day. The profit function, P( x ), can be modeled by the function P( x ) = 2 x 3 − 81x 2 + 840 x. Find the range of values of x for which the profit is decreasing. -R s es am br ev 8 It is given that f( x ) = -C C w C U b ge f( x ) = 3x 2 − 8x + 2 am R ie d a 4 A function f is defined as f( x ) = ev f( x ) = 2 x 2 − 4x + 7 2 Find the set of values of x for which each of the following is decreasing. 3 Find the set of values of x for which f( x ) = R b y ve rs ity a ni ev ie w C op 1 Find the set of values of x for which each of the following is increasing. Copyright Material - Review Only - Not for Redistribution 215 ve rs ity w ge 8.2 Stationary points C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 ev ie am br id Consider the following graph of the function y = f( x ). y y R C ve rs ity op dy =0 dx O dy >0 dx dy <0 dx Q P w dy =0 dx x C op y The red sections of the curve show where the gradient is negative (where f( x ) is a decreasing function) and the blue sections show where the gradient is positive (where f( x ) is an increasing function). The gradient of the curve is zero at the points P, Q and R. ge U ni ev ie R dy =0 dx dy >0 dx Pr es s -C dy <0 dx -R y = f(x) ie ev id br Maximum points w A point where the gradient is zero is called a stationary point or a turning point. -C -R am The stationary point Q is called a maximum point because the value of y at this point is greater than the value of y at other points close to Q. ity + – rs w y ve Minimum points – + -R am br ev the gradient is negative to the left of the minimum and positive to the right. ie id w ge C U At a minimum point: dy ● =0 dx ● op The stationary points P and R are called minimum points. ni ie ev R 0 the gradient is positive to the left of the maximum and negative to the right. C ● op 216 Pr y es s At a maximum point: dy ● =0 dx 0 es s -C Stationary points of inflexion Pr 0 op y – -R s es am br ev from positive to zero and then to positive again the gradient changes or from negative to zero and then to negative again. -C ● ie id g w e C U At a stationary point of inflexion: dy ● =0 dx – ni ve rs C w + ie ev R + 0 ity op y There is a third type of stationary point (turning point) called a point of inflexion. Copyright Material - Review Only - Not for Redistribution ve rs ity ge C U ni op y Chapter 8: Further differentiation am br id ev ie w WORKED EXAMPLE 8.4 Pr es s -C y y = x 3 − 12 x + 5 ve rs ity dy =0 dx 3x 2 − 12 = 0 ni ev ie For stationary points: y w C op dy = 3x 2 − 12 dx C op Answer -R Find the coordinates of the stationary points on the curve y = x 3 − 12 x + 5 and determine the nature of these points. Sketch the graph of y = x 3 − 12 x + 5. U R x2 − 4 = 0 id ie w ge ( x + 2)( x − 2) = 0 x = −2 or x = 2 br ev When x = −2, y = ( −2)3 − 12( −2) + 5 = 21 y = (2)3 − 12(2) + 5 = −11 -R am When x = 2, s es -C The stationary points are ( −2, 21) and (2, −11). dy dx 3( −2.1)2 − 12 = positive rs −2 −1.9 0 3( −1.9)2 − 12 = negative y C 0 3(2.1)2 − 12 = positive es op y Pr ity So ( −2, 21) is a maximum point and (2, −11) is a minimum point. ni ve rs y y = x3 – 12x + 5 C w ie 5 ev x O -R (2, –11) s -C am br id g e U R (–2, 21) y op ev The sketch graph of y = x 3 − 12 x + 5 is: es C shape of curve w 2.1 s direction of tangent ie ie 3(1.9)2 − 12 = negative ev 2 -R id 1.9 -C am br x dy dx w ge shape of curve TIP op ni U ev direction of tangent R Pr −2.1 ity x ve ie w C op y Now consider the gradient on either side of the points ( −2, 21) and (2, −11): Copyright Material - Review Only - Not for Redistribution This is called the First Derivative Test. 217 ve rs ity C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 w ge Second derivatives and stationary points – Pr es s -C + dy , starts as a positive value, decreases to zero at the maximum point and dx then decreases to a negative value. dy dy Since decreases as x increases, then the rate of change of is negative. dx dx d dy d 2 y dy The rate of change of is written as . = dx dx dx 2 dx d2 y is called the second derivative of y with respect x. dx 2 y w ie ev y es -C s d2 y dy , 0 , then the point is a maximum point. = 0 and dx dx 2 If -R br am KEY POINT 8.1 id This leads to the rule: ge U ni C op y ve rs ity op C w ev ie -R 0 The gradient, R ev ie am br id Consider moving from left to right along a curve, passing through a maximum point. y + 0 op ni R ev ve ie – rs w C ity op Pr Now, consider moving from left to right along a curve, passing through a minimum point. 218 -R am br ev id ie w ge C U dy , starts as a negative value, increases to zero at the minimum point and dx then increases to a positive value. dy dy Since increases as x increases, then the rate of change of is positive. dx dx The gradient, s -C This leads to the rule: op y es KEY POINT 8.2 ity Pr dy d2 y = 0 and . 0 , then the point is a minimum point. dx dx 2 -R s es -C am br ev ie id g w e C U op y dy d2 y = 0 and = 0, then the nature of the stationary point can be found using the dx dx 2 first derivative test. If R ev ie w ni ve rs C If Copyright Material - Review Only - Not for Redistribution REWIND In Chapter 7, Section 7.4 we learnt how to find a second derivative. Here we will look at how second derivatives can be used to determine the nature of a stationary point. ve rs ity ge C U ni op y Chapter 8: Further differentiation am br id ev ie w WORKED EXAMPLE 8.5 the nature of these points. y Pr es s -C Answer x2 + 9 y= = x + 9x −1 x dy 9 = 1 − 9x −2 = 1 − 2 dx x ve rs ity op U R ni ev ie y dy =0 dx 9 1− 2 = 0 x x2 − 9 = 0 For stationary points: C op C w x2 + 9 and use the second derivative to determine x -R Find the coordinates of the stationary points on the curve y = br ev id ie w ge ( x + 3)( x − 3) = 0 x = −3 or x = 3 ( −3)2 + 9 = −6 −3 32 + 9 =6 y= When x = 3, 3 The stationary points are ( −3, −6) and (3, 6). -R y= s es y op d 2 y 18 = 3 .0 dx 2 3 rs When x = 3, ve d2 y 18 ,0 2 = dx ( −3)3 br w -R am EXPLORE 8.2 ev id ie ge ∴ ( −3, −6) is a maximum point and (3, 6) is a minimum point. C U ni When x = −3, 219 ity d2 y 18 = 18x −3 = 3 x dx 2 C w ie ev R Pr op y -C am When x = −3, es y y = f ′ (x) ity O 1 2 3 4 5 6 x 1 2 3 4 5 6 x y op ie C U 1 Discuss the properties of these two graphs and the information that can be obtained from them. w id g e 2 Without finding the equation of the function y = f( x ), determine, giving reasons: ie -R s 3 Sketch the graph of the function y = f( x ). es am b the coordinates of the minimum point on the curve. ev br a the coordinates of the maximum point on the curve -C ev R y = f ″ (x) ni ve rs O Pr y w C op y The following graphs show y = f ′( x ) and y = f ′′( x ). s -C The graph of the function y = f( x ) passes through the point (1, −35) and the point (6, 90). Copyright Material - Review Only - Not for Redistribution ve rs ity ev ie am br id EXERCISE 8B w ge C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 c y = x − 12 x + 6 e y = x 4 + 4x − 1 3 y = (3 + x )(2 − x ) d y = 10 + 9x − 3x 2 − x 3 f y = (2 x − 3)3 − 6x ve rs ity y op y 2 Find the coordinates of the stationary points on each of the following curves and determine the nature of each stationary point. 9 8 b y = 4x 2 + a y= x+ x x ( x − 3)2 48 d y = x3 + +4 c y= x x 8 e y=4 x −x f y = 2x + 2 x x2 − 9 3 The equation of a curve is y = . x2 dy Find and, hence, explain why the curve does not have a stationary point. dx 4 A curve has equation y = 2 x 3 − 3x 2 − 36x + k. es s -C -R am br ev id ie w ge U R ni C op C w ev ie b Pr es s y = x 2 − 4x + 8 -C a -R 1 Find the coordinates of the stationary points on each of the following curves and determine the nature of each stationary point. Sketch the graph of each function and use graphing software to check your graphs. w Pr rs C b Hence, find the two values of k for which the curve has a stationary point on the x-axis. ity op y a Find the x-coordinates of the two stationary points on the curve. y op ev a Find the value of a. ve ie 5 The curve y = x 3 + ax 2 − 9x + 2 has a maximum point at x = −3. ni 220 ge C U R b Find the range of values of x for which the curve is a decreasing function. id ie w 6 The curve y = 2 x 3 + ax 2 + bx − 30 has a stationary point when x = 3. br ev The curve passes through the point (4, 2). -R am a Find the value of a and the value of b. es s -C b Find the coordinates of the other stationary point on the curve and determine the nature of this point. Pr op y 7 The curve y = 2 x 3 + ax 2 + bx − 30 has no stationary points. ity k2 , where k is a positive constant. Find, 2x − 3 in terms of k, the values of x for which the curve has stationary points and determine the nature of each stationary point. U op y ni ve rs 8 A curve has equation y = 1 + 2 x + ie id g w e C 9 Find the coordinates of the stationary points on the curve y = x 4 − 4x 3 + 4x 2 + 1 and determine the nature of each of these points. Sketch the graph of the curve. br ev 10 The curve y = x 3 + ax 2 + b has a stationary point at (4, −27). -R am a Find the value of a and the value of b. es s b Determine the nature of the stationary point (4, −27). -C R ev ie w C Show that a 2 , 6b. Copyright Material - Review Only - Not for Redistribution ve rs ity C ev ie w ge Find the coordinates of the other stationary point on the curve and determine the nature of this stationary point. am br id c U ni op y Chapter 8: Further differentiation -R d Find the coordinates of the point on the curve where the gradient is minimum and state the value of the minimum gradient. b has a stationary point at (2, 12). x2 a Find the value of a and the value of b. Pr es s -C 11 The curve y = ax + y C op ve rs ity ni ev ie w C op y b Determine the nature of the stationary point (2, 12). b c Find the range of values of x for which ax + 2 is increasing. x a 12 The curve y = x 2 + + b has a stationary point at (3, 5). x a Find the value of a and the value of b. id ie w ge U R b Determine the nature of the stationary point (3, 5). a c Find the range of values of x for which x 2 + + b is decreasing. x 3 2 13 The curve y = 2 x + ax + bx + 7 has a stationary point at the point (2, −13). br am -R b Find the coordinates of the second stationary point on the curve. Determine the nature of the two stationary points. s es ity rs w y op id ie w ge C U R ni ev ve ie There are many problems for which we need to find the maximum or minimum value of an expression. For example, the manufacturers of canned food and drinks often need to minimise the cost of their manufacturing. To do this they need to find the minimum amount of metal required to make a container for a given volume. Other situations might involve finding the maximum area that can be enclosed within a shape. ev br -R am -C The surface area of the solid cuboid is 100 cm 2 and the volume is V cm 3. h cm es s a Express h in terms of x. x cm 1 b Show that V = 25x − x 3 . x cm 2 c Given that x can vary, find the stationary value of V and determine whether this stationary value is a maximum or a minimum. Answer U 2 x 2 + 4xh = 100 op Surface area = 2 x 2 + 4xh C a y ni ve rs ity Pr op y -R s es am br ev ie id g w e 100 − 2 x 2 4x 25 1 h= − x x 2 h= -C C w ie ev 221 8.3 Practical maximum and minimum problems WORKED EXAMPLE 8.6 R cubics edition. Pr d Find the coordinates of the point on the curve where the gradient is minimum and state the value of the minimum gradient. C op y -C c Try the following resources on the Underground Mathematics website: • Floppy hair • Two-way calculus • Curvy cubics • Can you find… curvy ev a Find the value of a and the value of b. WEB LINK Copyright Material - Review Only - Not for Redistribution ve rs ity ev ie Substitute for h. am br id b V = x2h w ge C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 -C dV = 0: dx 3 25 − x 2 = 0 2 5 6 x= 3 C op U R ni ev ie y ve rs ity Stationary values occur when w C op y dV 3 = 25 − x 2 dx 2 Pr es s 1 3 x 2 = 25x − c -R 25 1 = x2 − x x 2 3 5 6 15 6 5 6 − = 68.04 (to 2 decimal places) , V = 25 3 3 2 3 ie br ev id d 2V = −3x dx 2 w ge When x = -R am 5 6 d 2V , = −5 6, which is , 0. dx 2 3 The stationary value of V is 68.04 and it is a maximum value. op Pr y es s -C When x = 222 rs The diagram shows a solid cylinder of radius r cm and height 2 h cm cut from a solid sphere of radius 5 cm. The volume of the cylinder is C cm 3. C 3 id ie c Find the value for h for which there is a stationary value of V . -R am br ev d Determine the nature of this stationary value. y U dV = 0: dh 2 50 π − 6 πh = 0 50 π h2 = 6π 5 3 h= 3 op dV = 50 π − 6 πh2 dh w ie ev -R s es am br id g e C Stationary values occur when -C c ni ve rs = 50 πh − 2 πh3 w ie ev R ) = π 25 − h2 (2 h ) Substitute for r. ity C b V = πr 2 (2 h ) ( Use Pythagoras’ theorem. es 25 − h 2 Pr op y r= s r 2 + h 2 = 52 -C a w ge b Show that V = 50 πh − 2 πh . Answer r y op U R a Express r in terms of h. ni ev ve ie w C ity WORKED EXAMPLE 8.7 Copyright Material - Review Only - Not for Redistribution 5 cm 2h ve rs ity op y ve rs ity C WORKED EXAMPLE 8.8 w w -R Pr es s -C The stationary value is a maximum value. ev ie ev ie ge d 2V = −12 πh dh2 5 3 d 2V 5 3 , = −12 π , which is , 0. When h = dx 2 3 3 am br id d C U ni op y Chapter 8: Further differentiation The diagram shows a hollow cone with base radius 12 cm and height 24 cm. 24 cm U R ni C op y A solid cylinder stands on the base of the cone and the upper edge touches the inside of the cone. w ge The cylinder has base radius r cm, height h cm and volume V cm 3. r br ev id ie a Express h in terms of r. h 12 cm -R am b Show that V = 24 πr 2 − 2 πr 3. es s -C c Find the volume of the largest cylinder that can stand inside the cone. Pr r 24 − h = 24 12 2 r = 24 − h h = 24 − 2 r ev br dV = 0: dr 48 πr − 6 πr 2 = 0 Pr es s 6 πr(8 − r ) = 0 r=8 -R am -C When r = 8, V = 24 π(8)2 − 2 π(8)3 = 512 π DID YOU KNOW? ity d 2V = 48 π − 12 πr dr 2 d 2V = 48 π − 12 π(8), which is , 0. When r = 8, dx 2 op y ni ve rs op y C U The stationary value is a maximum value. -R s es am br ev ie id g w e Volume of the largest cylinder is 512 π cm3. -C C w ie ev y op w ie id ge = 24 πr 2 − 2 πr 3 dV = 48 πr − 6 πr 2 dr C U = πr (24 − 2 r ) Stationary values occur when R 223 ity 2 R Substitute for h. ni ev b V = πr 2 h c Use similar triangles. rs ie w C op a ve y Answer Copyright Material - Review Only - Not for Redistribution Differentiation can be used in business to find how to maximise company profits and to find how to minimise production costs. ve rs ity ev ie am br id EXERCISE 8C w ge C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 1 The sum of two real numbers, x and y, is 9. c Given that Q = 3x 2 + 2 y2, write down an expression for Q, in terms of x. ve rs ity w i ii Find the minimum value of Q. C op y ii Find the maximum value of P. Pr es s Given that P = x 2 y, write down an expression for P, in terms of x. -C b i -R a Express y in terms of x. y ge U R a Express θ in terms of r. θ r id ie w b Show that A = 20 r − r 2. ev Find the value of r for which there is a stationary value of A. br c r C op ni ev ie 2 A piece of wire, of length 40 cm, is bent to form a sector of a circle with radius r cm and sector angle θ radians, as shown in the diagram. The total area enclosed by the shape is A cm 2 . -R am d Determine the magnitude and nature of this stationary value. s -C 3 The diagram shows a rectangular enclosure for keeping animals. ym The total length of the fence is 50 m and the area enclosed is A m 2. Pr xm a Express y in terms of x. 1 b Show that A = x(50 − x ). 2 c Find the maximum possible area enclosed and the value of x for which this occurs. y op ni ev ve ie w rs C 224 ity op y es There is a fence on three sides of the enclosure and a wall on its fourth side. w ge PQRS is a quadrilateral where PB = AS = 2 x cm, BQ = x cm and DR = 4x cm. id ie a Express the area of PQRS in terms of x. -R am br ev b Given that x can vary, find the value of x for which the area of PQRS is a minimum and find the magnitude of this minimum area. 2x Q x A P es R B Q Pr 3x + 2y = 30 e ev br y y op x R S y = 9 – x2 w ) b Show that A = 2 p 9 − p2 . P -R Find the value of p for which A has a stationary value. am P ie id g a Express QR in terms of p. ( O C U The points P and Q lie on the x-axis and the points R and S lie on the curve y = 9 − x 2. es s d Find this stationary value and determine its nature. -C 2x y ity ni ve rs PQRS is a rectangle with base length 2 p units and area A units2. c C s -C C op y 6 R S 5 The diagram shows the graph of 3x + 2 y = 30. OPQR is a rectangle with area A cm 2. The point O is the origin, P lies on the x-axis, R lies on the y-axis and Q has coordinates ( x, y ) and lies on the line 3x + 2 y = 30. 3 a Show that A = 15x − x 2. 2 b Given that x can vary, find the stationary value of A and determine its nature. w ie ev R 4x D C U R 4 The diagram shows a rectangle, ABCD, where AB = 20 cm and BC = 16 cm. Copyright Material - Review Only - Not for Redistribution O 2p units Q x ve rs ity x cm x Pr es s -R 15 cm -C 24 cm The diagram shows a 24 cm by 15 cm sheet of metal with a square of side x cm removed from each corner. The metal is then folded to make an open rectangular box of depth x cm and volume V cm3. ve rs ity y op C w ev ie am br id w ge 7 C U ni op y Chapter 8: Further differentiation a Show that V = 4x 3 − 78x 2 + 360 x. C op y Determine the nature of this stationary value. U R c ni ev ie b Find the stationary value of V and the value of x for which this occurs. y cm w ge 8 The volume of the solid cuboid shown in the diagram is 576 cm 3 and the surface area is A cm 2. br -R am es s -C Pr y 225 ity op C R xm rs w ve ie S PQST is a rectangle and QRS is a semicircle with diameter SQ. PT = x m and PQ = ST = y m. a Express y in terms of x. 1 1 b Show that A = x − x 2 − πx 2. 2 8 dA d2 A and . c Find dx dx 2 d Find the value for x for which there is a stationary value of A. P ym Q -R am br ev id ie w ge C U op ni y The total area enclosed by the shape is A m 2. ev 2x cm T 9 The diagram shows a piece of wire, of length 2 m, is bent to form the shape PQRST . R x cm ev id ie a Express y in terms of x. 1728 . b Show that A = 4x 2 + x c Find the maximum value of A and state the dimensions of the cuboid for which this occurs. s -C e Determine the magnitude and nature of this stationary value. ity Pr rm hm 1 2 πr . 2 e U Find op dA d2 A and . dr dr 2 d Find the value for r for which there is a stationary value of A. c C b Show that A = 5r − 2 r 2 − y ni ve rs a Express h in terms of r. -R s es am br ev ie id g w e Determine the magnitude and nature of this stationary value. -C R ev ie w C op y es 10 The diagram shows a window made from a rectangle with base 2 r m and height h m and a semicircle of radius r m. The perimeter of the window is 5 m and the area is A m 2. Copyright Material - Review Only - Not for Redistribution 2r m ve rs ity C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 w ge 11 A piece of wire, of length 100 cm, is cut into two pieces. am br id ev ie One piece is bent to make a square of side x cm and the other is bent to make a circle of radius r cm. The total area enclosed by the two shapes is A cm 2. ve rs ity 12 A solid cylinder has radius r cm and height h cm. ni w ge U R C op ev ie a Express h in terms of r. 864 π . b Show that A = 2 πr 2 + r c Find the value for r for which there is a stationary value of A. y The volume of this cylinder is 432π cm 3 and the surface area is A cm 2. w C op y Pr es s -C -R a Express r in terms of x. ( π + 4)x 2 − 200 x + 2500 . b Show that A = π c Find the value of x for which A has a stationary value and determine the nature and magnitude of this stationary value. 6x cm Pr es y cm The diagram shows an open water container in the shape of a triangular prism of length y cm. The vertical cross-section is an isosceles triangle with sides 5x cm, 5x cm and 6x cm. ity op C 226 s 5x cm 5x cm y -C am 13 -R br ev id ie d Determine the magnitude and nature of this stationary value. ve ie w rs The water container is made from 500 cm 2 of sheet metal and has a volume of V cm 3. br y es op y ni ve rs ity Pr r es s -R br ev d Determine the magnitude and nature of this stationary value. am h C ie id g w e U a Express r in terms of h. 1 b Show that V = πh2 (20 − h ). 3 c Find the value for h for which there is a stationary value of V . op y 15 The diagram shows a right circular cone of base radius r cm and height h cm cut from a solid sphere of radius 10 cm. The volume of the cone is V cm 3. -C C w ie ev c h 5 3 πr . 6 Find the exact value of r such that V is a maximum. b Show that V = 160 πr − R op s -C -R am 14 The diagram shows a solid formed by joining a hemisphere of radius r cm to a cylinder of radius r cm and height h cm. The surface area of the solid is 320π cm 2 and the volume is V cm 3. a Express h in terms of r. PS ie ev id d Show that the value in part c is a maximum value. PS C w ge U R ni ev a Express y in terms of x. 144 3 x. b Show that V = 600 x − 5 c Find the value of x for which V has a stationary value. Copyright Material - Review Only - Not for Redistribution r ve rs ity am br id ev ie w ge 8.4 Rates of change C U ni op y Chapter 8: Further differentiation ve rs ity w C op y Pr es s -C -R EXPLORE 8.3 A B C 3 −1 y U R 1 Discuss how the height of water in container A changes with time. w ge 2 Discuss how the height of water in container B changes with time. C op ni ev ie Consider pouring water at a constant rate of 10 cm s into each of these three large containers. id ie 3 Discuss how the height of water in container C changes with time. h h O t t O 227 t rs ity 5 What can you say about the gradients? You should have come to the conclusion that: the height of water in container A increases at a constant rate ● the height of water in containers B and C does not increase at a constant rate. op y ve ● ni R ev ie w C O Pr op y es s -C h -R am br ev 4 On copies of the following axes, sketch graphs to show how the height of water in a container ( h cm) varies with time (t seconds) for each container. w ge C U The (constant) rate of change of the height of the water in container A can be found by finding the gradient of the straight-line graph. es s -C -R am br ev id ie The rate of change of the height of the water in containers B and C at a particular time, t seconds, can be estimated by drawing a tangent to the curve and then finding the gradient of the tangent. A more accurate method is to use differentiation if we know the equation of the graph. Pr ity ie id g w e C U dh Differentiate to obtain (the rate of change dt of h with respect to t). es s -R br ev dh 2 4 = (2) = dt 5 5 am When t = 2, y 1 2 t 5 dh 2 = t dt 5 h= op Answer -C ie ev R 1 2 t , find the rate of change of h with respect to t when t = 2. 