Cambridge International AS & A Level Mathematics: Pure Mathematics 1 Coursebook 9781108407144
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Table of contents :
Cover
Title page
Copyright
Contents
Series introduction
How to use this book
Acknowledgements
1 Quadratics
2 Functions
3 Coordinate geometry
4 Circular measure
5 Trigonometry
6 Series
7 Differentiation
8 Further differentiation
9 Integration
Practice exam-style paper
Answers
Glossary
Index
Citation preview
Cambridge International AS & A Level Mathematics:
Pure Mathematics 1 Coursebook
Copyright Material - Review Only - Not for Redistribution
Copyright Material - Review Only - Not for Redistribution
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Cambridge International AS & A Level Mathematics: w
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Sue Pemberton Series Editor: Julian Gilbey
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Pure Mathematics 1 y op y op -R s es
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Coursebook
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University Printing House, Cambridge CB2 8BS, United Kingdom
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One Liberty Plaza, 20th Floor, New York, NY 10006, USA 477 Williamstown Road, Port Melbourne, VIC 3207, Australia
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314–321, 3rd Floor, Plot 3, Splendor Forum, Jasola District Centre, New Delhi – 110025, India
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79 Anson Road, #06–04/06, Singapore 079906
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Cambridge University Press is part of the University of Cambridge.
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© Cambridge University Press 2018
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www.cambridge.org Information on this title: www.cambridge.org/9781108407144
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It furthers the University’s mission by disseminating knowledge in the pursuit of education, learning and research at the highest international levels of excellence.
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First published 2018
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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.
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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
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A catalogue record for this publication is available from the British Library
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ISBN 978-1-108-40714-4 Paperback
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® IGCSE is a registered trademark
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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.
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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.
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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.
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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.
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1.2 Completing the square
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1.3 The quadratic formula
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1.1 Solving quadratic equations by factorisation
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1.4 Solving simultaneous equations (one linear and one quadratic)
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1.5 Solving more complex quadratic equations
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1.8 The number of roots of a quadratic equation
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2.3 Inverse functions
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2.4 The graph of a function and its inverse
2.7 Stretches
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3.2 Parallel and perpendicular lines
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3.3 Equations of straight lines
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3.1 Length of a line segment and midpoint
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3.4 The equation of a circle
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3.5 Problems involving intersections of lines and circles
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End-of-chapter review exercise 3
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2.8 Combined transformations
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2.5 Transformations of functions
3 Coordinate geometry
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2.1 Definition of a function 2.2 Composite functions
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End-of-chapter review exercise 1
2.6 Reflections
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1.9 Intersection of a line and a quadratic curve
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1.7 Solving quadratic inequalities
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1.6 Maximum and minimum values of a quadratic function
2 Functions
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1 Quadratics
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Acknowledgements
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How to use this book
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Series introduction
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Contents
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Contents
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Cambridge International AS & A Level Mathematics: Pure Mathematics 1
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Cross-topic review exercise 1
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5.2 The general definition of an angle
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5.3 Trigonometric ratios of general angles
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5.4 Graphs of trigonometric functions
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5.5 Inverse trigonometric functions
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5.6 Trigonometric equations
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5.7 Trigonometric identities
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6.3 Arithmetic progressions 6.4 Geometric progressions
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6.2 Binomial coefficients
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6.1 Binomial expansion of ( a + b )
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6 Series
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6.5 Infinite geometric series
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6.6 Further arithmetic and geometric series
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End-of-chapter review exercise 6
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7.1 Derivatives and gradient functions
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7.3 Tangents and normals 7.4 Second derivatives
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7.2 The chain rule
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End-of-chapter review exercise 7
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7 Differentiation
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Cross-topic review exercise 2
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End-of-chapter review exercise 5
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5.8 Further trigonometric equations
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5.1 Angles between 0° and 90°
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5 Trigonometry
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End-of-chapter review exercise 4
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4.3 Area of a sector
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4.2 Length of an arc
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4.1 Radians
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4 Circular measure
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Contents
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8 Further differentiation
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8.1 Increasing and decreasing functions 8.3 Practical maximum and minimum problems
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End-of-chapter review exercise 8
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9 Integration
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9.1 Integration as the reverse of differentiation
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9.3 Integration of expressions of the form ( ax + b )
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9.2 Finding the constant of integration
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9.4 Further indefinite integration
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9.5 Definite integration
9.8 Improper integrals
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Glossary Index
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Answers
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Cross-topic review exercise 3 Practice exam-style paper
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9.7 Area bounded by a curve and a line or by two curves
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9.6 Area under a curve
9.9 Volumes of revolution
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8.5 Practical applications of connected rates of change
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8.4 Rates of change
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8.2 Stationary points
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Cambridge International AS & A Level Mathematics: Pure Mathematics 1
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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.
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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.
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Series introduction
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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.
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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.
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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.
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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’.
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Series introduction
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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.
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To support you in your learning, these coursebooks have a variety of new features, for example:
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■ 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.
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We wish you every success as you embark on this course.
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Julian Gilbey London, 2018
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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.
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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.
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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.
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How to use this book
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Learning objectives indicate the important concepts within each chapter and help you to navigate through the coursebook.
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Key point boxes contain a summary of the most important methods, facts and formulae.
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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.
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Throughout this book you will notice particular features that are designed to help your learning. This section provides a brief overview of these features.
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completing the square
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It is important to remember to show appropriate calculations in coordinate geometry questions. Answers from scale drawings are not accepted.
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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.)
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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.
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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.
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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.
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These questions focus on problem-solving. These questions focus on proofs.
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Throughout each chapter there are multiple exercises containing practice questions. The questions are coded:
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You should not use a calculator for these questions.
These questions are taken from past examination papers.
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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.
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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.
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These questions focus on modelling.
You can use a calculator for these questions.
At the end of each chapter there is a Checklist of learning and understanding.
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Cross-topic review exercises appear after several chapters, and cover topics from across the preceding chapters.
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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.
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WEB LINK
Extension material goes beyond the syllabus. It is highlighted by a red line to the left of the text.
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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.
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FAST FORWARD
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REWIND
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How to use this book
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Cambridge International AS & A Level Mathematics: Pure Mathematics 1
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Acknowledgements
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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.
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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.
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Past examination questions throughout are reproduced by permission of Cambridge Assessment International Education.
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Thanks to the following for permission to reproduce images:
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Cover image iStock/Getty Images
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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
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Chapter 1 Quadratics
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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.
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■ ■ ■ ■ ■ ■
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In this chapter you will learn how to:
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Cambridge International AS & A Level Mathematics: Pure Mathematics 1
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PREREQUISITE KNOWLEDGE
What you should be able to do
IGCSE® / O Level Mathematics
Solve quadratic equations by factorising.
b x 2 − 6x + 9 = 0
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c 3x 2 − 17 x − 6 = 0 2 Solve:
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a 5x − 8 . 2
3 Solve:
Solve simultaneous linear equations.
a 2 x + 3 y = 13
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7 x − 5 y = −1 b 2 x − 7 y = 31 3x + 5 y = −31 4 Simplify: a
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b ( 5 )2
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Carry out simple manipulation of surds.
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a x 2 + x − 12 = 0
Solve linear inequalities.
IGCSE / O Level Additional Mathematics
Why do we study quadratics?
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1 Solve:
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IGCSE / O Level Mathematics
IGCSE / O Level Mathematics
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Check your skills
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Where it comes from
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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.
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You will have plotted graphs of quadratics such as y = 10 − x 2 before starting your 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.
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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.
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WEB LINK Try the Quadratics resource on the Underground Mathematics website (www.underground mathematics.org).
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Chapter 1: Quadratics
w
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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
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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
Use the fact that if pq = 0, then p = 0 or q = 0.
x=
-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
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2 or x = 5 3
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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
Factorise.
ie
6x 2 − 17 x + 5 = 0
br
ge
Write in the form ax 2 + bx + c = 0.
C
U
6x 2 + 5 = 17 x
a
w
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
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Pr es s
21 2 − =1 2x x + 3
2 x 2 − 11x − 63 = 0
(2 x + 7)( x − 9) = 0
w
U
R
ev ie
Factorise.
y
Expand brackets and rearrange.
ni
21( x + 3) − 4x = 2 x( x + 3)
C
Multiply both sides by 2 x( x + 3).
ve rs ity
op
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=−
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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.
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ev
3x 2 + 26 x + 35 =0 x2 + 8
Factorise.
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3x 2 + 26 x + 35 = 0
C
ie
w
C
Solve
ity
op
y
es
s
WORKED EXAMPLE 1.3 4
br
ev
id
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(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
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am
br
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ni ve rs
C
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Pr
op y
x=−
C op
WORKED EXAMPLE 1.2
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Cambridge International AS & A Level Mathematics: Pure Mathematics 1
Copyright Material - Review Only - Not for Redistribution
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Chapter 1: Quadratics
ev ie
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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 + x − 6)
=1
B
( x 2 − 3x + 1)6 = 1
rs
w
2
C
( x 2 − 3x + 1)(2 x
ev
ve
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4(2 x
5
ity
EXPLORE 1.2 A
-R
am
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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
2
+ x − 6)
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
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ni
op
1 Discuss with your classmates how you would solve each of these equations.
=1
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
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C
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Pr
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4 Discuss your results.
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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)
c
e
6x 2 + x − 2 =0 x2 + 7x + 4
ev f
2 x 2 + 9x − 5 =0 x4 + 1
c
2( x
e ( x 2 + 2 x − 14)5 = 1
f
( x 2 − 7 x + 11)8 = 1
op
d 3(2 x
− 11x + 15)
s
=
+ 9 x + 2)
2
=1
1 9
5 The diagram shows a right-angled triangle with sides 2 x cm, (2 x + 1) cm and 29 cm.
2
− 4 x + 6)
y
U
R
b Find the lengths of the sides of the triangle.
w
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C
2x + 1
ev
id
br
-R
am
s
= 1.
Pr
op y
7 Solve ( x − 11x + 29)
x
es
-C
x+3
x–1
(6 x 2 + x − 2)
C
ity
1.2 Completing the square
w ev
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br am
C
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( 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.
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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.
PS
TIP
op
ni
ev
a Show that 2 x 2 + x − 210 = 0.
2
=8
29
2x
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rs
C
6
b 4(2 x
Pr
2
y
=1
ity
-C
4 Find the real solutions of the following equations. + 2 x − 15)
x2 − 9 =0 7 x + 10
w
x2 + x − 6 =0 x2 + 5
-R
am
br
x2 − 2x − 8 =0 x 2 + 7 x + 10
b
es
id
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3 Solve: 3x 2 + x − 10 =0 a x2 − 7x + 6
2
x(10 x − 13) = 3
y
3 1 + =2 x + 1 x( x + 1)
a 8( x
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
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a x 2 + 3x − 10 = 0
-R
1 Solve by factorisation.
ie
EXERCISE 1A
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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.
