Cambridge International
AS & A Level Mathematics:
Pure Mathematics 1
Coursebook
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Copyright Material - Review Only - Not for Redistribution
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Cambridge International
AS & A Level Mathematics:
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Sue Pemberton
Series Editor: Julian Gilbey
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Pure Mathematics 1
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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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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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End-of-chapter review exercise 2
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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
End-of-chapter review exercise 9
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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.
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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.
P
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These questions focus on modelling.
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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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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You should not use a calculator for these questions.
You can use a calculator for these questions.
At the end of each chapter there is a Checklist of
learning and understanding.
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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
1
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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.
Pr
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■
■
■
■
■
■
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In this chapter you will learn how to:
Copyright Material - Review Only - Not for Redistribution
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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
ve
rs
ity
c 3x 2 − 17 x − 6 = 0
2 Solve:
y
b 3 − 2x ø 7
C
op
ni
a 5x − 8 . 2
3 Solve:
Solve simultaneous linear
equations.
a 2 x + 3 y = 13
w
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U
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7 x − 5 y = −1
b 2 x − 7 y = 31
3x + 5 y = −31
4 Simplify:
a
ity
op
c
rs
C
w
20
b ( 5 )2
Pr
y
es
s
-C
Carry out simple manipulation
of surds.
8
2
ni
op
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ve
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ev
a x 2 + x − 12 = 0
Solve linear inequalities.
IGCSE / O Level Additional
Mathematics
Why do we study quadratics?
C
U
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1 Solve:
Pr
es
s
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op
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IGCSE / O Level Mathematics
IGCSE / O Level Mathematics
2
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.
y
You will have plotted graphs of quadratics such as y = 10 − x 2 before starting your
-R
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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.
Copyright Material - Review Only - Not for Redistribution
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
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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)
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C
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y
EXPLORE 1.1
U
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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
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This is Rosa’s solution to the previous equation:
-R
am
br
Discuss her solution with your classmates and explain why her solution is not fully correct.
es
s
-C
Now solve the equation correctly.
Solve:
rs
y
ge
Factorise.
Use the fact that if pq = 0, then p = 0 or q = 0.
x=
w
ie
-R
s
1
3
es
or
Divide both sides by the common factor of 3.
3x 2 − 13x − 10 = 0
Factorise.
ity
9x 2 − 39x − 30 = 0
y
ni
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op
C
U
Solve.
id
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2
or x = 5
3
-R
s
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-C
am
br
x=−
w
e
3x + 2 = 0 or x − 5 = 0
TIP
Divide by a common
factor first, if possible.
(3x + 2)( x − 5) = 0
ev
R
5
2
Solve.
Pr
x=
w
C
op
y
-C
2 x − 5 = 0 or 3x − 1 = 0
ev
id
am
(2 x − 5)(3x − 1) = 0
b
op
6x 2 − 17 x + 5 = 0
a
C
Write in the form ax 2 + bx + c = 0.
U
6x 2 + 5 = 17 x
br
ni
Answer
ev
b 9x 2 − 39x − 30 = 0
ve
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w
a 6x 2 + 5 = 17 x
R
3
ity
C
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Pr
y
WORKED EXAMPLE 1.1
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Pr
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21
2
−
=1
2x x + 3
op
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2 x 2 − 11x − 63 = 0
Factorise.
U
R
ni
(2 x + 7)( x − 9) = 0
Expand brackets and rearrange.
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w
C
21( x + 3) − 4x = 2 x( x + 3)
Multiply both sides by 2 x( x + 3).
y
y
-C
Answer
-R
21
2
−
= 1.
2x x + 3
Solve
Solve.
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id
br
ev
7
or x = 9
2
-C
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am
x=−
w
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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.
U
R
ev
3x 2 + 26 x + 35
=0
x2 + 8
Factorise.
w
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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
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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
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-C
am
br
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id
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C
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C
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Pr
op
y
x=−
C
op
WORKED EXAMPLE 1.2
C
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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
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WORKED EXAMPLE 1.4
A rectangle has sides of length x cm and (6x − 7) cm.
Pr
es
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-C
Find the lengths of the sides of the rectangle.
y
Answer
C
w
6x 2 − 7 x − 90 = 0
Factorise.
ni
C
op
y
(2 x − 9)(3x + 10) = 0
Solve.
w
10
3
When x = 4 21 , 6x − 7 = 20.
op
Pr
y
es
-C
The rectangle has sides of length 4 21 cm and 20 cm.
C
2
4(2 x + x − 6) = 1
( x 2 − 3x + 1)6 = 1
C
2
( x 2 − 3x + 1)(2 x + x − 6) = 1
ev
ve
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B
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5
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EXPLORE 1.2
A
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Length is a positive quantity, so x = 4 21 .
s
x=−
or
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9
2
br
x=
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2 x − 9 = 0 or 3x + 10 = 0
w
Remember to check
each of your answers.
ev
br
-R
am
3 State how many values of x satisfy:
b equation B
c equation C
s
-C
a equation A
TIP
c equation C
ie
b equation B
id
a equation A
C
U
2 Solve:
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1 Discuss with your classmates how you would solve each of these equations.
y
ev
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6x – 7
Rearrange.
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op
Area = x(6x − 7) = 6x 2 − 7 x = 90
x
-R
The area of the rectangle is 90 cm 2.
y
op
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-C
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Pr
op
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4 Discuss your results.
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id
b x 2 − 7 x + 12 = 0
d 5x 2 + 19x + 12 = 0
e 20 − 7 x = 6x 2
b
2
3
+
=1
x x+2
d
5
3x
+
=2
x+3 x+4
f
3
1
1
+
=
x + 2 x − 1 ( x + 1)( x + 2)
x2 + x − 6
=0
x2 + 5
c
e
6x 2 + x − 2
=0
x2 + 7x + 4
ev
2 x 2 + 9x − 5
=0
x4 + 1
es
b 4(2 x − 11x + 15) = 1
c
2( x − 4 x + 6) = 8
e ( x 2 + 2 x − 14)5 = 1
f
( x 2 − 7 x + 11)8 = 1
2
1
9
5 The diagram shows a right-angled triangle with
sides 2 x cm, (2 x + 1) cm and 29 cm.
rs
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op
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a Show that 2 x 2 + x − 210 = 0.
U
R
b Find the lengths of the sides of the triangle.
w
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C
2x + 1
ev
id
br
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am
s
x
= 1.
Pr
es
-C
7 Solve ( x − 11x + 29)
op
y
PS
x+3
x–1
(6 x 2 + x − 2)
C
ity
1.2 Completing the square
w
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( x + d )2 = x 2 + 2 dx + d 2 and ( x − d )2 = x 2 − 2 dx + d 2
op
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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.
2
TIP
29
2x
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2
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y
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C
2
s
-C
2
-R
f
4 Find the real solutions of the following equations.
a 8( x + 2 x − 15) = 1
x2 − 9
=0
7 x + 10
w
b
Pr
ge
id
am
br
x2 − 2x − 8
=0
x 2 + 7 x + 10
d 3(2 x + 9 x + 2) =
x(10 x − 13) = 3
y
3
1
+
=2
x + 1 x( x + 1)
3 Solve:
3x 2 + x − 10
=0
a
x2 − 7x + 6
6
f
C
op
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ni
5x + 1 2 x − 1
−
= x2
4
2
ve
rs
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6
=0
x−5
c
d
x 2 − 6x − 16 = 0
U
R
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C
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y
2 Solve:
a x−
c
Pr
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s
-C
a x 2 + 3x − 10 = 0
-R
1 Solve by factorisation.
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EXERCISE 1A
w
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C
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y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
Copyright Material - Review Only - Not for Redistribution
Try the Factorisable
quadratics resource
on the Underground
Mathematics website.
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Chapter 1: Quadratics
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Rearranging these gives the following important results:
Pr
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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:
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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
w
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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
-R
am
To complete the square for 2 x 2 − 12 x + 5, we must first take a factor of 2 out of the first
s
-C
two terms, so:
w
es
Pr
ity
7
2 x 2 − 12 x + 5 = 2 [( x − 3)2 − 9] + 5 = 2( x − 3)2 − 13
ve
ie
6÷2 = 3
ւց
x 2 − 6x = ( x − 3)2 − 32 , giving
rs
C
op
y
2 x 2 − 12 x + 5 = 2( x 2 − 6x ) + 5
y
op
id
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w
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WORKED EXAMPLE 1.5
C
U
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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
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id
g
2 x 2 − 12 x + 3 = 2( x − 3)2 − 15
am
C
U
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Substituting p = 2 in equation (2) gives q = 3
(3)
y
2= p
ni
ve
rs
Comparing coefficients of x 2 , coefficients of x and the constant gives
-C
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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
C
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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
ie
w
C
8
ity
op
Pr
y
4x 2 + 20 x + 5 = ( −2 x − 5)2 − 20 = (2 x + 5)2 − 20
es
-C
Substituting b = −5 into equation (3) gives c = −20.
s
Substituting a = −2 into equation (2) gives b = −5.
-R
br
am
Alternatively:
ev
id
ie
4x 2 + 20 x + 5 = (2 x + 5)2 − 20
ie
id
Answer
w
ge
C
U
R
Leave your answers in surd form.
5
3
+
=1
x+2 x−5
-R
am
br
ev
Multiply both sides by ( x + 2)( x − 5).
-C
5( x − 5) + 3( x + 2) = ( x + 2)( x − 5)
s
2
Pr
ity
y
op
C
U
11
85
±
2
2
ie
ev
-R
1
(11 ± 85 )
2
s
-C
am
x =
es
br
id
g
x =
85
4
e
R
11
=±
2
ni
ve
rs
2
 x − 11  = 85

2 
4
ev
ie
w
C
 x − 11  −  11  + 9 = 0

 2 
2 
x−
Complete the square.
w
2
es
op
y
x 2 − 11x + 9 = 0
Expand brackets and collect terms.
Copyright Material - Review Only - Not for Redistribution
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EXERCISE 1B
w
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y
Chapter 1: Quadratics
1 Express each of the following in the form ( x + a )2 + b.
f
c
x 2 − 4x − 8
x 2 − 3x
g x2 + 7x + 1
d x 2 + 15x
h x 2 − 3x + 4
Pr
es
s
x 2 + 4x + 8
-C
e
b x 2 + 8x
-R
a x 2 − 6x
b 3x 2 − 12 x − 1
2 x 2 + 5x − 1
d 2x2 + 7x + 5
b 8x − x 2
c
4 − 3x − x 2
d 9 + 5x − x 2
y
a 4x − x 2
ev
ie
c
ve
rs
ity
a 2 x 2 − 12 x + 19
3 Express each of the following in the form a − ( x + b )2 .
w
C
op
y
2 Express each of the following in the form a ( x + b )2 + c.
U
c 13 + 4x − 2 x 2
4x 2 + 20 x + 30
c
25x 2 + 40 x − 4
ev
br
b
-R
9x 2 − 6x − 3
am
a
id
5 Express each of the following in the form ( ax + b )2 + c.
d 2 + 5x − 3x 2
w
b 3 − 12 x − 2 x 2
ge
a 7 − 8x − 2 x 2
ie
R
ni
C
op
4 Express each of the following in the form p − q ( x + r )2 .
b x 2 + 4x − 12 = 0
c
x 2 − 2 x − 35 = 0
d x 2 − 9x + 14 = 0
e x 2 + 3x − 18 = 0
f
x 2 + 9x − 10 = 0
op
Pr
y
s
-C
a x 2 + 8x − 9 = 0
es
6 Solve by completing the square.
d 9x 2 − 42 x + 61
9
b x 2 − 10 x + 2 = 0
c
x 2 + 8x − 1 = 0
d 2 x 2 − 4x − 5 = 0
rs
ity
a x 2 + 4x − 7 = 0
f
2 x 2 − 8x − 3 = 0
y
ve
op
5
3
+
= 2. Leave your answers in surd form.
x+2 x−4
w
ge
9 The diagram shows a right-angled triangle with
sides x m, (2 x + 5) m and 10 m.
br
ev
x
-R
Find the value of x. Leave your answer in surd form.
am
10
ie
id
PS
C
U
R
8 Solve
e 2 x 2 + 6x + 3 = 0
ni
ev
ie
w
C
7 Solve by completing the square. Leave your answers in surd form.
2x + 5
PS
10 Find the real solutions of the equation (3x + 5x − 7) = 1.
PS
49x 2
11 The path of a projectile is given by the equation y = ( 3 )x −
, where x and y
9000
are measured in metres.
es
s
4
Pr
op
y
-C
2
ity
op
y
(x, y)
Range
br
ev
a Find the range of this projectile.
am
x
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g
O
w
e
C
U
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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
b2 − 4ac
2a
rs
Write the right-hand side as a single fraction.
y
op
w
ie
ev
b
from both sides.
2a
−b ± b2 − 4ac
2a
ie
w
ge
id
WORKED EXAMPLE 1.8
C
U
R
x=
es
b
±
2a
Pr
x=−
Subtract
ity
b2 − 4ac
2a
C
b
=±
2a
ve
op
x+
Find the square root of both sides.
ni
y
-C
2
 x + b  = b − 4ac

2a 
4a 2
10
y
Complete the square.
ge
R
x2 +
Divide both sides by a.
ni
ax 2 + bx + c = 0
C
op
ev
ie
w
The quadratic formula can be proved by completing the square for the equation
ax 2 + bx + c = 0:
-R
am
br
ev
Solve the equation 6x 2 − 3x − 2 = 0.
-C
Write your answers correct to 3 significant figures.
es
s
Answer
3 + 57
3 − 57
or x =
12
12
y
op
-R
s
es
am
br
ev
ie
id
g
w
e
C
U
or x = −0.379 (to 3 significant figures)
-C
w
ie
ev
R
x = 0.879
Pr
x=
ni
ve
rs
−( −3) ± ( −3)2 − 4 × 6 × ( −2)
2×6
C
x=
ity
op
y
Using a = 6, b = −3 and c = −2 in the quadratic formula gives:
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
ev
ie
am
br
id
EXERCISE 1C
w
ge
C
U
ni
op
y
Chapter 1: Quadratics
1 Solve using the quadratic formula. Give your answer correct to 2 decimal places.
b x 2 + 6x + 4 = 0
d 2 x 2 + 5x − 6 = 0
e 4x 2 + 7 x + 2 = 0
c
x 2 + 3x − 5 = 0
f
5x 2 + 7 x − 2 = 0
Pr
es
s
-C
-R
a x 2 − 10 x − 3 = 0
The area of the rectangle is 63 cm 2.
ve
rs
ity
w
C
op
y
2 A rectangle has sides of length x cm and (3x − 2) cm.
Find the value of x, correct to 3 significant figures.
y
ev
ie
3 Rectangle A has sides of length x cm and (2 x − 4) cm.
U
R
ni
C
op
Rectangle B has sides of length ( x + 1) cm and (5 − x ) cm.
ge
Rectangle A and rectangle B have the same area.
id
ie
w
Find the value of x, correct to 3 significant figures.
br
ev
5
2
+
= 1.
x −3 x +1
Give your answers correct to 3 significant figures.
-C
-R
am
4 Solve the equation
WEB LINK
5 Solve the quadratic equation ax − bx + c = 0, giving your answers in terms
of a, b and c.
es
Pr
y
rs
ity
op
C
How do the solutions of this equation relate to the solutions of
the equation ax 2 + bx + c = 0?
w
ve
ie
s
2
y
op
ni
ev
1.4 Solving simultaneous equations (one linear and one quadratic)
C
ie
y = x2 – 4
y
op
w
ni
ve
rs
(–1, –3)
x
ity
C
O
Pr
op
y
es
s
-C
-R
am
ev
y = 2x – 1
br
id
y
(3, 5)
ie
C
U
The diagram shows the graphs of y = x 2 − 4 and y = 2 x − 1.
ie
id
g
w
e
The coordinates of the points of intersection of the two graphs are ( −1, −3) and (3, 5).
-R
s
es
am
br
ev
It follows that x = −1, y = −3 and x = 3, y = 5 are the solutions of the simultaneous
equations y = x 2 − 4 and y = 2 x − 1.
-C
ev
R
w
ge
U
R
In this section, we shall learn how to solve simultaneous equations where one equation is
linear and the second equation is quadratic.
Copyright Material - Review Only - Not for Redistribution
Try the Quadratic
solving sorter resource
on the Underground
Mathematics website.
11
ve
rs
ity
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
w
ev
ie
(1)
am
br
id
(2)
-R
y = x2 − 4
y = 2x − 1
ge
The solutions can also be found algebraically:
Substitute for y from equation (2) into equation (1):
-C
2x − 1 = x2 − 4
x2 − 2x − 3 = 0
( x + 1)( x − 3) = 0
x = −1 or x = 3
C
ve
rs
ity
op
y
Pr
es
s
Rearrange.
Factorise.
ni
The solutions are: x = −1, y = −3 and x = 3, y = 5.
id
ie
w
ge
In general, an equation in x and y is called quadratic if it has the form
ax 2 + bxy + cy2 + dx + ey + f = 0, where at least one of a, b and c is non-zero.
C
op
y
Substituting x = 3 into equation (2) gives y = 6 − 1 = 5.
U
R
ev
ie
w
Substituting x = −1 into equation (2) gives y = −2 − 1 = −3.
-C
-R
am
br
ev
Our technique for solving one linear and one quadratic equation will work for these more general
quadratics, too. (The graph of a general quadratic function such as this is called a conic.)
y
es
s
WORKED EXAMPLE 1.9
op
Pr
Solve the simultaneous equations.
12
ity
R
x − 4y = 8
y
(1)
2
ni
2
op
Answer
2x + 2 y = 7
ve
rs
x 2 − 4 y2 = 8
(2)
U
ev
ie
w
C
2x + 2 y = 7
br
2
ev
id
ie
w
ge
C
7 − 2y
.
2
Substitute for x in equation (2):
From equation (1), x =
-C
-R
am
 7 − 2 y  − 4 y2 = 8
 2 
s
49 − 28 y + 4 y2
− 4 y2 = 8
4
es
ity
Rearrange.
Factorise.
id
g
ev
ie
17
19
in equation (1) gives x =
.
6
3
-R
s
es
-C
am
br
Substituting y = −
w
e
C
U
1
17
or y =
2
6
op
(6 y + 17)(2 y − 1) = 0
y
ni
ve
rs
12 y2 + 28 y − 17 = 0
y=−
Multiply both sides by 4.
Pr
op
y
R
ev
ie
w
C
49 − 28 y + 4 y2 − 16 y2 = 32
Expand brackets.
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
Pr
es
s
From equation (1), 2 y = 7 − 2 x .
C
op
y
Rearrange.
ni
Factorise.
U
3x 2 − 28x + 57 = 0
ie
id
19
17
1
,y=−
and x = 3, y = .
3
6
2
op
Pr
y
es
s
-C
The solutions are: x =
ity
rs
y = 6−x
id
8x 2 − 2 xy = 4
y
op
x 2 + 3xy = 10
m 2 x + 3 y + 19 = 0
s
x 2 + y2 + 4xy = 24
2 x − y = 14
o x − 12 y = 30
Pr
n x + 2y = 5
2 y2 − xy = 20
x 2 + y2 = 10
ity
2x2 + 3 y = 5
l
x + 2y = 6
y 2 = 8x + 4
es
xy = 12
ev
i
k x − 4y = 2
2
4x − 3 y = 5
ni
ve
rs
op
y
2 The sum of two numbers is 26. The product of the two numbers is 153.
op
y
a What are the two numbers?
id
g
w
e
C
U
b If instead the product is 150 (and the sum is still 26), what would the two
numbers now be?
-R
s
es
am
br
ev
ie
3 The perimeter of a rectangle is 15.8 cm and its area is 13.5 cm 2. Find the lengths
of the sides of the rectangle.
-C
C
x 2 − 4xy = 20
-R
br
-C
5x − 2 y = 23
x − 5xy + y = 1
w
f
2 x 2 − 3 y2 = 15
am
xy = 8
ie
x − 2y = 6
h 2y − x = 5
g 2x + y = 8
2
x 2 + y2 = 100
w
ge
e
x 2 + 2 xy = 8
ie
ni
U
R
y = x2
j
c 3 y = x + 10
b x + 4y = 6
ve
ev
a
d y = 3x − 1
ev
13
1 Solve the simultaneous equations.
ie
w
C
EXERCISE 1D
R
-R
br
ev
19
or x = 3
3
am
x=
w
ge
(3x − 19)( x − 3) = 0
C
ev
ie
x 2 − 49 + 28x − 4x 2 = 8
R
Expand brackets.
ve
rs
ity
w
C
op
y
Substitute for 2 y in equation (2):
x 2 − (7 − 2 x )2 = 8
ev
ie
am
br
id
-C
Alternative method:
-R
1
into equation (1) gives x = 3.
2
19
17
1
The solutions are: x = , y = −
and x = 3, y = .
3
6
2
Substituting y =
w
ge
C
U
ni
op
y
Chapter 1: Quadratics
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
am
br
id
ev
ie
w
ge
4 The sum of the perimeters of two squares is 50 cm and the sum of the areas is
93.25 cm 2.
-R
Find the side length of each square.
Pr
es
s
-C
5 The sum of the circumferences of two circles is 36 π cm and the sum of the
areas is 170 π cm 2.
6 A cuboid has sides of length 5 cm, x cm and y cm. Given that x + y = 20.5 and
the volume of the cuboid is 360 cm 3, find the value of x and the value of y.
ve
rs
ity
w
C
op
y
Find the radius of each circle.
y
C
op
U
R
ni
ev
ie
7 The diagram shows a solid formed by joining a
hemisphere, of radius r cm, to a cylinder, of radius r cm
and height h cm.
h
ie
w
ge
The total height of the solid is 18 cm and the surface
area is 205 π cm 2.
18 cm
r
The surface area, A,
of a sphere with radius
r is A = 4 π r 2.
br
ev
id
Find the value of r and the value of h.
TIP
s
-C
a Find the coordinates of the points A and B.
-R
am
8 The line y = 2 − x cuts the curve 5x 2 − y2 = 20 at the points A and B.
Pr
y
es
b Find the length of the line AB.
rs
C
a Find the coordinates of the points A and B.
ity
op
9 The line 2 x + 5 y = 1 meets the curve x 2 + 5xy − 4 y2 + 10 = 0 at the points A and B.
14
ve
ie
w
b Find the midpoint of the line AB.
y
op
C
U
R
ni
ev
10 The line 7 x + 2 y = −20 intersects the curve x 2 + y2 + 4x + 6 y − 40 = 0 at the
points A and B. Find the length of the line AB.
id
ie
w
ge
11 The line 7 y − x = 25 cuts the curve x 2 + y2 = 25 at the points A and B.
br
ev
Find the equation of the perpendicular bisector of the line AB.
-R
am
12 The straight line y = x + 1 intersects the curve x 2 − y = 5 at the points A and B.
es
s
-C
Given that A lies below the x-axis and the point P lies on AB such that
AP : PB = 4 : 1, find the coordinates of P.
Pr
ity
ni
ve
rs
WEB LINK
op
y
14 a Split 10 into two parts so that the difference between the squares of the parts
is 60.
es
s
-R
br
ev
ie
id
g
w
e
C
U
b Split N into two parts so that the difference between the squares of the parts
is D.
am
R
ev
ie
PS
Find the equation of the perpendicular bisector of the line AB.
-C
w
C
op
y
13 The line x − 2 y = 1 intersects the curve x + y2 = 9 at two points, A and B.
Copyright Material - Review Only - Not for Redistribution
Try the Elliptical
crossings resource
on the Underground
Mathematics website.
ve
rs
ity
C
U
ni
op
y
Chapter 1: Quadratics
w
ge
1.5 Solving more complex quadratic equations
am
br
id
ev
ie
You may be asked to solve an equation that is quadratic in some function of x.
-R
WORKED EXAMPLE 1.10
Pr
es
s
-C
Solve the equation 4x 4 − 37 x 2 + 9 = 0.
op
y
Answer
ve
rs
ity
y
U
w
ge
1
or y = 9
4
id
es
s
-C
-R
am
br
ev
1
or x 2 = 9
4
1
or x = ± 3
x=±
2
Method 2: Factorise directly
x2 =
op
y
ve
rs
ity
15
ge
C
U
R
ni
ev
ie
w
C
op
Pr
y
4x 4 − 37 x 2 + 9 = 0
(4x 2 − 1)( x 2 − 9) = 0
1
or x 2 = 9
x2 =
4
1
x=±
or x = ± 3
2
ev
id
ie
w
WORKED EXAMPLE 1.11
-R
-C
Answer
am
br
Solve the equation x − 4 x − 12 = 0.
es
Pr
x.
ity
y2 − 4 y − 12 = 0
y = −2
Substitute
U
x = −2 has no solutions as x is never negative.
ev
-R
s
es
am
br
∴ x = 36
ie
id
g
w
e
x = 6 or x = −2
x for y.
C
y = 6 or
ni
ve
rs
( y − 6)( y + 2) = 0
-C
R
ev
ie
w
C
op
y
Let y =
s
x − 4 x − 12 = 0
y
R
(4 y − 1)( y − 9) = 0
y=
Substitute x 2 for y.
ni
4 y2 − 37 y + 9 = 0
ie
ev
ie
Let y = x 2 .
C
op
4x 4 − 37 x 2 + 9 = 0
op
w
C
Method 1: Substitution method
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
ev
ie
w
ge
am
br
id
WORKED EXAMPLE 1.12
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
y
ni
w
-R
es
s
-C
EXERCISE 1E
am
br
ev
id
ie
R
x = −1 or x = 2
1
= 3−1 and 9 = 32.
3
U
1
or 3x = 9
3
Substitute 3x for y.
ge
3x =
ev
ie
1
or y = 9
3
ve
rs
ity
(3 y − 1)( y − 9) = 0
w
C
op
y
3 y2 − 28 y + 9 = 0
y=
Let y = 3x.
Pr
es
s
-C
3(3x )2 − 28(3x ) + 9 = 0
C
op
Answer
-R
Solve the equation 3(9x ) − 28(3x ) + 9 = 0.
Pr
b x6 − 7x3 − 8 = 0
c
x 4 − 6x 2 + 5 = 0
e 3x 4 + x 2 − 4 = 0
f
8x 6 − 9x 3 + 1 = 0
h x 4 + 9x 2 + 14 = 0
i
x8 − 15x 4 − 16 = 0
l
8
7
+
=1
x6 x3
c
6x − 17 x + 5 = 0
f
3 x+
32 x10 − 31x5 − 1 = 0
k
9
5
+
=4
x 4 x2
b
x ( x + 1) = 6
C
ge
2 Solve:
op
j
br
ev
id
a 2 x − 9 x + 10 = 0
w
x 4 + 2 x 2 − 15 = 0
ie
g
ve
rs
d 2 x 4 − 11x 2 + 5 = 0
y
ity
a x 4 − 13x 2 + 36 = 0
ni
R
ev
ie
w
C
16
U
op
y
1 Find the real values of x that satisfy the following equations.
d 10 x + x − 2 = 0
-R
am
e 8x + 5 = 14 x
5
= 16
x
s
-C
3 The curve y = 2 x and the line 3 y = x + 8 intersect at the points A and B.
es
op
y
a Write down an equation satisfied by the x-coordinates of A and B.
Pr
ity
y
7
O
1
es
s
-R
br
ev
ie
id
g
w
e
C
U
op
ni
ve
rs
4 The graph shows y = ax + b x + c for x ù 0. The graph crosses the x-axis
49
at the points ( 1, 0 ) and  , 0  and it meets the y-axis at the point (0, 7).
 4 
Find the value of a, the value of b and the value of c.
y
Find the length of the line AB.
am
R
ev
ie
w
PS
c
-C
C
b Solve your equation in part a and, hence, find the coordinates of A and B.
Copyright Material - Review Only - Not for Redistribution
49
4
x
ve
rs
ity
y
ev
ie
w
ge
5 The graph shows y = a (22 x ) + b(2 x ) + c.
The graph crosses the axes at the points (2, 0), (4, 0) and (0, 90).
Find the value of a, the value of b and the value of c.
90
2
O
4
x
op
y
Pr
es
s
-C
-R
am
br
id
PS
C
U
ni
op
y
Chapter 1: Quadratics
C
op
y
The general form of a quadratic function is f( x ) = ax 2 + bx + c, where a, b and c are
constants and a ≠ 0.
ni
ev
ie
w
C
ve
rs
ity
1.6 Maximum and minimum values of a quadratic function
br
A point where the
gradient is zero is
called a stationary
point or a turning point.
-R
am
es
s
-C
Pr
y
ity
op
17
If a , 0, the curve has a maximum
point that occurs at the highest point
of the curve.
ge
C
U
R
ni
op
y
ve
rs
C
ie
w
If a . 0, the curve has a minimum
point that occurs at the lowest point
of the curve.
ev
TIP
ev
id
ie
w
ge
U
R
The shape of the graph of the function f( x ) = ax 2 + bx + c is called a parabola.
The orientation of the parabola depends on the value of a, the coefficient of x 2.
id
ie
w
In the case of a parabola, we also call this point the vertex of the parabola.
br
ev
Every parabola has a line of symmetry that passes through the vertex.
-R
am
One important skill that we will develop during this course is ‘graph sketching’.
es
s
-C
A sketch graph needs to show the key features and behaviour of a function.
C
●
Pr
WEB LINK
op
y
Depending on the context we should show some or all of these.
-R
s
es
am
br
ev
ie
id
g
w
e
C
U
The skills you developed earlier in this chapter should enable you to draw a clear sketch
graph for any quadratic function.
-C
R
ev
ie
w
●
the general shape of the graph
the axis intercepts
the coordinates of the vertex.
ni
ve
rs
●
ity
op
y
When we sketch the graph of a quadratic function, the key features are:
Copyright Material - Review Only - Not for Redistribution
Try the Quadratic
symmetry resource
on the Underground
Mathematics
website for a further
explanation of this.
ve
rs
ity
ev
ie
w
ge
am
br
id
DID YOU KNOW?
y
U
ni
C
op
ev
ie
w
C
ve
rs
ity
op
y
Pr
es
s
-C
-R
If we rotate a parabola about its axis of symmetry, we
obtain a three-dimensional shape called a paraboloid.
Satellite dishes are paraboloid shapes. They have the
special property that light rays are reflected to meet at a
single point, if they are parallel to the axis of symmetry
of the dish. This single point is called the focus of the
satellite dish. A receiver at the focus of the paraboloid
then picks up all the information entering the dish.
WORKED EXAMPLE 1.13
R
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
w
ge
For the function f( x ) = x 2 − 3x − 4 :
ie
id
a Find the axes crossing points for the graph of y = f( x ).
-R
Answer
am
br
ev
b Sketch the graph of y = f( x ) and find the coordinates of the vertex.
es
s
-C
a y = x 2 − 3x − 4
( x + 1)( x − 4) = 0
x = −1 or x = 4
w
Pr
ity
When y = 0, x 2 − 3x − 4 = 0
C
ev
ve
ie
Axes crossing points are: (0, −4), ( −1, 0) and (4, 0).
3
2
w
ge
y = x 2 – 3x – 4
ev
–1 O
x
-R
am
br
id
ie
x=
y
C
U
R
ni
op
b The line of symmetry cuts the x-axis midway between the axis intercepts
of −1 and 4.
s
es
ni
ve
rs
2
3
3
3
, y =   −3  − 4



2
2
2
25
y=−
4
Since a . 0, the curve is U-shaped.
3
25 
Minimum point =  , −
2
4 
-R
s
es
am
br
ev
ie
id
g
w
e
C
U
When x =
y
3
.
2
op
Hence, the line of symmetry is x =
-C
ev
ie
w
( 32 , –254
ity
Pr
–4
C
op
y
-C
4
(
R
y
18
rs
op
y
When x = 0, y = −4
Copyright Material - Review Only - Not for Redistribution
TIP
Write your answer in
fraction form.
ve
rs
ity
C
U
ni
op
y
Chapter 1: Quadratics
ev
ie
am
br
id
w
ge
Completing the square is an alternative method that can be used to help sketch the graph
of a quadratic function.
2
-R
Completing the square for x 2 − 3x − 4 gives:
2
3
25
=x−  −


2
4
2
C
This part of the expression is a square so it will be
at least zero. The smallest value it can be is 0. This
3
occurs when x = .
2
y
25
3
3
25
and this minimum occurs when x = .
is −
The minimum value of  x −  −


4
2
2
4
25 
3

2
So the function f( x ) = x − 3x − 4 has a minimum point at
.
,−
2
4 
3
The line of symmetry is x = .
2
Pr
19
ity
op
WORKED EXAMPLE 1.14
w
rs
C
es
s
-C
if a , 0, there is a maximum point at ( h, k ).
y
●
-R
am
br
ev
If f( x ) = ax 2 + bx + c is written in the form f( x ) = a ( x − h )2 + k, then:
b
● the line of symmetry is x = h = −
2a
● if a . 0, there is a minimum point at ( h, k )
y
C
U
Answer
ge
id
ie
w
This part of the expression is a square so ( x − 2)2 ù 0.
The smallest value it can be is 0. This occurs when x = 2.
br
ev
The maximum value of 9 − 4( x − 2)2 is 9 and this
maximum occurs when x = 2.
-R
am
Since this is being subtracted from 9, the whole expression
is greatest when x = 2.
s
-C
So the function f( x ) = 16x − 7 − 4x 2 has
a maximum point at (2, 9).
y
es
y
O
31
2
1
2
C
w
ev
s
-R
x=2
es
am
-C
x
ie
–7
br
id
g
e
U
R
3
x−2 = ±
2
1
x = 3 21 or x =
2
y = 16x – 7 – 4x2
ity
ni
ve
rs
ie
w
C
When x = 0, y = −7
When y = 0, 9 − 4( x − 2)2 = 0
9
( x − 2)2 =
4
(2, 9)
Pr
op
y
The line of symmetry is x = 2.
op
R
op
ni
ev
ve
ie
Sketch the graph of y = 16x − 7 − 4x 2 .
Completing the square gives:
16x − 7 − 4x 2 = 9 − 4( x − 2)2
ev
ie
id
KEY POINT 1.3
w
ge
U
R
ni
C
op
w
ev
ie
Pr
es
s
2
ve
rs
ity
op
y
-C
3
3
x 2 − 3x − 4 =  x −  −   − 4

 2
2
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EXERCISE 1F
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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
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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.
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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.
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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.
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4 a Express 2 x 2 + 9x + 4 in the form a ( x + b )2 + c, where a, b and c are constants.
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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.
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b Sketch the graph of y = 1 + x − 2 x 2 .
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6 a Write 1 + x − 2 x 2 in the form p − 2( x − q )2.
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5 Find the minimum value of x 2 − 7 x + 8 and the corresponding value of x.
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8 Find the equations of parabolas A, B and C.
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7 Prove that the graph of y = 4x 2 + 2 x + 5 does not intersect the x-axis.
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–2
8 x
6
C
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–4
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–6
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4
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2
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2
–2
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4
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8
6
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–4
10
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B
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12
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am
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–8
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9 The diagram shows eight parabolas.
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PS
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Chapter 1: Quadratics
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The equations of two of the parabolas are y = x 2 − 6x + 13 and y = − x 2 − 6x − 5.
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a Identify these two parabolas and find the equation of each of the other parabolas.
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E
F
A
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b Use graphing software to create your own parabola pattern.
B
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D
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11 A parabola passes through the points ( −2, −3), (2, 9) and (6, 5).
Find the equation of the parabola.
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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.
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Find the equation of the parabola.
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10 A parabola passes through the points (0, −24), ( −2, 0) and (4, 0).
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PS
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[This question is an adaptation of Which parabola? on the Underground Mathematics website
and was developed from an original idea from NRICH.]
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1.7 Solving quadratic inequalities
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We already know how to solve linear inequalities.
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The following text shows two examples.
Pr
Expand brackets.
2 x + 14 , − 4
Subtract 14 from both sides.
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Divide both sides by 2.
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Divide both sides by −2.
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Subtract 11 from both sides.
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xø3
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−2 x ù − 6
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Solve 11 − 2 x ù 5.
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x , −9
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2 x , −18
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Solve 2( x + 7) , − 4 .
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Cambridge International AS & A Level Mathematics: Pure Mathematics 1
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The second of the previous examples uses the important rule that:
KEY POINT 1.4
Pr
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If we multiply or divide both sides of an inequality by a negative number, then the inequality sign
must be reversed.
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y = x2 – 5x – 14
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Sketch the graph of y = x 2 − 5x − 14.
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–2
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Sketch the graph of y = 2 x 2 + 3x − 27.
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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).
–
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The solution is − 4 21 < x < 3.
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So the x-axis intercepts are −4 21 and 3.
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+
–4 1
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x = −4 21 or x = 3
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(2 x + 9)( x − 3) = 0
+
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When y = 0, 2 x 2 + 3x − 27 = 0
y = 2x2 + 3x – 27
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Pr
op
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Rearranging: 2 x 2 + 3x − 27 ø 0
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Answer
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Solve 2 x 2 + 3x ø 27.
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–
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The solution is x , −2 or x . 7.
x
7
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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.
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( x + 2)( x − 7) = 0
x = −2 or x = 7
WORKED EXAMPLE 1.16
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When y = 0, x 2 − 5x − 14 = 0
22
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Answer
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Solve x 2 − 5x − 14 . 0.
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WORKED EXAMPLE 1.15
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Quadratic inequalities can be solved by sketching a graph and considering when the graph
is above or below the x-axis.
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3
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Ivan is asked to solve the inequality
2x − 4
ù 7.
x
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EXPLORE 1.3
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Chapter 1: Quadratics
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This is his solution:
2x − 4 ù 7x
Multiply both sides by x:
x ø−
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y
−4 ù 5x
Subtract 2x from both sides:
Divide both sides by 5:
4
5
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She writes:
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Anika checks to see if x = −1 satisfies the original inequality.
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When x = −1: (2(−1) − 4) ÷ (−1) = 6
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Hence, x = −1is a value of x that does not satisfy the original inequality.
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So Ivan’s solution must be incorrect!
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Discuss Ivan’s solution with your classmates and explain Ivan’s error.
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Pr
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How could Ivan have approached this problem to obtain a correct solution?
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e 6x 2 − 23x + 20 , 0
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( x − 6)( x − 4) ø 0
f
(1 − 3x )(2 x + 1) , 0
c
x 2 + 6x − 7 . 0
f
4 − 7x − 2x2 , 0
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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 )
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a x 2 , 36 − 5x
Pr
2
g ( x + 4)2 ù 25
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c
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d 14x 2 + 17 x − 6 ø 0
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b x 2 + 7 x + 10 ø 0
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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:
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and ( x − 5)2 , 4
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and x 2 − 2 x − 3 ù 0
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2
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6 Find the range of values of x for which 2 x − 3x − 40 . 1.
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x2 + x − 2 . 0
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b x 2 + 4x − 21 ø 0 and x 2 − 9x + 8 . 0
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a x 2 − 25 ù 0
3 Solve:
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e (5 − x )( x + 6) ù 0
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d (2 x + 3)( x − 2) , 0
b ( x − 3)( x + 2) . 0
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a x( x − 3) ø 0
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1 Solve:
2 Solve:
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EXERCISE 1G
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x( x − 1)
.x
x +1
e
x 2 + 4x − 5
ø0
x2 − 4
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b
c
x2 − 9
ù4
x −1
f
x−3 x+2
ù
x+4 x−5
Pr
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x 2 − 2 x − 15
ù0
x−2
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7 Solve:
x
ù3
a
x −1
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Cambridge International AS & A Level Mathematics: Pure Mathematics 1
1.8 The number of roots of a quadratic equation
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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
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two distinct real roots
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x=
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2 ×1
−2 ± 36
x=
2
x = 2 or x = −4
x=
y
x 2 + 2x + 6 = 0
−2 ± 2 2 − 4 × 1 × ( −8 )
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x=
x 2 + 6x + 9 = 0
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x 2 + 2x − 8 = 0
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Consider solving the following three quadratic equations of the form ax 2 + bx + c = 0
−b ± b2 − 4ac
using the formula x =
.
2a
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The part of the quadratic formula underneath the square root sign is called the discriminant.
=0
,0
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Nature of roots
two distinct real roots
C
b 2 − 4 ac
.0
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The sign (positive, zero or negative) of the discriminant tells us how many roots there are
for a particular quadratic equation.
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two equal real roots (or 1 repeated real root)
no real roots
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The discriminant of ax 2 + bx + c = 0 is b 2 − 4ac.
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KEY POINT 1.5
24
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x
x
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or
a,0
x
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x
a.0
or
x
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a,0
The curve is entirely above or entirely below the x-axis.
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no real roots
or
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a.0
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,0
x
The curve touches the x-axis at one point.
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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)
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=0
Pr
two distinct real roots
C
.0
Nature of roots of ax 2 + bx +
+c = 0
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b 2 − 4 ac
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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.
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a,0
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Chapter 1: Quadratics
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WORKED EXAMPLE 1.17
b2 − 4ac = 0
For two equal roots:
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WORKED EXAMPLE 1.18
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k 2 = 16
k = −4 or k = 4
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k2 − 4 × 4 × 1 = 0
Pr
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Answer
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Find the values of k for which the equation 4x 2 + kx + 1 = 0 has two equal roots.
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Answer
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Find the values of k for which x 2 − 5x + 9 = k (5 − x ) has two equal roots.
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x 2 − 5x + 9 = k (5 − x )
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Rearrange the equation into the form ax 2 + bx + c = 0 .
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x 2 − 5x + 9 − 5 k + kx = 0
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x 2 + ( k − 5)x + 9 − 5 k = 0
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For two equal roots: b2 − 4ac = 0
Pr
2
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( k + 11)( k − 1) = 0
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k 2 + 10 k − 11 = 0
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k − 10 k + 25 − 36 + 20 k = 0
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( k − 5)2 − 4 × 1 × (9 − 5 k ) = 0
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k = −11 or k = 1
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WORKED EXAMPLE 1.19
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Pr
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b2 − 4ac . 0
2
4 k − 32 k . 0
4 k ( k − 8) . 0
+
0
k
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Note that the critical values are where 4 k ( k − 8) = 0 .
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Hence, k , 0 or k . 8.
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8
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Critical values are 0 and 8.
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( −2 k )2 − 4 × k × 8 . 0
-C
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For two distinct roots:
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kx 2 − 2 kx + 8 = 0
op
Answer
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Find the values of k for which kx 2 − 2 kx + 8 = 0 has two distinct roots.
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EXERCISE 1H
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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
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x 2 − 12 x + 36 = 0
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a
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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
.
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3 The equation x 2 + bx + c = 0 has roots −5 and 7.
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2 Use the discriminant to determine the nature of the roots of 2 − 5x =
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Find the value of b and the value of c.
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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
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x 2 + kx + 4 = 0
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4 Find the values of k for which the following equations have two equal roots.
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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
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kx 2 + (2 k − 5)x = 1 − k
Pr
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kx 2 + 2 kx = 4x − 6
d 2 x 2 + k = 3( x − 2)
f
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b 3x 2 + 5x + k + 1 = 0
kx 2 + kx = 3x − 2
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7 The equation kx 2 + px + 5 = 0 has repeated real roots.
op
2 x 2 + 8x − 5 = kx 2
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kx 2 − 4x + 8 = 0
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C
6 Find the values of k for which the following equations have no real roots.
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Find k in terms of p.
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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.
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11 Prove that x 2 + kx + 2 = 0 has real roots if k ù 2 2.
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For which other values of k does the equation have real roots?
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P
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WEB LINK
Try the Discriminating
resource on the
Underground
Mathematics website.
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Chapter 1: Quadratics
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1.9 Intersection of a line and a quadratic curve
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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.
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two points of intersection
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Pr
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Situation 1
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When considering the intersection of a straight line and a parabola, there are three
possible situations.
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We have already learnt that to find the points of intersection of a straight line and a
quadratic curve, we solve their equations simultaneously.
.0
two distinct real roots
=0
two equal real roots (repeated roots) one point of intersection (line is a tangent)
,0
no real roots
Pr
two distinct points of intersection
no points of intersection
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WORKED EXAMPLE 1.20
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s
Line and curve
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Nature of roots
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b 2 − 4 ac
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-C
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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.
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Answer
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Find the value of k for which y = x + k is a tangent to the curve y = x 2 + 5x + 2.
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x 2 + 5x + 2 = x + k
-R
am
x 2 + 4x + (2 − k ) = 0
-C
Since the line is a tangent to the curve, the discriminant of the quadratic must be zero, so:
es
s
b2 − 4ac = 0
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42 − 4 × 1 × (2 − k ) = 0
16 − 8 + 4 k = 0
4 k = −8
k = −2
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27
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WORKED EXAMPLE 1.21
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Cambridge International AS & A Level Mathematics: Pure Mathematics 1
-C
Answer
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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
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x 2 − 2 x = kx − 1
x − ( k + 2)x + 1 = 0
2
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+
–4
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Critical values are −4 and 0.
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+
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k 2 + 4k . 0
k ( k + 4) . 0
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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.
–
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Hence, k , −4 or k . 0.
k
0
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-C
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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.
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WORKED EXAMPLE 1.22
28
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Answer
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Find the set of values of k for which the line 2 x + y = k does not intersect the curve xy = 8 .
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2 x 2 − kx + 8 = 0
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x( k − 2 x ) = 8
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Substituting y = k − 2 x into xy = 8 gives:
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Since the line and curve do not intersect, we must have discriminant , 0.
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ev
b2 − 4ac , 0
( − k )2 − 4 × 2 × 8 , 0
+
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+
k 2 − 64 , 0
( k + 8)( k − 8) , 0
8
s
-C
–8
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es
Critical values are −8 and 8.
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s
es
-C
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Pr
Hence, −8 , k , 8 .
–
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EXERCISE 1I
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Chapter 1: Quadratics
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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
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4 The line y = k − 3x is a tangent to the curve x 2 + 2 xy − 20 = 0.
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3 Find the value of k for which the line x + ky = 12 is a tangent to the curve y =
y
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a Find the possible values of k.
U
R
ni
C
op
b For each of these values of k, find the coordinates of the point of contact of the tangent with the curve.
w
ge
5 Find the values of m for which the line y = mx + 6 is a tangent to the curve y = x 2 − 4x + 7.
id
ie
For each of these values of m, find the coordinates of the point where the line touches the curve.
-R
am
br
ev
6 Find the set of values of k for which the line y = 2 x − 1 intersects the curve y = x 2 + kx + 3 at two distinct
points.
es
s
-C
7 Find the set of values of k for which the line x + 2 y = k intersects the curve xy = 6 at two distinct points.
29
9 Find the set of values of m for which the line y = mx + 5 does not meet the curve y = x 2 − x + 6.
ity
C
op
Pr
y
8 Find the set of values of k for which the line y = k − x cuts the curve y = 5 − 3x − x 2 at two distinct points.
y
ve
ie
w
rs
10 Find the set of values of k for which the line y = 2 x − 10 does not meet the curve y = x 2 − 6x + k.
op
ie
id
br
ev
13 The line y = mx + c is a tangent to the curve ax 2 + by2 = c, where a, b, c and m are constants.
abc − a
.
Prove that m2 =
b
y
op
-R
s
es
-C
am
br
ev
ie
id
g
w
e
C
U
R
ev
ie
w
ni
ve
rs
C
ity
Pr
op
y
es
s
-C
-R
am
P
w
Prove that m2 + 8 m + 4c = 0.
C
12 The line y = mx + c is a tangent to the curve y = x 2 − 4x + 4.
ge
P
U
R
ni
ev
11 Find the value of k for which the line y = kx + 6 is a tangent to the curve x 2 + y2 − 10 x + 8 y = 84.
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
●
completing the square
●
using the quadratic formula x =
−b ± b 2 − 4ac
.
2a
Pr
es
s
-C
factorisation
-R
Quadratic equations can be solved by:
●
w
ev
ie
am
br
id
ge
Checklist of learning and understanding
op
y
Solving simultaneous equations where one is linear and one is quadratic
Substitute this for x or y in the quadratic equation and then solve.
Maximum and minimum points and lines of symmetry
y
ev
ie
w
●
Rearrange the linear equation to make either x or y the subject.
ve
rs
ity
C
●
●
if a , 0, there is a maximum point at ( h, k ).
ie
if a . 0, there is a minimum point at ( h, k )
br
ev
id
●
w
ge
U
R
ni
C
op
For a quadratic function f( x ) = ax 2 + bx + c that is written in the form f( x ) = a( x − h )2 + k :
b
● the line of symmetry is x = h = −
2a
-R
am
Quadratic equation ax 2 + bx + c = 0 and corresponding curve y = ax 2 + bx + c
Discriminant = b 2 − 4ac.
●
If b 2 − 4ac . 0, then the equation ax 2 + bx + c = 0 has two distinct real roots.
●
If b 2 − 4ac = 0, then the equation ax 2 + bx + c = 0 has two equal real roots.
op
Pr
y
es
s
-C
●
rs
The condition for a quadratic equation to have real roots is b 2 − 4ac ù 0.
y
ve
Intersection of a line and a general quadratic curve
If a line and a general quadratic curve intersect at one point, then the line is a tangent to the curve at that point.
●
Solving simultaneously the equations for the line and the curve gives an equation of the form ax 2 + bx + c = 0.
●
b 2 − 4ac gives information about the intersection of the line and the curve.
C
w
ie
ge
U
ni
op
●
Nature of roots
Line and parabola
two distinct real roots
two distinct points of intersection
=0
two equal real roots
one point of intersection (line is a tangent)
,0
no real roots
no points of intersection
-R
y
op
-R
s
es
-C
am
br
ev
ie
id
g
w
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U
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ev
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w
ni
ve
rs
C
ity
Pr
op
y
es
-C
am
.0
s
b 2 − 4 ac
ev
id
R
ev
ie
w
●
If b 2 − 4ac , 0, then the equation ax 2 + bx + c = 0 has no real roots.
ity
C
●
br
30
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ve
rs
ity
ge
C
U
ni
op
y
Chapter 1: Quadratics
ev
ie
am
br
id
1
w
END-OF-CHAPTER REVIEW EXERCISE 1
A curve has equation y = 2 xy + 5 and a line has equation 2 x + 5 y = 1.
[2]
b Find the set of values of x that satisfy the inequality 9x 2 − 15x , 6.
[2]
ve
rs
ity
36
25
+ 4 = 2.
x4
x
3
Find the real roots of the equation
4
Find the set of values of k for which the line y = kx − 3 intersects the curve y = x 2 − 9x at two
distinct points.
C
op
y
[4]
U
R
Find the set of values of the constant k for which the line y = 2 x + k meets the curve y = 1 + 2 kx − x 2
at two distinct points.
ie
id
[3]
br
ev
b Find the values of the constant k for which the line y = kx + 3 is a tangent to the curve
y = 4x 2 − 12 x + 7.
-R
s
Pr
y
op
For the case where the line is a tangent to the curve at a point C , find the value of k and the
coordinates of C.
[4]
C
U
ie
w
ge
Show that the curve lies above the x-axis.
b
Find the coordinates of the points of intersection of the line and the curve.
[3]
-R
am
br
ev
id
a
Write down the set of values of x that satisfy the inequality x 2 − 5x + 7 , 2 x − 3.
[3]
[1]
s
-C
es
A curve has equation y = 10 x − x 2 .
Express 10 x − x 2 in the form a − ( x + b )2 .
[3]
b
Write down the coordinates of the vertex of the curve.
[2]
c
Find the set of values of x for which y ø 9.
[3]
ity
Pr
a
ni
ve
rs
op
y
op
For the case where k = 2, the line and the curve intersect at points A and B.
C
U
i
y
10 A line has equation y = kx + 6 and a curve has equation y = x 2 + 3x + 2 k, where k is a constant.
id
g
ev
[5]
[4]
s
-R
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q9 November 2011
es
am
ie
Find the two values of k for which the line is a tangent to the curve.
br
ii
w
e
Find the distance AB and the coordinates of the mid-point of AB.
-C
C
w
ie
ev
R
[3]
A curve has equation y = x 2 − 5x + 7 and a line has equation y = 2 x − 3.
c
9
[1]
ity
rs
For one value of k, the line intersects the curve at two distinct points, A and B, where the
coordinates of A are ( −2, 13). Find the coordinates of B.
ve
w
ie
R
ev
c
Show that the x-coordinates of the points of intersection of the curve and the line are given by the
equation x 2 − 4x + (5 − k ) = 0.
ni
a
es
A curve has equation y = 5 − 2 x + x 2 and a line has equation y = 2 x + k, where k is a constant.
b
8
[5]
[4]
-C
C
op
y
7
[4]
a Find the coordinates of the vertex of the parabola y = 4x 2 − 12 x + 7.
am
6
w
ge
5
[4]
a Express 9x 2 − 15x in the form (3x − a )2 − b.
ni
ev
ie
w
C
op
y
2
Pr
es
s
-C
-R
The curve and the line intersect at the points A and B. Find the coordinates of the midpoint
of the line AB.
Copyright Material - Review Only - Not for Redistribution
31
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rs
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am
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id
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C
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Cambridge International AS & A Level Mathematics: Pure Mathematics 1
11 A curve has equation y = x 2 − 4x + 4 and a line has the equation y = mx, where m is a constant.
For the case where m = 1, the curve and the line intersect at the points A and B.
Find the non-zero value of m for which the line is a tangent to the curve, and find the coordinates
of the point where the tangent touches the curve.
op
y
ii
ve
rs
ity
C
Express 2 x 2 − 4x + 1 in the form a ( x + b )2 + c and hence state the coordinates of the minimum point, A,
on the curve y = 2 x 2 − 4x + 1.
[4]
y
w
[5]
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q7 June 2013
12 i
ev
ie
[4]
Pr
es
s
-C
Find the coordinates of the mid-point of AB.
-R
i
U
R
ni
C
op
The line x − y + 4 = 0 intersects the curve y = 2 x 2 − 4x + 1 at the points P and Q.
w
ge
It is given that the coordinates of P are (3, 7).
[3]
id
ie
ii Find the coordinates of Q.
iii Find the equation of the line joining Q to the mid-point of AP.
ev
br
ity
op
Pr
y
es
s
-C
-R
am
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q10 June 2011
y
op
y
op
-R
s
es
-C
am
br
ev
ie
id
g
w
e
C
U
R
ev
ie
w
ni
ve
rs
C
ity
Pr
op
y
es
s
-C
-R
am
br
ev
id
ie
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C
U
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ni
ev
ve
ie
w
rs
C
32
[3]
Copyright Material - Review Only - Not for Redistribution
op
y
ve
rs
ity
ni
C
U
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-R
am
br
id
Pr
es
s
-C
y
ni
C
op
y
ve
rs
ity
op
C
w
ev
ie
rs
op
y
ve
w
ge
C
U
ni
ie
w
C
ity
op
Pr
y
es
s
-C
-R
am
br
ev
id
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U
R
ev
R
Chapter 2
Functions
33
id
es
s
-C
-R
am
br
ev
understand the terms function, domain, range, one-one function, inverse function and
composition of functions
identify the range of a given function in simple cases, and find the composition of two given
functions
determine whether or not a given function is one-one, and find the inverse of a one-one
function in simple cases
illustrate in graphical terms the relation between a one-one function and its inverse
understand and use the transformations of the graph y = f( x ) given by y = f( x ) + a,
y = f( x + a ), y = a f( x ), y = f( ax ) and simple combinations of these.
Pr
ity
op
y
ni
ve
rs
-R
s
es
-C
am
br
ev
ie
id
g
w
e
C
U
R
ev
ie
w
C
op
y
■
■
■
■
■
ie
In this chapter you will learn how to:
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
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C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
am
br
id
ev
ie
w
PREREQUISITE KNOWLEDGE
What you should be able to do
IGCSE / O Level Mathematics
Find an output for a given
function.
Pr
es
s
-C
op
2 If f( x ) = 2 x + 1 and
g( x ) = 1 − x, find fg( x ).
Find the inverse of a simple
function.
3 If f( x ) = 5x + 4, find f −1( x ).
Complete the square.
4 Express 2 x 2 − 12 x + 5 in the
form a ( x + b )2 + c.
ni
C
op
y
ve
rs
ity
C
w
IGCSE / O Level Mathematics
Chapter 1
ev
ie
1 If f( x ) = 3x − 2 , find f(4).
Find a composite function.
y
IGCSE / O Level Mathematics
w
ge
U
R
Check your skills
-R
Where it comes from
id
ie
Why do we study functions?
-R
am
br
ev
At IGCSE / O Level, you learnt how to interpret expressions as functions with inputs and
outputs and find simple composite functions and simple inverse functions.
es
Pr
y
w
ve
ni
op
y
Modelling these situations using appropriate functions enables us to make predictions
about real-life situations, such as: How long will it take for the number of bacteria to
exceed 5 billion?
w
ge
C
U
R
ev
ie
●
WEB LINK
ity
●
rs
C
●
the temperature of a hot drink as it cools over time
the height of a valve on a bicycle tyre as the bicycle travels along a horizontal road
the depth of water in a conical container as it is filled from a tap
the number of bacteria present after the start of an experiment.
op
●
34
s
-C
There are many situations in the real world that can be modelled as functions. Some
examples are:
br
ev
id
ie
In this chapter we will develop a deeper understanding of functions and their special
properties.
Try the Thinking
about functions
and Combining
functions resources
on the Underground
Mathematics website.
-R
am
2.1 Definition of a function
es
An alternative name for a function is a mapping.
Pr
op
y
s
-C
A function is a relation that uniquely associates members of one set with members of
another set.
C
ity
A function can be either a one-one function or a many-one function.
ie
w
ni
ve
rs
The function x ֏ x + 2, where x ∈ ℝ is an example of a one-one function.
TIP
y
op
C
y = x+2
-R
x
s
O
es
-C
am
br
ev
ie
id
g
w
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U
R
ev
f(x)
Copyright Material - Review Only - Not for Redistribution
x ∈ ℝ means that x
belongs to the set of
real numbers.
ve
rs
ity
C
U
ni
op
y
Chapter 2: Functions
am
br
id
ev
ie
w
ge
A one-one function has one output value for each input value. Equally important is the
fact that for each output value appearing there is only one input value resulting in this
output value.
-R
We can write this function as f : x ֏ x + 2 for x ∈ ℝ or f( x ) = x + 2 for x ∈ ℝ.
y
Pr
es
s
-C
f : x ֏ x + 2 is read as ‘the function f is such that x is mapped to x + 2’ or ‘f maps x
to x + 2’.
The function x ֏ x 2 , where x ∈ ℝ is a many-one function.
ev
ie
w
C
ve
rs
ity
op
f( x ) is the output value of the function f when the input value is x. For example, when
f( x ) = x + 2, f(5) = 5 + 2 = 7.
y
C
op
y = x2
-R
am
br
ev
id
ie
w
ge
U
R
ni
f (x)
O
s
-C
x
op
Pr
y
es
A many-one function has one output value for each input value but each output value can
have more than one input value.
rs
f : x ֏ x 2 is read as ‘the function f is such that x is mapped to x 2 ’ or ‘f maps x to x 2 ’.
ve
ie
w
C
ity
We can write this function as f : x ֏ x 2 for x ∈ ℝ or f( x ) = x 2 for x ∈ ℝ.
y
op
U
y2 = x
x
Pr
op
y
es
s
-C
-R
ev
O
am
br
id
ie
w
ge
C
R
y
ni
ev
If we now consider the graph of y2 = x :
ni
ve
rs
y
The set of input values for a function is called the domain of the function.
U
op
When defining a function, it is important to also specify its domain.
-R
s
es
am
br
ev
ie
id
g
w
e
C
The set of output values for a function is called the range (or codomain) of the function.
-C
R
ev
ie
w
C
ity
We can see that the input value shown has two output values. This means that this relation
is not a function.
Copyright Material - Review Only - Not for Redistribution
35
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ev
ie
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am
br
id
WORKED EXAMPLE 2.1
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
-R
f( x ) = 5 − 2 x for x ∈ ℝ, −4 < x < 5.
b Sketch the graph of the function f .
op
y
c Write down the range of the function f .
a The domain is −4 < x < 5 .
ve
rs
ity
w
C
Answer
Pr
es
s
-C
a Write down the domain of the function f .
y
ev
ie
b The graph of y = 5 − 2 x is a straight line with gradient −2 and y-intercept 5.
ni
C
op
When x = −4, y = 5 − 2( − 4) = 13
f(x)
Pr
y
op
range
ity
y
id
br
ev
Sketch the graph of the function.
s
-C
-R
am
Find the range of f .
Answer
w
The function f is defined by f( x ) = ( x − 3)2 + 8 for −1 < x < 9.
ie
ge
C
U
WORKED EXAMPLE 2.2
R
op
ni
ev
ve
ie
rs
c The range is −5 < f( x ) < 13.
w
(5, –5)
domain
C
36
x
es
O
s
-C
-R
am
br
ev
id
ie
(–4, 13)
w
ge
U
R
When x = 5, y = 5 − 2(5) = −5
.
The circled part of the expression is a
square so it will always be > 0 .
The smallest value it can be is 0.
This occurs when x = 3.
ni
ve
rs
( x − 3)2 + 8
U
op
y
The minimum value of the expression is 0 + 8 = 8 and this
minimum occurs when x = 3.
ie
id
g
w
e
C
So the function f( x ) = ( x − 3)2 + 8 will have a minimum
point at the point (3, 8).
s
es
am
When x = 9, y = (9 − 3)2 + 8 = 44
-R
br
ev
When x = −1, y = ( −1 − 3)2 + 8 = 24
-C
R
ev
ie
w
C
ity
will be of the form
Pr
op
y
es
f( x ) = ( x − 3)2 + 8 is a positive quadratic function so the graph
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
y
op
y
Pr
es
s
-C
range
-R
am
br
id
ev
ie
(9, 44)
(–1, 24)
ve
rs
ity
(3, 8)
y
domain
U
R
ni
The range is 8 < f( x ) < 44.
x
C
op
C
O
w
ev
ie
w
ge
C
U
ni
op
y
Chapter 2: Functions
id
ie
w
ge
EXERCISE 2A
1 Which of these graphs represent functions? If the graph represents a function,
state whether it is a one-one function or a many-one function.
ev
for x ∈ ℝ
d
y = 2x
for x ∈ ℝ
s
for x ∈ ℝ, x > 0
x
y = x2 − 3
es
y
Pr
y=
ity
g
C
op
e
y = 2 x 3 − 1 for x ∈ ℝ
10
y=
for x ∈ ℝ, x > 0
x
b
-R
br
am
-C
a y = 2 x − 3 for x ∈ ℝ
c
TIP
x ∈ ℝ , x > 0 is
sometimes shortened
to just x > 0.
f
y = 3x 2 + 4 for x ∈ ℝ, x > 0
h
y 2 = 4x
37
for x ∈ ℝ
w
rs
2 a Represent on a graph the function:
y
op
U
R
ni
ev
ve
ie
2
 9 − x for x ∈ ℝ, −3 < x < 2
x֏
 2 x + 1 for x ∈ ℝ, 2 < x < 4
w
ge
C
b State the nature of the function.
id
ie
3 a Represent on a graph the relation:
-R
am
br
ev
2
 x + 1 for 0 < x < 2
y=
 2 x − 3 for 2 < x < 4
es
s
-C
b Explain why this relation is not a function.
Pr
y
(1, 8)
b
y
(–2, 20)
ity
a
C
op
y
4 State the domain and range for the functions represented by these two graphs.
U
y
(–3, 9)
x
O
-R
(1, –7)
s
es
am
-C
(2, 4)
x
w
ie
ev
(5, –8)
br
id
g
e
R
O
op
ev
(–1, 4)
y = 2x3 + 3x2 – 12x
C
ie
w
ni
ve
rs
y = 7 + 2x – x2
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
w
ge
5 Find the range for each of these functions.
b f( x ) = 2 x − 7 for −3 < x < 2
a
f( x ) = x + 4
c
f( x ) = 7 − 2 x for −1 < x < 4
e
f( x ) = 2 x
am
br
id
ev
ie
for x > 8
d f : x ֏ 2x2
12
f f( x ) =
x
b
f( x ) = 3 − 2 x 2 for x < 2
f : x ֏ x 2 + 3 for −2 < x < 5
d f( x ) = 7 − 3x 2 for −1 < x < 2
ve
rs
ity
y
op
c
f : x ֏ 8 − ( x − 5)2 for 4 < x < 10
1
2
for x > 4
b f( x ) = (2 x − 1)2 − 7 for x >
for x > 2
d
f( x ) = 1 + x − 4
2
y
f( x ) = ( x − 2)2 + 5
C
op
a
ni
C
f( x ) = x 2 − 2 for x ∈ ℝ
7 Find the range for each of these functions.
w
ev
ie
c
Pr
es
s
-C
6 Find the range for each of these functions.
a
for 1 < x < 8
-R
for −5 < x < 4
for 1 < x < 4
w
b f( x ) = 3x 2 − 10 x + 2 for x ∈ ℝ
ev
id
ie
f( x ) = x 2 + 6x − 11 for x ∈ ℝ
br
a
ge
U
R
8 Express each function in the form a ( x + b ) + c , where a, b and c are constants
and, hence, state the range of each function.
-R
f( x ) = 7 − 8x − x 2 for x ∈ ℝ
b f( x ) = 2 − 6x − 3x 2 for x ∈ ℝ
es
s
-C
a
am
9 Express each function in the form a − b( x + c )2 , where a, b and c are constants
and, hence, state the range of each function.
C
2
for
 3 − x
f( x ) = 
 3x − 7 for
0<x<2
Pr
38
2<x<4
ve
ie
w
rs
b Find the range of the function.
ity
op
y
10 a Represent, on a graph, the function:
y
op
ni
ev
11 The function f : x ֏ x 2 + 6x + k , where k is a constant, is defined for x ∈ ℝ.
C
U
R
Find the range of f in terms of k.
w
ge
12 The function g : x ֏ 5 − ax − 2 x 2 , where a is a constant, is defined for x ∈ ℝ.
br
ev
id
ie
Find the range of g in terms of a.
-R
am
13 f( x ) = x 2 − 2 x − 3 for x ∈ ℝ, − a < x < a
s
14 f( x ) = x 2 + x − 4 for x ∈ ℝ , a < x < a + 3
es
-C
If the range of the function f is −4 < f( x ) < 5, find the value of a.
Pr
ity
15 f( x ) = 2 x 2 − 8x + 5 for x ∈ ℝ, 0 < x < k
ni
ve
rs
a Express f( x ) in the form a ( x + b )2 + c .
op
For your value of k from part b, find the range of f .
C
U
c
y
b State the value of k for which the graph of y = f( x ) has a line of symmetry.
w
f( x ) = 2 x
ie
ev
c
f
f( x ) =
x−3−2
-R
br
am
b f( x ) = x 2 + 2
1
e f( x ) =
x−2
es
f( x ) = 3x − 1
1
d f( x ) =
x
a
s
id
g
e
16 Find the largest possible domain for each function and state the corresponding
range.
-C
R
ev
ie
w
C
op
y
If the range of the function f is −2 < f( x ) < 16, find the possible values of a.
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
C
U
ni
op
y
Chapter 2: Functions
w
ge
2.2 Composite functions
am
br
id
ev
ie
Most functions that we meet can be described as combinations of two or more functions.
(the function ‘multiply by 3’)
f :x֏x−7
(the function ‘subtract 7’)
y
Pr
es
s
-C
g : x ֏ 3x
-R
For example, the function x ֏ 3x − 7 is the function ‘multiply by 3 and then subtract 7’.
It is a combination of the two functions g and f, where:
x
f
fg(x)
U
R
ni
g(x)
y
ev
ie
w
C
ve
rs
ity
g
C
op
op
So, x ֏ 3x − 7 can be described as the function ‘first do g, then do f ’.
w
ge
fg
-R
am
br
ev
id
ie
When one function is followed by another function, the resulting function is called a
composite function.
s
-C
KEY POINT 2.1
op
Pr
y
es
fg( x ) means the function g acts on x first, then f acts on the result.
39
C
ity
There are three important points to remember about composite functions:
rs
ve
ie
w
KEY POINT 2.2
y
C
U
In general, fg( x ) ≠ gf( x ).
R
op
ni
ev
fg only exists if the range of g is contained within the domain of f .
br
-R
am
EXPLORE 2.1
ev
id
ie
w
ge
ff( x ) means you apply the function f twice.
g( x ) = 3x − 1 for x ∈ ℝ
es
s
-C
f( x ) = 2 x − 5 for x ∈ ℝ
ity
Student B
Student C
gf( x ) = 2(3x − 1) − 5
= 6x − 7
gf( x ) = 3(2x − 5) − 1
= 6x − 16
gf( x ) = (3x − 1)(2x − 5)
U
= 6x 2 − 17x + 5
op
ni
ve
rs
Student A
e
C
Discuss these solutions with your classmates.
-R
s
es
am
br
ev
ie
id
g
w
Which student is correct? What error has each of the other students made?
-C
R
ev
ie
w
C
Here are their solutions.
y
Pr
op
y
Three students are asked to find the composite function gf( x ).
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
y
Answer
ve
rs
ity
w
f is the function ‘subtract 4, square and then
subtract 1’.
ie
br
ev
id
WORKED EXAMPLE 2.4
w
ge
U
R
ni
= 1 14
g( x ) = x 2 − 1 for x ∈ ℝ
b
c
gf( x )
es
y
Answer
= 2( x − 1) + 3
ve
= (2 x + 3) − 1
f acts on x first and f( x ) = 2 x + 3 .
ni
g is the function ‘square and subtract 1’.
U
R
2
y
gf( x ) = g(2 x + 3)
ev
b
f is the function ‘double and add 3’.
rs
= 2x2 + 1
g acts on x first and g( x ) = x 2 − 1.
op
C
2
w
ie
Pr
fg( x ) = f( x 2 − 1)
ity
op
a
40
ff( x )
s
-C
fg( x )
-R
a
am
f( x ) = 2 x + 3 for x ∈ ℝ
Find:
ge
C
= 4x 2 + 12 x + 9 − 1
id
ie
w
= 4x 2 + 12 x + 8
ff( x ) = f(2 x + 3)
ev
f is the function ‘double and add 3’.
br
c
es
s
-C
-R
am
= 2(2 x + 3) + 3
= 4x + 9
Pr
ity
y
ff( x )
U
op
b
fg( x )
Answer
g acts on x first and g( x ) = 3 − x 2 .
ie
id
g
w
e
fg( x ) = f(3 − x 2 )
5
=
(3 − x 2 ) − 2
5
=
1 − x2
ev
f is the function ‘subtract 2 and then divide into 5’.
es
s
-R
br
am
-C
a
g( x ) = 3 − x 2 for x ∈ ℝ
C
w
ie
ev
R
a
Find:
5
for x ∈ ℝ, x ≠ 2
x−2
ni
ve
rs
C
op
y
WORKED EXAMPLE 2.5
f :x֏
2(4) + 3 11
= .
4−2
2
y
2
11
=
− 4 − 1
 2