5 ni ve rs Given that h = w C op y WORKED EXAMPLE 8.9 Copyright Material - Review Only - Not for Redistribution ve rs ity ev ie w ge am br id WORKED EXAMPLE 8.10 C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 -C Find the rate of change of V with respect to t when t = 4. Pr es s Answer y C op Connected rates of change ni dV = 4(4) − 3 = 13 dt U R ev ie w C dV = 4t − 3 dt dV Differentiate to obtain (the rate of change dt of V with respect to t). ve rs ity op y V = 2t 2 − 3t + 8 When t = 4, -R Variables V and t are connected by the equation V = 2t 2 − 3t + 8. es Pr op ni C U ie Pr op y WORKED EXAMPLE 8.11 1 C id g 1 dx = 0.06 dt w and e y = x + (2 x + 3) 2 op U Answer y ni ve rs ev ie Differentiate to find -R dy . dx s -C am br − dy 1 = 1 + (2 x + 3) 2 (2) dx 2 1 = 1+ 2x + 3 es w C ity A point with coordinates ( x, y ) moves along the curve y = x + 2 x + 3 in such a way that the rate of increase of x has the constant value 0.06 units per second. Find the rate of increase of y at the instant when x = 3. State whether the y-coordinate is increasing or decreasing. ie ev R -R es s -C am br ev id We can deduce this as: if we set t = y in the chain rule, we get dy dy dx = × dy dx dy dy Since = 1, the rule follows. dy w ge R ev dx 1 = y d dy dx y ve rs KEY POINT 8.4 ie w C ity op We may also need to use the rule: 228 In Chapter 7, Section 7.2 we learnt how to differentiate using the chain rule. Here we will look at how the chain rule can be used for problems involving connected rates of change. s dy dy dx = × dt dx dt y -C The chain rule states: REWIND -R br am KEY POINT 8.3 ev id ie w ge When two variables, x and y, both vary with a third variable, t, we can connect the three variables using the chain rule. Copyright Material - Review Only - Not for Redistribution ve rs ity w ge 1 4 = 3 2 (3) + 3 ev ie dy = 1+ dx am br id When x = 3, C U ni op y Chapter 8: Further differentiation Rate of change of y is 0.08 units per second. dy is a positive quantity). The y-coordinate is increasing (since dt C op ni U R EXERCISE 8D y ve rs ity ev ie w C op y Pr es s -C -R Using the chain rule: dy = dy × dx dt dx dt 4 = × 0.06 3 = 0.08 br ev id ie w ge 1 A point is moving along the curve y = 3x − 2 x 3 in such a way that the x-coordinate is increasing at 0.015 units per second. Find the rate at which the y-coordinate is changing when x = 2, stating whether the y-coordinate is increasing or decreasing. s -C -R am 2 A point with coordinates ( x, y ) moves along the curve y = 1 + 2 x in such a way that the rate of increase of x has the constant value 0.01 units per second. Find the rate of increase of y at the instant when x = 4. es 8 in such a way that the x-coordinate is increasing at a constant rate x2 − 2 of 0.005 units per second. Find the rate of change of the y-coordinate as the point passes through the point (2, 4). Pr ity 5 in such a way that the x-coordinate is increasing at a x constant rate of 0.02 units per second. Find the rate of change of the y-coordinate when x = 1. 4 A point is moving along the curve y = 3 x − y ve ie w rs C op y 3 A point is moving along the curve y = ev 1 in such a way that the x-coordinate of P is increasing at a x constant rate of 0.5 units per second. Find the rate at which the y-coordinate of P is changing when P is at the point (1, 4). ie w ge C U R ni op 5 A point, P, travels along the curve y = 3x + id 2 + 5x in such a way that the x-coordinate is increasing at a constant x rate of 0.02 units per second. Find the rate at which the y-coordinate is changing when x = 2, stating whether the y-coordinate is increasing or decreasing. -C -R am br ev 6 A point is moving along the curve y = s 8 . As it passes through the point P, the x-coordinate is increasing at 7 − 2x a rate of 0.125 units per second and the y-coordinate is increasing at a rate of 0.08 units per second. Find the possible x-coordinates of P. es Pr ity op y ni ve rs es s -R br ev ie id g w e C U 9 A point, P ( x, y ), travels along the curve y = x 3 − 5x 2 + 5x in such a way that the rate of change of x is constant. Find the values of x at the points where the rate of change of y is double the rate of change of x. am PS 8 A point, P, travels along the curve y = 3 2 x 2 − 3 in such a way that at time t minutes the x-coordinate of P is increasing at a constant rate of 0.012 units per minute. Find the rate at which the y-coordinate of P is changing when P is at the point (1, −1). -C R ev ie w C op y 7 A point moves along the curve y = Copyright Material - Review Only - Not for Redistribution 229 ve rs ity am br id w ev ie ge 8.5 Practical applications of connected rates of change C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 -R WORKED EXAMPLE 8.12 Oil is leaking from a pipeline under the sea and a circular patch is formed on the surface of the sea. Pr es s -C The radius of the patch increases at a rate of 2 metres per hour. op y Find the rate at which the area is increasing when the radius of the patch is 25 metres. C ve rs ity Answer w ev ie y dA when r = 25 . dt dr = 2. Radius increasing at a rate of 2 metres per hour, so dt We need to find U w id dA = 2 πr dr br ev dA = 50 π dr -R am When r = 25, Differentiate with respect to r. ge A = πr 2 ie R ni C op Let A = area of circular oil patch, in m 2 . op Pr y es s -C Using the chain rule, dA = dA × dr dt dr dt = 50 π × 2 = 100 π 230 rs op y ve WORKED EXAMPLE 8.13 ni ev ie w C ity The area is increasing at a rate of 100 π m 2 per hour. br -C C -R am b Find the rate of increase of the volume when r = 5. Answer w ev id ie ge U R A solid sphere has radius r cm, surface area A cm 2 and volume V cm 3. 1 cm s −1. The radius is increasing at a rate of 5π a Find the rate of increase of the surface area when r = 3. s dA when r = 3. dt 1 1 dr cm s −1, so = . Radius increasing at a rate of dt 5 π 5π Pr ity Differentiate with respect to r. op w ie ev -R s es am br id g e C U dA = 8 πr dr dA When r = 3, = 24 π dr dA dA dr Using the chain rule, = × dt dr dt 1 = 24 π × 5π = 4.8 The surface area is increasing at a rate of 4.8 cm 2 s −1. y ni ve rs A = 4 πr 2 -C R ev ie w C op y es a We need to find Copyright Material - Review Only - Not for Redistribution ve rs ity ev ie dV when r = 5. dt am br id b We need to find w ge C U ni op y Chapter 8: Further differentiation -R 4 3 πr 3 -C V = Pr es s dV = 100 π dr dV dV dr = × Using the chain rule, dt dr dt 1 = 100 π × 5π = 20 C op y ve rs ity When r = 5 , U R ni ev ie w C op y dV = 4 πr 2 dr Differentiate with respect to r. br ev id ie w ge The volume is increasing at a rate of 20 cm 3 s −2. -R am WORKED EXAMPLE 8.14 s -C Water is poured into the conical container shown, at a rate of 2π cm 3 s −1. op Pr y es 1 πh3, After t seconds, the volume of water in the container, V cm 3, is given by V = 12 where h cm is the height of the water in the container. 231 ity h b Given that the container has radius 10 cm and height 20 cm, find the rate of change of h when the container is half full. Give your answer correct to 3 significant figures. op ni dh when h = 5. dt C br 1 πh3 12 dV 1 = πh2 dh 4 s Differentiate with respect to h. op y ni ve rs ity Pr es dV 25 π = dh 4 dh dh dV = × Using the chain rule, dt dV dt 4 = × 2π 25 π = 0.32 When h = 5, -R s es am br ev ie id g w e C U The height is increasing at a rate of 0.32 cm s −1. -C R ev ie w C op y -C -R am V = w dV = 2 π. dt ev id Volume increasing at a rate of 2π cm 3 s −1, so ie ge a We need to find U R Answer y ve ev ie w rs C a Find the rate of change of h when h = 5. Copyright Material - Review Only - Not for Redistribution ve rs ity 1 1 1000 3 1 1 πh3 = π ( 20 ) = π 2 12 2 12 3 1 1 1000 Using V = πh3 , πh3 = π 12 12 3 h3 = 4000 h = 15.874 dV 1 = π(15.874)2 = 197.9 When h = 15.874 , dh 4 dh dh dV = × dt dV dt 1 = × 2π 197.9 = 0.0317 cm s −1 C op ie br ev id EXERCISE 8E w ge U R ni ev ie y ve rs ity Using the chain rule, w C op y Pr es s -C -R am br id ev ie b Volume when half full = w ge C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 -R am 1 A circle has radius r cm and area A cm 2. y Pr 2 A sphere has radius r cm and volume V cm 3. 1 The radius is increasing at a rate of cm s −1. 2π Find the rate of increase of the volume when V = 36 π. y ve ie w rs ity op C 232 s Find the rate of increase of A when r = 4. es -C The radius is increasing at a rate of 0.1cm s −1. op ni ev 3 A cone has base radius r cm and a fixed height of 30 cm. br 4 A square has side length x cm and area A cm 2. -R am The area is increasing at a constant rate of 0.03 cm 2 s −1. Find the rate of increase of x when A = 25. s -C ev id ie w ge Find the rate of change of the volume when r = 5. C U R The radius of the base is increasing at a rate of 0.01cm s −1. op y es 5 A cube has sides of length x cm and volume V cm 3. Pr ni ve rs ity Find the rate of increase of x when V = 8. C U Find the rate of increase of x when x = 2. op The cuboid is heated and the volume increases at a rate of 0.15 cm 3 s −1. y 6 A solid metal cuboid has dimensions x cm by x cm by 4x cm. e 400 π . r Given that the radius of the cylinder is increasing at a rate of 0.25 cm s −1, find the rate of change of A when r = 10. -R s es am br ev ie id g w 7 A closed circular cylinder has radius r cm and surface area A cm 2, where A = 2 πr 2 + -C R ev ie w C The volume is increasing at a rate of 1.5 cm 3 s −1. Copyright Material - Review Only - Not for Redistribution ve rs ity The vertical cross-section is an equilateral triangle. ev ie am br id w ge 8 The diagram shows a water container in the shape of a triangular prism of length 120 cm. C U ni op y Chapter 8: Further differentiation Water is poured into the container at a rate of 24 cm 3 s −1. -R h 120 cm 3 Pr es s b Find the rate of change of h when h = 12. ve rs ity 9 Water is poured into the hemispherical bowl of radius 5 cm at a rate of 3π cm 3 s −1. w C op y -C a Show that the volume of water in the container, V cm , is given by V = 40 3 h2, where h cm is the height of the water in the container. y h C op ge U R ni ev ie After t seconds, the volume of water in the bowl, V cm 3, is given by 1 V = 5 πh2 − πh3, where h cm is the height of the water in the bowl. 3 a Find the rate of change of h when h = 1. id ie w b Find the rate of change of h when h = 3. The cone is initially completely filled with water. 10 cm -R am br ev 10 The diagram shows a right circular cone with radius 10 cm and height 30 cm. s 30 cm es h Pr a Show that the volume of water in the cone, V cm 3, when the height of the πh3 water is h cm is given by the formula V = . 27 b Find the rate of change of h when h = 20 . ity 233 ve ie w rs C op y -C Water leaks out of the cone through a small hole at the vertex at a rate of 4 cm 3 s −1. y op ni ev 11 Oil is poured onto a flat surface and a circular patch is formed. C U R The radius of the patch increases at a rate of 2 r cm s −1. ie w ge Find the rate at which the area is increasing when the circumference is 8 π cm. -R am The area of the patch increases at a rate of 5 cm 2 s −1. ev br id 12 Paint is poured onto a flat surface and a circular patch is formed. -C a Find, in terms of π, the radius of the patch after 8 seconds. es Pr 13 A cylindrical container of radius 8 cm and height 25 cm is completely filled with water. ity The water is then poured at a constant rate from the cylinder into an empty inverted cone. ni ve rs The cone has radius 15 cm and height 24 cm and its axis is vertical. y It takes 40 seconds for all of the water to be transferred. U op a If V represents the volume of water, in cm 3, in the cone at time t seconds, find C w the rate of change of the height of the water in the cone ie id g i e b When the depth of the water in the cone is 10 cm, find: -R s es am br ev ii the rate of change of the horizontal surface area of the water in the cone. -C R ev ie w C op y PS s b Find, in terms of π, the rate of increase of the radius after 8 seconds. Copyright Material - Review Only - Not for Redistribution dV in terms of π. dt ve rs ity C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 Increasing and decreasing functions y = f( x ) is decreasing for a given interval of x if op y Stationary points ve rs ity Stationary points (turning points) of a function y = f( x ) occur when C y U ni dy = 0 dx w ie ev br -R am dy = 0 dx ge the gradient is positive to the left of the maximum and negative to the right. At a minimum point: es d2 y dy = 0 and 2 , 0, then the point is a maximum point. dx dx If d2 y dy = 0 and 2 . 0, then the point is a minimum point. dx dx If dy d2 y = 0 and 2 = 0, then the nature of the stationary point can be found using the dx dx Pr y If op C w ge first derivative test. id ie Connected rates of change ev s y op -R s es -C am br ev ie id g w e C U R ev ie w ni ve rs C ity dx 1 . = y d dy dx -R dy dy dx . = × dt dx dt es You may also need to use the rule: op y • -C be connected using the chain rule: Pr br When two variables, x and y, both vary with a third variable, t, the three variables can am • y rs ve ni U C w ie ev R ity op • • • s -C the gradient is negative to the left of the minimum and positive to the right. Second derivative test for maximum and minimum points 234 dy = 0. dx C op At a maximum point: id w ev ie dy , 0 throughout the interval. dx First derivative test for maximum and minimum points • • • • R -R dy . 0 throughout the interval. dx Pr es s y = f( x ) is increasing for a given interval of x if -C • • • w ev ie am br id ge Checklist of learning and understanding Copyright Material - Review Only - Not for Redistribution ve rs ity ge C U ni op y Chapter 8: Further differentiation ev ie am br id The volume of a spherical balloon is increasing at a constant rate of 40 cm 3 per second. Find the rate of increase of the radius of the balloon when the radius is 15 cm. 4 3 πr .] 3 An oil pipeline under the sea is leaking oil and a circular patch of oil has formed on the surface of the sea. At midday the radius of the patch of oil is 50 m and is increasing at a rate of 3 metres per hour. Find the rate at which the area of the oil is increasing at midday. ve rs ity y [4] y A watermelon is assumed to be spherical in shape while it is growing. Its mass, M kg, and radius, r cm, are related by the formula M = kr 3 , where k is a constant. It is also assumed that the radius is increasing at a constant rate of 0.1 centimetres per day. On a particular day the radius is 10 cm and the mass is 3.2 kg. Find the value of k and the rate at which the mass is increasing on this day. ie w ge br ev id [5] -R am Cambridge International AS & A Level Mathematics 9709 Paper 11 Q4 June 2012 y s -C 5 [5] C op 4 4 8 A curve has equation y = 27 x − . Show that the curve has a stationary point at x = − and 3 ( x + 2)2 determine its nature. ni ev ie 3 R [4] Cambridge International AS & A Level Mathematics 9709 Paper 11 Q3 November 2012 U w C op Pr es s -C [The volume, V , of a sphere with radius r is V = 2 -R 1 w END-OF-CHAPTER REVIEW EXERCISE 8 235 Q O x op P (p, 0) y X (–2, 0) ni ev ve ie w rs C ity op Pr y es y = 2x2 w [2] ev ie ge id Express the area, A, of triangle XPQ in terms of p. br i C U R The diagram shows the curve y = 2 x 2 and the points X ( −2, 0) and P ( p, 0). The point Q lies on the curve and PQ is parallel to the y-axis. -R Find the rate at which A is increasing when p = 2. op y es Cambridge International AS & A Level Mathematics 9709 Paper 11 Q2 June 2015 ym y op w ie e C U A farmer divides a rectangular piece of land into 8 equal-sized rectangular sheep pens as shown in the diagram. Each sheep pen measures x m by y m and is fully enclosed by metal fencing. The farmer uses 480 m of fencing. Show that the total area of land used for the sheep pens, A m 2, is given by A = 384x − 9.6x 2. ii Given that x and y can vary, find the dimensions of each sheep pen for which the value of A is a maximum. (There is no need to verify that the value of A is a maximum.) ev -R br [3] s Cambridge International AS & A Level Mathematics 9709 Paper 11 Q5 June 2016 es am [3] ie id g w i -C ev xm ni ve rs C ity Pr 6 R [3] s -C ii am The point P moves along the x-axis at a constant rate of 0.02 units per second and Q moves along the curve so that PQ remains parallel to the y-axis. Copyright Material - Review Only - Not for Redistribution ve rs ity am br id ev ie w ge C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 Pr es s -C -R The variables x, y and z can take only positive values and are such that z = 3x + 2 y and xy = 600. 1200 . i Show that z = 3x + x ii Find the stationary value of z and determine its nature. [1] [6] U R x 2y y 3y 3x ni ev ie w C ve rs ity 8 C op op y Cambridge International AS & A Level Mathematics 9709 Paper 11 Q6 June 2011 ge y w 7 id ie 4x br ev The diagram shows the dimensions in metres of an L-shaped garden. The perimeter of the garden is 48 m. Find an expression for y in terms of x. ii Given that the area of the garden is A m 2, show that A = 48x − 8x 2. -R [1] [2] s -C am i Pr Cambridge International AS & A Level Mathematics 9709 Paper 11 Q7 November 2011 8 A curve has equation y = + 2 x. x dy d2 y [3] and 2 . i Find dx dx ii Find the coordinates of the stationary points and state, with a reason, the nature of each stationary point. [5] ni op y ve ie ev C w ie Cambridge International AS & A Level Mathematics 9709 Paper 11 Q5 November 2015 -R am br 10 ev id ge U R [4] ity 9 w C 236 rs op y es iii Given that x can vary, find the maximum area of the garden, showing that this is a maximum value rather than a minimum value. s -C x cm x cm op y es y cm Pr ity i Express y in terms of x. ii Show that the area of the plate, A cm 2, is given by A = 30 x − x 2. [2] op y ni ve rs [2] U Given that x can vary, e C iii find the value of x at which A is stationary, w ie ev -R s es am [2] [2] Cambridge International AS & A Level Mathematics 9709 Paper 11 Q8 November 2010 br id g iv find this stationary value of A, and determine whether it is a maximum or a minimum value. -C R ev ie w C The diagram shows a metal plate consisting of a rectangle with sides x cm and y cm and a quarter-circle of radius x cm. The perimeter of the plate is 60 cm. Copyright Material - Review Only - Not for Redistribution ve rs ity am br id ev ie w ge C U ni op y Chapter 8: Further differentiation 11 A curve has equation y = x 3 + x 2 − 5x + 7. Find the coordinates of the two stationary points on the curve and determine the nature of each stationary point. [5] -R [4] -C b Find the set of values of x for which the gradient of the curve is less than 3. Pr es s a x metres C op y r metres ni ev ie w C ve rs ity op y 12 id ie w ge U R The inside lane of a school running track consists of two straight sections each of length x metres, and two semicircular sections each of radius r metres, as shown in the diagram. The straight sections are perpendicular to the diameters of the semicircular sections. The perimeter of the inside lane is 400 metres. i Show that the area, A m 2, of the region enclosed by the inside lane is given by A = 400 r − πr 2 . ii Given that x and r can vary, show that, when A has a stationary value, there are no straight sections in the track. Determine whether the stationary value is a maximum or a minimum. [5] -C -R am br ev [4] es s Cambridge International AS & A Level Mathematics 9709 Paper 11 Q8 November 2013 ii Find the nature of each of the stationary points. [3] ve rs w ie Pr Show that the origin is a stationary point on the curve and find the coordinates of the other stationary [4] point in terms of p. ity i C op y 13 The equation of a curve is y = x 3 + px 2, where p is a positive constant. 3 2 y op ni ev Another curve has equation y = x + px + px. C ie ev id br -R am es s -C ity Pr op y -R s es -C am br ev ie id g w e C U op y ni ve rs C w ie ev R [3] Cambridge International AS & A Level Mathematics 9709 Paper 11 Q9 June 2015 w ge U R iii Find the set of values of p for which this curve has no stationary points. Copyright Material - Review Only - Not for Redistribution 237 op y ve rs ity ni C U ev ie w ge -R am br id Pr es s -C y ni C op y ve rs ity op C w ev ie op y ve ni w ge C U R ev ie w Chapter 9 Integration rs C ity op Pr y es s -C -R am br ev id ie w ge U R 238 id es s -C -R am br ev understand integration as the reverse process of differentiation, and integrate ( ax + b ) n (for any rational n except −1 ), together with constant multiples, sums and differences solve problems involving the evaluation of a constant of integration evaluate definite integrals use definite integration to find the: area of a region bounded by a curve and lines parallel to the axes, or between a curve and a line, or between two curves volume of revolution about one of the axes. • • op y ni ve rs -R s es -C am br ev ie id g w e C U R ev ie w C ity Pr op y ■ ■ ■ ■ ie In this chapter you will learn how to: Copyright Material - Review Only - Not for Redistribution ve rs ity ge C U ni op y Chapter 9: Integration What you should be able to do IGCSE / O Level Mathematics Substitute values for x and y into equations of the form y = f( x ) + c and solve to find c . 