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Chapter 1: Quadratics
ev ie
am br id
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Rearranging these gives the following important results:
Pr es s
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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
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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
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8÷2 = 4 ւց x 2 + 8x − 7 = ( x + 4)2 − 42 − 7
br
ev
x 2 + 8x − 7 = ( x + 4)2 − 23
s
-C
-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 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
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WORKED EXAMPLE 1.5
C
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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
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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
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WORKED EXAMPLE 1.6
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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
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w
C
8
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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
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C
U
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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
am -C
1 (11 ± 85 ) 2
s
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
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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.
6 Solve by completing the square. a x 2 + 8x − 9 = 0
b x 2 + 4x − 12 = 0
d x 2 − 9x + 14 = 0
e x 2 + 3x − 18 = 0
d 9x 2 − 42 x + 61
c
x 2 − 2 x − 35 = 0
f
x 2 + 9x − 10 = 0
op
Pr
y
es
s
-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 .
9
ity
b x 2 − 10 x + 2 = 0
rs ve
f
2 x 2 − 8x − 3 = 0
5 3 + = 2. Leave your answers in surd form. x+2 x−4
w
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9 The diagram shows a right-angled triangle with sides x m, (2 x + 5) m and 10 m.
br
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x
-R
Find the value of x. Leave your answer in surd form.
am
10
ie
id
PS
C
U
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8 Solve
e 2 x 2 + 6x + 3 = 0
x 2 + 8x − 1 = 0
ni
ev
ie
d 2 x 2 − 4x − 5 = 0
c
y
w
a x 2 + 4x − 7 = 0
op
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.
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s
4
Pr
op y
-C
2
ity
op
y
(x, y)
Range
br
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a Find the range of this projectile.
am
x
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O
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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 .
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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
es
b from both sides. 2a
b2 − 4ac 2a
b ± 2a
ve
Write the right-hand side as a single fraction.
y
−b ± b2 − 4ac 2a
ni
ie
w
ge
id
WORKED EXAMPLE 1.8
C
U
R
x=
Subtract
op
C ev
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x=−
Pr
b2 − 4ac 2a
b =± 2a
ity
op
x+
Find the square root of both sides.
rs
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
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C
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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
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C
op
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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
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C
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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): x2 − 4 0 0 3
ve rs ity
Pr es s
Rearrange. Factorise.
C
op
y
-C
2x − 1 = x2 − 2x − 3 = ( x + 1)( x − 3) = x = −1 or x =
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
s
49 − 28 y + 4 y2 − 4 y2 = 8 4
es ity
12 y2 + 28 y − 17 = 0
Factorise.
y
id g
ev
ie
17 19 in equation (1) gives x = . 6 3
-R s es
am
br
Substituting y = −
w
e
C
U
1 17 or y = 2 6
op
(6 y + 17)(2 y − 1) = 0 y=−
Rearrange.
ni ve rs
-C
R
ev
ie
w
C
49 − 28 y + 4 y2 − 16 y2 = 32
Expand brackets.
Multiply both sides by 4.
Pr
op y
-R
am
7 − 2 y − 4 y2 = 8 2
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
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1 or y = 9 4
id
-R s es Pr
=0 =0
ity
15
rs
=9 = ±3
y op
ge
C
U
R
ni
ev
ie
w
C
op
y
4x 4 − 37 x 2 + 9 (4x 2 − 1)( x 2 − 9) 1 or x 2 x2 = 4 1 x=± or x 2
ve
-C
am
br
ev
1 or x 2 = 9 4 1 or x = ± 3 x=± 2 Method 2: Factorise directly x2 =
ev
id
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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 w ie -R es
s
-C
EXERCISE 1E
am
br
ev
id
x = −1 or x = 2
1 = 3−1 and 9 = 32. 3
U
1 or 3x = 9 3
Substitute 3x for y.
ni
1 or y = 9 3
ge
ev ie
R
3x =
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
C
ge
2 Solve:
9 5 + =4 x 4 x2
op
j
a 2 x − 9 x + 10 = 0
x ( x + 1) = 6
br
ev
id
b
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
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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. Since this is being subtracted from 9, the whole expression is greatest when x = 2.
-R
am
br
ev
The maximum value of 9 − 4( x − 2)2 is 9 and this maximum occurs when x = 2.
s
-C
So the function f( x ) = 16x − 7 − 4x 2 has a maximum point at (2, 9). y
es O
y
ni ve rs
ity
y = 16x – 7 – 4x2
31 2
1 2
w ev
ie
–7
s
-R
x=2
es
-C
am
br
id g
x = 3 21
x
C
U
R
3 x−2 = ± 2 1 or x = 2
e
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
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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
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op
12
y op -R s es
-C
am
br
ev
ie
id g
w
e
C
U
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ev
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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
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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
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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.
x = −4 21 or x = 3
+
–4 1
U
op
So the x-axis intercepts are −4 21 and 3.
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).
-R
–
s es
am
br
ev
ie
The solution is − 4 21 < x < 3.
-C
O
2
ni ve rs
C w ie ev
(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
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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
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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 . 1.
ev
− 3x − 40
s
2
es
am
6 Find the range of values of x for which 2 x
ie
and x 2 − 2 x − 3 ù 0
-R
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
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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
k 2 − 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.
-R
Line and curve
.0
two distinct real roots
=0
two equal real roots (repeated roots) one point of intersection (line is a tangent)
,0
no real roots
two distinct points of intersection
no points of intersection
op
ni
ev
WORKED EXAMPLE 1.20
y
ve
ie
w
rs
ity
Pr
y op C
Nature of roots
s
b 2 − 4 ac
es
-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
es ity
Pr
0 0 −8 −2
ev
-R s es
-C
am
br
ev
ie
id g
w
e
C
U
R
y
ni ve rs
= = = =
ie
w
C
op y
42 − 4 × 1 × (2 − k ) 16 − 8 + 4 k 4k k
s
b2 − 4ac = 0
op
-C
Since the line is a tangent to the curve, the discriminant of the quadratic must be zero, so:
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 .
–
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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. b 2 − 4 ac
C
w
ie
ge
U
ni
op
●
id
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
-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
am
-C
ev
Nature of roots
.0
es
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.
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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.
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Cambridge International AS & A Level Mathematics 9709 Paper 11 Q10 June 2011
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32
[3]
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Chapter 2 Functions
33
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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.
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■ ■ ■ ■ ■
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In this chapter you will learn how to:
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Cambridge International AS & A Level Mathematics: Pure Mathematics 1
am br id
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PREREQUISITE KNOWLEDGE
What you should be able to do
IGCSE / O Level Mathematics
Find an output for a given function.
Pr es s
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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
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1 If f( x ) = 3x − 2 , find f(4).
Find a composite function.
y
IGCSE / O Level Mathematics
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Check your skills
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Where it comes from
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Why do we study functions?
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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
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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?
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●
WEB LINK
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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.
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34
s
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There are many situations in the real world that can be modelled as functions. Some examples are:
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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.
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2.1 Definition of a function
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An alternative name for a function is a mapping.
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op y
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A function is a relation that uniquely associates members of one set with members of another set.
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A function can be either a one-one function or a many-one function.
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The function x ֏ x + 2, where x ∈ ℝ is an example of a one-one function.
TIP
y op C
y = x+2
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f(x)
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x ∈ ℝ means that x belongs to the set of real numbers.
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Chapter 2: Functions
am br id
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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
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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.
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ve rs ity
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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
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f (x)
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x
op
Pr
y
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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
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We can write this function as f : x ֏ x 2 for x ∈ ℝ or f( x ) = x 2 for x ∈ ℝ.
y op
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y2 = x
x
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op y
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If we now consider the graph of y2 = x :
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The set of input values for a function is called the domain of the function.
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When defining a function, it is important to also specify its domain.
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The set of output values for a function is called the range (or codomain) of the function.
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We can see that the input value shown has two output values. This means that this relation is not a function.
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35
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WORKED EXAMPLE 2.1
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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
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c Write down the range of the function f .
a The domain is −4 < x < 5 .
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Answer
Pr es s
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a Write down the domain of the function f .
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b The graph of y = 5 − 2 x is a straight line with gradient −2 and y-intercept 5.
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C op
When x = −4, y = 5 − 2( − 4) = 13
f(x)
Pr
y op
range
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Sketch the graph of the function.
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Find the range of f . Answer
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The function f is defined by f( x ) = ( x − 3)2 + 8 for −1 < x < 9.
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WORKED EXAMPLE 2.2
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c The range is −5 < f( x ) < 13.
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(5, –5)
domain
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36
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(–4, 13)
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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.
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( x − 3)2 + 8
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The minimum value of the expression is 0 + 8 = 8 and this minimum occurs when x = 3.
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So the function f( x ) = ( x − 3)2 + 8 will have a minimum point at the point (3, 8).
s es
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When x = 9, y = (9 − 3)2 + 8 = 44
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When x = −1, y = ( −1 − 3)2 + 8 = 24
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will be of the form
Pr
op y
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f( x ) = ( x − 3)2 + 8 is a positive quadratic function so the graph
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Pr es s
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range
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(9, 44)
(–1, 24)
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(3, 8)
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domain
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The range is 8 < f( x ) < 44.
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Chapter 2: Functions
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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
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y
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y = 2 x 3 − 1 for x ∈ ℝ 10 y= for x ∈ ℝ, x > 0 x
b
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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
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2 a Represent on a graph the function:
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2 9 − x for x ∈ ℝ, −3 < x < 2 x֏ 2 x + 1 for x ∈ ℝ, 2 < x < 4
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b State the nature of the function.
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3 a Represent on a graph the relation:
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2 x + 1 for 0 < x < 2 y= 2 x − 3 for 2 < x < 4
es
s
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b Explain why this relation is not a function.
Pr
y
(1, 8)
b
y (–2, 20)
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4 State the domain and range for the functions represented by these two graphs.
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y (–3, 9)
x
O
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(1, –7)
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(2, 4)
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(5, –8)
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(–1, 4)
y = 2x3 + 3x2 – 12x
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y = 7 + 2x – x2
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Cambridge International AS & A Level Mathematics: Pure Mathematics 1
w
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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 ֏ x 2 + 3 for −2 < x < 5
f( x ) = 3 − 2 x 2 for x < 2
d f( x ) = 7 − 3x 2 for −1 < x < 2
ve rs ity
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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
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f( x ) = ( x − 2)2 + 5
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f( x ) = x 2 − 2 for x ∈ ℝ
7 Find the range for each of these functions.
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c
Pr es s
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6 Find the range for each of these functions. a
for 1 < x < 8
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for −5 < x < 4
for 1 < x < 4
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b f( x ) = 3x 2 − 10 x + 2 for x ∈ ℝ
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f( x ) = x 2 + 6x − 11 for x ∈ ℝ
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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.