g acts on 4 first and g(4) =
C
op
C
op
11
fg(4) = f  
 2 
ev
ie
2x + 3
for x ∈ ℝ, x > 2
x−2
Pr
es
s
-C
Find fg(4).
g( x ) =
-R
f( x ) = ( x − 4)2 − 1 for x ∈ ℝ
ev
ie
am
br
id
WORKED EXAMPLE 2.3
w
ge
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
ge
ev
ie
w
5 
ff ( x ) = f 
 x−2
am
br
id
-R
=
5x − 10
9 − 2x
g( x ) = 3x − 1 for x ∈ ℝ
ge
f( x ) = x 2 + 4x for x ∈ ℝ
U
R
ni
y
ev
ie
WORKED EXAMPLE 2.6
C
op
5( x − 2)
5 − 2( x − 2)
Pr
es
s
op
Multiply numerator and denominator by ( x − 2).
w
C
5
5
−2
x−2
=
y
-C
=
ve
rs
ity
b
C
U
ni
op
y
Chapter 2: Functions
br
Answer
ev
id
ie
w
Find the values of k for which the equation fg( x ) = k has real solutions.
am
fg( x ) = (3x − 1)2 + 4(3x − 1)
-R
Expand brackets and simplify.
= 9x + 6x − 3
When fg( x ) = k,
s
es
ity
41
b2 − 4ac > 0
For real solutions:
w
rs
C
9x 2 + 6x + ( −3 − k ) = 0
y
op
k > −4
a
b
fg(6)
gf(4)
ff( −3)
c
x ֏ x + 10
x for x ∈ ℝ, x . 0
k:x֏
es
c
x֏
x +5
b
x+5
x֏
ni
ve
rs
3 f( x ) = ax + b for x ∈ ℝ
ity
a
Pr
Express each of the following in terms of h and/or k.
op
y
Given that f(5) = 3 and f(3) = −3:
C
U
a find the value of a and the value of b
w
-R
s
Solve the equation gf( x ) = 2 .
es
am
a Find gf( x ).
b
12
for x ∈ ℝ, x ≠ 1
1− x
ie
g:x֏
br
4 f : x ֏ 2 x + 3 for x ∈ ℝ
ev
id
g
e
b solve the equation ff( x ) = 4.
-C
ie
w
C
op
y
2 h : x ֏ x + 5 for x ∈ ℝ, x . 0
ev
x + 3 − 2 for x ∈ ℝ, x > −3
s
-C
Find:
g( x ) =
-R
am
1 f( x ) = x 2 + 6 for x ∈ ℝ
ev
br
id
ie
w
ge
C
U
R
ni
ev
ve
ie
62 − 4 × 9 × ( −3 − k ) > 0
144 + 36 k > 0
EXERCISE 2B
R
Rearrange and simplify.
Pr
9x 2 + 6x − 3 = k
op
y
-C
2
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
5 g( x ) = x 2 − 2 for x ∈ ℝ
ev
ie
w
ge
am
br
id
a Find gh( x ).
h( x ) = 2 x + 5 for x ∈ ℝ
g( x ) =
f( x ) = x 2 + 1 for x ∈ ℝ
-C
Pr
es
s
Solve the equation fg( x ) = 5 .
y
2
for x ∈ ℝ, x ≠ −1
x +1
Solve the equation hg( x ) = 11.
h( x ) = ( x + 2)2 − 5 for x ∈ ℝ
x +1
for x ∈ ℝ
2
Solve the equation gf( x ) = 1.
2x + 3
for x ∈ ℝ, x ≠ 1
x −1
ni
ev
ie
g:x֏
y
ve
rs
ity
w
C
op
7 g( x ) =
8 f:x֏
3
for x ∈ ℝ, x ≠ 2
x−2
C
op
6
-R
b Solve the equation gh( x ) = 14 .
R
x +1
for x ∈ ℝ, x . 0
2x + 5
Find an expression for ff( x ) , giving your answer as a single fraction in its simplest form.
id
ie
w
ge
U
9 f( x ) =
10 f : x ֏ x 2 for x ∈ ℝ
br
ev
g : x ֏ x + 1 for x ∈ ℝ
b
x ֏ x2 + 1
d
x ֏ x4
e
x ֏ x2 + 2x + 2
s
es
x ֏ x 4 + 2x2 + 1
ity
rs
U
13 f( x ) = x 2 − 3x for x ∈ ℝ
w
ge
C
g( x ) = 2 x − 5 for x ∈ ℝ
op
ni
ve
g( x ) =
y
C
w
ie
f
g( x ) = 2 x + 5 for x ∈ ℝ
2
for x ∈ ℝ, x ≠ 0
x
Find the values of k for which the equation fg( x ) = x has two equal roots.
ev
x֏x+2
Show that the equation gf( x ) = 0 has no real solutions.
12 f( x ) = k − 2 x for x ∈ ℝ
R
c
Pr
y
op
11 f( x ) = x 2 − 3x for x ∈ ℝ
42
-R
x ֏ ( x + 1)2
-C
a
am
Express each of the following as a composite function, using only f and/or g.
id
ie
Find the values of k for which the equation gf( x ) = k has real solutions.
-C
-R
am
br
ev
x+5
1
for x ∈ ℝ, x ≠
2x − 1
2
Show that ff( x ) = x.
14 f( x ) =
es
s
15 f( x ) = 2 x 2 + 4x − 8 for x ∈ ℝ, x > k
Pr
op
y
a Express 2 x 2 + 4x − 8 in the form a ( x + b )2 + c.
ity
ni
ve
rs
16 f( x ) = x 2 − 2 x + 4 for x ∈ ℝ
op
y
a Find the set of values of x for which f( x ) > 7 .
C
Write down the range of f .
w
ie
g( x ) = 2 x + 3 for x ∈ ℝ
br
17 f( x ) = x 2 − 5x for x ∈ ℝ
ev
id
g
e
c
U
b Express x 2 − 2 x + 4 in the form ( x − a )2 + b.
-R
s
b Find the range of the function fg( x ).
es
am
a Find fg( x ).
-C
R
ev
ie
w
C
b Find the least value of k for which the function is one-one.
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
am
br
id
Pr
es
s
1
for x ∈ ℝ, x ≠ 0
x
S( x ) =
R( x ) =
C
w
P( x ) = x 2 − 1 for x ∈ ℝ
w
Q( x ) = x + 2 for x ∈ ℝ
ve
rs
ity
y
19
op
PS
Find the values of x for which f( x ) = ff( x ).
-C
c
-R
b Show that if f( x ) = ff( x ) then x 2 + x − 2 = 0.
ev
ie
ge
2
for x ∈ ℝ , x ≠ −1
x +1
a Find ff( x ) and state the domain of this function.
18 f( x ) =
C
U
ni
op
y
Chapter 2: Functions
x + 1 − 1 for x ∈ ℝ, x > −1
y
ev
ie
Functions P, Q, R and S are composed in some way to make a new function, f( x ).
-R
am
br
ev
id
ie
w
ge
U
R
ni
C
op
For each of the following, write f( x ) in terms of the functions P, Q, R and/or S, and state the domain and
range for each composite function.
1
b f( x ) = x 2 + 1
c f( x ) = x
d f( x ) = 2 + 1
a f( x ) = x 2 + 4x + 3
x
1
f f( x ) = x − 2 x + 1 + 1
g f( x ) = x − 1
e f( x ) =
x+4
s
-C
2.3 Inverse functions
The inverse of a function f( x ) is the function that undoes what f( x ) has done.
es
ity
43
x
rs
KEY POINT 2.3
ff −1( x ) = f −1f( x ) = x
ve
ie
w
C
op
Pr
y
We write the inverse of the function f( x ) as f −1( x ) .
y
The range of f −1( x ) is the domain of f( x ).
C
U
w
f –1(x)
br
ev
ie
ge
id
It is important to remember that not every function has an inverse.
-R
am
KEY POINT 2.4
es
s
-C
An inverse function f −1( x ) exists if, and only if, the function f( x ) is a one-one mapping.
We want to find the function f −1( x ), so if we write y = f −1( x ), then f( y ) = f(f −1( x )) = x,
because f and f −1 are inverse functions. So if we write x = f( y ) and then rearrange it to get
y = … , then the right-hand side will be f −1( x ).
-R
s
es
-C
am
br
ev
ie
id
g
w
e
C
U
op
y
ni
ve
rs
ev
ie
w
C
ity
Pr
op
y
You should already know how to find the inverse function of some simple one-one
mappings.
R
y
op
ni
ev
The domain of f −1( x ) is the range of f( x ).
R
f(x)
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
x = 3y − 1
Step 3: Rearrange to make y the subject.
y=
Pr
es
s
-C
x +1
3
x +1
.
3
If f and f −1 are the same function, then f is called a self-inverse function.
1
1
for x ≠ 0 , then f −1( x ) =
for x ≠ 0 .
For example, if f( x ) =
x
x
1
for x ≠ 0 is a self-inverse function.
So f( x ) =
x
y
C
op
R
ni
w
C
ve
rs
ity
op
y
Hence, if f( x ) = 3x − 1, then f −1( x ) =
ev
ie
w
Step 2: Interchange the x and y variables.
-R
y = 3x − 1
am
br
id
Step 1: Write the function as y =
ev
ie
ge
We find the inverse of the function f( x ) = 3x − 1 by following these steps:
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
ie
f(x)
ev
f(x) = (x – 2)2 + 1
-R
am
br
id
The diagram shows the function f( x ) = ( x − 2)2 + 1 for x ∈ ℝ .
Discuss the following questions with your classmates.
w
ge
U
EXPLORE 2.2
-C
1 What type of mapping is this function?
es
s
2 What are the coordinates of the vertex of the parabola?
ve
y
op
br
ev
id
ie
w
ge
C
U
R
ni
x + 2 − 7 for x ∈ ℝ, x > −2
-R
am
f( x ) =
b Solve the equation f −1( x ) = f(62) .
s
-C
a Find an expression for f −1( x ) .
es
f( x ) =
x+2 −7
Step 1: Write the function as y =
x+2 −7
Step 2: Interchange the x and y variables.
x=
y+2 −7
Step 3: Rearrange to make y the subject.
x+7 =
U
( x + 7) = y + 2
e
2
id
g
w
y = ( x + 7)2 − 2
-R
s
es
am
br
ev
ie
f −1( x ) = ( x + 7)2 − 2
-C
C
R
y+2
op
ity
y=
ni
ve
rs
ie
w
C
a
Pr
op
y
Answer
y
ev
6 If f has an inverse, what is it? If not, then how could you change the domain of f so that the function does have
an inverse?
WORKED EXAMPLE 2.7
ev
x
rs
C
w
ie
5 Does this function have an inverse?
O
ity
4 What is the range of the function?
Pr
op
y
3 What is the domain of the function?
44
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
ge
w
62 + 2 − 7 = 1
ev
ie
f(62) =
am
br
id
b
C
U
ni
op
y
Chapter 2: Functions
( x + 7)2 − 2 = 1
-R
( x + 7)2 = 3
-C
x+7 = ± 3
ni
C
op
y
Hence, the only solution of f −1( x ) = f(62) is x = −7 + 3 .
w
ev
ie
The range of f is f( x ) > −7 so the domain of f −1 is x > −7.
ve
rs
ity
C
op
y
x = −7 − 3 or x = −7 + 3
ge
U
WORKED EXAMPLE 2.8
R
Pr
es
s
x = −7 ± 3
ie
id
w
f( x ) = 5 − ( x − 2)2 for x ∈ ℝ, k < x < 6
br
ev
a State the smallest value of k for which f has an inverse.
-C
-R
am
b For this value of k find an expression for f −1( x ) , and state the domain and range of f −1 .
y
(2, 5)
es
s
Answer
Pr
When x = 6, y = 5 − 42 = −11
y = 5 – (x – 2)2
ity
C
op
y
a The vertex of the graph of y = 5 − ( x − 2)2 is at the point (2, 5).
For the function f to have an inverse it must be a one-one function.
rs
w
ie
op
y
ve
ni
f( x ) = 5 − ( x − 2)2
C
U
ev
R
b
y = 5 − ( x − 2)2
ie
w
ge
id
am
br
ev
Step 2: Interchange the x and y variables.
-R
s
-C
y−2 = 5−x
Pr
es
y =2+ 5−x
Hence, f −1( x ) = 2 + 5 − x .
ity
op
y
ni
ve
rs
The domain of f −1 is the same as the range of f .
y
Hence, the domain of f −1 is −11 < x < 5 .
U
op
The range of f −1 is the same as the domain of f .
-R
s
es
am
br
ev
ie
id
g
w
e
C
Hence, the range of f −1 is 2 < f −1 ( x ) < 6.
-C
C
w
ie
x = 5 − ( y − 2)2
( y − 2)2 = 5 − x
Step 3: Rearrange to make y the subject.
ev
x
Hence, the smallest value of k is 2.
Step 1: Write the function as y =
R
O
Copyright Material - Review Only - Not for Redistribution
(6, –11)
45
ve
rs
ity
ev
ie
am
br
id
EXERCISE 2C
w
ge
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
c
f( x ) = ( x − 5)2 + 3 for x ∈ ℝ , x > 5
e
f( x ) =
y
f( x ) = ( x − 2)3 − 1 for x ∈ ℝ , x > 2
f
y
C
op
w
ge
R
ni
5
for x ∈ ℝ, x > 2
2x + 1
a Find an expression for f −1( x ).
f:x֏
U
ev
ie
b Find an expression for f −1( x ) .
3
br
ev
id
ie
b Find the domain of f −1.
f : x ֏ ( x + 1)3 − 4 for x ∈ ℝ, x > 0
-R
am
4
s
-C
a Find an expression for f −1( x ) .
C
g : x ֏ 2 x 2 − 8x + 10 for x ∈ ℝ, x > 3
w
rs
a Explain why g has an inverse.
Pr
5
ity
op
y
es
b Find the domain of f −1.
46
d
ve
rs
ity
op
f : x ֏ x 2 + 4x for x ∈ ℝ, x > −2
f( x ) = x 2 + 3 for x ∈ ℝ , x > 0
8
f( x ) =
for x ∈ ℝ , x ≠ 3
x−3
b
a State the domain and range of f −1.
w
C
2
x+7
for x ∈ ℝ , x ≠ −2
x+2
Pr
es
s
f( x ) = 5x − 8 for x ∈ ℝ
-C
a
-R
1 Find an expression for f −1( x ) for each of the following functions.
y
op
f : x ֏ 2 x 2 + 12 x − 14 for x ∈ ℝ, x > k
C
U
R
6
ni
ev
ve
ie
b Find an expression for g −1( x ).
ge
a Find the least value of k for which f is one-one.
br
f : x ֏ x 2 − 6x for x ∈ ℝ
-R
am
7
ev
id
ie
w
b Find an expression for f −1( x ).
-C
a Find the range of f.
Pr
op
y
f( x ) = 9 − ( x − 3)2 for x ∈ ℝ, k < x < 7
find an expression for f −1( x )
ii
state the domain and range of f −1.
-R
s
es
-C
am
br
ev
ie
id
g
w
e
C
U
op
i
y
ni
ve
rs
b For this value of k:
ity
a State the smallest value of k for which f has an inverse.
R
ev
ie
w
C
8
es
s
b State, with a reason, whether f has an inverse.
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
y
ge
Pr
es
s
-C
ve
rs
ity
y
op
C
5x − 1
for x ∈ ℝ, 0 , x < 3.
x
b State the domain of f.
g( x ) = b − 5x for x ∈ ℝ
y
w
The diagram shows the graph of y = f −1( x ) , where f −1( x ) =
10 f( x ) = 3x + a for x ∈ ℝ
ev
ie
x
O
a Find an expression for f( x ).
5x – 1
x
ev
ie
am
br
id
w
y=
-R
9
C
U
ni
op
y
Chapter 2: Functions
U
R
ni
C
op
Given that gf( −1) = 2 and g −1(7) = 1, find the value of a and the value of b.
11 f( x ) = 3x − 1 for x ∈ ℝ
3
for x ∈ ℝ, x ≠ 2
2x − 4
w
ge
g( x ) =
id
ie
a Find expressions for f −1( x ) and g −1( x ).
-R
am
br
ev
b Show that the equation f −1( x ) = g −1( x ) has two real roots.
12 f : x ֏ (2 x − 1)3 − 3 for x ∈ ℝ, 1 < x < 3
s
es
47
ity
C
13 f : x ֏ x 2 − 10 x for x ∈ ℝ, x > 5
Pr
b Find the domain of f −1.
op
y
-C
a Find an expression for f −1( x ) .
w
rs
a Express f( x ) in the form ( x − a )2 − b.
y
ve
ie
b Find an expression for f −1( x ) and state the domain of f −1.
ni
w
ge
C
U
R
op
ev
1
for x ∈ ℝ , x ≠ 1
x −1
a Find an expression for f −1( x ).
14 f( x ) =
es
s
-C
15 Determine which of the following functions are self-inverse functions.
1
2x + 1
a f( x ) =
for x ∈ ℝ , x ≠ 3
b f( x ) =
for x ∈ ℝ , x ≠ 2
x−2
3−x
3x + 5
3
for x ∈ ℝ , x ≠
c f( x ) =
4x − 3
4
g : x ֏ 4 − 2 x for x ∈ ℝ
ni
ve
rs
16 f : x ֏ 3x − 5 for x ∈ ℝ
ity
Pr
op
y
op
y
a Find an expression for (fg) −1( x ).
g −1 f −1( x ) .
C
ii
w
Comment on your results in part b.
ie
c
f −1 g −1( x )
e
i
U
R
b Find expressions for:
id
g
-R
s
-C
am
br
ev
Investigate if this is true for other functions.
es
C
w
ie
ie
-R
am
Give your answer in surd form.
ev
ev
Find the values of x for which f( x ) = f −1( x ) .
br
c
id
b Show that if f( x ) = f −1( x ) , then x 2 − x − 1 = 0 .
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
ge
2.4 The graph of a function and its inverse
am
br
id
ev
ie
Consider the function defined by f( x ) = 2 x + 1 for x ∈ ℝ , −4 < x < 2 .
w
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
-R
f( −4) = −7 and f(2) = 5.
y
op
y
Pr
es
s
-C
The domain of f is −4 < x < 2 and the range is −7 < f( x ) < 5.
x −1
.
The inverse of this function is f −1( x ) =
2
The domain of f −1 is the same as the range of f.
O
x
C
f–1
(–7, –4)
f
C
op
y
Hence, the range of f −1 is −4 < f −1( x ) < 2.
ni
ev
ie
w
The range of f −1 is the same as the domain of f.
The representation of f and f −1 on the same graph can be seen in the diagram opposite.
U
R
y=x
(5, 2)
ve
rs
ity
Hence, the domain of f −1 is −7 < x < 5.
(2, 5)
(–4, –7)
br
ev
id
ie
w
ge
It is important to note that the graphs of f and f −1 are reflections of each other in the line
y = x. This is true for each one-one function and its inverse functions.
-R
am
KEY POINT 2.5
-C
The graphs of f and f −1 are reflections of each other in the line y = x.
es
s
This is because ff −1( x ) = x = f −1 f( x )
C
ity
op
Pr
y
When a function f is self-inverse, the graph of f will be symmetrical about the line y = x.
48
ve
ie
w
rs
WORKED EXAMPLE 2.9
y
op
ni
ev
f( x ) = ( x − 1)2 − 2 for x ∈ ℝ, 1 < x < 4
ie
ev
es
s
-C
The function is one-one, so the inverse function exists.
Pr
f
f
Reflect f in y = x
2
C
8 x
ie
6
ev
4
O
–2
s
es
am
–2
–2
-R
br
2
w
e
id
g
O
-C
–2
f –1
4
U
2
6
y
4
y=x
op
w
ni
ve
rs
6
ie
y
8
ity
C
op
y
y
8
ev
-R
am
When x = 4, y = 7.
R
The circled part of the expression is a square so it
will always be > 0. The smallest value it can be is 0.
This occurs when x = 1. The vertex is at the
point (1, −2) .
br
y = ( x − 1)2 − 2
id
Answer
w
ge
C
U
R
On the same axes, draw the graph of f and the graph of f −1.
Copyright Material - Review Only - Not for Redistribution
2
4
6
8 x
ve
rs
ity
ge
C
U
ni
op
y
Chapter 2: Functions
am
br
id
ev
ie
w
WORKED EXAMPLE 2.10
-C
-R
2x + 7
for x ∈ ℝ, x ≠ 2
x−2
a Find an expression for f −1( x ) .
f:x֏
f :x֏
2x + 7
x−2
U
R
ni
ev
ie
Step 1: Write the function as y =
ie
-R
s
Pr
es
2x + 7
.
x−2
49
ity
rs
y
C
w
y
id
g–1
y=x
-R
x
x
ity
O
Pr
op
y
Ali states that:
-R
s
es
am
br
ev
ie
id
g
w
e
C
U
Explain your answer.
op
Is Ali correct?
y
ni
ve
rs
The diagrams show the functions f and g, together with their inverse functions f −1 and g−1.
-C
C
w
s
f–1
O
ie
g
es
-C
am
br
ev
y=x
ie
ge
U
EXPLORE 2.3
R
op
ni
ev
ve
ie
w
The graph of y = f( x ) is symmetrical about the line y = x.
f
ev
y( x − 2) = 2 x + 7
2x + 7
y=
x−2
f −1( x ) = f( x ) , so the function f is self-inverse.
y
R
2y + 7
y−2
ev
id
br
am
-C
y
op
C
b
x=
xy − 2 x = 2 y + 7
Step 3: Rearrange to make y the subject.
Hence f −1( x ) =
2x + 7
x−2
w
ge
Step 2: Interchange the x and y variables.
y=
y
w
C
a
ve
rs
ity
op
Answer
C
op
y
Pr
es
s
b State what your answer to part a tells you about the symmetry of the graph of y = f( x ).
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
ev
ie
am
br
id
EXERCISE 2D
w
ge
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
1 On a copy of each grid, draw the graph of f −1( x ) if it exists.
ev
ie
w
–2
U
ev
id
5
-R
1
1
2
3
4
6
O
x
6
x
w
ie
ev
-R
s
y
Pr
es
4
for x ∈ ℝ, x > 0.
x+2
ity
State the domain and range of f −1.
f
O
y
ni
ve
rs
d On a copy of the diagram, sketch the graph of y = f −1( x ) , making clear the
relationship between the graphs.
ie
id
g
3
3x − 1
for x ∈ ℝ, x ≠
2x − 3
2
ev
d
f( x ) =
4
4x + 5
for x ∈ ℝ, x ≠
3x − 4
3
es
s
-R
br
f( x ) =
am
c
w
e
C
U
op
4 For each of the following functions, find an expression for f −1( x ) and, hence, decide if the graph of y = f( x )
is symmetrical about the line y = x.
x+5
2x − 3
1
b f( x ) =
for x ∈ ℝ, x ≠
for x ∈ ℝ, x ≠ 5
a f( x ) =
2x − 1
x−5
2
-C
w
C
op
y
b Find an expression for f −1( x ) .
ie
5
C
ge
id
br
am
-C
a State the range of f .
ev
4
Sketch, on the same diagram, the graphs of y = f( x ) and y = f −1( x ) , making clear the relationship
between the graphs.
3 The diagram shows the graph of y = f( x ) , where f( x ) =
c
3
y
ni
U
a Find an expression for f −1( x ).
c
2
op
f : x ֏ 2 x − 1 for x ∈ ℝ, −1 < x < 3
b State the domain and range of f −1.
R
1
ve
5
rs
w
2
ity
1
f
3
s
2
ie
ev
R
2
4
es
f
O
y
6
C
50
d
Pr
op
y
3
6 x
4
–6
br
-C
4
2
–4
am
5
O
–2
ie
y
6
–4
–2
ge
R
–6
c
–6
ni
–4
6 x
4
y
2
C
op
O
–2
4
2
ve
rs
ity
–4
f
Pr
es
s
f
2
y
6
b
w
-C
4
y
op
C
–6
-R
y
6
a
Copyright Material - Review Only - Not for Redistribution
x
ve
rs
ity
w
ev
ie
f( x ) =
ge
x+a
1
for x ∈ ℝ, x ≠ , where a and b are constants.
bx − 1
b
Prove that this function is self-inverse.
ax + b
d
for x ∈ ℝ, x ≠ − , where a, b, c and d are constants.
b g( x ) =
cx + d
c
Find the condition for this function to be self-inverse.
5 a
Pr
es
s
-C
-R
am
br
id
P
C
U
ni
op
y
Chapter 2: Functions
op
y
2.5 Transformations of functions
y
w
ge
U
EXPLORE 2.4
R
C
op
ni
ev
ie
w
C
ve
rs
ity
At IGCSE / O Level you met various transformations that can be applied to two-dimensional
shapes. These included translations, reflections, rotations and enlargements. In this section
you will learn how translations, reflections and stretches (and combinations of these) can be
used to transform the graph of a function.
id
ie
1 a Use graphing software to draw the graphs of y = x 2, y = x 2 + 2 and y = x 2 − 3.
-R
12
12
12
,y=
+ 5 and y =
− 4.
x
x
x
es
y
Pr
d Can you generalise your results?
op
x − 2.
s
x, y =
c Repeat part a using the graphs y =
ity
2 a Use graphing software to draw the graphs of y = x 2, y = ( x + 2)2 and y = ( x − 5)2 .
Discuss your observations with your classmates and explain how the second and third graphs could be
obtained from the first graph.
op
ni
ev
b Repeat part a using the graphs y = x 3, y = ( x + 1)3 and y = ( x − 4)3 .
c Can you generalise your results?
C
U
R
y
ve
ie
w
rs
C
x + 1 and y =
b Repeat part a using the graphs y =
-C
am
br
ev
Discuss your observations with your classmates and explain how the second and third graphs could be
obtained from the first graph.
w
ge
3 a Use graphing software to draw the graphs of y = x 2 and y = − x 2 .
-R
am
b Repeat part a using the graphs y = x 3 and y = − x 3.
ev
br
id
ie
Discuss your observations with your classmates and explain how the second graph could be obtained from
the first graph.
-C
c Repeat part a using the graphs y = 2 x and y = −2 x .
es
s
d Can you generalise your results?
Pr
ity
Discuss your observations with your classmates and explain how the second graph could be obtained from
the first graph.
ni
ve
rs
b Repeat part a using the graphs y =
2 + x and y =
2 − x.
c Can you generalise your results?
op
y
5 a Use graphing software to draw the graphs of y = x 2 and y = 2 x 2 and y = (2 x )2.
C
x , y = 2 x and y =
w
id
g
b Repeat part a using the graphs y =
2x .
ie
e
U
Discuss your observations with your classmates and explain how the second graph could be obtained from
the first graph.
c Repeat part a using the graphs y = 3 , y = 2 × 3 and y = 32 x .
x
ev
x
es
s
-R
br
am
d Can you generalise your results?
-C
R
ev
ie
w
C
op
y
4 a Use graphing software to draw the graphs of y = 5 + x and y = 5 − x.
Copyright Material - Review Only - Not for Redistribution
51
ve
rs
ity
y
w
ge
Translations
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
12
am
br
id
ev
ie
The diagram shows the graphs of two functions that differ only by a constant.
y = x2 – 2x + 4
10
-R
y = x − 2x + 1
2
-C
y = x2 − 2x + 4
8
op
y
Pr
es
s
When the x-coordinates on the two graphs are the same ( x = x ) the
y-coordinates differ by 3 ( y = y + 3).
y = x2 – 2x + 1
6
4
Hence, the graph of y = x 2 − 2 x + 4 is a translation of the graph of y = x 2 − 2 x + 1
w
ie
s
-C
-R
am
br
ev
 0
The graph of y = f( x ) + a is a translation of the graph y = f( x ) by the vector   .
 a
es
Now consider the two functions:
op
Pr
y
y = x2 − 2x + 1
52
C
ity
y = ( x − 3)2 − 2( x − 3) + 1
ve
ni
op
y
The graphs of these two functions are:
ev
4
-R
am
br
6
es
s
-C
3
Pr
4
x
U
op
y
The curves have exactly the same shape but this time they are separated by 3 units in the
positive x-direction.
-R
s
es
am
br
ev
ie
id
g
w
e
C
You may be surprised that the curve has moved in the positive x-direction. Note, however,
that a way of obtaining y = y is to have x = x − 3 or equivalently x = x + 3. This means
that the two curves are at the same height when the red curve is 3 units to the right of the
blue curve.
-C
ie
ev
R
6
ni
ve
rs
2
ity
O
w
C
op
y
2
–2
ie
y = (x – 3)2 – 2 (x – 3) + 1
id
y = x2 – 2x + 1
w
ge
8
C
U
R
ev
ie
w
rs
We obtain the second function by replacing x by x − 3 in the first function.
y
3
y
–2
id
KEY POINT 2.6
2
C
op
U
ni
 0
by the vector   .
 3
ge
R
ev
ie
w
C
ve
rs
ity
This means that the two curves have exactly the same shape but that they are
separated by 3 units in the positive y direction.
Copyright Material - Review Only - Not for Redistribution
O
2
4
x
ve
rs
ity
C
U
ni
op
y
Chapter 2: Functions
w
ge
Hence, the graph of y = ( x − 3)2 − 2( x − 3) + 1 is a translation of the graph of
-R
am
br
id
ev
ie
 3
y = x 2 − 2 x + 1 by the vector   .
 0
Pr
es
s
-C
KEY POINT 2.7
y
C
op
R
ni
KEY POINT 2.8
ve
rs
ity
Combining these two results gives:
ev
ie
w
C
op
y
 a
The graph of y = f( x − a ) is a translation of the graph y = f( x ) by the vector   .
 0
br
ev
id
ie
w
ge
U
 a
The graph of y = f( x − a ) + b is a translation of the graph y = f( x ) by the vector   .
 b
-R
am
WORKED EXAMPLE 2.11
Pr
y
es
s
-C
The graph of y = x 2 + 5x is translated 2 units to the right. Find the equation of the resulting graph. Give your
answer in the form y = ax 2 + bx + c.
y = x 2 + 5x
Expand and simplify.
C
U
R
ni
op
y = x2 + x − 6
y
ve
y = ( x − 2) + 5( x − 2)
ie
Replace all occurrences of x by x − 2.
rs
w
2
ev
53
ity
C
op
Answer
id
ie
w
ge
WORKED EXAMPLE 2.12
br
ev
 −5 
2 x is translated by the vector   . Find the equation of the resulting graph.
 3
-R
y=
2 x + 10 + 3
es
2( x + 5) + 3
Pr
y=
Replace x by x + 5 , and add 3 to the resulting function.
-R
s
es
-C
am
br
ev
ie
id
g
w
e
C
U
op
ie
ev
R
y
ni
ve
rs
ity
2x
op
y
y=
w
C
s
-C
Answer
am
The graph of y =
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
ev
ie
am
br
id
EXERCISE 2E
w
ge
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
y = 2x2
-C
 0
after translation by  
 −2 
c
y = 7x2 − 2x
 0
after translation by  
 1
d
y = x2 − 1
 0
after translation by  
 2
e
y=
2
x
 −5 
after translation by  
 0
f
y=
x
x +1
 3
after translation by  
 0
g
y = x2 + x
C
op
w
ie
ev
Pr
es
-C
 2
after translation by  
 3
y = 3x 2 − 2
s
 −1 
after translation by  
 0
y
h
y
ve
rs
ity
ni
U
ge
id
br
am
ev
ie
w
C
op
y
Pr
es
s
y=5 x
b
R
 0
after translation by  
 4
-R
a
-R
1 Find the equation of each graph after the given transformation.
to the graph y = x 2 + 5x + 2
b
y = x3 + 2x2 + 1
to the graph y = x 3 + 2 x 2 − 4
c
y = x 2 − 3x
6
y=x+
x
y = 2x + 5
5
y = 2 − 3x
x
to the graph y = ( x + 1)2 − 3( x + 1)
6
to the graph y = x − 2 +
x−2
to the graph y = 2 x + 3
5
to the graph y =
− 3x + 10
( x − 2)2
rs
op
y
ve
ni
C
w
ie
-R
am
3 The diagram shows the graph of y = f( x ).
b
y = f ( x − 2)
c
y = f ( x + 1) − 5
y = f(x)
1 2 3 4 x
–4 –3 –2 –1 O
–1
–2
–3
–4
b
 0
y = 2 x can be transformed to y = 2 x + 2 by a translation of   .
 a
C
w
e
id
g
ie
Find the value of a.
es
s
-R
br
ev
 b
y = 2 x can be transformed to y = 2 x + 2 by a translation of   .
 0
Find the value of b.
am
c
op
On the same diagram, sketch the graphs of y = 2 x and y = 2 x + 2.
U
4 a
-C
R
ev
ie
w
ni
ve
rs
ity
Pr
es
y = f( x ) − 4
C
a
op
y
s
-C
Sketch the graphs of each of the following functions.
y
4
3
2
1
y
br
f
ev
e
U
R
d
ity
y = x 2 + 5x − 2
ge
C
w
ie
ev
a
id
op
2 Find the translation that transforms the graph.
54
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
w
ge
5 A cubic graph has equation y = ( x + 3 )( x − 2 )( x − 5 ).
C
U
ni
op
y
Chapter 2: Functions
am
br
id
ev
ie
 2
Write, in a similar form, the equation of the graph after a translation of   .
 0
-C
-R
 1
6 The graph of y = x 2 − 4x + 1 is translated by the vector   .
 2
WEB LINK
Try the Between the
lines resource on
the Underground
Mathematics website.
y
 2
7 The graph of y = ax 2 + bx + c is translated by the vector   .
 −5 
The resulting graph is y = 2 x 2 − 11x + 10 . Find the value of a, the value of b and
the value of c.
w
ge
The diagram shows the graphs of the two functions:
br
ev
y = −( x 2 − 2 x + 1)
2
s
-C
-R
am
When the x-coordinates on the two graphs are the same ( x = x ), the y-coordinates
are negative of each other ( y = − y ).
rs
Hence, the graph of y = −( x 2 − 2 x + 1) is a reflection of the graph of y = x 2 − 2 x + 1
in the x-axis.
y
C
U
R
op
ni
ev
ve
ie
w
C
ity
op
Pr
y
es
This means that, when the x-coordinates are the same, the red curve is the same
vertical distance from the x-axis as the blue curve but it is on the opposite side of the
x-axis.
KEY POINT 2.9
id
ie
w
ge
The graph of y = −f( x ) is a reflection of the graph y = f( x ) in the x-axis.
br
ev
Now consider the two functions:
-R
am
y = x2 − 2x + 1
-C
y = ( − x )2 − 2( − x ) + 1
es
s
We obtain the second function by replacing x by −x in the first function.
Pr
op
y
The graphs of these two functions are demonstrated in the diagram.
w
2
4
ie
O
C
U
id
g
–2
x
-R
s
es
-C
am
br
–4
e
2
1
op
y = x2 – 2x + 1
3
ev
ie
4
y = (–x)2 – 2(–x) + 1
y
ni
ve
rs
w
6
R
ev
ity
C
y
5
y = x2 – 2x + 1
4
ie
id
y = x2 − 2x + 1
y
6
C
op
U
2.6 Reflections
R
y
ve
rs
ity
ni
ev
ie
w
C
op
PS
Pr
es
s
Find, in the form y = ax 2 + bx + c, the equation of the resulting graph.
Copyright Material - Review Only - Not for Redistribution
–2
O
2
4 x
–2
–4
–6
y = –(x2 – 2x + 1)
55
ve
rs
ity
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
w
ge
The curves are at the same height ( y = y ) when x = − x or equivalently x = − x.
-R
am
br
id
ev
ie
This means that the heights of the two graphs are the same when the red graph has the
same horizontal displacement from the y-axis as the blue graph but is on the opposite side
of the y-axis.
op
y
Pr
es
s
-C
Hence, the graph of y = ( − x )2 − 2( − x ) + 1 is a reflection of the graph of y = x 2 − 2 x + 1 in
the y-axis.
U
R
ni
WORKED EXAMPLE 2.13
y
The graph of y = f( − x ) is a reflection of the graph y = f( x ) in the y-axis.
C
op
ev
ie
w
C
ve
rs
ity
KEY POINT 2.10
ie
-R
y
es
s
y = − f( x ) is a reflection of y = f( x ) in the x-axis.
Pr
b
The turning point is (5, 7). It is a maximum point.
y = f( − x ) is a reflection of y = f( x ) in the y-axis.
ity
op
C
56
y = f( − x )
ev
id
br
-C
Answer
a
b
am
y = − f( x )
a
w
ge
The quadratic graph y = f( x ) has a minimum at the point (5, −7). Find the coordinates of the vertex and state
whether it is a maximum or minimum of the graph for each of the following graphs.
y
ge
C
U
EXERCISE 2F
R
op
ni
ev
ve
ie
w
rs
The turning point is ( −5, −7) . It is a minimum point.
y
w
1 The diagram shows the graph of y = g( x ) .
am
b
y = g( − x )
s
es
ie
3
2
y = g(x)
1
–4 –3 –2 –1 O
–1
1
2
3
4 x
–2
Pr
op
y
–3
ity
–4
ni
ve
rs
C
2 Find the equation of each graph after the given transformation.
y = 5x 2 after reflection in the x-axis.
b
y = 2 x 4 after reflection in the y-axis.
c
y = 2 x 2 − 3x + 1 after reflection in the y-axis.
d
y = 5 + 2 x − 3x 2 after reflection in the x-axis.
-R
s
es
am
br
ev
ie
id
g
w
e
C
U
op
y
a
-C
w
ie
ev
R
-R
y = −g( x )
-C
a
br
Sketch the graphs of each of the following functions.
ev
id
4
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
C
U
ni
op
y
Chapter 2: Functions
c
w
ev
ie
y = 2 x − 5x 2 onto the graph y = 5x 2 − 2 x
y = x 3 + 2 x 2 − 3x + 1 onto the graph y = − x 3 − 2 x 2 + 3x − 1.
y
Pr
es
s
-C
d
y = x 2 − 3x + 4 onto the graph y = x 2 + 3x + 4
-R
b
y = x 2 + 7 x − 3 onto the graph y = − x 2 − 7 x + 3
am
br
id
a
ge
3 Describe the transformation that maps the graph:
op
2.7 Stretches
y
10
ni
y = 2( x 2 − 2 x − 3)
×2
y
ev
ie
w
y = x2 − 2x − 3
When the x-coordinates on the two graphs are the same ( x = x ), the
y-coordinate on the red graph is double the y-coordinate on the blue
graph ( y = 2 y ).
y = x 2 – 2x – 3
w
5
ev
id
ie
ge
U
R
C
op
C
ve
rs
ity
The diagram shows the graphs of the two functions:
-R
am
br
This means that, when the x-coordinates are the same, the red curve is twice the
distance of the blue graph from the x-axis.
–2
O
es
ity
rs
op
y
ve
ni
w
ie
id
KEY POINT 2.11
C
U
●
R
–5
a stretch with scale factor 2 with the line y = 0 invariant
a stretch with stretch factor 2 with the x-axis invariant
a stretch with stretch factor 2 relative to the x-axis
a vertical stretch with stretch factor 2.
ge
ev
●
6 x
Pr
y
op
C
w
ie
●
4
57
Note: there are alternative ways of expressing this transformation:
●
2
s
-C
–4
Hence, the graph of y = 2( x 2 − 2 x − 3) is a stretch of the graph of
y = x 2 − 2 x − 3 from the x-axis. We say that it has been stretched with stretch
factor 2 parallel to the y-axis.
-R
am
br
ev
The graph of y = a f( x ) is a stretch of the graph y = f( x ) with stretch factor a parallel to the
y-axis.
y
op
-R
s
es
-C
am
br
ev
ie
id
g
w
e
C
U
R
ev
ie
w
ni
ve
rs
C
ity
Pr
op
y
es
s
-C
Note: if a , 0, then y = a f( x ) can be considered to be a stretch of
y = f( x ) with a negative scale factor or as a stretch with positive scale
factor followed by a reflection in the x-axis.
Copyright Material - Review Only - Not for Redistribution
×2
y = 2(x 2 – 2x – 3)
ve
rs
ity
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
y = (2 x ) − 2(2 x ) − 3
2
×1
2
ev
ie
am
br
id
y = x2 − 2x − 3
w
ge
Now consider the two functions:
y = (2x)2 – 2(2x) – 3
y
10
-R
We obtain the second function by replacing x by 2x in the first function.
ni
y
O
2
4
6 x
ie
ev
–5
-R
s
es
1
parallel to the
a
op
ni
C
U
ie
w
ge
ev
A stretch parallel to the y-axis, factor 4, gives
the function 4f( x ).
Pr
op
y
es
The equation of the resulting graph is y = 20 − 2 x 2.
s
-C
-R
am
br
1 2
x
2
4f( x ) = 20 − 2 x 2
Let f( x ) = 5 −
id
Answer
y
rs
ve
1 2
x is stretched with stretch factor 4 parallel to the y-axis.
2
Find the equation of the resulting graph.
ni
ve
rs
C
ity
WORKED EXAMPLE 2.15
op
C
Express 4x 2 − 6x − 5 in terms of f( x ).
e
Let f( x ) = x 2 − 3x − 5
U
Answer
y
Describe the single transformation that maps the graph of y = x 2 − 3x − 5 to the graph of y = 4x 2 − 6x − 5.
ie
id
g
w
4x 2 − 6x − 5 = (2 x )2 − 3(2 x ) − 5
-R
s
es
am
br
ev
= f(2 x )
1
The transformation is a stretch parallel to the x-axis with stretch factor .
2
-C
w
–2
ity
op
C
w
ev
ie
WORKED EXAMPLE 2.14
The graph of y = 5 −
R
–4
Pr
y
x-axis.
58
ie
y = x2 – 2x – 3
w
ge
id
br
am
-C
The graph of y = f( ax ) is a stretch of the graph y = f( x ) with stretch factor
ev
5
C
op
y = x 2 − 2 x − 3 from the y-axis. We say that it has been stretched with stretch
1
factor parallel to the x-axis.
2
KEY POINT 2.12
R
×1
2
Hence, the graph of y = (2 x )2 − 2(2 x ) − 3 is a stretch of the graph of
U
R
ev
ie
w
C
ve
rs
ity
op
y
Pr
es
s
-C
The two curves are at the same height ( y = y ) when x = 2 x or equivalently
1
x = x.
2
This means that the heights of the two graphs are the same when the red graph
has half the horizontal displacement from the y-axis as the blue graph.
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
1 The diagram shows the graph of y = f( x ).
y
–6
–4
4
6 x
y = f(x)
2
–4
ni
y
–6
U
2 Find the equation of each graph after the given transformation.
y = 3x 2 after a stretch parallel to the y-axis with stretch factor 2.
b
y = x 3 − 1 after a stretch parallel to the y-axis with stretch factor 3.
1
y = 2 x + 4 after a stretch parallel to the y-axis with stretch factor .
2
y = 2 x 2 − 8x + 10 after a stretch parallel to the x-axis with stretch factor 2.
1
y = 6x 3 − 36x after a stretch parallel to the x-axis with stretch factor .
3
ie
ev
id
es
e
s
-C
-R
am
c
d
w
ge
a
br
R
–2 O
–2
ve
rs
ity
op
C
w
ev
ie
2
Pr
es
s
y = f(2 x )
b
4
C
op
y = 3f( x )
-C
a
y
6
-R
Sketch the graphs of each of the following functions.
ev
ie
am
br
id
EXERCISE 2G
w
ge
C
U
ni
op
y
Chapter 2: Functions
Pr
y = x 2 + 2 x − 5 onto the graph y = 4x 2 + 4x − 5
b
y = x 2 − 3x + 2 onto the graph y = 3x 2 − 9x + 6
c
y = 2 x + 1 onto the graph y = 2 x + 1 + 2
d
y=
59
ni
y
x − 6 onto the graph y =
3x − 6
ge
C
U
op
ve
rs
ity
a
R
ev
ie
w
C
op
y
3 Describe the single transformation that maps the graph:
id
ie
w
2.8 Combined transformations
br
ev
In this section you will learn how to apply simple combinations of transformations.
-R
am
The transformations of the graph of y = f( x ) that you have studied so far can each be
s
-C
categorised as either vertical or horizontal transformations.
Vertical transformations
es
 0
translation  
 a
y = −f( x )
reflection in the x-axis
y = a f( x )
vertical stretch, factor a
y = f( x + a )
 −a 
translation 

 0
y = f( − x )
reflection in the y-axis
y = f( ax )
horizontal stretch, factor
y
1
a
op
ni
ve
rs
ity
Pr
y = f( x ) + a
op
y
C
-R
s
es
am
br
ev
ie
id
g
w
e
C
U
When combining transformations care must be taken with the order in which the
transformations are applied.
-C
w
ie
ev
R
Horizontal transformations
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
Apply the transformations in the given order to triangle T
Pr
es
s
-C
-R
and for each question comment on whether the final images
are the same or different.
y
1 Combining two vertical transformations
1
T
O
1
y
 0
1
, then translate   .
2
 3
ni
C
op
b Investigate for other pairs of vertical transformations.
U
R
3 x
2
ve
rs
ity
w
C
op
 0
1
a i Translate   , then stretch vertically with factor .
2
 3
ii Stretch vertically with factor
ev
ie
y
2
ev
ie
am
br
id
EXPLORE 2.5
w
ge
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
ge
2 Combining one vertical and one horizontal transformation
-R
am
br
ev
id
ie
w
 −2 
a i Reflect in the x-axis, then translate   .
 0
 −2 
ii Translate   , then reflect in the x-axis.
 0
op
3 Combining two horizontal transformations
ity
 2
a i Stretch horizontally with factor 2, then translate   .
 0
 2
ii Translate   , then stretch horizontally with factor 2.
 0
y
op
ni
ev
ve
ie
w
rs
C
60
Pr
y
es
s
-C
b Investigate for other pairs of transformations where one is vertical and the
other is horizontal.
w
ge
C
U
R
b Investigate for other pairs of horizontal transformations.
br
-R
am
KEY POINT 2.13
ev
id
ie
From the Explore activity, you should have found that:
es
s
-C
• When two vertical transformations or two horizontal transformations are combined, the order
in which they are applied may affect the outcome.
C
ity
Pr
op
y
• When one horizontal and one vertical transformation are combined, the order in which they
are applied does not affect the outcome.
op
y
We will now consider how the graph of y = f( x ) is transformed to the graph y = a f( x ) + k.
w
ie
 0
translate  
 k
→ a f( x ) + k
add k to the function
-R
s
es
am
br
multiply function by a
→ a f( x ) →
ev
U
e
id
g
stretch vertically, factor a
-C
f( x ) →
C
This can be shown in a flow diagram as:
R
ev
ie
w
ni
ve
rs
Combining two vertical transformations
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
C
U
ni
op
y
Chapter 2: Functions
am
br
id
ev
ie
w
ge
This leads to the important result:
KEY POINT 2.14
Pr
es
s
-C
-R
Vertical transformations follow the ‘normal’ order of operations, as used in arithmetic.
Combining two horizontal transformations
ve
rs
ity
replace x with bx
id
ie
w
ge
U
R
This leads to the important result:
KEY POINT 2.15
1
b → f( bx + c )
y
→ f( x + c ) →
stretch horizontally, factor
C
op
f( x ) →
 −c 
translate  
 0
replace x with x + c
ni
ev
ie
w
C
op
y
Now consider how the graph of y = f( x ) is transformed to the graph y = f( bx + c ).
-C
-R
am
br
ev
Horizontal transformations follow the opposite order to the ‘normal’ order of operations, as used in
arithmetic.
es
ity
Sketch the graph of y = 2f( x ) − 3.
y
6
w
y = f(x)
2
ni
op
y
ve
ie
ev
U
R
–4
–2
ie
ev
id
br
4
6 x
-R
–6
y
6
es
s
-C
y = 2f(x)
4
Pr
op
y
2
–4
y = 2f( x ) − 3 is a combination of two vertical transformations of y = f( x ) ,
hence the transformations follow the ‘normal’ order of operations.
Step 1: Sketch the graph y = 2f( x ):
O
–2
w
ge
C
–6
am
Answer
61
4
rs
C
op
The diagram shows the graph of y = f( x ) .
Pr
y
s
WORKED EXAMPLE 2.16
Stretch y = f( x ) vertically with stretch factor 2 .
op
C
w
e
ev
ie
id
g
es
s
-R
br
am
-C
–4
y
–6
U
R
ev
ie
w
ni
ve
rs
C
ity
2
Copyright Material - Review Only - Not for Redistribution
–2
O
–2
–4
–6
2
4
6 x
ve
rs
ity
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
–6
–4
–2
O
–2
ve
rs
ity
–4
ni
C
op
y
–6
w
ge
U
WORKED EXAMPLE 2.17
ev
es
s
-C
y = g(x)
4
y = x2
Pr
op
y
3
ity
2
rs
C
62
-R
am
br
id
ie
The diagram shows the graph of y = x 2 and its image, y = g( x ), after a combination
of transformations.
y
y
2
op
1
x
3
C
O
–1
ni
R
–2
U
ev
ve
ie
w
1
w
ge
a Find two different ways of describing the combination of transformations.
br
-R
am
Answer
ev
id
ie
b Write down the equation of the graph y = g( x ).
es
y = g(x)
ity
U
R
ev
2
y
ni
ve
rs
ie
w
3
w
4
5
6
ie
3
x
ev
2
-R
1
s
am
-C
O
es
–1
br
–2
id
g
e
1
C
C
y = x2
Pr
y
op
op
y
s
-C
 4
1
a Translation of   followed by a horizontal stretch, stretch factor .
2
 0
4
y = 2f(x) – 3
2
Pr
es
s
-C
y
op
C
w
ev
ie
4
-R
am
br
id
ev
ie
 