1 a Given that the line y = 5x + c passes through the point (3, − 4), find the value of c. Pr es s -C y b Given that the curve y = x 2 − 2 x + c passes through the point ( −1, 2), find the value of c. ve rs ity op C 2 Find the x-coordinates of the points where the curve crosses the x-axis. Find the x-coordinates of the points where a curve crosses the x-axis. ni ev id br -R am a dy . dx y = 3x8 − 13x − 10 b y = 5x 2 − 4x + 10 x es s -C y=3 x −x Pr y op ity In Chapters 7 and 8 you studied differentiation, which is the first basic tool of calculus. In this chapter you will learn about integration, which is the second basic tool of calculus. We often refer to integration as the reverse process of differentiation. It has many applications; for example, planning spacecraft flight paths, or modelling real-world behaviour for computer games. ni op y ve rs C w ie b 3 Find ie Differentiate constant multiples, sums and differences of expressions containing terms of the form ax n. Why do we study integration? ev y = 3x 2 − 13x − 10 w ge U R Chapter 7 a C op w IGCSE / O Level Mathematics ev ie Check your skills -R Where it comes from y am br id ev ie w PREREQUISITE KNOWLEDGE br ie 9.1 Integration as the reverse of differentiation ev id w ge C U R Isaac Newton and Gottfried Wilhelm Leibniz formulated the principles of integration independently, in the 17th century, by thinking of an integral as an infinite sum of rectangles of infinitesimal width. Pr op y es You learnt the rule for differentiating power functions: ity op ie ev y = x3 – 0.1 y = x3 es s y = x3 – 59 dy = 3x2 dx -R id g -C am br y = x3 + 4 1 2 C y = x3 + e y = x3 – 7 w U Applying this rule to functions of the form y = x 3 + c, we obtain: R ev y ni ve rs dy n −1 . = nx dx ie w C KEY POINT 9.1 If y = x n , then s -C -R am dy In Chapter 7, you learnt about the process of obtaining when y is known. We call this dx process differentiation. Copyright Material - Review Only - Not for Redistribution WEB LINK Explore the Calculus meets functions station on the Underground Mathematics website. 239 ve rs ity C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 -R am br id ev ie w ge This shows that there are an infinite number of functions that when differentiated give the answer 3x 2. They are all of the form y = x 3 + c, where c is some constant. dy is known. In this chapter you will learn about the reverse process of obtaining y when dx We can call this reverse process antidifferentiation. C op ni U R ev ie w Because of this theorem, we do not need to use the term antidifferentiation. So from now on, we will only talk about integration, whether we are reversing the process of differentiation or finding the area under a graph. w ge EXPLORE 9.1 e y=− -R 1 −7 x + 0.2 7 c y= 1 15 x +1 15 f y= 2 2 5 x − 3 8 3 s 1 −2 x +3 2 Pr y d y=− es -C am br ev id ie dy for each of the following functions. dx 1 1 a y = x3 − 2 b y = x6 + 8 3 6 1 Find y C ve rs ity op y Pr es s -C There is a seemingly unrelated problem that you will study in Section 9.6: what is the area under the graph of y = 3x 2 ? The process used to answer that question is known as integration. There is a remarkable theorem due to both Newton and Leibniz that says that integration is essentially the same as antidifferentiation. This is now known as the Fundamental theorem of Calculus. op w ge -R am br ev id ie From the class discussion we can conclude that: KEY POINT 9.2 dy 1 = . dx x C U R ni 4 Discuss with your classmates whether your rule works for finding y when y rs 3 Describe your rule, in words. ve ev ie w C ity op 2 Discuss your results with those of your classmates and try to find a rule for dy obtaining y if = x n. dx 240 dy 1 n +1 = x n , then y = x + c (where c is an arbitrary constant and n ≠ −1). n+1 dx es You may find it easier to remember this in words: Pr op y s -C If op w e 1 n +1 x + c (where c is an arbitrary constant and n ≠ −1). n+1 ev es s -R br ∫ is used to denote integration. am The special symbol ie id g If f ′( x ) = x n, then f( x ) = C U KEY POINT 9.3 y Using function notation we write this rule as: -C R ev ie w ni ve rs C ity ‘Increase the power n by 1 to obtain the new power, then divide by the new power. Remember to add a constant c at the end.’ Copyright Material - Review Only - Not for Redistribution ve rs ity C U ni op y Chapter 9: Integration w am br id ∫ x dx = 4 x + c ev ie 1 ge When we need to integrate x 3, for example, we write: 4 -R 3 ∫ x dx is called the indefinite integral of x with respect to x. 3 Pr es s -C 3 We call it ‘indefinite’ because it has infinitely many solutions. + c (where c is a constant and n ≠ −1). y ev ie n +1 C op 1 ∫ x dx = n + 1 x n U R ve rs ity KEY POINT 9.4 ni w C op y Using this notation, we can write the rule for integrating powers as: ie ev id br KEY POINT 9.5 w ge We write the rule for integrating constant multiples of a function as: es s -C -R am ∫ kf(x ) dx = k ∫ f(x ) dx, where k is a constant. op Pr y We write the rule for integrating sums and differences of two functions as: ity ∫ f(x ) ± g(x ) dx = ∫ f(x ) dx ± ∫ g(x ) dx 241 y op w ge WORKED EXAMPLE 9.1 C U R ni ev ve ie w rs C KEY POINT 9.6 -R w ev ie id g -R s es Copyright Material - Review Only - Not for Redistribution 1 ( 52 ) 5 x2 + c 5 y op 1 +c x C =− = e U dy = x2 dx 3 +1 1 y= 3 x2 + c 1 + 2 = − x −1 + c br am c Pr ev R -C dy =x x dx 3 dy = x −2 dx 1 y= x −2 + 1 + c −2 + 1 ity 1 4 x +c 4 ni ve rs = b s dy = x3 dx 1 y= x3 + 1 + c 3+1 ie w C op y a c es -C Answer am br ev id ie Find y in terms of x for each of the following. 1 dy dy = x3 = 2 b a dx dx x = 2 2 x +c 5 ve rs ity ev ie am br id WORKED EXAMPLE 9.2 w ge C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 Find f( x ) in terms of x for each of the following. -R Pr es s -C C op c ve rs ity y f ′( x ) = 4x 3 − 2 x −2 + 4x 0 4 2 4 f( x ) = x 4 − x −1 + x1 + c 4 ( −1) 1 Write in index form ready for integration. U R ni ev ie a y w Answer C op b 2 +4 x2 1 f ′( x ) = 8x 2 − + 2x 2x 4 ( x + 3)( x − 1) f ′( x ) = x f ′( x ) = 4x 3 − a br ev id ie w ge = x 4 + 2 x −1 + 4x + c 2 = x 4 + + 4x + c x 1 −4 x + 2 x1 2 8 1 2 x −3 + x 2 + c f( x ) = x 3 − 3 2( −3) 2 8 3 1 −3 = x + x + x2 + c 3 6 8 3 1 = x + 3 + x2 + c 3 6x f ′( x ) = 8x 2 − Write in index form ready for integration. 3 1 − 1 Pr ity y Write in index form ready for integration. U R rs x2 + 2x − 3 x op f ′( x ) = ev c ve ie w C 242 ni op y es s -C -R am b ( 23 ) 5 C 3 ( 21 ) 1 w x2 − x2 + c ie 3 2 ev ( 52 ) 5 x2 + id 1 br f( x ) = ge = x 2 + 2 x 2 − 3x 2 3 -R am 2 2 4 2 x + x −6 x +c 5 3 es s -C = Pr op y WORKED EXAMPLE 9.3 Answer ∫ 4x 2 − 3 x dx x 2 C 3 w 4x 4 4x 3 3x 2 + − +c 4 3 2 4x 3 3x 2 = x4 + − +c 3 2 s -R ev ie = es br id g e U ∫ x(2x − 1)(2x + 3) dx = ∫ (4x + 4x − 3x ) dx am a b y x(2 x − 1)(2 x + 3) dx op ∫ ni ve rs a -C R ev ie w C ity Find: Copyright Material - Review Only - Not for Redistribution ve rs ity C 1 y = 2 x 2 − 6x 2 + c = 2x2 − 6 x + c y ve rs ity op -R ( ) C f ′( x ) = Pr 2 8 + + 6x x3 x2 f ′( x ) = 9 3 − −4 x7 x2 e dy = dx b ∫ 20x dx e ∫ 3 x dx 2 c ∫ 3x dx f ∫ x x dx w 2 −2 es 5 ∫ (x − 3) dx c ∫ (2 x − 1) dx e ∫ x2 − 1 dx 2x2 f ∫ x3 + 6 dx 2x3 ∫ x 4 − 10 dx x x ∫ 3 2 x − x 2 x dx 2 es s -R br ev ie id g w e h op y 2 C ∫ x2 + 2 x dx 3x x (x 2 + 1) dx am dy 5x 2 + 3x + 1 = dx x ie ev 3 Pr ∫ 3 -C f b ity d ni ve rs ∫ (x + 1)(x + 4) dx U op y a C w dy = x( x + 2)( x − 8) dx s 3 5 Find each of the following. g C x( x − 3) -R id br am 4 ∫ x dx -C d ∫ 12x dx 5 c y dy = 2 x 2 (3x + 1) dx ni b ge R rs dy x 4 − 2 x + 5 = dx 2x3 ve d U ie dy = x( x + 5) dx ev a a ie 243 3 Find y in terms of x for each of the following. 4 Find each of the following. ev dy 4 = dx x ity d f f ′( x ) = 3x5 + x 2 − 2 x es c b dy = 12 x 3 dx s -C y op f ′( x ) = 5x 4 − 2 x 3 + 2 w C a c ie dy 1 = dx 2 x 3 ev id br e am dy = 14x 6 dx -R dy 3 = dx x 2 b w ni d ge dy = 15x 2 dx U R a 2 Find f( x ) in terms of x for each of the following. R C op 1 Find y in terms of x for each of the following. op w w 1 4 2 3 x − 1 x2 + c 2 2 = EXERCISE 9A ev ie ev ie ∫ 1 − 2 dx x − 4 3 x Pr es s am br id ∫ 4x 2 − 3 x dx = x -C b ge U ni op y Chapter 9: Integration Copyright Material - Review Only - Not for Redistribution i 2 ve rs ity C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 w ge 9.2 Finding the constant of integration -R am br id ev ie The next two examples show how we can find the equation of a curve if we know the gradient function and the coordinates of a point on the curve. Pr es s -C WORKED EXAMPLE 9.4 dy 6x5 − 18 = , and (1, 6) is a point on the curve. dx x3 Find the equation of the curve. y Write in index form ready for integration. ie w ge U ni dy 6x5 − 18 = dx x3 = 6x 2 − 18x −3 y = 2 x 3 + 9x −2 + c 9 = 2x3 + 2 + c x C op Answer id R ev ie w C ve rs ity op y A curve is such that y ity op rs WORKED EXAMPLE 9.5 y ve ie w C 244 Pr 9 − 5. x2 The equation of the curve is y = 2 x 3 + es s -C -R am br ev When x = 1, y = 6. 9 6 = 2(1)3 + 2 + c (1) 6 = 2+9+c c = −5 op ie ev br f( x ) = 3x5 − 3x 2 + c id f ′( x ) = 15x 4 − 6x w ge Answer C U R ni ev The function f is such that f ′( x ) = 15x 4 − 6x and f( −1) = 1. Find f( x ). -R am Using f( −1) = 1 gives: s es Pr op y -C 1 = 3( −1)5 − 3( −1)2 + c 1 = −3 − 3 + c c=7 y op -R s es -C am br ev ie id g w e C U R ev ie w ni ve rs C ity ∴ f( x ) = 3x5 − 3x 2 + 7 Copyright Material - Review Only - Not for Redistribution ve rs ity ge C U ni op y Chapter 9: Integration am br id ev ie w WORKED EXAMPLE 9.6 dy = 6x + k , where k is a constant. The gradient of the normal to the curve at the dx 1 point (1, −3) is . Find the equation of the curve. 2 Pr es s -C -R A curve is such that Answer ni U (1) dy = 6(1) + k = 6 + k dx id 1 so gradient of tangent = −2 2 6 + k = −2 k = −8 w ge R When x = 1, y −3 = 3(1) + k (1) + c c + k = −6 C op y = −3. 2 ie When x = 1, rs ve op ni dy = 3x 2 + 1, P = (1, 4) dx c dy 4 = 2, dx x e dy 2 = − 1, P = (4, 6) dx x ev ie d -R am f P = (3, 7) dy (1 − x )2 = , P = (9, 5) x dx s dy k = − 2 , where k is a constant. Given that the curve passes through the points (6, 2.5) dx x and ( −3, 1), find the equation of the curve. es Pr ity op y op y ni ve rs dy = kx 2 − 12 x + 5, where k is a constant. Given that the curve passes through the points dx (1, −3) and (3, 11), find the equation of the curve. 3 A curve is such that dy 6 = kx 2 − 3 , where k is a constant. Given that the curve passes through the point dx x P (1, 6) and that the gradient of the curve at P is 9, find the equation of the curve. -R s es am br ev ie id g w e C U 4 A curve is such that -C C w ie dy 2x3 − 6 , = dx x2 w P = (4, 9) br id ge C U a dy and a point P on the curve. dx dy b = 2 x(3x − 1), P = ( −1, 2) dx -C R ev 1 Find the equation of the curve, given y C w ie EXERCISE 9B 2 A curve is such that ev 245 ity op Pr y The equation of the curve is y = 3x 2 − 8x + 2. es s Substituting for k into (1) gives c = 2. -R -C am br ev Gradient of normal = R Integrate. ve rs ity ev ie w C op y dy = 6x + k dx y = 3x 2 + kx + c Copyright Material - Review Only - Not for Redistribution ve rs ity C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 ge dy = 5x x + 2. Given that the curve passes through the point (1, 3), find: dx a the equation of the curve -R b the equation of the tangent to the curve when x = 4 . dy = kx + 3, where k is a constant. The gradient of the normal to the curve at the point dx 1 (1, −2) is − . Find the equation of the curve. 7 7 A function y = f( x ) has gradient function f ′( x ) = 8 − 2 x. The maximum value of the function is 20. Find f( x ) and sketch the graph of y = f( x ). ve rs ity C op y Pr es s -C 6 A curve is such that w ev ie am br id w 5 A curve is such that C op U R ni ev ie y dy = 3x 2 + x − 10 . Given that the curve passes through the point (2, −7) find: dx a the equation of the curve 8 A curve is such that w d2 y = 12 x + 12. The gradient of the curve at the point (0, 4) is 10. dx 2 a Express y in terms of x. ev id ie 9 A curve is such that am br PS ge b the set of values of x for which the gradient of the curve is positive. d2 y = −6x − 4. Given that the curve has a minimum point at ( −2, −6), find the equation dx 2 s 10 A curve is such that es -C -R b Show that the gradient of the curve is never less than 4. op Pr y of the curve. 11 A curve y = f( x ) has a stationary point at P (2, −13) and is such that f ′( x ) = 2 x 2 + 3x − k, where k is a ity C ve ie w constant. rs 246 y op ni ev a Find the x-coordinate of the other stationary point, Q, on the curve y = f( x ). dy = k + x, where k is a constant. dx a Given that the tangents to the curve at the points where x = 5 and x = 7 are perpendicular, find the value of k. ie w ge 12 A curve is such that br ev id PS C U R b Determine the nature of each of the stationary points P and Q. -C -R am b Given also that the curve passes through the point (10, −8), find the equation of the curve. 4 . Find f ′( x ) and f( x ). x3 15 A curve is such that dy = 3 − 2 x and (1, 11) is a point on the curve. dx op y ni ve rs ity Pr d2 y = 2 x + 8. Given that the curve has a minimum point at (3, −49), find the coordinates dx 2 of the maximum point. 14 A curve is such that -R am br ev ie id g w e C U a Find the equation of the curve. 1 b A line with gradient is a normal to the curve at the point (4, 5). Find the equation of this normal. 5 dy 16 A curve is such that = 3 x − 6 and the point P (1, 6) is a point on the curve. dx a Find the equation of the curve. es s b Find the coordinates of the stationary point on the curve and determine its nature. -C R ev ie w C op y es s 13 A curve y = f( x ) has a stationary point at (1, −1) and is such that f ′′( x ) = 2 + Copyright Material - Review Only - Not for Redistribution ve rs ity C U ni op y Chapter 9: Integration ge d2 y 12 = 2 − 3 . The curve has a stationary point at P where x = 1. Given that the curve dx 2 x passes through the point (2, 5), find the coordinates of the stationary point P and determine its nature. am br id ev ie w 17 A curve is such that -R dy = 2 x − 5 and the point P (3, −4) is a point on the curve. The normal to the curve dx at P meets the curve again at Q . b Find the equation of the normal to the curve at P. c Find the coordinates of Q. w ev ie Pr es s a Find the equation of the curve. ve rs ity C op y -C 18 A curve is such that 1 ev br id ie w ge d 1 (3x − 1)7 = (3x − 1)6 dx 3 × 7 ∫ (3x − 1) dx = 3 × 7 (3x − 1) + c 6 7 -R am Hence, es s -C This leads to the general rule: ve ie w rs C Pr ∫ If n ≠ −1 and a ≠ 0, then 247 1 ( ax + b ) dx = ( ax + b ) n + 1 + c a ( n + 1) n ity op y KEY POINT 9.7 y id ie w ge C U ∫ R op ni ev It is very important to note that this rule only works for powers of linear functions. 1 ( ax 2 + b )3 + c. (Try differentiating the latter For example, ( ax 2 + b )6 dx is not equal to 3a expression to see why.) b 20 ∫ (1 − 4x ) dx 1 (2 x − 3)5 + c 10 Pr +c y 20 ∫ (1 − 4x ) dx = 20∫ (1 − 4x ) dx 20 (1 − 4x )−6 + 1 + c ( −4)( −6 + 1) ie ev -R s = (1 − 4x )−5 + c 1 = +c (1 − 4x )5 es am br id g w e = op −6 6 C b -C ev ie w = 4+1 5 ∫ 2x + 7 dx ity 4 ni ve rs C 1 ∫ (2x − 3) dx = 2(4 + 1) (2x − 3) U op y Answer a c 6 s 4 es -C ∫ (2x − 3) dx -R br am Find: a ev WORKED EXAMPLE 9.7 R C op ni U R In Chapter 7 you learnt that: y 9.3 Integration of expressions of the form (ax + b)n Copyright Material - Review Only - Not for Redistribution ve rs ity C w 1 1 +1 Pr es s a ∫ (2x − 7) dx d ∫ 3(1 − 2x ) dx g 2 dx 3x − 2 U ni 8 2 Find the equation of the curve, given 3 dy = (2 x − 1)3 , P = , 4 2 dx ∫ 3 2 dx 2x + 1 dy and a point P on the curve. dx dy = b dx es 1 , P = (3, 7) x−2 3 A curve is such that 3 Pr dy = dx ∫ 5 − 4x dx ∫ 4(7 − 2x ) dx 5 5 2 x + 5, P = (2, 2) op ev br s 12 − 4x − 2. 3x + 1 es dy = dx -R b the equation of the curve. w ie id a the equation of the normal to the curve at P am i 3 C U ge Given that the curve passes through the point P(2, 1), find: -C ∫ ( 2x + 1) dx y ve ni ev f 8 dy 4 = , P = (2, 4) dx ( 3 − 2 x )2 d 1 . Find the equation of the curve. 12 dy 5 = . 4 A curve is such that dx 2x − 3 R ∫ 2(5x − 2) dx 3 dy = k ( x − 5) , where k is a constant. The gradient of the normal to the curve at the dx point (4, 2) is 5 A curve is such that c ity c e 5 rs ie w C 248 ∫ (3x + 1) dx s -C op y a b h br ∫ id ge 5 am R ev ie 1 Find: y w C ve rs ity EXERCISE 9C C op op y -C = 5 2x + 7 + c +c w ) − (2 x + 7) 2 ie − 21 + 1 ev 2( 5 -R = -R ∫ ev ie − 5 dx = 5 (2 x + 7) 2 dx 2x + 7 am br id ∫ c ge U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 Pr op y a Show that the curve has a stationary point when x = 1 and determine its nature. ity dy 4 = , where k is a constant. The point P(3, 2) lies on the curve and the normal dx 2x + k to the curve at P is x + 4 y = 11. Find the equation of the curve. y -R s es -C am br ev ie id g w e C U op 6 A curve is such that R ev ie PS ni ve rs w C b Given that the curve passes through the point (0, 13), find the equation of the curve. Copyright Material - Review Only - Not for Redistribution ve rs ity C U ni op y Chapter 9: Integration w ge 9.4 Further indefinite integration -R am br id ev ie In this section we use the concept that integration is the reverse process of differentiation to help us integrate some more complicated expressions. ) b Hence, find ge Answer ie ev s 249 rs C w es 1 (3x 2 − 4)8 + c 8 EXERCISE 9D y ve ie 7 ity op y = 2 Pr -C 7 op ∫ x(x + 2) dx. 3 w ge 2 C U R ni ev 1 a Differentiate ( x 2 + 2)4 with respect to x. b Hence, find b Hence, find ∫ ev s 1 with respect to x. 4 − 3x 2 3x dx . (4 − 3x 2 )2 op 2 5 C U ∫ 2(2x − 3)(x − 3x + 5) dx. e b Hence, find y 5 a Differentiate ( x 2 − 3x + 5)6 with respect to x. ev -R s ∫ ( x + 3)7 dx. x es am br b Hence, find ie id g w 6 a Differentiate ( x + 3)8 with respect to x. -C R ev ie w C 4 a Differentiate es ∫ Pr op y b Hence, find dy kx 1 , show that = , and state the value of k. dx ( x 2 − 5)2 x2 − 5 4x dx. ( x 2 − 5)2 ity -C 3 a Given that y = 4 -R 2 ni ve rs br am ∫ x(2x − 1) dx. ie id 2 a Differentiate (2 x 2 − 1)5 with respect to x. b Hence, find 7 -R br am 1 ∫ 6x(3x − 4) dx = 8 ∫ 48x(3x − 4) dx 2 2 Use the chain rule. dy = (6x )(8)(3x 2 − 4)8 − 1 dx = 48x(3x 2 − 4)7 b ∫ 6x ( 3x − 4 ) dx. w Let y = (3x 2 − 4)8 id a ( C op ) ni ( y 7 8 d 3x 2 − 4 = 48x 3x 2 − 4 . dx U R a Show that Pr es s op C ev ie w WORKED EXAMPLE 9.8 ve rs ity ∫ f(x ) dx = F(x ) + c d [ F( x ) ] = f( x ) , then dx y If -C KEY POINT 9.8 Copyright Material - Review Only - Not for Redistribution ve rs ity C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 ev ie ∫ 3 x (2x x − 1) dx. am br id 4 -R b Hence, find w ge 7 a Differentiate (2 x x − 1)5 with respect to x. -C 9.5 Definite integration 4 ni 3 y 1 ∫ x dx = 4 x + c, where c is an arbitrary constant, is called the indefinite integral of x 3 with respect to x. U R ev ie w Recall that C op C ve rs ity op y Pr es s In the remaining sections of this chapter, you will be learning how to find areas and volumes of various shapes. To do this, you will be using a technique known as definite integration, which is an extension of the indefinite integrals you have been using up to now. In this section, you will learn this technique, before going on to apply it in the next section. w ge We can integrate a function between two specified limits. br -R -C 2 x3 dx am ∫ 4 ev id ie We write the integral of the function x 3 with respect to x between the limits x = 2 and x = 4 as: es 4 4 1 x 3 d x = x 4 + c 4 2 2 Pr y ve = 60 ni 4 w ge ∫ U 1 x 3 dx = x 4 4 2 2 C 4 br ev id ie 1 1 = × 44 − × 2 4 4 4 -R am = 60 4 op Note that the ‘c’s cancel out, so the process can be simplified to: R ev ie w C ity 1 1 = × 44 + c − × 2 4 + c 4 4 rs op y ∫ 250 The limits of integration are always written either to the right of the integral sign, as printed, or directly below and above it. They should never be written to the left of the integral sign. s The method for evaluating this integral is: ∫ x dx is called the definite integral of x with respect to x between the limits 2 and 4. 3 es s -C 3 2 Pr op y Hence, we can write the evaluation of a definite integral as: ni ve rs b U op b a a y ∫ f(x ) dx = [ F(x ) ] = F(b) − F(a ) -R s es am br ev ie id g w e C The following rules for definite integrals may also be used. -C R ev ie w C ity KEY POINT 9.9 TIP Copyright Material - Review Only - Not for Redistribution ve rs ity ev ie b ∫ f(x ) dx, where k is a constant ∫ [ f(x ) ± g(x ) ] dx = ∫ f(x ) dx ± ∫ g(x ) dx ∫ f(x ) dx = − ∫ f(x ) dx a b a a a a b ve rs ity C ∫ b f( x ) d x + ∫ c ∫ f(x ) dx f( x ) d x = b a w ge U R a c ni ev ie w KEY POINT 9.11 y b op y -C a b C op a Pr es s b kf ( x ) d x = k -R ∫ w ge am br id KEY POINT 9.10 b C U ni op y Chapter 9: Integration s c es y 2 ∫ ( 6x − x ) d x −2 2 −2 w ni C U R ) op ve ie ev ) ( = 2(2)3 + (2)−1 − 2(1)3 + (1)−1 y 2 6 = x 3 + x −1 3 1 ( -R s 0 1 ( 5x + 1 ) 2 d x 3 w ni ve rs C ity Pr op y 3 1 = ( 5x + 1 ) 2 (5) 3 2 0 es ∫ 3 ev = 13 2 5x + 1 d x = 0 ie 1 id br am -C ∫ 3 w ge 1 = 16 + − (2 + 1) 2 b 3 y op ev ie 3 2 = ( 5x + 1 ) 2 15 0 id g w e C U R 3 3 2 2 × 12 = × 16 2 − 15 15 -R s es -C am br ev ie 128 2 = − 15 15 = 8 25 2 251 rs 1 8 ∫ ( 5 − 2x ) dx ity op C 6x 4 − 1 dx = x2 1 2 ∫ 1 5x + 1 d x 0 Answer a ∫ 3 Pr ∫ b -R br 6x 4 − 1 dx x2 1 2 -C a am Evaluate: ev id ie WORKED EXAMPLE 9.9 Copyright Material - Review Only - Not for Redistribution ve rs ity ∫ 8(5 − 2x ) dx −2 1 Pr es s 1 8 9 ni U 2 ∫0 ( 10 − x2 ) dx es f 2x ∫ ( x + 5 ) dx. ( ie 5 a Given that y = x 3 − 2 1 C w ) , find ddxy . 5 2 4 C U 3 5 1 x w ie ev ( x + 1 )4 d x . -R 4 dx s am -C ∫ 10 es br id g e ( x + 1) , find dy . 