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f( x ) = 7 − 8x − x 2 for x ∈ ℝ
b f( x ) = 2 − 6x − 3x 2 for x ∈ ℝ
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s
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a
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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 −3
s
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Find:
g( x ) =
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1 f( x ) = x 2 + 6 for x ∈ ℝ
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62 − 4 × 9 × ( −3 − k ) > 0 144 + 36 k > 0
EXERCISE 2B
R
Rearrange and simplify.
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9x 2 + 6x − 3 = k
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2
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Cambridge International AS & A Level Mathematics: Pure Mathematics 1
5 g( x ) = x 2 − 2 for x ∈ ℝ
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a Find gh( x ).
h( x ) = 2 x + 5 for x ∈ ℝ
g( x ) =
f( x ) = x 2 + 1 for x ∈ ℝ
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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
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g:x֏
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7 g( x ) =
8 f:x֏
3 for x ∈ ℝ, x ≠ 2 x−2
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6
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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.
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9 f( x ) =
10 f : x ֏ x 2 for x ∈ ℝ
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g : x ֏ x + 1 for x ∈ ℝ
b
x ֏ x2 + 1
d
x ֏ x4
e
x ֏ x2 + 2x + 2
s
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x ֏ x 4 + 2x2 + 1
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13 f( x ) = x 2 − 3x for x ∈ ℝ
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g( x ) = 2 x − 5 for x ∈ ℝ
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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
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11 f( x ) = x 2 − 3x for x ∈ ℝ
42
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x ֏ ( x + 1)2
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a
am
Express each of the following as a composite function, using only f and/or g.
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Find the values of k for which the equation gf( x ) = k has real solutions.
-C
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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
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16 f( x ) = x 2 − 2 x + 4 for x ∈ ℝ
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a Find the set of values of x for which f( x ) > 7 .
C
Write down the range of f .
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g( x ) = 2 x + 3 for x ∈ ℝ
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17 f( x ) = x 2 − 5x for x ∈ ℝ
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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
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a Find fg( x ).
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R
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b Find the least value of k for which the function is one-one.
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ve rs ity am br id
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Q( x ) = x + 2 for x ∈ ℝ
1 for x ∈ ℝ, x ≠ 0 x
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C
R( x ) =
w
Pr es s
P( x ) = x 2 − 1 for x ∈ ℝ
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.
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2 for x ∈ ℝ , x ≠ −1 x +1 a Find ff( x ) and state the domain of this function.
18 f( x ) =
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Chapter 2: Functions
S( x ) =
x + 1 − 1 for x ∈ ℝ, x > −1
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ev ie
Functions P, Q, R and S are composed in some way to make a new function, f( x ).
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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.
=
f −1f( x )
es
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x
rs
KEY POINT 2.3
ff −1( x )
43
=x
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Pr
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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
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f –1(x)
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id
It is important to remember that not every function has an inverse.
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KEY POINT 2.4
es
s
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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
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ni ve rs
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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)
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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
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We find the inverse of the function f( x ) = 3x − 1 by following these steps:
C
U
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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
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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?
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C
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R
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x + 2 − 7 for x ∈ ℝ, x > −2
-R
am
b Solve the equation f −1( x ) = f(62) .
s
-C
a Find an expression for f −1( x ) .
es
x+2 −7
Pr
Step 1: Write the function as y =
Step 2: Interchange the x and y variables.
y+2 −7
U
y+2
( x + 7) = y + 2
e id g
w
y = ( x + 7)2 − 2
ie ev es
s
-R
br am
x=
2
f −1( x ) = ( x + 7)2 − 2
-C
x+2 −7
x+7 =
Step 3: Rearrange to make y the subject.
R
y=
op
f( x ) =
ni ve rs
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C
a
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Answer
C
f( x ) =
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
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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
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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
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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
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R
b
y = 5 − ( x − 2)2
ie
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id
am
br
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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
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br
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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.
x − 2.
s
Pr
op
d Can you generalise your results?
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 =
12 12 12 ,y= + 5 and y = − 4. x x x
y
c Repeat part a using the graphs y =
x, y =
es
-C
b Repeat part a using the graphs y =
-R
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
C
op
0 y = 2 x can be transformed to y = 2 x + 2 by a translation of . a
w
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
1 2 3 4 x
On the same diagram, sketch the graphs of y = 2 x and y = 2 x + 2.
e
b
y = f(x)
–4 –3 –2 –1 O –1 –2 –3 –4
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.
s
-C
-R
am
The transformations of the graph of y = f( x ) that you have studied so far can each be categorised as either vertical or horizontal transformations.
es
0 translation a
Pr
y = f( x + a )
vertical stretch, factor a
y = f( − x )
reflection in the y-axis
y = f( ax )
horizontal stretch, factor
1 a
y
y = a f( x )
−a translation 0
op
reflection in the x-axis
ni ve rs
y = −f( x )
-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
R
ev
ie
w
C
y = f( x ) + a
Horizontal transformations
ity
op y
Vertical 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
0 1 a i Translate , then stretch vertically with factor . 2 3
1
T O
1
0 1 , then translate . 2 3 b Investigate for other pairs of vertical transformations.
ni
C op
y
ii Stretch vertically with factor
ev ie
U
R
3 x
2
ve rs ity
op C w
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
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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
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–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 = 2 − g( x )
d
y = 2g( − x ) + 1
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
e
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
ev
y = 2g( x ) + 1
y = g(x)
1
-R
b
-C
c
y = g( x + 2) + 3
am
a
br
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
3
2
2
1
1
–4 –3 –2 –1–1O
1
2
4 x
3
–2
U
–3 –4
ge
–4 –3 –2 –1–1O
w
–2 –3
y 4
c
y
4 x
x
C op
3
3
–2
Pr es s
-R
2
ni
1
2
id
ie
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.
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–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
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
w
ev
x onto the curve y = −2 3 x − 3 + 4
ie
3
id
the curve y =
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
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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
●
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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:
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The domain of f −1 ( x ) is the range of f( x ).
op
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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.
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If f is self-inverse then ff( x ) = x.
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The graph of a self-inverse function has y = x as a line of symmetry.
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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.
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The graph of y = f( − x ) is a reflection of the graph y = f( x ) in the y-axis.
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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.
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Vertical transformations follow the ‘normal’ order of operations, as used in arithmetic
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Horizontal transformations follow the opposite order to the ‘normal’ order of operations, as used in arithmetic.
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−a . 0
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Transformations of functions
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Inverse functions
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A function is a rule that maps each x value to just one y value for a defined set of input values.
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Composite functions
66
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Checklist of learning and understanding
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Chapter 2: Functions
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g : x ֏ 5x − x 2
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Functions f and g are defined for x ∈ ℝ by: f : x ֏ 3x − 1
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1
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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]
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x
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3
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2
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a Sketch the curve, showing the coordinates of any axes crossing points.
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[3]
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Express − x 2 + 6x − 5 in the form a ( x + b )2 + c, where a, b and c are constants.
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i
[3]
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b On the same diagram, sketch the graphs of f and f .
[4]
[3]
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−1
67
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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 .
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4
[3]
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A curve has equation y = x 2 + 6x + 8.
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3
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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 ) .
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The function f : x ֏ − x 2 + 6x − 5 is defined for x > m, where m is a constant.
s
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ii State the smallest possible value of m for which f is one-one.
[1] −1
−1
es
Pr
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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]
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6
[4]
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q9 November 2015
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[4]
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Cambridge International AS & A Level Mathematics 9709 Paper 11 Q8 June 2012
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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]
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iii State the smallest value of p for which f is one-one.
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iii For the case where m = 5, find an expression for f ( x ) and state the domain of f .
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x
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Cambridge International AS & A Level Mathematics: Pure Mathematics 1
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for − 1 < x < 1,
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for 1 < x < 4.
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3x − 2 f( x ) = 4 5 − x
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The diagram shows the function f defined for −1 < x < 4 , where
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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]
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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
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ii State the range of g.
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2
[1]
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[1]
Pr
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iii Find the range of f.
iv Find the expression for f −1( x ) and state the domain of f −1.
[5]
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Cambridge International AS & A Level Mathematics 9709 Paper 11 Q8 June 2013
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Express x 2 − 2 x − 15 in the form ( x + a )2 + b.
y
[2]
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The function f is defined for p < x < q, where p and q are positive constants, by
id g
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f : x ֏ x 2 − 2 x − 15 .
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The range of f is given by c < f( x ) < d, where c and d are constants.
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ii State the smallest possible value of c.
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[3]
s
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The value of k is now given to be 7.
ie
[4]
C
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id
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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
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iii Find an expression for g −1( x ) and state the domain of g −1.
9
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[4]
The function g is defined by g( x ) = 4x 2 − 24x + 11 , for x < 1.
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8
[1]
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[1]
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Chapter 2: Functions
For the case where c = 9 and d = 65, iv find an expression for f −1( x ).
[3]
Pr es s
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[4]
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iii find p and q,
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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]
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iii Find the set of values of x for which f( x ) < 21.
U
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The function g is defined by g : x ֏ 2 x + k for x ∈ ℝ .
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iv Find the value of the constant k for which the equation gf( x ) = 0 has two equal roots.
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id br
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g : x ֏ x 2 − 2.
Find and simplify expressions for fg( x ) and gf( x ).
Pr
[2] [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]
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ii Hence find the value of a for which fg( a ) = gf( a ).
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f : x ֏ 2 x + 1,
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The function h is defined by
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h : x ֏ x 2 − 2, for x < 0 .
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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]
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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]
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Cambridge International AS & A Level Mathematics 9709 Paper 11 Q11 November 2011
es
s
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v Find an expression for f −1( x ).
am
[3] [1]
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ii State the range of f.
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13 Functions f and g are defined by
i
[2]
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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]
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69
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Pr es s
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U
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Chapter 3 Coordinate geometry id
es
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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
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Pr
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■ ■ ■ ■ ■
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In this chapter you will learn how to:
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Chapter 3: Coordinate geometry
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PREREQUISITE KNOWLEDGE
What you should be able to do
IGCSE / O Level Mathematics
Find the midpoint and length of a line segment.
Pr es s
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Find the gradient of a line and state the gradient of a line that is perpendicular to the line.
w ie
3 The equation of a line is 2 y = x − 5. Write down: 3 a the gradient of the line
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c the x-intercept.
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4 a Complete the square for x 2 − 8x − 5. b Solve x 2 − 8x − 5 = 0..
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71
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Why do we study coordinate geometry?
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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.
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Chapter 1
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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.