Translate y = 2f ( x ) by the vector  0  .
 −3 
R
y
6
w
ge
Step 2: Sketch the graph y = 2f ( x ) − 3:
Copyright Material - Review Only - Not for Redistribution
2
4
6 x
ve
rs
ity
w
ev
ie
 2
1
, followed by a translation of   .
2
 0
y
y = g(x)
Pr
es
s
4
ev
ie
w
2
–2
O
–1
1
2
3
x
id
ie
w
ge
–3
U
R
ni
1
y
ve
rs
ity
C
op
3
C
op
y = x2
y
-C
Horizontal stretch, factor
-R
am
br
id
ge
OR
C
U
ni
op
y
Chapter 2: Functions
s
ity
y
ge
C
U
EXERCISE 2H
R
op
ni
ev
ve
ie
rs
C
1
means ‘replace x by 2x’.
2
y = ( x − 4)2 becomes y = (2 x − 4)2
Hence, g( x ) = (2 x − 4)2.
w
y = ( x − 4)2
Pr
Horizontal stretch, factor
op
y
-C
becomes
es
y = x2
The same answer will
be obtained when
using the second
combination of
transformations. You
may wish to check this
yourself.
-R
br
am
 4
Translation of   means ‘replace x by x − 4 ’.
 0
TIP
ev
b Using the first combination of transformations:
1 The diagram shows the graph of y = g( x ) .
id
ie
w
y
2
y = 2g( x ) + 1
y = 2 − g( x )
d
y = 2g( − x ) + 1
f
y = g(2 x ) + 3
g
y = g(2 x − 6)
h
y = g( − x + 1)
–3 –2 –1 O
–1
1
2
3 x
–2
es
s
y = −2g( x ) − 1
ity
Pr
–3
y
op
-R
s
es
-C
am
br
ev
ie
id
g
w
e
C
U
R
ev
ie
w
ni
ve
rs
C
op
y
e
y = g(x)
1
-R
br
b
-C
c
y = g( x + 2) + 3
am
a
ev
Sketch the graph of each of the following.
Copyright Material - Review Only - Not for Redistribution
63
ve
rs
ity
2 The diagram shows the graph of y = f( x ).
3
2
1
1
2
4 x
3
–3
–4
–4 –3 –2 –1–1O
id
ie
ge
y
–2
1
C
op
–4 –3 –2 –1–1O
U
–2
–3
y
4
c
2
4 x
3
x
w
2
3
–2
Pr
es
s
-R
1
2
3
ni
ev
ie
w
1
1
y = f(x)
y
4
ve
rs
ity
C
2
R
ev
ie
am
br
id
-C
y
op
b
3
–4
1
–4 –3 –2 –1 O
–1
y
4
–4 –3 –2 –1 O
–1
w
ge
y
2
Write down, in terms of f( x ), the equation of
the graph of each of the following diagrams.
a
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
-R
am
br
ev
3 Given that y = x 2 , find the image of the curve y = x 2 after each of the following
combinations of transformations.
es
s
-C
 1
a a stretch in the y-direction with factor 3 followed by a translation by the vector  
 0
 1
b a translation by the vector   followed by a stretch in the y-direction with
 0
factor 3
w
rs
ity
Pr
y
op
C
64
y
op
U
R
ni
ev
ve
ie
4 Find the equation of the image of the curve y = x 2 after each of the following
combinations of transformations and, in each case, sketch the graph of the
resulting curve.
-C
-R
am
br
ev
id
ie
w
ge
C
a a stretch in the x-direction with factor 2 followed by a translation by the
 5
vector  
 0
 5
b a translation by the vector   followed by a stretch in the x-direction with
 0
factor 2
s
On a graph show the curve y = x 2 and each of your answers to parts a and b.
es
c
Pr
ni
ve
rs
ity
 0
a translation   followed by a stretch parallel to the y-axis with stretch factor 2
 −5 
-R
s
es
-C
am
br
ev
ie
id
g
w
e
C
U
op
y
 2
b translation   followed by a reflection in the x-axis
 0
R
ev
ie
w
C
op
y
5 Given that f( x ) = x 2 + 1, find the image of y = f( x ) after each of the following
combinations of transformations.
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–2
–3
–4
1
2
3
4 x
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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 
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6 a
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the x-axis. Write down the equation of the resulting graph.
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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 ) =
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y
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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
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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
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the curve y = 3 x onto the curve y = −2 3 x − 3 + 4
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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
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10 Given that g( x ) = x 2 , write down the equation of the image of g( x ) after:
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 −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
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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 
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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 ).
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PS
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WEB LINK
Try the Transformers
resource on the
Underground
Mathematics website.
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Cambridge International AS & A Level Mathematics: Pure Mathematics 1
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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.
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●
fg only exists if the range of g is contained within the domain of f.
●
In general, fg( x ) ≠ gf( x ).
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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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●
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The domain of f −1 ( x ) is the range of f( x ).
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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.
●
If f is self-inverse then ff( x ) = x.
●
The graph of a self-inverse function has y = x as a line of symmetry.
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●
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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.
●
The graph of y = f( − x ) is a reflection of the graph y = f( x ) in the y-axis.
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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
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Combining transformations
When two vertical transformations or two horizontal transformations are combined, the order
in which they are applied may affect the outcome.
●
When one horizontal and one vertical transformation are combined, the order in which they are
applied does not affect the outcome.
●
Vertical transformations follow the ‘normal’ order of operations, as used in arithmetic
●
Horizontal transformations follow the opposite order to the ‘normal’ order of operations, as
used in arithmetic.
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●
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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]
O
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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]
s
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
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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.
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[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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O
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x
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Cambridge International AS & A Level Mathematics: Pure Mathematics 1
y
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 ) .
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-C
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q10 June 2014
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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 ) .
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ii State the range of g.
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2
[1]
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[1]
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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.
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[2]
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The function f is defined for p < x < q, where p and q are positive constants, by
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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]
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The value of k is now given to be 7.
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[4]
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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
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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
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i
Express f( x ) in the form a ( x − b )2 − c.
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ii State the range of f .
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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.
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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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g : x ֏ x 2 − 2.
[2]
ii Hence find the value of a for which fg( a ) = gf( a ).
[3]
iii Find the value of b ( b ≠ a ) for which g( b ) = b.
[2]
iv Find and simplify an expression for f −1g( x ) .
[2]
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Pr
Find and simplify expressions for fg( x ) and gf( x ).
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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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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.
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[4]
[3]
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Cambridge International AS & A Level Mathematics 9709 Paper 11 Q11 November 2011
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v Find an expression for f −1( x ).
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[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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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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Chapter 3
Coordinate geometry
id
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br
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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
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■
■
■
■
■
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In this chapter you will learn how to:
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Chapter 3: Coordinate geometry
What you should be able to do
IGCSE / O Level
Mathematics
Find the midpoint and length of a
line segment.
Find the gradient of a line and
state the gradient of a line that is
perpendicular to the line.
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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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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.
-C
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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U
IGCSE / O Level
Mathematics
1 Find the midpoint and length
of the line segment joining
( − 7, 4) and ( − 2, −8).
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IGCSE / O Level
Mathematics
R
Check your skills
-R
Where it comes from
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PREREQUISITE KNOWLEDGE
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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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br
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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
y
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
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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.
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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

y
es
s
Find the value of a and the value of b.
op
Pr
Answer
y
ve
rs
Decide which values to use for x1, y1, x2 , y2.
ni
C
U
x1 + x2 y1 + y2 
3
,
and midpoint =  , −11 

2

2
2
 −7 + a , 4 + b  =  3 , −11 

 2
2  2
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b = −26.
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and
R
Hence, a = 10
Pr
4+b
= −11
2
4 + b = −22
b = −26
Equating the y-coordinates:
-R
−7 + a 3
=
2
2
−7 + a = 3
a = 10
op
y
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Equating the x-coordinates:
ity
am
br
id
R
Using 

( a, b )
↑ ↑
( x2 , y2 )
op
( −7, 4)
↑ ↑
( x1, y1 )
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Method 1: Using algebra
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72
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Chapter 3: Coordinate geometry
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Method 2: Using vectors
P (−7, 4)
8 
PM =
1
2
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

 −15 
1
2
b = −11 + ( −15)
∴ a = 10
and
b = −26.
Q (a, b)
y
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WORKED EXAMPLE 3.2
w
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Three of the vertices of a parallelogram, ABCD, are A( −5, −1), B( −1, −4) and C(6, −2).
br
ev
id
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a Find the midpoint of AC.
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ity
Since ABCD is a parallelogram, the midpoint of BD is the same as the midpoint of AC.
−1 + m −4 + n   1
3
=
,
,− 


2
2
2
2
rs
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ni
op
ie
Midpoint of BD = 

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op
b Let the coordinates of D be ( m, n ).
es
s
−5 + 6 −1 + −2   1
3
=
,
,− 
 2
2
2
2
y
-C
a Midpoint of AC = 

Pr
Answer
am
b Find the coordinates of D.
−1 + m 1
=
2
2
−1 + m = 1
m=2
−4 + n
3
=−
Equating the y-coordinates:
2
2
−4 + n = −3
n=1
D is the point (2, 1).
Equating the x-coordinates:
w
ge
O
D (2, 1)
br
x
C (6, −2)
ev
id
ie
A (−5, −1)
-R
am
s
B (−1, −4)
Pr
es
-C
op
y
ni
ve
rs
ity
WORKED EXAMPLE 3.3
y
The distance between two points P ( −2, a ) and Q( a − 2, −7 ) is 17.
Answer
( a − 2, −7 )
↑ ↑
( x1, y1 )
↑ ↑
( x2 , y2 )
e
( −2, a )
C
U
op
Find the two possible values of a.
ev
ie
id
g
w
Decide which values to use for x1, y1, x2 , y2.
es
s
-R
br
am
-C
C
w
ie
y
C
U
R
ev
R
( 32 , −11)
ve
rs
ity
and
op
∴ a = 23 + 8 21
Pr
es
s
y


 −15 
M
C
op
-C
8 
∴ MQ =
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
( x2 − x1 )2 + ( y2 − y1 )2 = 17
am
br
id
ev
ie
w
ge
Using PQ =
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
( a − 2 + 2)2 + ( −7 − a )2 = 17
Square both sides.
2 a 2 + 14a − 240 = 0
op
-R
Collect terms on one side.
Divide both sides by 2.
Factorise.
ve
rs
ity
C
a 2 + 7 a − 120 = 0
Expand brackets.
Pr
es
s
a 2 + 49 + 14a + a 2 = 289
y
-C
a 2 + ( −7 − a )2 = 289
or
y
a=8
a + 15 = 0
a = −15
C
op
or
Solve.
ni
a−8= 0
ie
-C
Tamar says that this triangle is right angled.
2√7
es
C
ve
rs
w
EXERCISE 3A
ie
4√3
ity
op
Pr
y
Explain your reasoning.
74
5√3
s
Discuss whether he is correct.
-R
am
The triangle has sides of length 2 7 cm, 4 3 cm and 5 3 cm.
ev
br
id
EXPLORE 3.1
w
ge
U
R
ev
ie
w
( a − 8)( a + 15) = 0
y
op
ni
ev
1 Calculate the lengths of the sides of the triangle PQR.
P( −4, 6), Q(6, 1), R(2, 9)
w
ge
a
C
U
R
Use your answers to determine whether or not the triangle is right angled.
br
P(1, 6), Q( −2, 1) and R(3, −2).
-R
am
2
ev
id
ie
b P( −5, 2), Q(9, 3), R( −2, 8)
s
-C
Show that triangle PQR is a right-angled isosceles triangle and calculate the area of the triangle.
es
op
y
3 The distance between two points, P ( a, −1) and Q( −5, a ), is 4 5.
Pr
ni
ve
rs
Find the two possible values of b.
ity
4 The distance between two points, P( −3, −2) and Q( b, 2b ), is 10.
op
y
5 The point ( −2, −3) is the midpoint of the line segment joining P( −6, −5) and Q( a, b ).
C
U
Find the value of a and the value of b.
ie
id
g
w
e
6 Three of the vertices of a parallelogram, ABCD, are A( −7, 3), B( −3, −11) and C(3, −5).
br
ev
a Find the midpoint of AC.
-R
s
Find the length of the diagonals AC and BD.
es
c
am
b Find the coordinates of D.
-C
R
ev
ie
w
C
Find the two possible values of a.
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
am
br
id
w
ev
ie
ge
7 The point P ( k, 2 k ) is equidistant from A(8, 11) and B(1, 12).
Find the value of k.
C
U
ni
op
y
Chapter 3: Coordinate geometry
-R
8 Triangle ABC has vertices at A( −6, 3), B(3, 5) and C(1, −4).
Show that triangle ABC is isosceles and find the area of this triangle.
y
Pr
es
s
-C
9 Triangle ABC has vertices at A( −7, 8), B (3, k ) and C(8, 5).
Given that AB = 2 BC , find the value of k.
ve
rs
ity
w
C
op
5
10 The line x + y = 4 meets the curve y = 8 − at the points A and B.
x
Find the coordinates of the midpoint of AB.
ev
ie
11 The line y = x − 3 meets the curve y2 = 4x at the points A and B.
ni
C
op
y
a Find the coordinates of the midpoint of AB.
ie
w
12 In triangle ABC , the midpoints of the sides AB, BC and AC are (1, 4), (2, 0) and ( −4, 1), respectively.
Find the coordinates of points A, B and C .
am
br
ev
id
PS
ge
U
R
b Find the length of the line segment AB.
-R
3.2 Parallel and perpendicular lines
op
Pr
y
es
s
-C
At IGCSE / O Level you learnt how to find the gradient of the line joining the points
P ( x1, y1 ) and Q( x2 , y2 ) using the formula in Key point 3.2.
ity
Q(x2, y2)
rs
y2 − y1
x2 − x1
y
op
w
ge
C
U
R
ni
ev
ie
w
Gradient of PQ =
75
ve
C
KEY POINT 3.2
ie
-R
am
br
ev
id
P(x1, y1)
s
-C
You also learnt the following rules about parallel and perpendicular lines.
Parallel lines
gradient = m
y
op
U
R
ev
ie
w
ni
ve
rs
C
ity
Pr
op
y
es
Perpendicular lines
w
ie
If a line has gradient m, then every line
1
perpendicular to it has gradient − .
m
-R
s
es
-C
am
br
If two lines are parallel, then their gradients
are equal.
1
m
ev
id
g
e
C
gradient = −
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
am
br
id
ev
ie
w
ge
We can also write the rule for perpendicular lines as:
KEY POINT 3.3
Pr
es
s
-C
-R
If the gradients of two perpendicular lines are m1 and m2 , then m1 × m2 = −1.
C
ve
rs
ity
op
y
You need to know how to apply the rules for gradients to solve problems involving
parallel and perpendicular lines.
C
op
ni
U
R
Find the two possible values of k if A, B and C are collinear.
w
ge
id
ie
If A, B and C are collinear, then they lie on the same line.
-R
am
br
ev
gradient of AB = gradient of BC
s
es
k + 15 k
=
15 − k 2
k = 10
rs
C
or
U
∴k = 3
Solve.
ev
id
ie
w
ge
k − 10 = 0
Factorise.
-R
am
br
R
( k − 3)( k − 10) = 0
or
Collect terms on one side.
ni
ev
k 2 − 13k + 30 = 0
k−3=0
Expand brackets.
ve
ie
w
2 k + 30 = 15 k − k 2
Cross-multiply.
ity
2( k + 15) = k (15 − k )
C
76
Simplify.
Pr
op
y
-C
−k − k
k − ( −15)
=
10 − ( k − 5)
6 − 10
op
Answer
y
The coordinates of three points are A( k − 5, −15), B (10, k ) and C (6, − k ).
y
ev
ie
w
WORKED EXAMPLE 3.4
s
-C
WORKED EXAMPLE 3.5
op
y
es
The vertices of triangle ABC are A(11, 3), B (2 k, k ) and C( −1, −11).
Pr
ni
ve
rs
ity
b Draw diagrams to show the two possible triangles.
Answer
ie
id
g
w
e
Simplify the second fraction.
ev
+
-R
s
es
am
br
−
y
C
k−3
−11 − k
×
= −1
2 k − 11
−1 − 2 k
op
U
a Since angle ABC is 90°, gradient of AB × gradient of BC = −1.
-C
R
ev
ie
w
C
a Find the two possible values of k if angle ABC is 90°.
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
w
ge
C
U
ni
op
y
Chapter 3: Coordinate geometry
k−3
k + 11
×
= −1
2 k − 11
2k + 1
ev
ie
am
br
id
Multiply both sides by (2k − 11)(2k + 1).
( k − 3)( k + 11) = −(2 k − 11)(2 k + 1)
5k 2 − 12 k − 44 = 0
(5k − 22)( k + 2) = 0
y
or
k+2 = 0
∴ k = 4.4 or
k = −2
-R
Factorise.
ve
rs
ity
op
C
w
Collect terms on one side.
Pr
es
s
-C
k 2 + 8 k − 33 = −4 k 2 + 20 k + 11
5 k − 22 = 0
Expand brackets.
y
U
ge
The two possible triangles are:
ie
x
es
s
-C
Pr
C (−1, −11)
ity
y
rs
EXERCISE 3B
ev
ve
ie
w
C
op
C (−1, −11)
U
R
ni
1 The coordinates of three points are A( −6, 4), B(4, 6) and C(10, 7).
ge
C
a Find the gradient of AB and the gradient of BC.
id
ie
w
b Use your answer to part a to decide whether or not the points A, B and C are collinear.
br
ev
2 The midpoint of the line segment joining P( −4, 5) and Q(6, 1) is M.
-R
am
The point R has coordinates ( −3, −7).
s
-C
Show that RM is perpendicular to PQ.
es
Pr
Find the gradient of CD and the gradient of BC .
ni
ve
rs
ity
4 The coordinates of three of the vertices of a trapezium, ABCD, are A(3, 5), B( −5, 4) and C(1, −5).
AD is parallel to BC and angle ADC is 90° .
ie
id
g
w
Find the value of k if A, B and C are collinear.
C
e
U
5 The coordinates of three points are A(5, 8), B ( k, 5) and C ( − k, 4).
op
y
Find the coordinates of D.
-R
s
es
am
Find the two possible values of k if angle ABC is 90°.
ev
br
6 The vertices of triangle ABC are A( −9, 2 k − 8), B (6, k ) and C ( k, 12).
-C
ev
ie
w
C
op
y
3 Two vertices of a rectangle, ABCD, are A( −6, −4) and B(4, −8).
R
x
-R
am
B (−4, −2)
A (11, 3)
O
ev
id
br
O
B (8.8, 4.4)
w
A (11, 3)
y
y
y
R
ni
C
op
If k = −2, then B is the point ( −4, 2).
op
ev
ie
b If k = 4.4, then B is the point (8.8, 4.4).
Copyright Material - Review Only - Not for Redistribution
77
ve
rs
ity
7
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
w
ge
A is the point (0, 8) and B is the point (8, 6).
am
br
id
ev
ie
Find the point C on the y-axis such that angle ABC is 90°.
-R
8 Three points have coordinates A(7, 4), B(19, 8) and C ( k, 2 k ).
-C
Find the value of the constant k for which:
Pr
es
s
a C lies on the line that passes through the points A and B
x y
− = 1, where a and b are positive constants, meets the x-axis at P and the y-axis at Q.
a b
2
The gradient of the line PQ is and the length of the line PQ is 2 29.
5
Find the value of a and the value of b.
ve
rs
ity
ev
ie
w
C
9 The line
y
op
y
b angle CAB is 90° .
U
R
ni
C
op
10 P is the point ( a, a − 2) and Q is the point (4 − 3a, − a ).
w
ge
a Find the gradient of the line PQ.
id
br
ev
Given that the distance PQ is 10 5 , find the two possible values of a.
am
c
ie
b Find the gradient of a line perpendicular to PQ.
C
es
Pr
b Find the value of a, the value of b and the value of c.
c
Find the perimeter of the rhombus.
A (a, 1)
B (8, 2)
O
w
rs
d Find the area of the rhombus.
op
ni
C
U
3.3 Equations of straight lines
br
ev
id
ie
w
ge
At IGCSE / O Level you learnt the equation of a straight line is:
KEY POINT 3.4
x
y
ve
ie
ev
C (b, c)
M
ity
op
y
a Find the coordinates of M.
R
D (4, 10)
s
-C
M is the midpoint of BD.
78
y
-R
11 The diagram shows a rhombus ABCD.
es
s
-C
-R
am
y = mx + c, where m is the gradient and c is the y-intercept, when the line is non-vertical.
x = b when the line is vertical, where b is the x-intercept.
y
P (x, y)
C
ity
Pr
op
y
There is an alternative formula that we can use when we know the gradient of a straight
line and a point on the line.
A (x1, y1)
y
op
-R
s
es
-C
am
br
ev
ie
id
g
w
e
C
U
R
ev
ie
w
ni
ve
rs
Consider a line, with gradient m, that passes through the known point A( x1, y1 ) and whose
general point is P ( x, y ).
Copyright Material - Review Only - Not for Redistribution
O
x
ve
rs
ity
C
U
ni
op
y
Chapter 3: Coordinate geometry
y − y1
=m
x − x1
y − y1 = m( x − x1 )
ge
ev
ie
w
Multiply both sides by ( x − x1 ).
-R
am
br
id
Gradient of AP = m, hence
Pr
es
s
-C
KEY POINT 3.5
C
ve
rs
ity
op
y
y − y1 = m( x − x1 )
ev
ie
w
WORKED EXAMPLE 3.6
y
The equation of a straight line, with gradient m, that passes through the point ( x1, y1 ) is:
Answer
w
ge
U
R
ni
C
op
Find the equation of the straight line with gradient −2 that passes through
the point (4, 1).
y
es
s
-C
-R
am
br
y − 1 = −2( x − 4)
y − 1 = −2 x + 8
2x + y = 9
ity
Find the equation of the straight line passing through the points ( −4, 3) and (6, −2).
y
ve
op
Decide which values to use for x1, y1, x2 , y2.
ni
(6, −2)
↑ ↑
( x2 , y2 )
ie
y
op
-R
s
es
-C
am
br
ev
ie
id
g
w
e
C
U
R
ev
ie
w
ni
ve
rs
C
ity
Pr
op
y
es
s
-C
-R
am
br
ev
id
y2 − y1 ( −2) − 3
1
=
=−
x2 − x1 6 − ( −4)
2
1
Using y − y1 = m( x − x1 ) with m = − , x1 = −4 and y1 = 3:
2
1
y − 3 = − ( x + 4)
2
2 y − 6 = −x − 4
x + 2y = 2
Gradient = m =
w
C
U
( −4, 3)
↑ ↑
( x1, y1 )
ge
ie
Answer
ev
79
rs
w
C
op
Pr
WORKED EXAMPLE 3.7
R
ie
ev
id
Using y − y1 = m( x − x1 ) with m = −2, x1 = 4 and y1 = 1:
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
ev
ie
w
ge
am
br
id
WORKED EXAMPLE 3.8
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
-C
Gradient of AB =
1
−2 − 1
−3
=−
=
7 − ( −5) 12
4
op
y
Gradient of the perpendicular = 4
C
op
ni
w
ev
Expand brackets and simplify.
-R
es
s
Multiply both sides by 2.
Pr
1 Find the equation of the line with:
ity
op
C
ie
ge
id
br
am
-C
y
EXERCISE 3C
80
ve
ie
w
rs
a gradient 2 passing through the point (4, 9)
y
op
w
ge
2 Find the equation of the line passing through each pair of points.
br
ev
id
ie
( 1, 0 ) and (5, 6)
-R
am
b (3, −5) and ( −2, 4)
(3, −1) and ( −3, −5)
s
-C
c
C
U
R
ni
ev
b gradient −3 passing through the point (1, −4)
2
c gradient − passing through the point ( −4, 3).
3
a
es
3 Find the equation of the line:
Pr
ity
b parallel to the line x + 2 y = 6, passing through the point (4, −6)
ni
ve
rs
perpendicular to the line y = 2 x − 3, passing through the point (6, 1)
op
y
d perpendicular to the line 2 x − 3 y = 12, passing through the point (8, −3).
e
C
U
4 Find the equation of the perpendicular bisector of the line segment joining the
points:
ie
id
g
w
a (5, 2) and ( −3, 6)
es
s
-R
br
( −2, −7) and (5, −4).
am
c
ev
b ( −2, −5) and (8, 1)
-C
R
ev
ie
w
C
op
y
a parallel to the line y = 3x − 5, passing through the point (1, 7)
c
x1 + x2 y1 + y2 
,
.
2
2 
y
ve
rs
ity
Use midpoint = 

U
R
2 y = 8x − 9
y2 − y1
.
x2 − x1
Use m1 × m2 = −1.
 −5 + 7 1 + ( −2 )  
1
Midpoint of AB = 
,
 =  1, − 2 
2
 2
∴ The perpendicular bisector is the line with gradient 4 passing
1
through the point  1, −  .

2
1
Using y − y1 = m( x − x1 ) with x1 = 1, y1 = − and m = 4:
2
1
y + = 4( x − 1)
2
y = 4x − 4 21
C
w
ev
ie
Use gradient =
Pr
es
s
Answer
-R
Find the equation of the perpendicular bisector of the line segment joining A( −5, 1) and B(7, −2).
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
C
U
ni
op
y
Chapter 3: Coordinate geometry
ev
ie
P is the point ( −4, 2) and Q is the point (5, −4).
-R
am
br
id
6
w
ge
5 The line l1 passes through the points P( −10, 1) and Q(2, 10). The line l2 is parallel
to l1 and passes through the point (4, −1). The point R lies on l2 , such that QR is
perpendicular to l2 . Find the coordinates of R.
Pr
es
s
b Find the coordinates of the point R.
c
w
ev
ie
a Find the equation of the line l.
ve
rs
ity
C
op
y
-C
A line, l, is drawn through P and perpendicular to PQ to meet the y-axis at the
point R.
Find the area of triangle PQR.
U
R
ni
C
op
y
7 The line l1 has equation 3x − 2 y = 12 and the line l2 has equation y = 15 − 2 x .
The lines l1 and l2 intersect at the point A.
w
ge
a Find the coordinates of A.
id
ie
b Find the equation of the line through A that is perpendicular to the line l1.
-R
am
br
ev
8 The perpendicular bisector of the line joining A( −10, 5) and B( −2, −1) intersects
the x-axis at P and the y-axis at Q.
s
-C
a Find the equation of the line PQ.
es
Find the length of PQ.
Pr
c
81
9 The line l1 has equation 2 x + 5 y = 10.
ity
C
op
y
b Find the coordinates of P and Q.
ve
ie
w
rs
The line l2 passes through the point A( −9, −6) and is perpendicular to the line l1.
y
op
ni
ev
a Find the equation of the line l2 .
w
ge
C
U
R
b Given that the lines l1 and l2 intersect at the point B, find the area of triangle
ABO, where O is the origin.
-R
G
x
w
ni
ve
rs
C
ity
O
ie
op
y
H (5, −7)
id
g
w
e
C
U
11 The coordinates of three points are A( −4, −1), B(8, −9) and C ( k, 7).
M is the midpoint of AB and MC is perpendicular to AB. Find the value of k.
br
ev
ie
12 The point P is the reflection of the point ( −2, 10) in the line 4x − 3 y = 12.
-R
s
es
am
Find the coordinates of P.
-C
ev
R
Pr
op
y
F
s
E
-C
y
es
am
br
ev
id
ie
10 The diagram shows the points E , F and G lying on the line x + 2 y = 16. The
point G lies on the x-axis and EF = FG. The line FH is perpendicular to EG.
Find the coordinates of E and F .
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
am
br
id
ev
ie
w
ge
13 The coordinates of triangle ABC are A( −7, 3), B(3, −7) and C(8, 8).
P is the foot of the perpendicular from B to AC.
a Find the equation of the line BP .
-R
b Find the coordinates of P.
Find the lengths of AC and BP .
Pr
es
s
-C
c
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
y
d Use your answers to part c to find the area of triangle ABC .
ni
15 The equations of two of the sides of triangle ABC are x + 2 y = 8 and 2 x + y = 1.
Given that A is the point (2, −3) and that angle ABC = 90°, find:
w
ge
PS
C
op
b Find the coordinates of the point that is equidistant from P, Q and R.
ie
id
-R
am
br
ev
b the coordinates of the point B.
16 Find two straight lines whose x-intercepts differ by 7, whose y-intercepts
differ by 5 and whose gradients differ by 2.
Pr
ity
[This question is based upon Straight line pairs on the Underground
Mathematics website.]
y
ve
3.4 The equation of a circle
op
In this section you will learn about the equation of a circle. A circle is defined as the
locus of all the points in a plane that are a fixed distance (the radius) from a given point
(the centre).
-R
am
br
EXPLORE 3.2
ev
id
ie
w
ge
C
U
ni
ev
ie
w
C
82
rs
op
y
Is your solution unique? Investigate further.
es
s
-C
PS
y
ve
rs
ity
ii PR
PQ
U
ev
ie
R
i
a the equation of the third side
R
WEB LINK
a Find the equation of the perpendicular bisectors of:
w
C
op
14 The coordinates of triangle PQR are P(1, 1), Q(1, 8) and R(6, 6).
s
es
Centre
b
( x − 2)2 + ( y − 1)2 = 9
c
( x + 3)2 + ( y + 5)2 = 16
d
( x − 8)2 + ( y + 6)2 = 49
e
x 2 + ( y + 4)2 = 4
f
( x + 6)2 + y 2 = 64
ity
ni
ve
rs
y
x + y = 25
Radius
2
ie
id
g
w
e
C
U
op
a
2
Pr
Equation of circle
-R
s
es
am
br
ev
2 Discuss your results with your classmates and explain how you can find the
coordinates of the centre of a circle and the radius of a circle just by looking at the
equation of the circle.
-C
R
ev
ie
w
C
op
y
-C
1 Use graphing software to draw each of the following circles. From your graphs
find the coordinates of the centre and the radius of each circle, and copy and
complete the following table.
Copyright Material - Review Only - Not for Redistribution
Try the following
resources on the
Underground
Mathematics website:
• Lots of lines!
• Straight lines
• Simultaneous squares
• Straight line pairs.
ve
rs
ity
C
U
ni
op
y
Chapter 3: Coordinate geometry
ev
ie
am
br
id
w
ge
To find the equation of a circle, we let P ( x, y ) be any point on the circumference of a circle
with centre C ( a, b ) and radius r.
y
-R
P (x, y)
Pr
es
s
-C
r
C (a, b)
C
ve
rs
ity
op
y
Q
ev
ie
w
O
x
ni
C
op
y
Using Pythagoras’ theorem on triangle CQP gives CQ2 + PQ2 = r 2 .
U
R
Substituting CQ = x − a and PQ = y − b into CQ2 + PQ2 = r 2 gives:
ie
ev
id
-R
am
br
KEY POINT 3.6
w
ge
( x − a )2 + ( y − b )2 = r 2
s
-C
The equation of a circle with centre ( a, b ) and radius r can be written in completed
square form as:
ity
83
EXPLORE 3.3
ve
ie
w
rs
C
op
Pr
y
es
( x − a )2 + ( y − b )2 = r 2
y
op
b decreasing the value of a
c increasing the value of b
d decreasing the value of b.
-R
am
br
ev
id
ie
w
ge
a increasing the value of a
C
U
R
ni
ev
The completed square form for the equation of a circle with centre ( a, b ) and radius r
is ( x − a )2 + ( y − b )2 = r 2 . Use graphing software to investigate the effects of:
DID YOU KNOW?
s
-C
WORKED EXAMPLE 3.9
es
Write down the coordinates of the centre and the radius of each of these circles.
Pr
b
Centre = (2, 4), radius = 100 = 10
c
Centre = ( −1, 8), radius = 12 = 2 3
op
Centre = (0, 0), radius =
4 =2
-R
s
es
am
br
ev
ie
id
g
w
e
U
a
C
Answer
y
ni
ve
rs
c ( x + 1)2 + ( y − 8)2 = 12
ity
b ( x − 2)2 + ( y − 4)2 = 100
-C
R
ev
ie
w
C
op
y
a x 2 + y2 = 4
Copyright Material - Review Only - Not for Redistribution
In the 17th century, the
French philosopher
and mathematician
René Descartes
developed the idea
of using equations to
represent geometrical
shapes. The
Cartesian coordinate
system is named
after this famous
mathematician.
ve
rs
ity
-C
Answer
-R
Find the equation of the circle with centre ( −4, 3) and radius 6.
ev
ie
w
ge
am
br
id
WORKED EXAMPLE 3.10
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
( x − ( −4))2 + ( y − 3)2 = 62
C
op
y
ve
rs
ity
WORKED EXAMPLE 3.11
ni
ev
ie
w
C
op
( x + 4)2 + ( y − 3)2 = 36
Pr
es
s
y
Equation of circle is ( x − a )2 + ( y − b )2 = r 2 , where a = −4, b = 3 and r = 6.
U
R
A is the point (3, 0) and B is the point (7, −4).
ie
x
r
-C
C
rs
The centre of the circle, C , is the midpoint of AB.
ve
ie
w
ity
B (7, −4)
C
84
Pr
op
y
es
s
O
-R
am
A (3, 0)
br
y
ev
id
Answer
w
ge
Find the equation of the circle that has AB as a diameter.
y
op
C
U
R
ni
ev
3 + 7 0 + ( −4) 
C = 
,
 = (5, −2)
 2
2
w
ie
(5 − 3)2 + ( −2 − 0)2 = 8
ev
r=
br
id
ge
Radius of circle, r, is equal to AC.
es
Pr
ity
op
y
x 2 − 2 ax + a 2 + y2 − 2by + b2 = r 2
x 2 + y2 − 2 ax − 2by + ( a 2 + b2 − r 2 ) = 0
U
id
g
ie
ev
es
s
-R
br
am
●
the coefficients of x 2 and y2 are equal
there is no xy term.
-C
●
w
e
C
When we write the equation of a circle in this form, we can note some important
characteristics of the equation of a circle. For example:
op
y
ni
ve
rs
C
w
ie
( x − a )2 + ( y − b )2 = r 2 gives:
Rearranging gives:
ev
R
( 8 )2
( x − 5)2 + ( y + 2)2 = 8
Expanding the equation
-R
-C
( x − 5)2 + ( y + 2)2 =
s
am
Equation of circle is ( x − a )2 + ( y − b )2 = r 2 , where a = 5, b = −2 and r = 8.
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
C
U
ni
op
y
Chapter 3: Coordinate geometry
w
x 2 + y 2 + 2 gx + 2 fy + c = 0
Pr
es
s
-C
g 2 + f 2 − c is the radius.
You should not try to
memorise the formulae
for the centre and
radius of a circle in this
form, but rather work
them out if needed,
as shown in Worked
example 3.12.
-R
KEY POINT 3.7
where ( − g, − f ) is the centre and
TIP
ev
ie
am
br
id
ge
We often write the expanded form of a circle as:
ve
rs
ity
WORKED EXAMPLE 3.12
ev
ie
w
C
op
y
This is the equation of a circle in expanded general form.
C
op
ni
U
R
ge
Answer
id
ie
w
We answer this question by first completing the square.
x 2 + 10 x + y2 − 8 y − 40 = 0
br
ev
Complete the square.
-R
ity
Centre = ( −5, 4) and radius = 9.
85
rs
C
r 2 = 81
w
ve
ie
es
b=4
op
a = −5
Pr
y
( x + 5)2 + ( y − 4)2 = 81
y
op
ie
P
es
s
-C
Pr
op
y
The tangent to a circle at a point
is perpendicular to the radius at
that point.
The perpendicular from the
centre of a circle to a chord
bisects the chord.
w
ni
ve
rs
C
ity
The angle in a semicircle is a
right angle.
op
y
From these statements we can conclude that:
If triangle ABC is right angled at B, then the points A, B and C lie on the circumference
of a circle with AC as diameter.
●
The perpendicular bisector of a chord passes through the centre of the circle.
●
If a radius and a line at a point, P, on the circumference are at right angles, then the line
must be a tangent to the curve.
-R
s
es
am
br
ev
ie
id
g
w
e
C
U
●
-C
ie
O
-R
am
O
O
C
ev
A
br
id
w
ge
C
U
R
ni
ev
It is useful to remember the three following right angle facts for circles.
B
ev
Collect constant terms together.
Compare with ( x − a )2 + ( y − b )2 = r 2.
s
-C
am
( x + 5)2 − 52 + ( y − 4)2 − 42 − 40 = 0
R
y
Find the centre and the radius of the circle x 2 + y2 + 10 x − 8 y − 40 = 0.
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
A circle passes through the points P ( −1, 4), Q(1, 6) and R(5, 4).
Q (1, 6)
ve
rs
ity
P (−1, 4)
C
op
ge
U
R
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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
-C
-R
am
Midpoint of PQ = 

C
ity
Equation of perpendicular bisector of PQ is:
rs
(1)
y
ve
ev
ie
w
( y − 5) = −1( x − 0)
y = −x + 5
ni
C
U
id
ie
w
ge
R
op
1+ 5 6 + 4
,
= (3, 5)
2
2 
4−6
1
Gradient of QR =
=−
5−1
2
Midpoint of QR = 

br
ev
Gradient of perpendicular bisector of QR = 2
es
s
-C
-R
am
Equation of perpendicular bisector of QR is:
( y − 5) = 2( x − 3)
(2)
y = 2x − 1
Pr
ity
x = 2, y = 3
C
ni
ve
rs
Radius = CR =
(5 − 2)2 + (4 − 3)2 = 10
-R
s
es
-C
am
br
ev
ie
id
g
w
e
C
U
op
Hence, the equation of the circle is ( x − 2)2 + ( y − 3)2 = 10 .
R
ev
ie
w
Centre of circle = (2, 3)
y
op
y
Solving equations (1) and (2) gives:
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
ev
ie
w
ge
am
br
id
Alternative method:
C
U
ni
op
y
Chapter 3: Coordinate geometry
The equation of the circle is ( x − a )2 + ( y − b )2 = r 2.
-R
The points ( −1, 4), (1, 6) and (5, 4) lie on the circle, so substituting gives:
a 2 + 2 a + b2 − 8b + 17 = r 2
Pr
es
s
-C
( −1 − a )2 + (4 − b )2 = r 2
(1)
ve
rs
ity
Then subtracting (1) − (3) and (2) − (3) gives two simultaneous equations for a and b, which can then be solved.
U
R
EXERCISE 3D
2x 2 + 2 y2 = 9
c
x 2 + ( y − 2)2 = 25
d
( x − 5)2 + ( y + 3)2 = 4
( x + 7)2 + y2 = 18
f
-R
w
id
b
-C
e
x 2 + y2 − 8x + 20 y + 110 = 0
ev
2( x − 3)2 + 2( y + 4)2 = 45
2 x 2 + 2 y2 − 14x − 10 y − 163 = 0
s
h
es
g
ie
x 2 + y2 = 16
am
a
br
ge
1 Find the centre and the radius of each of the following circles.
C
op
y
Finally, substituting into (1) gives r 2 .
ni
ev
ie
w
C
op
y
and similar for the other two points, giving equations (2) and (3).
Pr
centre ( −1, 3), radius 7
1
3
5
d centre  , −  , radius
2
2
2
y
ev
ve
ie
c
b centre (5, −2), radius 4
ity
a centre (0, 0), radius 8
rs
w
C
op
y
2 Find the equation of each of the following circles.
w
ge
4 A diameter of a circle has its end points at A( −6, 8) and B(2, −4).
C
U
R
ni
op
3 Find the equation of the circle with centre (2, 5) passing through the point (6, 8).
br
ev
id
ie
Find the equation of the circle.
-R
am
5 Sketch the circle ( x − 3)2 + ( y + 2)2 = 9.
s
-C
6 Find the equation of the circle that touches the x-axis and whose centre is (6, −5).
es
ity
Pr
Show that the centre of the circle lies on the line 4x + 2 y = 15.
ni
ve
rs
8 A circle passes through the points (3, 2) and (7, 2) and has radius 2 2.
y
Find the two possible equations for this circle.
op
U
9 A circle passes through the points O(0, 0), A(8, 4) and B(6, 6).
w
e
C
Show that OA is a diameter of the circle and find the equation of this circle.
-R
s
es
am
br
ev
ie
id
g
10 Show that x 2 + y2 − 6x + 2 y = 6 can be written in the form ( x − a )2 + ( y − b )2 = r 2 ,
where a, b and r are constants to be found. Hence, write down the coordinates of
the centre of the circle and also the radius of the circle.
-C
R
ev
ie
w
C
op
y
7 The points P(1, −2) and Q(7, 1) lie on the circumference of a circle.
Copyright Material - Review Only - Not for Redistribution
87
ve
rs
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C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
am
br
id
ev
ie
w
ge
11 The equation of a circle is ( x − 3)2 + ( y + 2)2 = 25. Show that the point A(6, −6)
lies on the circle and find the equation of the tangent to the circle at the point A.
Pr
es
s
-C
-R
12 The line 2 x + 5 y = 20 cuts the x-axis at A and the y-axis at B. The point C is
the midpoint of the line AB. Find the equation of the circle that has centre C and
that passes through the points A and B. Show that this circle also passes through
the point O(0, 0).
b Find the equation of the circle that passes through the points P, Q and R.
w
ev
ie
a Show that angle PQR is a right angle.
ve
rs
ity
C
op
y
13 The points P ( −5, 6), Q( −3, 8) and R(3, 2) are joined to form a triangle.
U
R
ni
C
op
y
14 Find the equation of the circle that passes through the points (7, 3) and (11, −1)
and has its centre lying on the line 2 x + y = 7.
w
ge
15 A circle passes through the points O(0, 0), P (3, 9) and Q(11, 11).
br
ev
id
ie
Find the equation of the circle.
i
-R
s
Pr
The radius of each green circle is 1 unit.
Find the radius of the orange circle.
ity
op
C
w
rs
ii Use graphing software to draw the design.
ii
Use graphing software to draw this
extended design.
op
The radius of each green circle is 1 unit.
Find the radius of the blue circle.
y
ve
i
-C
-R
am
br
ev
id
ie
w
ge
C
U
R
ev
ie
b The design in part a is extended, as shown.
ni
88
Try the following
resources on the
Underground
Mathematics website:
• Olympic rings
• Teddy bear.
es
-C
17 a The design shown is made from four green
circles and one orange circle.
y
PS
am
16 A circle has radius 10 units and passes through the point (5, −16). The x-axis is
a tangent to the circle. Find the possible equations of the circle.
WEB LINK
es
s
3.5 Problems involving intersections of lines and circles
TIP
Line and parabola
y
Nature of roots
ni
ve
rs
b 2 − 4 ac
two distinct points of intersection
=0
two equal real roots
one point of intersection (line is a tangent)
,0
no real roots
C
ie
w
e
id
g
no points of intersection
op
two distinct real roots
U
.0
-R
s
es
am
br
ev
In this section you will solve problems involving the intersection of lines and circles.
-C
R
ev
ie
w
C
ity
Pr
op
y
In Chapter 1 you learnt that the points of intersection of a line and a curve can be found by
solving their equations simultaneously. You also learnt that if the resulting equation is of
the form ax 2 + bx + c = 0, then b2 − 4ac gives information about the line and the curve.
Copyright Material - Review Only - Not for Redistribution
We can also describe
an equation that has
‘two equal real roots’
as having ‘one repeated
(real) root’.
ve
rs
ity
ge
C
U
ni
op
y
Chapter 3: Coordinate geometry
am
br
id
ev
ie
w
WORKED EXAMPLE 3.14
-R
The line x = 3 y + 10 intersects the circle x 2 + y2 = 20 at the points A and B.
-C
a Find the coordinates of the points A and B.
Pr
es
s
b Find the equation of the perpendicular bisector of AB and show that it passes through the centre of the circle.
Factorise.
ge
y2 + 6 y + 8 = 0
Expand and simplify.
U
R
(3 y + 10)2 + y2 = 20
Substitute 3 y + 10 for x.
y
x 2 + y2 = 20
C
op
Answer
a
ve
rs
ity
Find the exact coordinates of P and Q.
ni
ev
ie
w
C
op
y
c The perpendicular bisector of AB intersects the circle at the points P and Q.
ie
y = −4
ev
id
or
am
When y = −2, x = 4 and when y = −4, x = −2.
A and B are the points ( −2, −4) and (4, −2).
−2 − ( −4) 1
=
4 − ( −2)
3
So the gradient of the perpendicular bisector = –3.
-C
Pr
es
s
Gradient of AB =
89
Use m = −3, x1 = 1 and y1 = −3.
ve
ie
w
y − y1 = m( x − x1 )
ity
−2 + 4 −4 + ( −2) 
Midpoint of AB = 
,
 = (1, −3)
 2
2
rs
C
op
y
b
-R
br
y = −2
w
( y + 2)( y + 4) = 0
y
ev
y − ( −3) = −3( x − 1)
C
U
R
ni
op
Perpendicular bisector is y = −3x.
br
x 2 + y2 = 20
s
2, y = −3 2 .
Pr
When x = − 2, y = 3 2 and when x =
es
x=± 2
(
)
ity
P and Q are the points − 2, 3 2 and
( 2, −3 2 ), respectively.
y
op
-R
s
es
-C
am
br
ev
ie
id
g
w
e
C
U
R
ev
ie
w
ni
ve
rs
op
y
-C
10 x 2 = 20
C
Substitute −3x for y.
-R
am
c
ev
id
ie
Hence, the perpendicular bisector of AB passes through the
point (0, 0), the centre of the circle x 2 + y2 = 20.
w
ge
When x = 0, y = −3(0) = 0.
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
ev
ie
w
ge
am
br
id
WORKED EXAMPLE 3.15
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
Answer
-R
Show that the line y = x − 13 is a tangent to the circle x 2 + y2 − 8x + 6 y + 7 = 0.
x + ( x − 13) − 8x + 6( x − 13) + 7 = 0
2
y
2
x − 14x + 49 = 0
op
2
Expand and simplify.
Factorise.
C
ve
rs
ity
( x − 7)( x − 7) = 0
x = 7 or x = 7
w
ev
ie
Substitute x − 13 for y.
Pr
es
s
-C
x 2 + y 2 − 8x + 6 y + 7 = 0
U
w
ge
EXERCISE 3E
ie
R
ni
C
op
y
The equation has one repeated root, hence y = x − 13 is a tangent.
br
ev
id
1 Find the points of intersection of the line y = x − 3 and the circle ( x − 3)2 + ( y + 2)2 = 20.
-R
am
2 The line 2 x − y + 3 = 0 intersects the circle x 2 + y2 − 4x + 6 y − 12 = 0 at two points, D and E.
s
-C
Find the length of DE .
Pr
y
es
3 Show that the line 3x + y = 6 is a tangent to the circle x 2 + y2 + 4x + 16 y + 28 = 0.
4 Find the set of values of m for which the line y = mx + 1 intersects the circle ( x − 7)2 + ( y − 5)2 = 20 at two
distinct points.
rs
ity
op
C
90
ve
ie
w
5 The line 2 y − x = 12 intersects the circle x 2 + y2 − 10 x − 12 y + 36 = 0 at the points A and B.
y
op
ni
ev
a Find the coordinates of the points A and B.
The perpendicular bisector of AB intersects the circle at the points P and Q.
w
ge
c
C
U
R
b Find the equation of the perpendicular bisector of AB.
br
ev
id
ie
Find the exact coordinates of P and Q.
-R
6 Show that the circles x 2 + y2 = 25 and x 2 + y2 − 24x − 18 y + 125 = 0 touch each other.
-C
PS
am
d Find the exact area of quadrilateral APBQ.
es
s
Find the coordinates of the point where they touch.
Pr
ity
ni
ve
rs
7 Two circles have the following properties:
the x-axis is a common tangent to the circles
●
the point (8, 2) lies on both circles
●
the centre of each circle lies on the line x + 2 y = 22.
ie
id
g
w
a Find the equation of each circle.
C
e
U
op
y
●
br
ev
b Prove that the line 4x + 3 y = 88 is a common tangent to these circles.
es
s
-R
[Inspired by Can we find the two circles that satisfy these three conditions? on the Underground
Mathematics website.]
am
R
ev
ie
w
PS
-C
C
op
y
[This question is taken from Can we show that these two circles touch? on the Underground
Mathematics website.]
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
Midpoint, gradient and length of line segment
Gradient of PQ is
●
Length of segment PQ is
-R
●
Pr
es
s
y2 − y1
.
x2 − x1
( x2 − x1 )2 + ( y2 − y1 )2
If the gradients of two parallel lines are m1 and m2 , then m1 = m2.
●
If the gradients of two perpendicular lines are m1 and m2 , then m1 × m2 = −1.
ni
C
op
●
ge
The equation of a straight line is:
ie
w
y − y1 = m( x − x1 ), where m is the gradient and ( x1, y1 ) is a point on the line.
id
●
y
Parallel and perpendicular lines
x + x2 y1 + y2 
,
Midpoint, M, of PQ is  1
 .

2
2
U
R
ev
ie
w
C
op
P (x1, y1)
●
ve
rs
ity
M
y
-C
Q (x2, y2)
w
ev
ie
am
br
id
ge
Checklist of learning and understanding
C
U
ni
op
y
Chapter 3: Coordinate geometry
br
ev
The equation of a circle is:
( x − a )2 + ( y − b )2 = r 2 , where ( a, b ) is the centre and r is the radius.
●
x 2 + y 2 + 2 gx + 2 fy + c = 0 , where ( − g, − f ) is the centre and
g 2 + f 2 − c is the radius.
91
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Cambridge International AS & A Level Mathematics: Pure Mathematics 1
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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
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s
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A line has equation 2 x + y = 20 and a curve has equation y = a +
1
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op
y
2
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END-OF-CHAPTER REVIEW EXERCISE 3
[4]
y = 6x + k
y = 7√x
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B
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A
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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
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Cambridge International AS & A Level Mathematics 9709 Paper 11 Q5 June 2012
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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
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3
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The curve and the line intersect at the points A and B.
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The diagram shows the curve y = 7 x and the line y = 6x + k , where k is a constant.
[6]
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op
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The curve y = 3 x − 2 and the line 3x − 4 y + 3 = 0 intersect at the points P and Q.
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Find the possible values of a and b.
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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
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b Find the coordinates of the midpoint of AB.
[3]
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5
[6]
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Find the length of PQ.
Find the area of triangle AOB in terms of t.
[3]
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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
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6
[3]
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c Find the equation of the perpendicular bisector of the line AB.
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Cambridge International AS & A Level Mathematics 9709 Paper 11 Q6 June 2015
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Find the coordinates of P. You must show all your working.
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The point P is the reflection of the point ( −7, 5) in the line 5x − 3 y = 18.
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7
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Show that the mid-point of PC lies on the line y = x.
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The line through P perpendicular to AB intersects the x-axis at C .
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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.
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The curve y = x + 2 −
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8
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Chapter 3: Coordinate geometry
b Find the perpendicular bisector of the line AB.
Pr
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[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.
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9
b Find the equation of the perpendicular bisector of the line PQ.
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[4]
[4]
[3]
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c Find the x-coordinates of the points where this perpendicular bisector intersects the circle.
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Give your answers in exact form.
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B (15, 22)
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x
A (3, −2)
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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
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iii Find the coordinates of D.
[4]
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The perpendicular bisector of AB meets BC at D.
[2]
[4]
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AC and BC are 2 m , −2 m and m respectively, where m is a positive constant.
s
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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]
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Cambridge International AS & A Level Mathematics 9709 Paper 11 Q7 November 2013
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11 The point A has coordinates ( −1, 6) and the point B has coordinates (7, 2).
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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).
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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
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b Use your answer to part a to find the value of p.
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[4]
[1]
[4]
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c Find the equation of the circle.
14
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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
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A (13, 17)
C (13, 4)
B (3, 2)
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[2]
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q9 June 2012
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[2]
iii Given that ABCD is a square, find the coordinates of D and the length of AD.
C
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[5]
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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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b Calculate the area of trapezium ABCD.
[7]
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15 The equation of a curve is xy = 12 and the equation of a line is 3x + y = k , where k is a constant.
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a In the case where k = 20, the line intersects the curve at the points A and B.
[4]
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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]
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Pr
a Find the equation of the line through A and B.
op
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17 The equation of a circle is x 2 + y2 − 8x + 4 y + 4 = 0.
[4]
C
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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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b Find the x-coordinates of the points where the circle crosses the x-axis, giving your answers in
exact form.
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3x + 3 y = 12 3 − 6.
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d Show that the equation of the tangent to the circle at A is
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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.
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Cross-topic review exercise 1
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4
17
+ 18 = 2 .
x4
x
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4 x
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–4
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A (–2, 21)
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2
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Solve the equation
1
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CROSS-TOPIC REVIEW EXERCISE 1
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The diagram shows the graph of y = f( x ) for −4 < x < 4.
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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
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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]
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es
1
–2
O
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1
3
2
–2
y = f(x)
–3
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Sketch the graph of y = 2 − f( x ).
C
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The diagram shows the graph of y = f( x ) for −3 < x < 3.
am
x
–1
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–1
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–3
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2
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6
95
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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
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[2]
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b y = −2f( x )
3
[2]
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a y = f( x ) + 5
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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]
C
8
The line x + ky + k 2 = 0, where k is a constant, is a tangent to the curve y2 = 4x at the point P.
Find, in terms of k, the coordinates of P.
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[6]
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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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am
A curve has equation xy = 12 + x and a line has equation y = kx − 9, where k is a constant.
-C
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.
Pr
[4]
The function f is such that f( x ) = 2 x − 3 for x > k, where k is a constant.
The function g is such that g( x ) = x 2 − 4 for x > −4.
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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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s
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i Find and simplify an expression for fg( x ) and state the range of fg.
Pr
[5]
ity
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q8 November 2016
ni
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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.
y
[2]
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b Using these values of b and c, find:
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g
ii the set of values of x for which x 2 + bx + c , 10.
C
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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.
14
[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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[3]
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op
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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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9
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[3]
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Find the two possible values of m.
Pr
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b The line y = mx − 11 is a tangent to the curve y = f( x ).
Copyright Material - Review Only - Not for Redistribution
[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
U
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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.
s
es
The function f is such that f( x ) = 2 x + 1 for x ∈ ℝ .
Pr
The function g is such that g( x ) = 8 − ax − bx 2 for x > k, where a, b and k are constants.
The function fg is such that fg( x ) = 17 − 24x − 4x for x > k.
ity
[3]
b Find the least possible value of k for which g has an inverse.
[4]
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op
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97
2
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A circle has centre (8, 3) and passes through the point P (13, 5) .
[2]
C
c For the value of k found in part b, find g −1( x ).
R
[3]
[2]
2
a Find the value of a and the value of b.
19
[3]
b Write down the coordinates of the vertex of the curve y = 3x + 12 x − 1.
-C
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18
[2]
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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.
ve
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y
b Find the coordinates of the point of intersection of lines L1 and L2 .
[4]
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id
a Find the equation of the circle.
[3]
b Find f −1( x ) and g −1( x ).
[3]
ni
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Pr
es
The function f is such that f( x ) = 3x − 7 for x ∈ ℝ .
18
for x ∈ ℝ , x ≠ 5.
The function g is such that g( x ) =
5−x
a Find the value of x for which fg( x ) = 5.
[3]
op
A curve has equation y = 2 − 3x − x 2.
C
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21
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c Show that the equation f −1( x ) = g −1( x ) has no real roots.
id
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a Express 2 − 3x − x 2 in the form a − ( x + b )2, where a and b are constants.
[2]
[1]
[3]
[3]
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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.
am
br
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b Write down the coordinates of the maximum point on the curve.
es
s
d For each value of m in part c, find the coordinates of the point where the line touches the curve.
-C
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20
[5]
s
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Give your answer in the form ax + by = c.
-R
am
b Find the equation of the tangent to the circle at the point P.
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Cambridge International AS & A Level Mathematics: Pure Mathematics 1
a Find the coordinates of the centre of the circle.
[2]
Pr
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b Find the radius of the circle.
[2]
-R
A circle, C , has equation x 2 + y2 − 16x − 36 = 0.
22
[3]
y
The function f is such that f( x ) = 3x − 2 for x > 0.
23
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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.
U
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op
The function g is such that g( x ) = 2 x 2 − 8 for x < k, where k is a constant.
ge
a Find the greatest value of k for which the composite function fg can be formed.
id
ie
w
b For the case where k = −3:
[2]
br
ev
i find the range of fg
[4]
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am
ii find (fg)−1( x ) and state the domain and range of (fg)−1.
A curve has equation xy = 20 and a line has equation x + 2 y = k, where k is a constant.
s
-C
24
[3]
es
Find:
i the coordinates of the points A and B
ity
[3]
ii the equation of the perpendicular bisector of the line AB.
[4]
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98
Pr
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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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op
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Pr
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b Find the values of k for which the line is a tangent to the curve.
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op
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Chapter 4
Circular measure
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am
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99
id
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am
br
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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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op
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am
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■
■
ie
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.
Pr
es
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y
5 cm
br
40°
w
-R
am
x cm
6 cm
ev
id
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.
es
s
-C
y cm
x°
12 cm
1
Area of triangle = ab sin C
2
Pr
y
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C
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op
At IGCSE / O Level, you will have always worked with angles that were measured in
degrees. Have you ever wondered why there are 360° in one complete revolution?
The original reason for choosing the degree as a unit of angular measure is unknown
but there are a number of different theories.
y
ve
w
●
The ancient Babylonians divided the circle into 6 equilateral triangles and then
subdivided each angle at O into 60 further parts, resulting in 360 divisions in one
complete revolution.
op
Ancient astronomers claimed that the Sun advanced in its path by one degree each day
and that a solar year consisted of 360 days.
O
360 has many factors that make division of the circle so much easier.
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●
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IGCSE / O Level Mathematics
Another measure for angles
ev
1 Find the perimeter and area of a
sector of a circle with radius 6 cm
and sector angle 30°.
Use Pythagoras’ theorem and
trigonometry on right-angled
triangles.
op
IGCSE / O Level Mathematics
100
Check your skills
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Where it comes from
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PREREQUISITE KNOWLEDGE
op
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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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O
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B
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ev
ie
In the diagram, the magnitude of angle AOB is 1 radian.
am
br
id
A
w
ge
4.1 Radians
C
U
ni
op
y
Chapter 4: Circular measure
y
KEY POINT 4.1
C
op
ni
U
br
ev
id
ie
w
ge
R
KEY POINT 4.2
2 π radians = 360°
π radians = 180°
y
It follows that the circumference (an arc of length 2 πr) subtends an angle of 2 π radians at
the centre, therefore:
ev
ie
w
C
ve
rs
ity
op
An arc equal in length to the radius of a circle subtends an angle of 1 radian at the centre.
-R
am
When an angle is written in terms of π, we usually omit the word radian (or rad).
s
-C
Hence, π = 180°.
es
Converting from degrees to radians
π
π
, 45° =
etc.
2
4
We can convert angles that are not simple fractions of 180° using the following rule.
y
π
.
180
C
w
ge
U
R
To change from degrees to radians, multiply by
op
ni
ve
KEY POINT 4.3
ie
ev
rs
w
C
ity
op
Pr
y
Since 180° = π, then 90° =
id
ie
Converting from radians to degrees
ni
ve
rs
C
y
op
-R
s
-C
am
br
ev
ie
id
g
w
e
C
U
R
180
≈ 57°.)
π
es
w
(It is useful to remember that 1 radian = 1 ×
ev
ie
180
.
π
ity
To change from radians to degrees, multiply by
Pr
op
y
KEY POINT 4.4
es
s
-C
-R
am
br
ev
π
π
= 30°,
= 18° etc.
6
10
We can convert angles that are not simple fractions of π using the following rule.
Since π = 180°,
Copyright Material - Review Only - Not for Redistribution
101
ve
rs
ity
Pr
es
s
-C
Method 2:
y
Method 2:
w
ge
Method 1:
ni
R
b
π
radians
6
5π
5 π 180  °
×
radians = 
 9
π 
9
br
ev
id
ie
π radians = 180°
5π
radians = 100°
9
-R
am
π
radians = 20°
9
op
Pr
y
es
s
-C
5π
radians = 100°
9
102
π
radians
6
30° =
U
ev
ie
w
 180  ° = π radians
 6 
6
30° =
π 
radians
30° =  30 ×