6 a Given that y = b Hence, evaluate f s ∫ ∫ 2 ∫ 2 1 ev 9 3 dx 2 ( 2x − 3 ) c ie 3 ∫ x ( x − 2 ) dx. 0 R 2x + 1 d x y C 0 b Hence, evaluate 2 2 2 ∫ 3 x + x dx 1 op op y 2 ∫ (x + 3)(7 − 2x ) dx 4 ni U br am -C dy 2 , find . dx x +5 2 b Hence, evaluate 2 1 y ∫ 4 6 op ∫ ∫ 2 − x dx 2 ni ve rs ev R (3 − x )(8 + x ) dx x4 1 2 0 e 4 a Given that y = w b 6 2 dx −1 ( x − 2 ) ∫ c es 1 ge −1 ( 2x + 3 )3 d x id ∫ 0 e −1 8 − x2 x 2 d x −2 ∫ −1 2 Pr ity x (1 − x ) d x 3 Evaluate: d ev b 0 a f Pr ∫ 1 3x 2 − 2 + 1 d x x2 1 rs ie d ∫ 2 ∫ (2x − 3) dx 4 ∫−1 ( 4x2 − 2x ) dx ity C w a ve op y 2 Evaluate: 252 1 c s -C e 3 -R br 3 d 4 ∫ x dx 1 am 1 b ev 3 2 -R 2 ∫ 3x dx ie w ge id 1 Evaluate: a y ve rs ity C 4 4 = − 3 9 C op op y 4 = 5 − 2 x −2 = -R -C 8 = ( 5 − 2x )−1 2 1 − − ( ) ( ) −2 EXERCISE 9E R −2 ev ie am br id ∫ w ev ie 1 w ge 1 8 2 dx = −2 ( 5 − 2 x ) c C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 Copyright Material - Review Only - Not for Redistribution −2 ( x − 1 )3 d x 4 dx 5 − 2x ve rs ity C w ge 9.6 Area under a curve U ni op y Chapter 9: Integration ev ie am br id Consider the area bounded by the curve y = x 2 , the x-axis and the lines x = 2 and x = 5. -R y = x2 y C op O 2 x 5 ge U R ni ev ie w C ve rs ity op y Pr es s -C y -C As δx → 0, then A → 5 ∫ y dx. ∑ yδx. y s The approximation for A is then -R am br ev id ie w The area, A, of the region can be approximated by a series of rectangular strips of thickness δx (corresponding to a small increase in x) and height y (corresponding to the height of the function). y = x2 253 y op O C 2 5 x w ge U R ni ev ve ie w rs C ity op Pr y es 2 br ev id ie This leads to the general rule: -R am KEY POINT 9.12 s -C If y = f( x ) is a function with y ù 0, then the area, A, bounded by the curve y = f( x ), the x-axis b ∫ y d x. es and the lines x = a and x = b is given by the formula A = y op -R s es -C am br ev ie id g w e C U R ev ie w ni ve rs C ity Pr op y a Copyright Material - Review Only - Not for Redistribution ve rs ity ev ie w ge am br id WORKED EXAMPLE 9.10 C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 U 6 w ev id -C -R am br = (216) − (64) ge 3 = x3 3 4 ie R y ni 2 4 6x 4 C op ∫ 3x dx = 152 units2 O ve rs ity 6 y = 3x2 Pr es s -C y op C w ev ie Answer Area = y -R Find the area of the shaded region. op ∫ f(x ) dx will have a negative value. This a Pr y b If the required area lies below the x-axis, then 254 es s In Worked example 9.10, the required area is above the x-axis. rs WORKED EXAMPLE 9.11 y op ie ev -R s es 0 op U Area is 36 units2. y ni ve rs = (72 − 108) − (0 − 0) = −36 w Pr 6 -R s es -C am br ev ie id g w e C C 6 1 ∫0 ( x2 − 6x ) dx = 3 x3 − 2 x2 ie O ity 6 ev y = x2 – 6x w ge id br am -C op y Answer R y C U R ni Find the area of the shaded region. ve ev ie w C ity is because the integral is summing the y values, and these are all negative. Copyright Material - Review Only - Not for Redistribution 6 x ve rs ity C U ni op y Chapter 9: Integration ev ie am br id w ge The required region could consist of a section above the x-axis and a section below the x-axis. If this happens we must evaluate each area separately. -C -R This is illustrated in Worked example 9.12. y Pr es s WORKED EXAMPLE 9.12 y 2 ge 3 2 0 2 es ( ) Pr y ev 1 8 1 8 = (2)4 − (2)3 + 6(2)2 − (0)4 − (0)3 + 6(0)2 4 4 3 3 s -C am br 1 8 = x 4 − x 3 + 6x 2 4 0 3 = 6 23 − (0) op 6 6 2 2 255 ity = 6 23 2 6 y 3 rs ∫ x(x − 2)(x − 6) dx = ∫ ( x − 8x + 12x ) dx ve C w ie w ∫ ( x − 8x + 12x ) dx id 0 ie x( x − 2)(x − 6) dx = op C U R ni ev 1 8 = x 4 − x 3 + 6x 2 4 2 3 w ( ) ie id ge 1 8 1 8 = (6)4 − (6)3 + 6(6)2 − (2)4 − (2)3 + 6(2)2 4 3 4 3 ev = −42 23 -R am br = ( −36) − 6 23 1 Pr op y es s -C Hence, the total area of the shaded regions = 6 23 + 42 23 = 49 3 units2. ev C U w x = f(y) ie x -R s es am br ev id g e a -C R A op b y ni ve rs ity y ie w C Area enclosed by a curve and the y-axis O 6 x -R ∫ 2 2 C op U R Answer y = x(x – 2)(x – 6) y O ni ev ie w C ve rs ity op Find the total area of the shaded regions. Copyright Material - Review Only - Not for Redistribution ve rs ity ev ie am br id KEY POINT 9.13 w ge C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 If x = f( y ) is a function with x ù 0, then the area, A, bounded by the curve x = f( y ), the y-axis -R U id ev br -R am s -C 1 1 = 2(4)2 − (4)3 − 2(0)2 − (0)3 3 3 y es = 10 23 rs ni op y ve EXERCISE 9F ev ie w C ity op Pr Area is 10 23 units2 . 1 Find the area of each shaded region. b y ie w ge y y = 2x + ev br id a C U R s x O Pr 4 es -C op y c d 1 3 w y op x O ev ie id g es s -R br am -C 4 w e 5 y = 4√x – 2x C U ev ie y = x(x – 5) O x y ni ve rs ity y R 3 –1 x2 -R am y = x3 – 8x2 + 16x O C x O 4 1 = y2 − y3 3 2 0 256 x = y(4 – y) w R 4 ge 2 ni ∫ x dy = ∫ ( 4y − y ) d y 4 y 4 y C op Answer 0 a ie ev ie w C Find the area of the shaded region. 0 4 b ve rs ity op y WORKED EXAMPLE 9.13 Area = ∫ x d y when x ù 0. Pr es s -C and the lines y = a and y = b is given by the formula A = Copyright Material - Review Only - Not for Redistribution x ve rs ity w y ev ie am br id ge R1 A O B x Pr es s -C R2 y The diagram shows the curve y = x( x − 2)( x − 4) that crosses the x-axis at the points O(0, 0), A(2, 0) and B(4, 0). ve rs ity op C w -R 2 C U ni op y Chapter 9: Integration y C op U R ni ev ie Show by integration that the area of the shaded region R1 is the same as the area of the shaded region R2. y = x( x − 3)( x + 1) c ( ) y = x x2 − 9 y = x(2 x − 1)( x + 2) d y = ( x − 1)( x + 1)( x − 4) ev -R am ie b br id a w ge 3 Sketch the curve and find the total area bounded by the curve and the x-axis for each of these functions. es Pr ity y = 3 x , the x-axis and the lines x = 1 and x = 4 e y= f y = 2 x + 3, the x-axis and the line x = 3 y ve d ni op 4 , the x-axis and the lines x = 1 and x = 9 x C U w R ev ie w c 257 rs b y = x 4 − 6x 2 + 9, the x-axis and the lines x = 0 and x = 1 5 y = 2 x + 2 , the x-axis and the lines x = 1 and x = 2 x 8 y = 5 + 3 , the x-axis and the lines x = 2 and x = 5 x ge C op y a s -C 4 Sketch the curve and find the enclosed area for each of the following. br y = √2x + 1 es y 6 x C U O op ie 1 y ni ve rs w C ity Pr op y 3 ev -R s es am br ev ie id g w e The diagram shows the curve y = 2 x + 1. The shaded region is bounded by the curve, the y-axis and the line y = 3. Find the area of the shaded region. -C R ie -R s x = y2 + 1 , the y-axis and the lines y = −1 and y = 2 -C b y = x 3 , the y-axis and the lines y = 8 and y = 27 am a ev id 5 Sketch the curve and find the enclosed area for each of the following. Copyright Material - Review Only - Not for Redistribution ve rs ity ge y = 2x2 + 1 ev ie am br id 7 w y O x ve rs ity C op y Pr es s -C -R 9 Find the area of the region bounded by the curve y = 2 x + 1, the line y = 9 and the y-axis. 2 y w ev ie C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 12 , the x-axis and the x2 R ni C op 8 a Find the area of the region enclosed by the curve y = ge U lines x = 1 and x = 4. br ev id ie w b The line x = p divides the region in part a into two parts of equal area. Find the value of p. 9 y es Pr y ity op rs C 258 x 2 ve ie w O op ev C id w y s -C -R am br ev y = 2√ x Pr es x+y=8 op y O x ni ve rs ity Find the shaded area enclosed by the curve y = 2 x , the line x + y = 8 and the x-axis. y 11 The tangent to the curve y = 8x − x 2 at the point (2, 12) cuts the x-axis at the point P. (2, 12) C U op y y = 8x – x2 id g w e a Find the coordinates of P. ie 8 x s -R ev O P es am br b Find the area of the shaded region. -C C w ie ev ie ge U R ) ni ( y d x x2 + 5 = . 2 dx x +5 b Use your result from part a to evaluate the area of the shaded region. a Show that 10 R x √ x2 + 5 s -C -R am y= Copyright Material - Review Only - Not for Redistribution ve rs ity C y ge 12 y = √ 2x + 1 Pr es s The diagram shows the curve y = 2 x + 1 that intersects the x-axis at A. The normal to the curve at B(4, 3) meets the x-axis at C. Find the area of the shaded region. y y = f(x) U R ni y ev br id ie w ge Q(7, 12) s x es O Pr The figure shows part of the curve y = f( x ). The points P(2, 4) and Q(7, 12) lie 7 on the curve. Given that ∫ y dx = 42, find the value of ∫ x dy. 4 op C U R ni y y ve rs 2 14 -R am br ev id ie w ge A (2, 8) s WEB y = g(x) x Pr es O on the curve. Given that ∫ 6 ity The figure shows part of the curve y = g(x ). The points A(2, 8) and B(6, 1) lie y d x = 16 , find the value of ni ve rs 2 ∫ x d y. y -R s -C am br ev ie id g w e C U op ev R 8 1 es C op y -C B (6, 1) ie w 259 12 ity C op y -C -R am P(2, 4) w ie ev PS C op 13 ve rs ity ev ie PS C x O A w C op y -C -R am br id ev ie B w U ni op y Chapter 9: Integration Copyright Material - Review Only - Not for Redistribution Try the following resources on the Underground Mathematics website: • What else do you know? • Slippery areas. ve rs ity w ge 9.7 Area bounded by a curve and a line or by two curves C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 -R am br id ev ie The following example shows a possible method for finding the area enclosed by a curve and a straight line. Pr es s -C WORKED EXAMPLE 9.14 op y The diagram shows the curve y = − x 2 + 8x − 5 and the line y = x + 1 that intersect at the points (1, 2) and (6, 7). y O U ni C op C w ev ie R ie 6 id Area = area under curve − area of trapezium ∫ ( −x + 8x − 5 ) dx − 2 × (2 + 7) × 5 ev am 1 1 -R 2 br = 1 w ge Answer y = –x2 + 8x – 5 y=x+1 ve rs ity Find the area of the shaded region. y 6 s -C 1 1 = − x 3 + 4x 2 − 5x − 22 2 3 1 5 ni op y ve There is an alternative method for finding the shaded area in Worked example 9.14. ev y = f(x) ie w ge id C U R y A am a b y = g(x) ev br O x -R ie w rs ity = 20 6 units 2 C 260 Pr op y es 1 1 3 2 1 = − ( 6 ) + 4 ( 6 ) − 5(6) − − (1)3 + 4(1)2 − 5(1) − 22 2 3 3 Pr op y es s -C If two functions, f( x ) and g( x ), intersect at x = a and x = b, then the area, A, enclosed between the two curves is given by: A= a a -R s es -C am br ev ie id g w e C U R b ∫ [ f(x ) − g(x ) ] dx y a or op b ity b ∫ f(x ) dx − ∫ g(x ) dx ni ve rs A= ev ie w C KEY POINT 9.14 Copyright Material - Review Only - Not for Redistribution 6 x ve rs ity C U ni op y Chapter 9: Integration op y Pr es s -C y=x+1 C 6 x C op 6 y Using f( x ) = −x 2 + 8x − 5 and g( x ) = x + 1 gives: ni w ev ie 1 ve rs ity O 6 2 am 1 w 1 br 1 6 6 ie id 2 ev 1 -R 1 6 ge f( x ) d x − U ∫ ∫ g(x ) dx = ∫ ( −x + 8x − 5 ) dx − ∫ (x + 1) dx = ∫ ( − x + 7x − 6 ) d x Area = R w y = –x2 + 8x – 5 ev ie y -R am br id ge So for the area enclosed by y = −x 2 + 8x − 5 and y = x + 1: 6 es s -C 1 7 = − x 3 + x 2 − 6x 3 1 2 261 rs ity = 20 56 units2 This alternative method is the easiest method to use in the next example. y ge C U WORKED EXAMPLE 9.15 R op ni ev ve ie w C op Pr y 7 1 7 1 = − (6)3 + (6)2 − 6(6) − − (1)3 + (1)2 − 6(1) 3 3 2 2 y br ev id ie w The diagram shows the curve y = x 2 − 6x − 2 and the line y = 2 x − 9, which intersect when x = 1 and x = 7. -R 2 2 7 ni ve rs 1 1 = − x 3 + 4x 2 − 7 x 3 1 y = x2 – 6x – 2 y 1 s 1 7 w e ev ie id g es s -R br am -C C U op 1 1 = − (7)3 + 4(7)2 − 7(7) − − (1)3 + 4(1)2 − 7(1) 3 3 = 36 units2 7 es 7 1 O ity op y C w ie ev R 7 ∫ (2x − 9) dx − ∫ ( x − 6x − 2 ) dx = ∫ ( − x + 8x − 7 ) d x Area = y = 2x – 9 Pr -C Answer am Find the area of the shaded region. Copyright Material - Review Only - Not for Redistribution x ve rs ity Pr es s y 4 O x Find the area of the region bounded by the curve y = 5 + 6x − x 2 , the line x = 4 and the line y = 5. ve rs ity op C y y = 2x – 3 B w ge U R ni y y = (x – 3)2 C op w y = 5 + 6x – x2 5 2 ev ie -R y -C 1 ev ie am br id EXERCISE 9G w ge C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 ie ev id A x -R am br O 2 3 Pr y = –x2 + 11x – 18 ity op y C ev ve ie w rs A x ge C U R ni op O y 262 es y s -C The diagram shows the curve y = ( x − 3 ) and the line y = 2 x − 3 that intersect at points A and B. Find the area of the shaded region. 2x + y = 12 br ev id ie w B -R am The diagram shows the curve y = −x 2 + 11x − 18 and the line 2 x + y = 12 . Find the area of the shaded region. c y = x 2 − 4x + 4 and 2 x + y = 12 Pr y = −x 2 + 12 x − 20 and y = 2 x + 1 ni ve rs ity b es y = x 2 − 3 and y = 6 op -R s es -C am br ev ie id g w e C U R y 5 Sketch the curves y = x 2 and y = x(2 − x ) and find the area enclosed between the two curves. ev ie w C op y a s -C 4 Sketch the following curves and lines and find the area enclosed between their graphs. Copyright Material - Review Only - Not for Redistribution ve rs ity y O –4 1 x + 2 meeting at the points ( −4, 0) and (0, 2). 2 U R Find the area of the shaded region. ge y = √ 2x + 3 es s -C -R am br P(3, 3) ie id w y ev 7 x y x + 4 and the line y = ni The diagram shows the curve y = 1 x +2 2 y= C op ev ie w C ve rs ity op y Pr es s -C -R y = √x + 4 ev ie am br id w ge 6 C U ni op y Chapter 9: Integration Pr op y R x w rs O 263 ity C Q y ev ve ie The curve y = 2 x + 3 meets the y-axis at the point Q. U R ni op The tangent at the point P(3, 3) to this curve meets the y-axis at the point R. w ge C a Find the equation of the tangent to the curve at P. y -R am 8 O Q R y ni ve rs C ity Pr op y y = 10 + 9x – x2 es s -C P(6, 28) w x C U op The diagram shows the curve y = 10 + 9x − x 2. Points P(6, 28) and Q(10, 0) lie on the curve. The tangent at P intersects the x-axis at R. w id g e a Find the equation of the tangent to the curve at P. -R s es am br ev ie b Find the area of the shaded region. -C ie ev R ev br id ie b Find the exact value of the area of the shaded region PQR. Copyright Material - Review Only - Not for Redistribution ve rs ity w y ge y Pr es s -C -R am br id Q ev ie 9 C ve rs ity op y = 4x – x3 w ev ie C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 P x O U R ni C op y The diagram shows the curve y = 4x − x 3. The point P has coordinates (2, 0) and the point Q has coordinates ( − 4, 48). ie w ge a Find the equation of the tangent to the curve at P. y -R am 10 ev br id b Find the area of the shaded region. es s -C Pr y ity op O op y x ni ev ve ie w rs C 264 P(9, 4) y = 5 – √ 10 – x C U R The diagram shows part of the curve y = 5 − 10 − x and the tangent to the curve at P(9, 4). w ge a Find the equation of the tangent to the curve at P. -R am br ev id ie b Find the area of the shaded region. Give your answer correct to 3 significant figures. 9.8 Improper integrals Pr op y es s -C In this section, we will consider what happens if some part of a definite integral becomes infinite. These are known as improper integrals, and we will look at two different types. These are definite integrals that have either one limit infinite or both limits infinite. ∞ −2 1 1 d x. Examples of these are 2 d x and 3 x x 1 −∞ ∫ op C ∞ w ∫ f(x ) dx by replacing the infinite limit with a finite a -R s es -C am br value, X , and then taking the limit as X → ∞, provided the limit exists. ie id g We can evaluate integrals of the form ev e U KEY POINT 9.15 y ni ve rs ∫ w ie ev R ity C Type 1 Copyright Material - Review Only - Not for Redistribution ve rs ity ev ie am br id b ∫ f(x ) dx by replacing the infinite limit with a finite value, X, We can evaluate integrals of the form −∞ -R KEY POINT 9.16 w ge C U ni op y Chapter 9: Integration Pr es s -C and then taking the limit as X → −∞, provided the limit exists. ∞ 1 ∫ x d x has a finite value and find this value. w X 1 R 2 1 ni X ∫ x dx = ∫ x dx −2 1 X w ge = − x −1 Write the integral with an upper limit X . U ev ie Answer y 2 1 C op Show that the improper integral ve rs ity C op y WORKED EXAMPLE 9.16 1 ∞ 1 C w Pr ∞ Hence, the improper integral 265 ity 2 1 1 ∫ x d x has a finite value of 1. 2 -C y w C am Show that as p → −∞, the shaded area tends to a finite value and find this value. Pr 3 0 1 (1 – x)3 es 1 p −3 ni ve rs p O p ity op y 0 op C U w ie id g es s -R br ev 1 1 − 2 2(1 − p)2 am = e 0 1 = 2 2(1 − x ) p y 1 = (1 − x )−2 ( −2)( −1) p -C C w ie ev R 0 ∫ (1 − x ) dx = ∫ (1 − x ) dx Area = y= s Answer y ie 1 . (1 − x )3 ev br id The diagram shows part of the curve y = -R ge U WORKED EXAMPLE 9.17 R op ni ev ve ie 1 rs y ∫ x dx = 1 − 0 = 1 op ∴ 1 →0 X s As X → ∞, 1 X es -C = 1− -R am br ev id ie 1 1 =− −− X 1 Copyright Material - Review Only - Not for Redistribution x ve rs ity 1 2 → 0. 2 (1 − p ) am br id As p → −∞ , Pr es s -C y Type 2 1 . 2 -R Hence, as p → −∞, the shaded area tends to a finite value of ev ie w ge C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 1 −1 2 1 1 2 ∫ x d x is an improper integral because x tends to infinity as x → 0 and it is 2 0 1 C op ni However, 1 ∫ x dx is an invalid integral because x is not defined when x = 0. 2 U R ev ie w 1 For example, y C ve rs ity op These are integrals where the function to be integrated approaches an infinite value (or approaches ± infinity) at either or both end points in the interval (of integration). w ge well-defined everywhere else in the interval of integration. am br ev id ie For this section we will consider only those improper integrals where the function is not defined at one end of the interval. -C -R KEY POINT 9.17 b es s ∫ f(x ) dx where f(x ) is not defined when x = a can be We can evaluate integrals of the form a rs op y ve KEY POINT 9.18 ni ev ie w C 266 ity op Pr y evaluated by replacing the limit a with an X and then taking the limit as X → a, provided the limit exists. b ∫ f(x ) dx where f(x ) is not defined when x = b by C U R We can evaluate integrals of the form a br ev id ie w ge replacing the limit b with an X and then taking the limit as X → b, provided the limit exists. 0 2 Pr op y -R 5 ∫ x dx. s -C 2 Find the value, if it exists, of es am WORKED EXAMPLE 9.18 Answer 5 is not defined when x = 0. x2 2 X ev ie id g 5 5 − X 2 es s -R br = w e 5 5 =− −− 2 X C = −5x −1 Write the integral with a lower limit X . y −2 X op ∫ 5x d x am ∫ 2 ni ve rs 5 2 dx = x X U 2 -C R ev ie w C ity The function f( x ) = Copyright Material - Review Only - Not for Redistribution ve rs ity w ge C U ni op y Chapter 9: Integration am br id ev ie 5 tends to infinity. X As X → 0 , 2 5 -R ∫ x dx is undefined. Hence, 2 Pr es s -C 0 ve rs ity p y ge Answer 3 ie id 1 2 am − 0 p ) ( s es y ve p 267 3 dx → 6 2 . 2−x ni U ie ev ∫ As p → 2, Pr ) =6 2 −6 2− p ge 0 br ev id ie w Hence, as p → 2 the shaded area tends to a finite value of 6 2. -R am EXERCISE 9H x op w = −6 2 − p − −6 2 2 ity ( p 0 p rs C op y -C 1 3 = (2 − x ) 2 1 ( −1) 2 0 = −6 2 − x O -R br p ev 0 3 √2 – x w ∫ 2 − x dx = ∫ 3(2 − x ) dx Area = y= C op ni U R find this value. R y 3 . 2−x Show that as p → 2 the shaded area tends to a finite value and The diagram shows part of the curve y = C ev ie w C op y WORKED EXAMPLE 9.19 ∫ 3 0 3 dx 3−x e s ∫ 25 ∫ ∞ 5 dx x f 1 dx (1 − x )2 i 0 2 -R s es -C am br ev ie id g w e C U h y 4 g ∫ op Pr ∫ xx dx ni ve rs ie 4 ity ∞ d R ev ∫ es ∫ w C op y -C 1 Show that each of the following improper integrals has a finite value and, in each case, find this value. −2 ∞ ∞ 2 4 10 dx dx dx a b c 2 5 3 x x −∞ x 1 4 Copyright Material - Review Only - Not for Redistribution 8 4 4 dx x−4 ∞ 2 ∫ 4 ∫ x + ( x + 2) d x 1 2 3 ve rs ity 20 (2x + 5)2 p x 20 . (2 x + 5)2 Show that as p → ∞, the shaded area tends to the value 2. w ev ie The diagram shows part of the curve y = ve rs ity C op y O Pr es s -C y= -R am br id ev ie w ge y 2 C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 y C op U 4 ∫ xx dx e ∫ 2 5 dx 1 (2 x − 1)2 2 9 c 25 s es y y = x2 ity rs op ni O ni ve rs C ity Pr op y es s -C -R am br ev id ie w ge y = x2 y es s -R br ev ie id g w e C U R op w We can approximate the volume, V , of the solid by a series of cylindrical discs of thickness δx (corresponding to a small increase in x) and radius y (corresponding to the height of the function). am x 5 y C U When this area is rotated about the x-axis through 360° a solid of revolution is formed. -C 2 y ve O The volume of this solid is called a volume of revolution. ie 2 Pr y op C w ie ev R 1 ∫ x + x dx 0 Consider the area bounded by the curve y = x 2, the x-axis, and the lines x = 2 and x = 5. ev 2 -R -C 9.9 Volumes of revolution 268 12 ∫ x x dx 0 f ie 2 dx x+4 w 0 am 5 ∞ ev d b ge ∫ ∞ 4 6 dx x id ∫ ∞ br R a ni 3 Show that none of the following improper integrals exists. Copyright Material - Review Only - Not for Redistribution 2 5 x ve rs ity C U ni op y Chapter 9: Integration y 5 w ev ie As δx → 0, then V → y = x2 ∫ πy dx. 2 -R 2 am br id ∑ πy δx. ge The volume of each cylindrical disc is πy2δ x. An approximation for V is then 2 op y KEY POINT 9.19 x O Pr es s -C This leads to a general formula: the boundary values x = a and x = b is given by the formula V = 2 C op y a w ge U WORKED EXAMPLE 9.20 R b ∫ πy dx. ni ev ie w C ve rs ity The volume, V, obtained when the function y = f( x ) is rotated through 360° about the x-axis between br op y 1 9 dx 1 3x + 2 2 C O ity −2 1 2 y ev ve ie w rs 81 =π (3x + 2)−1 1 3( −1) 2 ge C U R ni op −27 =π (3x + 2) 1 -R s -C am br ev id ie w −27 −27 = π − 8 5 81π units3 = 40 Pr op y es Sometimes a curve is rotated about the y-axis. In this case the general rule is: ity 2 a -R s es -C am br ev ie id g w e C U R b ∫ πx dy. y the boundary values y = a and y = b is given by the formula V = op ni ve rs The volume, V, obtained when the function x = f( y ) is rotated through 360° about the y-axis between ev ie w C KEY POINT 9.20 9 3x + 2 s 2 2 ∫ = π 81(3x + 2) d x ∫ y2 d x = π es ∫ 2 y= Pr Volume = π -R am -C Answer y ev id ie Find the volume obtained when the shaded region is rotated through 360° about the x-axis. Copyright Material - Review Only - Not for Redistribution 1 2 x 269 ve rs ity ev ie w ge am br id WORKED EXAMPLE 9.21 C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 y Find the volume obtained when the shaded region is rotated through 360° about the y-axis. -R 5 ve rs ity y Using the given y = x 2. id ie w ge y2 = π 2 2 ity op Pr y es s -C -R am br ev 25 4 = π − 2 2 21π units3 = 2 rs Find the volume of the solid obtained when the shaded region is rotated through 360° about the x-axis. op 2x2 + y2 = 9 P (2, 1) ie w ge O ev id br x -R am Answer y C U R ni ev ve ie w C WORKED EXAMPLE 9.22 y 270 x C op 2 O U 2 R 5 ∫ y dy x 2 dy = π 2 ni ∫ 5 y = x2 Pr es s -C y op C w ev ie Answer Volume = π 5 1 es s -C When the shaded region is rotated about the x-axis, a solid with a cylindrical hole is formed. 2 2 Pr op y The radius of the cylindrical hole is 1 unit and the length of the hole is 2 units. ∫ y dx − volume of cylinder = π ( 9 − 2x ) d x − π × r × h ∫ ity 2 2 0 2 y ie w 0 2 ni ve rs C Volume of solid = π ev 2 op C U R 2 = π 9x − x 3 − π × 12 × 2 3 0 -R s es -C am br ev ie id g w e 16 = π 18 − − (0 − 0) − 2 π 3 32 π units3 = 3 Copyright Material - Review Only - Not for Redistribution ve rs ity ev ie am br id EXERCISE 9I w ge C U ni op y Chapter 9: Integration 1 Find the volume obtained when the shaded region is rotated through 360° about the x-axis. -R y y= ve rs ity c w 2 x √x ev 1 -R br am y 5 3–x y= x 4 O –1 x 1 s -C O x 2 ie ge U R id y= 1 d ni y O x 2 4 2x – 1 C op 1 y y y op C ev ie w O b Pr es s 2 y = x2 + x -C a es a 271 3 ity 11 y b Pr y w rs C op y 2 Find the volume obtained when the shaded region is rotated through 360° about the y-axis. y = √2x + 1 y C O x br ev ie x id O w ge 2 op 1 U R ni ev ve ie y = x2 + 2 y a , where a . 