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IGCSE / O Level Mathematics
1 Find the midpoint and length of the line segment joining ( − 7, 4) and ( − 2, −8).
C op
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C
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op
IGCSE / O Level Mathematics
R
Check your skills
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Where it comes from
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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.
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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
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WEB LINK
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The Geometry of equations and Circles stations on the Underground Mathematics website have many useful resources for studying this topic.
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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.
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circle ellipse parabola hyperbola
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Cambridge International AS & A Level Mathematics: Pure Mathematics 1
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3.1 Length of a line segment and midpoint
TIP
Pr es s
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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
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KEY POINT 3.1
Q (x2, y2)
( x2 − x1 )2 + ( y2 − y1 )2
ni
P (x1, y1)
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WORKED EXAMPLE 3.1
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M
C op
To find the length of PQ: PQ =
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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.
y
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3 The point M , −11 is the midpoint of the line segment joining the points P( −7, 4) and Q( a, b ). 2 Find the value of a and the value of b.
op
Pr
Answer
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b = −26.
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4+b = −11 2 4 + b = −22 b = −26
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Equating the y-coordinates:
and
3 , −11 2 3 , −11 2
−7 + a 3 = 2 2 −7 + a = 3 a = 10
Equating the x-coordinates:
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x + x2 y1 + y2 Using 1 , and midpoint = 2 2 −7 + a 4 + b , = 2 2
Hence, a = 10
Decide which values to use for x1, y1, x2 , y2.
rs
( a, b ) ↑ ↑ ( x2 , y2 )
ie
( −7, 4) ↑ ↑ ( x1, y1 )
ni
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Method 1: Using algebra
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Chapter 3: Coordinate geometry
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Method 2: Using vectors
P (−7, 4)
PM =
-R
81 2 −15
∴ MQ =
b = −11 + ( −15)
and
b = −26.
ev ie
ni U
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WORKED EXAMPLE 3.2
Q (a, b)
y
∴ a = 10
and
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Three of the vertices of a parallelogram, ABCD, are A( −5, −1), B( −1, −4) and C(6, −2).
br
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a Find the midpoint of AC.
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s
−5 + 6 −1 + −2 1 3 = , , − 2 2 2 2
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a Midpoint of AC =
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Answer
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b Find the coordinates of D.
rs
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D (2, 1)
O
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A (−5, −1)
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C (6, −2)
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3 2 = −3 =1
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=−
x
B (−1, −4)
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WORKED EXAMPLE 3.3
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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
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Find the two possible values of a.
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id g
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Decide which values to use for x1, y1, x2 , y2.
es
s
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br am
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C w
1 2 =1 =2 =
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ni
−1 + m 2 −1 + m m −4 + n Equating the y-coordinates: 2 −4 + n n D is the point (2, 1).
Equating the x-coordinates:
y
ve
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Pr
Since ABCD is a parallelogram, the midpoint of BD is the same as the midpoint of AC. −1 + m −4 + n 1 3 = , , − Midpoint of BD = 2 2 2 2
R ev
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b Let the coordinates of D be ( m, n ).
R
( 32 , −11)
C op
+ 8 21
Pr es s
3 2
w
C
op
∴a =
M
ve rs ity
y
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81 2 −15
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ve rs ity
( x2 − x1 )2 + ( y2 − y1 )2 = 17
am br id
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Using PQ =
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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
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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
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Tamar says that this triangle is right angled. 2√7
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EXERCISE 3A
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4√3
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Explain your reasoning. 74
5√3
s
Discuss whether he is correct.
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am
The triangle has sides of length 2 7 cm, 4 3 cm and 5 3 cm.
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br
id
EXPLORE 3.1
w
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w
( a − 8)( a + 15) = 0
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1 Calculate the lengths of the sides of the triangle PQR.
P( −4, 6), Q(6, 1), R(2, 9)
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a
C
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Use your answers to determine whether or not the triangle is right angled.
br
P(1, 6), Q( −2, 1) and R(3, −2).
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2
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b P( −5, 2), Q(9, 3), R( −2, 8)
s
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Show that triangle PQR is a right-angled isosceles triangle and calculate the area of the triangle.
es
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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
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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.
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6 Three of the vertices of a parallelogram, ABCD, are A( −7, 3), B( −3, −11) and C(3, −5).
br
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a Find the midpoint of AC.
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s
Find the length of the diagonals AC and BD.
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b Find the coordinates of D.
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Find the two possible values of a.
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7 The point P ( k, 2 k ) is equidistant from A(8, 11) and B(1, 12). Find the value of k.
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Chapter 3: Coordinate geometry
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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
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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
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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
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a Find the coordinates of the midpoint of AB.
ie
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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
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b Find the length of the line segment AB.
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3.2 Parallel and perpendicular lines
op
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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
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Gradient of PQ =
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KEY POINT 3.2
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P(x1, y1)
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You also learnt the following rules about parallel and perpendicular lines. Parallel lines
gradient = m
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Perpendicular lines
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If a line has gradient m, then every line 1 perpendicular to it has gradient − . m
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am
br
If two lines are parallel, then their gradients are equal.
1 m
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id g
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gradient = −
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Cambridge International AS & A Level Mathematics: Pure Mathematics 1
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We can also write the rule for perpendicular lines as: KEY POINT 3.3
Pr es s
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If the gradients of two perpendicular lines are m1 and m2 , then m1 × m2 = −1.
C
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You need to know how to apply the rules for gradients to solve problems involving parallel and perpendicular lines.
C op
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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
rs
C
U
k − 10 = 0
w
ge
ie
k = 10
ev
or
Solve.
id
∴k = 3
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 2 k + 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 k+2 = 0
-R
Factorise.
ve rs ity
w
C
op
y
(5k − 22)( k + 2) = 0 or
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.
∴ k = 4.4 or
k = −2
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
ie
-R
s
es
Pr
ity
2y − 6 = x + 2y =
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
-C
y−3=
w
C
U
am
br
Using y − y1 =
y2 − y1 ( −2) − 3 1 = =− x2 − x1 6 − ( −4) 2 1 m( x − x1 ) with m = − , x1 = −4 and y1 = 3: 2 1 − ( x + 4) 2 −x − 4 2
ev
Gradient = m =
Decide which values to use for x1, y1, x2 , y2.
ni
(6, −2) ↑ ↑ ( x2 , y2 )
ge
( −4, 3) ↑ ↑ ( x1, y1 )
id
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
ni
C op
y
ve rs ity
x + x2 y1 + y2 , Use midpoint = 1 . 2 2
br
w
es
s
Multiply both sides by 2.
Pr
1 Find the equation of the line with:
ity
op C
ie
-R
am
-C y
EXERCISE 3C 80
Expand brackets and simplify.
ev
id
ge
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).
ve
ie
w
rs
a gradient 2 passing through the point (4, 9)
y op w
ev
id
ie
( 1, 0 ) and (5, 6)
br
a
ge
2 Find the equation of the line passing through each pair of points.
-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
es
3 Find the equation of the line:
Pr
perpendicular to the line y = 2 x − 3, passing through the point (6, 1)
ni ve rs
c
ity
b parallel to the line x + 2 y = 6, passing through the point (4, −6)
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)
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
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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
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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
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U
R
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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
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Midpoint of PQ =
C
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Equation of perpendicular bisector of PQ is:
rs
(1)
y
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( y − 5) = −1( x − 0) y = −x + 5
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1+ 5 6 + 4 Midpoint of QR = , = (3, 5) 2 2 4−6 1 Gradient of QR = =− 5−1 2
br
ev
Gradient of perpendicular bisector of QR = 2
es
s
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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
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br
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w
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C
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Hence, the equation of the circle is ( x − 2)2 + ( y − 3)2 = 10 .
R
ev
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Centre of circle = (2, 3)
y
op y
Solving equations (1) and (2) gives:
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ve rs ity
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w
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am br id
Alternative method:
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op
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Chapter 3: Coordinate geometry
The equation of the circle is ( x − a )2 + ( y − b )2 = r 2.
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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
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( −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
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EXERCISE 3D
br
am
( x + 7)2 + y2 = 18
( x − 5)2 + ( y + 3)2 = 4
x 2 + y2 − 8x + 20 y + 110 = 0
2( x − 3)2 + 2( y + 4)2 = 45 2 x 2 + 2 y2 − 14x − 10 y − 163 = 0
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h
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d f
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x 2 + ( y − 2)2 = 25
2x 2 + 2 y2 = 9
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c
b
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x 2 + y2 = 16
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a
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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
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c
b centre (5, −2), radius 4
ity
a centre (0, 0), radius 8
rs
w
C
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2 Find the equation of each of the following circles.
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4 A diameter of a circle has its end points at A( −6, 8) and B(2, −4).
C
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3 Find the equation of the circle with centre (2, 5) passing through the point (6, 8).
br
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Find the equation of the circle.
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am
5 Sketch the circle ( x − 3)2 + ( y + 2)2 = 9.
s
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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.
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am
br
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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.
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C
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7 The points P(1, −2) and Q(7, 1) lie on the circumference of a circle.
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C
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Cambridge International AS & A Level Mathematics: Pure Mathematics 1
am br id
ev ie
w
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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
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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.
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ni
C op
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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
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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
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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
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no points of intersection
op
two distinct real roots
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.0
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br
ev
In this section you will solve problems involving the intersection of lines and circles.
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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.
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We can also describe an equation that has ‘two equal real roots’ as having ‘one repeated (real) root’.
ve rs ity ge
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Chapter 3: Coordinate geometry
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WORKED EXAMPLE 3.14
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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
ity
−2 + 4 −4 + ( −2) Midpoint of AB = , = (1, −3) 2 2 y − y1 = m( x − x1 )
Use m = −3, x1 = 1 and y1 = −3.
ve
ie
w
rs
C
op
y
b
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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 es
x=± 2
2, y = −3 2 .
Pr
When x = − 2, y = 3 2 and when x =
(
)
ity
P and Q are the points − 2, 3 2 and
(
)
2, −3 2 , respectively.
y op -R s es
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am
br
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C
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ev
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w
ni ve rs
op y
-C
10 x 2 = 20
C
Substitute −3x for y.
-R
am
c
ev
id
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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.
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ve rs ity
ev ie
w
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am br id
WORKED EXAMPLE 3.15
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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
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x 2 + y 2 − 8x + 6 y + 7 = 0
U
w
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EXERCISE 3E
ie
R
ni
C op
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The equation has one repeated root, hence y = x − 13 is a tangent.
br
ev
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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
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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
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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
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b Find the equation of the perpendicular bisector of AB.
br
ev
id
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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.
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id g
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a Find the equation of each circle.