180 
ve
rs
ity
op
C
180° = π radians
C
op
Method 1:
y
a
-R
a Change 30° to radians, giving your answer in terms of π.
5π
radians to degrees.
b Change
9
Answer
ev
ie
w
ge
am
br
id
WORKED EXAMPLE 4.1
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
π
radians.
6
There are other angles, which you should learn, that can be written as simple multiples of π.
45°
60°
90°
180°
270°
Radians
0
π
6
π
4
π
3
π
2
π
3π
2
360°
ie
w
2π
br
ev
id
C
30°
op
y
ve
0°
ge
Degrees
U
ni
There are:
R
ev
ie
w
rs
C
ity
In Worked example 4.1, we found that 30° =
-R
es
s
-C
EXERCISE 4A
am
We can quickly find other angles, such as 120°, using these known angles.
b 40°
c
150°
g 135°
ity
Pr
a 20°
i
m 9°
n 35°
j
300°
op
y
o 600°
π
12
9π
20
7π
12
i
l
4π
15
m
5π
4
n
w
h
7π
3
-R
3π
10
e
4π
3
j
9π
2
o
9π
8
C
d
s
7π
5
π
6
g
id
g
k
br
4π
9
am
f
c
e
U
2 Change these angles to degrees.
π
π
b
a
2
3
225°
ie
540°
e 5°
ev
l
d 50°
es
k 65°
25°
h 210°
ni
ve
rs
f
-C
R
ev
ie
w
C
op
y
1 Change these angles to radians, giving your answers in terms of π.
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
C
U
ni
op
y
Chapter 4: Circular measure
47°
c
d 200°
w
b 32°
am
br
id
a 28°
ev
ie
ge
3 Write each of these angles in radians, correct to 3 significant figures.
e
320°
d 1.52 rad
e
0.79 rad
b 0.8 rad
c 1.34 rad
-C
a 1.2 rad
-R
4 Write each of these angles in degrees, correct to 1 decimal place.
Degrees
0
Radians
0
30
90
135 180
225 270 315 360
π
2π
60
90
120 150 180
210 240 270 300 330 360
π
2π
y
0
45
U
cos(0.9)
π
tan
5
w
π
sin
3
f
ie
e
id
c
br
π
d cos
2
b tan(1.5)
ge
a sin(0.7)
ev
R
ni
6 Use your calculator to find:
C
op
C
ev
ie
w
b
0
Radians
Pr
es
s
Degrees
ve
rs
ity
a
op
y
5 Copy and complete the tables, giving your answers in terms of π.
-R
am
7 Calculate the length of QR.
es
Pr
y
103
ity
op
ve
op
Q
5 cm
ie
Robert is told the size of angle BAC in degrees and he is then asked to calculate
the length of the line BC. He uses his calculator but forgets that his calculator
is in radian mode. Luckily he still manages to obtain the correct answer. Given
that angle BAC is between 10° and 15°, use graphing software to help you find
the size of angle BAC, correct to 2 decimal places.
-R
s
es
-C
am
br
ev
ie
id
g
w
e
C
U
op
ie
ev
R
y
ni
ve
rs
ity
Pr
es
s
-C
op
y
ev
am
B
-R
br
id
6 cm
A
w
C
w
C
ge
8
C
U
R
ni
1 rad
y
rs
C
w
ie
ev
P
PS
You do not need to
change the angle to
degrees. You should
set the angle mode
on your calculator to
radians.
s
-C
R
TIP
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
ev
ie
am
br
id
EXPLORE 4.1
w
ge
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
chord
arc
sector
Pr
es
s
-C
Explain what is meant by:
minor arc and major arc
minor sector and major sector
minor segment and major segment.
●
A
r
r
O
C
op
y
Given that the radius of a circle is r cm and that the angle
subtended at the centre of the circle by the chord AB is θ °,
discuss and write down an expression, in terms of r and θ , for
finding each of the following:
●
w
ie
●
length of chord AB
area of minor sector AOB
area of minor segment AOB.
ev
id
●
-R
am
●
length of minor arc AB
perimeter of minor sector AOB
perimeter of minor segment AOB
br
●
ge
U
R
●
B
θ°
ni
ev
ie
w
C
●
segment
ve
rs
ity
op
y
●
-R
Discuss and explain, with the aid of diagrams, the meaning of each of these words.
es
s
-C
What would the answers be if the angle θ was measured in radians instead?
WEB LINK
op
Pr
y
DID YOU KNOW?
ity
A geographical coordinate system is used to describe the location of any
point on the Earth’s surface. The coordinates used are longitude and latitude.
‘Horizontal’ circles and ‘vertical’ circles form the ‘grid’. The horizontal circles
are perpendicular to the axis of rotation of the Earth and are known as lines
of latitude. The vertical circles pass through the North and South poles and
are known as lines of longitude.
y
op
w
ge
-R
am
br
ev
From the definition of a radian, an arc that subtends an angle of 1 radian at the centre of the circle is
of length r. Hence, if an arc subtends an angle of θ radians at the centre, the length of the arc is rθ .
es
s
-C
KEY POINT 4.5
ie
ev
-R
s
es
am
br
π
3
= 5 π cm
= 15 ×
id
g
Arc length = rθ
w
e
Answer
C
U
op
π
radians at the centre of a circle with radius 15 cm.
3
Find the length of the arc in terms of π.
An arc subtends an angle of
y
ni
ve
rs
WORKED EXAMPLE 4.2
-C
R
ev
ie
w
C
ity
Pr
op
y
Arc length = rθ
rθ
B
ie
id
4.2 Length of an arc
Try the Where are
you? resource on
the Underground
Mathematics website.
C
U
R
ni
ev
ve
ie
w
rs
C
104
Copyright Material - Review Only - Not for Redistribution
r
θ
O
r
A
ve
rs
ity
ge
C
U
ni
op
y
Chapter 4: Circular measure
am
br
id
ev
ie
w
WORKED EXAMPLE 4.3
A sector has an angle of 1.5 radians and an arc length of 12 cm.
Pr
es
s
-C
ni
WORKED EXAMPLE 4.4
y
ev
ie
w
C
ve
rs
ity
op
y
Arc length = rθ
12 = r × 1.5
r = 8 cm
ev
br
-R
am
a the length of arc CD
b the length of AD
es
Pr
b
AB = 2 × 8 cos 0.9 = 9.9457…
AD = AB − DB
= 9.9457… − 8
c
= 17.1cm (to 3 significant
figures)
br
b radius 7 cm and angle
es
s
-R
am
-C
Pr
b radius 3.5 cm and angle 0.65 radians.
ity
a radius 10 cm and angle 1.3 radians
ni
ve
rs
3 Find, in radians, the angle of a sector of:
U
op
4 The High Roller Ferris wheel in the USA has a diameter of 158.5 metres.
y
b radius 12 cm and arc length 9.6 cm.
a radius 10 cm and arc length 5 cm
ev
ie
id
g
es
s
-R
br
am
π
radians.
16
w
e
C
Calculate the distance travelled by a capsule as the wheel rotates through
-C
C
op
y
2 Find the arc length of a sector of:
3π
7
7π
d radius 24 cm and angle
.
6
ev
id
ie
w
ge
C
U
ni
op
y
ve
ie
1 Find, in terms of π, the arc length of a sector of:
π
a radius 8 cm and angle
4
3π
c radius 16 cm and angle
8
w
Perimeter = DC + CA + AD
= 7.2 + 8 + 1.945…
= 1.95 cm (to 3 significant figures)
EXERCISE 4B
ie
B
D
ity
C
w
Arc length = rθ
= 8 × 0.9
= 7.2 cm
ev
R
8 cm
0.9 rad
A
rs
op
y
Answer
ev
8 cm
s
-C
c the perimeter of the shaded region.
a
C
ie
ge
id
Find:
w
U
R
Triangle ABC is isosceles with AC = CB = 8 cm.
CD is an arc of a circle, centre B, and angle ABC = 0.9 radians.
R
C
op
Answer
-R
Find the radius of the sector.
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105
ve
rs
ity
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
w
2.1 rad
-R
Pr
es
s
6 cm
ve
rs
ity
y
w
ie
7 cm
O
es
Pr
ABCD is a rectangle with AB = 5 cm and BC = 24 cm.
rs
y
ve
O is the midpoint of BC .
E
C
U
ge
R
a the length of AO
w
B
ev
-R
am
es
s
-C
A
θ
10 cm
B
O
ity
Pr
op
y
C
e
U
op
 x− y
.
Show that P = 4 xy + π( x + y ) + 2( x − y ) sin −1 
 x + y 
x
y
ni
ve
rs
10 The diagram shows the cross-section of two cylindrical metal rods of radii
x cm and y cm. A thin band, of length P cm, holds the two rods tightly
together.
-R
s
es
am
br
ev
ie
id
g
w
[This question is based upon Belt on the Underground Mathematics website.]
-C
C
w
ie
C
O
C
9 The diagram shows a semicircle with radius 10 cm and centre O.
Angle BOC = θ radians. The perimeter of sector AOC is twice the
perimeter of sector BOC.
π−2
.
a Show that θ =
3
b Find the perimeter of triangle ABC .
ev
D
ie
id
the perimeter of the shaded region.
br
c
A
op
ni
ev
OAED is a sector of a circle, centre O. Find:
b angle AOD, in radians
R
7 cm
ity
op
y
the perimeter of the shaded segment.
B
2 rad
s
b the length of chord AB
C
A
-R
-C
a the length of arc AB
PS
O
ev
id
br
am
AB is a chord and angle AOB = 2 radians. Find:
8
6 cm
C
op
the perimeter of the shaded area.
7 The circle has radius 7 cm and centre O.
c
P
R
U
c
8 cm
Q
ge
R
b the length of QR
w
8 cm
5 cm
ni
a angle POQ, in radians
ie
4.3 rad
6 The circle has radius 6 cm and centre O.
PQ is a tangent to the circle at the point P.
QRO is a straight line. Find:
ev
ie
w
C
op
y
-C
1.2 rad
106
c
ev
ie
b
am
br
id
a
ge
5 Find the perimeter of each of these sectors.
Copyright Material - Review Only - Not for Redistribution
y
ve
rs
ity
w
ge
ev
ie
am
br
id
B
r
θ
O
A
r
op
y
Pr
es
s
-C
-R
4.3 Area of a sector
C
U
ni
op
y
Chapter 4: Circular measure
ve
rs
ity
y
area of sector
angle in the sector
=
area of circle
complete angle at the centre
ni
U
w
ge
θ
× πr 2
2π
-R
am
br
ev
id
ie
R
area of sector
θ
=
2π
πr 2
area of sector =
C
op
When θ is measured in radians, the ratio becomes:
s
1 2
rθ
2
ity
107
WORKED EXAMPLE 4.5
ni
ev
Find the area of a sector of a circle with radius 9 cm and angle
w
ge
Answer
C
U
R
Give your answer in terms of π.
-R
s
es
Pr
ni
ve
rs
ity
WORKED EXAMPLE 4.6
op
y
The circle has radius 6 cm and centre O. AB is a chord and angle AOB = 1.2 radians.
Find:
id
g
w
e
C
U
a the area of sector AOB
b the area of triangle AOB
-R
s
es
am
br
ev
ie
c the area of the shaded segment.
-C
ev
ie
w
C
op
y
-C
am
br
ev
id
ie
1 2
rθ
2
1
π
= × 92 ×
2
6
27 π
cm 2
=
4
Area of sector =
R
π
radians.
6
y
ve
ie
w
rs
C
op
Pr
y
Area of sector =
es
-C
KEY POINT 4.6
op
ev
ie
w
C
To find the formula for the area of a sector, we use the ratio:
Copyright Material - Review Only - Not for Redistribution
A
B
1.2 rad
6 cm
6 cm
O
ve
rs
ity
ev
ie
w
ge
am
br
id
Answer
1 2
rθ
2
1
= × 62 × 1.2
2
= 21.6 cm 2
Pr
es
s
-C
1
ab sin C
2
1
= × 6 × 6 × sin1.2
2
= 16.7767…
ve
rs
ity
y
Area of triangle AOB =
w
C
op
b
-R
Area of sector AOB =
a
ev
ie
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
C
op
ni
Area of shaded segment = area of sector AOB − area of triangle AOB
= 21.6 − 16.7767…
w
ie
= 4.82 cm 2 (to 3 significant figures)
-R
am
br
ev
id
ge
U
c
R
y
= 16.8 cm 2 (to 3 significant figures)
es
s
-C
WORKED EXAMPLE 4.7
op
Pr
y
The diagram shows a circle inscribed inside a square of side length 10 cm.
108
C
ity
A quarter circle, of radius 10 cm, is drawn with the vertex of the square as centre.
y
op
ie
br
C
-R
β
( 10 + 10 ) = 5 2 cm
2
5
es
s
2
10
C
U
Shaded area = area of segment PQR − area of segment PQS
ie
id
g
w
e
1
1
1
1
=  × 52 × β − × 52 × sin β  −  × 102 × 2θ − × 102 × sin 2θ 
2
 2

2
2
s
es
am
-R
br
ev
= 21.968 − 7.3296
= 14.6 cm 2 (to 3 significant figures)
5√2
θ
θ
O
y
sin θ
sin α
=
5
10
θ = 0.4867 rad
α
α
Q
ni
ve
rs
Sine rule:
-C
R
ev
ie
w
Hence, β = 2 π − 2α = 2.4189 rad
S
Pr
op
y
-C
1
1
Pythagoras: (diagonal of square) =
2
2
52 + (5 2 )2 − 102
Cosine rule: cos α =
2×5×5 2
α = 1.932 rad
ity
am
Radius of inscribed circle = 5 cm
P
R
op
OQ = 10 cm
ev
id
Answer
w
ge
C
U
R
ni
ev
ve
ie
w
rs
Find the shaded area.
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
ev
ie
2π
radians
5
4π
d radius 9 cm and angle
radians.
3
b radius 10 cm and angle
b radius 2.6 cm and angle 0.9 radians.
ve
rs
ity
a radius 34 cm and angle 1.5 radian
C
op
y
2 Find the area of a sector of:
Pr
es
s
-C
1 Find, in terms of π, the area of a sector of:
π
a radius 12 cm and angle radians
6
2π
c radius 4.5 cm and angle
radians
9
-R
am
br
id
EXERCISE 4C
w
ge
C
U
ni
op
y
Chapter 4: Circular measure
ev
ie
w
3 Find, in radians, the angle of a sector of:
C
op
ni
AOB is a sector of a circle, centre O, with radius 8 cm.
U
R
4
b radius 6 cm and area 27 cm 2 .
y
a radius 4 cm and area 9 cm 2
w
ge
The length of arc AB is 10 cm. Find:
b the area of the sector AOB.
br
ev
id
ie
a angle AOB, in radians
Q
s
es
O
b Find the length of PX .
109
Find the area of the shaded region.
ni
op
ve
ie
ev
R
8 cm
C
U
O
ie
id
P
π
3
w
ge
Find the exact area of the shaded region.
Q
y
w
rs
P
6 The diagram shows a sector, POR, of a circle, centre O, with radius 8 cm
π
and sector angle radians. The lines OR and QR are perpendicular
3
and OPQ is a straight line.
R
X
Pr
a Find angle POQ, in radians.
c
4 cm
ity
C
op
y
-C
-R
am
5 The diagram shows a sector, POQ, of a circle, centre O, with radius
4 cm. The length of arc PQ is 7 cm. The lines PX and QX are
tangents to the circle at P and Q, respectively.
es
s
-C
P
Pr
π
3
b Find the exact area of the shaded region.
O
A
ity
5 cm
ni
ve
rs
8 The diagram shows three touching circles with radii 6 cm, 4 cm and 2 cm.
2
op
-R
s
es
-C
am
br
ev
ie
id
g
w
e
C
U
R
y
Find the area of the shaded region.
ev
ie
w
C
op
y
a Find the exact length of AP.
PS
B
-R
am
br
ev
7 The diagram shows a sector, AOB, of a circle, centre O, with radius
π
5 cm and sector angle radians. The lines AP and BP are tangents
3
to the circle at A and B, respectively.
Copyright Material - Review Only - Not for Redistribution
2
6
4
6
4
ve
rs
ity
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
am
br
id
w
FH is the arc of a circle, centre E. Find the area of:
F
ev
ie
ge
9 The diagram shows a semicircle, centre O, with radius 8 cm.
b sector FOG
c
d the shaded region.
2 rad
-R
a triangle EOF
E
Pr
es
s
-C
sector FEH
a The perimeter of the shaded region is P cm.
r
Show that P = (3 + 3 3 + π ).
3
b The area of the shaded region is A cm 2.
r2
(3 3 − π ).
Show that A =
6
E
ve
rs
ity
F
C
op
y
G
U
R
H G
O
r cm
ni
ev
ie
w
C
op
y
10 The diagram shows a sector, EOG, of a circle, centre O, with radius r cm.
The line GF is a tangent to the circle at G, and E is the midpoint of OF .
O
w
ge
11 The diagram shows two circles with radius r cm.
ie
es
ity
A quarter circle, of radius 10 cm, is drawn from each vertex of the square.
Find the exact area of the shaded region.
y
op
13 The diagram shows a circle with radius 1cm, centre O.
w
ge
PS
C
U
R
ni
ev
ve
ie
w
rs
C
Pr
y
12 The diagram shows a square of side length 10 cm.
op
PS
110
s
-C
-R
am
br
Find, in terms of r, the exact area of the shaded region.
ev
id
The centre of each circle lies on the circumference of the other circle.
O
br
ev
id
ie
Triangle AOB is right angled and its hypotenuse AB is a tangent to the circle at P.
-R
am
Angle BAO = x radians.
-C
a Find an expression for the length of AB in terms of tan x.
x
es
P
B
B
Pr
ity
op
area of inner circle 2
= .
area of sector
3
-R
s
es
am
br
ev
ie
id
g
w
e
C
U
b Show that
y
ni
ve
rs
a Show that R = 3r.
-C
C
w
ie
ev
R
A
14 The diagram shows a sector, AOB, of a circle, centre O, with radius R cm and
π
sector angle radians.
3
An inner circle of radius r cm touches the three sides of the sector.
op
y
P
s
b Find the value of x for which the two shaded areas are equal.
Copyright Material - Review Only - Not for Redistribution
π
3
O
A
ve
rs
ity
am
br
id
r
Pr
es
s
-C
r
-R
Radians and degrees
1 rad
r
w
C
ve
rs
ity
op
y
O
One radian is the size of the angle subtended at the centre of a circle, radius r, by an arc of length r.
●
π radians = 180°
●
To change from degrees to radians, multiply by
●
To change from radians to degrees, multiply by
w
π
.
180
ie
180
.
π
ev
br
id
ge
U
ni
C
op
y
●
R
ev
ie
w
ev
ie
ge
Checklist of learning and understanding
C
U
ni
op
y
Chapter 4: Circular measure
-R
am
Arc length and area of a sector
θ
r
A
111
y
op
ni
ev
ve
ie
w
rs
C
ity
O
Pr
op
y
r
es
s
-C
B
When θ is measured in radians, the length of arc AB is rθ .
1 2
● When θ is measured in radians, the area of sector AOB is
r θ.
2
y
op
-R
s
es
-C
am
br
ev
ie
id
g
w
e
C
U
R
ev
ie
w
ni
ve
rs
C
ity
Pr
op
y
es
s
-C
-R
am
br
ev
id
ie
w
ge
C
U
R
●
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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
Copyright Material - Review Only - Not for Redistribution
115
op
y
ve
rs
ity
ni
C
U
ev
ie
w
ge
-R
am
br
id
Pr
es
s
-C
y
ni
C
op
y
ve
rs
ity
op
C
w
ev
ie
op
y
ve
ni
w
ge
C
U
R
ev
ie
w
Chapter 5
Trigonometry
rs
C
ity
op
Pr
y
es
s
-C
-R
am
br
ev
id
ie
w
ge
U
R
116
id
es
s
-C
-R
am
br
ev
sketch and use graphs of the sine, cosine and tangent functions (for angles of any size, and using
either degrees or radians)
use the exact values of the sine, cosine and tangent of 30°, 45°, 60°, and related angles
use the notations sin −1 x, cos −1 x, tan −1 x to denote the principal values of the inverse trigonometric
relations
sin θ
= tan θ and sin2 θ + cos2 θ = 1
use the identities
cos θ
find all the solutions of simple trigonometrical equations lying in a specified interval.
y
op
-R
s
es
-C
am
br
ev
ie
id
g
w
e
C
U
R
ev
ie
w
ni
ve
rs
C
ity
Pr
op
y
■
■
■
■
■
ie
In this section you will learn how to:
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
ge
C
U
ni
op
y
Chapter 5: Trigonometry
am
br
id
ev
ie
w
PREREQUISITE KNOWLEDGE
What you should be able to do
IGCSE / O Level Mathematics
Use Pythagoras’ theorem and
trigonometry on right-angled
triangles.
Check your skills
1
y
Find each of the following in
terms of r.
y
b sin θ
ni
C
op
c cos θ
U
ge
2 a Convert to radians.
es
ity
rs
ii
720°
op
y
ve
ni
45°
b Convert to degrees.
π
i
6
7π
ii
2
13π
iii
12
Pr
y
op
C
w
ie
ev
i
iii 150°
s
-C
-R
am
br
ev
id
ie
w
Convert between degrees and
radians.
Solve quadratic equations.
b Solve 2 x 2 + 7 x − 15 = 0.
ev
id
ie
w
ge
C
U
3 a Solve x 2 − 5x = 0.
br
-R
am
FAST FORWARD
Pr
op
y
es
s
-C
You should already know how to calculate lengths and angles using the sine, cosine and
tangent ratios. In this chapter you shall learn about some of the special rules connecting
these trigonometric functions together with the special properties of their graphs. The
graphs of y = sin x and y = cos x are sometimes referred to as waves.
ity
Oscillations and waves occur in many situations in real life. A few examples of these
are musical sound waves, light waves, water waves, electricity, vibrations of an aircraft
wing and microwaves. Scientists and engineers represent these oscillations/waves using
trigonometric functions.
w
e
C
U
op
y
ni
ve
rs
C
-R
s
es
am
br
ev
ie
id
g
Try the Trigonometry: Triangles to functions resource on the Underground Mathematics website.
-C
w
ie
ev
R
B
a BC
Why do we study trigonometry?
WEB LINK
r cm
d tan θ
IGCSE / O Level Mathematics
R
θ°
A
ve
rs
ity
op
C
w
ev
ie
R
Chapter 4
C
1 cm
Pr
es
s
-C
-R
Where it comes from
Copyright Material - Review Only - Not for Redistribution
In the Pure
Mathematics 2 & 3
Coursebook, Chapter 3
you shall learn about
the secant, cosecant and
cotangent functions,
which are closely
connected to the sine,
cosine and tangent
functions. You shall
also learn many more
rules involving these six
trigonometric functions.
117
ve
rs
ity
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
am
br
id
You should already know the following trigonometric ratios.
cos θ =
y
ni
y
sin θ
iii tan θ
br
ev
id
ie
w
ge
1 − tan2 θ
3 5 −6 .
b Show that
=
cos θ + sin θ
5
Answer
-R
Pr
ity
rs
w
ve
ie
3
C
w
ev
s
es
y
op
C
w
ity
Pr
) =5−2 5 +2 5 −4 =1
es
s
3 5 −6
=
5
( 5 + 2 )( 5 − 2 )
ie
3
Multiply the numerator and
denominator by 5 − 2.
ev
e
( 5 − 2)
5 ( 5 + 2 )( 5 − 2 )
(
id
g
am
br
=
)
Multiply the numerator and
denominator by 15.
-R
(
3
5 +2
U
5
Simplify.
ni
ve
rs
-C
=
R
-R
2
 2 
1− 
 5 
1 − tan2 θ
=
cos θ + sin θ
5
2
+
3
3
 1
 5
=
 5 + 2

3 
-C
√5
2
.
5
br
am
w
C
op
y
b
θ°
ie
id
2
3
x
y
ni
32 − ( 5 )2 = 2
U
R
x=
ge
ev
Using Pythagoras’ theorem:
iii From the triangle, tan θ =
ie
cos2 θ means
(cos θ )2
5
5
×
3
3
5
=
9
ii A right-angled triangle with angle θ is shown in this diagram.
∴ sin θ =
ev
TIP
s
2
=
y
op
C
118
 5
=
 3 
es
-C
am
a i cos2 θ = cos θ × cos θ
op
R
ii
U
cos2 θ
i
ve
rs
ity
5
, where 0° < θ < 90°.
3
a Find the exact values of:
cos θ =
ev
ie
w
C
op
WORKED EXAMPLE 5.1
C
op
x
opposite
adjacent
y
tan θ =
x
tan θ =
-R
-C
θ°
adjacent
hypotenuse
x
cos θ =
r
opposite
hypotenuse
y
sin θ =
r
sin θ =
y
Pr
es
s
r
ev
ie
w
ge
5.1 Angles between 0° and 90°
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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°
The value
1
can be
2
1
2
1
2
op
ity
1
3
written as
2
.
2
3
2
1
2
3
The value
w
C
1
ev
-C
-R
am
WORKED EXAMPLE 5.2
ev
es
Rationalise the denominator.
-R
s
es
-C
am
br
ev
ie
id
g
w
e
C
U
R
y
1
1
×
2
2
1
=
2 2
1× 2
=
2 2 × 2
2
=
4
sin 30° cos 45° =
ni
ve
rs
a
op
Answer
c
3
Pr
sin
π
π
sin
4
6
2 π
2 π
+ sin
cos
3
3
2 cos
2 π
ity
op
y
C
b
sin 30° cos 45°
ie
w
s
Find the exact value of:
a
1
can be
3
3
.
written as
3
ie
ve
rs
3
2
1
2
U
π
3
TIP
ge
θ = 60° =
tan θ
id
π
4
cos θ
ni
sin θ
119
br
R
ev
ie
w
C
op
These two triangles give the important results:
θ = 45° =
-R
s
y
es
3
π
6
1
Pr
-C
am
We can find the height of the triangle using Pythagoras’
theorem:
θ = 30° =
√3
w
ge
The perpendicular bisector to the base splits the
equilateral triangle into two congruent right-angled
triangles.
22 − 12 =
1
y
Triangle 1
w
ge
We can obtain exact values of the sine, cosine and tangent of 30°, 45° and 60°
 or π , π and π  from the following two triangles.

6 4
3
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
ev
ie
-R
Pr
es
s
The denominator simplifies to
1 3
+ = 1.
4 4
y
Rationalise the denominator.
C
op
ve
rs
ity
ev
ie
2
=
2
ni
1
2
=
U
s
e
c
sin2 θ + cos2 θ
f
5
1 + cos θ
c
1 − sin2 θ
f
5−
rs
op
e
2
sin θ + 1
1
and that θ is acute, find the exact value of:
4
b tan θ
ie
d
sin θ cos θ
tan θ
am
1
1
+
tan θ sin θ
-R
cos θ
br
a
ev
id
3 Given that sin θ =
-C
e
es
sin 45° + cos 30°
e
sin2 45°
2 + tan 60°
f
sin2 30° + cos2 30°
2 sin 45° cos 45°
c
1 − 2 sin2
ni
ve
rs
π
4
y
cos
π
3
cos
f
-R
s
es
am
-C
C
U
op
−
w
tan
1
br
e
1
ev
d
π
π
− tan
6
3
π
sin
4
e
sin
id
g
C
w
ie
c
5 Find the exact value of each of the following.
π
π
π
a sin cos
b cos2
4
4
3
R
ev
sin2 45°
Pr
sin 60°
sin 30°
b
ity
d
op
y
sin 30° cos 60°
tan θ
sin θ
s
4 Find the exact value of each of the following.
a
y
ve
ni
cos θ
sin θ
3 − sin θ
3 + cos θ
ge
d
U
w
ie
ev
R
sin θ
f
2
and that θ is acute, find the exact value of:
5
b cos θ
2 Given that tan θ =
a
1 − sin2 θ
cos θ
C
5
tan θ
2 sin θ cos θ
w
d
c
Pr
sin θ
ie
C
120
4
and that θ is acute, find the exact value of:
5
b tan θ
ity
a
op
y
1 Given that cos θ =
es
-C
-R
am
br
ev
id
ie
w
ge
R
2 2
2
=
2
EXERCISE 5A
2
π
π
means  sin  .
sin

3
3
am
br
id
-C
y
w
C
op
c
w
ge
π
π
π
= sin × sin
3
3
3
3
3
=
×
2
2
3
=
4
1
1
π
π
2 ×
×
2 cos sin
2
2
4
6
=
2
2
2 π
2 π
+ sin
cos
 1 +  3 
3
3


 2
 2 
sin2
b
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
Copyright Material - Review Only - Not for Redistribution
π
6
π
π
+ tan
3
6
π
sin
3
ve
rs
ity
C
w
π
1
1
and the missing function is from the list sin θ , tan θ ,
and
.
cos θ
2
tan θ
ev
ie
6 In the table, 0 ø θ ø
am
br
id
PS
ge
U
ni
op
y
Chapter 5: Trigonometry
……
1
cos θ
1
2
-R
ve
rs
ity
θ =…
3
1
3
1
2
……
……
2
y
……
ge
U
R
ni
C
op
op
C
w
1
sin θ
ev
ie
θ =…
Pr
es
s
θ =…
y
-C
Without using a calculator, copy and complete the table.
w
5.2 The general definition of an angle
id
ie
second
quadrant
br
P
-R
am
To do this we need a general definition for an angle:
θ
es
op
Pr
y
third
quadrant
rs
C
w
y
ve
ni
op
ie
WORKED EXAMPLE 5.3
ev
fourth
quadrant
ity
The Cartesian plane is divided into four quadrants, and the angle θ is said to be in the
quadrant where OP lies. In the previous diagram, θ is in the first quadrant.
C
-R
br
am
Answer
-C
a 120° is an anticlockwise rotation.
b
y
x
P
430°
70°
op
y
ni
ve
rs
w
ie
Acute angle made with x-axis = 60°
ev
ie
id
g
w
e
C
U
Acute angle made with x-axis = 70°
es
s
-R
br
am
x
O
ity
O
y
Pr
60°
-C
−
430° is an anticlockwise rotation.
es
120°
op
y
d
s
P
C
3π
4
w
c
430°
ev
id
b
ie
ge
U
R
Draw a diagram showing the quadrant in which the rotating line OP lies for each of the following angles. In each
case, find the acute angle that the line OP makes with the x-axis.
a 120°
ev
x
O
s
-C
An angle is a measure of the rotation of a line segment OP about a fixed point O.
The angle is measured from the positive x-direction. An anticlockwise rotation is taken
as positive and a clockwise rotation is taken as negative.
R
first
quadrant
ev
We need to be able to use the three basic trigonometric functions for any angle.
y
Copyright Material - Review Only - Not for Redistribution
2π
3
121
ve
rs
ity
am
br
id
d
−
w
3π
is an anticlockwise rotation.
4
c
2 π is a clockwise rotation.
3
ev
ie
ge
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
y
3π
4
x
ni
2π
3
y
P
π
4
Acute angle made with x-axis =
π
3
w
ge
U
x
O
C
op
C
w
ev
ie
π
3
–
ve
rs
ity
op
O
Acute angle made with x-axis =
R
Pr
es
s
π
4
y
-C
-R
y
P
-R
s
-C
EXERCISE 5B
am
br
ev
id
ie
The acute angle made with the x-axis is sometimes called the basic angle or the
reference angle.
Pr
y
w
y
op
ni
U
d
ie
ev
id
br
x
θ = –500°
es
s
-C
x
O
-R
am
O
y
w
ge
y
x
O
ve
ie
ev
R
θ = –320°
x
O
θ = 200°
y
rs
θ = 110°
c
b
C
a
C
122
ity
op
y
es
1 For each of the following diagrams, find the basic angle of θ .
Pr
am
-C
f
2π
3
h
−
C
op
y
−150°
w
5π
3
17 π
−
8
ie
j
ev
13π
9
d
-R
i
−100°
s
7π
6
es
g
ity
400°
ni
ve
rs
e
b
U
310°
e
c
id
g
100°
br
a
R
ev
ie
w
C
op
y
2 Draw a diagram showing the quadrant in which the rotating line OP lies for each of the following angles. On
each diagram, indicate clearly the direction of rotation and state the acute angle that the line OP makes with
the x-axis.
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
C
U
ni
op
y
Chapter 5: Trigonometry
am
br
id
ev
ie
w
ge
3 In each part of this question you are given the basic angle, b, the quadrant in which θ lies and the range in
which θ lies. Find the value of θ .
a b = 55°, second quadrant, 0° , θ , 360°
b = 32°, fourth quadrant, 360° , θ , 720°
Pr
es
s
π
, third quadrant, 0 , θ , 2 π
4
π
e b = , second quadrant, 2 π , θ , 4 π
3
π
f b = , fourth quadrant, −4 π , θ , − 2 π
6
d b=
y
C
op
U
R
ni
ev
ie
w
C
ve
rs
ity
op
y
-C
c
-R
b b = 20°, third quadrant, −180° , θ , 0°
5.3 Trigonometric ratios of general angles
ie
y
P(x, y)
ev
id
r
-R
y
x
y
, cos θ = , tan θ = , when x ≠ 0
r
r
x
y
θ
O
x
x
y
es
s
-C
sin θ =
am
br
KEY POINT 5.1
w
ge
In general, trigonometric ratios of any angle θ in any quadrant are defined as:
x 2 + y2 .
op
Pr
Where x and y are the coordinates of the point P and r is the length of OP, where r =
123
rs
cos θ =
op
y
ve
ni
x
r
tan θ =
y
x
w
y
r
U
sin θ =
ge
R
ev
ie
EXPLORE 5.1
C
w
C
ity
You need to know the signs of the three trigonometric ratios in each of the four quadrants.
-R
cos θ
y
+
=
=+
r
+
x
+
=
=+
r
+
y
+
=
=+
x
+
x
−
=
=−
r
+
y
=
x
x
=
r
y
−
=
=+
x
−
x
=
r
y
=
x
s
C
U
y
w
e
On a copy of the diagram, record which ratios are positive in each quadrant.
ie
id
g
The first quadrant has been completed for you.
-R
s
es
am
br
ev
(All three ratios are positive in the first quadrant.)
sin
cos
tan
y
4th quadrant
y
−
=
=−
r
+
op
y
=
r
es
3rd quadrant
tan θ
Pr
y
=
r
ni
ve
rs
2nd quadrant
-C
R
ev
ie
w
C
op
y
1st quadrant
sin θ
ity
-C
am
br
ev
id
ie
By considering whether x and y are positive or negative ( + or − ) in each of the four quadrants, copy and
complete the table. (r is positive in all four quadrants.)
Copyright Material - Review Only - Not for Redistribution
O
x
ve
rs
ity
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
90°
am
br
id
ev
ie
w
ge
The diagram shows which trigonometric functions are
positive in each quadrant.
Sin
All
-R
180°
y
C
ve
rs
ity
op
WORKED EXAMPLE 5.4
0°, 360°
O
Tan
Pr
es
s
-C
We can memorise this diagram using a mnemonic such as
‘All Students Trust Cambridge’.
Cos
270°
y
cos( −130°)
U
Answer
y
S
br
ev
id
ie
In the second quadrant, sin is positive.
es
Pr
A
ity
ve
y
−130°
ni
ev
T
ie
w
ge
id
ev
-R
br
am
es
s
-C
3
and that 180° ø θ ø 270°, find the value of sin θ and the value of tan θ .
5
Pr
op
y
Answer
θ is in the third quadrant.
sin is negative and tan is positive in this quadrant.
y
ity
S
y2 = 25 − 9 = 16
−3
x
C
w
e
ev
ie
id
g
es
s
-R
br
am
5
T
−4 4
4
−4
= .
∴ sin θ =
= − and tan θ =
−3 3
5
5
-C
O
y
U
Since y < 0, y = −4.
A
θ
ni
ve
rs
y2 + ( −3)2 = 52
C
C
w
ie
C
C
U
R
Given that cos θ = −
x
50° O
op
C
w
C
S
WORKED EXAMPLE 5.5
ev
x
y
In the third quadrant only tan is positive, so cos is
negative.
cos( −130°) = − cos 50°
ie
40°
O
T
rs
op
y
b The acute angle made with the x-axis is 50°.
R
A
140°
s
-C
-R
am
sin 140° = sin 40°
124
w
ge
a The acute angle made with the x-axis is 40°.
y
R
b
C
op
sin 140°
op
a
ni
ev
ie
w
Express in terms of trigonometric ratios of acute angles:
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
ge
C
U
ni
op
y
Chapter 5: Trigonometry
ev
ie
am
br
id
w
WORKED EXAMPLE 5.6
b
sin 120°
-C
a
-R
Without using a calculator, find the exact values of:
Pr
es
s
Answer
∴ sin 120° is positive.
y
S
ve
rs
ity
C
op
y
a 120° lies in the second quadrant.
7π
6
cos
A
T
w
ge
U
R
7π
lies in the third quadrant.
6
7π
∴ cos
is negative.
6
7π
π
−π=
Basic acute angle =
6
6
3
7π
π
∴ cos
= − cos = −
6
6
2
ie
-R
s
es
C
Pr
T
ity
125
rs
y
ve
ev
b
c
cos 50°
op
w
-R
am
230° lies in the third quadrant.
S
A
230°
es
Basic acute angle = 230° − 180° = 50°
x
O
50°
Pr
∴ sin 230° = − sin 50° = −b
y
s
-C
∴ sin 230° is negative.
C
ity
T
y
op
-R
s
es
-C
am
br
ev
ie
id
g
w
e
C
U
R
ev
ie
w
ni
ve
rs
C
op
y
d
ie
br
ev
id
Answer
tan 40°
C
ni
U
sin 230°
ge
R
x
O
π
6
Given that sin 50° = b, express each of the following in terms of b.
a
A
7π
6
ev
id
am
-C
y
op
C
w
ie
WORKED EXAMPLE 5.7
a
C
y
S
br
b
x
O
y
3
2
ni
ev
ie
∴ sin 120° = sin 60° =
120°
60°
C
op
w
Basic acute angle = 180° − 120° = 60°
Copyright Material - Review Only - Not for Redistribution
tan 140°
ve
rs
ity
am
br
id
1 − b2
= 1 − b2
1
-R
1
50°
b
50°
√1 – b2
C
op
U
R
ni
ev
ie
40°
1
d 140° lies in the second quadrant.
-R
1 − b2
b
es
ity
tan 125°
ni
4π
5
f
9π
8
g
7π 
cos  −
 10 
sin
d
cos( −245°)
h
tan
sin 225°
d
tan( −300°)
π
tan  − 
 6 
h
cos
d
cos 245°
y
c
op
cos
cos 305°
C
e
b
ve
sin 190°
R
a
rs
1 Express the following as trigonometric ratios of acute angles.
U
C
w
C
Pr
y
op
EXERCISE 5C
ie
x
O
T
s
-C
∴ tan 140° = − tan 40° = −
40°
ev
id
br
am
Basic acute angle = 180° − 140° = 40°
ev
A
140°
ie
S
w
ge
y
∴ tan 140° is negative.
126
y
1 − b2
b
ve
rs
ity
∴ tan 40° =
w
C
op
y
c Show 40° on the triangle:
b
√ 1 – b2
Pr
es
s
-C
∴ cos 50° =
w
b Draw the right-angled triangle showing the angle 50°:
ev
ie
ge
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
11π
9
am
br
4π
3
w
c
f
cos
7π
3
ie
sin
tan 330°
ev
e
b
g
-R
cos 120°
id
a
ge
2 Without using a calculator, find the exact values of each of the following.
10 π
3
s
y
ni
ve
rs
U
sin θ
5
and that 180° ø θ ø 360°, find the value of:
12
b cos θ
e
a
w
ie
c
ev
sin 25°
cos 65°
s
-R
b
es
br
tan 205°
am
a
id
g
7 Given that tan 25° = a, express each of the following in terms of a.
-C
C
w
ie
ev
R
6 Given that tan θ = −
op
sin θ
1
and that 180° ø θ ø 270°, find the value of:
3
b tan θ
5 Given that cos θ = −
C
a
Pr
cos θ
a
2
and that θ is obtuse, find the value of:
5
b tan θ
ity
op
y
4 Given that sin θ =
es
-C
3 Given that sin θ < 0 and tan θ < 0, name the quadrant in which angle θ lies.
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
C
U
ni
op
y
Chapter 5: Trigonometry
cos A
-C
a
y
w
ev
ie
ve
rs
ity
sin A
−1
sin θ
……
1
2
1
cos θ
−2
− 2
ie
ev
1
3
−
1
2
s
……
es
id
br
am
op
Pr
y
-C
θ = 210°
w
……
ge
θ = ……
-R
U
R
ni
C
op
Without using a calculator, copy and complete the table.
θ = 120°
1
1
and
.
sin θ
tan θ
y
ev
ie
w
PS 11 In the table, 0° ø θ ø 360° and the missing function is from the list cos θ , tan θ ,
……
127
rs
y
ve
ni
ev
ie
EXPLORE 5.2
op
w
C
ity
5.4 Graphs of trigonometric functions
-C
C
w
-R
am
br
ev
id
ie
ge
U
R
Consider taking a ride on a Ferris wheel with radius 50 metres
that rotates at a constant speed.
You enter the ride from a platform that is level with the centre
of the wheel and the wheel turns in an anticlockwise direction
through one complete turn.
es
s
1 Sketch the following two graphs and discuss their properties.
Pr
ni
ve
rs
ity
b The graph of your horizontal displacement from the centre of the wheel plotted
against the angle turned through.
-R
s
es
am
br
ev
ie
id
g
w
e
C
U
op
y
2 Discuss with your classmates what the two graphs would be like if you turned
through two complete turns.
-C
ev
ie
w
C
op
y
a The graph of your vertical displacement from the centre of the wheel plotted
against the angle turned through.
R
cos 347°
2
3
and cos B = , where A and B are in the same quadrant, find the value of:
3
4
b cos A
c sin B
d tan B
C
op
d
sin 257°
5
4
and cos B = − , where A and B are in the same quadrant, find the value of:
13
5
b tan A
c sin B
d tan B
10 Given that tan A = −
a
c
tan13°
-R
9 Given that sin A =
b
Pr
es
s
sin 77°
am
br
id
a
ge
8 Given that cos 77° = b, express each of the following in terms of b.
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
The graphs of y = sin x and y = cos x
w
ge
y
am
br
id
ev
ie
Suppose that OP makes an angle of x with the positive
horizontal axis and that P moves around the unit circle,
through one complete revolution.
P(cos x, sin x)
1
-R
x
The coordinates of P will be (cos x, sin x ).
x
O
op
y
Pr
es
s
-C
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
C
ve
rs
ity
The height of P above the horizontal axis changes from 0 → 1 → 0 → −1 → 0.
360 x
270
-R
–1
-C
am
br
ev
y = sin x
w
180
ie
90
id
O
ge
U
R
ni
1
y
ev
ie
y
C
op
w
The graph of sin x against x for 0° < x < 360° is therefore:
es
s
The displacement of P from the vertical axis changes from 1 → 0 → −1 → 0 → 1.
Pr
op
y
The graph of cos x against x for 0° ø x ø 360° is therefore:
y = cos x
ity
y
C
128
O
y
360 x
270
180
ie
w
ge
br
ev
id
–1
C
U
R
90
op
ni
ev
ve
ie
w
rs
1
s
es
Pr
O
90
–1
270
360 x
–360
–270
–90
O
C
w
e
ev
ie
id
g
es
s
-R
br
am
-C
–180
–1
U
R
ev
ie
180
y
–90
y = cos x
op
–180
ity
–270
y
1
ni
ve
rs
op
y
-C
y
1
y = sin x
w
C
–360
-R
am
The graphs of y = sin x and y = cos x can be continued beyond 0° ø x ø 360° :
Copyright Material - Review Only - Not for Redistribution
90
180
270
360 x
ve
rs
ity
C
U
ni
op
y
Chapter 5: Trigonometry
am
br
id
ev
ie
w
ge
The sine and cosine functions are called periodic functions because they repeat themselves
over and over again.
-R
The period of a periodic function is defined as the length of one repetition or cycle.
-C
The sine and cosine functions repeat every 360° .
Pr
es
s
We say they have a period of 360° (or 2π radians).
C
ve
rs
ity
op
y
The amplitude of a periodic function is defined as the distance between a maximum
(or minimum) point and the principal axis.
C
op
ni
w
-R
am
br
●
ie
●
ev
●
sin( − x ) = − sin x
sin(180° − x ) = sin x
sin(180° + x ) = − sin x
sin(360° − x ) = − sin x
sin(360° + x ) = sin x
U
R
●
ge
●
y
The symmetry of the curve y = sin x shows these important relationships:
id
ev
ie
w
The functions y = sin x and y = cos x both have amplitude 1.
s
-C
EXPLORE 5.3
2
cos(180° − x ) =
4
cos(360° − x ) =
5
cos(360° + x ) =
3
rs
y
op
ni
C
U
-R
s
–90
90
O
es
–180
180
270
360
450
540 x
w
ie
U
op
The tangent function behaves very differently to the sine and cosine functions.
y
y = tan x
ni
ve
rs
C
ity
Pr
op
y
-C
am
br
ev
id
ie
w
ge
y
w
e
C
The tangent function repeats its cycle every 180° so its period is 180° (or π radians).
s
es
am
The tangent function does not have an amplitude.
-R
br
ev
ie
id
g
The red dashed lines at x = ± 90°, x = 270° and x = 450° are called asymptotes. The
branches of the graph get closer and closer to the asymptotes without ever reaching them.
-C
ev
R
129
cos(180° + x ) =
ve
ev
The graph of y = tan x
R
Pr
cos( − x ) =
ity
1
ie
w
C
op
y
es
By considering the shape of the cosine curve, complete the following statements,
giving your answers in terms of cos x.
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
ev
ie
am
br
id
EXPLORE 5.4
w
ge
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
2
tan(180° − x ) =
4
tan(360° − x ) =
5
tan(360° + x ) =
3
tan(180° + x ) =
op
y
Pr
es
s
tan( − x ) =
-C
1
-R
By considering the shape of the tangent curve, complete the following statements,
giving your answers in terms of tan x.
Transformations of trigonometric functions
ve
rs
ity
C
ev
br
am
x
360
s
270
es
-C
180
y = sin x
op
Pr
y
90
-R
y = 2 sin x
1
rs
C
ity
–2
130
y
ie
w
ge
id
y
2
–1
U
The graph of y = a sin x
O
C
op
ni
w
ev
ie
R
REWIND
These rules for the transformations of the graph y = f( x ) can be used to transform
graphs of trigonometric functions. These transformations include y = a f( x ), y = f( ax ),
y = f(x ) + a and y = f( x + a ) and simple combinations of these.
ve
ni
op
y
It is a stretch, stretch factor 2, parallel to the y-axis.
The amplitude of y = 2 sin x is 2 and the period is 360°.
ge
C
U
R
ev
ie
w
The graph of y = 2 sin x is a stretch of the graph of y = sin x.
ie
ev
id
br
y
am
s
180
360
270
es
-C
y = sin 2x
x
Pr
op
y
90
-R
y = sin x
1
O
w
The graph of y = sin ax
C
ity
–1
y
op
C
U
w
e
ev
ie
id
g
es
s
-R
br
am
-C
R
ev
ie
w
ni
ve
rs
The graph of y = sin 2 x is a stretch of the graph of y = sin x.
1
It is a stretch, stretch factor , parallel to the x-axis.
2
The amplitude of y = sin 2 x is 1 and the period is 180°.
Copyright Material - Review Only - Not for Redistribution
In Section 2.6, you
learnt some rules for
the transformation of
the graph y = f( x ).
Here we will look
at how these rules
can be used to
transform graphs
of trigonometric
functions.
ve
rs
ity
C
U
ni
op
y
Chapter 5: Trigonometry
ev
ie
-R
y = 1 + sin x
-C
1
y
O
270
180
y = sin x
op
90
C
y
U
ni
C
op
w
ev
ie
The graph of y = 1 + sin x is a translation of the graph of y = sin x.
 0
It is a translation of   .
 1
R
x
ve
rs
ity
–1
360
Pr
es
s
am
br
id
y
2
w
ge
The graph of y =
= a + sin x
w
ge
The amplitude of y = 1 + sin x is 1 and the period is 360°.
am
y
y = sin(x + 90)
O
C
y = sin x
rs
w
–1
ve
ie
131
360 x
270
180
ity
90
Pr
op
y
es
s
-C
1
-R
br
ev
id
ie
The graph of y = sin(x + a )
y
w
ge
C
U
 −90 
.
It is a translation of 
 0 
R
op
ni
ev
The graph of y = sin( x + 90) is a translation of the graph of y = sin x.
br
ie
ev
id
The amplitude of y = sin( x + 90) is 1 and the period is 360°.
-C
-R
am
WORKED EXAMPLE 5.8
es
s
On the same grid, sketch the graphs of y = sin x and y = sin( x − 90) for 0° < x < 360°.
ity
Pr
 90 
y = sin( x − 90) is a translation of the graph y = sin x by the vector   .
 0
ni
ve
rs
y
op
y
y = sin x
-R
s
es
am
-C
360 x
270
C
180
ie
–1
90
ev
O
y = sin(x – 90)
br
id
g
e
U
R
ev
ie
1
w
w
C
op
y
Answer
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
am
br
id
y
Step 2: Sketch the graph of y = cos( x + 90).
op
O
y = cos(x + 90)
ni
C
op
O
U
ge
w
ie
-R
s
es
90
180
270
360 x
Pr
ity
–2
y
y = 2 cos(x + 90) + 1
y
ve
3
op
ni
C
U
2
O
90
–1
-R
am
br
ev
id
ie
w
ge
1
–2
es
s
-C
ni
ve
rs
ity
Pr
op
y
a Write down the period and amplitude of f.
op
y
b Write down the coordinates of the maximum and minimum points on the curve y = f( x ).
w
-R
s
es
am
br
ev
ie
id
g
e
d Use your answer to part c to sketch the graph of y = 1 + 3 cos 2 x.
C
U
c Sketch the graph of y = f( x ).
-C
w
C
f( x ) = 3 cos 2 x for 0° < x < 360°.
ie
y = 2 cos(x + 90)
–1
rs
w
ie
ev
R
360 x
y
O
 0
Translate y = 2 cos( x + 90) by the vector   .
 1
Period = 360°
WORKED EXAMPLE 5.9
ev
270
1
Step 4: Sketch the graph of y = 2 cos( x + 90) + 1.
Amplitude = 2.
R
180
2
ev
id
br
am
-C
y
op
C
132
90
–1
Stretch y = cos( x + 90) with stretch factor 2, parallel to the y-axis.
Amplitude = 2
360 x
1
Step 3: Sketch the graph of y = 2 cos( x + 90).
Period = 360°
270
y
ve
rs
ity
C
w
ev
ie
R
Amplitude = 1
180
2
 −90 
Translate y = cos x by the vector  0  .


Period = 360°
90
–1
Pr
es
s
Amplitude = 1
y = cos x
y
-C
Period = 360°
y
2
1
-R
Step 1: Start with a sketch of y = cos x.
w
for 0° < x < 360°, we can build up the transformation in steps.
ev
ie
ge
To sketch the graph of a trigonometric function, such as y = 2 cos( x + 90) + 1
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
Copyright Material - Review Only - Not for Redistribution
180
270
360
x
ve
rs
ity
Answer
w
ge
C
U
ni
op
y
Chapter 5: Trigonometry
am
br
id
ev
ie
360°
= 180°
2
Amplitude = 3
Period =
b
y = cos x has its maximum and minimum points at:
Pr
es
s
-C
-R
a
Hence, f( x ) = 3 cos 2 x has its maximum and minimum points at:
y
4
3
2
1
270
180
y
w
ie
ev
s
-R
am
Pr
es
y = 3 cos 2x + 1
ity
360 x
270
rs
180
op
y
ve
90
133
ge
C
U
R
ni
y
-C
y
5
4
3
2
1
op
C
w
ie
x
360
 0
y = 1 + 3 cos 2 x is a translation of the graph y = 3 cos 2 x by the vector   .
 1
O
–1
–2
–3
–4
ev
C
op
ni
U
90
br
O
–1
–2
–3
–4
y = 3 cos 2x
ge
R
ev
ie
w
c
d
ve
rs
ity
(0°, 3), (90°, −3), (180°, 3), (270°, −3) and (360°, 3)
id
C
op
y
(0°, 1), (180°, −1), (360°, 1), (540°, −1) and ( 720°, 1 )
br
ev
id
ie
w
WORKED EXAMPLE 5.10
-R
am
a On the same grid, sketch the graphs of y = sin 2 x and y = 1 + 3 cos 2 x for 0° < x < 360°.
-C
b State the number of solutions of the equation sin 2 x = 1 + 3 cos 2 x for 0° < x < 360°.
es
ity
y = 1 + 3 cos 2x
360
y = sin 2x
x
y
270
op
180
w
e
C
U
90
ni
ve
rs
O
–1
–2
–3
–4
Pr
y
5
4
3
2
1
ie
id
g
b The graphs of y = sin 2 x and y = 1 + 3 cos 2 x intersect each other at four points in the interval.
-R
s
es
am
br
ev
Hence, the number of solutions of the equation sin 2 x = 1 + 3 cos 2 x is four.
-C
R
ev
ie
w
C
op
y
a
s
Answer
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
1 Write down the period of each of these functions.
ev
ie
am
br
id
EXERCISE 5D
w
ge
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
f
1
x°
2
y = 5 cos(2 x + 45)°
y = 5 cos 2 x°
c
y = 7 sin
y = 4 sin(2 x + 60)°
f
1
x°
2
y = 2 sin(3x + 10)° + 5
c
y = tan 3x
f
y = 2 sin 3x − 1
i
y = tan( x − 90)
b
y = sin 2 x°
d
y = 1 + 2 sin 3x°
e
y = tan( x − 30)°
c
Pr
es
s
-R
y = cos x°
-C
a
y = 3tan
a
y = sin x°
d
y = 2 − 3 cos 4x°
b
ve
rs
ity
w
C
op
y
2 Write down the amplitude of each of these functions.
e
y
C
op
ie
y = 2 cos( x + 60)
w
h
y = sin( x − 45)
id
g
ge
U
R
ni
ev
ie
3 Sketch the graph of each of these functions for 0° < x < 360°.
1
a y = 2 cos x
b y = sin x
2
d y = 3 cos 2 x
e y = 1 + 3 cos x
ii
ev
am
y = 2 sin x
-C
i
π
iii y = sin  2 x + 

4
π
y = cos  x − 

2
-R
br
4 a Sketch the graph of each of these functions for 0 < x < 2 π .
es
s
b Write down the coordinates of the turning points for your graph for part a iii.
Pr
6 a On the same diagram, sketch the graphs of y = 2 sin x and y = 2 + cos 3x for 0 < x < 2 π.
ve
ie
w
rs
C
b State the number of solutions of the equation sin 2 x = 1 + cos 2 x for 0° < x < 360°.
ity
op
y
5 a On the same diagram, sketch the graphs of y = sin 2 x and y = 1 + cos 2 x for 0° < x < 360°.
134
y
op
ni
ev
b Hence, state the number of solutions, in the interval 0 < x < 2 π, of the equation 2 sin x = 2 + cos 3x.
C
U
R
7 a On the same diagram, sketch and label the graphs of y = 3sin x and y = cos 2 x for the interval 0 < x < 2 π.
ie
-R
s
-C
es
6
Pr
op
y
5
O
π
–
π
3π
–
2π
x
2
e
U
2
y
1
op
2
ni
ve
rs
ity
3
R
ie
-R
s
-C
am
br
Find the value of a, the value of b and the value of c.
ev
id
g
w
Part of the graph y = a sin bx + c is shown above.
es
ev
ie
w
C
4
C
7
am
9
8
ev
id
y
br
8
w
ge
b State the number of solutions of the equation 3sin x = cos 2 x in the interval 0 < x < 2 π.
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
-C
1
w
ev
ie
-R
2
O
120
180
240
300
360 x
Part of the graph of y = a + b cos cx is shown above.
ve
rs
ity
Write down the value of a, the value of b and the value of c.
ev
ie
10 a Sketch the graph of y = 2 sin x for −π < x < π.
ni
U
R
w
br
y
ev
id
ie
State the coordinates of the other points where the line intersects the curve.
-R
s
es
P
π
2
π
135
2π x
y
ev
ve
ie
w
rs
3π
2
ity
O
Pr
5
C
op
y
-C
am
11
ge
b Find the value of k. Give your answer in terms of π.
c
w
s
es
Pr
ity
13 f( x ) = a − b cos x for 0° < x < 360°, where a and b are positive constants.
ni
ve
rs
The maximum value of f( x ) is 8 and the minimum value is −2.
op
y
a Find the value of a and the value of b.
C
U
b Sketch the graph of y = f( x ).
id
g
w
e
14 f( x ) = a + b sin cx for 0° < x < 360°, where a and b are positive constants.
-R
s
es
am
Find the value of a, the value of b and the value of c.
ev
br
ie
The maximum value of f( x ) is 9, the minimum value of f( x ) is 1 and the period is 120°.
-C
ie
w
C
op
y
a the value of a and the value of b
ev
-R
br
am
-C
7π 
Given that f(0) = 3 and that f 
= 2, find:
 6 
b the range of f .
ie
ev
id
ge
C
U
R
ni
op
Part of the graph of y = a tan bx + c is shown above.
π
The graph passes through the point P  , 8  .
4

Find the value of a, the value of b and the value of c.
12 f( x ) = a + b sin x for 0 < x < 2 π
R
C
op
The straight line y = kx intersects this curve at the maximum point.
y
w
C
op
y
60
Pr
es
s
3
am
br
id
5
4
C
y
ge
9
U
ni
op
y
Chapter 5: Trigonometry
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
am
br
id
-R
a Write down the value of A and the value of B.
-C
b Write down the amplitude of f( x ).
y
16 The graph of y = sin x is reflected in the line x = π and then in the line y = 1.
17 The graph of y = cos x is reflected in the line x =
π
and then in the line y = 3.
2
ni
C
op
y
Find the equation of the resulting function.
U
R
ev
ie
w
PS
Find the equation of the resulting function.
ve
rs
ity
C
op
PS
Sketch the graph of f( x ).
Pr
es
s
c
w
The maximum value of f( x ) is 7 and the period is 60°.
ev
ie
ge
15 f( x ) = A + 5 cos Bx for 0° < x < 120°
ge
5.5 Inverse trigonometric functions
-R
am
br
ev
id
ie
w
The functions y = sin x, y = cos x and y = tan x for x ∈ R are many-one functions. If,
however, we suitably restrict the domain of each of these functions, it is possible to make
the function one-one and hence we can define each inverse function.
y
1
C
136
π
–
2
y = sin –1x
π
–
y
op
ni
x
O
–1
1
-R
y = sin x
π
π
<x<
2
2
range: −1 < sin x < 1
s
es
π
––
2
y = sin −1 x
op
y
domain: − 1 < x < 1
π
π
range: − < sin −1 x <
2
2
-R
s
es
-C
am
br
ev
ie
id
g
w
e
C
U
R
ev
ie
w
ni
ve
rs
C
ity
Pr
op
y
-C
domain: −
REWIND
In Chapter 2 you
learnt about functions
and that only one-one
functions can have
an inverse function.
You also learnt that if
f and f −1 are inverse
functions, then the
graph of f −1 is a
reflection of the graph
of f in the line y = x .
x
ev
id
br
am
w
C
2
ie
R
–1
U
O
π
––
2
ge
ev
ve
ie
w
rs
y = sin x
In Section 2.5 you learnt
about the inverse of a
function. Here we will
look at the particular
case of the inverse of a
trigonometric function.
y
ity
op
Pr
y
es
s
-C
The graphs of the suitably restricted functions y = sin x, y = cos x and y = tan x and their
inverse functions y = sin −1 x, y = cos −1 x and y = tan −1 x, together with their domains and
ranges are:
REWIND
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
ge
C
U
ni
op
y
Chapter 5: Trigonometry
ev
ie
w
y
π
am
br
id
π
x
y
Pr
es
s
π
2
w
y = cos x
ev
ie
domain: 0 < x < π
–1
x
w
ge
id
ie
range: 0 < cos −1 x < π
ev
y = tan x
y
y = tan –1 x
π
–
2
es
s
-C
-R
am
br
1
domain: −1 < x < 1
π x
2
137
O
op
y
x
C
U
R
ni
ev
ve
ie
w
rs
O
ity
–π
2
Pr
y
op
C
O
y = cos −1 x
U
R
y
y = cos –1 x
2
ni
range: −1 < cos x < 1
ve
rs
ity
C
op
–1
π
–
C
op
-C
O
-R
y = cos x
y
y
1
br
ev
id
ie
w
ge
π
––
2
am
y = tan x
domain: x ∈ R
π
π
range: − < tan −1 x <
2
2
-R
π
π
<x<
2
2
range: tan x ∈ R
y = tan −1 x
Pr
op
y
es
s
-C
domain: −
C
U
op
y
ni
ve
rs
x = sin−1 0.5
π
Using a calculator gives x = .
6
The angle that the calculator gives is the one that lies in the range of the function sin −1.
(This is sometimes called the principal angle.)
-R
s
es
am
br
ev
ie
id
g
w
e
The principal angle is the angle that lies in the range of the inverse trigonometric function.
5π
, that satisfies sin x = 0.5 with 0 < x < π. We can
There is a second angle, x =
6
find this second angle either by using skills learnt earlier in this chapter or by using the
symmetry of the curve y = sin x.
-C
R
ev
ie
w
C
ity
When solving the equation sin x = 0.5 for 0 < x < π, we can find one solution using the
inverse functions:
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
ev
ie
w
ge
am
br
id
WORKED EXAMPLE 5.11
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
b
 3
cos−1 

 2 
ev
ie
ve
rs
ity
a sin −1 0 means the angle whose sine is 0, where −90° < angle < 90°.
Hence, sin −1 0 = 0°.
w
C
op
y
Answer
tan −1( −1)
c
Pr
es
s
sin −1 0
-C
a
-R
The output of the sin −1, cos −1 and tan −1 functions can be given in degrees if that is needed.
Without using a calculator, write down, in degrees, the value of:
U
R
ni
C
op
y
 3
3
b cos −1 
, where 0° < angle < 180°.
means the angle whose cosine is
2
 2 
id
ie
w
ge
 3
= 30°.
Hence, cos −1 
 2 
br
ev
c tan −1( −1) means the angle whose tangent is −1, where −90° < angle < 90°.
es
s
-C
-R
am
Hence, tan −1 ( −1) = −45°.
op
Pr
y
WORKED EXAMPLE 5.12
ity
x
The function f( x ) = 3sin   − 1 is defined for the domain −π < x < π.
 2
a Sketch the graph of y = f( x ) and explain why f has an inverse function.
rs
ie
ev
-R
Pr
ni
ve
rs
w
C
-R
s
es
am
br
ev
ie
id
g
w
e
x
f( x ) = 3sin   − 1
 2
-C
c
U
b Range is −4 < f( x ) < 2.
op
f has an inverse function because f is a one-one function with this domain.
y
ity
C
–2
–4
ie
π x
π
2
O
s
π
2
f
es
am
-C
–
y
2
op
y
–π
w
ge
id
br
a
ev
C
U
R
c Find f −1( x ) and state its domain.
Answer
R
op
ni
ev
b Find the range of f .
y
ve
ie
w
C
138
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
C
U
ni
op
y
Chapter 5: Trigonometry
ge
x
y = 3sin   − 1
 2
am
br
id
ev
ie
w
Step 1: Write the function as y =
ve
rs
ity
y = 2 sin −1

x + 1
for −4 < x < 2 .
3 
U
R
ni
C
op
The inverse function is f −1( x ) = 2 sin −1

id
ie
w
ge
EXERCISE 5E
x + 1
3 
y
y
op
C
w
x+
+1
y
= sin  

3
2
y
x + 1
= sin −1

2
3 
Pr
es
s
Step 3: Rearrange to make y the subject.
ev
ie
y
y
x = 3sin   − 1
 2
-R
-C
Step 2: Interchange the x and y variables.
-R
am
br
ev
1 Without using a calculator, write down, in degrees, the value of:
1
a cos −1 1
b sin −1
2
tan −1 − 3
(
)
tan −1 3
f
 1 
cos −1  −


2 
es
e
s
sin −1 ( −1 )
-C
d
c
Pr
sin −1 0
d
 1 
tan −1  −

3 
tan −1 1
c
 1 
cos −1 
 2 
e
1
cos −1  − 
 2
f

3
sin −1  −
 2 
y
op
rs
ni
ve
w
ie
ev
b
ity
a
C
op
y
2 Without using a calculator, write down, in terms of π, the value of:
w
π
π
<x< .
2
2
br
ev
4 The function f( x ) = 3sin x − 4 is defined for the domain −
ie
id
ge
C
U
R
3
3 Given that θ = cos−1   , find the exact value of:
5
b tan2 θ
a sin2 θ
b Find f −1( x ) .
-R
am
a Find the range of f .
s
es
Pr
a Find the range of f and sketch the graph of y = f( x ).
b Explain why f has an inverse and find the equation of this inverse.
6 The function f( x ) = 5 − 2 sin x is defined for the domain
π
< x < p.
2
C
U
b For this value of p, find f −1( x ) and state the domain of f −1.
op
a Find the largest value of p for which f has an inverse.
y
ity
Sketch the graph of y = f −1( x ) on your graph for part a.
ni
ve
rs
c
w
ie
Find f −1( x ) and state its range.
s
-R
b
es
am
br
a Find the range of f .
ev
id
g
e
x
7 The function f( x ) = 4 cos   − 5 is defined for the domain 0 < x < 2 π.
 2
-C
R
ev
ie
w
C
op
y
-C
5 The function f( x ) = 4 − 2 cos x is defined for the domain 0 < x < π.
Copyright Material - Review Only - Not for Redistribution
139
ve
rs
ity
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
One solution is given by x = sin −1(0.5) =
π
(or 30°).
6
w
-R
am
br
id
Consider solving the equation sin x = 0.5 for −360° < x < 360°.
ev
ie
ge
5.6 Trigonometric equations
ve
rs
ity
C
w
O
−90
ni
ev
ie
90
30
360 x
270
180
150
U
R
−360 −270 −180
–330
–210
y = 0.5
y
op
y
1
C
op
y
The graph of y = sin x for −360° < x < 360° is:
Pr
es
s
-C
There are, however, many more values of x for which sin x = 0.5.
y = sin x
w
id
ie
ge
−1
-R
am
br
ev
The graph shows there are four values of x, between −360° and 360°, for which sin x = 0.5.
es
s
-C
We can use the calculator value of x = 30°, together with the symmetry of the curve to find
the remaining answers.
C
ve
ie
w
rs
WORKED EXAMPLE 5.13
y
op
w
ge
Answer
C
U
R
ni
ev
Solve cos x = −0.7 for 0° < x < 360°.
Use a calculator to find cos−1( −0.7), correct to
1 decimal place.
id
ie
cos x = −0.7
ev
br
One solution is x = 134.4°
-R
am
y
es
Pr
op
y
134.4
O
225.6
180
270
360
w
x
y = –0.7
ni
ve
rs
C
ity
90
y
–1
ie
U
The sketch graph shows there are two values of x, between − 0° and 360°, for which cos x = − 0.7 .
w
e
C
Using the symmetry of the curve, the second value is (360° − 134.4°) = 225.6°.
ie
ev
(correct to 1 decimal place)
s
-R
x = 134.4° or 225.6°
es
am
br
id
g
Hence, the solution of cos x = −0.7 for 0° < x < 360° is:
-C
ev
R
y = cos x
s
-C
1
op
140
ity
op
Pr
y
Hence, the solution of sin x = 0.5 for − 360° < x < 360° is:
x = −330°, −210°, 30° or 150°
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
ge
C
U
ni
op
y
Chapter 5: Trigonometry
am
br
id
ev
ie
w
WORKED EXAMPLE 5.14
Pr
es
s
-C
Answer
-R
Solve tan 2 A = −2.1 for 0° < A < 180°.
tan x = −2.1
A solution is x = −64.54°.
Let 2A = x.
Use a calculator to find tan −1( −2.1).
ve
rs
ity
C
op
y
tan 2 A = −2.1
115.46
90
-R
Pr
y
ity
2 A = x:
2 A = −64.54°
A = −32.3°
2 A = 115.46°
A = 57.7°
y
ve
ni
op
Hence, the solution of tan 2 A = −2.1 for 0° < A < 180° is:
ev
141
2 A = 295.46°
A = 147.7°
rs
op
C
w
ie
x = (115.46° + 180°)
= 295.46°
es
x = ( −64.54° + 180°)
= 115.46°
s
x = −64.54°
id
ie
w
ge
C
U
R
A = 57.7° or 147.7° (correct to 1 decimal place)
-R
am
br
ev
WORKED EXAMPLE 5.15
es
s
-C
π
Solve sin  2 A +  = 0.6 for 0 < A < π.