0. The volume x obtained when the shaded region is rotated through 360° about the x-axis is 18 π. Find the value of a. -R ity x 2 y op C U w e y = √x3 + 4x2 + 3x + 2 ev ie id g es s -R br am -C 1 y ni ve rs C w ie O 4 The diagram shows part of the curve y = x 3 + 4x 2 + 3x + 2 . Find the volume obtained when the shaded region is rotated through 360° about the x-axis. R ev a y= x Pr op y es s -C am 3 The diagram shows part of the curve y = Copyright Material - Review Only - Not for Redistribution –2 O 1 x ve rs ity y am br id ev ie w ge 5 The diagram shows part of the line 3x + 8 y = 24 . Rotating the shaded region through 360° about the x-axis would give a cone of base radius 3 and perpendicular height 8. C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 -R Find the volume of the cone using: 3x + 8y = 24 x 8 O Pr es s -C a integration 3 y b the formula for the volume of a cone. b Find the volume of the solid formed when the enclosed region bounded by the curve, the x-axis and the y-axis is rotated through 360° about the x-axis. ve rs ity ev ie w C op 6 a Sketch the graph of y = ( x − 2)2 . y ni C op y 7 The diagram shows part of the curve y = 5 x − x. U R The curve meets the x-axis at O and P. br 9 − 1 that intercepts the y2 y-axis at the point P. The shaded region is bounded by the curve, the y-axis and the line y = 1 . -R P es y b Find the volume obtained when the shaded region is rotated through 360° about the y-axis. rs w y 2 . x The line y = 7 intersects the curve at the points P and Q. ni op y ve ie ev ie w ge 2 . The shaded area is 2x + 1 rotated through 360° about the x-axis between x = 0 and x = p. y br ev id O es y= Pr p x y y = √ 25 – x2 y ity ni ve rs 25 − x 2 . The point op C w ie ev O es s -R br P(4, 3) 3 U id g e b Find the volume obtained when the shaded region is rotated through 360° about the x-axis. am 2 2x + 1 O a Find the volume obtained when the shaded region is rotated through 360° about the y-axis. -C w C op y Show that as p → ∞, the volume approaches the value 2 π. 11 The diagram shows part of the curve y = P (4, 3) lies on the curve. x -R s -C am 10 The diagram shows part of the curve y = ie y=7 Q C U R 2 y = 3x + x P a Find the coordinates of P and Q. ev x O 9 The diagram shows part of the curve y = 3x + b Find the volume obtained when the shaded region is rotated through 360° about the x-axis. R 9 –1 y2 1 Pr a Find the coordinates of P. x= ity op C 272 y s -C am 8 The diagram shows part of the curve x = x P O ev id ie b Find the volume obtained when the shaded region is rotated through 360° about the x-axis. w ge a Find the coordinates of P. y = 5√x – x Copyright Material - Review Only - Not for Redistribution 4 x ve rs ity C U ni op y Chapter 9: Integration am br id ev ie w ge 12 The diagram shows the curve y = 4 − x and the line x + 2 y = 4 that intersect at the points (4, 0) and (0, 2). y 2 -R a Find the volume obtained when the shaded region is rotated through 360° about the x-axis. op y Pr es s -C b Find the volume obtained when the shaded region is rotated through 360° about the y-axis. x + 2y = 4 ve rs ity C w ev ie y x2 + y2 = 100 w ge b Find the volume of water in the bowl. –8 op Pr y es s -C -R am br ev 14 Use integration to prove that the volume, V cm , of a sphere with radius 4 r cm is given by the formula V = πr 3. 3 273 ity rs C w y ve ∫ f(x ) dx = F(x ) + c. op d [ F( x ) ] = f( x ), then dx If ∫ x dx = n + 1 x ● ∫ (ax + b) dx = a( n + 1) (ax + b) + c (where c is a constant and n ≠ −1) ev n +1 am 1 n +1 + c ( n ≠ −1 and a ≠ 0) s -C n -R br n id ● ie w ge 1 C U R ● ni ev ie Integration as the reverse of differentiation Integration formulae y w ∫ f(x ) dx, where k is a constant. ev a es s -R br a C b a ie k f( x ) d x = k op ni ve rs ity b a e U ∫ f(x ) dx = [ F(x ) ] = F(b) − F(a ). id g ∫ b b f( x ) dx = F( x ) + c, then am ● ∫ If -C ie Rules for definite integration ● Pr ∫ k f(x ) dx = k ∫ f(x ) dx, where k is a constant ∫ [ f(x ) ± g(x ) ] dx = ∫ f(x ) dx ± ∫ g(x ) dx op y ● w C ● es Rules for indefinite integration ev x ie id 3 Checklist of learning and understanding R x y C op ni U R The bowl is filled with water to a depth of 3 cm. P 4 O 13 A mathematical model for the inside of a bowl is obtained by rotating the curve x 2 + y2 = 100 through 360° about the y-axis between y = −8 and y = 0. Each unit of x and y represents 1cm. a Find the volume of the bowl. y =√ 4 – x Copyright Material - Review Only - Not for Redistribution ve rs ity b a a a Pr es s b y ve rs ity y C op Area under a curve ev ie w y = f(x) w x id b ie a O ev The area, A, bounded by the curve y = f( x ), the x-axis and the lines x = a and x = b is given by the formula: ∫ y dx when y ù 0 A= b ∫ f(x ) dx when f(x ) ù 0 ). a es s -C a (or A = -R b am br ● ge U R ni A y a -C ∫ f(x ) dx = − ∫ f(x ) dx C op b -R ∫ [ f(x ) ± g(x ) ] dx = ∫ f(x ) dx ± ∫ g(x ) dx a ● b ev ie am br id b ● w ge C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 Pr b C 274 ity op y y rs ie w A ve a y C U x ie b ∫ f( y) dy when f( y) ù 0). a es s -C Pr ity ni ve rs A a b y = g(x) x y O -R s es -C am br ev ie id g w e C U R ev ie w C op y y = f(x) op a y (or A = ev id b ∫ x dy when x ù 0 -R A= w ge The area, A, bounded by the curve x = f( y ), the y-axis and the lines y = a and y = b is given by the formula: br ● am R O op ni ev x = f(y) Copyright Material - Review Only - Not for Redistribution ve rs ity ev ie am br id The area, A, enclosed between y = f( x ) and y = g( x ) is given by the formula: A= b -R ● w ge C U ni op y Chapter 9: Integration ∫ [ f(x ) − g(x ) ] dx Pr es s -C a op y where a and b are the x-coordinates of the points of intersection of the functions f and g. ve rs ity ∞ ● w C Improper integrals Integrals of the form ∫ f(x ) dx can be evaluated by replacing the infinite limit with a finite a Integrals of the form y ∫ f(x ) dx can be evaluated by replacing the infinite limit with a finite U R b ● C op ni ev ie value, X , and then taking the limit as X → ∞, provided the limit exists. ie ev id b ∫ f(x ) dx where f(x ) is not defined when x = a can be evaluated by -R br Integrals of the form am ● w ge −∞ value, X , and then taking the limit as X → −∞, provided the limit exists. a Pr ∫ f(x ) dx where f(x ) is not defined when x = b can be evaluated by op a rs ity replacing the limit b with an X and then taking the limit as X → b, provided the limit exists. U R between the boundary values x = a and x = b is given by the formula V = ∫ πy dx. 2 w ge a ev id ie The volume, V , obtained when the function x = f( y ) is rotated through 360° about the y-axis br ● b op y The volume, V , obtained when the function y = f( x ) is rotated through 360° about the x-axis ni ev ● ve ie Volume of revolution C C w es b Integrals of the form y ● s -C replacing the limit a with an X and then taking the limit as X → a , provided the limit exists. b ∫ πx d y. 2 a y op -R s es -C am br ev ie id g w e C U R ev ie w ni ve rs C ity Pr op y es s -C -R am between the boundary values y = a and y = b is given by the formula V = Copyright Material - Review Only - Not for Redistribution 275 ve rs ity ge C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 am br id ev ie w END-OF-CHAPTER REVIEW EXERCISE 9 The function f is such that f ′( x ) = 12 x 3 + 10 x and f ( −1) = 1. 1 2 ∫ 5x − x2 dx. 4 dy 6 = − 5x and the point (3, 5.5) lies on the curve. Find the equation of the curve. dx x 2 8 3 − 3 and that f(2) = 3. Find f( x ). A curve has equation y = f( x ). It is given that f ′( x ) = x+2 x A curve is such that ve rs ity 5 y w ge 1 ie O br id [5] C op x = 62 + 1 y 3 U R ni y [4] x ev ev ie w C op 3 [3] Pr es s -C Find y 2 [3] -R Find f( x ). 6 + 1. The shaded region is bounded by the curve, the y-axis, and y2 the lines y = 1 and y = 3. Find the volume, in terms of π, when this shaded region is rotated through 360° about the y-axis. [5] y es s -C -R am The diagram shows part of the curve x = Pr ity a State the value of x for which f( x ) has a stationary value. [1] b Find an expression for f( x ) in terms of x. [4] y C op y = 6x – x2 U R ni 7 y rs C w ie ev A function is defined for x ∈ and is such that f ′( x ) = 6x − 6. The range of the function is given by f( x ) ù 5. ve op 6 276 ev br id ie w ge y=5 x -R am O s a Sketch the curve y = ( x − 3)2 + 2. b The region enclosed by the curve, the x-axis, the y-axis, the line x = 3 is rotated through 360° about the x-axis. Find the volume obtained, giving your answer in terms of π. [6] op y ni ve rs ity Pr [1] -R s es -C am br ev ie id g w e C U R ev ie w C 8 [6] Cambridge International AS & A Level Mathematics 9709 Paper 11 Q4 June 2010 es op y -C The diagram shows the curve y = 6x − x 2 and the line y = 5. Find the area of the shaded region. Copyright Material - Review Only - Not for Redistribution ve rs ity y -R O Pr es s U C ev id s es O A x Pr y 277 ity rs op C w Find the coordinates of B and C. [2] ve ie -R br am -C B y op ni w U ge ie ev 2 √x + 1 y=1 Pr op y es s -C am y= -R br id y x ity C ni ve rs 2 . x +1 2 4 op 2 -R br ev iii The shaded region is rotated through 360° about the y-axis. Find the exact value of the volume of revolution obtained. [5] [5] s Cambridge International AS & A Level Mathematics 9709 Paper 11 Q11 June 2012 es am [1] ie 1 w e ∫ y − 1 d y. Hence find the area of the shaded region. id g ii Find 2 4 can be written in the form x = 2 − 1. x +1 y C U Show that the equation y = y The diagram shows the line y = 1 and part of the curve y = -C [5] Cambridge International AS & A Level Mathematics 9709 Paper 11 Q10 November 2011 O i [4] C iii Find the volume obtained when the shaded region is rotated through 360° about the y-axis. w ie y = √ 1 + 2x ie ge y ii Find the equation of the normal to the curve at C. ev [6] The diagram shows the curve y = 1 + 2 x meeting the x-axis at A and the y-axis at B. The y -coordinate of the point C on the curve is 3. i R [2] Cambridge International AS & A Level Mathematics 9709 Paper 11 Q8 November 2012 w R ni C op ii Find, showing all necessary working, the area of the shaded region. y ve rs ity The diagram shows the curve y2 = 2 x − 1 and the straight line 3 y = 2 x − 1. 1 The curve and straight line intersect at x = and x = a , where a is a constant. 2 i Show that a = 5. 11 ev x 2 10 R a 1 – y op C w ev ie 3y = 2x – 1 y2 = 2x – 1 -C 9 ev ie am br id w ge C U ni op y Chapter 9: Integration Copyright Material - Review Only - Not for Redistribution ve rs ity ev ie am br id w ge C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 1 − 1 12 A curve has equation y = f( x ) and is such that f ′( x ) = 3x 2 + 3x 2 − 10. Pr es s [3] iii It is given that the curve y = f( x ) passes through the point (4, −7). Find f( x ). [4] ve rs ity y op [4] ii Find f ′′( x ) and hence, or otherwise, determine the nature of each stationary point. C Cambridge International AS & A Level Mathematics 9709 Paper 11 Q9 June 2013 y x=4 y= ev br id P x 8 . The curve intersects the y-axis at A(0, 4) . The normal to 3x + 4 s -C -R am O The diagram shows part of the curve y = 8 √ 3x + 4 w ge Q C op y B A (0, 4) U R ni ev ie 13 ie w -R 1 By using the substitution u = x 2 , or otherwise, find the values of x for which the curve y = f( x ) has stationary points. -C i C es i Find the coordinates of B. [5] Pr ii Show, with all necessary working, that the areas of the regions P and Q are equal. [6] y U C w ge y = (3 – 2x)3 O x es s -C ie -R am br id ( 12 , 8) ev R ni ev 14 y ve ie w rs Cambridge International AS & A Level Mathematics 9709 Paper 11 Q10 June 2015 op 278 ity op y the curve at A intersects the line x = 4 at the point B. Pr ity ii Find the area of the shaded region. [5] [6] ni ve rs w y op -R s -C am br ev ie id g w e C U R ev ie Cambridge International AS & A Level Mathematics 9709 Paper 11 Q10 November 2013 es C op y 1 The diagram shows the curve y = (3 − 2 x )3 and the tangent to the curve at the point , 8 . 2 i Find the equation of this tangent, giving your answer in the form y = mx + c. Copyright Material - Review Only - Not for Redistribution ve rs ity -R y y Pr es s -C 15 ev ie am br id w ge C U ni op y Chapter 9: Integration Q (2, 3) 1 2 y = 1 x2 + 1 2 C op y P (0, 1) U and y = br ev id The diagram shows parts of the curves x 1 y = (4x + 1) 2 w ge O 1 2 x + 1 intersecting at points P (0, 1) and 2 ie R ni ev ie w C ve rs ity op y = (4x + 1) α -R Find α , giving your answer in degrees correct to 3 significant figures. s -C i am Q(2, 3) . The angle between the tangents to the curves at Q is α . ii Find by integration the area of the shaded region. [6] es [6] 279 y op y op -R s es -C am br ev ie id g w e C U R ev ie w ni ve rs C ity Pr op y es s -C -R am br ev id ie w ge C U R ni ev ve ie w rs C ity op Pr y Cambridge International AS & A Level Mathematics 9709 Paper 11 Q11 November 2014 Copyright Material - Review Only - Not for Redistribution ve rs ity ge C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 ev ie am br id 1 dy = 2 x 2 − 3. Given that the curve passes through the point ( −3, −2) , find the equation dx of the curve. [4] 1 − dy = 2 − 8(3x + 4) 2 . A curve is such that dx i A point P moves along the curve in such a way that the x-coordinate is increasing at a constant rate of 0.3 units per second. Find the rate of change of the y-coordinate as P crosses the y-axis. [2] 4 The curve intersects the y-axis where y = . 3 ii Find the equation of the curve. [4] Pr es s -C -R A curve is such that y ve rs ity ev ie w C op y 2 w CROSS-TOPIC REVIEW EXERCISE 3 3 1 dy = 3x 2 − 6 and the point (9, 2) lies on the curve. dx i Find the equation of the curve. [4] ii Find the x-coordinate of the stationary point on the curve and determine the nature of the stationary point. [3] U R ni C op Cambridge International AS & A Level Mathematics 9709 Paper 11 Q4 June 2016 s -C Cambridge International AS & A Level Mathematics 9709 Paper 11 Q6 June 2010 es i Find the equation of the curve. [4] ity op 1 ii Find the set of values of x for which the gradient of the curve is less than . 3 w ie y y C U w ge y = (2x – 1)2 y2 = 1 – 2x ity Pr op y ni ve rs C U i State the coordinates of A. w ie ev -R s es am [1] [6] Cambridge International AS & A Level Mathematics 9709 Paper 11 Q7 November 2016 br id g e C ii Find, showing all necessary working, the area of the shaded region. op y The diagram shows parts of the curves y = (2 x − 1)2 and y2 = 1 − 2 x , intersecting at points A and B. -C w ie x A es O s -C -R am br ev id B ie 5 op ni ev ve Cambridge International AS & A Level Mathematics 9709 Paper 11 Q7 June 2011 R ev R [3] rs C 280 dy 3 1 = and the point 1, lies on the curve. 2 dx (1 + 2 x )2 Pr A curve is such that y 4 -R am br ev id ie w ge A curve is such that Copyright Material - Review Only - Not for Redistribution ve rs ity ev ie am br id 1 Pr es s U id -R am br ev 2r cm The diagram shows a metal plate. The plate has a perimeter of 50 cm and consists of a rectangle of width 2 r cm and height x cm, and a semicircle of radius r cm. 1 [4] a Show that the area, A cm 2, of the plate is given by A = 50 r − 2 r 2 − πr 2. 2 Given that x and r can vary: 50 [4] b show that A has a stationary value when r = 4+π c find this stationary value of A and determine the nature of this stationary value. [2] y ve op A line has equation y = 2 x + c and a curve has equation y = 8 − 2 x − x 2. i For the case where the line is a tangent to the curve, find the value of the constant c. C U R 8 ni ev ie w rs ity Pr es s -C y op y w ge x cm ie R ni r cm C op 7 C [3] Cambridge International AS & A Level Mathematics 9709 Paper 11 Q10 November 2016 ve rs ity w 1 ii Given that the curve also passes through the point (4, 10), find the y-coordinate of A, giving your answer as a fraction. [6] C op y -C i Find the x-coordinate of A. ev ie − A curve has equation y = f( x ) and it is given that f ′( x ) = 3x 2 − 2 x 2 . The point A is the only point on the curve at which the gradient is −1. -R 6 w ge C U ni op y Cross-topic review exercise 3 [3] br ev id ie w ge ii For the case where c = 11, find the x-coordinates of the points of intersection of the line and the curve. Find also, by integration, the area of the region between the line and the curve. [7] s 9 . 2−x es dy and determine, with a reason, whether the curve has any stationary points. [3] dx ii Find the volume obtained when the region bounded by the curve, the coordinate axes and the line x = 1 is rotated through 360° about the x-axis. [4] Pr ity iii Find the set of values of k for which the line y = x + k intersects the curve at two distinct points. ie ni ve rs [4] y op -R s -C am br ev ie id g w e C U R ev Cambridge International AS & A Level Mathematics 9709 Paper 11 Q11 November 2010 es C op y i Find an expression for w -R am The equation of a curve is y = -C 9 Cambridge International AS & A Level Mathematics 9709 Paper 11 Q11 June 2014 Copyright Material - Review Only - Not for Redistribution 281 ve rs ity 4 for x > 0 . 2x + 1 ev ie 10 A function f is defined as f( x ) = -R am br id w ge C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 Pr es s -C a Find an expression, in terms of x, for f ′( x ) and explain how your answer shows that f is a decreasing function. [3] [4] c On a diagram, sketch the graph of y = f( x ) and the graph of y = f −1( x ) , making clear the relationship between the two graphs. [4] ve rs ity C op y b Find an expression, in terms of x, for f −1( x ) and find the domain of f −1. 1 1 y [4] [2] [5] id ie w ge U R ni ev ie C op w − dy 2 = x 2 − x 2 . The curve passes through the point 4, . 3 dx i Find the equation of the curve. d2 y ii Find 2 . dx iii Find the coordinates of the stationary point and determine its nature. 11 A curve is such that br ev Cambridge International AS & A Level Mathematics 9709 Paper 11 Q12 June 2014 es s -C -R am 2 12 The function f is defined for x . 0 and is such that f ′( x ) = 2 x − 2 . The curve y = f( x ) passes through the x point P (2, 6). i Find the equation of the normal to the curve at P. [3] Pr iii Find the x-coordinate of the stationary point and state with a reason whether this point is a maximum or a minimum. [4] w rs C 282 [4] ity op y ii Find the equation of the curve. ve ie Cambridge International AS & A Level Mathematics 9709 Paper 11 Q9 November 2014 y 1 9 − . x −1 x −5 i Find the x-coordinate of the point where the normal to the curve at P intersects the x-axis. op ge C U R ni ev 13 The point P (3, 5) lies on the curve y = am Cambridge International AS & A Level Mathematics 9709 Paper 11 Q11 November 2016 s es y = (x – 2)4 A (1, 1) O x B y ni ve rs C ity Pr op y C w C U op The diagram shows part of the curve y = ( x − 2)4 and the point A(1, 1) on the curve. The tangent at A cuts the x-axis at B and the normal at A cuts the y-axis at C . br w -R iii Find the area of the shaded region. [6] [2] [4] s Cambridge International AS & A Level Mathematics 9709 Paper 11 Q10 June 2013 es am a , where a and b are integers. b ev ii Find the distance AC, giving your answer in the form ie id g e i Find the coordinates of B and C. -C ie ev R -R y -C 14 [6] ev br id ie w ii Find the x-coordinate of each of the stationary points on the curve and determine the nature of each stationary point, justifying your answers. [5] Copyright Material - Review Only - Not for Redistribution ve rs ity P (6, 5) y op -R ve rs ity O 1 2 Pr es s y = (1 + 4x) C w ev ie am br id y -C 15 w ge C U ni op y Cross-topic review exercise 3 Q (8, 0) x 1 y C op ni ev ie The diagram shows part of the curve y = (1 + 4x ) 2 and a point P (6, 5) lying on the curve. The line PQ intersects the x-axis at Q(8, 0). U R i Show that PQ is a normal to the curve. [7] id ie w ge ii Find, showing all necessary working, the exact volume of revolution obtained when the shaded region is rotated through 360° about the x-axis. [5] br ev [In part ii you may find it useful to apply the fact that the volume, V , of a cone of base radius r and vertical -R am 1 2 πr h.] 3 Cambridge International AS & A Level Mathematics 9709 Paper 11 Q11 November 2015 283 y op y op -R s es -C am br ev ie id g w e C U R ev ie w ni ve rs C ity Pr op y es s -C -R am br ev id ie w ge C U R ni ev ve ie w rs C ity op Pr y es s -C height h, is given by V = Copyright Material - Review Only - Not for Redistribution ve rs ity C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 am br id ev ie w ge PRACTICE EXAM-STYLE PAPER Time allowed is 1 hour 50 minutes (75 marks). 5 , for x . 0. Show that f is an increasing function. x3 2 The graph of y = x 3 − 3 is transformed by applying a translation of followed by a reflection 0 in the x -axis. Pr es s -C ve rs ity Find the equation of the resulting graph in the form y = ax 3 + bx 2 + cx + d . 3 Prove the identity 4 a Find the first three terms in the expansion of (3 − 2 x )7, in ascending powers of x. b Find the coefficient of x 2 in the expansion of (1 + 5x )(3 − 2 x )7 . [4] y w ie -R Pr 3 ity The point X lies on the line OA and BX is perpendicular to OA. a Find the exact area of the shaded region. b Find the exact perimeter of the shaded region. y ve rs [4] ni C U a Find the equation of the circle. b Find the equation of the tangent to the circle at the point P, giving your answer in the form ax + by = c. [3] [4] br ev id ie w ge [3] op A circle has centre (3, −2) and passes through the point P (5, −6). -R am a The sum, Sn , of the first n terms of an arithmetic progression is given by Sn = 11n − 4 n2. Find the first term and the common difference. 1 b The first term of a geometric progression is 2 14 and the fourth term is . Find: 12 i the common ratio Pr es s [3] ii the sum to infinity. [3] [2] ity op y ni ve rs The equation of a curve is y = 3 + 12 x − 2 x 2. Express 3 + 12 x − 2 x 2 in the form a − 2( x + b )2 , where a and b are constants to be found. b Find the coordinates of the stationary point on the curve. c Find the set of values of x for which y ø − 5. -R s es am br ev ie id g w e C U op y a -C C w ie ev R 8 A The diagram shows sector OAB of a circle with centre O, radius 6 cm and sector angle π radians. -C 7 X es O s π 3 -C R 6 [3] ev id br am 6 cm y op ev ie w C 284 [3] C op ni B ge 5 [2] [3] 1 − tan x ≡ 2 cos2 x − 1. 