C
e
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op
y
●
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b Prove that the line 4x + 3 y = 88 is a common tangent to these circles.
es
s
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[Inspired by Can we find the two circles that satisfy these three conditions? on the Underground Mathematics website.]
am
R
ev
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PS
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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
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ev ie
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C
op
P (x1, y1)
●
ve rs ity
M
y
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Q (x2, y2)
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ev ie
am br id
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Checklist of learning and understanding
C
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op
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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
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C
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ni ve rs
C
ity
Pr
op y
es
s
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-R
am
br
ev
id
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C
U
R
ni
ev
ve
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w
rs
C
ity
op
Pr
y
es
s
-C
-R
am
●
Copyright Material - Review Only - Not for Redistribution
ve rs ity
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C
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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
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END-OF-CHAPTER REVIEW EXERCISE 3
[4]
y = 6x + k y = 7√x
w
C
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B
y C op
O
x
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U
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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
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am
i
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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
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The curve and the line intersect at the points A and B.
ev
id
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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]
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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]
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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
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Find the coordinates of P. You must show all your working.
ie
id g
w
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The point P is the reflection of the point ( −7, 5) in the line 5x − 3 y = 18.
-C
7
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Show that the mid-point of PC lies on the line y = x.
R
ev
ii
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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
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C
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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.
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C op
Give your answers in exact form.
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10
[4]
s
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B (15, 22)
op
es Pr
y
C
93 x
A (3, −2)
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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
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id
The perpendicular bisector of AB meets BC at D.
[2] [4]
w
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R
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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
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y
op -R s
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br
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Cambridge International AS & A Level Mathematics 9709 Paper 11 Q7 November 2013
es
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C
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11 The point A has coordinates ( −1, 6) and the point B has coordinates (7, 2).
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ve rs ity
am br id
ev ie
w
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C
U
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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
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b Use your answer to part a to find the value of p.
y
[4] [1] [4]
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c Find the equation of the circle. 14
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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
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am
br
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A (13, 17)
C (13, 4)
B (3, 2)
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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
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i
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ABCD is a trapezium with AB parallel to DC and angle BAD = 90°.
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a Calculate the coordinates of D.
[2]
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C
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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
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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]
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ev
br
id g
w
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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
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16 A is the point ( −3, 6) and B is the point (9, −10).
[4]
s
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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]
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Cross-topic review exercise 1
w
4 17 + 18 = 2 . x4 x
[4]
y
ve rs ity ni
4 x
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U
C op
op C w ev ie
R
O
–4
y
Pr es s
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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 Pr
O
–1
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
–2
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
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R
ni
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4
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ie
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rs
C
[2]
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op
y
b y = −2f( x )
3
[2]
s
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a y = f( x ) + 5
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[4]
ve rs ity
am br id
ev ie
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C
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Cambridge International AS & A Level Mathematics: Pure Mathematics 1
a Find the set of values of x for which f( x ) < x.
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The function f is such that f( x ) = x 2 − 5x + 5 for x ∈ ℝ.
7
[3]
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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.
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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.
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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]
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A curve has equation xy = 12 + x and a line has equation y = kx − 9, where k is a constant.
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11
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a In the case where k = 2, find the coordinates of the points of intersection of the curve and the line.
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[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.
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12
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a Find the smallest value of k for which the composite function gf can be formed.
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4 − 2 for x . 0 , x 4 g( x ) = for x > 0. 5x + 2
[3]
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i Find and simplify an expression for fg( x ) and state the range of fg.
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Cambridge International AS & A Level Mathematics 9709 Paper 11 Q8 November 2016
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The equation x 2 + bx + c = 0 has roots −2 and 7. a Find the value of b and the value of c.
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b Using these values of b and c, find:
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ii the set of values of x for which x 2 + bx + c , 10.
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i the coordinates of the vertex of the curve y = x 2 + bx + c
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ii Find an expression for g −1( x ) and find the domain of g −1.
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[4]
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The functions f and g are defined by f( x ) =
[3]
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b Solve the inequality gf( x ) . 45. 13
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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
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[3]
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Find the two possible values of m.
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b The line y = mx − 11 is a tangent to the curve y = f( x ).
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[3] [3]
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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).
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15
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Cross-topic review exercise 1
[4]
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a Find the equation of the line L2 .
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16
A curve has equation y = 12 x − x 2.
[3]
b State the maximum value of 12 x − x 2.
[1]
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The function g is defined as g: x ֏ 12 x − x 2, for x ù 6. −1
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c State the domain and range of g −1.
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a Express 3x 2 + 12 x − 1 in the form a ( x + b )2 + c, where a, b and c are constants.
[4]
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c Find the set of values of k for which 3x 2 + 12 x − 1 = kx − 4 has no real solutions.
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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.
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[3]
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97
2
b Find the least possible value of k for which g has an inverse.
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A circle has centre (8, 3) and passes through the point P (13, 5) .
[4]
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c For the value of k found in part b, find g −1( x ).
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[3] [2]
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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.
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18
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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.
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b Find the coordinates of the point of intersection of lines L1 and L2 .
[4]
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a Find the equation of the circle.
[3]
b Find f −1( x ) and g −1( x ).
[3]
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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.
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a Express 2 − 3x − x 2 in the form a − ( x + b )2, where a and b are constants.
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b Write down the coordinates of the maximum point on the curve.
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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.
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A curve has equation y = 2 − 3x − x 2.
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c Show that the equation f −1( x ) = g −1( x ) has no real roots.
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d For each value of m in part c, find the coordinates of the point where the line touches the curve.
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20
[5]
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Give your answer in the form ax + by = c.
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b Find the equation of the tangent to the circle at the point P.
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Cambridge International AS & A Level Mathematics: Pure Mathematics 1
a Find the coordinates of the centre of the circle.
[2]
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b Find the radius of the circle.
[2]
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A circle, C , has equation x 2 + y2 − 16x − 36 = 0.
22
[3]
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The function f is such that f( x ) = 3x − 2 for x > 0.
23
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[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
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c Find the coordinates of the points where the circle meets the x-axis.
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The function g is such that g( x ) = 2 x 2 − 8 for x < k, where k is a constant.
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a Find the greatest value of k for which the composite function fg can be formed.
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b For the case where k = −3:
[2]
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i find the range of fg
[4]
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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.
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24
[3]
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Find:
i the coordinates of the points A and B
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[3]
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98
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a In the case where k = 14, the line intersects the curve at the points A and B.
[4]
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ii the equation of the perpendicular bisector of the line AB.
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b Find the values of k for which the line is a tangent to the curve.
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[4]
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Chapter 4 Circular measure
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99
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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.
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In this section you will learn how to:
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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.
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5 cm
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40°
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x cm
6 cm
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3
8 cm
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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.
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y cm
x°
12 cm
1 Area of triangle = ab sin C 2
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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.
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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.
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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.
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2
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IGCSE / O Level Mathematics
Another measure for angles
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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.
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IGCSE / O Level Mathematics
100
Check your skills
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Where it comes from
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PREREQUISITE KNOWLEDGE
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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
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angular measure and we use it extensively in mathematics because it can simplify many formulae and calculations.
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1 rad
1 radian is sometimes written as 1 rad, but often no symbol at all is used for angles measured in radians.
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B
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In the diagram, the magnitude of angle AOB is 1 radian.
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4.1 Radians
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Chapter 4: Circular measure
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KEY POINT 4.1
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KEY POINT 4.2 2 π radians = 360° π radians = 180°
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It follows that the circumference (an arc of length 2 πr) subtends an angle of 2 π radians at the centre, therefore:
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An arc equal in length to the radius of a circle subtends an angle of 1 radian at the centre.
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When an angle is written in terms of π, we usually omit the word radian (or rad).
s
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Hence, π = 180°.
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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.
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π . 180
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To change from degrees to radians, multiply by
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KEY POINT 4.3
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Since 180° = π, then 90° =
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Converting from radians to degrees
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180 ≈ 57°.) π
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(It is useful to remember that 1 radian = 1 ×
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180 . π
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To change from radians to degrees, multiply by
Pr
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KEY POINT 4.4
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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°,
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101
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Pr es s
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Method 2:
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Method 1:
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b
π radians 6
5π 5 π 180 ° × radians = 9 π 9
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π radians = 180°
5π radians = 100° 9
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π radians = 20° 9
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5π radians = 100° 9
102
π radians 6
30° =
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180 ° = π radians 6 6 30° =
π radians 30° = 30 × 180
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180° = π radians
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Method 1:
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a Change 30° to radians, giving your answer in terms of π. 5π radians to degrees. b Change 9 Answer
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WORKED EXAMPLE 4.1
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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.
180°
270°
π 4
π 3
π 2
π
3π 2
360° 2π
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90°
C
π 6
60°
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0
45°
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Radians
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Degrees
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There are:
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In Worked example 4.1, we found that 30° =
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EXERCISE 4A
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We can quickly find other angles, such as 120°, using these known angles.
300°
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o 600°
π 12 9π 20
3π 10
h
7π 12
i
l
4π 15
m
5π 4
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4π 3
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9π 2
o
9π 8
C
d
7π 3
s
7π 5
π 6
g
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k
br
4π 9
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f
j
n 35°
c
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2 Change these angles to degrees. π π b a 2 3
m 9°
225°
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540°
i
e 5°
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h 210°
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k 65°
d 50°
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g 135°
25°
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150°
c
Pr
b 40°
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a 20°
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1 Change these angles to radians, giving your answers in terms of π.
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Chapter 4: Circular measure
47°
c
d 200°
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b 32°
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a 28°
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3 Write each of these angles in radians, correct to 3 significant figures.
e
320°
e
0.79 rad
b 0.8 rad
c 1.34 rad
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a 1.2 rad
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4 Write each of these angles in degrees, correct to 1 decimal place. d 1.52 rad
Degrees
0
Radians
0
90
135 180
225 270 315 360
π 30
60
90
120 150 180
2π 210 240 270 300 330 360
π
2π
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0
45
U
cos(0.9)
π tan 5
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π sin 3
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br
π d cos 2
b tan(1.5)
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a sin(0.7)
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6 Use your calculator to find:
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b
0
Radians
Pr es s
Degrees
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5 Copy and complete the tables, giving your answers in terms of π.
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7 Calculate the length of QR.
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103
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Q
5 cm
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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.
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6 cm
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8
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1 rad
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P
PS
You do not need to change the angle to degrees. You should set the angle mode on your calculator to radians.
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TIP
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EXPLORE 4.1
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Cambridge International AS & A Level Mathematics: Pure Mathematics 1
chord
arc
sector
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Explain what is meant by:
minor arc and major arc minor sector and major sector minor segment and major segment.
●
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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:
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length of chord AB area of minor sector AOB area of minor segment AOB.
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length of minor arc AB perimeter of minor sector AOB perimeter of minor segment AOB
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B
θ°
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segment
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Discuss and explain, with the aid of diagrams, the meaning of each of these words.
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What would the answers be if the angle θ was measured in radians instead?
WEB LINK
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DID YOU KNOW?