6
Pr
y
Use a calculator to find sin−1 ( 0.6 ).
x = 0.6435 radians
-R
s
es
-C
am
br
ev
ie
id
g
w
e
C
U
R
π
= x.
6
op
sin x = 0.6
ev
Let 2 A +
ity
π
sin  2 A +  = 0.6

6
ni
ve
rs
w
C
op
y
Answer
ie
y = –2.1
ev
br
am
-C
Using the symmetry of the curve:
Using
360 x
270
180
ie
–90
295.46
w
O
id
ge
U
R
–64.54
C
op
y
y = tan x
ni
ev
ie
w
y
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
2.498 π
2π x
3π
–
2
U
R
y
= 2.498
C
op
π
= x:
6
π
2 A + = 0.6435
6
1
π
A =  0.6435 − 
2
6
2A +
x = π − 0.6435
ni
ev
ie
w
x = 0.6435
π
= 2.498
6
1
π
A =  2.498 − 
2
6
br
ev
id
ie
ge
2A +
w
C
Using the symmetry of the curve:
ve
rs
ity
op
y
y = sin x
–1
Using
-R
y = 0.6
Pr
es
s
2
ev
ie
w
ge
π
–
0.6435
-C
O
am
br
id
y
1
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
A = 0.0600
-R
am
A = 0.987
es
s
-C
π
Hence, the solution of sin  2 A +  = .0.6 for 0 < A < π is:

6
ity
op
Pr
y
A = 0.0600 or 0.987 radians (correct to 3significant figures)
es
Pr
ity
ni
ve
rs
C
w
x 2 + y2 = r 2
op
C
U
x
y
and sin θ = .
r
r
-R
s
es
am
br
ev
ie
id
g
e
Use cos θ =
w
2
y
Divide both sides by r 2.
 x + y =1
 r
 r
-C
ie
ev
C
w
-R
s
-C
op
y
sin θ
for all θ with cos θ ≠ 0.
cos θ
Rule 2
2
ie
y
x
and cos θ = .
r
r
ev
Use sin θ =
am
br
id
ge
Divide numerator and denominator by r.
KEY POINT 5.2
tan θ =
y
ve
x
U
y
tan θ =
x
y
θ°
ni
Rule 1
 y
 r
=
 x
 r
R
r
Two very important rules can be found using this triangle.
R
ev
ie
w
rs
Consider a right-angled triangle.
op
C
142
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
ev
ie
am
br
id
KEY POINT 5.3
w
ge
C
U
ni
op
y
Chapter 5: Trigonometry
-R
cos2 θ + sin2 θ = 1 for all θ .
op
y
Pr
es
s
-C
If we use the unit circle definition of the trigonometric functions, we discover that these
two important rules are true for all valid values of θ . We can use them to help solve more
complicated trigonometric equations.
w
C
ve
rs
ity
WORKED EXAMPLE 5.16
y
ni
R
Answer
C
op
ev
ie
Solve 3 cos2 x − sin x cos x = 0 for 0° < x < 360°.
U
3 cos2 x − sin x cos x = 0
Factorise.
w
ge
cos x(3 cos x − sin x ) = 0
ev
-R
s
es
tan x = 3
x = 71.6 or 180 + 71.6
x = 71.6° or 251.6°
y
Pr
op
y
-C
am
br
id
ie
cos x = 0 or 3 cos x − sin x = 0
x = 90°, 270°
sin x = 3 cos x
3
143
y
y = tan x
ge
U
R
360 x
270
op
251.6
C
180
s
-C
-R
am
br
ev
id
The solution of 3 cos2 x − sin x cos x = 0 for 0° < x < 360° is:
x = 71.6°, 90°, 251.6° or 270°
w
90
ni
ev
71.6
ie
O
ve
ie
w
rs
C
ity
y=3
Pr
op
y
es
WORKED EXAMPLE 5.17
ity
ni
ve
rs
Answer
2 sin2 x + 3 cos x − 3 = 0
op
y
Replace sin2 x with 1 − cos2 x.
2(1 − cos2 x ) + 3 cos x − 3 = 0
C
U
Expand brackets and collect terms.
id
g
w
e
2 cos2 x − 3 cos x + 1 = 0
ie
Factorise.
-R
s
es
am
br
ev
(2 cos x − 1)(cos x − 1) = 0
-C
R
ev
ie
w
C
Solve 2 sin2 x + 3 cos x − 3 = 0 for 0 < x < 2 π.
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
w
ge
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
Pr
es
s
-C
-R
am
br
id
ev
ie
1
cos x = 1
or
2
π
π π
x=
or 2 π − x x = 0 or 2 π
3 3
3
π
5π
x=
or
3
3
cos x =
op
y
y
1
π
3
π
2
π
2π x
C
op
y = cos x
-C
ie
ev
-R
am
br
id
The solution of 2 sin2 x + 3 cos x − 1 = 0 for 0 < x < 2 π is:
π 5π
x = 0, ,
or 2 π
3 3
w
ge
U
–1
R
3π 5π
2 3
ni
ev
ie
w
O
y
C
ve
rs
ity
y = 0.5
y
es
s
EXERCISE 5F
Pr
tan x = 1.5
b
sin x = 0.4
c
cos x = 0.7
d
sin x = −0.3
e
cos x = −0.6
f
tan x = −2
g
2 cos x − 1 = 0
h
5 sin x + 3 = 0
d
sin x = −0.7
h
5 tan x + 7 = 0
ity
C
w
ve
ie
a
rs
op
1 Solve each of these equations for 0° < x < 360°.
144
y
c
tan x = 3
f
cos x = −0.5
g
4 sin x = 3
cos 2 x = 0.6
b
sin 3x = 0.8
e
3 cos 2 x = 2
f
5 sin 2 x = −4
w
c
tan 2 x = 4
d
sin 2 x = −0.5
g
4 + 2 tan 2 x = 0
h
1 − 5 sin 2 x = 0
b
π
cos x +  = −0.5 for 0 < x < 2 π

6
3sin(2 x − 4) = 2
x π
2 sin +  = 1 for 0 < x < 4 π
 3 4
s
-C
ev
a
-R
br
3 Solve each of these equations for 0° < x < 180°.
am
op
cos x = 0.5
C
b
ie
tan x = −3
ni
e
U
sin x = 0.3
ge
a
id
R
ev
2 Solve each of the these equations for 0 < x < 2 π.
sin( x − 60°) = 0.5
c
ity
es
a
cos(2 x + 45°) = 0.8 for 0° < x < 180°
d
e
x
2 tan  + 3 = 0 for 0° < x < 540°
 2
f
-R
s
es
-C
am
br
ev
ie
id
g
w
e
C
U
op
y
ni
ve
rs
Pr
for 0° < x < 360°
R
ev
ie
w
C
op
y
4 Solve each of these equations for the given domains.
Copyright Material - Review Only - Not for Redistribution
for 0 < x < π
ve
rs
ity
C
U
ni
op
y
Chapter 5: Trigonometry
w
2 sin x − 3 cos x = 0
d
3 cos 2 x − 4 sin 2 x = 0
b
5 sin2 x − 3sin x = 0
d
sin2 x + 2 sin x cos x = 0
f
sin x tan x = 4 sin x
b
4 tan2 x = 9
ev
ie
b
4 sin x + 7 cos x = 0
-R
c
2 sin x = cos x
am
br
id
a
ge
5 Solve each of these equations for 0° < x < 360°.
Pr
es
s
-C
6 Solve 4 sin(2 x + 0.3) − 5 cos(2 x + 0.3) = 0 for 0 < x < π.
a
sin x cos( x − 60) = 0
c
tan2 x = 5 tan x
e
2 sin x cos x = sin x
ve
rs
ity
ev
ie
w
C
op
y
7 Solve each of these equations for 0° < x < 360°.
3 cos2 x − 2 cos x − 1 = 0
br
d
2 sin2 x − cos x − 1 = 0
f
cos x + 5 = 6 sin2 x
h
1 + tan x cos x = 2 cos2 x
b
2 cos2 x + 5 sin x = 4
s
2 cos2 x − sin2 x − 2 sin x − 1 = 0
es
g
tan2 x + 2 tan x − 3 = 0
-R
am
3 cos2 x − 3 = sin x
-C
e
ie
c
b
ev
2 sin2 x + sin x − 1 = 0
id
a
w
ge
9 Solve each of these equations for 0° < x < 360°.
C
op
ni
4 cos2 x = 1
U
R
a
y
8 Solve each of these equations for 0° < x < 360°.
Pr
a
4 tan x = 3 cos x
ity
C
op
y
10 Solve each of these equations for 0 < x < 2 π.
y
ev
ve
ie
w
rs
11 Solve sin2 x + 3sin x cos x + 2 cos2 x = 0 for 0 < x < 2 π.
U
R
ni
op
5.7 Trigonometric identities
C
ge
x + x = 2 x is called an identity because it is true for all values of x.
br
ev
id
ie
w
When writing an identity, we often replace the = symbol with a ≡ symbol to emphasise
that it is an identity.
Pr
op
y
es
s
-C
-R
am
Two commonly used trigonometric identities are:
sin x
sin2 x + cos2 x ≡ 1 and tan x ≡
cos x
In this section you will learn how to use these two identities to simplify expressions and to
prove other more complicated identities that involve sin x, cos x and tan x.
op
y
ni
ve
rs
-R
s
es
-C
am
br
ev
ie
id
g
w
e
C
U
R
ev
ie
w
C
ity
When proving an identity, it is usual to start with the more complicated side of the identity
and prove that it simplifies to the less complicated side.
Copyright Material - Review Only - Not for Redistribution
145
ve
rs
ity
ev
ie
w
ge
am
br
id
WORKED EXAMPLE 5.18
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
≡ 4 cos2 x − 3 + 3 cos2 x
ve
rs
ity
TIP
y
WORKED EXAMPLE 5.19
ni
LHS ≡
1 + sin x
cos x
+
cos x
1 + sin x
≡
(1 + sin x )2 + cos2 x
cos x(1 + sin x )
Expand the brackets in the numerator.
≡
1 + 2 sin x + sin2 x + cos2 x
cos x(1 + sin x )
Use sin2 x + cos2 x ≡ 1 .
≡
2 + 2 sin x
cos x(1 + sin x )
≡
2(1 + sin x )
cos x( 1 + sin x )
≡
2
cos x
es
s
-C
-R
am
br
ev
Add the two fractions.
ity
y
C
op
Divide numerator and denominator by
1 + sin x.
br
ev
id
ie
w
ge
U
ni
ve
rs
Factorise the numerator.
-R
am
≡ RHS
Pr
y
op
R
ev
ie
w
C
146
C
ity
op
y
Pr
2
1 + sin x 
1 
≡ tan x +
.
Prove the identity
1 − sin x 
cos x 
Add the two fractions.
w
ie
ev
br
-R
s
(1 + sin )2
2
es
-C
am
 sin x + 1 
≡
 cos x 
sin x
.
cos x
C
2
id
g
 sin x
1 
≡
+
 cos x cos x 
Use tan x =
U
2
e

1 
RHS ≡  tan x +
cos x 

op
w
ni
ve
rs
Answer
ie
es
s
-C
WORKED EXAMPLE 5.20
ev
ie
id
w
ge
U
R
1 + sin x
cos x
2
Prove the identity
+
≡
.
cos x
1 + sin x
cos x
y
ev
ie
w
C
op
≡ 7 cos2 x − 3
Answer
R
Replace sin2 x with 1 − cos2 x.
Pr
es
s
y
-C
4 cos2 x − 3sin2 x ≡ 4 cos2 x − 3(1 − cos2 x )
C
op
Answer
-R
Express 4 cos2 x − 3sin2 x in terms of cos x.
Copyright Material - Review Only - Not for Redistribution
LHS means left-hand
side and RHS means
right-hand side.
ve
rs
ity
w
(1 + sin x )2
1 − sin2 x
Replace cos2 x with 1 − sin2 x in the
denominator.
ev
ie
≡
ge
(1 + sin x )2
cos2 x
(1 + sin x )2
(1 + sin x )(1 − sin x )
≡
1 + sin x
1 − sin x
Pr
es
s
≡
Divide numerator and denominator by
1 + sin x.
y
U
w
ge
EXPLORE 5.5
id
ie
R
ni
C
op
≡ LHS
ev
ie
Use 1 − sin2 x = (1 + sin x )(1 − sin x ).
ve
rs
ity
w
C
op
y
-C
-R
am
br
id
≡
C
U
ni
op
y
Chapter 5: Trigonometry
-R
s
cos x
es
tan 2 x cos 3 x
cos 3 x + cos x sin 2 x
147
ity
op
C
sin 2 x
tan 2 x cos x
Pr
y
-C
am
1
sin x tan x + cos x
1 – sin 2 x
cos x
sin 2 x
sin x
tan x
cos 3 x
sin 2 x cos x
(1 – cos x)(1 + cos x)
rs
1 – sin 2 x
y
op
ni
ev
ve
ie
w
ev
br
Equivalent trigonometric expressions:
id
ie
Create trigonometric expressions of your own that simplify to sin x.
w
ge
C
U
R
Discuss why each of the trigonometric expressions in the coloured boxes simplifies to
cos x.
es
s
-C
Compare your answers with those of your classmates.
-R
am
br
ev
(Your expressions must contain at least two different trigonometric ratios.)
Pr
op
y
EXERCISE 5G
ity
1 + sin x − sin2 x
≡ cos x + tan x
cos x
op
y
d
w
e
id
g
ie
ev
cos 4 x + sin2 x cos2 x ≡ cos2 x
-R
s
es
am
1 − cos2 x
≡ tan x
sin x cos x
f
br
e
b
C
cos2 x
≡ 1 + sin x
1 − sin x
cos2 x − sin2 x
+ sin x ≡ cos x
cos x + sin x
-C
R
c
cos x tan x ≡ sin x
U
a
ni
ve
rs
2 Prove each of these identities.
ev
ie
w
C
1 Express 2 sin2 x − 7 cos2 x + 4 in terms of sin x.
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
c
2 − (sin x + cos x )2 ≡ (sin x − cos x )2
b
cos2 x − sin2 x ≡ 1 − 2 sin2 x
d
cos 4 x + sin2 x ≡ sin 4 x + cos2 x
b
sin 4 x − cos 4 x ≡ 2 sin2 x − 1
ve
rs
ity
cos 4 x − sin4 x
≡ 1 − tan2 x
cos2 x
e
sin x − cos x tan x − 1
≡
sin x + cos x tan x + 1
g
tan x + 1
≡ sin x + cos x
sin x tan x + cos x
d
cos x
1
≡ 1+
tan x(1 − sin x )
sin x
f
 1
1 − cos x
1 
 sin x − tan x  ≡ 1 + cos x
ie
sin2 x (1 − cos2 x )
≡ tan 4 x
cos2 x (1 − sin2 x )
b
tan x +
-R
s
es
1
− cos x ≡ sin x tan x
cos x
e
sin x
sin x
2 tan x
+
≡
1 − sin x 1 + sin x
cos x
Pr
c
1
1
≡
tan x sin x cos x
sin x
1 + cos x
2
+
≡
1 + cos x
sin x
sin x
ity
d
1 + cos x
1 − cos x
4
−
≡
1 − cos x 1 + cos x
sin x tan x
f
y
ve
rs
C
w
ie
h
ev
id
br
am
op
y
-C
6 Prove each of these identities.
cos x
1
a
−
≡ tan x
cos x 1 + sin x
148
2
w
ge
U
ni
c
y
tan2 x − sin2 x ≡ tan2 x sin2 x
op
C
w
(cos2 x − 2)2 − 3sin2 x ≡ cos 4 x + sin2 x
C
op
c
5 Prove each of these identities.
cos2 x − sin2 x
≡ cos x + sin x
a
cos x − sin x
ev
ie
d
2(1 + cos x ) − (1 + cos x )2 ≡ sin2 x
-R
cos2 x − sin2 x ≡ 2 cos2 x − 1
y
a
Pr
es
s
-C
4 Prove each of these identities.
R
b
w
(sin x + cos x )2 ≡ 1 + 2 sin x cos x
am
br
id
a
ev
ie
ge
3 Prove each of these identities.
op
U
R
ni
ev
7 Show that (1 + cos x )2 + (1 − cos x )2 + 2 sin2 x has a constant value for all x and state this value.
w
ge
C
8 a Express 7 sin2 x + 4 cos2 x in the form a + b sin2 x.
-R
am
br
9 a Express 4 sin θ − cos2 θ in the form (sin θ + a )2 + b.
ev
id
ie
b State the range of the function f( x ) = 7 sin2 x + 4 cos2 x, for the domain 0 < x < 2 π.
s
1 − sin θ
1 2(1 + sin θ )
, show that
=
.
2 cos θ
a
cos θ
es
10 a Given that a =
op
y
PS P
-C
b Hence, state the maximum and minimum values of 4 sin θ − cos2 θ , for the domain 0 < θ < 2 π.
y
op
-R
s
es
-C
am
br
ev
ie
id
g
w
e
C
U
R
ev
ie
w
ni
ve
rs
C
ity
Pr
b Hence, find sin θ and cos θ in terms of a.
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
C
U
ni
op
y
Chapter 5: Trigonometry
w
ge
5.8 Further trigonometric equations
-R
am
br
id
ev
ie
This section uses trigonometric identities to help solve some more complex trigonometric
equations.
Pr
es
s
1 − tan2 θ
≡ 2 cos2 θ − 1.
1 + tan2 θ
ni
U
Use tan θ =
 sin θ 
1− 
 cos θ 
2
 sin θ 
1+ 
 cos θ 
2
w
ie
ev
-R
am
Simplify.
rs
1 − tan2 θ
= 5 cos θ − 3
1 + tan2 θ
Use the result from part a.
2 cos2 θ − 1 = 5 cos θ − 3
Rearrange.
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b
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≡ RHS
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149
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≡ cos2 θ − (1 − cos2 θ )
≡ 2 cos2 θ − 1
Replace sin2 θ with 1 − cos2 θ .
Pr
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≡ cos2 θ − sin2 θ
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am
2 cos2 θ − 5 cos θ + 2 = 0
s
-C
(2 cos θ − 1)(cos θ − 2) = 0
1
or cos θ = 2
2
1
θ = cos −1  
 2
Pr
es
cos θ = 2 has no solutions.
ity
θ = 60°
Factorise.
θ = 360° − 60°
or
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Solution is θ = 60° or θ = 300°.
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cos θ =
ie
Use sin2 θ + cos2 θ = 1.
s
cos2 θ − sin2 θ
cos2 θ + sin2 θ
es
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≡
sin θ
.
cos θ
Multiply numerator and denominator by
cos2 θ .
br
≡
1 − tan2 θ
1 + tan2 θ
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a LHS ≡
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Answer
1 − tan2 θ
= 5 cos θ − 3 for 0° < θ < 360°.
1 + tan2 θ
y
b Hence, solve the equation
C
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a Prove the identity
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WORKED EXAMPLE 5.21
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EXERCISE 5H
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Cambridge International AS & A Level Mathematics: Pure Mathematics 1
1 a Show that the equation cos θ + sin θ = 5 cos θ can be written in the form tan θ = k.
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b Hence, solve the equation cos θ + sin θ = 5 cos θ for 0° < θ < 360°.
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b Hence, solve the equation 3sin2 θ + 5 sin θ cos θ = 2 cos2 θ for 0° < θ < 180°.
3 a Show that the equation 8 sin2 θ + 2 cos2 θ − cos θ = 6 can be written in the form 6 cos2 θ + cos θ − 2 = 0.
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s
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2 a Show that the equation 3sin2 θ + 5 sin θ cos θ = 2 cos2 θ can be written in the form
3tan2 θ + 5 tan θ − 2 = 0.
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b Hence, solve the equation 8 sin2 θ + 2 cos2 θ − cos θ = 6 for 0° < θ < 360°.
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4 a Show that the equation 4 sin 4 θ + 14 = 19 cos2 θ can be written in the form 4x 2 + 19x − 5 = 0, where
x = sin2 θ .
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b Hence, solve the equation 4 sin 4 θ + 14 = 19 cos2 θ for 0° < θ < 360°.
br
ev
5 a Show that the equation sin θ tan θ = 3 can be written in the form cos2 θ + 3 cos θ − 1 = 0.
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b Hence, solve the equation sin θ tan θ = 3 for 0° < θ < 360°.
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s
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6 a Show that the equation 5(2 sin θ − cos θ ) = 4(sin θ + 2 cos θ ) can be written in the form tan θ =
op
Pr
y
b Hence, solve the equation 5(2 sin θ − cos θ ) = 4(sin θ + 2 cos θ ) for 0° < θ < 360°.
sin θ
1 + cos θ
2
+
≡
.
1 + cos θ
sin θ
sin θ
ity
7 a Prove the identity
sin θ
1 + cos θ
+
= 1 + 3 sin θ for 0° < θ < 360°.
1 + cos θ
sin θ
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cos θ
1
≡
− 1.
tan θ (1 + sin θ ) sin θ
cos θ
= 1 for 0° < θ < 360°.
b Hence, solve the equation
tan θ (1 + sin θ )
8 a Prove the identity
op
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b Hence, solve the equation
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C
150
1
1
2
.
+
≡
1 + sin θ 1 − sin θ
cos2 θ


1
1
b Hence, solve the equation cos θ 
+
= 5 for 0° < θ < 360°.
 1 + sin θ 1 − sin θ 
es
1 + cos θ .
1 − cos θ
Pr
≡
2
= 2 for 0° < θ < 360°.
y
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 1
1 
+
b Hence, solve the equation 
 sin θ tan θ 
op
2
ity
10 a Prove the identity  1 + 1 
 sin θ tan θ 
C
1
for 0° < θ < 360°.
2
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b Hence, solve the equation cos 4 θ − sin 4 θ =
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11 a Prove the identity cos 4 θ − sin 4 θ ≡ 2 cos2 θ − 1.
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9 a Prove the identity
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.
6
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Chapter 5: Trigonometry
Exact values of trigonometric functions
1
2
3
2
1
3
θ = 45° =
π
4
1
2
1
2
1
θ = 60° =
π
3
1
2
3
Pr
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Angles measured anticlockwise from the positive x-direction are positive.
●
Angles measured clockwise from the positive x -direction are negative.
180°
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Cos
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0°, 360°
O
Tan
151
270°
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Useful mnemonic: ‘A ll Students Trust Cambridge’.
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All
s
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Sin
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90°
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●
●
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C
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3
2
Diagram showing where sin, cos and tan are positive
ev
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π
6
Positive and negative angles
tan θ
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cos θ
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sin θ
θ =
= 30° =
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am
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Checklist of learning and understanding
op
1
180
270
360 x
270
360 x
s
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am
–1
90
ie
–90 O
br
–360 –270 –180
ev
id
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y = sin x
es
y
Pr
op
y
1
y = cos x
90
180
op
y
–1
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s
es
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C
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C
ity
–360 –270 –180 –90 O
-C
C
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U
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Graphs of trigonometric functions
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Pr
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s
–180
–90
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The graph of y = sin( ax ) is a stretch of y = sin x , stretch factor
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1
, parallel to the x -axis.
a
0
The graph of y = a + sin x is a translation of y = sin x by the vector   .
a
 −a 
.
 0 
es
s
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The graph of y = sin( x + a ) is a translation of y = sin x by the vector 
ity
y
π
–
2
y
op
2
y = sin −1 x
domain: −1 < x < 1
π
π
range: − < sin −1 x <
2
2
s
1
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y
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Pr
y = tan −1 x
domain: x ∈ R
π
π
range: − < tan −1 x <
2
2
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id
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sin2 x + cos2 x ≡ 1
C
U
Trigonometric identities
sin x
● tan x ≡
cos x
-C
π
––
2
x
y = cos −1 x
domain: −1 < x < 1
range: 0 < cos −1 x < π
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y
C
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O
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–1
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am
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π
––
2
●
O
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1
y = cos –1 x
ie
x
y = tan –1 x
C
π
–
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O
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y=
y
π
sin –1x
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2
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Pr
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C
y
π
–
–1
ie
360 x
C
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C
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R
270
●
Inverse trigonometric functions
ev
180
The graph of y = a sin x is a stretch of y = sin x , stretch factor a, parallel to the y -axis.
●
R
90
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–270
●
●
152
y = tan x
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–360
y
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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 5: Trigonometry
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3
Pr
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y
1
–π
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3π
––
2
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y = a + b sin x
2
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END-OF-CHAPTER REVIEW EXERCISE 5
1π
––
2
O
1 π
–
2
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State the values of the constants a and b.
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Find the value of x satisfying the equation sin−1( x − 1) = tan−1(3).
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br
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am
Given that θ is an acute angle measured in radians and that cos θ = k , find, in terms of k, an expression for:
s
tan θ
c
cos( π − θ )
es
b
Pr
[1]
[1]
Solve the equation sin 2 x = 5 cos 2 x, for 0° < x < 180°.
6
Solve the equation
rs
5
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[4]
id
br
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am
[4]
1
for 0 < θ < 2 π.
2
ii Write down the number of roots of the equation 2 cos 2θ − 1 = 0 in the interval 0 < θ < 2 π.
Sketch, on a single diagram, the graphs of y = cos 2θ and y =
es
s
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iii Deduce the number of roots of the equation 2 cos 2θ − 1 = 0 in the interval 10π < θ < 20 π.
[3]
[1]
[1]
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Cambridge International AS & A Level Mathematics 9709 Paper 11 Q3 November 2014
C
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q3 November 2011
Show that the equation 2 tan2 θ sin2 θ = 1 can be written in the form 2 sin 4 θ + sin2 θ − 1 = 0.
C
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[2]
[4]
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q5 June 2011
e
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ii Hence solve the equation 2 tan2 θ sin2 θ = 1 for 0° < θ < 360°.
y
i
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9
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[4]
Solve the equation 2 cos2 x = 5 sin x − 1 for 0° < x < 360°.
i
[4]
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ge
13sin2 θ
+ cos θ = 2 for 0° < θ < 180°.
2 + cos θ
C
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[1]
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sin θ
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a
Solve the equation cos −1(8x 4 + 14x 2 − 16) = π.
8
[3]
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q2 November 2014
4
7
[2]
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q1 June 2014
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3
2π x
–1
The diagram shows part of the graph of y = a + b sin x.
2
ev
3π
–
2
π
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10 i
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Cambridge International AS & A Level Mathematics: Pure Mathematics 1
Solve the equation 4 sin2 x + 8 cos x − 7 = 0 for 0° < x < 360°.
[4]
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1
1
ii Hence find the solution of the equation 4 sin2  θ  + 8 cos  θ  − 7 = 0 for 0° < θ < 360°.
2 
2 
Pr
es
s
y
C
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2 − sin x
3
for 0 < x < 2 π.
=
1 + 2 sin x
4
U
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12 a Solve the equation
y
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Prove the identity
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Cambridge International AS & A Level Mathematics 9709 Paper 11 Q4 November 2013
sin x tan x
1
≡ 1+
.
[3]
1 − cos x
cos x
sin x tan x
ii Hence solve the equation
+ 2 = 0, for 0° < x < 360°.
[3]
1 − cos x
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q4 November 2010
11 i
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[1]
s
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[1]
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[2]
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14 i
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Cambridge International AS & A Level Mathematics 9709 Paper 11 Q7 November 2010
Solve the equation 2 cos2 θ = 3sin θ , for 0° < θ < 360°.
[4]
rs
C
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[3]
Pr
y
iv Obtain an expression, in terms of x, for f −1( x ).
ni
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ii The smallest positive solution of the equation 2 cos2 ( nθ ) = 3sin( nθ ), where n is a positive integer,
is 10°. State the value of n and hence find the largest solution of this equation in the interval
0° < θ < 360°.
ev
[4]
ie
1
13 A function f is defined by f : x → 3 − 2 tan  x  for 0 < x < π.
2 
i State the range of f.
ii State the exact value of f  2 π  .
3 
iii Sketch the graph of y = f(x ).
C
[3]
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q7 November 2012
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[3]
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b Solve the equation sin x − 2 cos x = 2 ( 2 sin x − 3 cos x ) for −π < x < π.
154
[2]
s
-C
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am
br
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id
Show that
ie
sin θ
cos θ
1
+
≡
.
[3]
sin θ + cos θ sin θ − cos θ
sin2 θ − cos2 θ
sin θ
cos θ
+
= 3, for 0° < θ < 360°.
[4]
ii Hence solve the equation
sin θ + cos θ sin θ − cos θ
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q5 June 2013
15 i
es
4 cos θ
+ 15 = 0 can be expressed as 4 sin2 θ − 15 sin θ − 4 = 0.
tan θ
4 cos θ
+ 15 = 0 for 0° < θ < 360°.
ii Hence solve the equation
tan θ
Show that the equation
[3]
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Pr
[3]
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Cambridge International AS & A Level Mathematics 9709 Paper 11 Q4 November 2015
C
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17 The function f : x → 5 + 3 cos 1 x  is defined for 0 < x < 2 π.
2 
i Solve the equation f( x ) = 7, giving your answer correct to 2 decimal places.
id
g
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e
ii Sketch the graph of y = f(x ).
ie
iii Explain why f has an inverse.
ev
[3]
[2]
[1]
[3]
s
-R
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q8 June 2015
es
am
br
iv Obtain an expression for f −1( x ).
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R
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C
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16 i
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id
Pr
es
s
-C
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ni
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y
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op
C
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C
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am
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Chapter 6
Series
155
id
es
s
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am
br
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use the expansion of ( a + b ) n, where n is a positive integer
recognise arithmetic and geometric progressions
use the formulae for the nth term and for the sum of the first n terms to solve problems involving
arithmetic or geometric progressions
use the condition for the convergence of a geometric progression, and the formula for the sum to
infinity of a convergent geometric progression.
y
op
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am
br
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id
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C
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C
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Pr
op
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■
■
■
■
ie
In this chapter you will learn how to:
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Cambridge International AS & A Level Mathematics: Pure Mathematics 1
w
IGCSE / O Level
Mathematics
Expand brackets.
b (1 − 3x )(1 + 2 x − 3x 2 )
2 Simplify:
b ( −2 x 3 )5
3 Find the nth term of these linear sequences.
ni
y
Find the nth term of a linear
sequence.
a
U
R
(2 x + 3)2
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IGCSE / O Level
Mathematics
a
a (5x 2 )3
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1 Expand:
Pr
es
s
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op
Simplify indices.
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IGCSE / O Level
Mathematics
Check your skills
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What you should be able to do
-C
Where it comes from
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PREREQUISITE KNOWLEDGE
5, 7, 9, 11, 13, …
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b 8, 5, 2, −1, −4, …
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am
Why study series?
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es
s
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At IGCSE / O Level you learnt how to expand expressions such as (1 + x )2 . In this chapter
you will learn how to expand expressions of the form (1 + x ) n, where n can be any positive
integer. Expansions of this type are called binomial expansions.
ve
op
ni
Binomial means ‘two terms’.
br
The expansion of ( a + b )2 can be used to expand ( a + b )3:
-R
am
C
ev
id
ie
You should already know that ( a + b )2 = a 2 + 2 ab + b2.
( a + b )3 = ( a + b )( a 2 + 2 ab + b2 )
s
-C
= a 3 + 2 a 2b + ab2 + a 2b + 2 ab2 + b3
Pr
op
y
es
= a 3 + 3a 2b + 3ab2 + b3
Similarly, it can be shown that ( a + b )4 = a 4 + 4a 3b + 6a 2b2 + 4ab3 + b 4.
WEB LINK
1a + 1b
1a 2 + 2 ab + 1b2
( a + b )3 =
3
1a 3 + 3a 2b + 3ab2 + 1b3
( a + b )4 =
4
1a 4 + 4a 3b + 6a 2b2 + 4ab3 + 1b 4
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2
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id
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( a + b )2 =
y
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( a + b )1 =
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Writing the expansions of ( a + b ) n in full in order:
1
FAST FORWARD
Properties of binomial
expansions are also used
in probability theory,
which you will learn
about if you go on to
study the Probability
and Statistics 1
Coursebook, Chapter 7.
w
ge
The word is used in algebra for expressions such as x + 3 and 5x − 2 y.
( a + b )0 = 1
In the Pure Mathematics
2 and 3 Coursebook,
Chapter 7, you will
learn how to expand
these expressions
for any real value
of n.
y
6.1 Binomial exapansion of ( a + b ) n
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rs
C
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Pr
This chapter also covers arithmetic and geometric progressions. Both the mathematical and
the real world are full of number sequences that have particular special properties. You will
learn how to find the sum of the numbers in these progressions. Some fractal patterns can
generate these types of sequences.
156
FAST FORWARD
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Try the Sequences
and Counting and
binomials resources
on the Underground
Mathematics website.
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C
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Chapter 6: Series
ev
ie
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The first term is a 4 and then the power of a decreases by 1 while the power of b
increases by 1 in each successive term.
All of the terms have a total index of 4 ( a 4, a 3b, a 2b2 , ab3 and b 4 ).
Pr
es
s
-C
●
am
br
id
●
w
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If you look at the expansion of ( a + b )4, you should notice that the powers of a and b form
a pattern.
There is a similar pattern in the other expansions.
ve
rs
ity
ց ւ
6
The next row is then:
10
10
1
5
id
1
br
s
es
DID YOU KNOW?
ni
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1
6
15
6
4
10
10
20
1
5
15
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5
1
op
4
1
1
1
3
1
6
1
ge
There are many number patterns to be found in Pascal’s triangle.
4
s
1
10
20
Pr
ni
ve
rs
ity
1 What do you notice if you find the total of each row in Pascal’s triangle?
Can you explain your findings?
op
y
2 Can you find the Fibonacci sequence (1, 1, 2, 3, 5, 8, 13, …) in Pascal’s triangle?
You may want to add terms together.
-R
s
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3 Pascal’s triangle has many other number patterns.
Which number patterns can you find?
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These numbers are called tetrahedral numbers.
es
-C
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am
br
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id
For example, the numbers 1, 4, 10 and 20 have been highlighted.
C
3
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ie
2
rs
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1
1
1
ity
C
1
Pascal’s triangle is
named after the French
mathematician Blaise
Pascal (1623–1662).
Pr
y
op
1
Each number is
the sum of the two
numbers in the row
above it.
-R
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am
( a + b )5 = 1a 5 + 5a 4b + 10 a 3b2 + 10 a 2b3 + 5ab 4 + 1b5
EXPLORE 6.1
Each row always starts
and finishes with a 1.
ev
This row can then be used to write down the expansion of ( a + b )5:
w
n = 5: 1 5
4
w
4
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1
y
1 + 2
1
ց ւ
1
3 + 3
1
TIP
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1
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1
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n = 1:
n = 2:
n = 3:
n = 4:
1
ie
n = 0:
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The coefficients also form a pattern that is known as Pascal’s triangle.
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WORKED EXAMPLE 6.1
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Cambridge International AS & A Level Mathematics: Pure Mathematics 1
Use Pascal’s triangle to find the expansion of:
(5 − 2 x )4
-C
b
(3x + 2)3
(3x + 2)3 = 1(3x )3 + 3(3x )2 (2) + 3(3x )(2)2 + 1(2)3
4
= 27 x 3 + 54x 2 + 36x + 8
y
(5 − 2 x )
b
ev
ie
The index is 3 so use the row for n = 3 in Pascal’s triangle (1, 3, 3, 1).
ve
rs
ity
w
C
op
y
a
Pr
es
s
Answer
-R
(3x + 2)3
a
U
R
(5 − 2 x )4 = 1(5)4 + 4(5)3 ( −2 x ) + 6(5)2 ( −2 x )2 + 4(5)( −2 x )3 + 1( −2 x )4
br
ev
id
ie
w
ge
= 625 − 1000 x + 600 x 2 − 160 x 3 + 16x 4
C
op
ni
The index is 4 so use the row for n = 4 in Pascal’s triangle (1, 4, 6, 4, 1).
-R
am
WORKED EXAMPLE 6.2
s
es
-C
a Use Pascal’s triangle to expand (1 − 2 x )5.
op
Pr
y
b Find the coefficient of x 3 in the expansion of (3 + 5x )(1 − 2 x )5.
158
ity
a (1 − 2 x )5
rs
w
C
Answer
y
ev
ve
ie
The index is 5 so use the row for n = 5 in Pascal’s triangle (1, 5, 10, 10, 5, 1).
U
R
= 1 − 10 x + 40 x 2 − 80 x 3 + 80 x 4 − 32 x5
w
ge
b (3 + 5x )(1 − 2 x )5 = (3 + 5x )(1 − 10 x + 40 x 2 − 80 x 3 + 80 x 4 − 32 x5 )
br
ev
id
ie
The term in x 3 comes from the products:
-R
am
(3 + 5x )(1 − 10 x + 40 x 2 − 80 x 3 + 80 x 4 − 32 x5 )
-C
3 × ( −80 x 3 ) = −240 x 3 and 5x × 40 x 2 = 200 x 3
ni
ve
rs
s
b
(1 − x )4
c
( x + y )3
d
e
( x − y )4
f
(2 x + 3 y )3
g
(2 x − 3)4
C
 x2 + 3 

2 x3 
w
e
ev
ie
id
g
es
s
-R
br
am
h
(2 − x )3
y
( x + 2)3
U
a
op
1 Use Pascal’s triangle to find the expansions of:
-C
ie
w
C
ity
EXERCISE 6A
ev
es
Pr
op
y
Coefficient of x 3 = −240 + 200 = −40.
R
C
ni
op
(1 − 2 x )5 = 1(1)5 + 5(1)4 ( −2 x ) + 10(1)3 ( −2 x )2 + 10(1)2 ( −2 x )3 + 5(1)( −2 x )4 + 1( −2 x )5
Copyright Material - Review Only - Not for Redistribution
3
ve
rs
ity
C
U
ni
op
y
Chapter 6: Series
(2 x − 1)4
(3 + x )5 + (3 − x )5 ≡ A + Bx 2 + Cx 4
w
f
c
(3 − x )5
d
(4 + x )4
g
(4x + 3)4
h
5− x

2
ev
ie
(1 + x )5
4
Pr
es
s
-C
3
( x − 2)5
b
-R
e
( x + 3)4
am
br
id
a
ge
2 Find the coefficient of x 3 in the expansions of:
y
Find the value of A, the value of B and the value of C .
ve
rs
ity
Find the possible values of the constant a.
ev
ie
5 a Expand (2 + x )4 .
U
R
ni
C
op
b Use your answer to part a to express (2 + 3 )4 in the form a + b 3 .
w
ge
6 a Expand (1 + x )3 .
id
br
ev
(1 + 5 )3 in the form a + b 5
(1 − 5 )3 in the form c + d 5 .
-R
am
ii
Use your answers to part b to simplify (1 + 5 )3 + (1 − 5 )3.
es
s
-C
c
op
Pr
y
7 Expand (1 + x )(2 + 3x )4.
159
ity
8 a Expand ( x 2 − 1)4 .
b Find the coefficient of x 6 in the expansion of (1 − 2 x 2 )( x 2 − 1)4.
4
ve
ie
w
rs
C
ie
b Use your answer to part a to express:
i
y
op
ni
4
w
ge
U
3
10 Find the term independent of x in the expansion of  x 2 − 2  .

x 
C
ev
2
9 Find the coefficient of x 2 in the expansion of  3x −  .

x
R
y
w
C
op
4 The coefficient of x 2 in the expansion of (3 + ax )4 is 216.
id
ie
11 a Find the first three terms, in ascending powers of y, in the expansion of (1 + y )4 .
-R
am
br
ev
b By replacing y with 5x − 2 x 2, find the coefficient of x 2 in the expansion of (1 + 5x − 2 x 2 )4.
-C
12 The coefficient of x 2 in the expansion of (1 + ax )4 is 30 times the coefficient of x in the expansion of
3
s
es
Pr
4
ity
1
13 Find the power of x that has the greatest coefficient in the expansion of  3x 4 +  .

x
ni
ve
rs
14 a Write down the expansion of ( x + y )5 .
5
b Without using a calculator and using your result from part a, find the value of  10  , correct to

4
y
1
op
U
C
the nearest hundred.
ie
id
g
w
e
4
4
q
1
1
15 a Given that  x 2 +  −  x 2 −  = px5 + , find the value of p and the value of q.


x
x
x
4
4
-R
s
es
am
br
ev
1 
1 


b Hence, without using a calculator, find the exact value of  2 +
 −  2 −
 .

2
2
-C
R
ev
ie
w
C
op
y
 1 + ax  . Find the value of a.

3 
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
C
U
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op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
1
x
1
a Express x 3 + 3 in terms of y.
x
1
b Express x5 + 5 in terms of y.
x
Pr
es
s
-C
-R
am
br
id
ev
ie
w
ge
16 y = x +
PS
y
6.2 Binomial coefficients
y
U
R
ni
EXPLORE 6.2
C
op
ev
ie
w
C
ve
rs
ity
op
Pascal’s triangle can be used to expand ( a + b ) n for any positive integer n, but if n is large
it can take a long time to write out all the rows in the triangle. Hence, we need a more
efficient method to find the coefficients in the expansions. The coefficients in the binomial
expansion of (1 + x ) n are known as binomial coefficients.
w
ge
Consider the expansion:
10
10
5
1
op
ie
es
1 Use your calculator to find the values of:
Pr
y
s
-C
Find the nCr function on your calculator. On some calculators this may be n C r
n
or   .
 r 
ity
5
 5 5 5 5 5
 0  ,  1  ,  2  ,  3  ,  4  and  5  .
       
 
op
C
w
ie
-R
am
br
ev
id
ge
U
R
ni
ev
2 What do you notice about your answers to question 1?
3 Complete the following four statements.
 5 
The coefficient of x 2 in the expansion of (1 + x )5 is 
.
…
…
The coefficient of x r in the expansion of (1 + x ) n is 
.
…
y
ve
ie
w
rs
C
160
C
ity
Pr
op
y
es
s
-C
 5 
5
The coefficient of the 4th term in the expansion of ( 1 + x ) is 
.
…
…
n
The coefficient of the ( r + 1)th term in the expansion of ( 1 + x ) is 
.
…
y
op
w
ev
ie
id
g
es
s
-R
br
am
C
U
e
 n  n
 n
 n
If n is a positive integer, then (1 + x ) n =   +   x +   x 2 + … +   x n .
 2
 0  1
 n
-C
R
ev
ie
w
ni
ve
rs
We write the binomial expansion of (1 + x ) n , where n is a positive integer as:
KEY POINT 6.1
 5
To find   , key in
 2
5 nCr 2 .
-R
am
The coefficients are: 1 5
ev
br
id
(1+ x )5 = 1+ 5x + 10x 2 + 10x 3 + 5x 4 + x 5
TIP
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
C
U
ni
op
y
Chapter 6: Series
am
br
id
ev
ie
w
ge
We can therefore write the expansion of (1 + x ) n using binomial coefficients; the result is
known as the Binomial theorem.
n
-C
-R
b
We can use the Binomial theorem to expand ( a + b ) n, too. We can write ( a + b ) n = a n  1 + 

a
(assuming that a ≠ 0 ).
Pr
es
s
KEY POINT 6.2
y
ni
ev
ie
WORKED EXAMPLE 6.3
(2 − 3x )10
ie
ev
-R
am
 15   15 
 15  2  15  3
(1 + x )15 = 
x+
x +
x +…
+


0
1

 2 
 
 3 

= 1 + 15x + 105x 2 + 455x 3 + …
es
s
-C
ity
Pr
 10  7
 10  10  10  9
 10  8
(2 − 3x )10 = 
2 +
2 ( −3x )1 + 
2 ( −3x )2 + 
2 ( −3x )3 + …



 1 
 2 
 3 
 0 
= 1024 − 15 360 x + 103680 x 2 − 414 720 x 3 + …
ve
ie
w
rs
C
op
y
b
w
b
br
Answer
a
U
(1 + x )15
ge
a
id
R
Find, in ascending powers of x, the first four terms in the expansion of:
y
ev
You should also know how to work out the binomial coefficients without using a calculator.
w
ge
C
U
R
ni
op
5
5
From Pascal’s triangle, we know that   = 1 and   = 1.
0
5
br
ev
id
ie
In general, we can write this as:
-R
am
KEY POINT 6.3
Pr
y
 5  5×4×3×2
=5
 =
 4  4× 3× 2 ×1
C
 5  5×4×3
= 10
  =
 3  3× 2 ×1
op
ni
ve
rs
U
 5  5×4
= 10
 =
 2  2 ×1
-R
s
es
am
br
ev
ie
id
g
w
e
5 5
 = =5
1 1
-C
ev
ie
w
C
5 5 5
5
We write   ,   ,   and   as:
1 2 3
4
ity
op
y
es
s
-C
 n
 n
 0  = 1 and  n  = 1
 
 
R
C
op
w
C
ve
rs
ity
op
y
 n
 n
 n
 n
( a + b ) n =   a n +   a n −1b1 +   a n − 2 b 2 + … +   b n
0
1
2
 
 
 
 n
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161
ve
rs
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C
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Cambridge International AS & A Level Mathematics: Pure Mathematics 1
am
br
id
ev
ie
w
ge
In general, if r is a positive integer less than n, then:
KEY POINT 6.4
y
Pr
es
s
-C
-R
 n
n × ( n − 1) × ( n − 2) × … × ( n − r + 1)
 r  = r × ( r − 1) × ( r − 2) × … × 3 × 2 × 1
ve
rs
ity
C
op
y
8
a Without using a calculator, find the value of   .
4
n
b Find an expression, in terms of n, for   .
4
U
R
ni
ev
ie
w
C
op
WORKED EXAMPLE 6.4
w
ge
Answer
 8  8×7×6×5
 4  = 4 × 3 × 2 × 1 = 70
b
 n
n × ( n − 1) × ( n − 2) × ( n − 3) n( n − 1)( n − 2)( n − 3)
=
 4  =
4×3×2 ×1
24
op
Pr
y
es
s
-C
-R
am
br
ev
id
ie
a
162
rs
n
x
When  1 −  is expanded in ascending powers of x, the coefficient of x 2 is 4. Given that n is the positive

3
integer, find the value of n.
y
op
ie
-R
s
Pr
-R
s
es
-C
am
br
ev
ie
id
g
w
e
C
U
op
ie
ev
R
y
ni
ve
rs
As n is a positive integer, n = 9.
ity
op
y
( n − 9)( n + 8) = 0
n = 9 or n = −8
w
C
n2 − n − 72 = 0
es
-C
am
n × ( n − 1)
=4
18
n( n − 1) = 72
ev
br
id
2
 n
x
n × ( n − 1) x 2
n × ( n − 1) 2
x
Term in x 2 =    −  =
×
=


3
2 ×1
9
18
 2
w
ge
Answer
C
U
R
ni
ev
ve
ie
w
C
ity
WORKED EXAMPLE 6.5
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
ge
C
U
ni
op
y
Chapter 6: Series
am
br
id
ev
ie
w
WORKED EXAMPLE 6.6
 8
Term in x 4 =   (2)4 ( kx )4 = 1120 k 4 x 4
 4
 
ve
rs
ity
ev
ie
w
C
op
y
 8
Term in x5 =   (2)3 ( kx )5 = 448 k 5 x5
 5
 
Pr
es
s
-C
Answer
-R
When (2 + kx )8 is expanded, the coefficient of x5 is two times the coefficient of x 4 . Given that k > 0, find the
value of k.
ni
C
op
y
Coefficient of x5 = 2 × coefficient of x 4
U
w
id
448 k 4 ( k − 5) = 0
ge
448 k 5 − 2240 k 4 = 0
ie
R
448 k 5 = 2 × 1120 k 4
br
ev
k = 0 or kk==50 or k = 5
s
-C
-R
am
As k is a positive integer, k = 5 .
op
Pr
y
es
WORKED EXAMPLE 6.7
163
ity
b Use your answer to part a to estimate the value of 1.99 × 1.029.
s
Let x = 0.01.
es
(2 − x )(1 + 2 x )9 = 2 + 35x + 270 x 2 + …
ity
ni
ve
rs
1.99 × 1.029 ≈ 2.377
Pr
1.99 × 1.029 ≈ 2 + 35(0.01) + 270(0.01)2
w
w
e
C
U
op
y
n
There is an alternative formula for calculating   . To be able to understand and apply
 r 
the alternative formula, we need to first know about factorial notation.
-R
s
es
am
br
ev
ie
id
g
We write 6! to mean 6 × 5 × 4 × 3 × 2 × 1, and call it ‘6 factorial’.
-C
ie
ev
R
op
ev
= 2 + (2 × 18 − 1)x + (2 × 144 − 18)x 2 + …
= 2 + 35x + 270 x 2 + …
-R
am
br
id
ie
w
= (2 − x )(1 + 18x + 144x 2 + …)
= 2(1 + 18x + 144x 2 + …) − x(1 + 18x + 144x 2 + …)
-C
C
op
y
b
C
ni
 9   9 

 9
(2 − x )(1 + 2 x )9 = (2 − x )    +   (2 x )1 +   (2 x )2 + … 
 2
  0   1 

ge
R
a
U
ev
Answer
y
ve
ie
w
rs
C
a Obtain the first three terms in the expansion of (2 − x )(1 + 2 x )9.
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
C
U
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op
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Cambridge International AS & A Level Mathematics: Pure Mathematics 1
am
br
id
ev
ie
w
ge
In general, if n is a positive integer, then:
KEY POINT 6.5
-C
w
ie
am
 n
n!
to find the value of:
Use the formula   =
 r  r! ( n − r )!
8
a  
4
164
s
es
Pr
op
y
9
b  
3
ity
rs
ve
ie
w
C
Answer
 8
8!
8!
=
= 70
a   =
−
4!(8
4)!
4!
4!
4
 
-R
br
ev
id
WORKED EXAMPLE 6.8
ge
U
R
ni
 n
n!
 r  = r! ( n − r )!
y
w
KEY POINT 6.6
C
op
C
ve
rs
ity
op
y
n
The formula for   then becomes:
 r 
ev
ie
-R
Pr
es
s
-C
n! = n × ( n − 1) × ( n − 2) × ( n − 3) × … × 3 × 2 × 1
y
op
id
ie
w
ge
C
U
R
ni
ev
 9
9!
9!
b   =
=
= 84
3!(9 − 3)!
3! 6!
 3
-R
am
br
ev
WORKED EXAMPLE 6.9
9
s
-C
5
Find the term independent of x in the expansion of  x + 2  .

x 
op
y
es
Answer
op
y
ni
ve
rs
The term that is independent of x is the term that when simplified does not involve x.
5
The x terms cancel each other out when the power of x is double the power of 2 .
x
Also, the sum of these powers must be 9.
ev
-R
s
es
am
br
 9  6  5 3
125
6
 3  x  x 2  = 84 × x × x 6 = 10 500
ie
id
g
w
e
C
U
 9
Hence, we are looking for powers of 6 and 3, respectively, and the corresponding binomial coefficient is   .
 3
The term independent of x is:
-C
R
ev
ie
w
C
ity
Pr
3
2
1
9
 x + 5  =  9  x 9 +  9  x8  5  +  9  x 7  5  +  9  x 6  5  + …
 3   x 2 
 0 
 2   x 2 
 1   x 2 

x2 
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
ev
ie
am
br
id
EXERCISE 6B
w
ge
C
U
ni
op
y
Chapter 6: Series
1 Without using a calculator, find the value of each of the following.
9
 
6
-R
b
Pr
es
s
7
 
3
-C
a
c
 12 


 4 
c
n
 
3
d
 15 


 6 
d
 12 


 7 
n
 
2
b
n
 
1
ve
rs
ity
a
b
8
 
5
c
 14 


 3 
w
C
op
ni
 10 


 2 
ge
R
a
y
 n
n!
3 Use the formula   =
to find the value of of each of the following.
 r  r! ( n − r )!
U
ev
ie
w
C
op
y
2 Express each of the following in terms of n.
7
f
-R
3+ x

2
(2 − x )13
g
(2 + x 2 )8
(1 + x 2 )12
h

x2 
 2 + 2 
d
3− x

3
9
s
-C
e
am
br
ev
id
ie
4 Find, in ascending powers of x, the first three terms in each of the following expansions.
7
x
d
b (1 − 3x )10
c 1+ 
a (1 + 2 x )8

2
(1 + 3x )12
c
2+ x

4
12
C
es
b
Pr
(1 − x )9
7
10
ity
a
6 Find the coefficient of x 4 in the expansion of (2 x + 1)12.
w
rs
C
op
y
5 Find the coefficient of x 3 in each of the following expansions.
y
op
ni
ev
ve
ie
7 Find the term in x5 in the expansion of (5 − 2 x )8.
U
R
8 Find the coefficient of x8 y5 in the expansion of ( x − 2 y )13.
id
ie
w
ge
3
9 Find the term independent of x in the expansion of  x − 2  .

x 
ev
b
(1 + 2 x )(1 − 3x )10
x
(1 + x )  1 − 

2
c
-R
(1 − x )(2 + x )7
am
a
br
10 Find, in ascending powers of x, the first three terms of each of the following expansions.
8
s
-C
11 a Find, in ascending powers of x, the first three terms in the expansion of (2 + x )10.
8
ity
Pr
x
12 a Find, in ascending powers of x, the first three terms in the expansion of  1 −  .

2
8
x
b Hence, obtain the coefficient of x 2 in the expansion of (2 + 3x − x 2 )  1 −  .

2
ie
w
ni
ve
rs
C
op
y
es
b By replacing x with 2 y − 3 y2, find the first three terms in the expansion of (2 + 2 y − 3 y2 )10.
y
op
ev
13 Find the first three terms, in ascending powers of x, in the expansion of (2 − 3x )4 (1 + 2 x )10.
id
g
n
-R
s
es
am
br
ev
ie
x
15 The first two terms, in ascending powers of x, in the expansion of (1 + x )  2 −  are p + qx 2.