1 + tan2 x 2 U R ev ie w C op y 2 -R It is given that f( x ) = 2 x − 1 Copyright Material - Review Only - Not for Redistribution [3] [2] [3] ve rs ity ev ie am br id The function f : x ֏ 6 − 5 cos x is defined for the domain 0 ø x ø 2 π. a Sketch the graph of y = f( x ). [2] Solve the equation f( x ) = 3. [3] The function g : x ֏ 6 − 5 cos x is defined for the domain 0 ø x ø π. [2] C op 6 and A(3, 2) is a point on the curve. 10 A curve has equation y = 9 − 2x a Find the equation of the normal to the curve at the point A. b y Find g−1( x ) . ve rs ity d ni ev ie w C op y c [1] Pr es s -C b Find the range of f . -R 9 w ge C U ni op y Practice exam-style paper U ie ev id br -R am -C [4] es Pr rs op y ve ni C U ie w ge ev id br -R am es s -C ity Pr op y -R s es am br ev ie id g w e C U op y ni ve rs C w ie [5] 285 ity op C w ie ev R ev R -C [3] s Find the volume of the solid formed when the region enclosed by the curve, the x -axis, and the lines x = 1 and x = 2 is rotated about the x -axis. y c [5] w ge R A point P ( x, y ) moves along the curve in such a way that the y -coordinate is increasing at a constant rate of 0.05 units per second. Find the rate of increase of the x -coordinate when x = 4. 16 − x 2. 11 A curve has equation y = x dy d2 y and in terms of x. a Find dx dx 2 b Find the coordinates of the stationary point on the curve and determine its nature. [5] Copyright Material - Review Only - Not for Redistribution ve rs ity am br id a −4, 3 a x.2 a x = 2, y = 3 a 2 5 Pr es s -C c (5x + 4)2 − 20 d (3x − 7)2 + 12 a −9, 1 b −6, 2 c −5, 7 d 2, 7 e −6, 3 f a − 2 ± 11 b 5 ± 23 c − 4 ± 17 −10, 1 id y w ie c −4.19, 1.19 d −3.39, 0.89 e − 1.39, − 0.36 f x= y 5 op 4.93 3.19 −0.217, 9.22 b ± b2 − 4ac b ; the solutions each increase by . 2a a ev -R a ( − 3, 9), (2, 4) 7 b −8, , (2, 1) 2 c ( − 10, 0), (8, 6) d ( − 2, − 7), (1, 2) e (2, − 2), (10, 2) f g (2, 4) h ( − 3, 1), (9, 7) 2 7 45 g x + − 2 4 2 3 7 h x − + 2 4 ni ve rs ( ) ), (4, −9) 2 a 9 and 17 3 2 21 cm and 5 25 cm 4 ev U e 3 21 cm and 9 cm s -R 2 7 9 d 2 x + − 4 8 ( −3, −4 13 ( − 1, − 3), (2, 1) j ( − 5, − 24), (5, 1) l (4, − 6), (12, 10) n ( − 1, 3), (3, 1) o (6, − 2), (18, − 1) es br am m b 3( x − 2)2 − 13 id g 2 5 33 c 2 x + − 4 8 k ( −6, −2), ( x − 2)2 − 12 8, 1 21 y f (2, 2), (10, − 2) op e ( x + 2)2 + 4 i C 2 15 225 d x + − 2 4 w 2 3 9 c x − − 2 4 ity b ( x + 4)2 − 16 ie op y Pr es s -C 1 a ( x − 3)2 − 9 a 2( x − 3)2 + 1 −1.64, 0.24 Exercise 1D am br b − 5.24, − 0.76 2 3 4 b 20 cm, 21cm, 29 cm Exercise 1B -C a −0.29, 10.29 C U ge 1 2 2, 3, 4, 5 f 11 2 w d −4, − 2± ie c 1, 3 ni a −5, 3 5 b ,3 2 f -R s es Pr rs ve 1 2 −3 ± 3 2 Exercise 1C 1 ity d 4 −5, e ev U ge id -C 11 b −3, 2 f 7 2 C op ni 7 d 1± 1 − ,0 2 f 2 1 − , , 4, 6, 7 3 2 C b (2 x + 5)2 + 5 2 10 7 w a (3x − 1)2 − 4 49 5 − 3 x − 12 6 d ±2 6 ie d 19 − 2 8 1 1 − , 1, ( −5 − 97 ), ( 97 − 5) 3 6 6 9000 3 9000 3 a ≈ 318 m b ≈ 159 m 49 98 5 21 ev c 15 − 2( x − 1)2 9 a Proof R b 21 − 2( x + 3)2 3 ± 10 5 2 a 15 − 2( x + 2)2 8 5 e − , 1 13 2 a −1, 6 3 c − ,1 4 1 e − ,1 2 5 a −2, 3 c ±3 2 1 e − , 3 2 e −5, 3 1 2 4 5 1 3 f − , 5 2 b −1, 4 d −3, − y C w ie ev R 4 6 61 5 − x− 4 2 c −2, 8 op 3 286 d b 3, 4 am 2 2 a −5, 2 br R ev ie w Exercise 1A 1 5 ve rs ity op y b x > −2 b x = − 2, y = − 5 b 5 c 4 2 C 2 3 4 1 c − ,6 3 b 3 25 3 − x+ 4 2 -R 4 Prerequisite knowledge 1 w c 1 Quadratics b 16 − ( x − 4)2 a 4 − ( x − 2)2 ev ie 3 ge Answers C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 Copyright Material - Review Only - Not for Redistribution b 13 − 19 and 13 + 19 ve rs ity x = 4 21 and y = 16 or x = 16 and y = 4 21 7 r = 5, h = 13 8 a ( − 3, 5) and (2, 0) b 5 2 -C 1 b , 0 2 14 a 2, 8 w ev ie -R w ev -R s es 8 A: y = ( x − 4)2 + 2 or x 2 − 8x + 18 y op a y = x 2 − 6x + 13 B y = x 2 − 6x + 5 C y = − x 2 + 6x − 5 D y = − x 2 + 6x − 13 E y = x 2 + 6x + 13 F y = x 2 + 6x + 5 G y = − x 2 − 6x − 5 H y = − x 2 − 6x − 13 s es Pr op y ity ni ve rs a ∪-shaped curve, minimum point: (3, − 1), axes crossing points: (2, 0), (4, 0), (0, 8) br ( y = 5 + 3x − 12 Proof w ), 1 2 x 2 ) -R s am 11 ie −1 43 , −2181 axes crossing points: ( −5, 0), 1 21 , 0 , (0, −15) -C y = 3x 2 − 6x − 24 ev id g e ( − 7, 0), (2, 0), (0, − 14) ( 10 C U ( −2 21 , −20 14 ), axes crossing points: op y b Student’s own answers b ∪-shaped curve, minimum point: c ∪-shaped curve, minimum point: 1 1 ( x − 2)2 or 6 + 2 x − x 2 2 2 A -R a = 2, b = − 40, c = 128 C w ie ev Proof ve 5 Exercise 1F R b ∩ -shaped curve, maximum point: 1 , 1 1 , axes crossing points: 4 8 7 ni a = 2, b = − 9, c = 7 2 − 1 , 0 , (1, 0), (0, 1) 2 9 -C 4 1 9 1 − 2 x − 8 4 C: y = 8 − am c 4 10 a ie a 6 B: y = 4( x + 2)2 − 6 or 4x 2 + 16x + 10 b 4 4 d 25 1 f , 25 9 b (4, 4), (16, 8) es 3 −4 14 when x = 3 21 ev e 5 U R c −1, 2 ge a l id 2 3 ± 2 4, 6 14 1 1 ,6 9 4 1 9 ,1 4 16 x−6 x +8 = 0 br C w ie k ±2 ( −2 14 , −6 81 ) , minimum C op y g ± 3 i b ge -C e ±1 ( 2 21 , 13 14 ) , maximum 2 N D N D + , − 2 2N 2 2N b −1, 2 2 d ± ,± 5 2 1 f 1, 2 h No solutions 1 j − ,1 2 am c ± 5, ±1 id a ±2, ±3 br 1 b 9 49 a 2 x + − 4 8 4 U b 53 5 − x− 4 2 y y = −2x − 3 2 a ie 13 3 Pr (2, 3) ity 12 rs 7x + y = 0 a 2( x − 2)2 − 3 b x=2 ni 11 ve rs ity op 2 53 ( − 3, 0), (4, 0), (0, 12) w 10 2 Pr es s a ( − 2, 1) and (3, − 1) y am br id 9 1 , 12 1 , axes crossing points: 4 2 C op 6 ge 7 cm and 11cm C w ev ie R d ∩ -shaped curve, maximum point: 5 Exercise 1E ev C U ni op y Answers Copyright Material - Review Only - Not for Redistribution 287 ve rs ity Pr es s U w -R id 8 k,6 9 −3 , m , 1 10 k.6 y op Proof y 1 , 0 2 C op 5 25 1 a 3x − − b − ,x,2 3 2 4 3 x = ±2, x = ± 2 x , − 9 − 2 3 or x . − 9 + 2 3 e 3 4 k , 1 or k . 2 es s -R 5 w 57 8 1 d k, 2 25 f k, 16 1 2 Proof s 13 2 id g am 3 2 k , − 4 3 or k . 4 3 2 k = − 10 or k = 14 f b k, br c k,2 e k. es Pr 8 9 7 1 d k = 0 or k = 2 a k . −13 -C R 5 ni ve rs b k = 4 or k = 1 e k = 0 or k = − k , − 2 or k . 6 End-of-chapter review exercise 1 ity b = − 2, c = − 35 a k = ±4 1 c k= 4 6 U ev ie w 4 − 6, − 2, ( − 1, 12), (1, 4) b (2, 4), ( − 2, − 4) ie am -C C 3 No real roots op y 2 5 12 Two distinct roots f a ±10 11 b Two distinct roots c Two distinct roots d Two equal roots e No real roots 4 -R br a Two equal roots 1 5 C ge U R e −5 < x , − 2 or 1 < x , 2 1 f x , − 4 or < x , 5 2 Exercise 1H 3 w ni ev d − 3 < x , 2 or x > 5 −1, 7 ie es ve ie w rs ity C 7 Pr y x , −5 or x . 8 3 a 1, x < b −1 , x , 0 2 c − 1 < x , 1 or x > 5 op 6 − 5, − 9 2 s b −7 < x , 1 c x , − 2 or x > 3 288 k < −2 2 ie br 5 11 1 am -C 4 Proof Exercise 1I id g x < − 9 or x > 1 7 5 i − ,x, 2 3 5 −3 , x , 2 a 5<x,7 ge R e x , − 4 or x . 1 10 y d −3 , x , 2 1 3 f − ,x, 2 5 h x , − 2 or x . 5 ni c − 12 < x < 1 9 8 k= ev 3 4 5 ,x, 3 2 a −9 , x , 4 e 7 − 2 10 , k , 7 + 2 10 p2 20 25 k< 8 Proof 7 ve rs ity w C op y c x < − 7 or x . 1 f C op a x < − 5 or x > 5 2 ev -C e −6 < x < 5 -R am br id c 4<x<6 ev ie ev ie b x , − 2 or x . 3 3 d − ,x,2 2 1 1 f x , − or x . 2 3 b −5 < x < −2 3 2 d − <x< 2 7 1 f x , − 4 or x . 2 b x , 7 or x . 8 1 13 b k. 2 12 39 26 c k. d k.− 8 5 e 5 − 21 , k , 5 + 21 a k. ev a 0<x<3 1 6 w ge Exercise 1G C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 Copyright Material - Review Only - Not for Redistribution ve rs ity b (2, 1), (5, 7), b (5, 25) 1 3 5 , − , 5 2 ii k = 3 or 11 11 i ( 2 21 , 2 21 ) ii m = −8, ( − 2, 16) 12 i 1 2( x − 1)2 − 1, (1, − 1) ii − , 3 21 2 w y C op ni b domain: x ∈ R, − 3 < x < 2 range: f( x ) ∈ R, − 7 < f( x ) < 20 b − 13 < f( x ) < − 3 ie ge a f( x ) . 12 c − 1 < f( x ) < 9 1 e < f( x ) < 16 32 a f( x ) > − 2 d 2 < f( x ) < 32 3 f < f( x ) < 12 2 b 3 < f( x ) < 28 c f( x ) < 3 d − 5 < f( x ) < 7 a f( x ) > 5 b f( x ) > − 7 c −17 < f( x ) < 8 d f( x ) > 1 8 a f( x ) > − 20 b f( x ) > −6 13 9 a f( x ) < 23 b f( x ) < 5 10 a s Pr rs ve 2( x − 3)2 − 13 ni 4 ity x−4 5 f −1( x ) = U 6 br w 4 function, one-one y s es Pr a 2( x − 2 ) − 3 y 15 2 -R c x ∈ R, −3 < x < 5 s es am b Many-one -C a = 1 or a = − 5 ev 4 x br 2 14 12 e id g O –2 13 f( x ) > k − 9 a2 g( x ) < +5 8 a=2 11 U 2 R b − 1 < f( x ) < 5 op w ie ev 4 4 2 –2 ni ve rs 6 –4 O ity C 8 C op y 10 2 w -C -R am g function, one-one h not a function a 289 ie e function, one-one f ev id c function, one-one d function, one-one 2 y a function, one-one b function, many-one ge R Exercise 2A 7 ie C w ie ev es y op 3 − 2x 3 1 6 -C 2 w -R am Prerequisite knowledge y br ev id 5 2 Functions 10 x 4 a domain: x ∈ R, − 1 < x < 5 range: f( x ) ∈ R, − 8 < f( x ) < 8 4 1 iii y − 3 = − ( x − 2) 5 1 2 b each input does not have a unique output U w ev ie ve rs ity i C 10 O op op c x < 1 or x > 9 R 2 Pr es s -C 2 y a 25 − ( x − 5) -R 4 a Proof c 2,x,5 9 y 6 b (6, 29) c k = 1, C = (2, 5) 8 a ev ie a Proof 3 ge 7 b k = − 4 or k = − 20 am br id a C ( 121 , −2 ) 6 C U ni op y Answers Copyright Material - Review Only - Not for Redistribution b k=4 x ve rs ity C c RR( x ), domain is x ∈ R, x ≠ 0, range is f( x ) ∈ R, f( x ) ≠ 0 ev ie w ge d QPR( x ), domain is x ∈ R, x ≠ 0, range is f( x ) ∈ R, f( x ) . 1 -R b domain: x ∈ R range: f( x ) ∈ R, f( x ) > 2 Pr es s 1 2 C c f( x ) > 3 w 5 a g is one-one for x > 3, since vertex = ( 2, 2 ) y op a −3 7 a f( x ) > − 9 b f −1( x ) = − 3 + w x + 32 2 a k=3 10 11 a f −1( x ) = s es Pr ity f −1( x ) = 3 + 9 − x ii Domain is x < 9, range is 3 < f −1( x ) < 7 1 b Domain is x < 4 23 a f( x ) = 5−x a = 5, b = 12 9 12 4x + 3 x + 1 −1 , g (x) = 3 2x 1 (1 + 3 x + 3 ) 2 b Domain is − 2 < x < 122 a f −1( x ) = C U e 13 id g a PQ( x ), domain is x ∈ R, range is f( x ) ∈ R, f( x ) > − 1 a f( x ) = ( x − 5)2 − 25 ev ie b f −1( x ) = 5 + x + 25 , domain is x > − 25 es s -R br am x−2 2 b Proof 2( x + 1) a ff ( x ) = for x ∈ R, x ≠ − 3 x+3 b Proof c −2 or 1 -C b x > −3 C 6 b fg( x ) > −6 14 b QP( x ), domain is x ∈ R, range is f( x ) ∈ R, f( x ) > 1 y a f −1( x ) = − 1 + 3 x + 4 b x <1 b No inverse since it is not one-one ni ve rs op y w R a 4x 2 + 2 x − 6 5−x 2x y b ( x − 1)2 + 3 19 b f −1( x ) = − 2 + x + 4 -R a x < − 1 or x > 3 18 ie Pr ity 16 f −1( x ) = 2 + 3 x + 1 4 rs b −1 f a f −1( x ) = ve ni U a 2( x + 1) − 10 -C 15 2 7 − 2x x −1 a Domain is x > − 4, range is f −1( x ) > − 2 b i 14 17 3x + 8 x 3 8 19 k>− 2 Proof 13 c f −1( x ) = 5 + x − 3 d f −1( x ) = op ±4 ge 12 c gg f fgf id Proof b gf e gfg br 11 x−3 b g −1( x ) = 2 + am 9 b f −1( x ) = ev 2 y 1 or 3 21 2 4 − or 0 3 −9 x+2 4x + 9 a fg d ff x+8 5 ie id br am b −4 op C 8 10 ie b a f −1( x ) = e f −1( x ) = 5 79 b −4 21 or − -C a (2 x + 5)2 − 2 w ie ev c hh g SP( x ), domain is x ∈ R, x > − 1, range is f( x ) ∈ R, f( x ) > −1 w a a = 3, b = −12 6 a − x +1 b kh PS( x ), domain is x ∈ R, x > − 1, range is f( x ) ∈ R, f( x ) > − 1 -R 3 c 231 s a hk b 3 es 2 7 ev U a 7 f Exercise 2C 1 ge 1 6 ve rs ity w ev ie R Exercise 2B 5 R domain: x ∈ R, x > 3, range: f( x ) ∈ R, f( x ) > − 2 f 4 290 e domain: x ∈ R, x ≠ 2, range: f( x ) ∈ R, f( x ) ≠ 0 ni C op y d domain: x ∈ R, x ≠ 0, range: f( x ) ∈ R, f( x ) ≠ 0 e RQQ( x ), domain is x ∈ R, x ≠ − 4, range is f( x ) ∈ R, f( x ) ≠ 0 C op -C c domain: x ∈ R range: f( x ) ∈ R, f( x ) . 0 ev a domain: x ∈ R range: f( x ) ∈ R am br id 16 U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 Copyright Material - Review Only - Not for Redistribution ve rs ity y O 2 4 6 x 291 w ie ity rs y 4 − 2x x −1 c Domain is 0 , x < 2, range is f ( x ) > 0 b f −1( x ) = op C y f –1 y=x ie d w ve ev s x a Symmetrical about y = x 4 b Not symmetrical about y = x ni ve rs c Symmetrical about y = x es s -R br ev ie id g w e C op a Proof 5 y d Symmetrical about y = x U R f O x ity f –1 6 es 4 Pr 2 –6 am –2 -R br am -C O C w ie f –1 a 0 , f( x ) < 2 3 4 –4 ev f –4 2 –2 y=x 4 –4 y=x –2 -C 6 –2 ni id ge 6 f x y ev x U y b 6 2 Pr y op C w ev ie 6 5 4 C op c -R 4 3 s 2 –4 op y C w ev ie -R ge id br am -C O –6 R 2 1 d f −1 does not exist since f is not one-one x +1 a f −1( x ) = 2 b Domain is − 3 < x < 5, range is − 1 < x < 3 2 f 2 –2 –4 f O 4 –2 –6 1 ni U R –4 2 y=x f –1 f–1 3 es ev ie y 6 –6 4 Pr es s op c (fg)−1( x ) = g−1 f −1( x ) w C ve rs ity 7−x ii 6 a y=x 6 -C 7−x a 6 14 − x b i 6 Exercise 2D 1 y 5 y 16 c b Proof 1± 5 2 b and c c 15 x +1 x ge a f −1( x ) = am br id 14 U ni op y Answers Copyright Material - Review Only - Not for Redistribution b d = −a ve rs ity −1 c Translation 0 2 d Translation 0 ni U 7 a = 2, b = − 3, c = 1 ev -R a s x 3 2 ity 1 rs –4 –3 –2 –1 O –1 4 s 3 w y op d y = 3x 2 − 2 x − 5 w a Reflection in the x-axis b Reflection in the y-axis ie ev c Reflection in the x-axis s -R d Reflection in the x-axis es am -C y = 2 x 2 + 3x + 1 b y = 2x 4 C 3 a y = − 5x 2 c 4 x U 2 e 1 2 br –4 –4 id g –3 –3 ni ve rs w ie 1 1 –2 ity C 4 –2 1 y y –4 –3 –2 –1 O –1 2 –4 –3 –2 –1 O –1 es –4 2 4 x 3 Pr -C op y –3 3 3 C x –2 c 2 4 ev 3 4 x y ie b -R 2 3 op ni ge 1 2 –4 id br am –4 –3 –2 –1 O –1 –3 U R 4 1 –2 ve w ie ev y 1 y 4 Pr op C y = x 2 − 6x + 8 es 4 –4 3 ev 6 ie ge 3 –3 2 R y = ( x + 1)( x − 4)( x − 7) br am -C y 2 –2 b c b = −1 5 1 1 4 x 2 –3 1 –4 –3 –2 –1 O –1 O –1 –2 Exercise 2F 2 292 –2 b a=2 id 4 3 ve rs ity op C w ev ie R 3 –4 –4 2 Translation 4 f 1 y 0 b Translation −5 y 0 a Translation 4 1 e Translation 0 y a 2 C op -C g y = ( x + 1)2 + x + 1 h y = 3( x − 2)2 + 1 2 w x−3 y= x−2 f 4 3 -R e 2 y= x+5 d y = x2 + 1 Pr es s y = 7x2 − 2x + 1 y w am br id c a ev ie b y =5 x −2 a y = 2x2 + 4 1 4 ge Exercise 2E C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 Copyright Material - Review Only - Not for Redistribution ve rs ity O 2 6 –4 –3 –2 –1 O –1 w ie s es –3 Pr –4 293 c ity 4 6 x y ve op 2 C U 1 ge –4 –3 –2 –1 O –1 w ie ev d y 4 x 1 2 3 4 x -R 4 3 ity 2 ni ve rs 1 y –4 –3 –2 –1 O –1 op c Stretch parallel to the y-axis with stretch factor 2 C U –2 –3 –4 es s -R br ev ie id g w e d Stretch parallel to the x-axis with stretch 1 factor 3 am 3 –4 Pr a Stretch parallel to the x-axis with stretch 1 factor 2 b Stretch parallel to the y-axis with stretch factor 3 -C R ev ie w C 3 2 –2 –3 es y = 162 x 3 − 108x 1 s br id -C +2 op y e y=2 x −1 y 4 3 b y = 3x 3 − 3 1 d y = x 2 − 4x + 10 2 am a y = 6x 2 x –2 –6 c 1 ni ie ev R –4 2 2 rs w 2 –2 4 3 –4 –3 –2 –1 O –1 -R am -C y op C O 3 y U y 6 –2 2 4 4 –4 1 y b 2 –6 4 x –4 ev id br b 3 –3 ge –6 2 –2 x ni –4 1 C op ev ie w –2 R 4 1 ve rs ity –2 3 -R 2 –4 y 4 2 Pr es s am br id 4 y op C –6 a w 1 6 ev ie a y -C 1 Exercise 2H ge Exercise 2G C U ni op y Answers Copyright Material - Review Only - Not for Redistribution ve rs ity 3 x –2 b y = − x 2 + 4x − 5 6 a y = 2g( − x ) b y = 3 − f( x − 2) 7 a Stretch parallel to the y-axis with stretch 0 1 factor followed by a translation 2 3 3 w 4 0 translation 2 x ev 2 br 1 b Reflection in the x-axis followed by a ie id ge 1 -R am –2 s ity 4 rs ve 1 3 4 x parallel to the y-axis with stretch factor -R s Pr 2 1 2 3 0 x-axis, translation 4 U e b y = 2 − f( x ) id g 1 x −1 + 3 2 a y=− 10 a y = 3[( − x + 4)2 + 2] = 3(4 − x )2 + 6 ev b y = 3( x − 1)2 b y=− 1 ( x − 1) − 3 2 9 -R b y = 3[( −( x + 4))2 ] + 2 = 3( x + 4)2 + 2 s am a y = 3( x − 1)2 es br y = 2f( x − 1) + 1 -C 3 y-axis with stretch factor 2, reflection in the y ev R –4 c x op ie –2 a y = 2f( − x ) 4 ni ve rs w –4 –3 –2 –1 O –1 3 c Translation , stretch parallel to the 0 ity 1 w op y C 3 es -C y 4 –3 2 −1 b Translation , stretch parallel to the 0 1 y-axis with stretch factor , reflection in the 2 0 x-axis, translation −2 ie br am h 1 2 ev id –3 w ge –2 –4 2 −5 a Translation followed by a stretch 0 U R –4 –3 –2 –1 O –1 8 ni ev ie 1 C w 2 op C 3 6 c Translation followed by a stretch parallel 0 1 to the x-axis with stretch factor 2 d Stretch parallel to the y-axis with stretch 0 factor 2 followed by a translation −8 y op Pr y g 294 –4 es y -C –3 C –4 –3 –2 –1 O –1 a y = 2x2 − 8 U 2 x 5 ni R 3 10 C op ve rs ity 4 5 O b ie op y a y = x2 y 4 y Pr es s -C y 3 –4 C w 2 –3 f ev ie 1 -R 1 2 1 b y = x − 5 2 1 ( x − 5)2 4 c 2 –4 –3 –2 –1 O –1 a y= ev ie 4 am br id 4 w ge y e C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 Copyright Material - Review Only - Not for Redistribution ve rs ity -R 1 –2 i ( x − 2)2 − 4 + k ie ii f( x ) > k − 4 ev -R iv f −1( x ) = 2 + x + 4 − k , domain is x > k − 4 − 5 < f( x ) < 4 ii y es s i Pr 295 ity f C U op x 6 w ge ie -R s es 4( x − 3)2 − 25, vertex is (3, − 25) Pr ii g( x ) > − 9 i 1 2 x + 25 , domain is x > − 9 2( x − 3)2 − 5 iii f( x ) > 27 ii 3 y 9 op x+5 , domain is x > 27 2 ( x − 1)2 − 16 ii −16 id g i C e 10 w U iv f −1( x ) = 3 + ev ie iii p = 6, q = 10 es s -R br am i ity x O b y = 3x 2 + 6x -C 8 iii g−1( x ) = 3 − ni ve rs C w –2 1 ( x + 2 ) for − 5 < x < 1 3 iii f −1( x ) = 5− 4 for 1 , x < 4 x ev id br am -C op y y 8 x y ve ni ev R ie y=x f–1 O 3 by translation 0 ev ii 3 iii p = 2 the y-axis or reflection in the y-axis followed a x w 6 rs y 4 iii f ( x ) = 3 + 4 − x , domain is x < 0 U -C y op C 2 ie w −( x − 3)2 + 4 −1 −3 b Translation followed by a reflection in 0 R y i –2 –4 2 C op ni 5 7 O 3 O –1 –3 1 2 End-of-chapter review exercise 2 f–1 2 –2 ge id br am −5 followed by translation 0 2 3 Pr es s ve rs ity y op C parallel to the x-axis with stretch factor f 4 x −10 Translation followed by a stretch parallel 0 1 to the x-axis with stretch factor or stretch 2 25 7 − 9 x − 4 6 a y=x y y = f(x) O –2 w ev ie w ev ie ge -C y y = g(x) 1 x + 2 for x > − 2 b −2 translation 0 R a f −1 : x ֏ 4 y-axis or reflection in the y-axis followed by 12 C 2 Translation followed by reflection in the 0 am br id 11 U ni op y Answers Copyright Material - Review Only - Not for Redistribution iv f −1( x ) = 1 + x + 16 ve rs ity br Prerequisite knowledge am ity rs ev ie C b x + 2 y = −8 c x + 2y = 8 d 3x + 2 y = 18 a y = 2x + 2 b 5x + 3 y = 9 c 7x + 3 y = −6 6 a y= 3 x+8 2 b (0, 8) a y= y 8 op a (6, 3) 4 x + 10 3 2 b y=− x+7 3 b ( −7 21 , 0 ), (0, 10) w C U 7 a 2 y = 5x + 33 -R s ( − 2, 6) a y = 3x + 4 (8, 2) 9 es 10 b 9x + 5 y = 2 a 2 y = 3x − 3 c 12 21 br k=2 -C 9 b y = − 3x − 1 c 39 id g units a y = 2x + 1 5 e 2 am 8 k=4 38 21 d 100 op s es Pr b ( − 1, 9) ni ve rs a ( − 2, − 1) ity a = 2, b = − 1 c 2 41, 2 101 7 c 4 145 w 4 op y C R ev ie w 6 b a = − 4, b = 16, c = 11 ie b = 3 or b = −5 54 a (6, 6) -R 4 b −2 ie id 3 br a = 3 or a = − 9 1 2 c a = 6 or a = − 4 c 2x − 3 y = 9 am 3 a y ve ni ge 2 -C 17 units2 a = 10, b = 4 c 2x + 3 y = 1 b PQ = 197, QR = 146, PR = 3 5 , not right angled 5 11 1 a PQ = 5 5, QR = 4 5, PR = 3 5 , right-angled triangle 2 b 5 Exercise 3C b 4 − 21, 4 + 21 Exercise 3A 1 a 1 -R Pr b −5 a ( x − 4)2 − 21 4 es b 6 U R ev ie w C 3 8 s -C 1 a − 6 2 a 3 c 7 21 op 296 (0, − 26) 10 y 2 7 9 ( −4 21 , −2 ), 13 1 6 k= C op ge id 3 Coordinate geometry (7, − 1) 5 7 k = 2 or k = 3 5 ni 1 ( x − 2) 2 4 w iv f( x ): half parabola from (0, 10) to (2, 2); g( x ): line through O at 45°; f −1( x ): reflection of f( x ) in g( x ) b Not collinear ie ve rs ity 3 1 1 , 5 6 Proof 2 5 − , 5 2 a y 2 U C w ev ie 1 ev -C ii 2 < f( x ) < 10 y 2( x − 2)2 + 2 v f −1 ( x ) = 2 − R Exercise 3B v h −1( x ) = − x + 2 iii 2 < x < 10 w A( − 5, 5), B(7, 3), C( − 3, − 3) 2 iii b = 2 op i 12 fg( x ) = 2 x − 3, gf( x ) = 4x + 4x − 1 ii a = − 1 1 iv ( x 2 − 3) 2 13 iv k = 22 2 b 8 2 ev i a (5, 2) -R 12 11 am br id iii − 1 , x , 7 ii f > − 11 ev 2( x − 3)2 − 11 Pr es s i ge 11 C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 Copyright Material - Review Only - Not for Redistribution b 33 ve rs ity E (4, 6), F (10, 3) 11 10 12 (14, −2) 13 a y = −3x + 2 b ( −1, 5) c 5 10, 4 10 d 100 ge a Proof 14 ( x − 5)2 + ( y + 3)2 = 40 15 ( x − 9)2 + ( y − 2)2 = 85 16 ( x + 3)2 + ( y + 10)2 = 100 , ( x − 13)2 + ( y + 10)2 = 100 17 a i 1+ 2 b i 3+2 2 3 2 b (0, 0), 2 d (5, − 3), 2 Exercise 3E ( 4 25 , 154 ) 15 a y = 2x − 7 16 x + y = 8, 3x + y = 3. Other solutions possible. U a x + y = 64 b ( x − 5) + ( y + 2)2 = 16 ) Pr 2 2 5 y 2 6 (4, 3) 7 a ( x − 12)2 + ( y − 5)2 = 25 and w ge ie ev -R End-of-chapter review exercise 3 1 s es Pr ity 9 ( x − 4) + ( y − 2) = 20 5 6 a a = 5, b = −2 2 c y = − x − 3 25 5 2 i 16t 7 (13, −7) 8 a ( −2, 2), (4, 5) e id g es ev s -R br 2 am b (4, −5) op ( x − 5)2 + y2 = 8 and ( x − 5)2 + ( y − 4)2 = 8 10 C 8 4 i ii Proof w Proof 3 4 1 49 and ii 24 9 4 a = −4, b = −1 or a = 12, b = 7 2 ie 7 U ( x − 6)2 + ( y + 5)2 = 25 -C ie ev R 6 2 , a , 26 y ni ve rs –6 w C op y –4 2 ( 5 − 5, 6 + 2 5 ), ( 5 + 5, 6 − 2 5 ) b Proof 6 x 4 b y = −2 x + 16 ( x − 2)2 + ( y − 10)2 = 100 -C am 2 br O –2 ie ity rs ( x + 2)2 + ( y − 2)2 = 52 2 ,m,2 29 a (0, 6), (8, 10) d 20 5 ve 4 − c ni ( x − 2)2 + ( y − 5)2 = 25 4 5 U 3 Proof id R ev ie w C op y c ( x + 1)2 + ( y − 3)2 = 7 1 3 25 d x − + y + = 2 2 4 3 s 2 2 5 y ( 2 es 2 2 -R am -C 2 w g (4, − 10), 6 e ( − 7, 0), 3 2 ( −1, −4), (5, 2) 1 3 10 f (3, − 4), 2 1 1 h 3 2 , 2 2 , 10 C br id c (0, 2), 5 ii Student’s own answer ev a (0, 0), 4 ge 1 ii Student’s own answer op b b ( x + 1)2 + ( y − 4)2 = 20 y ve rs ity ) y= C op -R Pr es s ii x + y = 7 Exercise 3D R 13 ev ie am br id -C y ( 12 ni ev ie w C b 2 21 , 4 21 ( x − 3)2 + ( y + 1)2 = 16, (3, − 1), 4 3 21 x− 4 2 ( x − 5)2 + ( y − 2)2 = 29 11 y = 4 21 a i op 14 10 w 10 C U ni op y Answers Copyright Material - Review Only - Not for Redistribution b y = −2 x + 5 21 297 ve rs ity iii ( −1, 8), 2 10 2 b p = −1 a y=− x+3 3 c ( x − 6)2 + ( y + 1)2 = 26 5 U ni C op y 4 a (19, 13) b 104 10 , 10 b k , −12, k . 