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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.
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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 .
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KEY POINT 4.5
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π 3 = 5 π cm = 15 ×
id g
Arc length = rθ
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Answer
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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
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WORKED EXAMPLE 4.2
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Arc length = rθ
rθ
B
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4.2 Length of an arc
Try the Where are you? resource on the Underground Mathematics website.
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θ O
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Chapter 4: Circular measure
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WORKED EXAMPLE 4.3
A sector has an angle of 1.5 radians and an arc length of 12 cm.
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WORKED EXAMPLE 4.4
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Arc length = rθ 12 = r × 1.5 r = 8 cm
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a the length of arc CD b the length of AD
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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
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b radius 3.5 cm and angle 0.65 radians.
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a radius 10 cm and angle 1.3 radians
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3 Find, in radians, the angle of a sector of:
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4 The High Roller Ferris wheel in the USA has a diameter of 158.5 metres.
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b radius 12 cm and arc length 9.6 cm.
a radius 10 cm and arc length 5 cm
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π radians. 16
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Calculate the distance travelled by a capsule as the wheel rotates through
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2 Find the arc length of a sector of:
3π 7 7π d radius 24 cm and angle . 6
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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
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Perimeter = DC + CA + AD = 7.2 + 8 + 1.945…
= 1.95 cm (to 3 significant figures)
EXERCISE 4B
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D
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Arc length = rθ = 8 × 0.9 = 7.2 cm
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8 cm 0.9 rad
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Answer
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8 cm
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c the perimeter of the shaded region.
a
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Find:
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Triangle ABC is isosceles with AC = CB = 8 cm. CD is an arc of a circle, centre B, and angle ABC = 0.9 radians.
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Answer
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Find the radius of the sector.
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Cambridge International AS & A Level Mathematics: Pure Mathematics 1
w 2.1 rad
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6 cm
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7 cm O
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ABCD is a rectangle with AB = 5 cm and BC = 24 cm.
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O is the midpoint of BC .
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a the length of AO
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A
θ 10 cm
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x− y . Show that P = 4 xy + π( x + y ) + 2( x − y ) sin −1 x + y
x
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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.
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[This question is based upon Belt on the Underground Mathematics website.]
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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 .
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the perimeter of the shaded region.
br
c
A
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OAED is a sector of a circle, centre O. Find:
b angle AOD, in radians
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7 cm
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the perimeter of the shaded segment.
B
2 rad
s
b the length of chord AB
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A
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a the length of arc AB
PS
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AB is a chord and angle AOB = 2 radians. Find:
8
6 cm
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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.
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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:
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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
C
)=5
β
2 cm
5
es
s
102 + 102
10
y C
U
Shaded area = area of segment PQR − area of segment PQS
s es
am
-R
br
ev
ie
id g
w
e
1 1 1 1 = × 52 × β − × 52 × sin β − × 102 × 2θ − × 102 × sin 2θ 2 2 2 2 = 21.968 − 7.3296 = 14.6 cm 2 (to 3 significant figures)
5√2
θ
θ O
op
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
-R
br
OQ = 10 cm
ev
id
Answer
w
ge
C
U
R
ni
ev
ve
ie
w
rs
Find the shaded area.
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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.
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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.
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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
●
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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
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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.
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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.
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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
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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:
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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
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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
√5
C w ie -R
ev
2 . 5
2
(
s es Pr
5 + 2
()
5 − 2
y
(
5 +2
) =5−2
)(
5 −2
)
5 +2 5 −4 =1
s
3 5 − 6 = 5
)(
)
op
ity
5 − 2
C
(
es
5
3
Multiply the numerator and denominator by 5 − 2.
w
id g -C
am
br
=
)
Multiply the numerator and denominator by 15.
-R
(
U
5
3 5 + 2
e
R
=
Simplify.
ni ve rs
-C
2 1 − 5 1 − tan2 θ = cos θ + sin θ 5 2 + 3 3 1 5 = 5 + 2 3
ie
br
am
w
C
op y
b
θ°
ev
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°
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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°
1 3
3 2
1 2
1
C
1 2
w
ve
ni 1 2
op
3 2
s
es
Pr
y
ev
-R s es
-C
am
br
ev
ie
id g
w
e
C
U
R
Rationalise the denominator.
op
1 1 × 2 2 1 = 2 2 1 × 2 = 2 2 × 2 2 = 4
sin 30° cos 45° =
ni ve rs
a
c
ity
op y C
Answer
ie
w
b
sin 30° cos 45°
π π sin 4 6 2 π 2 π cos + sin 3 3 2 cos
π sin 3 2
1 can be 2
written as
2 . 2
The value
ev
-C
Find the exact value of:
The value
1 can be 3 3 . written as 3
3
-R
am
WORKED EXAMPLE 5.2
a
TIP
ie
ity
tan θ
U
π 3
ge
θ = 60° =
id
π 4
1 2
cos θ
rs
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
C op
y
Rationalise the denominator.
-R
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
-C
e
1 1 + tan θ sin θ
-R
cos θ
br
a
ev
id
3 Given that sin θ =
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
EXERCISE 5A
am
br
ev
id
ie
w
ge
U
R
ni
ev ie
2
ve rs ity
1 2 2 = 2 2 2 = 2 =
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 3 1 cos + sin + 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.
sin θ
4th quadrant
y − = =− r +
y = x
x = r
y − = =+ x −
x = r
y = x
y
C
U
op 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.)
-C
C w ie ev
R
x − = =− r +
sin cos tan
y
y = r
op
s
3rd quadrant
y + = =+ x +
es
y = r
Pr
ity
2nd quadrant
tan θ
x + = =+ r +
ni ve rs
y + = =+ r +
-C
1st quadrant
cos θ
-R
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
ni f
sin
tan 125°
d
cos( −245°)
g
7π cos − 10
h
tan
sin 225°
d
tan( −300°)
π tan − 6
h
cos
d
cos 245°
y
4π 5
c
9π 8
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
cos
w
f
c
ie
sin
tan 330°
ev
e
b
7π 3
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
U
R
ni
C op
Without using a calculator, copy and complete the table.
……
−1
sin θ
……
1 2
1 cos θ
−2
ie ev
-R
1 2
s
……
es
id
br am
1 3 −
− 2
op
Pr
y
-C
θ = 210°
w
θ = ……
ge
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
ev ie
w
ge
To sketch the graph of a trigonometric function, such as y = 2 cos( x + 90) + 1 for 0° < x < 360°, we can build up the transformation in steps.
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
y
op
Step 2: Sketch the graph of y = cos( x + 90).
y = cos x
y
-C
Amplitude = 1
y 2 1
-R
Step 1: Start with a sketch of y = cos x. Period = 360°
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
br
ev
4 a Sketch the graph of each of these functions for 0 < x < 2 π .
π π iii y = sin 2 x + y = cos x − 4 2 b Write down the coordinates of the turning points for your graph for part a iii.
am
y = 2 sin x
es
s
-R
ii
-C
i
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
es
s
-R
br
am
Pr
b the range of f .
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
-C
7π Given that f(0) = 3 and that f = 2, find: 6 a the value of a and the value of b
ev
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 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 = 7 u6 × 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
WORKED EXAMPLE 7.7
ax + b passes through the point (12, 4) and has gradient
ni
The curve y =
1 at this point. 4
ie
(1)
and
ity
op
y
ni
C
U
ev
id
ie
w
ge
Substitute x = 12 and
(2)
-R
am
br
R
ev
=
op
1
1 −2 u × a 2 a dy = dx 2 ax + b a 1 = 4 2 12 a + b 2 a = 12 a + b
Use the chain rule.
rs
dy dy du = × dx du dx
ve
C w ie
dy 1 − 2 = u du 2
Pr
y
1
du =a dx
200
y = u2
so
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 .
s
1
Let u = ax + b
1
Write
es
-C
y = ( ax +
1 b) 2
-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
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
-C y op
ve rs ity
C w
g
2(4 − 7 x )4
k
( x 2 + 4 x )3
g
br
5 − 2x
2x + 3
c
f
2 3x + 1
8 3 − 2x 8 x 2 + 2 x
2 x2 − 1 1 2 x − 5
w
b
ie
3
e
id
x−5
a
g
ev
ge
U
R
(3 − 4x )5
c
ni
ev ie
2 Differentiate with respect to x: 1 3 a b x+2 x−5 4 3 f e (3x + 1)6 2(3x + 1)5 3 Differentiate with respect to x:
9
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 1 h (3x − 1)7 5 5 5 l x2 − x d
d h
16 x2 + 2 7 (2 x 2 − 5 x )7
x 3 − 5x 6 h 3 2 − 3x
d
-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
id g
)
(
)2
(
)2
y op
)2
3 4− 9 dy 1 =− =− dx 2 9 2 Copyright Material - Review Only - Not for Redistribution
s
-C
When x = 9,
= −3
es
am
br
3 4− 4 dy =− When x = 4, dx 2 4
(
3 4− x 1 −1 2 −2 x = − 2 x
C
e
(
2
w
dy = 3 4− x dx
ie
U
)3
ev
(
y = 4− x
-R
Answer a
ity
b Find the area of triangle PQR.
R
ev
ie
w
C
a Find the coordinates of R.
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1 3
am br id
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Normal at P: passes through the point (4, 8) and gradient =
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ni
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y
Chapter 7: Differentiation
1 ( x − 4) 3 3 y = x + 20
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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
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Normal at Q: passes through the point (9, 1) and gradient = 2
U
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When x = 14.2, y = 2(14.2) − 17 = 11.4
w 10.4
Q(9, 1) 5.2
5
203
x
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O
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7
y op
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P(4, 8)
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3.4
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R(14.2, 11.4)
10.2
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br
b
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Hence, R is the point (14.2, 11.4).
ve
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w
rs
Area of triangle PQR = area of rectangle − sum of areas of outside triangles
y
op
C
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w
= 44.2 units2
br
-C
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am
EXERCISE 7C
ev
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ni
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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)
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ni ve rs
c
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y = x 2 − 3x + 2 at the point (3, 2)
Pr
a
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d
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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
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2 Find the equation of the normal to each curve at the given point.
-C
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1 Find the equation of the tangent to each curve at the given point.
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ve rs ity
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ni
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Cambridge International AS & A Level Mathematics: Pure Mathematics 1
am br id
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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.
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b Find the equation of the normal to the curve at the point A.
Pr es s
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4 The equation of a curve is y = 5 − 3x − 2 x 2 .
C op
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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.
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b Find the coordinates of the point at which the normal meets the curve again.
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C
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a Show that the equation of the normal to the curve at the point ( −2, 3) is x + 5 y = 13.
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6 The tangents to the curve y = 5 − 3x − x 2 at the points ( −1, 7) and ( −4, 1) meet at the point Q.
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Find the coordinates of Q.