4
Find the values of n , p and q.
-C
PS
w
e
C
U
R
14 The first four terms, in ascending powers of x, in the expansion of (1 + ax + bx 2 )7 are 1 − 14x + 91x 2 + px 3.
Find the values of a, b and p.
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y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
w
ge
6.3 Arithmetic progressions
am
br
id
ev
ie
At IGCSE / O Level you learnt that a number sequence is a list of numbers and that the
numbers in the sequence are called the terms of the sequence.
The notation used for arithmetic progressions is:
d = common difference
op
y
a = first term
Pr
es
s
-C
-R
A linear sequence such as 5, 8, 11, 14, 17, … is also called an arithmetic progression. Each term
differs from the term before by a constant. This constant is called the common difference.
l = last term
y
C
op
a + 3d
term 4
U
a + 2d
term 3
a + 4d
term 5
w
ge
a+d
term 2
a
term 1
ni
The first five terms of an arithmetic progression whose first term is a and whose common
difference is d are:
R
ev
ie
w
C
ve
rs
ity
The common difference is also allowed to be zero or negative. For example, 10, 6, 2, −2, …
and 5, 5, 5, 5, … are both arithmetic progressions.
br
-R
am
KEY POINT 6.7
ev
id
ie
From this pattern, you can see that the formula for the nth term is given by:
y
es
s
-C
nth term = a + ( n − 1)d
C
ity
op
Pr
WORKED EXAMPLE 6.10
166
rs
y
ni
op
Use a = −3, d = 4 and nth term = 237.
U
R
nth term = a + ( n − 1)d
Solve.
-R
am
br
ev
id
ie
w
ge
237 = −3 + 4( n − 1)
n − 1 = 60
n = 61
C
Answer
ve
ev
ie
w
Find the number of terms in the arithmetic progression −3, 1, 5, 9, 13, … , 237.
es
s
-C
WORKED EXAMPLE 6.11
ity
Pr
op
y
The fourth term of an arithmetic progression is 7 and the tenth term is 16. Find the first term and the
common difference.
ni
ve
rs
10th term = 16
⇒ a + 9d = 16
(2)
e
d = 1.5
C
U
(2) − (1) gives 6d = 9
(1)
y
⇒ a + 3d = 7
op
4th term = 7
R
id
g
w
Substituting into (1) gives a + 4.5 = 7
-R
s
-C
am
br
ev
ie
a = 2.5
First term = 2.5, common difference = 1.5
es
ev
ie
w
C
Answer
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rs
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ge
C
U
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op
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Chapter 6: Series
am
br
id
ev
ie
w
WORKED EXAMPLE 6.12
Pr
es
s
-C
Answer
-R
The nth term of an arithmetic progression is 5 − 6n. Find the first term and the common difference.
y
1st term = 5 − 6(1) = −1
ve
rs
ity
y
ev
ie
Substitute n = 2 into nth term = 5 − 6 n.
Common difference = 2nd term − 1st term = −6
w
C
op
2nd term = 5 − 6(2) = −7
Substitute n = 1 into nth term = 5 − 6 n.
ni
C
op
The sum of an arithmetic progression
ie
br
ev
id
EXPLORE 6.3
w
ge
U
R
When the terms in a sequence are added together we call the resulting sum a series.
-R
am
1+ 2 + 3 + 4 + … + 97 + 98 + 99 + 100 = ?
ity
1 Can you complete Gauss’s method to find the answer?
w
rs
C
op
Pr
y
es
s
-C
It is said that, at the age of seven or eight, the famous mathematician Carl Gauss
was asked to find the sum of the numbers from 1 to 100. His teacher expected this
task to keep him occupied for some time but Gauss surprised him by writing down
the correct answer almost immediately. His method involved adding the numbers in
pairs: 1 + 100 = 101 , 2 + 99 = 101, 3 + 98 = 101, …
y
b
5 + 10 + 15 + 20 + … + 185 + 190 + 195 + 200
c
6 + 9 + 12 + 15 + … + 93 + 96 + 99 + 102
op
2 + 4 + 6 + 8 + … + 494 + 496 + 498 + 500
ev
id
ie
w
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C
U
ni
a
br
R
ev
ve
ie
2 Use Gauss’s method to find the sum of:
es
s
-C
-R
am
3 Use Gauss’s method to find an expression, in terms of n, for the sum:
1 + 2 + 3 + 4 + … + ( n − 3) + ( n − 2) + ( n − 1) + n
Pr
op
y
The sum of an arithmetic progression, Sn, can be written as:
-R
s
es
-C
am
br
ev
ie
id
g
w
e
C
U
R
n
[2 a + ( n − 1)d ]
2
y
Sn =
op
or
ity
n
(a + l )
2
ni
ve
rs
Sn =
ev
ie
w
C
KEY POINT 6.8
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167
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Cambridge International AS & A Level Mathematics: Pure Mathematics 1
w
ev
ie
+ ( a + d ) + ( a + 2d ) + … + (l − 2d ) + (l − d ) +
+ (l − d ) + (l − 2d ) + … + ( a + 2d ) + ( a + d ) +
a
l
am
br
id
Reversing:
Sn =
Sn =
ge
We can prove this result as follows, by writing out the series in full.
l
a
ve
rs
ity
op
y
Pr
es
s
-C
-R
2Sn = ( a + l ) + ( a + l ) + ( a + l ) + … + ( a + l ) + ( a + l ) + ( a + l )
2Sn = n( a + l ), as there are n terms in the series
n
So Sn = ( a + l ).
2
n
Using l = a + ( n − 1)d , this can be rewritten as Sn = [2a + ( n − 1)d ].
2
Adding:
y
C
op
KEY POINT 6.9
ni
ev
ie
w
C
It is useful to remember the following rule that applies for all sequences.
w
ie
br
ev
id
WORKED EXAMPLE 6.13
ge
U
R
nth term = Sn − Sn − 1
s
-C
-R
am
In an arithmetic progression, the 1st term is −12, the 17th term is 12 and the last term is 45.
Find the sum of all the terms in the progression.
ity
Use nth term = 12 when n = 17 and a = −12.
Solve.
op
y
ve
rs
12 = −12 + 16d
3
d =
2
ni
ev
ie
w
C
nth term = a + ( n − 1)d
Pr
op
y
We start by working out the common difference.
168
es
Answer
C
U
R
We now determine the number of terms in the whole sequence.
Use nth term = 45 when a = −12 and d =
id
ie
w
ge
nth term = a + ( n − 1)d
3
45 = −12 + ( n − 1)
2
n − 1 = 38
n = 39
s
es
ity
Pr
Use a = −12 , l = 45 and n = 39.
-R
s
es
-C
am
br
ev
ie
id
g
w
e
C
U
op
ev
R
y
ni
ve
rs
= 643 21
ie
w
C
op
y
-C
Finally, we can work out the sum of all the terms.
n
Sn = ( a + l )
2
39
( −12 + 45)
S39 =
2
-R
am
br
ev
Solve.
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3
.
2
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C
U
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op
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Chapter 6: Series
am
br
id
ev
ie
w
WORKED EXAMPLE 6.14
The 10th term in an arithmetic progression is 14 and the sum of the first 7 terms is 42.
-C
-R
Find the first term of the progression and the common difference.
Pr
es
s
Answer
ni
U
br
4
into equation (1) gives a = 2.
3
4
First term = 2, common difference =
3
ev
id
ie
w
ge
w
(1) − (2) gives 6d = 8
4
d =
3
Use n = 7 and S7 = 42.
y
n
[2 a + ( n − 1)d ]
2
7
42 = (2 a + 6d )
2
(2)
6 = a + 3d
Sn =
ev
ie
R
(1)
C
op
14 = a + 9d
op
-R
s
Pr
y
es
-C
am
Substituting d =
169
C
ity
WORKED EXAMPLE 6.15
ve
ie
w
rs
The sum of the first n terms, Sn, of a particular arithmetic progression is given by Sn = 4 n2 + n .
y
op
ni
ev
a Find the first term and the common difference.
id
ie
w
ge
C
U
R
b Find an expression for the nth term.
Answer
a S1 = 4(1)2 + 1 = 5
br
ev
First term = 5
-R
am
S2 = 4(2)2 + 2 = 18
C
es
Pr
U
op
y
Method 2:
nth term = Sn − Sn − 1 = 4 n2 + n − [4( n − 1)2 + ( n − 1)]
-R
s
es
am
br
ev
ie
id
g
w
e
C
= 4 n2 + n − (4 n2 − 8 n + 4 + n − 1)
= 8n − 3
-C
ev
ie
w
ni
ve
rs
= 5 + 8( n − 1)
= 8n − 3
Use a = 5, d = 8.
ity
op
y
First term = 5, common difference = 8
b Method 1:
nth term = a + ( n − 1)d
First term + second term = 18
s
-C
Second term = 18 − 5 = 13
R
Use nth term = 14 when n = 10.
ve
rs
ity
C
op
y
nth term = a + ( n − 1)d
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
ev
ie
am
br
id
EXERCISE 6C
w
ge
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
1 The first term in an arithmetic progression is a and the common difference is d .
-R
Write down expressions, in terms of a and d , for the seventh term and the 19th term.
13 + 17 + 21 + … + 97
op
y
a
Pr
es
s
-C
2 Find the number of terms and the sum of each of these arithmetic series.
b
152 + 149 + 146 + … + 50
b
4 + 1 + ( −2) + … (38 terms)
c
1 1 2 …
(20 terms)
+ + +
3 2 3
d
− x − 5x − 9x − … (40 terms)
C
op
y
5 + 12 + 19 + … (17 terms)
R
ni
ev
ie
ve
rs
ity
a
w
C
3 Find the sum of each of these arithmetic series.
w
ge
U
4 The first term of an arithmetic progression is 15 and the sum of the first 20 terms is 1630.
Find the common difference.
br
ev
id
ie
5 In an arithmetic progression, the first term is −27, the 16th term is 78 and the last term is 169.
s
-C
b Find the sum of the terms in this progression.
-R
am
a Find the common difference and the number of terms.
op
Pr
y
es
6 The first two terms in an arithmetic progression are 146 and 139. The last term is −43.
Find the sum of all the terms in this progression.
ity
7 The first two terms in an arithmetic progression are 2 and 9. The last term in the progression is the only
number that is greater than 150. Find the sum of all the terms in the progression.
ve
ie
w
rs
C
170
y
op
C
U
R
ni
ev
8 The first term of an arithmetic progression is 15 and the last term is 27. The sum of the first five terms
is 79. Find the number of terms in this progression.
w
ge
9 Find the sum of all the integers between 100 and 300 that are multiples of 7.
-R
am
br
ev
id
ie
10 The first term of an arithmetic progression is 2 and the 11th term is 17. The sum of all the terms in the
progression is 500. Find the number of terms in the progression.
op
y
es
s
-C
11 Robert buys a car for $8000 in total (including interest). He pays for the car by making monthly payments that
are in arithmetic progression. The first payment that he makes is $200 and the debt is fully repaid after
16 payments. Find the fifth payment.
Pr
ity
a Find the first term and the common difference.
ni
ve
rs
b Given that the nth term of this progression is −59, find the value of n.
C
U
op
y
13 The sum of the first n terms, Sn, of a particular arithmetic progression is given by Sn = 4 n2 + 3n.
Find the first term and the common difference.
-R
s
es
am
br
ev
ie
id
g
w
e
14 The sum of the first n terms, Sn, of a particular arithmetic progression is given by Sn = 12 n − 2 n2 .
Find the first term and the common difference.
-C
R
ev
ie
w
C
12 The sixth term of an arithmetic progression is −3 and the sum of the first ten terms is −10.
Copyright Material - Review Only - Not for Redistribution
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rs
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C
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op
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Chapter 6: Series
am
br
id
ev
ie
w
ge
15 The sum of the first n terms, Sn, of a particular arithmetic progression is given by Sn =
Find an expression for the nth term.
1
(5 n2 − 17 n ).
4
Pr
es
s
-C
-R
16 A circle is divided into ten sectors. The sizes of the angles of the sectors are in arithmetic progression. The
angle of the largest sector is seven times the angle of the smallest sector. Find the angle of the smallest sector.
b Find the 65th term in terms of a.
y
18 The tenth term in an arithmetic progression is three times the third term. Show that the sum of the first ten
terms is eight times the sum of the first three terms.
w
19 The first term of an arithmetic progression is sin2 x and the second term is 1.
U
R
ni
P
C
op
ev
ie
a Find d in terms of a.
ve
rs
ity
C
op
y
17 An arithmetic sequence has first term a and common difference d. The sum of the first 20 terms is seven times
the sum of the first five terms.
ge
a Write down an expression, in terms of sin x , for the fifth term of this progression.
ie
id
br
ev
20 The sum of the digits in the number 67 is 13 (as 6 + 7 = 13).
-R
am
PS
w
b Show that the sum of the first ten terms of this progression is 10 + 35 cos2 x.
a Show that the sum of the digits of the integers from 19 to 21 is 15.
y
es
s
-C
b Find the sum of the digits of the integers from 1 to 99.
op
Pr
6.4 Geometric progressions
171
rs
ni
op
r = common ratio
U
R
a = first term
y
ve
The notation used for a geometric progression is:
ev
ie
w
C
ity
The sequence 2, 6, 18, 54, … is called a geometric progression. Each term is three times the
preceding term. The constant multiplier, 3, is called the common ratio.
ar 2
term 3
ar 3
term 4
ar 4
term 5
-R
am
br
ar
term 2
ie
a
term 1
ev
id
w
ge
C
The first five terms of a geometric progression whose first term is a and whose common
ratio is r are:
s
-C
This leads to the formula for the nth term of a geometric progression:
op
y
es
KEY POINT 6.10
op
-R
s
es
am
C
ie
ev
Use nth term = 1 when n = 5 and r =
br
nth term = ar n − 1
id
g
Answer
w
e
U
1
The fifth term of a geometric progression is 1 and the common ratio is .
2
Find the eighth term and an expression for the nth term.
y
ni
ve
rs
WORKED EXAMPLE 6.16
-C
R
ev
ie
w
C
ity
Pr
nth term = ar n − 1
Copyright Material - Review Only - Not for Redistribution
1
.
2
ve
rs
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ev
ie
w
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4
-R
am
br
id
1
1= a 
 2
a = 16
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
7
n −1
WORKED EXAMPLE 6.17
y
ev
ie
w
C
ve
rs
ity
op
y
1
nth term = ar n − 1 = 16  
 2
Pr
es
s
-C
1
1
8th term = 16   =
 2
8
ie
br
4
ar 4
40.5
=
ar
12
27
r3 =
8
3
r=
2
3
Substituting r =
into equation (1) gives a = 8.
2
n −1
3
3
First term = 8, common ratio = , nth term = 8   .
 2
2
rs
y
op
ie
w
ge
ev
br
id
WORKED EXAMPLE 6.18
C
U
R
ni
ev
ve
ie
w
C
172
ity
op
Pr
y
es
s
-C
(2) ÷ (1) gives
-R
(2)
am
40.5 = ar
(1)
ev
12 = ar
id
Answer
w
ge
U
R
ni
C
op
The second and fifth terms in a geometric progression are 12 and 40.5, respectively. Find the first term and the
common ratio. Hence, write down an expression for the nth term.
n
-R
-C
am
2
The nth term of a geometric progression is 9  −  . Find the first term and the common ratio.
 3
es
s
Answer
1
Pr
op
y
2
1st term = 9  −  = −6
 3
2
ni
ve
rs
y
2nd term
4
2
=
=−
1st term
−6
3
op
Common ratio =
e
C
U
2
This is also clear from the formula directly: each term is  −  times the previous one.
 3
-R
s
es
am
br
ev
ie
id
g
w
2
First term = −6, common ratio = − .
3
-C
R
ev
ie
w
C
ity
2
2nd term = 9  −  = 4
 3
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
ev
ie
am
br
id
EXPLORE 6.4
w
ge
C
U
ni
op
y
Chapter 6: Series
-R
In this Explore activity you are not allowed to use a calculator.
Pr
es
s
-C
1 Consider the sum of the first 10 terms, S10 , of a geometric progression with
a = 1 and r = 3.
a Multiply both sides of the previous equation by the common ratio, 3, and
complete the following statement.
ve
rs
ity
C
op
y
S10 = 1+ 3 + 32 + 33 + … + 37 + 38 + 39
w
3S10 = 3 + 32 + 3… + 3… + … + 3… + 3… + 3…
y
C
op
ni
ev
ie
b How does this compare to the original expression? Can you use this to find
a simpler way of expressing the sum S10?
b
a + ar + ar 2 + …
ev
br
(10 terms)
(10 terms)
-R
am
a + ar + ar 2 + …
(n terms)
es
s
-C
c
ie
1+ r + r 2 + …
id
a
w
ge
U
R
2 Use the method from question 1 to find an alternative way of expressing each of
the following.
C
ity
op
Pr
y
You will have discovered in Explore 6.4 that the sum of a geometric progression, Sn, can be
written as:
rs
y
a ( r n − 1)
r −1
op
Sn =
C
U
ge
ie
id
ev
-R
br
●
Use the first formula when −1 < r < 1.
Use the second formula when r > 1 or when r ø −1.
am
●
w
Either formula can be used but it is usually easier to:
rSn =
ar + ar + … + ar n − 3 + ar n − 2 + ar n − 1 + ar n
s
n
ni
ve
rs
a ( r n − 1)
r −1
w
e
C
U
op
Multiplying the numerator and the denominator by −1 gives the alternative formula
a (1 − r n )
Sn =
.
1− r
-R
s
es
am
br
ev
ie
id
g
Can you see why this formula does not work when r = 1 ?
-C
ie
ev
Sn =
(2)
ity
( r − 1)Sn = a ( r n − 1)
w
C
(2) − (1): rSn − Sn = ar − a
R
(1)
es
op
y
r × (1):
2
Pr
-C
This is the proof of the formulae in Key point 6.11.
Sn = a + ar + ar 2 + … + ar n − 3 + ar n − 2 + ar n − 1
y
or
ve
a (1 − r n )
1− r
R
ev
Sn =
TIP
ni
ie
w
KEY POINT 6.11
173
Copyright Material - Review Only - Not for Redistribution
These formulae are not
defined when r = 1 .
ve
rs
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am
br
id
WORKED EXAMPLE 6.19
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Cambridge International AS & A Level Mathematics: Pure Mathematics 1
Simplify.
ni
WORKED EXAMPLE 6.20
y
C
w
ev
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Use a = 3, r = 2 and n = 12.
ve
rs
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op
y
Pr
es
s
-C
a ( r n − 1)
r −1
3(212 − 1)
S12 =
2 −1
= 12 285
Sn =
C
op
Answer
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Find the sum of the first 12 terms of the geometric series 3 + 6 + 12 + 24 + … .
ie
id
Answer
w
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U
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The third term of a geometric progression is nine times the first term. The sum of the first six terms is k times the
sum of the first two terms. Find the value of k.
br
ev
3rd term = 9 × first term
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id
EXERCISE 6D
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op
y
ve
When r = 3 , k = 91 and when r = −3, k = 91.
Hence, k = 91.
ev
Rearrange to make k the subject.
Pr
y
a ( r 6 − 1) ka ( r 2 − 1)
=
r −1
r −1
6
r −1
k= 2
r −1
op
ie
w
C
174
Divide both sides by a (which we assume is
non-zero) and solve.
s
-C
Use
-R
am
ar 2 = 9a
r = ±3
S6 = kS2
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-R
2, 4, 8, 14, …
b
7, 21, 63, 189, …
d
1 2 4 7
, , , ,…
9 9 9 9
e
1, 0.4, 0.16, 0.64, …
c
81, −27, 9, −3, …
f
1, −1, 1, −1, …
Pr
es
s
a
op
y
ni
ve
rs
ity
2 The first term in a geometric progression is a and the common ratio is r. Write down expressions,
in terms of a and r, for the sixth term and the 15th term.
op
y
3 The first term of a geometric progression is 270 and the fourth term is 80. Find the common ratio.
C
U
4 The first term of a geometric progression is 50 and the second term is −30. Find the fourth term.
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id
g
w
e
5 The second term of a geometric progression is 12 and the fourth term is 27. Given that all the terms are
positive, find the common ratio and the first term.
-R
s
es
am
6 The sum of the second and third terms in a geometric progression is 84. The second term is 16 less than the
first term. Given that all the terms in the progression are positive, find the first term.
-C
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R
am
1 Identify whether the following sequences are geometric.
If they are geometric, write down the common ratio and the eighth term.
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Chapter 6: Series
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7 Three consecutive terms of a geometric progression are x, 4 and x + 6 . Find the possible values of x.
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3 + 6 + 12 + 24 + …
1− 2 + 4 − 8 +…
b
128 + 64 + 32 + 16 + …
d
243 + 162 + 108 + 72 + …
Pr
es
s
-C
c
am
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id
a
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8 Find the sum of the first eight terms of each of these geometric series.
10 A ball is thrown vertically upwards from the ground. The ball rises to a height of 8 m and then falls and
3
bounces. After each bounce it rises to of the height of the previous bounce.
4
a Write down an expression for the height that the ball rises after the nth impact with the ground.
y
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op
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9 The first four terms of a geometric progression are 0.5, 1, 2 and 4. Find the smallest number of terms that
will give a sum greater than 1000 000.
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op
b Find the total distance that the ball travels from the first throw to the fifth impact with the ground.
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11 The second term of a geometric progression is 24 and the third term is 12( x + 1).
id
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a Find, in terms of x, the first term of the progression.
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am
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ev
b Given that the sum of the first three terms is 76, find the possible values of x.
s
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12 The third term of a geometric progression is nine times the first term. The sum of the first four terms is
k times the first term. Find the possible values of k.
Pr
rs
b Find the total value of the donations made during the years 2010 to 2016, inclusive.
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ni
op
S3 n − S2 n
= r 2 n.
Sn
w
ge
R
Show that
y
14 A geometric progression has first term a, common ratio r and sum to n terms Sn.
ev
P
ve
w
1
1
1
1
, 9, , 27,
, 81,
,….
3
9
27
81
Show that the sum of the first 2n terms of the sequence is 1 (2 + 3n − 31 − n ).
2
15 Consider the sequence 1, 1, 3,
P
16 Let Sn = 1 + 11 + 111 + 1111 + 11111 + … to n terms.
id
ie
P
s
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ni
ve
rs
1
, so it begins
2
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es
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s
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These sums are getting closer and closer to 4.
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1 1 1
2, 1, , , , … . We can work out the sum of the first n terms of this:
2 4 8
1
3
7
S1 = 2, S2 = 3, S3 = 3 , S4 = 3 , S5 = 3 , and so on.
2
4
8
op
Consider the infinite geometric progression where a = 2 and r =
y
An infinite sequence is a sequence whose terms continue forever.
-C
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ev
ie
w
6.5 Infinite geometric series
es
10 n + 1 − 10 − 9 n
.
81
ity
Show that Sn =
C
op
y
-C
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175
a Find the value of the donation in 2016.
ity
C
op
y
es
13 A company makes a donation to charity each year. The value of the donation increases exponentially
by 10% each year. The value of the donation in 2010 was $10 000.
Copyright Material - Review Only - Not for Redistribution
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Cambridge International AS & A Level Mathematics: Pure Mathematics 1
am
br
id
ev
ie
w
ge
The diagram of the 2 by 2 square is a visual representation of this series. If the pattern of
rectangles inside the square is continued, the total area of the rectangles approximates the
area of the whole square (which is 4) increasingly well as more rectangles are included.
1 1 1 …
+ + + is 4,
2 4 8
because the sum of the first n terms gets as close to 4 as we like as n gets larger. We write
1 1 1
2 + 1 + + + + … = 4. We also say that the sum to infinity of this series is 4, and that
2 4 8
the series converges to 4. A series that converges is also known as a convergent series.
2
-R
Pr
es
s
-C
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op
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op
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C
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to be very useful, and gives us answers that work consistently when we try to do more
mathematics with them.
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id
DID YOU KNOW?
w
ge
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w
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C
176
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op
Pr
y
es
s
-C
-R
am
You are probably also familiar with a very important example of an infinite geometric
series without realising it! What do we mean by the recurring decimal 0.3333…?
3
3
3
+
+
+ … . If we work out the sum
We can write this as a series: 0.3333… =
10 100 1000
3
33
of the first n terms of this geometric series, we find S1 =
= 0.3, S2 =
= 0.33,
10
100
333
1
S3 =
= 0.333 and so on. These sums are getting as close as we like to , so we say that
1000
3
1
1
the sum of the infinite series is equal to , and we write = 0.3333… . This justifies what
3
3
you have been writing for many years. Using the formula we will be working out shortly,
we can easily write any recurring decimal as an exact fraction.
s
-C
-R
am
br
ev
The first person to introduce infinite decimal numbers was Simon Stevin in 1585. He was an
influential mathematician who popularised the use of decimals more generally as well, through a
publication called De Thiende (‘The tenth’).
Pr
op
y
es
EXPLORE 6.5
y
a=−
1
, r = −2
2
op
d
1
5
C
a = 3, r = −
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2
3
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a = 6, r =
e
c
3
, r = −2
5
id
g
a a=
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ev
2 Find other convergent geometric series of your own.
In each case, find the sum to infinity.
-R
s
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am
3 Can you find a condition for r for which a geometric series is convergent?
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1 Investigate whether these infinite geometric series converge or not. You could use
a spreadsheet to help with the calculations. If they converge, state their sum to
infinity.
Copyright Material - Review Only - Not for Redistribution
1
4
2
We therefore say that the sum of the infinite geometric series 2 + 1 +
You might be wondering why we can say this, as no matter how many terms we add up,
the answer is always less than 4. The simplest answer is because it works. Mathematicians
and philosophers have struggled with the idea of infinity for thousands of years, and
1 1 1
whether something like ‘2 + 1 + + + + …’ even makes sense. But over the past few
2 4 8
1 1 1
hundred years, we have worked out that writing ‘2 + 1 + + + + … = 4’ turns out
2 4 8
1
8
1
2
1
2
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Chapter 6: Series
-R
am
br
id
ev
ie
w
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Consider the geometric series a + ar + ar 2 + ar 3 + … + ar n − 1.
a (1 − r n )
.
The sum, Sn, is given by the formula Sn =
1− r
n
If −1 < r < 1 , then as n gets larger and larger, r gets closer and closer to 0.
We say that as n tends to infinity, r n tends to zero, and we write ‘as n → ∞, r n → 0’.
y
Pr
es
s
-C
a (1 − r n )
a (1 − 0)
a
→
=
.
1− r
1− r
1− r
Hence, as n → ∞,
ve
rs
ity
KEY POINT 6.12
y
a
provided that −1 < r < 1.
1− r
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ni
S∞ =
C
op
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w
C
op
This gives the result:
br
ev
id
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If r ù 1 or r ø −1, then r n does not converge, and so the series itself does not converge.
So an infinite geometric series converges when and only when −1 < r < 1.
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am
WORKED EXAMPLE 6.21
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-C
The first four terms of a geometric progression are 5, 4, 3.2 and 2.56.
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es
a Write down the common ratio.
rs
y
ni
op
Use a = 5 and r =
C
w
-R
am
br
ev
id
4
1−  
 5
= 25
a (1 − r 3 )
1− r
3

2 
a 1−  −  
 3 

y
2
Use S3 = 63 and r = − .
3
C
U
S3 =
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id
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ev
Simplify.
es
s
-R
br
2
1−  − 
 3
am
63 =
w
e
a
op
Answer
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ve
rs
b Find the sum to infinity.
ity
Pr
A geometric progression has a common ratio of − 2 and the sum of the first three terms is 63.
3
a Find the first term of the progression.
-C
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es
s
-C
WORKED EXAMPLE 6.22
R
4
.
5
ie
=
a
1− r
5
U
ev
R
b S∞ =
second term
4
=
first term
5
ve
Common ratio =
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w
a
ity
Answer
177
ge
C
op
Pr
b Find the sum to infinity.
Copyright Material - Review Only - Not for Redistribution
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35
27
am
br
id
Solve.
5
3
-R
63 =
a×
C
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Cambridge International AS & A Level Mathematics: Pure Mathematics 1
-C
a = 81
a
1− r
81
=
2
1−  − 
 3
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es
s
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op
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w
= 48 53
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EXERCISE 6E
R
2
Use a = 81 and r = − .
3
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b S∞ =
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1 Find the sum to infinity of each of the following geometric series.
2 …
2 2
a 2+ + +
+
b 1 + 0.1 + 0.01 + 0.001 + …
3 9 27
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am
40 − 20 + 10 − 5 + …
-C
c
d
−64 + 48 − 36 + 27 − …
es
s
2 The first four terms of a geometric progression are 1, 0.52, 0.54 and 0.56. Find the sum to infinity.
op
Pr
y
3 The first term of a geometric progression is 8 and the second term is 6. Find the sum to infinity.
ity
4 The first term of a geometric progression is 270 and the fourth term is 80. Find the common ratio and the sum
to infinity.
w
rs
C
178
y
op
C
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ɺ ɺ as the sum of a geometric progression.
5 a Write the recurring decimal 0.57
ɺ ɺ can be written as 19 .
b Use your answer to part a to show that 0.57
33
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id
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6 The first term of a geometric progression is 150 and the sum to infinity is 200. Find the common ratio and the
sum of the first four terms.
-R
am
br
7 The second term of a geometric progression is 4.5 and the sum to infinity is 18. Find the common ratio and
the first term.
es
s
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8 Write the recurring decimal 0.315151515… as a fraction.
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ity
e
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10 The third term of a geometric progression is 16 and the sixth term is − 1 .
4
a Find the common ratio and the first term.
es
s
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b Find the sum to infinity.
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y
b the sum to infinity.
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a the common ratio and the first term
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9 The second term of a geometric progression is 9 and the fourth term is 4. Given that the common ratio is
positive, find:
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Chapter 6: Series
am
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id
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11 The first three terms of a geometric progression are 135, k and 60. Given that all the terms in the progression
are positive, find:
a the value of k
-R
b the sum to infinity.
Pr
es
s
a Find the value of k.
b Find the sum to infinity.
ve
rs
ity
13 The fourth term of a geometric progression is 48 and the sum to infinity is five times the first term.
Find the first term.
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12 The first three terms of a geometric progression are k + 12 , k and k − 9, respectively.
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op
14 A geometric progression has first term a and common ratio r. The sum of the first three terms is 3.92 and the
sum to infinity is 5. Find the value of a and the value of r.
id
ie
15 The first term of a geometric progression is 1 and the second term is 2 cos x, where 0 < x <
16 A circle of radius 1cm is drawn touching the three edges of an equilateral triangle.
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PS
am
br
ev
Find the set of values of x for which this progression is convergent.
π
.
2
s
es
Pr
This process is then repeated an infinite number of times, as shown in the diagram.
179
a Find the sum of the circumferences of all the circles.
op
w
pattern 2
pattern 3
pattern 4
br
ev
pattern 1
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id
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C
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17
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ve
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b Find the sum of the areas of all the circles.
w
ie
P
ev
1 cm
ity
C
op
y
-C
Three smaller circles are then drawn at each corner to touch the original circle and
two edges of the triangle.
-R
am
We can construct a Koch snowflake as follows.
Starting with an equilateral triangle (pattern 1), we perform the following steps to produce pattern 2.
es
s
-C
Step 1: Divide each line segment into three equal segments.
Pr
Step 3: Remove the line segments that were used as the base of the equilateral triangles in step 2.
ni
ve
rs
ity
These three steps are then repeated to produce the next pattern.
-R
s
es
am
br
ev
ie
id
g
w
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C
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op
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a Let pn be the perimeter of pattern n. Show that the sequence p1, p2 , p3 , … tends to infinity.
8
b Let An be the area of pattern n. Show that the sequence A1, A2 , A3 , … tends to times the area of the original
5
triangle.
8
c The Koch snowflake is the limit of the patterns. It has infinite perimeter but an area of of the original
5
triangle, as you have shown. This snowflake pattern is an example of a fractal. Use the internet to find out
about the Sierpinski triangle fractal.
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Step 2: Draw an equilateral triangle, pointing outwards, that has the middle segment from step 1 as its base.
Copyright Material - Review Only - Not for Redistribution
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Cambridge International AS & A Level Mathematics: Pure Mathematics 1
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6.6 Further arithmetic and geometric series
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EXPLORE 6.6
Pr
es
s
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a, b, c,…
op
y
1 Given that a, b and c are in arithmetic progression, find an equation connecting
a, b and c.
y
U
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ni
WORKED EXAMPLE 6.23
C
op
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C
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rs
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2 Given that a, b and c are in geometric progression, find an equation connecting
a, b and c.
id
ie
w
ge
The first, second and third terms of an arithmetic series are x, y and x 2 . The first,
second and third terms of a geometric series are x, x 2 and y. Given that x < 0 , find:
br
ev
a the value of x and the value of y
-R
am
b the sum to infinity of the geometric series
es
s
-C
c the sum of the first 20 terms of the arithmetic series.
op
a Arithmetic series is: x + y + x 2 + …
rs
(1)
Geometric series is: x + x + y + …
y
x2
2 =
x
x
y = x3
(2)
Use common ratios.
(1) and (2) give 2 x 3 = x 2 + x
Divide by x (since x ≠ 0) and rearrange.
y
ev
id
ie
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ge
C
U
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op
2
ve
ie
w
2 y = x2 + x
ev
Use common differences.
ity
y − x = x2 − y
C
180
Pr
y
Answer
2x − x − 1 = 0
(2 x + 1)( x − 1) = 0
br
2
1
or x = 1
2
1
1
Hence, x = − and y = − .
2
8
a
b S∞ =
1− r
1
−
1
2
=−
S∞ =
3
1
1−  − 
 2
-R
am
Factorise and solve.
s
Pr
es
x ≠ 1 since x < 0.
C
U
op
y
ni
ve
rs
ity
Use a = − 1 and r = − 1 .
2
2
1
1
3
Use n = 20 , a = − 1 , d = − −  −  = .
8  2 8
2
n
[2 a + ( n − 1)d ]
2
20 
3 
−1 + 19   
S20 =
 8
2 
s
es
= 61.25
-R
br
ev
ie
id
g
w
e
Sn =
am
c
-C
R
ev
ie
w
C
op
y
-C
x=−
Copyright Material - Review Only - Not for Redistribution
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rs
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ev
ie
am
br
id
DID YOU KNOW?
w
ge
C
U
ni
op
y
Chapter 6: Series
op
y
Pr
es
s
-C
-R
Georg Cantor (1845–1918) was a German mathematician who is famous for his work on set theory
and for formalising many ideas about infinity. He developed the theory that there are infinite sets
of different sizes. He showed that the set of natural numbers (1, 2, 3, ...) and the set of rational
numbers (all fractions) are actually the same size, whereas the set of real numbers is actually larger
than either of them.
ve
rs
ity
ni
b geometric
U
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a arithmetic
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1 The first term of a progression is 16 and the second term is 24. Find the sum of the first eight terms given that
the progression is:
ge
2 The first term of a progression is 20 and the second term is 16.
C
op
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ie
w
C
EXERCISE 6F
id
ie
w
a Given that the progression is geometric, find the sum to infinity.
-R
am
br
ev
b Given that the progression is arithmetic, find the number of terms in the progression if the sum of all the
terms is −160.
s
es
Pr
a the value of r
181
b the sixth term of each progression.
ity
C
op
y
-C
3 The first, second and third terms of a geometric progression are the first, fourth and tenth terms, respectively,
of an arithmetic progression. Given that the first term in each progression is 12 and the common ratio of the
geometric progression is r, where r ≠ 1 , find:
y
op
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C
U
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ev
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w
rs
1
4 A geometric progression has eight terms. The first term is 256 and the common ratio is .
2
1
An arithmetic progression has 51 terms and common difference .
2
The sum of all the terms in the geometric progression is equal to the sum of all the terms in the arithmetic
progression. Find the first term and the last term in the arithmetic progression.
-R
am
br
ev
id
ie
5 The first, second and third terms of a geometric progression are the first, sixth and ninth terms, respectively,
of an arithmetic progression. Given that the first term in each progression is 100 and the common ratio of
the geometric progression is r, where r ≠ 1 , find:
s
-C
a the value of r
es
op
y
b the fifth term of each progression.
Pr
ity
a Find the common difference of this progression.
y
ni
ve
rs
The first, third and nth terms of this arithmetic progression are the first, second and third terms, respectively,
of a geometric progression.
op
w
e
7 The first term of a progression is 2x and the second term is x 2 .
C
U
b Find the common ratio of the geometric progression and the value of n.
-R
s
es
am
br
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ie
id
g
a For the case where the progression is arithmetic with a common difference of 15, find the two possible
values of x and corresponding values of the third term.
1
b For the case where the progression is geometric with a third term of − , find the sum to infinity.
16
-C
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ie
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C
6 The first term of an arithmetic progression is 16 and the sum of the first 20 terms is 1080.
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Cambridge International AS & A Level Mathematics: Pure Mathematics 1
Binomial expansions
w
Pr
es
s
-C
 n
 n
n!
n × ( n − 1) × ( n − 2) × … × ( n − r + 1)
or   =
the formulae   =
.
r × ( r − 1) × ( r − 2) × … × 3 × 2 × 1
 r  r! ( n − r )!
 r 
op
y
●
-R
 n
Binomial coefficients, denoted by n C r or   , can be found using:
 r 
● Pascal’s triangle
ev
ie
am
br
id
ge
Checklist of learning and understanding
ni
We can extend this rule to give:
y
 n
 n  n
 n
 n
(1 + x ) n =   +   x +   x 2 + … +   x n , where the ( r + 1)th term =   x r.
 r 
 2
 0  1
 n
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If n is a positive integer, the Binomial theorem states that:
w
ge
U
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 n  n−r r
 n
 n
 n
 n
b.
( a + b ) n =   a n +   a n − 1b1 +   a n − 2 b 2 + … +   b n , where the ( r + 1)th term =   a
 r 
 2
 1
 0
 n
s
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Arithmetic series
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br
n( n − 1) 2 n( n − 1)( n − 2) 3
x +
x + … + xn
2!
3!
am
(1 + x ) n = 1 + nx +
ev
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We can also write the expansion of (1 + x ) n as:
y
Pr
the kth term is a + ( k − 1)d
op
●
es
For an arithmetic progression with first term a, common difference d and n terms:
the last term is l = a + ( n − 1)d
n
n
( a + l ) = [2 a + ( n − 1)d ].
● the sum of the terms is Sn =
2
2
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Geometric series
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●
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182
●
the last term is ar n − 1
●
sum of the terms is Sn =
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the kth term is ar k − 1
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●
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a
.
1− r
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When an infinite geometric series converges, S∞ =
s
am
The condition for an infinite geometric series to converge is −1 < r < 1.
ev
ie
a (1 − r n ) a ( r n − 1)
=
.
1− r
r −1
id
br
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For a geometric progression with first term a, common ratio r and n terms:
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Chapter 6: Series
ev
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5
3
Find the coefficient of x 2 in the expansion of  2 x + 2  .

x 
-R
[3]

x
In the expansion of  1 −  (5 + x )6, the coefficient of x 2 is zero.

a
Find the value of a.
[3]
Pr
es
s
2 
.
Find the term independent of x in the expansion of  3x −

5x 
y
U
1 
.
Find the term independent of x in the expansion of  3x 2 −

2 x3 
[4]
8
a
Find the first three terms in the expansion of ( x − 3x 2 )8, in descending powers of x.
am
[3]
b
Find the coefficient of x in the expansion of (1 − x )( x − 3x ) .
[2]
a
Find the first three terms in the expansion of (1 + px )8, in ascending powers of x.
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es
s
2 8
Pr
2
[4]
ii
(3 − x )
y
(1 + 2 x )5
op
i
ve
Find the first three terms, in ascending powers of x, in the expansion of:
ni
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5
w
Find the coefficient of x 2 in the expansion of [(1 + 2 x )(3 − x )]5.
[2]
[2]
[3]
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b
183
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Given that the coefficient of x in the expansion of (1 − 2 x )(1 + px ) is 204, find the possible
values of p.
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10 a
[3]
8
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b
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15
[4]
5
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7
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[4]
2
Find the coefficient of x5 in the expansion of  x 3 + 2  .

x 
9
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5
6
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[3]
In the expansion of (2 + ax )7 , where a is a constant, the coefficient of x is −2240.
Find the coefficient of x 2 .
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5
6
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4
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3
[3]
In the expansion of ( a + 2 x )6, the coefficient of x is equal to the coefficient of x 2.
Find the value of the constant a.
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2
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END-OF-CHAPTER REVIEW EXERCISE 6
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11 The first term of an arithmetic progression is 1.75 and the second term is 1.5. The sum of the
first n terms is −n . Find the value of n.
[4]
s
b
the first term
c
the sum to infinity.
es
the common ratio
[3]
Pr
a
[1]
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[2]
Find d in terms of a.
b
Write down an expression, in terms of a, for the 50th term.
[3]
[2]
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a
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13 An arithmetic progression has first term a and common difference d . The sum of the first 100 terms
is 25 times the sum of the first 20 terms.
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s
Given that the nth term of the progression is −19, find the value of n.
es
b
Find the first term of the progression and the common difference.
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14 The tenth term of an arithmetic progression is 17 and the sum of the first five terms is 190.
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12 The second term of a geometric progression is −1458 and the fifth term is 432. Find:
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[4]
[2]
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[3]
The seventh term of an arithmetic progression is 19 and the sum of the first twelve terms is 224.
Find the fourth term.
[4]
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The fifth term of an arithmetic progression is 18 and the sum of the first eight terms is 186.
Find the first term and the common difference.
1
The first term of a geometric progression is 32 and the fourth term is . Find the
2
sum to infinity of the progression.
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y
C
op
[3]
[3]
An arithmetic progression has first term −4. The nth term is 8 and the (2 n )th term is 20.8.
Find the value of n.
[4]
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A second geometric progression has first term 5a, common ratio 3r and sum to infinity 10S .
Find the value of r.
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b
[4]
A geometric progression has first term a, common ratio r and sum to infinity S.
ge
17 a
A geometric progression has first term 3 and common ratio r. A second geometric progression has
1
first term 2 and common ratio r. The two progressions have the same sum to infinity, S. Find the
5
value of r and the value of S.
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16 a
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Cambridge International AS & A Level Mathematics: Pure Mathematics 1
Pr
Model 1: Increase the prize money by $1000 each day.
Model 2: Increase the prize money by 10% each day.
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184
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18 A television quiz show takes place every day. On day 1 the prize money is $1000. If this is not won
the prize money is increased for day 2. The prize money is increased in a similar way every day until it is
won. The television company considered the following two different models for increasing the prize money.
if Model 1 is used,
ii
if Model 2 is used.
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[3]
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Cambridge International AS & A Level Mathematics 9709 Paper 11 Q8 June 2011
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s
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19 a The first two terms of an arithmetic progression are 1 and cos2 x respectively. Show that the sum of the first
ten terms can be expressed in the form a − b sin2 x, where a and b are constants to be found.
[3]
b The first two terms of a geometric progression are 1 and 1 tan2 θ respectively,
3
1
where 0 < θ < π .
2
[2]
i Find the set of values of θ for which the progression is convergent.
1
ii Find the exact value of the sum to infinity when θ = π .
[2]
6
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q7 June 2012
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For the case where the progression is geometric with a sum to infinity of 8, find the third term.
[4]
[4]
s
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q8 November 2015
es
am
ii
For the case where the progression is arithmetic with a common difference of 12, find the possible
values of x and the corresponding values of the third term.
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20 The first term of a progression is 4x and the second term is x 2.
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[4]
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On each day that the prize money is not won the television company makes a donation to charity.
The amount donated is 5% of the value of the prize on that day. After 40 days the prize money
has still not been won. Calculate the total amount donated to charity
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Chapter 6: Series
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1
2
21 a The third and fourth terms of a geometric progression are and respectively. Find the
3
9
sum to infinity of the progression.
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b A geometric progression has first term a, common ratio r and sum to infinity 6. A second
geometric progression has first term 2a, common ratio r 2 and sum to infinity 7. Find the
values of a and r.
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[5]
185
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Cambridge International AS & A Level Mathematics 9709 Paper 11 Q9 November 2013
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[5]
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22 a In an arithmetic progression the sum of the first ten terms is 400 and the sum of the
next ten terms is 1000. Find the common difference and the first term.
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[4]
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q7 June 2015
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b A circle is divided into 5 sectors in such a way that the angles of the sectors are in arithmetic
progression. Given that the angle of the largest sector is 4 times the angle of the smallest
sector, find the angle of the largest sector.
[4]
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Cambridge International AS & A Level Mathematics: Pure Mathematics 1
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CROSS-TOPIC REVIEW EXERCISE 2
3
6

2 
a Find the first three terms in the expansion of  3x − 2  , in descending powers of x.

x 
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op
[2]
a Find the first three terms when (1 − 2 x )5 is expanded, in ascending powers of x.
[3]
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op
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b Find the sum to infinity.
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[3]
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op
b Find the sum to infinity.
[2]
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The sum of the 1st and 2nd terms of a geometric progression is 50 and the sum of the 2nd and 3rd
terms is 30. Find the sum to infinity.
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a Find the value of k.
7
[2]
The first three terms of a geometric progression are 3k + 14, k + 14 and k, respectively.
All the terms in the progression are positive.
186
[3]
ie
a Find the fourth term.
6
[2]
The first term of a geometric progression is 50 and the second term is −40.
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Find the value of a.
y
b In the expansion of (3 + ax )(1 − 2 x )5, the coefficient of x 2 is zero.
5
[3]
6

2 
2
b Hence, find the coefficient of x 2 in the expansion of  1 +   3x −  .

x 
x
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C
[3]
Pr
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
1 
Find the term independent of x in the expansion of  4x − 2  .

x 
4
[2]
6
2
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4
Find the highest power of x in the expansion of  (5x 4 + 3)8 + (1 − 3x 3 )5 (4x 2 − 5x5 )6  .
[6]
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op
i Show that cos x ≡ 1 − 2 sin2 x + sin 4 x .
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Cambridge International AS & A Level Mathematics 9709 Paper 11 Q5 November 2016
4
[5]
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ii Hence, or otherwise, solve the equation 8 sin 4 x + cos 4 x = 2 cos2 x for 0° < x < 360°.
[1]
ie
a Show that the area, A cm 2, of the sector is given by A = 30 r − r 2.
[2]
b Express 30 r − r 2 in the form a − ( r − b )2, where a and b are constants.
[2]
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[1]
Pr
c find the value of r at which A is a maximum
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s
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Given that r can vary:
ity
[1]
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d find this stationary value of A.
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A sector of a circle, radius r cm, has a perimeter of 60 cm.
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9
id
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q6 November 2016
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r cm
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The diagram shows a metal plate consisting of a rectangle with sides x cm and r cm and two identical sectors of
a circle of radius r cm. The perimeter of the plate is 100 cm.
[2]
b Express 50 r − r 2 in the form a − ( r − b )2, where a and b are constants.
[2]
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c find the value of r at which A is a maximum
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am
rm
lm
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The diagram shows a running track. The track has a perimeter of 400 m and consists of two straight sections of
length l m and two semicircular sections of radius r m.
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[1]
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[1]
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d find this stationary value of A.
ev
[3]
[2]
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c show that A has a maximum value when l = 0
R
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a Show that the area, A m2, of the region enclosed by the track is given by A = 400 r − π r 2 .
2
a
b
b Express 400 r − πr 2 in the form − π  r −  , where a and b are constants.

π
π
Given that l and r can vary:
R
[1]
ie
d find this stationary value of A.
11
y
a Show that the area, A cm2, of the plate is given by A = 50 r − r 2.
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x cm
Given that r can vary:
R
r cm
Pr
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s
10
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Cross-topic review exercise 2
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187
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C1
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P
8 cm
T
Q
2 cm
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12
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Cambridge International AS & A Level Mathematics: Pure Mathematics 1
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The diagram shows two circles, C1 and C2 , touching at the point T . Circle C1 has centre P and radius 8 cm;
circle C2 has centre Q and radius 2 cm. Points R and S lie on C1 and C2 respectively, and RS is a
tangent to both circles.
[2]
es
s
i Show that RS = 8 cm.
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13
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Cambridge International AS & A Level Mathematics 9709 Paper 11 Q9 November 2010
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[4]
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iii Find the area of the shaded region.
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[2]
Pr
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ii Find angle RPQ in radians correct to 4 significant figures.
188
s
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am
In the diagram, OAB is an isosceles triangle with OA = OB and angle AOB = 2θ radians.
Arc PST has centre O and radius r, and the line ASB is a tangent to the arc PST at S.
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[5]
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The function f is such that f( x ) = 2 sin2 x − 3 cos2 x for 0 < x < π.
e
C
ii State the greatest and least values of f( x ).
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[2]
[2]
[3]
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Cambridge International AS & A Level Mathematics 9709 Paper 11 Q5 June 2010
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iii Solve the equation f( x ) + 1 = 0.
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i Express f( x ) in the form a + b cos2 x , stating the values of a and b.
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14
[4]
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q9 June 2011
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i Find the total area of the shaded regions in terms of r and θ .
1
ii In the case where θ = π and r = 6 , find the total perimeter of the shaded regions, leaving your
3
answer in terms of 3 and π.
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Cross-topic review exercise 2
sin θ
1
1
−
≡
.
1 − cos θ sin θ
tan θ
sin θ
1
−
= 4 tan θ for 0° , θ , 180°.
ii Hence solve the equation
1 − cos θ sin θ
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The function f is defined by f : x ֏ 4 sin x − 1 for −
i State the range of f .
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[3]
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q9 June 2014
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16
[4]
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i Prove the identity
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15
1
1
π < x < π.
2
2
[2]
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ii Find the coordinates of the points at which the curve y = f( x ) intersects the coordinate axes.
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iii Sketch the graph of y = f( x ) .
−1
−1
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U
iv Obtain an expression for f ( x ), stating both the domain and range of f .
[4]
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ie
id
ev
a The first term of a geometric progression in which all the terms are positive is 50. The third term is 32.
Find the sum to infinity of the progression.
-R
am
[3]
Pr
y
es
s
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b The first three terms of an arithmetic progression are 2 sin x , 3 cos x and (sin x + 2 cos x ) respectively, where
x is an acute angle.
4
[3]
i Show that tan x = .
3
ii Find the sum of the first twenty terms of the progression.
[3]
ity
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q9 June 2016
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[2]
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q11 June 2016
br
17
[3]
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189
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Chapter 7
Differentiation
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190
id
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understand that the gradient of a curve at a point is the limit of the gradients of a suitable
sequence of chords
dy
d2 y
and
use the notations f ′( x ) , f ′′( x ) ,
for the first and second derivatives
dx
dx 2
use the derivative of x n (for any rational n), together with constant multiples, sums, differences of
functions, and of composite functions using the chain rule
apply differentiation to gradients, tangents and normals.
y
op
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■
■
■
■
ie
In this chapter you will learn how to:
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Chapter 7: Differentiation
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PREREQUISITE KNOWLEDGE
What you should be able to do
IGCSE / O Level Mathematics
Use the rules of indices to simplify
expressions to the form ax n .
a 3x x
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b 5 3 x2
x
c
2 x
1
d
2x
3
e
x2
2
f − 2x
3
5 x
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id
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k
in the form
( ax + b ) n
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U
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Write
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am
br
k ( ax + b )− n .
Find the gradient of a
perpendicular line.
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Pr
y
ity
op
Find the equation of a line with a
given gradient and a given point
on the line.
y
4 Find the equation of the line
with gradient 2 that passes
through the point (2, 5).
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ni
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Chapter 3
2 Write in the form k ( ax + b )− n :
4
a
( x − 2)3
2
b
(3x + 1)5
3 The gradient of a line is 2 .
3
Write down the gradient of a
line that is perpendicular to it.
s
Chapter 3
C
U
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1 Write in the form ax n :
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IGCSE / O Level Mathematics
C
Check your skills
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Where it comes from
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Why do we study differentiation?
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Calculus is the mathematical study of change. Calculus has two basic tools, differentiation
and integration, and it has widespread uses in science, medicine, engineering and
economics. A few examples where calculus is used are:
designing effective aircraft wings
● the study of radioactive decay
● the study of population change
● modelling the financial world.
In this chapter you will be studying the first of the two basic tools of calculus. You will
learn the rules of differentiation and how to apply these to problems involving gradients,
tangents and normals. In Chapter 8 you will then learn how to apply these rules of
differentiation to more practical problems.
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7.1 Derivatives and gradient functions
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C
U
At IGCSE / O Level you learnt how to estimate the gradient of a curve at a point by
drawing a suitable tangent and then calculating the gradient of the tangent. This method
only gives an approximate answer (because of the inaccuracy of drawing the tangent) and
it is also very time consuming.
-R
s
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am
In this chapter you will learn a method for finding the exact gradient of the graph of a function
(which does not involve drawing the graph). This exact method is called differentiation.
-C
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●
Copyright Material - Review Only - Not for Redistribution
WEB LINK
Try the Calculus
resources on the
Underground
Mathematics website.
191
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EXPLORE 7.1
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Cambridge International AS & A Level Mathematics: Pure Mathematics 1
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Consider the quadratic function y = x 2 and a point P ( x, x 2 ) on the curve.
y = x2
y
Pr
es
s
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1 Let P be the point (2, 4).
op
y
The points A(2.2, 4.84), B (2.1, 4.41) and C (2.01, 4.0401) also lie on the
curve and are close to the point P (2, 4).
ii the chord PB
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iii the chord PC.
x
y
the chord PA
C
op
w
i
P (2, 4)
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C
a Calculate the gradient of:
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R
b Discuss your results with those of your classmates and make suggestions as to what is happening.
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c Suggest a value for the gradient of the curve y = x 2 at the point (2, 4).
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2 Let P be the point (3, 9).
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The points A(3.2, 10.24), B (3.1, 9.61) and C (3.01, 9.0601) also lie on the curve and are close to the
point P (3, 9).
es
Pr
ii the chord PB
iii the chord PC.
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rs
C
the chord PA
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op
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i
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c Suggest a value for the gradient of the curve y = x 2 at the point (3, 9).
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b Discuss your results with those of your classmates and make suggestions as to what is happening.
op
192
s
a Calculate the gradient of:
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C
3 Use a spreadsheet to investigate the value of the gradient at other points on the curve y = x 2.
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4 Can you suggest a general formula for the gradient of the curve y = x 2 at the point ( a, a 2 )?
What would be the gradient at ( x, x 2 )?
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Pr
e
y = x2
op
y
δy
δx
x
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id
g
O
A
C
P
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ity
y
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op
y
Take a point P ( x, y ) on the curve y = x 2 and a point A that is close to the point P.
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am
br
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The coordinates of A are ( x + δx, y + δy ), where δx is a small increase in the value of x
and δy is the corresponding small increase in the value of y.
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WEB LINK
s
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The general formula for the gradient of the curve y = x 2 at the point (x, x 2 ) can be proved
algebraically.
Copyright Material - Review Only - Not for Redistribution
There are other ways
of thinking about the
gradient of a curve. Try
the following resources
on the Underground
Mathematics website
Zooming in and
Mapping a derivative.
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Chapter 7: Differentiation
(
2
y
op
x 2 + 2 x δx + ( δx ) − x 2
δx
2
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C
We use the Greek
symbol delta, δ, to
denote a very small
change in a quantity.
-R
( x + δx ) − x
2 x δx + ( δx )
=
δx
= 2 x + δx
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( x + δx )2 − x2
2
=
TIP
Pr
es
s
-C
=
).
ev
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y2 − y1
x2 − x1
am
br
id
Gradient of chord PA =
w
ge
We can also write the coordinates of P and A as ( x, x 2 ) and x + δx, ( x + δx )
U
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C
op
y
As δx tends towards 0, A tends to P and the gradient of the chord PA tends to a value.
We call this value the gradient of the curve at P.
w
ge
In this case, therefore, the gradient of the curve at P is 2x.
id
ie
This process of finding the gradient of a curve at any point is called differentiation.
s
-C
-R
am
br
ev
Later in this chapter, you will learn some rules for differentiating functions without having to
calculate the gradients of chords as we have done here. The process of calculating gradients
using the limit of gradients of chords is sometimes called differentiation from first principles.
es
Notation
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op
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op
Pr
y
There are three different notations that are used to describe the previous rule.
dy
= 2 x.
1. If y = x 2, then
dx
2. If f( x ) = x 2, then f ′( x ) = 2 x.
d
( x2 ) = 2x
3.
dx
dy
is called the derivative of y with respect to x.
If y is a function of x, then
dx
Likewise f ′( x ) is called the derivative of f(x).
dy
or f ′( x ) is sometimes also called the
If y = f( x ) is the graph of a function, then
dx
gradient function of this curve.
d
( x 2 ) = 2 x means ‘if we differentiate x 2 with respect to x, the result is 2 x’.
dx
Pr
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EXPLORE 7.2
y
1 Use a spreadsheet to investigate the gradient of the curve y = x 3.
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2 Can you suggest a general formula for the gradient of the curve y = x 3 at
the point ( x, x 3 )?
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id
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3 Differentiate y = x 3 from first principles to confirm your answer to
question 2.
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You do not need to be able to differentiate from first principles but you are expected to
understand that the gradient of a curve at a point is the limit of a suitable sequence of chords.
Copyright Material - Review Only - Not for Redistribution
DID YOU KNOW?
Gottfried Wilhelm
Leibniz and Isaac
Newton are both
credited with
developing the modern
calculus that we
use today. Leibniz’s
notation for derivatives
dy
was
. Newton’s
dx
dy
notation for
dx
was yɺ . The notation
f ′( x ) is known as
Lagrange’s notation.
193
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C
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Cambridge International AS & A Level Mathematics: Pure Mathematics 1
w
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Differentiation of power functions
d
( x 6 ) = 6x 5
dx
Pr
es
s
d
( x 5 ) = 5x 4
dx
-C
d
( x 4 ) = 4x 3
dx
-R
am
br
id
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ie
d
d
( x 3 ) = 3x 2.
( x 2 ) = 2 x and that
dx
dx
Investigating the gradient of the curves y = x 4 , y = x5 and y = x 6 would give the results:
We now know that
op
y
This leads to the general rule for differentiating power functions:
d
( x n ) = nx n – 1
dx
y
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rs
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U
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ni
C
op
This is true for any real power n, not only for positive integer values of n.
id
op
Pr
y
es
s
-C
-R
am
br
ev
So for the earlier example where y = x 2:
dy
= 2 × x 2 −1
dx
= 2 x1
= 2x
y
op
U
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ev
s
-R
br
am
-C
id
g
1
1 −2
x
2
1
=
2 x
e
1
1 2 −1
x
2
1
Multiply by the power
and then subtract
2
one from the power.
C
1
=
1
Write x as x 2 .
x
f( x ) = x 2
f ′( x ) =
ev
es
Pr
ni
ve
rs
f( x ) =
Multiply by the power −2 and then subtract
one from the power.
−3
ity
op
y
= −2 x
2
=− 3
x
1
as x −2.
x2
s
am
Write
-R
br
id
ie
w
Multiply by the power 7 and then subtract one
from the power.
= −2 x −2 − 1
C
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ev
c
y=2
d
x
C
ge
d
( x7 ) = 7x7 − 1
dx
= 7x6
d  1 
d
=
( x −2 )
dx  x 2  dx
-C
b
f( x ) =
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Answer
a
R
c
y
rs
Find the derivative of each of the following.
1
b
a x7
x2
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WORKED EXAMPLE 7.1
es
194
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‘Multiply by the power n and then subtract one from the power.’
ie
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You may find it easier to remember this rule as:
op
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C
KEY POINT 7.1
Copyright Material - Review Only - Not for Redistribution
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Multiply by the power and then subtract one
from the power.
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y = 2x
0
TIP
w
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Write 2 as 2 x 0 .
am
br
id
y=2
d
C
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Chapter 7: Differentiation
op
y
Pr
es
s
-C
dy
= 0x 0 − 1
dx
=0
It is worth
remembering that
when you differentiate
a constant, the answer
is always 0.
Scalar multiple rule
C
op
ge
U
KEY POINT 7.2
R
y
If k is a constant and f( x ) is a function then:
ni
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You need to know and be able to use the following two rules.
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am
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id
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w
d
d
[ kf( x )] = k
[f( x )]
dx
dx
Addition/subtraction rule
es
s
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If f( x ) and g( x ) are functions then
ity
195
y
C
U
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1
4
+
+ 5 with respect to x.
x
2x2
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ev
id
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R
WORKED EXAMPLE 7.2
Differentiate 3x 4 −
op
ni
ev
ve
rs
d
d
d
[f( x ) ± g( x )] =
[f( x )] ±
[g( x )]
dx
dx
dx
ie
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C
op
Pr
y
KEY POINT 7.3
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am
Answer
Pr
1
d
1 d
d  −2 
d
(x4 ) −
( x −2 ) + 4
x
+5
(x0 )


dx
2 dx
dx 
dx

 1 −3 
1
( −2 x −3 ) + 4  − x 2  + 5(0 x −1 )
2
 2

ie
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ve
rs
= 3(4x 3 ) −
ity
C
op
y
=3
es
s
-C
1