12 a 3 4 a y=− x+2 b Proof 3 c ( x − 15)2 + ( y − 7)2 = 325 w ie -R ity rs ve a 1< x < 5 8 ( k , −2 k ) 9 6 5 10 a k = 14 11 a ( −1, −11), (6, 3) 12 a k=− 13 i C 1 x+4 3 25 b k,− 12 b y= w 1 b x.5 2 fg( x ) = 5x, range is fg( x ) > 0 ii g −1 ( x ) = 4 − 2 x , domain is 0 , x < 2 5x a b = −5, c = −14 ity a 2 y = 3x + 25 b ( −3, 8) 16 a 36 − ( x − 6) b 36 2 a 3( x + 2)2 − 13 a a = 12, b = 2 C 18 op c 6 , x , 18 x b ( −2, −13) y 17 b −3 26 − x 2 2 2 a ( x − 8) + ( y − 3) = 29 id g w c g−1( x ) = −3 + 19 ie U ii −3 , x , 8 15 e 4 (2.5, −20.25) b i c x < 36, g ( x ) > 6 d g−1( x ) = 6 + 36 − x es s -R br am 2 -R s es Pr ie (–2, –42) -C 3 −1 ni ve rs (2, 22) O –4 2 ie ge id br am -C y w C (2, –6) b 1 b −13, 3 7 14 4 x O op y –4 O ev 30 –1 –2 ev (–2, 26) ni 2 2 2 x = ± ,x = ± 3 2 y a R ev ie w 1 U C Cross-topic review exercise 1 –3 op op 298 1 s d Proof es c Proof Pr b 4 − 2 3, 4 + 2 3 3 2 ev id br a 4, (4, −2) y 17 -C am 16 ev y ge 14 15 R C 6 ve rs ity C ii Proof y = −x 2 + 6 x − 8 y y y = −2 x + 6, (3, 0) i w ev ie R 5 Translation , vertical stretch with stretch 0 factor 2 Pr es s 8 6 ii (0, −2), , 5 5 y = 2x − 2 12 13 4 5 op i -C iii (5, 12) 11 a = 5, b = −2 w 10 19 19 c − 113, + 113 2 2 ii ( −1, 6) i 2, m = 1 3 ev ie a ( −2, −3) am br id b y=− 1 x + 4 34 2 9 -R ge U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 Copyright Material - Review Only - Not for Redistribution b 5x + 2 y = 75 x ve rs ity w -R Degrees 0 45 90 135 180 225 270 315 360 w ev -R 7 8 12.79° Pr es s C w a 13 cm b 2.275 cm a 0.5 rad b 0.8 rad ie d 28 π cm 4 15.6 m 5 a 19.2 cm b 20.5 cm c 50.4 cm a 0.927 rad ity 6 b 4 cm ni ve rs c 17.6 cm a 14 cm 7 b 11.8 cm c 25.8 cm a 13 cm b 2.35 rad C U 8 f 80° g 54° h 105° 10 Proof j 810° k 252° l n 420° o 202.5° ie ev -R es s 48° w e 240° id g c 56.6 cm a Proof br 0.727 c 6 π cm ev -R 3 f b 3π cm 9 am 5π 6 a 2 π cm d 15° -C 7π 4 2π π 7π 6 5 π 11π 2π 3 6 d 0 c 30° m 225° 3π 2 c 0.622 3 e 2 7.79 cm b 60° 81° 2π 3 b 14.1 2 n 7π 36 e Radians 4 π 3 a 90° i π 2 a 0.644 1 ge id -C op y C w ie ev R 2 3π π 3 y s es ni b l 3π 2 Degrees 240 270 300 330 360 Exercise 4B 2π 9 5π d 18 5π f 6 h 7π 6 j 5π 3 U π 9 5π c 36 π e 36 3π g 4 5π i 4 k 13π 36 π m 20 o 10 π 3 a am 1 br R Exercise 4A π 6 6 Pr rs 5.14, 15.4 cm 2 ve w ie ev ity op 3 C 13, 67.4 5π 4 π ie id br y Prerequisite knowledge 2 3π 4 y U Radians 0 am -C 4 Circular measure (12 + π ) cm, 3π cm 2 π 2 b. Degrees 0 30 60 90 120 150 180 210 b k = ±4 10 1 π 4 Radians 0 ii 4x − 2 y = 21 (4, 5), (10, 2) a i d 87.1° a. ge R 5 range is (fg)−1 ( x ) < −3 24 c 76.8° e 45.3° x + 26 , domain is x > 28, 6 ii (fg)−1( x ) = − b 45.8° y fg( x ) > 28 b i a 68.8° C op a k = −2 d 3.49 op c ( −2, 0), (18, 0) 3 d y=− x+6 4 c 0.820 op b 10 ve rs ity a (8, 0) 4 ni 23 ev ie w C 22 b 0.559 e 5.59 Pr es s op y c a 0.489 ev ie ge -C c a 3 x + 7 −1 18 , g (x) = 5 − f (x) = x 3 Proof 2 3 17 17 3 − x + b − , 2 4 4 2 −5 and −1 d (1, −2), ( −1, 4) −1 b 21 1 2 a x= am br id 20 C U ni op y Answers Copyright Material - Review Only - Not for Redistribution b 43.4 cm 299 ve rs ity a 1.25 rad b 40 cm 2 5 a 1.75 rad y b 4.79 cm am br id U 11 2 πr 2 3 − 12 π 100 1 + − 3 cm 2 3 d 13.9 cm 2 2 b Proof c 1 d 6 5 f b Proof 15 4 15 c 16 a i id -R s es Pr ity U ii e −1 f θ = y π 4 π 3 cos θ 3r 2α ii + πr 2 2 -R 1 sin θ π 3 θ = 3 1 3 1 2 1 2 3 2 2 2 3 2 1 s 2 π 5 id g e iii α = br 2 πr + rα + 2 r am i -C 10 4α cos α + 4α + 8 − 8 cos α c C i tan θ w 9 ii 6.31 ie i 1 4 b θ = ev 8 Proof ii r 2 π r cos θ + 1 − sin θ + − θ 2 es i 3 f 1 6 ni ve rs 7 5 f 24 25 12 19 5 3 5 2 15(3 − 5 ) 4 15 15 15 16 75 − 4 15 15 1 2 d ev br iii A = 36, θ = 2 b 3+ 2 2 2− 3 e 2 1 a 2 2 −2 6 d 2 c am -C op y i R ev ie w C 6 1 4 C 5 ii 8 + 5 π 4 ii + 4 tan α + 2α cos α r(1 + θ + cos θ + sin θ ) ii 55.2 7 π AC = r − r cos θ ii +2 3−2 3 Proof ii 36 − ( r − 6)2 a f c op i d w U 4 4 25 3 25 π b − 4 8 ge 3 2 b e 4 + 15 End-of-chapter review exercise 4 1 e y ity rs ve a Proof 3 3 2 3 4 4 5 b b e ni ie ev R 3 5 20 d 3 2 a 3 a op es Pr y 14 w 13 1 a tan x + cm b 0.219 rad tan x a 15 − 5 3 + 5 3 π 6 π i α = 8 i 8 tan α − 2α b −5, a 0, 5 s -C 3r 2 cm 2 2 op C 300 ii 630° ie id a Proof am 10 iii w ge b 36.5 cm 2 br c 51.7 cm 2 1 5π 6 iii 195° ii 4 π ie a 29.1cm 2 d r b i 30° Exercise 5A 25 ( 2 3 − π ) cm2 6 b 1 1 + r2 π a i 4 -R 9 a 3 ni 8 -R ve rs ity 32 3 − 32 π cm 2 3 5 3 cm 3 1.86 cm 2 7 2 r 1 + r2 b c Pr es s -C op C w ev ie b 3.042 cm 2 c 5.16 cm 2 1 + r2 a y 4 1 C op b 1.5 rad w a 1.125 rad Prerequisite knowledge d 54 π cm 2 ev ie 3 6 R b 20 π cm 2 ev 2 a 12 π cm 2 9π c cm 2 4 a 867 cm 2 1 5 Trigonometry ge Exercise 4C C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 Copyright Material - Review Only - Not for Redistribution π 6 1 2 2+ 3 3 ve rs ity d 40° Pr es s ve rs ity −2 − 2 − 2 3 y 1 2 ev -R f 180° a 1 b 5 c 7 d 3 e 4 f 301 y 2 y 2 C op s es Pr ity e 180° w ie ev -R O 90 180 270 360 x 180 270 360 x s –1 –2 y 2 b a 1 + a2 a d − 1 + a2 b b 1 − b2 y 1 O C w 90 ie –1 1 − b2 ev d − es s -R id g br am c − 1 − b2 1 2 d 120° U 1− b 3 2 c 360° a e a 2 1 3 w ie 12 13 −1 b 180° es -C a 1 + a2 ve b b -C 8 2 − 3 1 2 21 b θ = 210° a 360° 1 2 b − a a c 3 Pr 21 5 a − 2 3 5 a − 13 op y C w ie R ev 7 3 ity id br a − 2 1 h − 2 am 4 1 rs ge f θ = 135° Exercise 5D U d θ = 120° C op ni U y op C w ie ev R 4th quadrant 6 1 cos θ id d − cos 65° π f − sin 8 2π h tan 9 3 b − 3 -C c − tan 55° π e − cos 5 3π g − cos 10 1 a − 2 2 c − 2 3 e − 2 3 g − 3 3 5 sin θ b cos 55° am a − sin 10° br Exercise 5C 2 2 13 c − 7 4 a − 11 ge 8π 3 3 5 tan θ ni C w ev ie R c 688° 5 12 3 d − 4 3 b 13 d − 7 3 b − op -C y op π 6 4π i 3rd quadrant, 9 a 125° g 3rd quadrant, 1 10 d 3rd quadrant, 30° π f 2nd quadrant, 3 π h 1st quadrant, 3 π j 4th quadrant, 8 b −160° 5π d 4 13π f − 6 e 1st quadrant, 40° e -R a 2nd quadrant, 80° b 3rd quadrant, 80° c 4th quadrant, 50° 3 c ni ve rs 2 12 13 w b 40° c 20° a − 9 ev ie a 70° am br id 1 ge Exercise 5B C U ni op y Answers Copyright Material - Review Only - Not for Redistribution ve rs ity br –1 180 -R y op C -R s O es Pr 2π x π 3π 2 2π x y 2 y ii π 2 ie id g w e O op 1 ev –1 –2 es s -R br am 3π 2 –3 360 x U R –4 π –2 ni ve rs –3 π 2 –1 ity 270 180 –2 -C 1 C -C op y O C y 3 2 1 –1 a i ev id 4 y 2 90 360 x 270 w 360 x 270 am –3 180 ie 180 br –2 w C s 90 ge U 1 O ie 360 x es ve ni ev 2 R 90 rs w ie 3 –1 ev 300 y ity y 5 4 f 240 Pr y op C e 180 –3 i –4 302 120 –2 360 x 270 60 ie 90 C op O -C –3 1 am –2 2 ev O –1 360 x 270 y 3 h U 1 180 –2 ge 2 id R 3 90 –1 ni y 4 O y 360 x 300 ve rs ity op C w ev ie d w ev ie 240 w 180 -R 120 y 60 y 2 1 Pr es s -C O g ge y am br id c U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 Copyright Material - Review Only - Not for Redistribution ve rs ity 1 3π 2 C w ve rs ity 1 a a = 3, b = 5 ge y w 90 14 a = 5, b = 4, c = 3 15 a A = 2, B = 6 y op 2π x ev s es -C am –4 -R br –2 ie id g y = cos 2x 3π 2 C e π w U ev π 2 180 –4 s es Pr O –2 ni ve rs ity 2 y = 3 sinx O f ie -R ev id br 2π x y = 2 sin x 2 b 2 y 10 8 b 1 < f( x ) < 5 op ve b 3π 2 op y C 4 y 13 303 4 π w 7 C op ie -R a a = 3, b = 2 –3 b 2 a y s Pr ity 12 rs a = 3, b = 1, c = 5 6 -C am π 2 4 π 11 U ie R O –2 b k= y = 2 + cos 3x 3 1 –2 –3 y = sin 2x ni y 4 2 –1 π c (0, 0), − , −2 2 es y op C w –2 –1 2π x 3π 2 π –1 a π x ev br π 2 b 4 6 π 2 w ge id y = 1 + cos 2x am -C O O –π 2 –π U y 3 y 3 2 ni a 1 ie a C y op –2 2π x -R -C π –1 C w ev ie R a = 3, b = 2, c = 3 π 5π 9 π 13π , −1 , 1 , , −1 , b , 1 , 8 8 8 8 5 ev 9 10 π 2 2 R a = 4, b = 2, c = 5 Pr es s O 8 ev ie ge y 2 am br id iii U ni op y Answers Copyright Material - Review Only - Not for Redistribution b 5 270 360 x ve rs ity d 197.5°, 342.5° e 126.9°, 233.1° f 116.6°, 296.6° g 60°, 300° h 216.9°, 323.1° π 5π b , 3 3 d 3.92, 5.51 2 π 4π f , 3 3 h 2.19, 5.33 2 c 1.25, 4.39 ni e 1.89, 5.03 ge 3 id b 17.7°, 42.3°, 137.7°, 162.3° -R es y op ni C U x + 4 3 0.298, 1.87 7 a 0°, 150°, 180°, 330°, 360° -R b 0°, 36.9°, 143.1°, 180°, 360° s es e 0°, 60°, 180°, 300°, 360° Pr 8 –1 b 56.3°, 123.7°, 236.3°, 303.7° a 30°, 150°, 270° y 9 2π x b 45°, 108.4°, 225°, 288.4° c 0°, 109.5°, 250.5°, 360° d 60°, 180°, 300° id g w e 3π 2 a 60°, 120°, 240°, 300° op π 0°, 76.0°, 180°, 256.0°, 360° C U π 2 d 0°, 116.6°, 180°, 296.6°, 360° f ity w ev ie e 0°, 180°, 199.5°, 340.5°, 360° br f 70.5°, 120°, 240°, 289.5° s -R g 19.5°, 160.5°, 270° es am a 26.6°, 206.6° 6 f 4−x b f is one-one, f −1 ( x ) = cos −1 2 3π a 2 5−x b f −1 ( x ) = sin −1 ,3<x<7 2 -C a 90°, 210° c 139.1°, 175.9° ge br am -C op y C 2 ie h 5.77°, 84.2° π 7π b , 2 6 d 0.0643, 2.36, 3.21, 5.51 3π f 0, 2 b 56.3°, 236.3° d 18.4°, 108.4°, 198.4°, 288.4° f 4 ev R g 58.3°, 148.3° c 0°, 78.7°, 180°, 258.7°, 360° O 6 f 116.6°, 153.4° e 278.2° y 6 a 2 < f( x ) < 6 5 e 24.1°, 155.9° c 119.7°, 299.7° b f −1 ( x ) = sin −1 a −7 < f( x ) < −1 4 4 5 id ie ev R 3 d 105°, 165° w π 4 2π e 3 16 a 25 c c 38.0°, 128.0° ie a 0 ni ve rs w C 2 Pr f 135° π b 4 π d − 6 π f − 3 16 b 9 ity e −60° op 304 rs d −90° ve c 60° y 1 s b 30° -C a 0° am Exercise 5E a 26.6°, 153.4° ev y = 6 + cos x g 0.848, 2.29 br 17 a 0.305, 2.84 C op 90 y 120 x U R –4 y = 2 + sin x C c 45.6°, 314.4° ev w b 23.6°, 156.4° w 60 –2 ev ie a 56.3°, 236.3° ve rs ity C O 1 ie op y 2 16 w ev ie -R f 30 x +5 , 0 < f −1( x ) < 2 π 4 Exercise 5F -C 4 a −9 < f( x ) < −1 b f −1 ( x ) = 2 cos −1 am br id 6 7 ge y 8 Pr es s c U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 h 30°, 150°, 270° Copyright Material - Review Only - Not for Redistribution ve rs ity op π 5π , 6 6 End-of-chapter review exercise 5 a 0.565, 2.58 11 3π 7π 2.03, , 5.18, 4 4 am br id b C 1 2 a Proof e Proof f Proof 5 a Proof b Proof 6 30° or 150° c Proof d Proof a Proof b Proof 7 30° or 150° 8 i e Proof f a 4 + 3 sin2 x b 4 < f( x ) < 7 9 a (sin θ + 2) − 5 b 4, −4 10 a Proof ii 45°, 135°, 225°, 315° 10 i 60° or 300° ii 120° 11 Proof π 5π a , 6 6 i f( x ) < 3 rs ve ni 13 w s es b 30°, 150°, 210°, 330° 5 a Proof b 72.4°, 287.6° 6 a Proof b 65.2°, 245.2° 7 a Proof b 41.8°, 138.2°, 270° 8 a Proof b 30°, 150° 9 a Proof b 66.4°, 293.6° 10 a Proof b 70.5°, 289.5° 11 a Proof 3−x iv f −1( x ) = 2 tan−1 2 i 30° or 150° 15 i Proof ii n = 3, θ = 290° i 17 i 1.68 C op ni ve rs ii 194.5° or 345.5° Proof 16 w e ie b 30°, 150°, 210°, 330° ev id g π x ii 54.7°, 125.3°, 234.7°, 305.3° es s -R br am -C π 2 14 ity Pr a Proof U op y 4 C w ie ev f ie id -C b 18.4°, 116.6° b 60°, 131.8°, 228.2°, 300° R ii 3 − 2 3 ev a Proof y 3 b −2.21, 0.927 O -R 3 ii 109.5° or 250.5° i iii ge a Proof br 2 iii 20 Proof 12 b 76.0°, 256.0° am a Proof ii 4 i U R 1 2π θ 9 1 − 4a 2 4a , cos θ = b sin θ = 2 1 + 4a 1 + 4a 2 Exercise 5H y w ie ev 8 ity 4 2 y= 1 2 –1 y op 7 C w ie ev Proof π y d Proof y = cos 2θ op c Proof O C -C b Proof -R a Proof s h Proof es g Proof y 6 Proof Pr f y 1 C op ni ge br d Proof am e Proof x=± 4 b Proof id a Proof c Proof ve rs ity y d Proof 1 − k2 k c −k d Proof c Proof 5 b c Proof U R 4 1 − k2 a 3 2 39.3° or 129.3° b Proof op C w ev ie 3 Pr es s -C 9 sin2 x − 3 1.95 3 Exercise 5G 1 a = 1, b = 2 -R 2 w 10 ev ie ge U ni y Answers Copyright Material - Review Only - Not for Redistribution 305 ve rs ity b −32 x15 3 a 2n + 3 14 a x5 + 5x 4 y + 10 x 3 y 2 + 10 x 2 y3 + 5xy 4 + y5 16 a y − 3y Pr ity 3 rs ie a 1 + 16x + 112 x b 1 − 30 x + 405x 2 7 21 2 c 1+ x + x d 1 + 12 x 2 + 66x 4 2 4 5103 5103 2 e 2187 + x+ x 2 4 f 8192 − 53248x + 159 744x 2 2 C ev -R s es a −84 35 c 4 b 5940 d −9720 7 −224 000 8 −41184 9 40 095 10 a 128 + 320 x + 224x 2 b 1 − 28x + 345x 2 op y 7920 C Pr ity ni ve rs w ii 16 − 8 5 11 a 1024 + 5120 x + 11520 x 2 s -R b 1024 + 10 240 y + 30 720 y 2 es -C am c 32 d 792 b 56 c 1 − 3x + 3x 2 br 16 + 8 5 b i 3 id g a 1 + 3x + 3x + x 2 c 364 b n 6 e b 97 + 56 3 6 5 U a 16 + 32 x + 24x 2 + 8x 3 + x 4 d 5005 w id br am -C op y C 5 c 495 n( n − 1) a 2 n( n − 1)( n − 2) c 6 a 45 h 512 + 1152 x 2 + 1152 x 4 f ±2 b 84 g 256 + 1024x 2 + 1792 x 4 d 16 −32 5 g 768 h − 2 A = 486, B = 540, C = 30 a 35 ie g 16x 4 − 96x 3 + 216x 2 − 216x + 81 9x 27 27 + + h x6 + 2 4x 4 8x 9 a 12 b 10 4 R ev ie w 3 b y5 − 5 y 3 + 5 y op ni U ge R 8x 3 + 36x 2 y + 54xy2 + 27 y3 e 40 b 36 2 y ve 4 e x 4 − 4x 3 y + 6x 2 y2 − 4xy3 + y 4 c −90 3 ie b 1 − 4x + 6x 2 − 4x 3 + x 4 c x 3 + 3x 2 y + 3xy2 + y3 2 y a p = 8, q = 8 C op 15 s y op a x 3 + 6x 2 + 12 x + 8 f C x11 es -C b 11 − 3n d 8 − 12 x + 6x 2 − x 3 w 13 2 Exercise 6A C 5 w a 125x 6 w 12 b 142 -R 2 ie a 1 + 4 y + 6 y2 ev id br b 9x 3 − 9x 2 − x + 1 am a 4x 2 + 12 x + 9 ev 11 1 1 1 54 Exercise 6B Prerequisite knowledge 306 10 Pr es s U ge 6 Series −216 b 113 100 ni iii f is one-one R 2π x 3π 2 π x−5 iv f −1( x ) = 2 cos−1 3 a x8 − 4x 6 + 6x 4 − 4x 2 + 1 b −16 ev C ev ie w π 2 ve rs ity op y 2 O 8 9 f -C 4 16 + 112 x + 312 x 2 + 432 x 3 + 297 x 4 + 81x5 -R 6 7 ev ie ge y 8 am br id ii U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 Copyright Material - Review Only - Not for Redistribution ve rs ity 13 16 + 224x + 1176x 14 a = −2, b = 1, p = −364 15 n = 8, p = 256, q = −144 ge b 1 am br id 2 -R Pr es s -C 13 a $17 715.61 14 Proof 15 Proof 16 Proof -R s es ity Proof 19 a 4 − 3 sin2 x 20 a Proof C 0.25, 199.21875 7 0.5, 9 ie w 6 8 b 900 -R 9 s es a −0.25, 256 b 204.8 11 a 90 b 405 12 a 36 b 192 13 93.75 14 a = 2, r = e id g 3 5 π π ,x, 3 2 a 5π 17 a Proof -R ev 16 c Proof es s br am b 40.5 10 15 U ar5 , ar14 −1, −1 ni ve rs f ity d No Pr b 3, 15 309 52 165 2 a , 13.5 3 ev U ge a No 1 1 c − ,− 3 27 e No -C y 57 ɺ ɺ = 57 + 57 + +… a 0.57 100 10 000 1000 000 b Proof b Proof -C am br id b 9a op y C w ie ev R 2 , 810 3 5 Exercise 6D 6 4 307 y 18 5 32 y a a = 8d c 26 23 4 3 op 17 b 1 91 d −36 74 a 3 3 ni 16 1 ( 5n − 11) 2 9° 2 3 −10.8 3 ,8 2 64 2 rs 10, −4 ve 14 Pr b 20 op 7, 8 C w ie ev R ie id br am -C a 17, −4 13 C op $360 1 op 11 Exercise 6E C 25 b $94 871.71 w 10 b 2, ie 5586 1 3 a ev 9 ve rs ity 31 ni 8 U 1817 ge 7 b 2059 11 w y d −3160 x 1442 4 b 48.8125 m 12 6 3 n b −1957 a 7, 29 2 3 a 8 4 a 1037 c 38 13 5 1 10 21 b 35, 3535 7 15 b 255 d 700 59 a 22, 1210 4 12 a 765 48 x +1 40, −20 y R ev ie w 3 8 9 a + 6d , a + 18d op C 2 −8, 2 c −85 Exercise 6C 1 7 w a 1 − 4x + 7 x 2 ev ie 12 C U ni op y Answers Copyright Material - Review Only - Not for Redistribution 11π 8 b Proof b ve rs ity -R es Pr ity rs ve c 15 d 225, maximum a Proof b 625 − ( r − 25)2 c 25 d 625, maximum 2 40 000 200 b −πr− π π 40 000 , maximum d π ii 0.9273 a Proof i Proof ni U i r 2 (tan θ − θ ) ii 12 + 12 3 + 4 π 14 i 2 − 5 cos2 x ii −3 and 2 16 i -R b 2187 i Proof ii 26.6°, 153.4° −5 < f( x ) < 3 ii (0.253, 0), (0, −1) y 4 es s -C iii a d = 2a op y b 99a 15 a a = 60, d = −10.5 b 42 23 16 a 17 b r = − 5, S = 7 4 7 17 a 18 i 19 a a = 10, b = 45 π b i 0 ,θ , ii 1.125 3 i x = −2 or 6, 3rd term = 16 or 48 Pr b n = 22 ity 2 –π 2 ni ve rs –π 4 π 4 op U ii 22100 w e C –4 -R 16 27 s ii –6 ev ie id g br O –2 b 16 am y = f(x) y 1 5 41000 -C y b 225 − ( r − 15)2 13 15 a a = 44, d = −3 C a Proof iii 0.685, 2.46 es ie Proof iii 5.90 cm 2 b − 17 , 3 7 ii 243 − 405x + 270 x 2 14 w ie ev 11 12 id am 2 a − 3 c 1312.2 br 23 i b 112 c Proof ge 1 + 10 x + 40 x 2 a i b 5940 R a 14 625 8 s -C y C w 135 2 a 6561x16 − 17 496x15 + 20 412 x14 a 1 + 8 px + 28 p2 x 2 20 6 10 b −37 908 13 a −25.6 y 7 12 5 b 12 b 27 79 -R 40 11 a 1 − 10 x + 40 x 2 9 br 6 10 4 ii 35.3°, 144.7°, 215.3°, 324.7° am 5 864 25 16800 − 9 a 729x 6 − 2916x 3 + 4860 b −5832 op 2 8 ev ve rs ity ni 3 3 U 5 3840 8 ge 2 2 7 id 240 x180 C op op 1 4 R b End-of-chapter review exercise 6 op R ev ie w 7 308 3 ,n=6 2 a x = −3 or 5, 3rd term = 24 or 40 4 b − 5 a 4 C 6 1 Pr es s -C b 12.96, 68 y 5 Cross-topic review exercise 2 C −2.5, 22.5 3 a 5 w 4 b 384, 32 ie a 2 a d = 6, a = 13 w 3 22 b 16 ie a 100 ev 2 b 115.2° b a = 12 , r = 5 7 7 w b 788.125 a 2 14 ev a 352 am br id 1 21 ev ie ge Exercise 6F C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 Copyright Material - Review Only - Not for Redistribution π x 2 ve rs ity 7 3 4 − 3 2 x x 1 3 i 6 x+ + 2 x x3 a 3 5 c − 4 15 8 −3 9 −8 10 ( −2, −10), (2, − 6) g 14x + 17 −2 , x , 3 18 x < −1 and x > 19 Proof op b 9x 8 1 d − 2 x 2 f 3 3 x e 10(5x − 2)7 C U ie -R ev g −56(4 − 7 x )3 1 ( x + 2)2 16 c (3 − 2 x )2 72 e − (3x + 1)7 50(2 x − 1)4 21 (3x − 1)6 h 5 j −16x(2 − x 2 )7 f l d y f 16( x + 1) x 2 ( x + 2)2 1 a 2 x −5 2x c 2x2 − 1 h b es s -R ev ie 3 b 16(2 x + 3)7 8 91 d x + 1 22 b w U e id g br am k 6x 2 ( x + 2)( x + 4)2 g − b 6x 2 + 8 e 16x 3 − 24x 10 x( x 2 + 3)4 a − 2 0 d 2x + 1 10 2 f − 2 3 x x i op Pr es s 2 1 h − 5 x c 10 x − 3 309 C f w ge id c 3x5 1 2 y a 6( x + 4)5 1 c −20(3 − 4x )4 b 15x 4 3 d − 2 x a 10 x − 1 y a = −10.5, b = 18 ev −2 a 8x 3 4 -C 16 ni ve rs C w ie ev R a = 4, b = −6 −3 d 10 e − 3 3x 2 g x 5 15 b −8, 2 Exercise 7B ve b op y 4 2.95 2.99 h 3x g 5x a = −5, b = 2 -R s 2.8 -C e 0 14 ni 4 1 c 2 a 5x 4 4 c − 5 x a br R 3 Pr es 2.5 EF ity 2 DF rs Gradient CF b 3 2 ev AF BF Chord am ie w C op a a = 2, b = −7 ity y Exercise 7A 1 13 b 0.5 C op ni ge id br am -C 4 a 4( x − 2)−3 3 − 2 y = 2x + 1 12 5 4 a ( −2, 7), (3, −8) 11 U R 1 −1 d x 2 5 2 f − x3 5 b 2(3x + 1)−5 e 3x −2 3 w b 5x 3 1 2 x 2 c ev ie 2 3 a 3x 2 1 2 ve rs ity 1 ev ie w C Prerequisite knowledge 5 1 + 2 x 4 x3 w op y 7 Differentiation h 3− ie -C ii 70 -R Proof b i 6 Pr es s 17 am br id ge x +1 iv f −1( x ) = sin−1 , domain is −5 < x < 3, 4 1 1 range is − π < f −1( x ) < π 2 2 a 250 C U ni op y Answers Copyright Material - Review Only - Not for Redistribution d 5( x 3 − 5)4 (2 x 3 + 5) x6 3 − ( x − 5)2 32 x − 2 ( x + 2)2 45 − 6 2 ( 3x + 1 ) 49(4x − 5) − (2 x − 5)8 x8 1 2x + 3 3x 2 − 5 2 x 3 − 5x ve rs ity d 2y = x − 1 ni a 4y = x + 4 b y = x −1 7 ( −2.5, 8.5) 7 9 a y = 4x − 68 20 a 2+ 3 x ( −7.5, 2.5) 10 ( −6.2, 6.6) 11 a ( −32.6, 28.4) 12 a (2 3, 8 3 ), ( −2 3, −8 3 ) C w − + + + + ev 11 a Proof 12 a = −4, b = 4 Pr ity rs ve ni U b −8, 8 C 1 3x 3 + 2 −3 59 7 4x 2 -R Proof 4 −15(3 − 5x )2 − 2, 150(3 − 5x ) es 27 2 (3x + 1)5 10 a y = −4x + 18 h 24x 2 − 36x + 20 11 a Proof ni ve rs 7 ii Proof w y=− ( −0.8, 15.5) 1 x 2 es s 12 b y= 1 x +1 4 ie ev b i -R id g ii (16.5, 0) op ity 6 − C Pr (4.25, −7.5) 5 y s 3 9 f br am − Proof U 4 (4x − 9)3 4x − 30 g x4 5x + 12 i − 4 x5 − 10 d 48(2 x − 3)2 e − + e 36 x4 -C w ie ev R c − + Proof b 30 x − 14 a 2 0 8 b (0.4, −5.48) Exercise 7D 1 − 9 ge a −7 − 8 15 i 4x + 5 y = 66 24 i 5− 3 x 37.4 C 16 0 x.5 id -C 73 op y 15 + 8 br y = 0.6x + 1.6, 30.96° dy dx End-of-chapter review exercise 7 am 14 7 − es -C y op C R ev ie w (4, −1) 6 d2 y dx 2 b 317.2 units2 13 5 + -R b (0.6, 2.48) b Proof 4 s am b y = 4x − 7.5 b (17, 0) 3 op 6 2 ie (0, 7.5) 1 w 5 0 ie a Proof b x + 4y = 0 2 81 x id 4 br a x + 4y = 4 8 c 10 48 − (2 x − 1)9 d 5x − 6 y = 3 3 b 4 y y = 3x + 9 6 U b 8x + y = 17 c x − 6y = 9 310 a −1 a y = 3x − 7 c 2 4 5 ge 1 R ev ie Exercise 7C 4 − 8(2 x − 1)3 , −48(2 x − 1)2 y ve rs ity a = 5, b = 3 w 8 f C op y op C 7 9 4 (1 − 3x )3 80 3 3 (2 x + 1)7 d − 3 w 4 3 (5, 1) 4, 6 3 x3 ev 12 b − 15 x −6 4 e Pr es s 5 30 45 − x 4 x7 4 45 − 3 x 4 x7 a c -C 10 2 ev ie h f -R 1 (2 x − 5)3 g − 4 3 3x + 1 6 3 (2 − 3x ) 4 ge 2 3 3 (5 − 2 x )2 am br id e − U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 Copyright Material - Review Only - Not for Redistribution ii (2.1, 8.25) ve rs ity ii 10 34 Proof 14 i Proof 15 1 iii , 4 34 2 1 i y − 2 = ( x − 1), y − 2 = −2( x − 1) 2 ii 2 21 c ( −3, −12) maximum; (3, 0) minimum b (2, − 2), maximum -R a a = −15, b = 36 Proof d x , 0 and x . 8 e x , 1 and x . 4 f a x , 1 13 b x . 4.5 c 2,x,5 d −1 , x , 3 ity rs ve 1 3 x 2 a a = −6, b = 5 b minimum c (0, 5), maximum d (2, − 11), −12 a a = 4, b = 16 b minimum y op ni a a = 54, b = −22 b minimum C b ( −1, 14) ie w ge a a = −3, b = −12 br ev id c (2, −13) = minimum, ( −1, 14) = maximum d (0.5, 0.5), −13.5 -R Exercise 8C s ii 97.2 Q = 5x − 36x + 162 40 − 2 r b Proof θ= r r = 10 d A = 100, maximum 50 − x y= b Proof 2 A = 312.5, x = 25 c 3 a c y a U b ( −0.5, 6.25) maximum 2 ii 108 2 op a (2, 4) minimum P = 9x 2 − x 3 C ity Exercise 8B b i c i ni ve rs 7 , x , 20 a y = 9−x 1 es Proof op y 8 a 3x 2 − 10 x + 160 d ( −3, −17) minimum; (1, 15) maximum b x = 1 23 , 151 23 cm 2 5 a Proof b A = 37.5, maximum ie -R a QR = 9 − p c 2 p= 3 s am (1, − 7) maximum; (2, −11) minimum -C f 6 ev id g br e ( −1, − 4) minimum w 4 e c ( −2, 22) maximum; (2, −10) minimum es C O c x , 0 and 0 , x , 3 13 1.5 , x , 3.5 8 , increasing (1 − 2 x )2 2x − 6 , neither ( x + 2)3 Proof 8x + 20, 8x + 20 ù 0, if x ù 0 1 12 x , − 4 and x . 2 7 9 1 c x , 0 and x . 2 Pr 6 11 am 5 10 −2 23 , x , 2 -C 4 f 2 –1 U R 3 y w ie ev c x , −1.75 x , 2 and x . 6 23 3 -R b x .1 Pr a x.4 e 4 s 18 (1 − 3x )3 es -C b y 2 C op ve rs ity ni id am a 10(2 x − 1)4 (2, 6) minimum y br 9 9 x , 4 x 2 f 3+ k 3−k , minimum; x = , maximum 2 2 (0, 1), minimum; (1, 2), maximum; (2, 1), minimum x= b −2 , x , 3 3 9 b − 3 , 4 x x op C w 6 7 U a 6x − 1, 6 ev ie b −3 , x , 1 ge 2 1 w a a=3 9 Exercise 8A ie 5 8 3 ev b −44, 81 8 Further differentiation c R 4 e (4, 4) maximum 18 18 , ≠0 x3 x3 a −2, 3 3 Pr es s y op C w ev ie R ev ie d ( −2, −28) maximum; (2, 36) minimum -C am br id 9 9 ii − , 2 4 6 12 iii , , E not midpoint of OA 11 11 a x , −1 and x . 3 b (1, 12) minimum w ge i 1 a (9, 6) minimum 2 13 Prerequisite knowledge C U ni op y Answers Copyright Material - Review Only - Not for Redistribution b Proof d A = 12 3, maximum 311 ve rs ity b V = 486, x = 3 3 0.09 units per second, increasing 7 1, 6 8 0.016 units per second 1 ,3 3 w ev ie y C op w ie b 1.024 cm s −1 5 4π π cm s −1 10 ii 8 π cm2 s −1 3 Maximum 4 k = 0.0032 kg cm 3 , 0.096 kg day -1 5 ev s ity 8 op C w ie i A = 2 p2 + p3 ii 0.4 i Proof ii 20 m by 24 m i Proof ii 120, minimum 4(6 − x ) ii Proof 3 iii A = 72 d 2 y 16 dy 8 i = − 2 + 2, 2 = 3 dx x dx x d2 y ii (2, 8), minimum since 2 . 0 when x = 2 dx d2 y ( −2, −8), maximum since 2 , 0 when x = −2 dx πx i y = 30 − x − ii Proof 4 iv Maximum iii x = 15 i y= id g w e C U op y ni ve rs 9 y 2 2 cm s −1 45 π 300 π m 2 hr −1 10 es s -R br am -C C 10 cm π a 40 π cm3 s −1 a 2 ie 6 ie 32 π cm2 s −1 -R 1.25 units per second es 5 Pr 0.08 units per second 1 cm s −1 3 -R s es Pr ve ni U 4 ev R id w op y 3 9 13 7 1 units per second 300 −0.04 units per second 2 a Proof 6 −0.315 units per second, decreasing 1 10 1 d 1241, maximum -C Exercise 8D a 3 cm s −1 120 1 b cm s −1 7 9 b − cm s −1 100 π b End-of-chapter review exercise 8 ge c 13 13 9 b i b Proof br a r = 20 h − h2 am 15 a Proof ev w ie ev R 14 8 12 rs 13 7 0.125 cm s −1 1 cm s −1 320 9 π cm2 s −1 11 ity op C 312 5 ev id br am -C y 12 π cm3 s −1 0.003 cm s −1 Pr es s ge U R 11 4 π cm 2 s −1 5 18 cm3 s −1 4 6 ni 10 2 ve rs ity w C op y 9 1 -R am br id c Maximum 288 b Proof a y= 2 x c 432, 12 cm by 6 cm by 8 cm 1 1 a y = 1 − x − πx b Proof 2 4 dA 1 d2 A 1 c = 1 − x − πx , = −1 − π 2 dx 4 4 dx 2 4 d e A= , maximum 4+π 4+π 5 − 2 r − πr a h= b Proof 2 dA d2 A c = 5 − 4r − πr, 2 = −4 − π dr dr 5 25 e , maximum d 4+π 8 + 2π 50 − 2 x a r= b Proof π 100 2500 = 14.0, = 350, minimum c ( π + 4) (4 + π) 432 a h= 2 b Proof r d A = 216 π, minimum c r=6 2 500 − 24x b Proof a y= 10 x 5 10 d Proof c 6 160 3 a h= − r b Proof 2 r c r=8 -C 8 ev ie Exercise 8E ge a Proof 7 C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 Copyright Material - Review Only - Not for Redistribution ve rs ity C 5 364 b Maximum at − , , minimum at (1, 4) 3 27 Pr es s y f 5 3 y = 2x 2 + 2x 2 + 2 f( x ) = 4 + 8x − x 2 y C w ie ev -R s 4 x 1 2 x − 10 x + 3 2 b x , − 2 and x . 