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7 The normal to the curve y = 4 − 2 x at the point P (16, −4) meets the x-axis at the point Q.
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b Find the coordinates of Q.
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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).
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op
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6 at the point (3, 6) meets the x-axis at P and x−2
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9 The normal to the curve y = the y-axis at Q.
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204
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a Find the equation of the normal PQ.
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Find the midpoint of PQ.
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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
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a Find the coordinates of R.
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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
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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.
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12 A curve has equation y = 3x +
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a Find the coordinates of the points C and D. Give your answer in exact form.
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s es
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b Find the equation of the perpendicular bisector of CD.
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R
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C
op y
Find the coordinates of R.
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ve rs ity
C
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ni
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Chapter 7: Differentiation
am br id
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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
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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.
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C
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15 The curve y =
y
C op
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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.
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a Find the value of k.
Try the Tangent or normal resource on the Underground Mathematics website.
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b Find the coordinates of the point where the normal meets the curve again.
br
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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
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2 5 , find d y . 3 (2 x − 3) dx 2
Answer
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Use the chain rule.
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Use the chain rule.
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d2 y = 120(2 x − 3)−5 × 2 dx 2 240 = (2 x − 3)5
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dy = −15(2 x − 3)−4 × 2 dx = −30(2 x − 3)−4
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y = 5(2 x − 3)−3
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WORKED EXAMPLE 7.10 Given that y =
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f ′( x ) = 3x 2 + 10 x − 3
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dy = 3x 2 + 10 x − 3 dx d2 y = 6x + 10 dx 2
f( x ) = x 3 + 5x 2 − 3x + 2
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So for y = x 3 + 5x 2 − 3x + 2
ev
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If we differentiate y with respect to x we obtain
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WEB LINK
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205
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A curve has equation y = x 3 + 3x 2 − 9x + 2.
y = x 3 + 3x 2 − 9x + 2
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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
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∴ −3 , x , − 1
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(2)
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x , −1
Combining (1) and (2) on a number line:
−4
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2 d 2 y dy . ≠ dx 2 dx
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EXERCISE 7D
d
y = (2 x − 3)4
g
y=
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y=
f
h
y = 2 x 2 (5 − 3x + x 2 )
y
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2x − 5 x2
4x − 9
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x −3
)
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f( x ) = x 2
2x − 3 x x2 15 f( x ) = 3 2x + 1 f( x ) =
f
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c
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f( x ) = 1 − 3x
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d
6 x2 2 y= 3x + 1 5x − 4 y= x
y=2−
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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
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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,
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d2 y dy = 6 x + 6 and = (3x 2 + 6 x − 9)2 . 2 dx dx
br
b
1
Pr
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6x + 6 , 0
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+
−
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(1)
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d2 y = 6x + 6 dx 2 d2 y , 0 when: dx 2
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dy , 0 when: dx
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dy = 3x 2 + 6x − 9 dx
y
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Answer a
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2
d 2 y dy ≠ . dx 2 dx
y
b Show that
dy d2 y and are negative. dx dx 2
Pr es s
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a Find the range of values of x for which
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WORKED EXAMPLE 7.11
C
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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
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a
C
dy d2 y and . dx dx 2
w
3 Given that y = 4x − (2 x − 1)4 , find
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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
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Pr
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s
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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
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5
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d2 y dx 2
2
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dy dx
1
C op
0
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x
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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
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C
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Pr es s
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5 Given that f ′( x ) =
y op C w y
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br
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C
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C
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Pr
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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.
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U
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ni
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ve
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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
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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
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br
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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
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ni ve rs
C
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Pr
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208
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●
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Tangents and normals
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ve rs ity
Scalar multiple rule:
C
●
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ev ie
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dy represents the gradient of the curve y = f( x ). dx
Pr es s
●
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Checklist of learning and understanding
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ve rs ity ge
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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
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b
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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]
s
[3]
es
Pr
Find the length of PQ, correct to 3 significant figures.
[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.
ity
9
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op
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C
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Find the equation of the normal to the curve at the point A.
[5] [2]
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b
[6]
y
ve
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[4] [1]
w
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Find the equation of the normal PQ.
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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
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[3]
4
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Pr es s
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3
[3]
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Differentiate
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2
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1
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END-OF-CHAPTER REVIEW EXERCISE 7
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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]
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i
[1]
op
C w ev
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s
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br am -C
[7]
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q8 June 2016
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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
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C
op y
b
Pr
-C
The normal meets the curve again at the point Q.
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209
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y C
C
y=2 −
A
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Pr es s
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13
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Cambridge International AS & A Level Mathematics: Pure Mathematics 1
x
y
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B
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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
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br
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The diagram shows part of the curve y = 2 −
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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
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[2] [4]
y
op
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1
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y = (6x + 2) 3
B
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E
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A(1, 2)
C x
1
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C
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Cambridge International AS & A Level Mathematics 9709 Paper 11 Q11 November 2012
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C
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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]
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Cambridge International AS & A Level Mathematics 9709 Paper 11 Q10 November 2010 15
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[6] [2]
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q7 June 2010
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210
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18 2x + 3
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op
y
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C
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Pr es s
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ni
C op
y
ve rs ity
op C w ev ie
Chapter 8 Further differentiation
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211
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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
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ni ve rs
C
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Pr
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■ ■
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In this chapter you will learn how to:
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Cambridge International AS & A Level Mathematics: Pure Mathematics 1
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Chapter 1
Solve quadratic inequalities.
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What you should be able to do
Check your skills
y
Pr es s
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Where it comes from
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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
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C w ev ie
a
dy for the following. dx y = (2x − 1)5
b
y=
3 Find
3 (1 − 3x )2
s
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br
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x2 − 2x − 3 . 0
a
b 6 + x − x2 . 0
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Chapter 7
1 Solve:
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Why do we study differentiation?
FAST FORWARD
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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.
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212
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op
Pr
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In Chapter 7, you learnt how to differentiate functions and how to use differentiation to find gradients, tangents and normals.
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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
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• • • •
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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.
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WEB LINK Explore the Calculus meets functions station on the Underground Mathematics website.
ve rs ity
ev ie
am br id
EXPLORE 8.1
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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
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‘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 ).
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ni
Consider the graph of y = g( x ).
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Section B: Decreasing functions
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‘As the value of x increases the value of y …’
O
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These types of functions are called decreasing functions.
Pr
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8.1 Increasing and decreasing functions
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2 Sketch other graphs that satisfy these statements.
-C
Try the Choose your families resource on the Underground Mathematics website.
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‘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.
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213
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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.
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C
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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.
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(a, b) O
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-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
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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
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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 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
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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
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w
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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
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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 − 2 x 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
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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
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b
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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
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8.2 Stationary points
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op
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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
Pr es s
-C
dy 0 . 2x + 1
ev ie
10 A function f is defined as f( x ) =
-R
am br id
w
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C
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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
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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]
283
y op y op -R s es
-C
am
br
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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
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ev
ve
ie
w
rs
C
ity
op
Pr
y
es
s
-C
-R
am
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 1 height h, is given by V = πr 2 h.] 3 Cambridge International AS & A Level Mathematics 9709 Paper 11 Q11 November 2015
Copyright Material - Review Only - Not for Redistribution
ve rs ity
C
U
ni
op
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Cambridge International AS & A Level Mathematics: Pure Mathematics 1
am br id
ev ie
w
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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
b −8,
c ( − 10, 0), (8, 6)
7 , (2, 1) 2 d ( − 2, − 7), (1, 2)
e (2, − 2), (10, 2)
f
g (2, 4)
h ( − 3, 1), (9, 7)
a ( − 3, 9), (2, 4)
2 3 7 h x − + 2 4
ni ve rs
U
e
2
a 9 and 17
3
2 21
cm and
5 25
cm
3 21 cm and 9 cm
s
-R
4
es
br
am
2
7 9 d 2 x + − 4 8
( −3,
( ) ), (4, −9)
( − 1, − 3), (2, 1)
j
( − 5, − 24), (5, 1)
l
(4, − 6), (12, 10)
n ( − 1, 3), (3, 1)
o (6, − 2), (18, − 1)
b 3( x − 2)2 − 13
id g
2
5 33 c 2 x + − 4 8
m
−4 13
y
2 7 45 g x + − 2 4
k ( −6, −2),
( x − 2)2 − 12
8, 1 21
op
f
(2, 2), (10, − 2)
C
e ( x + 2)2 + 4
i
w
2 15 225 d x + − 2 4
ie
2 3 9 c x − − 2 4
ity
b ( x + 4)2 − 16
ev
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
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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)
w
ev ie
y
a
ev -R s es
b ∩ -shaped curve, maximum point: 1 , 1 1 , axes crossing points: 4 8
Pr 7
Proof
8
A: y = ( x − 4)2 + 2 or x 2 − 8x + 18
ve
y
op
a
ity
ni ve rs
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
(
y
op
y = 3x 2 − 6x − 24
,
−2181
11
y = 5 + 3x −
),
12
Proof
w
−1 43
ie
(
10
1 2 x 2
ev
id g
e
( − 7, 0), (2, 0), (0, − 14)
br
B
C
U
( −2 21 , −20 14 ), axes crossing points:
)
es
s
-R
axes crossing points: ( −5, 0), 1 21 , 0 , (0, −15)
am
y = x 2 − 6x + 13
b Student’s own answers
b ∪-shaped curve, minimum point:
-C
A
s es Pr
op y
a ∪-shaped curve, minimum point: (3, − 1), axes crossing points: (2, 0), (4, 0), (0, 8)
c ∪-shaped curve, minimum point:
1 1 ( x − 2)2 or 6 + 2 x − x 2 2 2
C
9
-R
a = 2, b = − 40, c = 128
2
ie
9 1 − 2 x − 8 4
− 1 , 0 , (1, 0), (0, 1) 2
ni
5
C w ie ev
6
C: y = 8 −
-C
a = 2, b = − 9, c = 7
Exercise 1F
R
−4 14 when x = 3 21
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)
4
1
5
w −1, 2
am
c 4 10
( −2 14 , −6 81 ) , minimum
ie
a
( 2 21 , 13 14 ) , maximum
ev
e
b
9 49 a 2 x + − 4 8
U
R
c
l
ge
a
3 ± 2 4, 6 14 1 1 ,6 9 4 1 9 ,1 4 16 x−6 x +8 = 0
id
k
±2
br
ie
w
C
i
ity
op
y
g ± 3
53 5 − x− 4 2
b
rs
-C
e ±1
am
c ± 5, ±1
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
br
a ±2, ±3
id
Exercise 1E
2
a
2
U
b
a 2( x − 2)2 − 3
w
a 2, 8
ve rs ity
14
3
ni
y = −2x − 3
( − 3, 0), (4, 0), (0, 12)
b x=2
4
ge
w ev ie
13
3
Pr es s
-C
y
op
(2, 3)
C
12
2
2
1 b , 0 2
2 53 7x + y = 0
1
-R
am br id
b 5 2
a ( − 2, 1) and (3, − 1)
11
1 , 12 1 , axes crossing points: 4 2
C op
6
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7 cm and 11cm
10
R
d ∩ -shaped curve, maximum point:
5
9
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
1 2 Proof
13
Proof
1 , 0 2
2
a 3x −
e id g
3 4
op
5 25 1 b − ,x,2 − 3 2 4 3 x = ±2, x = ± 2 x , − 9 − 2 3 or x . − 9 + 2 3 k , 1 or k . 2
es
s
-R
5
2
y
1
C
57 8 1 d k, 2 25 f k, 16 b k,
am
3 2
k , − 4 3 or k . 4 3
s
k = − 10 or k = 14
f
7
w
8 9
br
c k,2 e k.