−
d  4
1
4
d  4 1 −2

3x −
3x − x + 4x 2 + 5x 0 
+
+ 5 =

2

 dx 
dx 
2
x
2x

−
3
y
op
2
x3
C
1
−
x3
-R
s
es
-C
am
br
ev
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id
g
e
U
R
= 12 x 3 +
w
ev
= 12 x 3 + x −3 − 2 x 2
Copyright Material - Review Only - Not for Redistribution
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WORKED EXAMPLE 7.3
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Cambridge International AS & A Level Mathematics: Pure Mathematics 1
-C
Answer
-R
Find the gradient of the tangent to the curve y = x(2 x − 1)( x + 3) at the point (1, 4).
y = 2 x + 5x − 3x
2
y
3
y
C
op
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ev
id
WORKED EXAMPLE 7.4
ge
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op
dy
= 6x 2 + 10 x − 3
dx
dy
= 6(1)2 + 10(1) − 3
When x = 1,
dx
dx
= 13
Gradient of curve at (1, 4) is 13.
R
Expand brackets and simplify.
Pr
es
s
y = x(2 x − 1)( x + 3)
-R
am
The curve y = ax 4 + bx 2 + x has gradient 3 when x = 1 and gradient −51 when x = −2.
es
s
-C
Find the value of a and the value of b.
Pr
y
Answer
op
y = ax 4 + bx 2 + x
ity
dy
= 4ax 3 + 2bx + 1
dx
dy
Since
= 3 when x = 1:
dx
rs
y
ve
op
4a (1)3 + 2b(1) + 1 = 3
4a + 2b = 2
2a + b = 1
-R
s
es
am
op
y
-C
4a ( −2)3 + 2b( −2) + 1 = −51
−32 a − 4b = −52
8a + b = 13
ev
id
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dy
= −51 when x = −2:
dx
br
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ni
ve
rs
C
∴a = 2
Substitute a = 2 into (1): 4 + b = 1
(2)
Pr
(2) − (1) gives 6a = 12
y
op
-R
s
-C
am
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ie
id
g
w
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C
U
R
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ie
∴ b = −3
es
Since
w
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(1)
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C
196
Copyright Material - Review Only - Not for Redistribution
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EXERCISE 7A
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Chapter 7: Differentiation
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1 The points A(0, 0), B(0.5, 0.75), C(0.8, 1.44), D(0.95, 1.8525), E(0.99, 1.9701) and
F(1, 2) lie on the curve y = f(x ).
AF
BF
Gradient
2
2.5
CF
DF
b Use the values in the table to predict the value of
w
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Pr
es
s
Chord
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C
op
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-C
a Copy and complete the table to show the gradients of the chords CF,
DF and EF.
y
C
op
d
y=
c
ev
f
3
g
es
-C
x2
b
f( x ) = 3x5
e
f( x ) =
5
3x 2
f
f( x ) = −2
Pr
c
ve
g
6
x
2
4x
f( x ) =
x
f( x ) =
a
d
y = ( x + 5)( x − 4)
h
w
)2
y = 7 − 3x + 5x 2
ev
f
br
2x − 5
x2
y=
4x 2 + 3x − 2
x
es
s
-C
-R
am
y = 7 x2 −
y=
ie
(
y = 2x2 − 3
c
C
U
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id
e
op
C
dy
when x = 2.
dx
id
g
e
8 Given that xy = 12, find the value of
w
U
7 Find the gradient of the curve y = (2 x − 5)( x + 4) at the point (3, 7) .
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op
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-R
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9 Find the gradient of the curve y = 5x 2 − 8x + 3 at the point where the curve
crosses the y-axis.
-C
C
w
3
x
2x x
f( x ) =
3x 3
f( x ) =
d
5
1
3
2
+
h y = 3x + −
i
x x2
x 2 x
dy
for each curve at the given point.
6 Find the value of
dx
a y = x 2 + x − 4 at the point (1, −2)
2
b y =5−
at the point (2, 4)
x
3x − 2
at the point ( −2, −2)
c y=
x2
g
ie
h
197
dy
for each of the following.
dx
y = 5x 2 − x + 1
b y = 2 x 3 + 8x − 4
5 Find
R
ity
f( x ) = 2 x 4
rs
a
x3 × x2
ni
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w
C
op
4 Find f ′( x ) for each of the following.
1
x
x5
x2
d
y
8
y
e
x −4
op
x9
12
at (2, 6)
x
s
am
b
x5
w
ge
id
y = x 2 − 2 x + 3 at (0, 3)
ie
y = 2 x at (4, 4)
b
-R
ni
c
U
y = x 4 at (1, 1)
br
R
a
a
ev
dy
when x = 1.
dx
2 By considering the gradient of a suitable sequence of chords, find a value for the
gradient of the curve at the given point.
3 Differentiate with respect to x:
R
EF
Copyright Material - Review Only - Not for Redistribution
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Cambridge International AS & A Level Mathematics: Pure Mathematics 1
ev
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11 Find the gradient of the curve y =
the x-axis.
5 x − 10
at the point where the curve crosses
x2
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am
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id
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10 Find the coordinates of the points on the curve y = x 3 − 3x − 8 where the
gradient is 9.
Pr
es
s
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12 The curve y = x 2 − 4x − 5 and the line y = 1 − 3x meet at the points A and B.
b Find the gradient of the curve at each of the points A and B.
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13 The gradient of the curve y = ax 2 + bx at the point (3, −3) is 5. Find the value of
a and the value of b.
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a Find the coordinates of the points A and B.
U
R
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C
op
14 The gradient of the curve y = x 3 + ax 2 + bx + 7 at the point (1, 5) is −5. Find the
value of a and the value of b.
ge
b
has gradient 16 when x = 1 and gradient −8 when
x2
x = −1. Find the value of a and the value of b.
br
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15 The curve y = ax +
-C
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am
16 Given that the gradient of the curve y = x 3 + ax 2 + bx + 3 is zero when x = 1
and when x = 6, find the value of a and the value of b.
s
dy
, 0.
dx
dy
18 Given that y = 4x 3 + 3x 2 − 6x − 9, find the range of values of x for which
> 0.
dx
19 A curve has equation y = 3x 3 + 6x 2 + 4x − 5. Show that the gradient of the curve
is never negative.
Try the following
resources on the
Underground
Mathematics website:
• Slippery slopes
• Gradient match.
op
y
ve
ni
7.2 The chain rule
U
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w
rs
C
198
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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
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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
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rs
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C
U
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op
y
Chapter 7: Differentiation
am
br
id
ev
ie
w
WORKED EXAMPLE 7.5
y = (3x − 2)7
Let u = 3x − 2
so
y = u7
and
dy
= 7u6
du
ve
rs
ity
y
y
ni
U
R
= 7(3x − 2)6 × 3
Use the chain rule.
C
op
ev
ie
w
C
op
du
=3
dx
dy
dy du
=
×
dx du dx
= 7u6 × 3
Pr
es
s
-C
Answer
-R
Find the derivative of y = (3x − 2)7.
id
ie
w
ge
= 21(3x − 2)6
br
ev
With practice you will be able to do this mentally.
-R
am
Consider the ‘inside’ of (3x − 2)7 to be 3x − 2.
y
op
Pr
3
Step 2: Differentiate the ‘inside’:
s
7(3x − 2)6
Step 1: Differentiate the ‘outside’:
es
-C
To differentiate (3x − 2)7:
199
rs
op
es
y
ni
ve
rs
-R
s
-C
am
br
ev
ie
id
g
w
e
C
U
R
= −10(3x 2 + 1)−6 × 6x
60 x
=−
(3x 2 + 1)6
ity
Pr
Use the chain rule.
op
am
C
w
ie
C
w
dy
= −10 u −6
du
ev
and
op
y
-C
du
= 6x
dx
dy
dy du
=
×
dx du dx
= −10 u −6 × 6x
-R
y = 2 u −5
s
so
Let u = 3x 2 + 1
ev
y
ve
ni
id
2
(3x 2 + 1)5
br
y=
ie
Answer
2
.
(3x + 1)5
2
ge
R
Find the derivative of y =
U
ev
ie
WORKED EXAMPLE 7.6
es
w
C
ity
Step 3: Multiply these two expressions: 21(3x − 2)6
Copyright Material - Review Only - Not for Redistribution
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rs
ity
w
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C
U
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op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
6x
ev
ie
Step 2: Differentiate the ‘inside’:
Pr
es
s
−10(3x 2 + 1)−6
60 x
(3x 2 + 1)6
The curve y =
ax + b passes through the point (12, 4) and has gradient
1
at this point.
4
ie
(1)
-C
1
y = ( ax + b ) 2
and
dy 1 − 2
= u
du 2
ity
rs
y
op
ni
C
U
ie
w
ge
Substitute x = 12 and
ev
(2)
-R
am
br
id
ev
Use the chain rule.
ve
1
1 −2
u ×a
2
a
dy
=
dx 2 ax + b
a
1
=
4 2 12 a + b
2 a = 12 a + b
=
R
es
y = u2
dy dy du
=
×
dx du dx
s
es
Pr
ity
4 = 24 + b
16 = 24 + b
b = −8
∴ a = 2, b = −8
y
op
-R
s
es
-C
am
br
ev
ie
id
g
w
e
C
U
R
ni
ve
rs
ie
w
C
op
y
-C
(1) and (2) give 2 a = 4
a=2
Substituting a = 2 into (1) gives:
ev
ax + b in the form ( ax + b ) 2 .
Pr
y
op
C
w
ie
so
1
du
=a
dx
1
Write
s
1
Let u = ax + b
200
-R
am
4 = 12 a + b
Substitute x = 12 and y = 4.
ev
ax + b
br
y=
id
Answer
w
ge
U
R
Find the value of a and the value of b.
y
WORKED EXAMPLE 7.7
ni
ev
ie
w
C
ve
rs
ity
op
y
Step 3: Multiply the two expressions: −60 x(3x 2 + 1)−6 = −
C
op
-C
Step 1: Differentiate the ‘outside’:
-R
am
br
id
Alternatively, to differentiate the expression mentally:
2
as 2(3x 2 + 1)−5.
Write
2
(3x + 1)5
Copyright Material - Review Only - Not for Redistribution
dy
1
= .
dx
4
ve
rs
ity
ev
ie
(2 x + 3)8
e
(5x − 2)8
4
f
5(2 x − 1)5
i
( x 2 + 3)5
j
(2 − x 2 )8
2 Differentiate with respect to x:
1
a
b
x+2
4
f
e
(3x + 1)6
3 Differentiate with respect to x:
3
x−5
3
2(3x + 1)5
ni
ve
rs
ity
-C
y
op
C
w
ev
ie
(3 − 4x )5
d
g
2(4 − 7 x )4
h
k
( x 2 + 4 x )3
l
c
g
2x + 3
c
f
2 3x + 1
8
3 − 2x
8
x2 + 2 x
2 x2 − 1
1
2x − 5
d
h
w
br
5 − 2x
b
ie
3
e
id
x−5
a
d
g
h
ev
ge
U
R
c
y
b
Pr
es
s
( x + 4)6
a
-R
1 Differentiate with respect to x:
C
op
am
br
id
EXERCISE 7B
w
ge
C
U
ni
op
y
Chapter 7: Differentiation
 1 x + 1
2

9
1
(3x − 1)7
5
5
 x2 − 5 

x
16
x2 + 2
7
(2 x 2 − 5 x )7
x 3 − 5x
6
3
2 − 3x
-R
am
4 Find the gradient of the curve y = (2 x − 3)5 at the point (2, 1).
6
at the point where the curve crosses the y-axis.
( x − 1)2
3
6 Find the gradient of the curve y = x −
at the points where the curve crosses the x-axis.
x+2
es
Pr
y
op
ity
7 Find the coordinates of the point on the curve y =
y
ve
ie
C
U
O
w
x
ni
ve
rs
ity
Pr
op
y
es
normal
C
dy
at the point A( x1, y1 ) is m, then the equation of the tangent
dx
at A is given by:
ev
ie
id
g
br
KEY POINT 7.5
w
e
C
U
If the value of
es
s
-R
y − y1 = m( x − x1 )
am
y
The line perpendicular to the tangent at the point A is called the normal at A.
-C
w
ie
-R
s
-C
A(x1, y1)
ie
id
br
am
tangent
ev
ge
y = f(x)
y
op
R
ni
ev
op
w
a
3
passes through the point (2, 1) and has gradient − at this point.
bx − 1
5
Find the value of a and the value of b.
8 The curve y =
7.3 Tangents and normals
ev
R
( x − 10 x + 26) where the gradient is 0.
2
rs
C
s
-C
5 Find the gradient of the curve y =
Copyright Material - Review Only - Not for Redistribution
TIP
We use the numerical
form for m in this
formula (not the
derivative formula).
201
ve
rs
ity
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
am
br
id
ev
ie
w
ge
The normal at the point ( x1, y1 ) is perpendicular to the tangent, so the gradient of the
1
normal is −
and the equation of the normal is given by:
m
-R
KEY POINT 7.6
Pr
es
s
-C
1
( x − x1 )
m
op
y
y − y1 = −
y
ie
id
Answer
8
− 9 at the point where x = 2 .
x2
w
ge
U
R
Find the equation of the tangent and the normal to the curve y = 2 x 2 +
C
op
WORKED EXAMPLE 7.8
ni
ev
ie
w
C
ve
rs
ity
This formula only makes sense when m ≠ 0. If m = 0, it means that the tangent is
horizontal and the normal is vertical, so it has equation x = x1 instead.
s
-C
-R
am
br
ev
y = 2 x 2 + 8x −2 − 9
dy
= 4x − 16x −3
dx
When x = 2, y = 2(2)2 + 8(2)−2 − 9 = 1
es
dy
= 4(2) − 16(2)−3 = 6
dx
Tangent: passes through the point (2, 1) and gradient = 6
Pr
y
ve
ni
C
br
ev
id
ie
w
ge
R
1
y − 1 = − ( x − 2)
6
x + 6y = 8
1
6
U
ev
Normal: passes through the point (2, 1) and gradient = −
op
w
ie
rs
y − 1 = 6( x − 2)
y = 6x − 11
C
202
ity
op
y
and
-R
am
WORKED EXAMPLE 7.9
)3
(
s
es
-C
A curve has equation y = 4 − x .
Pr
op
y
The normal at the point P (4, 8) and the normal at the point Q(9, 1) intersect at the point R.
ni
ve
rs
y
(
(
)2 = −3
(
)2
3 4− 9
dy
1
=−
=−
dx
2 9
2
Copyright Material - Review Only - Not for Redistribution
s
-C
When x = 9,
es
am
br
3 4− 4
dy
=−
When x = 4,
dx
2 4
ev
)
-R
id
g
e
(
)2
ie
1
3 4− x
2  1 − 
dy
= 3 4− x − x 2 = −
dx
2 x
 2

op
U
)3
C
(
y = 4− x
w
Answer
a
ity
b Find the area of triangle PQR.
R
ev
ie
w
C
a Find the coordinates of R.
ve
rs
ity
1
3
am
br
id
ev
ie
Normal at P: passes through the point (4, 8) and gradient =
w
ge
C
U
ni
op
y
Chapter 7: Differentiation
1
( x − 4)
3
3 y = x + 20
-R
y−8=
(1)
y = 2 x − 17
(2)
Solving equations (1) and (2) gives:
y
ni
C
op
w
3(2 x − 17) = x + 20
x = 14.2
ev
ie
Pr
es
s
y − 1 = 2( x − 9)
ve
rs
ity
C
op
y
-C
Normal at Q: passes through the point (9, 1) and gradient = 2
U
R
When x = 14.2, y = 2(14.2) − 17 = 11.4
w
10.4
Q(9, 1)
5.2
5
203
x
ity
O
Pr
7
y
op
es
s
-C
P(4, 8)
-R
am
3.4
C
ie
R(14.2, 11.4)
10.2
ev
y
br
b
id
ge
Hence, R is the point (14.2, 11.4).
ve
ie
w
rs
Area of triangle PQR = area of rectangle − sum of areas of outside triangles
y
op
C
ie
w
= 44.2 units2
br
-C
-R
am
EXERCISE 7C
ev
id
ge
U
R
ni
ev
1
1

 1
= 10.2 × 10.4 −   × 5 × 7  +  × 5.2 ×10.4  +  × 10.2 × 3.4  






2
2
2


= 106.08 − [ 17.5 + 27.04 + 17.34 ]
s
b
y = (2 x − 5)4 at the point (2, 1)
x3 − 5
y=
at the point ( −1, 6)
x
y = 2 x − 5 at the point (9, 4)
ity
y
op
d
ni
ve
rs
c
es
y = x 2 − 3x + 2 at the point (3, 2)
Pr
a
C
w
e
ie
ev
-R
s
es
d
id
g
c
br
b
y = 3x 3 + x 2 − 4x + 1 at the point (0, 1)
3
y= 3
at the point ( −2, −3)
x +1
y = (5 − 2 x )3 at the point (3, −1)
20
y= 2
at the point (3, 2)
x +1
am
a
U
2 Find the equation of the normal to each curve at the given point.
-C
R
ev
ie
w
C
op
y
1 Find the equation of the tangent to each curve at the given point.
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
am
br
id
ev
ie
w
ge
8
1
.
3 A curve passes through the point A  2,  and has equation y =
 2
( x + 2)2
a Find the equation of the tangent to the curve at the point A.
-R
b Find the equation of the normal to the curve at the point A.
Pr
es
s
-C
4 The equation of a curve is y = 5 − 3x − 2 x 2 .
C
op
ni
ev
ie
5 The normal to the curve y = x 3 − 5x + 3 at the point ( −1, 7) intersects the
y-axis at the point P.
Find the coordinates of P.
U
R
y
b Find the coordinates of the point at which the normal meets the curve again.
ve
rs
ity
w
C
op
y
a Show that the equation of the normal to the curve at the point ( −2, 3) is
x + 5 y = 13.
id
ie
w
ge
6 The tangents to the curve y = 5 − 3x − x 2 at the points ( −1, 7) and ( −4, 1) meet at
the point Q.
-R
am
br
ev
Find the coordinates of Q.
s
-C
7 The normal to the curve y = 4 − 2 x at the point P (16, −4) meets the x-axis at
the point Q.
es
b Find the coordinates of Q.
ity
10
8 The equation of a curve is y = 2 x − 2 + 8.
x
dy
.
a Find
dx
5
b Show that the normal to the curve at the point  −4, −  meets the y-axis at

8
the point (0, −3).
y
op
C
w
6
at the point (3, 6) meets the x-axis at P and
x−2
id
ge
9 The normal to the curve y =
the y-axis at Q.
ie
U
R
ni
ev
ve
ie
w
rs
C
204
Pr
op
y
a Find the equation of the normal PQ.
-R
am
br
ev
Find the midpoint of PQ.
es
s
-C
10 A curve has equation y = x5 − 8x 3 + 16x . The normal at the point P (1, 9) and
the tangent at the point Q( −1, −9) intersect at the point R.
)3
Pr
(
ni
ve
rs
a Find the coordinates of R.
ity
11 A curve has equation y = 2 x − 1 + 2 . The normal at the point P (4, 4) and
the normal at the point Q(9, 18) intersect at the point R.
op
y
b Find the area of triangle PQR.
U
12
and passes through the points A(2, 12) and
x
B (6, 20) . At each of the points C and D on the curve, the tangent is parallel to AB.
ie
id
g
w
e
C
12 A curve has equation y = 3x +
br
ev
a Find the coordinates of the points C and D. Give your answer in exact form.
-R
s
es
am
b Find the equation of the perpendicular bisector of CD.
-C
R
ev
ie
w
C
op
y
Find the coordinates of R.
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
C
U
ni
op
y
Chapter 7: Differentiation
am
br
id
ev
ie
w
ge
13 The curve y = x( x − 1)( x + 2) crosses the x-axis at the points O(0, 0), A(1, 0)
and B ( −2, 0) . The normals to the curve at the points A and B meet at the point C.
Find the coordinates of the point C .
5
and passes through the points P ( −1, 1) . Find
2 − 3x
the equation of the tangent to the curve at P and find the angle that this tangent
Pr
es
s
-C
-R
14 A curve has equation y =
y
makes with the x-axis.
12
− 4 intersects the x-axis at P. The tangent to the curve at
2x − 3
P intersects the y-axis at Q. Find the distance PQ.
ve
rs
ity
w
C
op
15 The curve y =
y
C
op
ni
ev
ie
16 The normal to the curve y = 2 x 2 + kx − 3 at the point (3, −6) is parallel to the
line x + 5 y = 10.
U
R
a Find the value of k.
Try the Tangent or
normal resource on
the Underground
Mathematics website.
id
ie
w
ge
b Find the coordinates of the point where the normal meets the curve again.
br
ev
7.4 Second derivatives
dy
.
dx
dy
is called the first derivative of y with respect to x.
dx
dy
d  dy 
with respect to x we obtain
If we then differentiate
 , which is usually

dx
dx  dx 
2
d y
written as
.
dx 2
d2 y
is called the second derivative of y with respect to x.
dx 2
f ′′( x ) = 6x + 10
-R
s
es
Pr
op
y
2
5
, find d y .
3
(2 x − 3)
dx 2
ity
y = 5(2 x − 3)−3
op
y
Use the chain rule.
C
U
Use the chain rule.
-R
s
es
am
br
ev
ie
id
g
w
e
d2 y
= 120(2 x − 3)−5 × 2
dx 2
240
=
(2 x − 3)5
ni
ve
rs
dy
= −15(2 x − 3)−4 × 2
dx
= −30(2 x − 3)−4
-C
ie
w
C
Answer
ev
y
ev
br
-C
am
WORKED EXAMPLE 7.10
Given that y =
op
or
w
f ′( x ) = 3x 2 + 10 x − 3
U
or
id
ge
R
dy
= 3x 2 + 10 x − 3
dx
d2 y
= 6x + 10
dx 2
f( x ) = x 3 + 5x 2 − 3x + 2
C
ni
or
ie
So for y = x 3 + 5x 2 − 3x + 2
ev
-R
rs
ve
ie
w
C
ity
op
Pr
y
es
s
-C
am
If we differentiate y with respect to x we obtain
R
WEB LINK
Copyright Material - Review Only - Not for Redistribution
205
ve
rs
ity
ev
ie
w
ge
A curve has equation y = x 3 + 3x 2 − 9x + 2.
y = x 3 + 3x 2 − 9x + 2
ni
U
3x 2 + 6x − 9 , 0
x2 + 2x − 3 , 0
( x + 3)( x − 1) , 0
−3 , x , 1
−3
-R
s
2
y
1
op
0
rs
−1
ve
−2
ni
w
ie
ev
−3
U
∴ −3 , x , − 1
R
es
(2)
ity
op
y
x , −1
Combining (1) and (2) on a number line:
−4
C
2
ie
ev
id
2
d 2 y  dy 
.
≠


dx 2  dx 
-C
-R
am
es
s
EXERCISE 7D
d
y = (2 x − 3)4
g
y=
Pr
4x − 9
f
h
y = 2 x 2 (5 − 3x + x 2 )
i
e
id
g
ie
ev
( x − 3)
-R
f( x ) = x 2
2x − 3 x
x2
15
f( x ) = 3
2x + 1
f( x ) =
f
s
e
c
es
br
f( x ) = 1 − 3x
am
d
op
U
2x − 5
x2
6
x2
2
y=
3x + 1
5x − 4
y=
x
y=2−
y
ity
y=
ni
ve
rs
e
2 Find f ′′( x ) for each of the following functions.
5
3
4x 2 − 3
b
f(
x
)
=
a f( x ) = 2 −
2x
x
2 x5
-C
C
w
ie
ev
R
c
C
op
y
a
d2 y
for each of the following functions.
dx 2
y = x 2 + 8x − 4
b y = 5x 3 − 7 x 2 + 5
1 Find
w
Hence,
w
ge
d2 y
 dy 
= 6 x + 6 and 
= (3x 2 + 6 x − 9)2 .
2
 dx 
dx
br
b
1
Pr
-C
6x + 6 , 0
C
206
+
−
ev
id
(1)
br
am
d2 y
= 6x + 6
dx 2
d2 y
, 0 when:
dx 2
w
+
ge
R
dy
, 0 when:
dx
ie
ev
ie
w
dy
= 3x 2 + 6x − 9
dx
y
ve
rs
ity
C
op
Answer
a
-R
2
d 2 y  dy 
≠
 .
dx 2  dx 
y
b Show that
dy
d2 y
and
are negative.
dx
dx 2
Pr
es
s
-C
a Find the range of values of x for which
C
op
am
br
id
WORKED EXAMPLE 7.11
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
Copyright Material - Review Only - Not for Redistribution
x
ve
rs
ity
am
br
id
4 Given that f( x ) = x 3 + 2 x 2 − 3x − 1, find:
b
f(1)
f ′(1)
c
-R
a
C
dy
d2 y
and
.
dx
dx 2
w
3 Given that y = 4x − (2 x − 1)4 , find
ev
ie
ge
U
ni
op
y
Chapter 7: Differentiation
f ′′(1)
3
, find f ′′( x ) .
(2 x − 1)8
2
, find the value of f ′′( −4).
6 Given that f( x ) =
1 − 2x
y
3
4
7
w
6
-R
Pr
y
es
s
-C
8 A curve has equation y = x 3 − 6x 2 − 15x − 7. Find the range of values of x for
dy
d2 y
and
which both
are positive.
dx
dx 2
d2 y
dy
= 2 y.
9 Given that y = x 2 − 2 x + 5, show that 4 2 + ( x − 1)
dx
dx
d2 y
dy
10 Given that y = 4 x , show that 4 x 2
+ 4x
= y.
2
dx
dx
207
w
rs
ity
op
C
5
ie
id
br
am
d2 y
dx 2
2
ge
dy
dx
1
C
op
0
ev
R
x
ni
ve
rs
ity
7 A curve has equation y = 2 x 3 − 21x 2 + 60 x + 5. Copy and complete the table
dy
d2 y
and
to show whether
are positive ( + ), negative ( − ) or zero (0) for the
dx
dx 2
given values of x.
U
ev
ie
w
C
op
y
Pr
es
s
-C
5 Given that f ′( x ) =
y
op
C
w
y
-R
s
es
-C
am
br
ev
ie
id
g
w
e
C
U
op
ev
ie
w
ni
ve
rs
C
ity
Pr
op
y
es
s
-C
-R
am
br
ev
id
dy
ax + b
. Given that
= 0 and
12 A curve has equation y =
2
dx
x
d2 y
1
=
when x = 2, find the value of a and the value of b.
2
dx 2
R
WEB LINK
Try the Gradients of
gradients resource
on the Underground
Mathematics website.
ie
ge
U
R
ni
ev
ve
ie
11 A curve has equation y = x 3 + 2 x 2 − 4x + 6 .
dy
2
= 0 when x = −2 and when x = .
a Show that
dx
3
2
d2 y
b Find the value of
when x = −2 and when x = .
2
3
dx
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
am
br
id
Gradient of a curve
d
( x n ) = nx n − 1
dx
Power rule:
op
y
●
-C
The four rules of differentiation
d
d
[ kf( x )] = k
[f( x )]
dx
dx
●
Addition/subtraction rule:
d
d
d
[f( x ) ± g( x )] =
[f( x )] ±
[g( x )]
dx
dx
dx
●
Chain rule:
dy
dy
du
=
×
dx
du
dx
y
C
op
s
-C
-R
am
br
ev
id
ie
dy
at the point ( x1, y1 ) is m, then:
dx
● the equation of the tangent at that point is given by y − y1 = m( x − x1 )
1
● the equation of the normal at that point is given by y − y1 = −
( x − x1 ) .
m
If the value of
es
Second derivatives
Pr
y
d  dy 
d2 y
=
dx  dx 
dx 2
y
op
y
op
-R
s
es
-C
am
br
ev
ie
id
g
w
e
C
U
R
ev
ie
w
ni
ve
rs
C
ity
Pr
op
y
es
s
-C
-R
am
br
ev
id
ie
w
ge
C
U
R
ni
ev
ve
ie
w
rs
C
208
ity
op
●
w
Tangents and normals
ge
U
ni
w
ve
rs
ity
Scalar multiple rule:
C
●
R
ev
ie
-R
dy
represents the gradient of the curve y = f( x ).
dx
Pr
es
s
●
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Checklist of learning and understanding
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Chapter 7: Differentiation
ev
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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
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Find the gradient of the curve y =
Find the coordinates of Q.
y
[4]
C
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br
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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]
Pr
es
s
[3]
[6]
The curve y = x( x − 3)( x − 5) crosses the x-axis at the points O(0, 0), A(3, 0) and B (5, 0) .
The tangents to the curve at the points A and B meet at the point C .
Find the coordinates of the point C .
2
.
10 A curve passes through the point A(4, 2) and has equation y =
( x − 3)2
a Find the equation of the tangent to the curve at the point A.
[6]
op
Find the length of PQ, correct to 3 significant figures.
ity
9
rs
w
ni
op
y
ve
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ev
w
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Find the equation of the normal to the curve at the point A.
[5]
[2]
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b
C
U
R
[4]
[1]
w
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Find the equation of the normal PQ.
-C
y
8
[3]
15
at the point where x = 5.
x2 − 2 x
am
7
C
dy
d y
and
.
dx
dx 2
5
a
[3]
2
The equation of a curve is y = (3 − 5x )3 − 2 x. Find
C
w
[3]
4
R
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Pr
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y
3
[3]
-R
Differentiate
-C
2
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1
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END-OF-CHAPTER REVIEW EXERCISE 7
-R
am
br
ev
10
11 A curve passes through the point P (5, 1) and has equation y = 3 −
.
x
a Show that the equation of the normal to the curve at the point P is 5x + 2 y = 27.
[4]
ii
Find the midpoint of PQ.
s
Find the coordinates of Q.
[3]
es
i
[1]
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am
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[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
ev
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w
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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-R
y
C
C
y=2 −
A
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y
Pr
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s
-C
13
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Cambridge International AS & A Level Mathematics: Pure Mathematics 1
x
y
U
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C
op
B
ni
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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
am
br
ev
id
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The diagram shows part of the curve y = 2 −
es
s
-C
Pr
y
14 The equation of a curve is y = 3 + 4x − x 2.
i Show that the equation of the normal to the curve at the point (3, 6) is 2 y = x + 9.
ii Given that the normal meets the coordinate axes at points A and B, find the coordinates of the
mid-point of AB.
iii Find the coordinates of the point at which the normal meets the curve again.
ity
rs
w
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[2]
[4]
y
op
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C
R
1
br
ev
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y = (6x + 2) 3
B
Pr
op
y
es
s
-C
E
-R
am
A(1, 2)
C
x
1
y
ni
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C
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O
w
-R
s
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q11 November 2012
es
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id
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op
The diagram shows the curve y = (6x + 2) 3 and the point A(1, 2) which lies on the curve. The tangent to the
curve at A cuts the y-axis at B and the normal to the curve at A cuts the x-axis at C .
[5]
i Find the equation of the tangent AB and the equation of the normal AC.
[3]
ii Find the distance BC .
iii Find the coordinates of the point of intersection, E , of OA and BC , and determine whether E is
the mid-point of OA.
[4]
-C
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[4]
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q10 November 2010
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[6]
[2]
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q7 June 2010
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210
R
18
2x + 3
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Chapter 8
Further differentiation
y
op
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C
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Pr
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-C
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211
id
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br
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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
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es
-C
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br
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ev
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C
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Pr
op
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s
-C
■
■
ie
In this chapter you will learn how to:
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Cambridge International AS & A Level Mathematics: Pure Mathematics 1
w
Chapter 1
Solve quadratic inequalities.
ev
ie
What you should be able to do
Check your skills
y
Pr
es
s
-C
Where it comes from
-R
am
br
id
PREREQUISITE KNOWLEDGE
Find the first and second derivatives
of x n.
Chapter 7
Differentiate composite functions.
2
2 Find dy and d y2 for the following.
dx
dx
a y = 3x 2 − x + 2
3
b y=
2x2
c y = 3x x
y
U
ni
C
op
C
w
ev
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a
dy
for the following.
dx
y = (2x − 1)5
b
y=
3 Find
3
(1 − 3x )2
s
-C
-R
am
br
ev
id
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w
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R
x2 − 2x − 3 . 0
a
b 6 + x − x2 . 0
ve
rs
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op
Chapter 7
1 Solve:
es
Why do we study differentiation?
FAST FORWARD
C
y
ve
ni
R
ev
ie
w
rs
In this chapter you will build on this knowledge and learn how to apply differentiation
to problems that involve finding when a function is increasing (or decreasing) or when a
function is at a maximum (or minimum) value. You will also learn how to solve practical
problems involving rates of change.
op
212
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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.
C
w
es
s
-C
-R
am
br
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id
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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
-C
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br
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id
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ev
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C
ity
Pr
op
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•
•
•
•
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U
There are many situations in real life where these skills are needed. Some
examples are:
In the Mechanics
Coursebook,
Chapter 6 you will
learn to apply these
skills to problems
concerning
displacement, velocity
and time.
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WEB LINK
Explore the Calculus
meets functions station
on the Underground
Mathematics website.
ve
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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
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w
C
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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 ).
U
R
y
y
ni
Consider the graph of y = g( x ).
C
op
ev
ie
Section B: Decreasing functions
w
ge
‘As the value of x increases the value of y …’
O
ie
br
These types of functions are called decreasing functions.
Pr
y
8.1 Increasing and decreasing functions
es
s
-R
am
2 Sketch other graphs that satisfy these statements.
-C
Try the Choose your
families resource on
the Underground
Mathematics website.
ev
id
‘The sign of the gradient at any point is always …’
WEB LINK
x
y = g(x)
rs
C
w
Likewise, a decreasing function f( x ) is one where the f( x ) values decrease whenever the
x value increases, or f ( a ) > f ( b ) whenever a < b.
ni
op
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ev
213
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op
As you probably worked out from Explore 8.1, an increasing function f( x ) is one
where the f( x ) values increase whenever the x value increases. More precisely, this means
that f ( a ) < f ( b ) whenever a < b.
id
ie
w
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C
U
R
Sometimes we talk about a function increasing at a point, meaning that the function values
are increasing around that point. If the gradient of the function is positive at a point, then
the function is increasing there.
-C
Now consider the function y = h( x ), shown on the graph.
Pr
ity
(a, b)
O
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op
C
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id
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br
am
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dy
>0
dx
dy
dx < 0
ni
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C
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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.
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WORKED EXAMPLE 8.1
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Cambridge International AS & A Level Mathematics: Pure Mathematics 1
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WORKED EXAMPLE 8.2
ge
U
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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
<x<4
2
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op
-R
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es
-C
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c When f ′( x ) < 0, f(x) is decreasing.
w
–
s
am
(2 x + 3)( x − 4) > 0
∴−
4
–3
2
ev
id
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2 x 2 − 5x − 12 > 0
Critical values are −
+
w
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12 x 2 − 30 x − 72 > 0
C
214
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Chapter 8: Further differentiation
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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
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f ′( x ) = −5(2 x − 3)−2 (2)
10
=−
(2 x − 3)2
R
Write in a form ready for differentiating.
ve
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C
op
f( x ) =
id
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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
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f( x ) = 10 + 9x − x 2
-C
f( x ) = 11 + 24x − 3x 2 − x 3
-R
br
f
f( x ) = x 3 − 3x 2 − 9x + 5
1
( 5 − 2x )3 + 4x is increasing.
6
s
4
for x > 1. Find an expression for f ′( x ) and determine whether f is
1 − 2x
an increasing function, a decreasing function or neither.
5
2
for x > 0. Find an expression for f ′( x ) and determine
5 A function f is defined as f( x ) =
2 −
x+2
( x + 2)
whether f is an increasing function, a decreasing function or neither.
es
op
y
x2 − 4
is an increasing function.
x
U
6 Show that f( x ) =
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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
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f( x ) = 3x 2 − 8x + 2
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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
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a
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1 Find the set of values of x for which each of the following is increasing.
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215
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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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Consider the following graph of the function y = f( x ).
y
y
R
C
ve
rs
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op
dy
=0
dx
O
dy
>0
dx
dy
<0
dx
Q
P
w
dy
=0
dx
x
C
op
y
The red sections of the curve show where the gradient is negative (where f( x ) is a
decreasing function) and the blue sections show where the gradient is positive (where f( x )
is an increasing function). The gradient of the curve is zero at the points P, Q and R.
ge
U
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R
dy
=0
dx
dy
>0
dx
Pr
es
s
-C
dy
<0
dx
-R
y = f(x)
ie
ev
id
br
Maximum points
w
A point where the gradient is zero is called a stationary point or a turning point.
-C
-R
am
The stationary point Q is called a maximum point because the value of y at this point is
greater than the value of y at other points close to Q.
ity
+
–
rs
w
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Minimum points
–
+
-R
am
br
ev
the gradient is negative to the left of the minimum and positive to
the right.
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id
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C
U
At a minimum point:
dy
●
=0
dx
●
op
The stationary points P and R are called minimum points.
ni
ie
ev
R
0
the gradient is positive to the left of the maximum and negative
to the right.
C
●
op
216
Pr
y
es
s
At a maximum point:
dy
●
=0
dx
0
es
s
-C
Stationary points of inflexion
Pr
0
op
y
–
-R
s
es
am
br
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 from positive to zero and then to positive again

the gradient changes  or
 from negative to zero and then to negative again.

-C
●
ie
id
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w
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C
U
At a stationary point of inflexion:
dy
●
=0
dx
–
ni
ve
rs
C
w
+
ie
ev
R
+
0
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op
y
There is a third type of stationary point (turning point) called a point of inflexion.
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Chapter 8: Further differentiation
am
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WORKED EXAMPLE 8.4
Pr
es
s
-C
y
y = x 3 − 12 x + 5
ve
rs
ity
dy
=0
dx
3x 2 − 12 = 0
ni
ev
ie
For stationary points:
y
w
C
op
dy
= 3x 2 − 12
dx
C
op
Answer
-R
Find the coordinates of the stationary points on the curve y = x 3 − 12 x + 5 and
determine the nature of these points. Sketch the graph of y = x 3 − 12 x + 5.
U
R
x2 − 4 = 0
id
ie
w
ge
( x + 2)( x − 2) = 0
x = −2 or x = 2
br
ev
When x = −2, y = ( −2)3 − 12( −2) + 5 = 21
y = (2)3 − 12(2) + 5 = −11
-R
am
When x = 2,
s
es
-C
The stationary points are ( −2, 21) and (2, −11).
dy
dx
3( −2.1)2 − 12 = positive
rs
−2
−1.9
0
3( −1.9)2 − 12 = negative
y
C
0
3(2.1)2 − 12 = positive
es
op
y
Pr
ity
So ( −2, 21) is a maximum point and (2, −11) is a minimum point.
ni
ve
rs
y
y = x3 – 12x + 5
C
w
ie
5
ev
x
O
-R
(2, –11)
s
-C
am
br
id
g
e
U
R
(–2, 21) y
op
ev
The sketch graph of y = x 3 − 12 x + 5 is:
es
C
shape of curve
w
2.1
s
direction of tangent
ie
ie
3(1.9)2 − 12 = negative
ev
2
-R
id
1.9
-C
am
br
x
dy
dx
w
ge
shape of curve
TIP
op
ni
U
ev
direction of tangent
R
Pr
−2.1
ity
x
ve
ie
w
C
op
y
Now consider the gradient on either side of the points ( −2, 21) and (2, −11):
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This is called the First
Derivative Test.
217
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C
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Cambridge International AS & A Level Mathematics: Pure Mathematics 1
w
ge
Second derivatives and stationary points
–
Pr
es
s
-C
+
dy
, starts as a positive value, decreases to zero at the maximum point and
dx
then decreases to a negative value.
dy
dy
Since
decreases as x increases, then the rate of change of
is negative.
dx
dx
d  dy  d 2 y
dy
The rate of change of
is written as
.
 =

dx  dx  dx 2
dx
d2 y
is called the second derivative of y with respect x.
dx 2
y
w
ie
ev
y
es
-C
s
d2 y
dy
, 0 , then the point is a maximum point.
= 0 and
dx
dx 2
If
-R
br
am
KEY POINT 8.1
id
This leads to the rule:
ge
U
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C
op
y
ve
rs
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op
C
w
ev
ie
-R
0
The gradient,
R
ev
ie
am
br
id
Consider moving from left to right along a curve, passing through a maximum point.
y
+
0
op
ni
R
ev
ve
ie
–
rs
w
C
ity
op
Pr
Now, consider moving from left to right along a curve, passing through
a minimum point.
218
-R
am
br
ev
id
ie
w
ge
C
U
dy
, starts as a negative value, increases to zero at the minimum point and
dx
then increases to a positive value.
dy
dy
Since
increases as x increases, then the rate of change of
is positive.
dx
dx
The gradient,
s
-C
This leads to the rule:
op
y
es
KEY POINT 8.2
ity
Pr
dy
d2 y
= 0 and
. 0 , then the point is a minimum point.
dx
dx 2
-R
s
es
-C
am
br
ev
ie
id
g
w
e
C
U
op
y
dy
d2 y
= 0 and
= 0, then the nature of the stationary point can be found using the
dx
dx 2
first derivative test.
If
R
ev
ie
w
ni
ve
rs
C
If
Copyright Material - Review Only - Not for Redistribution
REWIND
In Chapter 7, Section 7.4
we learnt how to find a
second derivative. Here
we will look at how
second derivatives can
be used to determine
the nature of a
stationary point.
ve
rs
ity
ge
C
U
ni
op
y
Chapter 8: Further differentiation
am
br
id
ev
ie
w
WORKED EXAMPLE 8.5
the nature of these points.
y
Pr
es
s
-C
Answer
x2 + 9
y=
= x + 9x −1
x
dy
9
= 1 − 9x −2 = 1 − 2
dx
x
ve
rs
ity
op
U
R
ni
ev
ie
y
dy
=0
dx
9
1− 2 = 0
x
x2 − 9 = 0
For stationary points:
C
op
C
w
x2 + 9
and use the second derivative to determine
x
-R
Find the coordinates of the stationary points on the curve y =
br
ev
id
ie
w
ge
( x + 3)( x − 3) = 0
x = −3 or x = 3
( −3)2 + 9
= −6
−3
32 + 9
=6
y=
When x = 3,
3
The stationary points are ( −3, −6) and (3, 6).
-R
y=
s
es
y
op
d 2 y 18
= 3 .0
dx 2
3
rs
When x = 3,
ve
d2 y
18
,0
2 =
dx
( −3)3
br
w
-R
am
EXPLORE 8.2
ev
id
ie
ge
∴ ( −3, −6) is a maximum point and (3, 6) is a minimum point.
C
U
ni
When x = −3,
219
ity
d2 y
18
= 18x −3 = 3
x
dx 2
C
w
ie
ev
R
Pr
op
y
-C
am
When x = −3,
es
y
y = f ′ (x)
ity
O
1 2 3 4 5 6 x
1 2 3 4 5 6 x
y
op
ie
C
U
1 Discuss the properties of these two graphs and the information that can be obtained from them.
w
id
g
e
2 Without finding the equation of the function y = f( x ), determine, giving reasons:
ie
-R
s
3 Sketch the graph of the function y = f( x ).
es
am
b the coordinates of the minimum point on the curve.
ev
br
a the coordinates of the maximum point on the curve
-C
ev
R
y = f ″ (x)
ni
ve
rs
O
Pr
y
w
C
op
y
The following graphs show y = f ′( x ) and y = f ′′( x ).
s
-C
The graph of the function y = f( x ) passes through the point (1, −35) and the point (6, 90).
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
ev
ie
am
br
id
EXERCISE 8B
w
ge
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
c
y = x − 12 x + 6
e
y = x 4 + 4x − 1
3
y = (3 + x )(2 − x )
d
y = 10 + 9x − 3x 2 − x 3
f
y = (2 x − 3)3 − 6x
ve
rs
ity
y
op
y
2 Find the coordinates of the stationary points on each of the following curves and
determine the nature of each stationary point.
9
8
b y = 4x 2 +
a y= x+
x
x
( x − 3)2
48
d y = x3 +
+4
c y=
x
x
8
e y=4 x −x
f y = 2x + 2
x
x2 − 9
3 The equation of a curve is y =
.
x2
dy
Find
and, hence, explain why the curve does not have a stationary point.
dx
4 A curve has equation y = 2 x 3 − 3x 2 − 36x + k.
es
s
-C
-R
am
br
ev
id
ie
w
ge
U
R
ni
C
op
C
w
ev
ie
b
Pr
es
s
y = x 2 − 4x + 8
-C
a
-R
1 Find the coordinates of the stationary points on each of the following curves and
determine the nature of each stationary point. Sketch the graph of each function
and use graphing software to check your graphs.
w
Pr
rs
C
b Hence, find the two values of k for which the curve has a stationary point on
the x-axis.
ity
op
y
a Find the x-coordinates of the two stationary points on the curve.
y
op
ev
a Find the value of a.
ve
ie
5 The curve y = x 3 + ax 2 − 9x + 2 has a maximum point at x = −3.
ni
220
ge
C
U
R
b Find the range of values of x for which the curve is a decreasing function.
id
ie
w
6 The curve y = 2 x 3 + ax 2 + bx − 30 has a stationary point when x = 3.
br
ev
The curve passes through the point (4, 2).
-R
am
a Find the value of a and the value of b.
es
s
-C
b Find the coordinates of the other stationary point on the curve and determine
the nature of this point.
Pr
op
y
7 The curve y = 2 x 3 + ax 2 + bx − 30 has no stationary points.
ity
k2
, where k is a positive constant. Find,
2x − 3
in terms of k, the values of x for which the curve has stationary points and
determine the nature of each stationary point.
U
op
y
ni
ve
rs
8 A curve has equation y = 1 + 2 x +
ie
id
g
w
e
C
9 Find the coordinates of the stationary points on the curve y = x 4 − 4x 3 + 4x 2 + 1
and determine the nature of each of these points. Sketch the graph of the curve.
br
ev
10 The curve y = x 3 + ax 2 + b has a stationary point at (4, −27).
-R
am
a Find the value of a and the value of b.
es
s
b Determine the nature of the stationary point (4, −27).
-C
R
ev
ie
w
C
Show that a 2 , 6b.
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
C
ev
ie
w
ge
Find the coordinates of the other stationary point on the curve and determine
the nature of this stationary point.
am
br
id
c
U
ni
op
y
Chapter 8: Further differentiation
-R
d Find the coordinates of the point on the curve where the gradient is minimum
and state the value of the minimum gradient.
b
has a stationary point at (2, 12).
x2
a Find the value of a and the value of b.
Pr
es
s
-C
11 The curve y = ax +
y
C
op
ve
rs
ity
ni
ev
ie
w
C
op
y
b Determine the nature of the stationary point (2, 12).
b
c Find the range of values of x for which ax + 2 is increasing.
x
a
12 The curve y = x 2 + + b has a stationary point at (3, 5).
x
a Find the value of a and the value of b.
id
ie
w
ge
U
R
b Determine the nature of the stationary point (3, 5).
a
c Find the range of values of x for which x 2 + + b is decreasing.
x
3
2
13 The curve y = 2 x + ax + bx + 7 has a stationary point at the point (2, −13).
br
am
-R
b Find the coordinates of the second stationary point on the curve.
Determine the nature of the two stationary points.
s
es
ity
rs
w
y
op
id
ie
w
ge
C
U
R
ni
ev
ve
ie
There are many problems for which we need to find the maximum or minimum value
of an expression. For example, the manufacturers of canned food and drinks often need
to minimise the cost of their manufacturing. To do this they need to find the minimum
amount of metal required to make a container for a given volume. Other situations might
involve finding the maximum area that can be enclosed within a shape.
ev
br
-R
am
-C
The surface area of the solid cuboid is 100 cm 2 and the volume is V cm 3.
h cm
es
s
a Express h in terms of x.
x cm
1
b Show that V = 25x − x 3 .
x cm
2
c Given that x can vary, find the stationary value of V and determine whether this stationary value is a
maximum or a minimum.
Answer
U
2 x 2 + 4xh = 100
op
Surface area = 2 x 2 + 4xh
C
a
y
ni
ve
rs
ity
Pr
op
y
-R
s
es
am
br
ev
ie
id
g
w
e
100 − 2 x 2
4x
25 1
h=
− x
x
2
h=
-C
C
w
ie
ev
221
8.3 Practical maximum and minimum problems
WORKED EXAMPLE 8.6
R
cubics edition.
Pr
d Find the coordinates of the point on the curve where the gradient is minimum
and state the value of the minimum gradient.
C
op
y
-C
c
Try the following
resources on the
Underground
Mathematics website:
• Floppy hair
• Two-way calculus
• Curvy cubics
• Can you find… curvy
ev
a Find the value of a and the value of b.
WEB LINK
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
ev
ie
Substitute for h.
am
br
id
b V = x2h
w
ge
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
-C
dV
= 0:
dx
3
25 − x 2 = 0
2
5 6
x=
3
C
op
U
R
ni
ev
ie
y
ve
rs
ity
Stationary values occur when
w
C
op
y
dV
3
= 25 − x 2
dx
2
Pr
es
s
1 3
x
2
= 25x −
c
-R
25 1 
= x2 
− x
 x
2 
3
5 6 15 6
5 6
−
= 68.04 (to 2 decimal places)
, V = 25 
3
 3  2  3 
ie
br
ev
id
d 2V
= −3x
dx 2
w
ge
When x =
-R
am
5 6 d 2V
,
= −5 6, which is , 0.
dx 2
3
The stationary value of V is 68.04 and it is a maximum value.
op
Pr
y
es
s
-C
When x =
222
rs
The diagram shows a solid cylinder of radius r cm and height 2 h cm cut from a solid
sphere of radius 5 cm. The volume of the cylinder is C cm 3.
C
3
id
ie
c Find the value for h for which there is a stationary value of V .
-R
am
br
ev
d Determine the nature of this stationary value.
y
U
dV
= 0:
dh
2
50 π − 6 πh = 0
50 π
h2 =
6π
5 3
h=
3
op
dV
= 50 π − 6 πh2
dh
w
ie
ev
-R
s
es
am
br
id
g
e
C
Stationary values occur when
-C
c
ni
ve
rs
= 50 πh − 2 πh3
w
ie
ev
R
)
= π 25 − h2 (2 h )
Substitute for r.
ity
C
b V = πr 2 (2 h )
(
Use Pythagoras’ theorem.
es
25 − h
2
Pr
op
y
r=
s
r 2 + h 2 = 52
-C
a
w
ge
b Show that V = 50 πh − 2 πh .
Answer
r
y
op
U
R
a Express r in terms of h.
ni
ev
ve
ie
w
C
ity
WORKED EXAMPLE 8.7
Copyright Material - Review Only - Not for Redistribution
5 cm
2h
ve
rs
ity
op
y
ve
rs
ity
C
WORKED EXAMPLE 8.8
w
w
-R
Pr
es
s
-C
The stationary value is a maximum value.
ev
ie
ev
ie
ge
d 2V
= −12 πh
dh2
5 3
d 2V
5 3
,
= −12 π 
, which is , 0.
When h =
dx 2
3
 3 
am
br
id
d
C
U
ni
op
y
Chapter 8: Further differentiation
The diagram shows a hollow cone with base radius 12 cm and height 24 cm.
24 cm
U
R
ni
C
op
y
A solid cylinder stands on the base of the cone and the upper
edge touches the inside of the cone.
w
ge
The cylinder has base radius r cm, height h cm and volume
V cm 3.
r
br
ev
id
ie
a Express h in terms of r.
h
12 cm
-R
am
b Show that V = 24 πr 2 − 2 πr 3.
es
s
-C
c Find the volume of the largest cylinder that can stand
inside the cone.
Pr
r
24 − h
=
24
12
2 r = 24 − h
h = 24 − 2 r
ev
br
dV
= 0:
dr
48 πr − 6 πr 2 = 0
Pr
es
s
6 πr(8 − r ) = 0
r=8
-R
am
-C
When r = 8, V = 24 π(8)2 − 2 π(8)3 = 512 π
DID YOU KNOW?
ity
d 2V
= 48 π − 12 πr
dr 2
d 2V
= 48 π − 12 π(8), which is , 0.
When r = 8,
dx 2
op
y
ni
ve
rs
op
y
C
U
The stationary value is a maximum value.
-R
s
es
am
br
ev
ie
id
g
w
e
Volume of the largest cylinder is 512 π cm3.
-C
C
w
ie
ev
y
op
w
ie
id
ge
= 24 πr 2 − 2 πr 3
dV
= 48 πr − 6 πr 2
dr
C
U
= πr (24 − 2 r )
Stationary values occur when
R
223
ity
2
R
Substitute for h.
ni
ev
b V = πr 2 h
c
Use similar triangles.
rs
ie
w
C
op
a
ve
y
Answer
Copyright Material - Review Only - Not for Redistribution
Differentiation can
be used in business to
find how to maximise
company profits and to
find how to minimise
production costs.
ve
rs
ity
ev
ie
am
br
id
EXERCISE 8C
w
ge
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
1 The sum of two real numbers, x and y, is 9.
c
Given that Q = 3x 2 + 2 y2, write down an expression for Q, in terms of x.
ve
rs
ity
w
i
ii Find the minimum value of Q.
C
op
y
ii Find the maximum value of P.
Pr
es
s
Given that P = x 2 y, write down an expression for P, in terms of x.
-C
b i
-R
a Express y in terms of x.
y
ge
U
R
a Express θ in terms of r.
θ
r
id
ie
w
b Show that A = 20 r − r 2.
ev
Find the value of r for which there is a stationary value of A.
br
c
r
C
op
ni
ev
ie
2 A piece of wire, of length 40 cm, is bent to form a sector of a circle with radius r cm
and sector angle θ radians, as shown in the diagram. The total area enclosed by the
shape is A cm 2 .
-R
am
d Determine the magnitude and nature of this stationary value.
s
-C
3 The diagram shows a rectangular enclosure for keeping animals.
ym
The total length of the fence is 50 m and the area enclosed is A m 2.
Pr
xm
a Express y in terms of x.
1
b Show that A = x(50 − x ).
2
c Find the maximum possible area enclosed and the value of x for which this occurs.
y
op
ni
ev
ve
ie
w
rs
C
224
ity
op
y
es
There is a fence on three sides of the enclosure and a wall on its fourth side.
w
ge
PQRS is a quadrilateral where PB = AS = 2 x cm, BQ = x cm and DR = 4x cm.
id
ie
a Express the area of PQRS in terms of x.
-R
am
br
ev
b Given that x can vary, find the value of x for which the area of PQRS is a
minimum and find the magnitude of this minimum area.
2x
Q
x
A
P
es
R
B
Q
Pr
3x + 2y = 30
e
ev
br
y
y
op
x
R
S
y = 9 – x2
w
)
b Show that A = 2 p 9 − p2 .
P
-R
Find the value of p for which A has a stationary value.
am
P
ie
id
g
a Express QR in terms of p.
(
O
C
U
The points P and Q lie on the x-axis and the points R and S lie on the
curve y = 9 − x 2.
es
s
d Find this stationary value and determine its nature.
-C
2x
y
ity
ni
ve
rs
PQRS is a rectangle with base length 2 p units and area A units2.
c
C
s
-C
C
op
y
6
R
S
5 The diagram shows the graph of 3x + 2 y = 30. OPQR is a rectangle with area A cm 2.
The point O is the origin, P lies on the x-axis, R lies on the y-axis and Q has
coordinates ( x, y ) and lies on the line 3x + 2 y = 30.
3
a Show that A = 15x − x 2.
2
b Given that x can vary, find the stationary value of A and determine its nature.
w
ie
ev
R
4x
D
C
U
R
4 The diagram shows a rectangle, ABCD, where AB = 20 cm and BC = 16 cm.
Copyright Material - Review Only - Not for Redistribution
O
2p units
Q
x
ve
rs
ity
x cm
x
Pr
es
s
-R
15 cm
-C
24 cm
The diagram shows a 24 cm by 15 cm sheet of metal with a square of side x cm removed from each corner.
The metal is then folded to make an open rectangular box of depth x cm and volume V cm3.
ve
rs
ity
y
op
C
w
ev
ie
am
br
id
w
ge
7
C
U
ni
op
y
Chapter 8: Further differentiation
a Show that V = 4x 3 − 78x 2 + 360 x.
C
op
y
Determine the nature of this stationary value.
U
R
c
ni
ev
ie
b Find the stationary value of V and the value of x for which this occurs.
y cm
w
ge
8 The volume of the solid cuboid shown in the diagram is 576 cm 3 and the surface area
is A cm 2.
br
-R
am
es
s
-C
Pr
y
225
ity
op
C
R
xm
rs
w
ve
ie
S
PQST is a rectangle and QRS is a semicircle with diameter SQ.
PT = x m and PQ = ST = y m.
a Express y in terms of x.
1
1
b Show that A = x − x 2 − πx 2.
2
8
dA
d2 A
and
.
c Find
dx
dx 2
d Find the value for x for which there is a stationary value of A.
P
ym
Q
-R
am
br
ev
id
ie
w
ge
C
U
op
ni
y
The total area enclosed by the shape is A m 2.
ev
2x cm
T
9 The diagram shows a piece of wire, of length 2 m, is bent to form the
shape PQRST .
R
x cm
ev
id
ie
a Express y in terms of x.
1728
.
b Show that A = 4x 2 +
x
c Find the maximum value of A and state the dimensions of the cuboid for which
this occurs.
s
-C
e Determine the magnitude and nature of this stationary value.
ity
Pr
rm
hm
1 2
πr .
2
e
U
Find
op
dA
d2 A
and
.
dr
dr 2
d Find the value for r for which there is a stationary value of A.
c
C
b Show that A = 5r − 2 r 2 −
y
ni
ve
rs
a Express h in terms of r.
-R
s
es
am
br
ev
ie
id
g
w
e Determine the magnitude and nature of this stationary value.
-C
R
ev
ie
w
C
op
y
es
10 The diagram shows a window made from a rectangle with base 2 r m and height
h m and a semicircle of radius r m. The perimeter of the window is 5 m and the area
is A m 2.
Copyright Material - Review Only - Not for Redistribution
2r m
ve
rs
ity
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
w
ge
11 A piece of wire, of length 100 cm, is cut into two pieces.
am
br
id
ev
ie
One piece is bent to make a square of side x cm and the other is bent to make a circle of radius r cm.
The total area enclosed by the two shapes is A cm 2.
ve
rs
ity
12 A solid cylinder has radius r cm and height h cm.
ni
w
ge
U
R
C
op
ev
ie
a Express h in terms of r.
864 π
.
b Show that A = 2 πr 2 +
r
c Find the value for r for which there is a stationary value of A.
y
The volume of this cylinder is 432π cm 3 and the surface area is A cm 2.
w
C
op
y
Pr
es
s
-C
-R
a Express r in terms of x.
( π + 4)x 2 − 200 x + 2500
.
b Show that A =
π
c Find the value of x for which A has a stationary value and determine the nature and magnitude of this
stationary value.
6x cm
Pr
es
y cm
The diagram shows an open water container in the shape of a triangular prism of length y cm.
The vertical cross-section is an isosceles triangle with sides 5x cm, 5x cm and 6x cm.
ity
op
C
226
s
5x cm
5x cm
y
-C
am
13
-R
br
ev
id
ie
d Determine the magnitude and nature of this stationary value.
ve
ie
w
rs
The water container is made from 500 cm 2 of sheet metal and has a volume of V cm 3.
br
y
es
op
y
ni
ve
rs
ity
Pr
r
es
s
-R
br
ev
d Determine the magnitude and nature of this stationary value.
am
h
C
ie
id
g
w
e
U
a Express r in terms of h.
1
b Show that V = πh2 (20 − h ).
3
c Find the value for h for which there is a stationary value of V .
op
y
15 The diagram shows a right circular cone of base radius r cm and height h cm cut
from a solid sphere of radius 10 cm. The volume of the cone is V cm 3.
-C
C
w
ie
ev
c
h
5 3
πr .
6
Find the exact value of r such that V is a maximum.
b Show that V = 160 πr −
R
op
s
-C
-R
am
14 The diagram shows a solid formed by joining a hemisphere of radius r cm to a
cylinder of radius r cm and height h cm. The surface area of the solid is 320π cm 2
and the volume is V cm 3.
a Express h in terms of r.
PS
ie
ev
id
d Show that the value in part c is a maximum value.
PS
C
w
ge
U
R
ni
ev
a Express y in terms of x.
144 3
x.
b Show that V = 600 x −
5
c Find the value of x for which V has a stationary value.
Copyright Material - Review Only - Not for Redistribution
r
ve
rs
ity
am
br
id
ev
ie
w
ge
8.4 Rates of change
C
U
ni
op
y
Chapter 8: Further differentiation
ve
rs
ity
w
C
op
y
Pr
es
s
-C
-R
EXPLORE 8.3
A
B
C
3 −1
y
U
R
1 Discuss how the height of water in container A changes with time.
w
ge
2 Discuss how the height of water in container B changes with time.
C
op
ni
ev
ie
Consider pouring water at a constant rate of 10 cm s into each of these three large
containers.
id
ie
3 Discuss how the height of water in container C changes with time.
h
h
O
t
t
O
227
t
rs
ity
5 What can you say about the gradients?
You should have come to the conclusion that:
the height of water in container A increases at a constant rate
●
the height of water in containers B and C does not increase at a constant rate.
op
y
ve
●
ni
R
ev
ie
w
C
O
Pr
op
y
es
s
-C
h
-R
am
br
ev
4 On copies of the following axes, sketch graphs to show how the height of water in
a container ( h cm) varies with time (t seconds) for each container.
w
ge
C
U
The (constant) rate of change of the height of the water in container A can be found
by finding the gradient of the straight-line graph.
es
s
-C
-R
am
br
ev
id
ie
The rate of change of the height of the water in containers B and C at a particular
time, t seconds, can be estimated by drawing a tangent to the curve and then finding
the gradient of the tangent. A more accurate method is to use differentiation if we
know the equation of the graph.
Pr
ity
ie
id
g
w
e
C
U
dh
Differentiate to obtain
(the rate of change
dt
of h with respect to t).
es
s
-R
br
ev
dh 2
4
= (2) =
dt
5
5
am
When t = 2,
y
1 2
t
5
dh 2
= t
dt
5
h=
op
Answer
-C
ie
ev
R
1 2
t , find the rate of change of h with respect to t when t = 2.
5
ni
ve
rs
Given that h =
w
C
op
y
WORKED EXAMPLE 8.9
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
ev
ie
w
ge
am
br
id
WORKED EXAMPLE 8.10
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
-C
Find the rate of change of V with respect to t when t = 4.
Pr
es
s
Answer
y
C
op
Connected rates of change
ni
dV
= 4(4) − 3 = 13
dt
U
R
ev
ie
w
C
dV
= 4t − 3
dt
dV
Differentiate to obtain
(the rate of change
dt
of V with respect to t).
ve
rs
ity
op
y
V = 2t 2 − 3t + 8
When t = 4,
-R
Variables V and t are connected by the equation V = 2t 2 − 3t + 8.
es
Pr
op
ni
C
U
ie
Pr
op
y
WORKED EXAMPLE 8.11
1
C
id
g
1
dx
= 0.06
dt
w
and
e
y = x + (2 x + 3) 2
op
U
Answer
y
ni
ve
rs
ev
ie
Differentiate to find
-R
dy
.
dx
s
-C
am
br
−
dy
1
= 1 + (2 x + 3) 2 (2)
dx
2
1
= 1+
2x + 3
es
w
C
ity
A point with coordinates ( x, y ) moves along the curve y = x + 2 x + 3 in such a way that the rate of increase
of x has the constant value 0.06 units per second. Find the rate of increase of y at the instant when x = 3. State
whether the y-coordinate is increasing or decreasing.
ie
ev
R
-R
es
s
-C
am
br
ev
id
We can deduce this as: if we set t = y in the chain rule, we get
dy dy dx
=
×
dy dx dy
dy
Since
= 1, the rule follows.
dy
w
ge
R
ev
dx
1
=
y
d
dy
dx
y
ve
rs
KEY POINT 8.4
ie
w
C
ity
op
We may also need to use the rule:
228
In Chapter 7, Section 7.2
we learnt how to
differentiate using the
chain rule. Here we
will look at how the
chain rule can be used
for problems involving
connected rates of
change.
s
dy
dy
dx
=
×
dt
dx
dt
y
-C
The chain rule states:
REWIND
-R
br
am
KEY POINT 8.3
ev
id
ie
w
ge
When two variables, x and y, both vary with a third variable, t, we can connect the three
variables using the chain rule.
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
w
ge
1
4
=
3
2 (3) + 3
ev
ie
dy
= 1+
dx
am
br
id
When x = 3,
C
U
ni
op
y
Chapter 8: Further differentiation
Rate of change of y is 0.08 units per second.
dy
is a positive quantity).
The y-coordinate is increasing (since
dt
C
op
ni
U
R
EXERCISE 8D
y
ve
rs
ity
ev
ie
w
C
op
y
Pr
es
s
-C
-R
Using the chain rule: dy = dy × dx
dt
dx
dt
4
= × 0.06
3
= 0.08
br
ev
id
ie
w
ge
1 A point is moving along the curve y = 3x − 2 x 3 in such a way that the x-coordinate is increasing at
0.015 units per second. Find the rate at which the y-coordinate is changing when x = 2, stating whether
the y-coordinate is increasing or decreasing.
s
-C
-R
am
2 A point with coordinates ( x, y ) moves along the curve y = 1 + 2 x in such a way that the rate of increase of
x has the constant value 0.01 units per second. Find the rate of increase of y at the instant when x = 4.
es
8
in such a way that the x-coordinate is increasing at a constant rate
x2 − 2
of 0.005 units per second. Find the rate of change of the y-coordinate as the point passes through the point (2, 4).
Pr
ity
5
in such a way that the x-coordinate is increasing at a
x
constant rate of 0.02 units per second. Find the rate of change of the y-coordinate when x = 1.
4 A point is moving along the curve y = 3 x −
y
ve
ie
w
rs
C
op
y
3 A point is moving along the curve y =
ev
1
in such a way that the x-coordinate of P is increasing at a
x
constant rate of 0.5 units per second. Find the rate at which the y-coordinate of P is changing when P is at
the point (1, 4).
ie
w
ge
C
U
R
ni
op
5 A point, P, travels along the curve y = 3x +
id
2
+ 5x in such a way that the x-coordinate is increasing at a constant
x
rate of 0.02 units per second. Find the rate at which the y-coordinate is changing when x = 2, stating whether
the y-coordinate is increasing or decreasing.
-C
-R
am
br
ev
6 A point is moving along the curve y =
s
8
. As it passes through the point P, the x-coordinate is increasing at
7 − 2x
a rate of 0.125 units per second and the y-coordinate is increasing at a rate of 0.08 units per second. Find the
possible x-coordinates of P.
es
Pr
ity
op
y
ni
ve
rs
es
s
-R
br
ev
ie
id
g
w
e
C
U
9 A point, P ( x, y ), travels along the curve y = x 3 − 5x 2 + 5x in such a way that the rate of change of x is
constant. Find the values of x at the points where the rate of change of y is double the rate of change of x.
am
PS
8 A point, P, travels along the curve y = 3 2 x 2 − 3 in such a way that at time t minutes the x-coordinate of
P is increasing at a constant rate of 0.012 units per minute. Find the rate at which the y-coordinate of P is
changing when P is at the point (1, −1).
-C
R
ev
ie
w
C
op
y
7 A point moves along the curve y =
Copyright Material - Review Only - Not for Redistribution
229
ve
rs
ity
am
br
id
w
ev
ie
ge
8.5 Practical applications of connected rates of change
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
-R
WORKED EXAMPLE 8.12
Oil is leaking from a pipeline under the sea and a circular patch is formed on the surface of the sea.
Pr
es
s
-C
The radius of the patch increases at a rate of 2 metres per hour.
op
y
Find the rate at which the area is increasing when the radius of the patch is 25 metres.
C
ve
rs
ity
Answer
w
ev
ie
y
dA
when r = 25 .
dt
dr
= 2.
Radius increasing at a rate of 2 metres per hour, so
dt
We need to find
U
w
id
dA
= 2 πr
dr
br
ev
dA
= 50 π
dr
-R
am
When r = 25,
Differentiate with respect to r.
ge
A = πr 2
ie
R
ni
C
op
Let A = area of circular oil patch, in m 2 .
op
Pr
y
es
s
-C
Using the chain rule, dA = dA × dr
dt
dr dt
= 50 π × 2
= 100 π
230
rs
op
y
ve
WORKED EXAMPLE 8.13
ni
ev
ie
w
C
ity
The area is increasing at a rate of 100 π m 2 per hour.
br
-C
C
-R
am
b Find the rate of increase of the volume when r = 5.
Answer
w
ev
id
ie
ge
U
R
A solid sphere has radius r cm, surface area A cm 2 and volume V cm 3.
1
cm s −1.
The radius is increasing at a rate of
5π
a Find the rate of increase of the surface area when r = 3.
s
dA
when r = 3.
dt
1
1
dr
cm s −1, so
=
.
Radius increasing at a rate of
dt 5 π
5π
Pr
ity
Differentiate with respect to r.
op
w
ie
ev
-R
s
es
am
br
id
g
e
C
U
dA
= 8 πr
dr
dA
When r = 3,
= 24 π
dr
dA dA dr
Using the chain rule,
=
×
dt
dr dt
1
= 24 π ×
5π
= 4.8
The surface area is increasing at a rate of 4.8 cm 2 s −1.
y
ni
ve
rs
A = 4 πr 2
-C
R
ev
ie
w
C
op
y
es
a We need to find
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
ev
ie
dV
when r = 5.
dt
am
br
id
b We need to find
w
ge
C
U
ni
op
y
Chapter 8: Further differentiation
-R
4 3
πr
3
-C
V =
Pr
es
s
dV
= 100 π
dr
dV
dV dr
=
×
Using the chain rule,
dt
dr
dt
1
= 100 π ×
5π
= 20
C
op
y
ve
rs
ity
When r = 5 ,
U
R
ni
ev
ie
w
C
op
y
dV
= 4 πr 2
dr
Differentiate with respect to r.
br
ev
id
ie
w
ge
The volume is increasing at a rate of 20 cm 3 s −2.
-R
am
WORKED EXAMPLE 8.14
s
-C
Water is poured into the conical container shown, at a rate of 2π cm 3 s −1.
op
Pr
y
es
1
πh3,
After t seconds, the volume of water in the container, V cm 3, is given by V =
12
where h cm is the height of the water in the container.
231
ity
h
b Given that the container has radius 10 cm and height 20 cm, find the rate of change of
h when the container is half full. Give your answer correct to 3 significant figures.
op
ni
dh
when h = 5.
dt
C
br
1
πh3
12
dV
1
= πh2
dh
4
s
Differentiate with respect to h.
op
y
ni
ve
rs
ity
Pr
es
dV
25 π
=
dh
4
dh
dh dV
=
×
Using the chain rule,
dt
dV
dt
4
=
× 2π
25 π
= 0.32
When h = 5,
-R
s
es
am
br
ev
ie
id
g
w
e
C
U
The height is increasing at a rate of 0.32 cm s −1.
-C
R
ev
ie
w
C
op
y
-C
-R
am
V =
w
dV
= 2 π.
dt
ev
id
Volume increasing at a rate of 2π cm 3 s −1, so
ie
ge
a We need to find
U
R
Answer
y
ve
ev
ie
w
rs
C
a Find the rate of change of h when h = 5.
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
1 1
1000
3
 1 1
πh3  =  π ( 20 )  =
π
2  12
2
12
3