1 23 a y = 2 x 3 + 6x 2 + 10 x + 4 10 y = 2 + 4x − 2 x − x 11 a −3.5 y 9 2 w -R 13 Proof b P = minimum, Q = maximum 1 a −6 b y = x 2 − 6x + 2 2 2 2 f ′( x ) = 2 x − 2, f( x ) = x 2 + − 4 x x ev 12 b 3 s b 5x 4 + c 2 d − 2 +c x b y = 42 x − 97 a y = x3 + x +c id g br a 2x6 + c 3 c − +c x 313 (4, 20) O e 5 w y s 7 es ni ve rs 3 + 6x 2 + c y= 7 − 5 12 x 2 e am 4 7 2x 2 -C R ev ie w d y = 2 x + 3x − 7 8 U C c es ity rs ve ni U ge op y b es a Pr 3 6 2 y ity d 2 2 x −1 3 5 4 id c y = 2 x − 2x + 3 +2 x y = 2 x 3 − 6x 2 + 5x − 4 3 y = 5x 3 + 2 − 2 x a y = 2x2 x + 2x − 1 br R b 3 f y= 2 3 am a 1 y=− 2 +c 4x x4 f( x ) = x5 − + 2x + c 2 x6 x3 f( x ) = + − x2 + c 2 3 1 8 f( x ) = 3x 2 − 2 − + c x x 3 3 f( x ) = − 6 + − 4 x + c x 2x x 3 5x 2 y= + +c 3 2 3x 4 2 x 3 y= + +c 2 3 x4 − 2 x 3 − 8x 2 + c y= 4 x2 1 5 y= − + +c 2 x 4 4x Pr -C e y = 3x 4 + c -C 2 ev ie w C op c w ie 5 x b y = 2x7 + c 3 d y=− +c x f y =8 x +c y a y = 5x 3 + c b y = 2x3 − x2 + 5 6 d y = x2 + − 4 x a y = x3 + x + 2 4 c y = 10 − x e y = 4 x −x+2 ev b 10 x − 4 + Exercise 9A 1 4 C op ni U ge 1 b (0, 9) a 24x 7 − 13 10 Exercise 9B -R 3 3x 3 3x 3 + +c d 10 4 x 3 − +c 2 2 x2 3 2x 2 20 x2 4 x g + +c h + +c 7 x 6 3 12 9 − +c i 2 x2 + x 4x4 b −1 id 2 a −19 2 a − , 5 3 br 1 am R Prerequisite knowledge 8x 2 c − + 2x2 + x + c 3 x 1 + +c e f 2 2x 7 ve rs ity ev ie w C op 2 p 4 p3 ii (0, 0) minimum, − , maximum 3 27 iii 0 , p , 3 9 Integration x3 − 3x 2 + 9x + c 3 op 2 p 4 p3 − 3 , 27 b op i − C 13 ii Maximum 10 +c x f ie Proof 5 -R i -C 12 4 x +c 3 x 3 5x 2 a + + 4x + c 3 2 e ev ie 4 3 ge a −2 , x , am br id 11 U ni op y Answers Copyright Material - Review Only - Not for Redistribution ve rs ity a y = 9 + 3x − x 2 16 a y = 2 x x − 6x + 10 b (4, 2), minimum 17 (1, 7), maximum 18 a y = x 2 − 5x + 2 1 y w ev b 84 25 ie -R es s b 8 20 56 d 5 13 op a 11 56 C 3 b 40 21 d 14 e 16 f 5 b 6 6 a 48 3 4 3 13 7 10 23 8 a 9 b 1.6 9 a Proof b 3− 5 10 18 23 11 a ( −1, 0) 12 10 21 13 34 op C w s es 9 y ev 26 -R 14 ie s es Pr U 1 ( x + 3)8 + c 4 1 b (2 x x − 1)5 + c 5 b ie -R w Proof y a 21 13 c 15.84 ity a 15 x (2 x x − 1)4 ( x + 1)4 4 x a b 5 21 id g br 7 am 6 b 2 151 f a 7 51 e b d 4 a 15x 2 ( x 3 − 2)4 Pr ity U ( x − 3x + 5) +c 3 4( x + 3)7 a x 6 26 3 d 21121 ni ve rs a 6(2 x − 3)( x 2 − 3x + 5)5 -C ie 6x (4 − 3x 2 )2 2 R ev 5 1 2 ( x + 2)4 + c 8 1 b (2 x 2 − 1)5 + c 20 2 b − 2 +c x −5 1 +c b 8 − 6x 2 b 3 c 4 32 id -C op y C a w 4 a k = −2 f 18 8 4 b 45 2 4 b a 20 x(2 x 2 − 1)4 2 3 br am a 8x( x 2 + 2)3 1 4 15 4x a − 2 ( x + 5)2 c ge y = 4 2x − 5 − 2 Exercise 9D 6 1 b y = 5 2x − 3 − 4 b y = 8 3x + 1 − 2 x 2 − 2 x + 5 6 d ev a Maximum b 3 Exercise 9F rs 5 5 ve a x + 5y = 7 4 ni C y = 3( x − 5)4 − 1 4 R ev ie w 3 ve rs ity ni U y = 2 x−2 +5 op c 314 2 5 e 2 3 -C a a 10 c 1 b y = (2 x + 5) 2 − 7 3 2 d y= +6 3 − 2x y 2 am i 3 ge br g 1 (3x + 1)6 + c 18 1 d − (1 − 2 x )6 + c 4 5 1 (2 x + 1) 2 + c f 5 2 h − +c (2 x + 1)2 b id e 1 (2 x − 7)9 + c 18 2 (5x − 2)9 + c 45 4 3 − (5 − 4x ) 3 + c 16 4 3x − 2 + c 3 5 +c 32(7 − 2 x )4 1 y = (2 x − 1)4 + 2 8 11 2 107 c 6 37 e 8 a C op Pr es s y op C w ev ie R c c −6 2 Exercise 9C a 16 9 d 21 5 f 2 b e 9 b x + y = −1 c (1, −2) 1 a 7 -R am br id -C b 5 y = x + 21 ev ie 15 ge ( −11, 408 13 ) w Exercise 9E 14 C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 Copyright Material - Review Only - Not for Redistribution b 90 ve rs ity 3 57 61 4 y a y= 9 a y = −8x + 16 b 108 1 10 a 2y = x − 1 b 8.83 2 ve br -C 16 π 3 25 π d 4 124 π b 15 y s es Pr ity ni ve rs op 3 C i w 20 4, 3 13 ii P and Q are both 14 i y = −24x + 20 -R e b 24 π a ∪-shaped curve, y-intercept = 4, vertex = (2, 0) 32 π b 5 1 iii f( x ) = 2 x 2 + 6x 2 − 10 x + 5 U 39 π 4 a 24 π i 29.7° 9 8 ii 1 ev id g s 15 es 6 11 12 br 5 C op C -R am 10 b am 4 -C 3 w w 9 Proof ie f a ∪-shaped curve, y-intercept (0, 11), vertex at (3, 2) 483π b 5 9 i Proof ii 4 i B (0, 1), C (4, 3) ii y = −3x + 15 2π iii 15 5π iii i Proof ii 1 3 1 i or 9 9 1 3 3 − 3 − 1 ii f ′′( x ) = x 2 − x 2 at x = max, at 2 2 9 x = 9 min ev id d Proof op y C 2 ie U ni 8 b Proof 71π 5 15 π c 8 81π a 2 6 a ev 7 ge y op C w ie 6 Exercise 9I 1 5 y a Proof 4 op 3 25 3 4 x − 20 x − + c x 3 6 5 2 y = 30 − − x x 2 4 f( x ) = 6 x + 2 + 2 − 10 x 194 π 9 a 1 b f( x ) = 3x 2 − 6x + 8 10 23 ie h 1 2 g 6 3 20 i 9 Proof es f 16 Pr e 50 ity d 4 rs 5 4 s 1 b 256 -C a 2 3 f( x ) = 3x 4 + 5x 2 − 7 -R am br id ge ni b y End-of-chapter review exercise 9 e Proof w b 171π cm 3 8 c Proof ie C 13 128 π 3 32 π b 5 b U C w ev ie R ev 12 Proof 52 π a 3 8π a 3 1888 π cm 3 a 3 Proof 250 π 9 14 c − ev 11 b 1 (2 3 − 3) 2 b 64 1 R 3125 π 6 b 16 π b 1 x+2 3 a y = −3x + 46 Exercise 9H R ve rs ity 7 a (0, 3) 1 a , 7 , (2, 7) 3 -R Pr es s b 10 23 op 6 8 10 a 36 1 3 1 13 a (25, 0) 9 c 36 5 7 w 2 10 23 ge 26 23 am br id 1 -C Exercise 9G ev ie U ni op y Answers Copyright Material - Review Only - Not for Redistribution ii 32 3 315 ve rs ity C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 iii -R 1 15 7 a Proof b Proof 1250 = 175 (3 s.f.), maximum c ( π + 4) i 12 ii x = −1 or x = −3, 1 13 3 Proof 4 a 2187 − 10 206x + 20 412 x 2 3 a (4 π − 3 3 ) cm 2 2 b (2 π + 3 + 3 3 ) cm b −30 618 a ( x − 3)2 + ( y + 2)2 = 20 b x − 2 y = 17 U ge br id ie 5 -R 6 7 Pr y es s am -C dy 9 dy i = , no turning points since ≠0 2 dx (2 − x ) dx 81π iii k , −8, k . 4 ii 2 8 8 <0 a f ′( x ) = − 2, (2 x + 1) (2 x + 1)2 4−x b f −1( x ) = ,0,x<4 2x c y 8 rs 9 ni ge 8 id ie 6 -R 1 3 1 −2 1 −2 x + x 2 2 y= i 2 2 y − 6 = − ( x − 2) ii y = x 2 + + 1 7 x 10 a 3x + 4 y = 17 11 a − ii x = −1 (max), x = 2 (min) 16 32 − 2 x, 3 − 2 x2 x 431π 5 ev ie id g c es s -R br am 3π 2 b C e i 13 π y ni ve rs U iii x = 1, minimum since f ′′(1) . 0 13 π 2 2π x c 0.927 rad, 5.36 rad 6−x d g−1( x ) = cos−1 5 ity ii O op 1 es 3 2 s x 4 i -C R f Pr -C am br ev 4 2 2 2 x − 2x 2 − 3 3 iii (1, −2), minimum op y 12 ev ie w C 11 C 10 U R O y 12 y=x 4 27 8 b (3, 21) ii a 1 < f( x ) < 11 b ve w ie ev f–1 a 7, −8 1 b i 3 a 21 − 2( x − 3)2 c x < 3 − 13, x > 3 + 13 ity op C 10 y i 2 f ′( x ) = 2 + y 6 316 284 π 3 C op i 9 ii 15 . 0 for all x x4 y = − x 3 + 6x 2 − 12 x + 11 1 ni 5 8 Proof Practice exam-style paper ve rs ity 3 y=− + 1 ii x . 1, x , − 2 2(1 + 2 x ) 1 1 ii , 0 2 6 4 2 ii − 9 27 3 40 w op w ev ie R i ii x = 4, minimum y 4 C i i 17 4 op 3 y = 2 x 2 − 6x + 2 3 ii w 16 (3x + 4) 2 + 12 3 Pr es s -C ii y = 2 x − 5 3 B , 0 , C 0, 4 4 w 2 i ev ie am br id 1 14 ev 2 y = x 3 − 3x + 7 3 i −0.6 w ge Cross-topic review exercise 3 Copyright Material - Review Only - Not for Redistribution 1 units per second 240 b ( −2, −12), maximum op y ve rs ity ni C U ev ie -R am br id F Pr es s -C G C op y General form of a circle: the equation x 2 + y2 + 2 gx + 2 fy + c = 0 , where ( − g, − f ) is the centre ni B and g 2 + f 2 − c is the radius of the circle Geometric progression: each term in the progression is a constant multiple of the preceding term Gradient function: the derivative f ′( x ) is also known as the gradient function of the curve y = f( x ) -R br am I ity op Pr y es s -C Chain rule: the rule for computing the derivative of the composition of two functions Common difference: the difference between successive terms in an arithmetic progression Common ratio: the constant ratio of successive terms in a geometric progression Completed square form: the equation ( x − a )2 + ( y − b )2 = r 2, where ( a, b ) is the centre and r is the radius of the circle Completing the square: writing the expression ax 2 + bx + c in the form d ( x + e )2 + f Composite function: a function obtained from two given functions by applying first one function and then applying the second function to the result Convergent series: a sequence that tends to a finite number y op ie w ge ev id s ity Pr op y es Decreasing function: a function whose value decreases as x increases Definite integral: an integral between limits whose result does not contain a constant of integration dy Derivative: denoted by of f ( x ); gives the gradient of a dx curve op id g e C U Differentiation: the process of finding the gradient of a curve Differentiation from first principles: the process of finding the gradient of a curve using small increments Discriminant: the part of the quadratic formula underneath the square root sign Domain: the set of input values for a function N ev Normal: the line perpendicular to the tangent at a point on a curve es s -R br am -C Many-one: a function which has one output value for each input value but each output value can have more than one input value Mapping: a diagram to show how the numbers in the domain and range are paired Maximum point: a point, P, on a curve where the value of y at this point is greater than the value of y at other points close to P Minimum point: a point, Q, on a curve where the value of y at this point is less than the value of y at other points close to Q y ni ve rs C w ie ev R -R br am M -C D Identity: a mathematical relationship equating one quantity to another Improper integrals: a definite integral that has either one limit or both limits are infinite, or a definite integral where the function to be integrated approaches an infinite value at either or both endpoints in the interval (of integration) Increasing function: a function whose value increases as x increases Indefinite integral: an integral without limits whose result contains a constant of integration Integration: the reverse process of differentiation Inverse function: the inverse of a function, f −1( x ) , is the function that undoes what f( x ) has done C U R ni ev ve rs C w ie ev id ie w ge U R Basic angle: the acute angle made with the x-axis Binomial: a polynomial with two terms Binomial coefficients: the coefficients in a binomial expansion Binomial theorem: the rule for expanding (1 + x ) n or ( a + b ) n C Factorial: 6 × 5 × 4 × 3 × 2 × 1 = 6 ! (read as ‘6 factorial’) First derivative: see Derivative Function: a rule that maps each x value to just one y value for a defined set of input (x) values w ev ie w C ve rs ity op y Amplitude: the distance between a maximum (or minimum) point and the principal axis of a sinusoidal function Arithmetic progression: each term in the progression differs from the term before by a constant Asymptote: a straight line such that the distance between a curve and the line approaches zero as they tend to infinity ie A w ge Glossary Copyright Material - Review Only - Not for Redistribution 317 ve rs ity Range: the set of output values for a function w ge O C U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 Reference angle: the acute angle made with the x-axis d2 y or f ″ (x ) and is used to dx 2 determine the nature of stationary points on a curve Second derivative: denoted by Series: the sum of the terms in a sequence ve rs ity Self-inverse function: a function f where f −1( x ) = f( x ) for all x ni C op y Solid of revolution: the solid formed when an area is rotated through 360° about an axis ge U Stationary point: a point on a curve where the gradient is zero, also known as a turning point ev Tangent: a straight line that touches a curve at a point Term: a number in a sequence Turning point: a point on a curve where the gradient is zero, also known as a stationary point op 318 -R s y op -R s es -C am br ev ie id g w e C U R ev ie w ni ve rs C ity Pr op y es s -C -R am br ev id ie w ge C U R ni op y ve w rs C Radian: one radian is the angle subtended at the centre of a circle by an arc that is equal in length to the radius of a circle ev V Vertex: the vertex of a parabola is the maximum or minimum point Volume of revolution: the volume of the solid formed when an area is rotated through 360° about an axis ity R ie es Pr y -C am Quadrant: the Cartesian plane is divided into four quadrants −b ± b2 − 4ac Quadratic formula: the formula x = , 2a which is used to solve the equation ax 2 + bx + c = 0 ie id T br Q w y op C w ev ie R S Pr es s -C Parabola: the graph of a quadratic function Pascal’s triangle: a triangular array of the binomial coefficients, where each number is the sum of the two numbers above Period: the length of one repetition or cycle of a periodic function Periodic functions: a function that repeats its values in regular intervals or periods Point of inflexion: a point on a curve at which the direction of curvature changes Principal angle: the angle that the calculator gives is called the principal angle ev ie P Roots: if f( x ) is a function, then the solutions to the equation f( x ) = 0 are called the roots of the equation -R am br id One-one: a function where exactly one input value gives rise to each value in the range Copyright Material - Review Only - Not for Redistribution ve rs ity am br id ev ie w ge completing the square 6–8 graph sketching 19 proof of the quadratic formula 10 composite functions 39–41 conic sections 71 connected rates of change 228–9 practical applications 230–2 constant of integration 244–5 convergent series 176 cosine angles between 0° and 90° 118–20 general angles 123–6 graph of 128–9 inverse of 137, 138 transformations of 132 trigonometric equations 140–4 trigonometric identities 145–7, 149 basic angle (reference angle) 121–2 binomial coefficients 160–4 binomial expansions 156–8 binomial theorem 161 decreasing functions 213–15 definite integration 250–2 area bounded by a curve and a line 260–1 area bounded by two curves 260–1 area enclosed by a curve and the y-axis 255–6 area under a curve 253–5 improper integrals 264–7 volumes of revolution 268–70 degrees 100 converting to and from radians 101–2 derivatives 193 first and second 205–6 see also differentiation Descartes, René 83 differentiation 191–6 addition/subtraction rule 195 chain rule 198–200 constants 195 increasing and decreasing functions 213–15 notation 193 power rule 194–5 practical applications of connected rates of change 230–2 practical maximum and minimum problems 221–3 rates of change 227–9 real life uses of 212 Pr es s -C ie w C op ve rs ity ni U ge ev id br -R am es s -C Pr y ity op y y C w ie Galileo Galilei 2 Gauss, Carl 167 geometric progressions 171–2, 180 infinite geometric series 175–8 sum of 173–4 gradients 75, 192–3 and equation of a straight line 78–80 of tangents and normals 201–3 see also differentiation s -R ev factorial notation 163–4 factorisation, quadratic equations 3–5 first derivative test 217 first derivatives 205 functions composite 39–41 definitions 34 domain and range 35–7 graph of a function and its inverse 48–9 increasing and decreasing 213–15 inverse of 43–5, 48–9 many-one 35 one-one 34–5 quadratic see quadratic functions self-inverse 44 transformations of 51 combined transformations 59–63 reflections 55–6 stretches 57–8 translations 52–3 trigonometric see cosine; sine; tangent; trigonometry use in modelling 34 op w ie ev -R s es Pr es -C am br id g e U ni ve rs ity ni U ge id br am -C op y C C ve calculus 191 see also differentiation; integration Cantor, Georg 181 Cartesian coordinate system 83 quadrants of 122 chain rule, differentiation 198–200 connected rates of change 228–9 circles area of a sector 107–8 equation of 82–7 intersection with a line 88–90 length of an arc 104–5 right-angle facts 85 codomain (range) of a function 35–7 inverse functions 43, 45 combined transformations of functions 59–60 trigonometric functions 132 two horizontal transformations 61–3 two vertical transformations 60–1 common difference, arithmetic progressions 166 common ratio, geometric progressions 171 communication vi completed square form, equation of a circle 83 rs C w ie w ie equation of a circle 82–3, 86–7 completed square form 83–4 expanded general form 84–5 equation of a straight line 78–80 tangents and normals 202 y y op C w ev ie R ev R ev R scalar multiple rule 195 second derivatives 205–6 stationary points 216–19 tangents and normals 201–3 differentiation from first principles 193 discriminant 24 domain of a function 35–7 inverse functions 43, 45 -R addition/subtraction rule, differentiation 195 amplitude of a periodic function 129 angles degrees 100 general definition of 121–2 radians 101–2 arcs, length of 104–5 area bounded by a curve and a line 260–1 area bounded by two curves 260–1 area enclosed by a curve and the y-axis 255–6 area under a curve 253–5 arithmetic progressions 166–7, 180 sum of 167–9 asymptotes 129 op Index C U ni op y Index Copyright Material - Review Only - Not for Redistribution 319 ve rs ity C w ev ie -R y y y op C w ie ev scalar multiple rule, differentiation 195 second derivatives 205–6 and stationary points 218–19 sectors, area of 107–8 self-inverse functions 44 graphs of 48 sequences arithmetic progressions 166–9 geometric progressions 171–4 infinite geometric series 175–8 series 156, 167 binomial coefficients 160–4 binomial expansion of (a + b)n 156–8 simultaneous equations 11–13 sine angles between 0° and 90° 118–20 general angles 123–6 graph of 128–9 inverse of 136, 138–9 transformations of 130–1 trigonometric equations 140–4 trigonometric identities 145–7, 149 sketching a graph 17–19 quadratic inequalities 22 solids of revolution 268 sphere, surface area of 14 C w ie ev -R s es Pr ity quadrants of the Cartesian plane 121 quadratic curves, intersection with a line 27–8 quadratic equations completing the square 6–8 functions of x 15–16 number of roots 24–5 simultaneous equations 11–13 solution by factorisation 3–5 es s -R radians 101 area of a sector 107–8 converting to and from degrees 101–2 length of an arc 104–5 simple multiples of π 102 range (codomain) of a function 35–7 inverse functions 43, 45 rates of change 227–8 connected 228–9 practical applications 230–2 recurring decimals 176 reference angle (basic angle) 121–2 reflections 55–6 revolution, volumes of 268–70 right-angle facts for circles 85 roots of a quadratic equation 24–5 op parabolas 17–19 intersection with a line 27–8 lines of symmetry 17–19 maximum and minimum values 17–19 paraboloids 18 parallel lines 75 Pascal’s triangle 157–8 periodic functions 129 perpendicular lines 75 points of inflexion 216 power rule, differentiation 194–5 principal angle 137, 138 problem solving vi ni ve rs e U -C am br id g Lagrange’s notation 193 latitude 104 Leibnitz, Gottfried 193, 239 length of a line segment 72–4 limits of integration 250 quadratic formula 10 quadratic functions 2 graph sketching 17–19 lines of symmetry 17–19 maximum and minimum values 17–19 quadratic inequalities 21–3 C op w ie ev -R s es Pr ity rs one-one functions 34–5 oscillations 117 ni U ge id br am -C op y C w ie R ev Newton, Isaac 193, 239 normals, gradient 201–3 nth term of a geometric progression 171–2 nth term of an arithmetic progression 166–7 ve ge id br am -C y op R ev ie w C 320 many-one functions 35 mappings see functions maximum points 17–19, 216–17 practical problems 221–3 and second derivatives 218–19 mid-point of a line segment 72–4 minimum points 17–19, 216–17 practical problems 221–3 and second derivatives 218–19 modelling vi–vii ni identities, trigonometric 145–7, 149 image set of a function see range (codomain) of a function improper integrals type 1 264–6 type 2 266–7 increasing functions 213–15 indefinite integrals 241 inequalities, quadratic 21–3 infinite geometric series 175–8 infinity 181 improper integrals 264–7 inflexion, points of 216 integration 239, 249 area bounded by a curve and a line 260–1 area bounded by two curves 260–1 area enclosed by a curve and the y-axis 255–6 area under a curve 253–5 definite 250–2 of expressions of the form (ax + b)n 247–8 finding the constant of 244–5 formulae for 241–3 improper integrals 264–7 notation 240 as the reverse of differentiation 239–40 volumes of revolution 268–70 intersection of a line and a parabola 27–8 inverse functions 43–5 graphs of 48–9 trigonometric functions 136–9 U R ev ie w C horizontal transformations, combination of 61–3 line segments gradient 75 length and mid-point 72–4 parallel and perpendicular 75–7 see also straight lines lines of symmetry, quadratic functions 17–19 longitude 104 Pr es s op y -C am br id ge gradients of parallel and perpendicular lines 75–7 graph of a function and its inverse 48–9 graph sketching 17–19 quadratic inequalities 22 graphs of trigonometric functions transformations 130–3 sin x and cos x 128–9 tan x 129 ve rs ity U ni op y Cambridge International AS & A Level Mathematics: Pure Mathematics 1 Copyright Material - Review Only - Not for Redistribution ve rs ity C U ni op y Index ev ie -R am br id Pr es s -C Underground Mathematics vii vertex of a parabola 17–19 vertical transformations, combination of 60–1 volumes of revolution 268–9 rotation around the x-axis 269–70 rotation around the y-axis 268–9 ie w C op ve rs ity ni U ge ev id br waves 117 321 y op y op -R s es -C am br ev ie id g w e C U R ev ie w ni ve rs C ity Pr op y es s -C -R am br ev id ie w ge C U R ni ev ve ie w rs C ity op Pr y es s -C -R am trigonometric identities 145–7, 149 trigonometry 117 angles between 0° and 90° 118–20 general definition of an angle 121–2 graphs of functions 127–33 inverse functions 136–9 ratios of general angles 123–6 transformations of functions 130–3 turning points 17–19, 216 see also stationary points y y op C w ev ie R tangent (trigonometric ratio) angles between 0° and 90° 118–20 general angles 123–6 graph of 129 inverse of 137 trigonometric equations 141–3 trigonometric identities 145–7, 149 tangents to a curve 27 gradient 201–3 terms of a sequence 166 trajectories 2 transformations of functions 51 combined transformations 59–63 reflections 55–6 stretches 57–8 translations 52–3 trigonometric functions 130–3 translations 52–3 trigonometric functions 131 trigonometric equations 140–4, 149 w ge stationary points 17–19, 216–17 practical maximum and minimum problems 221–3 and second derivatives 218–19 Stevin, Simon 176 straight lines equation of 78–80 intersection with a circle 88–90 intersection with a quadratic curve 27–8 see also line segments stretch factors 57–8 stretches 57–8 trigonometric functions 130 sum of an arithmetic progression 167–9 sum of a geometric progression 173–4 sum of an infinite geometric series 175–8 sum to infinity of a series 176–8 Copyright Material - 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