es
Pr 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 5
2
ev
-C
e −6 < x < 5
-R
am br id
c 4 − 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
w
5
a g is one-one for x > 3, since vertex = ( 2, 2 )
y 7
a f( x ) > − 9
w
x + 32 2
a k=3
f −1( x ) = 3 + 9 − x
10
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
11
a f −1( x ) =
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
b f −1( x ) = − 3 +
b Proof
2( x + 1) a ff ( x ) = for x ∈ R, x ≠ − 3 x+3 b Proof c −2 or 1
-C
x−2 2
op
a −3
b fg( x ) > −6 14
b QP( x ), domain is x ∈ R, range is f( x ) ∈ R, f( x ) > 1
b x > −3
C
6
s es Pr
ni ve rs
a 4x 2 + 2 x − 6
y
a f −1( x ) = − 1 + 3 x + 4
b x 3
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
6
y = x 2 − 6x + 8
7
a = 2, b = − 3, c = 1
ev a
x
3 2
ity
1 –4 –3 –2 –1 O –1
4
–4
3
d y = 3x 2 − 2 x − 5
y
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
c
4 x
U 3
e
2
a y = − 5x 2
op
2
br
–4
–3
id g
–3
1
1
–2
ni ve rs
w ie
–2
1
ity
C
4
–4 –3 –2 –1 O –1
y
op s
y
1
4 x
2
–4 –3 –2 –1 O –1
es
–4
2
3
3
Pr
-C op y
–3
3
2
C x
–2
c
4 x
4
ev
3
3
y
-R
2
w
b
ie
ge 1
2
–4
id
br
am
–4 –3 –2 –1 O –1
–3
U
R
4
1
–2
ve
ev
y
ni
ie
w
rs
–4
1
y 4
Pr
op C
y = ( x + 1)( x − 4)( x − 7)
-R 4
–3
b
c b = −1
s
3
3
ev
–3
es
2
–2
2
R
–2
ie
ge
br am -C
y
1
4 x
2
5
1
1
292
O –1
Exercise 2F
2
–4 –3 –2 –1 O –1
–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
3
4
x
1
2
3
4
x
ev
–3
-R
–4
d
y 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
2
–2
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
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
es
s
i
− 5 < f( x ) < 4
Pr
f–1
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
y
ii
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
16
x + y = 8, 3x + y = 3. Other solutions possible. 17
a i
1+ 2
b i
3+2 2
)
(4, 3)
7
a ( x − 12)2 + ( y − 5)2 = 25 and
-R
End-of-chapter review exercise 3 1
s es Pr ity
9
( x − 4) + ( y − 2) = 20
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
-R s es
am
ev
2
br
2
5
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
6
2 , a , 26
y
ni ve rs
–6
w
C
op y
–4
R
C
w
ev
id
ie
b Proof
6 x
4
5, 6 + 2 5 ), ( 5 + 5, 6 − 2 5 )
( x − 2)2 + ( y − 10)2 = 100
-C
–2
am
2
br
O
(5 −
b y = −2 x + 16
6
ge
y 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
y
2
2 5
3
s
es
Pr
y
2
( −1, −4), (5, 2)
-R
am
2
2
1 3 25 d x − + y + = 2 2 4
ii Student’s own answer
w
b ( x − 5) + ( y + 2)2 = 16
(
2
ii Student’s own answer
ev
id br
a x + y = 64
-C
( x + 3)2 + ( y + 10)2 = 100 , ( x − 13)2 + ( y + 10)2 = 100
1
g (4, − 10), 6
op C w ie ev
16
Exercise 3E
3 10 f (3, − 4), 2 1 1 h 3 2 , 2 2 , 10
c ( x + 1)2 + ( y − 3)2 = 7
R
( x − 9)2 + ( y − 2)2 = 85
U ge
3 2 b (0, 0), 2 d (5, − 3), 2
e ( − 7, 0), 3 2
5
15
op
b
a (0, 0), 4 c (0, 2), 5
2
( x − 5)2 + ( y + 3)2 = 40
b ( x + 1)2 + ( y − 4)2 = 20
y
a y = 2x − 7
2
14
y=
C op
( 4 25 , 154 )
15
1
a Proof
-R
)
Exercise 3D
R
Pr es s
ii x + y = 7
ve rs ity
,
13
ev ie
ge
am br id
-C
y
(
4 21
12
ni
ev ie
w
C
b
2 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
C op
y
4
b 104
w -R
d Proof
ity rs ve
ni
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
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
19
ie
id g
w
c g−1( x ) = −3 +
es
s
-R
ev
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
br am
1 x+4 3 25 b k,− 12 b y=
-R s es Pr
ie
(–2, –42)
-C
3
−1
ni ve rs
(2, 22)
O
–4
2
2
C
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
ge
R
(–2, 26)
U
C w ie
2
2 2 x = ± ,x = ± 3 2 y a
–1
–2
op
c Proof
–3
s
b 4 − 2 3, 4 + 2 3
Pr
a 4, (4, −2)
es
-C
= 325
Cross-topic review exercise 1 1
1
ev
id
br
b Proof
op
298
2
ie
b k , −12, k . 12
3
ie
a (19, 13) 10 , 10 a 3 4 a y=− x+2 3 c ( x − 15)2 + ( y − 7)2
y
17
ev
C
ve rs ity
5
iii ( −1, 8), 2 10 2 b p = −1 a y=− x+3 3 c ( x − 6)2 + ( y + 1)2 = 26
am
16
ev
y
y
y
6
U
14 15
R
y = −x 2 + 6 x − 8
ni
13
y = −2 x + 6, (3, 0)
Pr es s
8 6 ii (0, −2), , 5 5 ii Proof
y = 2x − 2
i
5 Translation , vertical stretch with stretch 0 factor 2
ge
R
ev ie
w
C
12
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 s es
ni
8
12.79°
w
Pr
es
s
d 28 π cm
a 13 cm
b 2.275 cm
a 0.5 rad
b 0.8 rad
ie
-R
3
0.727
b 3π cm
ev
2 4
15.6 m
5
a 19.2 cm
b 20.5 cm
c 50.4 cm a 0.927 rad
6
b 4 cm
a 14 cm
7
b 11.8 cm
c 25.8 cm a 13 cm
b 2.35 rad
C
U
8
y
ni ve rs
c 17.6 cm
c 56.6 cm a Proof
e 240°
f
80°
g 54°
h 105°
10
Proof
j
810°
k 252°
l
ie
ev
-R
es
s
48°
w
9
o 202.5°
5π 6
c 6 π cm
d 15°
n 420°
f
a 2 π cm
c 30°
m 225°
7π 4
2π
π
7π 6
5 π 11π 2π 3 6
7
b 60°
81°
3π 2
d 0
a 90° i
2π 3
c 0.622 3 e 2 7.79 cm
id g
2
3π
π 2
b 14.1
1
n 7π 36
am
R
o
-C
ev
ie
m
l
e
k
br
w
C
i
Radians 4 π 3
C
U
-C
op y
g
b
am
e
3π 2
Degrees 240 270 300 330 360
Exercise 4B
2π 9 5π d 18 5π f 6 h 7π 6 j 5π 3
π 3
a 0.644
ity
c
π 9 5π 36 π 36 3π 4 5π 4 13π 36 π 20 10 π 3
ge
a
id
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°
C op
fg( x ) > 28
b i
a 68.8°
op
a k = −2
d 3.49
y
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
am br id
12
π 100 1 + − 3 cm 2 3
-R
ve rs ity
b Proof
c 1
d
6 5
f
id
-R s es
Pr
ity
U ii
θ =
y
π 4
π 3
cos θ
3r 2α ii + πr 2 2
-R
1 sin θ
c
π 3
f
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
1 4
e −1
C
i
tan θ
w
9
ii 6.31
ie
i
3
b
θ =
ev
8
Proof ii r 2 π r cos θ + 1 − sin θ + − θ 2
es
i
f
24 25 12 19
f 1
6
ni ve rs
7
5
c
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
C
i
1 4
op
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
ie
i
4
25 3 25 π b − 4 8
f
w
U
4
d
e 4 + 15
ge
3
b
y
b Proof
15 4 15 c 16 a
End-of-chapter review exercise 4
2
e
op
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
s es Pr
y
14
w
13
1 a tan x + cm b 0.219 rad tan x
1
b −5,
a 0, 5
w
2
2 cm
a 15 − 5 3 + 5 3 π 6 π i α = 8 i 8 tan α − 2α
iii
ii 630°
b i 30°
a
5π 6 iii 195°
ii 4 π
ie
d 13.9 cm 2
op
C
300
d r
-R
2 πr 2 3 −
-C
ni
id br
11
am
a Proof
3r 2 2
1
b 36.5 cm 2
10
1 1 + r2 π a i 4
Exercise 5A
25 ( 2 3 − π ) cm2 6
b
U
a 29.1cm 2
3
ge
9
c 51.7 cm 2
Pr es s
-C
y
op
C 8
7
2
r 1 + r2
b
c
b 4.79 cm
32 3 − 32 π cm 2 3 5 3 a cm 3 1.86 cm 2
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
y − 2
−
ev -R
−
1 2 2 3
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
−2
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
y 10 8
w Pr ity
2
O 90
–2
a = 5, b = 4, c = 3
15
a A = 2, B = 6
ni ve rs
14
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
y = 3 sinx
π 2
180
–4
s es
op y
4
O
f
ie ev -R
2π x
3π 2 y = 2 sin x
2
b 1 < f( x ) < 5
op
ve
b
id br
π
w
C
y 13
303
4
b 2 a y
b 2
C op ie -R
a a = 3, b = 2
–3
7
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 3 + 13
rs
C
10
7
Pr
y
op
316
6
s
am
dy 9 dy i = , no turning points since ≠0 2 dx (2 − x ) dx 81π iii k , −8, k . 4 ii 2 8 8