1
1
1000
Using V =
πh3 ,
πh3 =
π
12
12
3
h3 = 4000
h = 15.874
dV
1
= π(15.874)2 = 197.9
When h = 15.874 ,
dh
4
dh
dh dV
=
×
dt
dV
dt
1
=
× 2π
197.9
= 0.0317 cm s −1
C
op
ie
br
ev
id
EXERCISE 8E
w
ge
U
R
ni
ev
ie
y
ve
rs
ity
Using the chain rule,
w
C
op
y
Pr
es
s
-C
-R
am
br
id
ev
ie
b Volume when half full =
w
ge
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
-R
am
1 A circle has radius r cm and area A cm 2.
y
Pr
2 A sphere has radius r cm and volume V cm 3.
1
The radius is increasing at a rate of
cm s −1.
2π
Find the rate of increase of the volume when V = 36 π.
y
ve
ie
w
rs
ity
op
C
232
s
Find the rate of increase of A when r = 4.
es
-C
The radius is increasing at a rate of 0.1cm s −1.
op
ni
ev
3 A cone has base radius r cm and a fixed height of 30 cm.
br
4 A square has side length x cm and area A cm 2.
-R
am
The area is increasing at a constant rate of 0.03 cm 2 s −1.
Find the rate of increase of x when A = 25.
s
-C
ev
id
ie
w
ge
Find the rate of change of the volume when r = 5.
C
U
R
The radius of the base is increasing at a rate of 0.01cm s −1.
op
y
es
5 A cube has sides of length x cm and volume V cm 3.
Pr
ni
ve
rs
ity
Find the rate of increase of x when V = 8.
C
U
Find the rate of increase of x when x = 2.
op
The cuboid is heated and the volume increases at a rate of 0.15 cm 3 s −1.
y
6 A solid metal cuboid has dimensions x cm by x cm by 4x cm.
e
400 π
.
r
Given that the radius of the cylinder is increasing at a rate of 0.25 cm s −1, find the rate of change
of A when r = 10.
-R
s
es
am
br
ev
ie
id
g
w
7 A closed circular cylinder has radius r cm and surface area A cm 2, where A = 2 πr 2 +
-C
R
ev
ie
w
C
The volume is increasing at a rate of 1.5 cm 3 s −1.
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
The vertical cross-section is an equilateral triangle.
ev
ie
am
br
id
w
ge
8 The diagram shows a water container in the shape of a triangular
prism of length 120 cm.
C
U
ni
op
y
Chapter 8: Further differentiation
Water is poured into the container at a rate of 24 cm 3 s −1.
-R
h
120 cm
3
Pr
es
s
b Find the rate of change of h when h = 12.
ve
rs
ity
9 Water is poured into the hemispherical bowl of radius 5 cm at a rate of
3π cm 3 s −1.
w
C
op
y
-C
a Show that the volume of water in the container, V cm , is given by
V = 40 3 h2, where h cm is the height of the water in the container.
y
h
C
op
ge
U
R
ni
ev
ie
After t seconds, the volume of water in the bowl, V cm 3, is given by
1
V = 5 πh2 − πh3, where h cm is the height of the water in the bowl.
3
a Find the rate of change of h when h = 1.
id
ie
w
b Find the rate of change of h when h = 3.
The cone is initially completely filled with water.
10 cm
-R
am
br
ev
10 The diagram shows a right circular cone with radius 10 cm and height 30 cm.
s
30 cm
es
h
Pr
a Show that the volume of water in the cone, V cm 3, when the height of the
πh3
water is h cm is given by the formula V =
.
27
b Find the rate of change of h when h = 20 .
ity
233
ve
ie
w
rs
C
op
y
-C
Water leaks out of the cone through a small hole at the vertex at a rate
of 4 cm 3 s −1.
y
op
ni
ev
11 Oil is poured onto a flat surface and a circular patch is formed.
C
U
R
The radius of the patch increases at a rate of 2 r cm s −1.
ie
w
ge
Find the rate at which the area is increasing when the circumference is 8 π cm.
-R
am
The area of the patch increases at a rate of 5 cm 2 s −1.
ev
br
id
12 Paint is poured onto a flat surface and a circular patch is formed.
-C
a Find, in terms of π, the radius of the patch after 8 seconds.
es
Pr
13 A cylindrical container of radius 8 cm and height 25 cm is completely filled with water.
ity
The water is then poured at a constant rate from the cylinder into an empty inverted cone.
ni
ve
rs
The cone has radius 15 cm and height 24 cm and its axis is vertical.
y
It takes 40 seconds for all of the water to be transferred.
U
op
a If V represents the volume of water, in cm 3, in the cone at time t seconds, find
C
w
the rate of change of the height of the water in the cone
ie
id
g
i
e
b When the depth of the water in the cone is 10 cm, find:
-R
s
es
am
br
ev
ii the rate of change of the horizontal surface area of the water in the cone.
-C
R
ev
ie
w
C
op
y
PS
s
b Find, in terms of π, the rate of increase of the radius after 8 seconds.
Copyright Material - Review Only - Not for Redistribution
dV
in terms of π.
dt
ve
rs
ity
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
Increasing and decreasing functions
y = f( x ) is decreasing for a given interval of x if
op
y
Stationary points
ve
rs
ity
Stationary points (turning points) of a function y = f( x ) occur when
C
y
U
ni
dy
= 0
dx
w
ie
ev
br
-R
am
dy
= 0
dx
ge
the gradient is positive to the left of the maximum and negative to the right.
At a minimum point:
es
d2 y
dy
= 0 and 2 , 0, then the point is a maximum point.
dx
dx
If
d2 y
dy
= 0 and 2 . 0, then the point is a minimum point.
dx
dx
If
dy
d2 y
= 0 and 2 = 0, then the nature of the stationary point can be found using the
dx
dx
Pr
y
If
op
C
w
ge
first derivative test.
id
ie
Connected rates of change
ev
s
y
op
-R
s
es
-C
am
br
ev
ie
id
g
w
e
C
U
R
ev
ie
w
ni
ve
rs
C
ity
dx
1
.
=
y
d
dy
dx
-R
dy
dy
dx
.
=
×
dt
dx
dt
es
You may also need to use the rule:
op
y
•
-C
be connected using the chain rule:
Pr
br
When two variables, x and y, both vary with a third variable, t, the three variables can
am
•
y
rs
ve
ni
U
C
w
ie
ev
R
ity
op
•
•
•
s
-C
the gradient is negative to the left of the minimum and positive to the right.
Second derivative test for maximum and minimum points
234
dy
= 0.
dx
C
op
At a maximum point:
id
w
ev
ie
dy
, 0 throughout the interval.
dx
First derivative test for maximum and minimum points
•
•
•
•
R
-R
dy
. 0 throughout the interval.
dx
Pr
es
s
y = f( x ) is increasing for a given interval of x if
-C
•
•
•
w
ev
ie
am
br
id
ge
Checklist of learning and understanding
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
ge
C
U
ni
op
y
Chapter 8: Further differentiation
ev
ie
am
br
id
The volume of a spherical balloon is increasing at a constant rate of 40 cm 3 per second.
Find the rate of increase of the radius of the balloon when the radius is 15 cm.
4 3
πr .]
3
An oil pipeline under the sea is leaking oil and a circular patch of oil has formed on the surface of the sea.
At midday the radius of the patch of oil is 50 m and is increasing at a rate of 3 metres per hour.
Find the rate at which the area of the oil is increasing at midday.
ve
rs
ity
y
[4]
y
A watermelon is assumed to be spherical in shape while it is growing. Its mass, M kg, and radius, r cm, are
related by the formula M = kr 3 , where k is a constant. It is also assumed that the radius is increasing at a
constant rate of 0.1 centimetres per day. On a particular day the radius is 10 cm and the mass is 3.2 kg.
Find the value of k and the rate at which the mass is increasing on this day.
ie
w
ge
br
ev
id
[5]
-R
am
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q4 June 2012
y
s
-C
5
[5]
C
op
4
4
8
A curve has equation y = 27 x −
. Show that the curve has a stationary point at x = − and
3
( x + 2)2
determine its nature.
ni
ev
ie
3
R
[4]
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q3 November 2012
U
w
C
op
Pr
es
s
-C
[The volume, V , of a sphere with radius r is V =
2
-R
1
w
END-OF-CHAPTER REVIEW EXERCISE 8
235
Q
O
x
op
P (p, 0)
y
X (–2, 0)
ni
ev
ve
ie
w
rs
C
ity
op
Pr
y
es
y = 2x2
w
[2]
ev
ie
ge
id
Express the area, A, of triangle XPQ in terms of p.
br
i
C
U
R
The diagram shows the curve y = 2 x 2 and the points X ( −2, 0) and P ( p, 0). The point Q lies on the curve
and PQ is parallel to the y-axis.
-R
Find the rate at which A is increasing when p = 2.
op
y
es
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q2 June 2015
ym
y
op
w
ie
e
C
U
A farmer divides a rectangular piece of land into 8 equal-sized rectangular sheep pens as shown in the diagram.
Each sheep pen measures x m by y m and is fully enclosed by metal fencing. The farmer uses 480 m of fencing.
Show that the total area of land used for the sheep pens, A m 2, is given by A = 384x − 9.6x 2.
ii
Given that x and y can vary, find the dimensions of each sheep pen for which the value of A is a
maximum. (There is no need to verify that the value of A is a maximum.)
ev
-R
br
[3]
s
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q5 June 2016
es
am
[3]
ie
id
g
w
i
-C
ev
xm
ni
ve
rs
C
ity
Pr
6
R
[3]
s
-C
ii
am
The point P moves along the x-axis at a constant rate of 0.02 units per second and Q moves along the curve
so that PQ remains parallel to the y-axis.
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
am
br
id
ev
ie
w
ge
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
Pr
es
s
-C
-R
The variables x, y and z can take only positive values and are such that z = 3x + 2 y and xy = 600.
1200
.
i Show that z = 3x +
x
ii Find the stationary value of z and determine its nature.
[1]
[6]
U
R
x
2y
y
3y
3x
ni
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rs
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8
C
op
op
y
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q6 June 2011
ge
y
w
7
id
ie
4x
br
ev
The diagram shows the dimensions in metres of an L-shaped garden. The perimeter of the garden is 48 m.
Find an expression for y in terms of x.
ii
Given that the area of the garden is A m 2, show that A = 48x − 8x 2.
-R
[1]
[2]
s
-C
am
i
Pr
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q7 November 2011
8
A curve has equation y = + 2 x.
x
dy
d2 y
[3]
and 2 .
i Find
dx
dx
ii Find the coordinates of the stationary points and state, with a reason, the nature of each
stationary point.
[5]
ni
op
y
ve
ie
ev
C
w
ie
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q5 November 2015
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am
br
10
ev
id
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U
R
[4]
ity
9
w
C
236
rs
op
y
es
iii Given that x can vary, find the maximum area of the garden, showing that this is a maximum
value rather than a minimum value.
s
-C
x cm
x cm
op
y
es
y cm
Pr
ity
i
Express y in terms of x.
ii
Show that the area of the plate, A cm 2, is given by A = 30 x − x 2.
[2]
op
y
ni
ve
rs
[2]
U
Given that x can vary,
e
C
iii find the value of x at which A is stationary,
w
ie
ev
-R
s
es
am
[2]
[2]
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q8 November 2010
br
id
g
iv find this stationary value of A, and determine whether it is a maximum or a minimum value.
-C
R
ev
ie
w
C
The diagram shows a metal plate consisting of a rectangle with sides x cm and y cm and a quarter-circle of
radius x cm. The perimeter of the plate is 60 cm.
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ve
rs
ity
am
br
id
ev
ie
w
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C
U
ni
op
y
Chapter 8: Further differentiation
11 A curve has equation y = x 3 + x 2 − 5x + 7.
Find the coordinates of the two stationary points on the curve and determine the nature of each
stationary point.
[5]
-R
[4]
-C
b
Find the set of values of x for which the gradient of the curve is less than 3.
Pr
es
s
a
x metres
C
op
y
r metres
ni
ev
ie
w
C
ve
rs
ity
op
y
12
id
ie
w
ge
U
R
The inside lane of a school running track consists of two straight sections each of length x metres, and two
semicircular sections each of radius r metres, as shown in the diagram. The straight sections are perpendicular
to the diameters of the semicircular sections. The perimeter of the inside lane is 400 metres.
i
Show that the area, A m 2, of the region enclosed by the inside lane is given by A = 400 r − πr 2 .
ii
Given that x and r can vary, show that, when A has a stationary value, there are no straight sections in
the track. Determine whether the stationary value is a maximum or a minimum.
[5]
-C
-R
am
br
ev
[4]
es
s
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q8 November 2013
ii
Find the nature of each of the stationary points.
[3]
ve
rs
w
ie
Pr
Show that the origin is a stationary point on the curve and find the coordinates of the other stationary
[4]
point in terms of p.
ity
i
C
op
y
13 The equation of a curve is y = x 3 + px 2, where p is a positive constant.
3
2
y
op
ni
ev
Another curve has equation y = x + px + px.
C
ie
ev
id
br
-R
am
es
s
-C
ity
Pr
op
y
-R
s
es
-C
am
br
ev
ie
id
g
w
e
C
U
op
y
ni
ve
rs
C
w
ie
ev
R
[3]
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q9 June 2015
w
ge
U
R
iii Find the set of values of p for which this curve has no stationary points.
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237
op
y
ve
rs
ity
ni
C
U
ev
ie
w
ge
-R
am
br
id
Pr
es
s
-C
y
ni
C
op
y
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rs
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op
C
w
ev
ie
op
y
ve
ni
w
ge
C
U
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ev
ie
w
Chapter 9
Integration
rs
C
ity
op
Pr
y
es
s
-C
-R
am
br
ev
id
ie
w
ge
U
R
238
id
es
s
-C
-R
am
br
ev
understand integration as the reverse process of differentiation, and integrate ( ax + b ) n (for any
rational n except −1 ), together with constant multiples, sums and differences
solve problems involving the evaluation of a constant of integration
evaluate definite integrals
use definite integration to find the:
area of a region bounded by a curve and lines parallel to the axes, or between a curve and a
line, or between two curves
volume of revolution about one of the axes.
•
•
op
y
ni
ve
rs
-R
s
es
-C
am
br
ev
ie
id
g
w
e
C
U
R
ev
ie
w
C
ity
Pr
op
y
■
■
■
■
ie
In this chapter you will learn how to:
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rs
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C
U
ni
op
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Chapter 9: Integration
What you should be able to do
IGCSE / O Level
Mathematics
Substitute values for x and y into
equations of the form y = f( x ) + c
and solve to find c .
1 a Given that the line y = 5x + c
passes through the point (3, − 4),
find the value of c.
Pr
es
s
-C
y
b Given that the curve y = x 2 − 2 x + c
passes through the point ( −1, 2), find
the value of c.
ve
rs
ity
op
C
2 Find the x-coordinates of the points
where the curve crosses the x-axis.
Find the x-coordinates of the
points where a curve crosses the
x-axis.
ni
ev
id
br
-R
am
a
dy
.
dx
y = 3x8 − 13x − 10
b
y = 5x 2 − 4x + 10 x
es
s
-C
y=3 x −x
Pr
y
op
ity
In Chapters 7 and 8 you studied differentiation, which is the first basic tool of calculus.
In this chapter you will learn about integration, which is the second basic tool of
calculus. We often refer to integration as the reverse process of differentiation. It has
many applications; for example, planning spacecraft flight paths, or modelling real-world
behaviour for computer games.
ni
op
y
ve
rs
C
w
ie
b
3 Find
ie
Differentiate constant multiples,
sums and differences of expressions
containing terms of the form ax n.
Why do we study integration?
ev
y = 3x 2 − 13x − 10
w
ge
U
R
Chapter 7
a
C
op
w
IGCSE / O Level
Mathematics
ev
ie
Check your skills
-R
Where it comes from
y
am
br
id
ev
ie
w
PREREQUISITE KNOWLEDGE
br
ie
9.1 Integration as the reverse of differentiation
ev
id
w
ge
C
U
R
Isaac Newton and Gottfried Wilhelm Leibniz formulated the principles of integration
independently, in the 17th century, by thinking of an integral as an infinite sum of
rectangles of infinitesimal width.
Pr
op
y
es
You learnt the rule for differentiating power functions:
ity
op
ie
ev
y = x3 – 0.1
y = x3
es
s
y = x3 – 59
dy
= 3x2
dx
-R
id
g
-C
am
br
y = x3 + 4
1
2
C
y = x3 +
e
y = x3 – 7
w
U
Applying this rule to functions of the form y = x 3 + c, we obtain:
R
ev
y
ni
ve
rs
dy
n −1
.
= nx
dx
ie
w
C
KEY POINT 9.1
If y = x n , then
s
-C
-R
am
dy
In Chapter 7, you learnt about the process of obtaining
when y is known. We call this
dx
process differentiation.
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WEB LINK
Explore the Calculus
meets functions station
on the Underground
Mathematics website.
239
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rs
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Cambridge International AS & A Level Mathematics: Pure Mathematics 1
-R
am
br
id
ev
ie
w
ge
This shows that there are an infinite number of functions that when differentiated give the
answer 3x 2. They are all of the form y = x 3 + c, where c is some constant.
dy
is known.
In this chapter you will learn about the reverse process of obtaining y when
dx
We can call this reverse process antidifferentiation.
C
op
ni
U
R
ev
ie
w
Because of this theorem, we do not need to use the term antidifferentiation. So from
now on, we will only talk about integration, whether we are reversing the process of
differentiation or finding the area under a graph.
w
ge
EXPLORE 9.1
e y=−
-R
1 −7
x + 0.2
7
c
y=
1 15
x +1
15
f
y=
2 2 5
x −
3
8
3
s
1 −2
x +3
2
Pr
y
d y=−
es
-C
am
br
ev
id
ie
dy
for each of the following functions.
dx
1
1
a y = x3 − 2
b y = x6 + 8
3
6
1 Find
y
C
ve
rs
ity
op
y
Pr
es
s
-C
There is a seemingly unrelated problem that you will study in Section 9.6: what is the
area under the graph of y = 3x 2 ? The process used to answer that question is known
as integration. There is a remarkable theorem due to both Newton and Leibniz that says
that integration is essentially the same as antidifferentiation. This is now known as the
Fundamental theorem of Calculus.
op
w
ge
-R
am
br
ev
id
ie
From the class discussion we can conclude that:
KEY POINT 9.2
dy
1
= .
dx
x
C
U
R
ni
4 Discuss with your classmates whether your rule works for finding y when
y
rs
3 Describe your rule, in words.
ve
ev
ie
w
C
ity
op
2 Discuss your results with those of your classmates and try to find a rule for
dy
obtaining y if
= x n.
dx
240
dy
1
n +1
= x n , then y =
x
+ c (where c is an arbitrary constant and n ≠ −1).
n+1
dx
es
You may find it easier to remember this in words:
Pr
op
y
s
-C
If
op
w
e
1
n +1
x
+ c (where c is an arbitrary constant and n ≠ −1).
n+1
ev
es
s
-R
br
∫ is used to denote integration.
am
The special symbol
ie
id
g
If f ′( x ) = x n, then f( x ) =
C
U
KEY POINT 9.3
y
Using function notation we write this rule as:
-C
R
ev
ie
w
ni
ve
rs
C
ity
‘Increase the power n by 1 to obtain the new power, then divide by the new power.
Remember to add a constant c at the end.’
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rs
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C
U
ni
op
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Chapter 9: Integration
w
am
br
id
∫ x dx = 4 x + c
ev
ie
1
ge
When we need to integrate x 3, for example, we write:
4
-R
3
∫ x dx is called the indefinite integral of x with respect to x.
3
Pr
es
s
-C
3
We call it ‘indefinite’ because it has infinitely many solutions.
+ c (where c is a constant and n ≠ −1).
y
ev
ie
n +1
C
op
1
∫ x dx = n + 1 x
n
U
R
ve
rs
ity
KEY POINT 9.4
ni
w
C
op
y
Using this notation, we can write the rule for integrating powers as:
ie
ev
id
br
KEY POINT 9.5
w
ge
We write the rule for integrating constant multiples of a function as:
es
s
-C
-R
am
∫ kf(x ) dx = k ∫ f(x ) dx, where k is a constant.
op
Pr
y
We write the rule for integrating sums and differences of two functions as:
ity
∫  f(x ) ± g(x )  dx = ∫ f(x ) dx ± ∫ g(x ) dx
241
y
op
w
ge
WORKED EXAMPLE 9.1
C
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ni
ev
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ie
w
rs
C
KEY POINT 9.6
-R
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ev
ie
id
g
-R
s
es
Copyright Material - Review Only - Not for Redistribution
1
( 52 )
5
x2 + c
5
y
op
1
+c
x
C
=−
=
e
U
dy
= x2
dx
3
+1
1
y= 3
x2 + c
1
+
2
= − x −1 + c
br
am
c
Pr
ev
R
-C
dy
=x x
dx
3
dy
= x −2
dx
1
y=
x −2 + 1 + c
−2 + 1
ity
1 4
x +c
4
ni
ve
rs
=
b
s
dy
= x3
dx
1
y=
x3 + 1 + c
3+1
ie
w
C
op
y
a
c
es
-C
Answer
am
br
ev
id
ie
Find y in terms of x for each of the following.
1
dy
dy
= x3
= 2
b
a
dx
dx
x
=
2 2
x +c
5
ve
rs
ity
ev
ie
am
br
id
WORKED EXAMPLE 9.2
w
ge
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
Find f( x ) in terms of x for each of the following.
-R
Pr
es
s
-C
C
op
c
ve
rs
ity
y
f ′( x ) = 4x 3 − 2 x −2 + 4x 0
4
2
4
f( x ) = x 4 −
x −1 + x1 + c
4
( −1)
1
Write in index form ready for integration.
U
R
ni
ev
ie
a
y
w
Answer
C
op
b
2
+4
x2
1
f ′( x ) = 8x 2 −
+ 2x
2x 4
( x + 3)( x − 1)
f ′( x ) =
x
f ′( x ) = 4x 3 −
a
br
ev
id
ie
w
ge
= x 4 + 2 x −1 + 4x + c
2
= x 4 + + 4x + c
x
1 −4
x + 2 x1
2
8
1
2
x −3 + x 2 + c
f( x ) = x 3 −
3
2( −3)
2
8 3 1 −3
= x + x + x2 + c
3
6
8 3
1
= x + 3 + x2 + c
3
6x
f ′( x ) = 8x 2 −
Write in index form ready for integration.
3
1
−
1
Pr
ity
y
Write in index form ready for integration.
U
R
rs
x2 + 2x − 3
x
op
f ′( x ) =
ev
c
ve
ie
w
C
242
ni
op
y
es
s
-C
-R
am
b
( 23 )
5
C
3
( 21 )
1
w
x2 −
x2 + c
ie
3
2
ev
( 52 )
5
x2 +
id
1
br
f( x ) =
ge
= x 2 + 2 x 2 − 3x 2
3
-R
am
2 2 4 2
x + x −6 x +c
5
3
es
s
-C
=
Pr
op
y
WORKED EXAMPLE 9.3
Answer
∫
4x 2 − 3 x
dx
x
2
C
3
w
4x 4 4x 3 3x 2
+
−
+c
4
3
2
4x 3 3x 2
= x4 +
−
+c
3
2
s
-R
ev
ie
=
es
br
id
g
e
U
∫ x(2x − 1)(2x + 3) dx = ∫ (4x + 4x − 3x ) dx
am
a
b
y
x(2 x − 1)(2 x + 3) dx
op
∫
ni
ve
rs
a
-C
R
ev
ie
w
C
ity
Find:
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
C
1
y
= 2 x 2 − 6x 2 + c
= 2x2 − 6 x + c
y
ve
rs
ity
op
-R
( )
C
f ′( x ) =
Pr
2
8
+
+ 6x
x3 x2
f ′( x ) =
9
3
−
−4
x7 x2
e
dy
=
dx
b
∫ 20x dx
e
∫ 3 x dx
2
c
∫ 3x dx
f
∫ x x dx
w
2
−2
es
5
∫ (x − 3) dx
c
∫ (2 x − 1) dx
e
∫
x2 − 1
dx
2x2
f
∫
x3 + 6
dx
2x3
∫
x 4 − 10
dx
x x
∫
3 

 2 x − x 2 x  dx
2
es
s
-R
br
ev
ie
id
g
w
e
h
op
y
2
C
∫
x2 + 2 x
dx
3x
x (x 2 + 1) dx
am
dy 5x 2 + 3x + 1
=
dx
x
ie
ev
3
Pr
∫
3
-C
f
b
ity
d
ni
ve
rs
∫ (x + 1)(x + 4) dx
U
op
y
a
C
w
dy
= x( x + 2)( x − 8)
dx
s
3
5 Find each of the following.
g
C
x( x − 3)
-R
id
br
am
4
∫ x dx
-C
d
∫ 12x dx
5
c
y
dy
= 2 x 2 (3x + 1)
dx
ni
b
ge
R
rs
dy x 4 − 2 x + 5
=
dx
2x3
ve
d
U
ie
dy
= x( x + 5)
dx
ev
a
a
ie
243
3 Find y in terms of x for each of the following.
4 Find each of the following.
ev
dy
4
=
dx
x
ity
d
f
f ′( x ) = 3x5 + x 2 − 2 x
es
c
b
dy
= 12 x 3
dx
s
-C
y
op
f ′( x ) = 5x 4 − 2 x 3 + 2
w
C
a
c
ie
dy
1
=
dx 2 x 3
ev
id
br
e
am
dy
= 14x 6
dx
-R
dy
3
=
dx x 2
b
w
ni
d
ge
dy
= 15x 2
dx
U
R
a
2 Find f( x ) in terms of x for each of the following.
R
C
op
1 Find y in terms of x for each of the following.
op
w
w
1
4 2
3
x − 1 x2 + c
2
2
=
EXERCISE 9A
ev
ie
ev
ie
∫
1

− 
2 dx
x
−
4
3
x




Pr
es
s
am
br
id
∫
4x 2 − 3 x
dx =
x
-C
b
ge
U
ni
op
y
Chapter 9: Integration
Copyright Material - Review Only - Not for Redistribution
i
2
ve
rs
ity
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
w
ge
9.2 Finding the constant of integration
-R
am
br
id
ev
ie
The next two examples show how we can find the equation of a curve if we know the
gradient function and the coordinates of a point on the curve.
Pr
es
s
-C
WORKED EXAMPLE 9.4
dy 6x5 − 18
=
, and (1, 6) is a point on the curve.
dx
x3
Find the equation of the curve.
y
Write in index form ready for integration.
ie
w
ge
U
ni
dy
6x5 − 18
=
dx
x3
= 6x 2 − 18x −3
y = 2 x 3 + 9x −2 + c
9
= 2x3 + 2 + c
x
C
op
Answer
id
R
ev
ie
w
C
ve
rs
ity
op
y
A curve is such that
y
ity
op
rs
WORKED EXAMPLE 9.5
y
ve
ie
w
C
244
Pr
9
− 5.
x2
The equation of the curve is y = 2 x 3 +
es
s
-C
-R
am
br
ev
When x = 1, y = 6.
9
6 = 2(1)3 + 2 + c
(1)
6 = 2+9+c
c = −5
op
ie
ev
br
f( x ) = 3x5 − 3x 2 + c
id
f ′( x ) = 15x 4 − 6x
w
ge
Answer
C
U
R
ni
ev
The function f is such that f ′( x ) = 15x 4 − 6x and f( −1) = 1. Find f( x ).
-R
am
Using f( −1) = 1 gives:
s
es
Pr
op
y
-C
1 = 3( −1)5 − 3( −1)2 + c
1 = −3 − 3 + c
c=7
y
op
-R
s
es
-C
am
br
ev
ie
id
g
w
e
C
U
R
ev
ie
w
ni
ve
rs
C
ity
∴ f( x ) = 3x5 − 3x 2 + 7
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
ge
C
U
ni
op
y
Chapter 9: Integration
am
br
id
ev
ie
w
WORKED EXAMPLE 9.6
dy
= 6x + k , where k is a constant. The gradient of the normal to the curve at the
dx
1
point (1, −3) is . Find the equation of the curve.
2
Pr
es
s
-C
-R
A curve is such that
Answer
ni
U
(1)
dy
= 6(1) + k = 6 + k
dx
id
1
so gradient of tangent = −2
2
6 + k = −2
k = −8
w
ge
R
When x = 1,
y
−3 = 3(1) + k (1) + c
c + k = −6
C
op
y = −3.
2
ie
When x = 1,
rs
ve
op
ni
dy
= 3x 2 + 1, P = (1, 4)
dx
c
dy
4
= 2,
dx
x
e
dy
2
=
− 1, P = (4, 6)
dx
x
ev
ie
d
-R
am
f
P = (3, 7)
dy
(1 − x )2
=
, P = (9, 5)
x
dx
s
dy
k
= − 2 , where k is a constant. Given that the curve passes through the points (6, 2.5)
dx
x
and ( −3, 1), find the equation of the curve.
es
Pr
ity
op
y
op
y
ni
ve
rs
dy
= kx 2 − 12 x + 5, where k is a constant. Given that the curve passes through the points
dx
(1, −3) and (3, 11), find the equation of the curve.
3 A curve is such that
dy
6
= kx 2 − 3 , where k is a constant. Given that the curve passes through the point
dx
x
P (1, 6) and that the gradient of the curve at P is 9, find the equation of the curve.
-R
s
es
am
br
ev
ie
id
g
w
e
C
U
4 A curve is such that
-C
C
w
ie
dy
2x3 − 6
,
=
dx
x2
w
P = (4, 9)
br
id
ge
C
U
a
dy
and a point P on the curve.
dx
dy
b
= 2 x(3x − 1), P = ( −1, 2)
dx
-C
R
ev
1 Find the equation of the curve, given
y
C
w
ie
EXERCISE 9B
2 A curve is such that
ev
245
ity
op
Pr
y
The equation of the curve is y = 3x 2 − 8x + 2.
es
s
Substituting for k into (1) gives c = 2.
-R
-C
am
br
ev
Gradient of normal =
R
Integrate.
ve
rs
ity
ev
ie
w
C
op
y
dy
= 6x + k
dx
y = 3x 2 + kx + c
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
ge
dy
= 5x x + 2. Given that the curve passes through the point (1, 3), find:
dx
a the equation of the curve
-R
b the equation of the tangent to the curve when x = 4 .
dy
= kx + 3, where k is a constant. The gradient of the normal to the curve at the point
dx
1
(1, −2) is − . Find the equation of the curve.
7
7 A function y = f( x ) has gradient function f ′( x ) = 8 − 2 x. The maximum value of the function is 20. Find f( x )
and sketch the graph of y = f( x ).
ve
rs
ity
C
op
y
Pr
es
s
-C
6 A curve is such that
w
ev
ie
am
br
id
w
5 A curve is such that
C
op
U
R
ni
ev
ie
y
dy
= 3x 2 + x − 10 . Given that the curve passes through the point (2, −7) find:
dx
a the equation of the curve
8 A curve is such that
w
d2 y
= 12 x + 12. The gradient of the curve at the point (0, 4) is 10.
dx 2
a Express y in terms of x.
ev
id
ie
9 A curve is such that
am
br
PS
ge
b the set of values of x for which the gradient of the curve is positive.
d2 y
= −6x − 4. Given that the curve has a minimum point at ( −2, −6), find the equation
dx 2
s
10 A curve is such that
es
-C
-R
b Show that the gradient of the curve is never less than 4.
op
Pr
y
of the curve.
11 A curve y = f( x ) has a stationary point at P (2, −13) and is such that f ′( x ) = 2 x 2 + 3x − k, where k is a
ity
C
ve
ie
w
constant.
rs
246
y
op
ni
ev
a Find the x-coordinate of the other stationary point, Q, on the curve y = f( x ).
dy
= k + x, where k is a constant.
dx
a Given that the tangents to the curve at the points where x = 5 and x = 7 are perpendicular, find the
value of k.
ie
w
ge
12 A curve is such that
br
ev
id
PS
C
U
R
b Determine the nature of each of the stationary points P and Q.
-C
-R
am
b Given also that the curve passes through the point (10, −8), find the equation of the curve.
4
. Find f ′( x ) and f( x ).
x3
15 A curve is such that
dy
= 3 − 2 x and (1, 11) is a point on the curve.
dx
op
y
ni
ve
rs
ity
Pr
d2 y
= 2 x + 8. Given that the curve has a minimum point at (3, −49), find the coordinates
dx 2
of the maximum point.
14 A curve is such that
-R
am
br
ev
ie
id
g
w
e
C
U
a Find the equation of the curve.
1
b A line with gradient is a normal to the curve at the point (4, 5). Find the equation of this normal.
5
dy
16 A curve is such that
= 3 x − 6 and the point P (1, 6) is a point on the curve.
dx
a Find the equation of the curve.
es
s
b Find the coordinates of the stationary point on the curve and determine its nature.
-C
R
ev
ie
w
C
op
y
es
s
13 A curve y = f( x ) has a stationary point at (1, −1) and is such that f ′′( x ) = 2 +
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
C
U
ni
op
y
Chapter 9: Integration
ge
d2 y
12
= 2 − 3 . The curve has a stationary point at P where x = 1. Given that the curve
dx 2
x
passes through the point (2, 5), find the coordinates of the stationary point P and determine its nature.
am
br
id
ev
ie
w
17 A curve is such that
-R
dy
= 2 x − 5 and the point P (3, −4) is a point on the curve. The normal to the curve
dx
at P meets the curve again at Q .
b Find the equation of the normal to the curve at P.
c
Find the coordinates of Q.
w
ev
ie
Pr
es
s
a Find the equation of the curve.
ve
rs
ity
C
op
y
-C
18 A curve is such that
1
ev
br
id
ie
w
ge

d  1
(3x − 1)7  = (3x − 1)6

dx  3 × 7
∫ (3x − 1) dx = 3 × 7 (3x − 1) + c
6
7
-R
am
Hence,
es
s
-C
This leads to the general rule:
ve
ie
w
rs
C
Pr
∫
If n ≠ −1 and a ≠ 0, then
247
1
( ax + b ) dx =
( ax + b ) n + 1 + c
a ( n + 1)
n
ity
op
y
KEY POINT 9.7
y
id
ie
w
ge
C
U
∫
R
op
ni
ev
It is very important to note that this rule only works for powers of linear functions.
1
( ax 2 + b )3 + c. (Try differentiating the latter
For example, ( ax 2 + b )6 dx is not equal to
3a
expression to see why.)
b
20
∫ (1 − 4x ) dx
1
(2 x − 3)5 + c
10
Pr
+c
y
20
∫ (1 − 4x ) dx = 20∫ (1 − 4x ) dx
20
(1 − 4x )−6 + 1 + c
( −4)( −6 + 1)
ie
ev
-R
s
= (1 − 4x )−5 + c
1
=
+c
(1 − 4x )5
es
am
br
id
g
w
e
=
op
−6
6
C
b
-C
ev
ie
w
=
4+1
5
∫ 2x + 7 dx
ity
4
ni
ve
rs
C
1
∫ (2x − 3) dx = 2(4 + 1) (2x − 3)
U
op
y
Answer
a
c
6
s
4
es
-C
∫ (2x − 3) dx
-R
br
am
Find:
a
ev
WORKED EXAMPLE 9.7
R
C
op
ni
U
R
In Chapter 7 you learnt that:
y
9.3 Integration of expressions of the form (ax + b)n
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
C
w
1
1
+1
Pr
es
s
a
∫ (2x − 7) dx
d
∫ 3(1 − 2x ) dx
g
2
dx
3x − 2
U
ni
8
2 Find the equation of the curve, given
3

dy
= (2 x − 1)3 , P =  , 4 
2

dx
∫
3
 2  dx
 2x + 1 
dy
and a point P on the curve.
dx
dy
=
b
dx
es
1
, P = (3, 7)
x−2
3 A curve is such that
3
Pr
dy
=
dx
∫ 5 − 4x dx
∫ 4(7 − 2x ) dx
5
5
2 x + 5,
P = (2, 2)
op
ev
br
s
12
− 4x − 2.
3x + 1
es
dy
=
dx
-R
b the equation of the curve.
w
ie
id
a the equation of the normal to the curve at P
am
i
3
C
U
ge
Given that the curve passes through the point P(2, 1), find:
-C
∫ ( 2x + 1) dx
y
ve
ni
ev
f
8
dy
4
=
, P = (2, 4)
dx ( 3 − 2 x )2
d
1
. Find the equation of the curve.
12
dy
5
=
.
4 A curve is such that
dx
2x − 3
R
∫ 2(5x − 2) dx
3
dy
= k ( x − 5) , where k is a constant. The gradient of the normal to the curve at the
dx
point (4, 2) is
5 A curve is such that
c
ity
c
e
5
rs
ie
w
C
248
∫ (3x + 1) dx
s
-C
op
y
a
b
h
br
∫
id
ge
5
am
R
ev
ie
1 Find:
y
w
C
ve
rs
ity
EXERCISE 9C
C
op
op
y
-C
= 5 2x + 7 + c
+c
w
)
−
(2 x + 7) 2
ie
− 21 + 1
ev
2(
5
-R
=
-R
∫
ev
ie
−
5
dx = 5 (2 x + 7) 2 dx
2x + 7
am
br
id
∫
c
ge
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
Pr
op
y
a Show that the curve has a stationary point when x = 1 and determine its nature.
ity
dy
4
=
, where k is a constant. The point P(3, 2) lies on the curve and the normal
dx
2x + k
to the curve at P is x + 4 y = 11. Find the equation of the curve.
y
-R
s
es
-C
am
br
ev
ie
id
g
w
e
C
U
op
6 A curve is such that
R
ev
ie
PS
ni
ve
rs
w
C
b Given that the curve passes through the point (0, 13), find the equation of the curve.
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
C
U
ni
op
y
Chapter 9: Integration
w
ge
9.4 Further indefinite integration
-R
am
br
id
ev
ie
In this section we use the concept that integration is the reverse process of differentiation
to help us integrate some more complicated expressions.
)
b Hence, find
ge
Answer
ie
ev
s
249
rs
C
w
es
1
(3x 2 − 4)8 + c
8
EXERCISE 9D
y
ve
ie
7
ity
op
y
=
2
Pr
-C
7
op
∫ x(x + 2) dx.
3
w
ge
2
C
U
R
ni
ev
1 a Differentiate ( x 2 + 2)4 with respect to x.
b Hence, find
b Hence, find
∫
ev
s
1
with respect to x.
4 − 3x 2
3x
dx .
(4 − 3x 2 )2
op
2
5
C
U
∫ 2(2x − 3)(x − 3x + 5) dx.
e
b Hence, find
y
5 a Differentiate ( x 2 − 3x + 5)6 with respect to x.
ev
-R
s
∫
( x + 3)7
dx.
x
es
am
br
b Hence, find
ie
id
g
w
6 a Differentiate ( x + 3)8 with respect to x.
-C
R
ev
ie
w
C
4 a Differentiate
es
∫
Pr
op
y
b Hence, find
dy
kx
1
, show that
=
, and state the value of k.
dx ( x 2 − 5)2
x2 − 5
4x
dx.
( x 2 − 5)2
ity
-C
3 a Given that y =
4
-R
2
ni
ve
rs
br
am
∫ x(2x − 1) dx.
ie
id
2 a Differentiate (2 x 2 − 1)5 with respect to x.
b Hence, find
7
-R
br
am
1
∫ 6x(3x − 4) dx = 8 ∫ 48x(3x − 4) dx
2
2
Use the chain rule.
dy
= (6x )(8)(3x 2 − 4)8 − 1
dx
= 48x(3x 2 − 4)7
b
∫ 6x ( 3x − 4 ) dx.
w
Let y = (3x 2 − 4)8
id
a
(
C
op
)
ni
(
y
7
8
d 
3x 2 − 4  = 48x 3x 2 − 4 .


dx 
U
R
a Show that
Pr
es
s
op
C
ev
ie
w
WORKED EXAMPLE 9.8
ve
rs
ity
∫ f(x ) dx = F(x ) + c
d
[ F( x ) ] = f( x ) , then
dx
y
If
-C
KEY POINT 9.8
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
ev
ie
∫ 3 x (2x x − 1) dx.
am
br
id
4
-R
b Hence, find
w
ge
7 a Differentiate (2 x x − 1)5 with respect to x.
-C
9.5 Definite integration
4
ni
3
y
1
∫ x dx = 4 x + c,
where c is an arbitrary constant, is called the indefinite integral of x 3 with respect to x.
U
R
ev
ie
w
Recall that
C
op
C
ve
rs
ity
op
y
Pr
es
s
In the remaining sections of this chapter, you will be learning how to find areas and
volumes of various shapes. To do this, you will be using a technique known as definite
integration, which is an extension of the indefinite integrals you have been using up to now.
In this section, you will learn this technique, before going on to apply it in the next section.
w
ge
We can integrate a function between two specified limits.
br
-R
-C
2
x3 dx
am
∫
4
ev
id
ie
We write the integral of the function x 3 with respect to x between the limits x = 2 and
x = 4 as:
es
4
4
1
x 3 d x =  x 4 + c 
4
2
2
Pr
y
ve
= 60
ni
4
w
ge
∫
U
1
x 3 dx =  x 4 
4
2
2
C
4
br
ev
id
ie
1
1
=  × 44  −  × 2 4 

4
 4
-R
am
= 60
4
op
Note that the ‘c’s cancel out, so the process can be simplified to:
R
ev
ie
w
C
ity
1
1
=  × 44 + c  −  × 2 4 + c 
4
 4

rs
op
y
∫
250
The limits of
integration are always
written either to the
right of the integral
sign, as printed, or
directly below and
above it. They should
never be written to the
left of the integral sign.
s
The method for evaluating this integral is:
∫ x dx is called the definite integral of x with respect to x between the limits 2 and 4.
3
es
s
-C
3
2
Pr
op
y
Hence, we can write the evaluation of a definite integral as:
ni
ve
rs
b
U
op
b
a
a
y
∫ f(x ) dx = [ F(x ) ] = F(b) − F(a )
-R
s
es
am
br
ev
ie
id
g
w
e
C
The following rules for definite integrals may also be used.
-C
R
ev
ie
w
C
ity
KEY POINT 9.9
TIP
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
ev
ie
b
∫ f(x ) dx, where k is a constant
∫ [ f(x ) ± g(x ) ] dx = ∫ f(x ) dx ± ∫ g(x ) dx
∫ f(x ) dx = − ∫ f(x ) dx
a
b
a
a
a
a
b
ve
rs
ity
C
∫
b
f( x ) d x +
∫
c
∫ f(x ) dx
f( x ) d x =
b
a
w
ge
U
R
a
c
ni
ev
ie
w
KEY POINT 9.11
y
b
op
y
-C
a
b
C
op
a
Pr
es
s
b
kf ( x ) d x = k
-R
∫
w
ge
am
br
id
KEY POINT 9.10
b
C
U
ni
op
y
Chapter 9: Integration
s
c
es
y
2
∫ ( 6x − x ) d x
−2
2
−2
w
ni
C
U
R
)
op
ve
ie
ev
) (
= 2(2)3 + (2)−1 − 2(1)3 + (1)−1
y
2
6
=  x 3 + x −1 
3
1
(
-R
s
0
1
( 5x + 1 ) 2 d x
3
w
ni
ve
rs
C
ity
Pr
op
y


3 

1
=
( 5x + 1 ) 2 
 (5)  3 

 2

 0
es
∫
3
ev
= 13 2
5x + 1 d x =
0
ie
1
id
br
am
-C
∫
3
w
ge
1
=  16 +  − (2 + 1)

2
b
3
y
op
ev
ie
3 
 2
=  ( 5x + 1 ) 2 
 15
0
id
g
w
e
C
U
R
3
3
 2
 2
× 12 
=
× 16 2  − 

  15
 15
-R
s
es
-C
am
br
ev
ie
128   2 
=
−
 15   15 
= 8 25
2
251
rs
1
8
∫ ( 5 − 2x ) dx
ity
op
C
6x 4 − 1
dx =
x2
1
2
∫
1
5x + 1 d x
0
Answer
a
∫
3
Pr
∫
b
-R
br
6x 4 − 1
dx
x2
1
2
-C
a
am
Evaluate:
ev
id
ie
WORKED EXAMPLE 9.9
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
∫ 8(5 − 2x ) dx
−2
1
Pr
es
s
1
8
9
ni
U
2
∫0 ( 10 − x2 ) dx
es
f
2x
∫ ( x + 5 ) dx.
(
ie
5 a Given that y = x 3 − 2
1
C
w
) , find ddxy .
5
2
4
C
U
3
5
1
x
w
ie
ev
( x + 1 )4 d x .
-R
4
dx
s
am
-C
∫
10
es
br
id
g
e
( x + 1) , find dy .
6 a Given that y =
b Hence, evaluate
f
s
∫
∫
2
∫
2
1
ev
9
3 dx
2 ( 2x − 3 )
c
ie
3
∫ x ( x − 2 ) dx.
0
R
2x + 1 d x
y
C
0
b Hence, evaluate
2
2

2 
∫  3 x + x  dx
1
op
op
y
2
∫ (x + 3)(7 − 2x ) dx
4
ni
U
br
am
-C
dy
2
, find
.
dx
x +5
2
b Hence, evaluate
2
1
y
∫
4
6 
op
∫

∫  2 − x  dx
2
ni
ve
rs
ev
R
(3 − x )(8 + x )
dx
x4
1
2
0
e
4 a Given that y =
w
b
6
2 dx
−1 ( x − 2 )
∫
c
es
1
ge
−1
( 2x + 3 )3 d x
id
∫
0
e
−1
 8 − x2 
 x 2  d x
−2 
∫
−1
2
Pr
ity
x (1 − x ) d x
3 Evaluate:
d
ev
b
0
a
f
Pr
∫
1
 3x 2 − 2 + 1  d x
x2 
1 
rs
ie
d
∫
2
∫ (2x − 3) dx
4
∫−1 ( 4x2 − 2x ) dx
ity
C
w
a
ve
op
y
2 Evaluate:
252
1
c
s
-C
e
3
-R
br
3
d
4
∫ x dx
1
am
1
b
ev
3
2
-R
2
∫ 3x dx
ie
w
ge
id
1 Evaluate:
a
y
ve
rs
ity
C
4
4
=  − 
 3  9
C
op
op
y
4 
=
 5 − 2 x  −2
=
-R
-C


8
=
( 5 − 2x )−1 
2
1
−
−
(
)
(
)

 −2
EXERCISE 9E
R
−2
ev
ie
am
br
id
∫
w
ev
ie
1
w
ge
1
8
2 dx =
−2 ( 5 − 2 x )
c
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
Copyright Material - Review Only - Not for Redistribution
−2
( x − 1 )3 d x
4
dx
5 − 2x
ve
rs
ity
C
w
ge
9.6 Area under a curve
U
ni
op
y
Chapter 9: Integration
ev
ie
am
br
id
Consider the area bounded by the curve y = x 2 , the x-axis and the lines x = 2 and x = 5.
-R
y = x2
y
C
op
O
2
x
5
ge
U
R
ni
ev
ie
w
C
ve
rs
ity
op
y
Pr
es
s
-C
y
-C
As δx → 0, then A →
5
∫ y dx.
∑ yδx.
y
s
The approximation for A is then
-R
am
br
ev
id
ie
w
The area, A, of the region can be approximated by a series of rectangular strips of
thickness δx (corresponding to a small increase in x) and height y (corresponding to the
height of the function).
y = x2
253
y
op
O
C
2
5
x
w
ge
U
R
ni
ev
ve
ie
w
rs
C
ity
op
Pr
y
es
2
br
ev
id
ie
This leads to the general rule:
-R
am
KEY POINT 9.12
s
-C
If y = f( x ) is a function with y ù 0, then the area, A, bounded by the curve y = f( x ), the x-axis
b
∫ y d x.
es
and the lines x = a and x = b is given by the formula A =
y
op
-R
s
es
-C
am
br
ev
ie
id
g
w
e
C
U
R
ev
ie
w
ni
ve
rs
C
ity
Pr
op
y
a
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
ev
ie
w
ge
am
br
id
WORKED EXAMPLE 9.10
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
U
6
w
ev
id
-C
-R
am
br
= (216) − (64)
ge
3
=  x3 
 3
 4
ie
R
y
ni
2
4
6x
4
C
op
∫ 3x dx
= 152 units2
O
ve
rs
ity
6
y = 3x2
Pr
es
s
-C
y
op
C
w
ev
ie
Answer
Area =
y
-R
Find the area of the shaded region.
op
∫ f(x ) dx will have a negative value. This
a
Pr
y
b
If the required area lies below the x-axis, then
254
es
s
In Worked example 9.10, the required area is above the x-axis.
rs
WORKED EXAMPLE 9.11
y
op
ie
ev
-R
s
es
0
op
U
Area is 36 units2.
y
ni
ve
rs
= (72 − 108) − (0 − 0)
= −36
w
Pr
6
-R
s
es
-C
am
br
ev
ie
id
g
w
e
C
C
6
1

∫0 ( x2 − 6x ) dx =  3 x3 − 2 x2 
ie
O
ity
6
ev
y = x2 – 6x
w
ge
id
br
am
-C
op
y
Answer
R
y
C
U
R
ni
Find the area of the shaded region.
ve
ev
ie
w
C
ity
is because the integral is summing the y values, and these are all negative.
Copyright Material - Review Only - Not for Redistribution
6
x
ve
rs
ity
C
U
ni
op
y
Chapter 9: Integration
ev
ie
am
br
id
w
ge
The required region could consist of a section above the x-axis and a section below the
x-axis.
If this happens we must evaluate each area separately.
-C
-R
This is illustrated in Worked example 9.12.
y
Pr
es
s
WORKED EXAMPLE 9.12
y
2
ge
3
2
0
2
es
( )
Pr
y
ev
1
8
1
8
=  (2)4 − (2)3 + 6(2)2  −  (0)4 − (0)3 + 6(0)2 
4
 4

3
3
s
-C
am
br
1
8
=  x 4 − x 3 + 6x 2 
 4
 0
3
= 6 23 − (0)
op
6
6
2
2
255
ity
= 6 23
2
6
y
3
rs
∫ x(x − 2)(x − 6) dx = ∫ ( x − 8x + 12x ) dx
ve
C
w
ie
w
∫ ( x − 8x + 12x ) dx
id
0
ie
x( x − 2)(x − 6) dx =
op
C
U
R
ni
ev
1
8
=  x 4 − x 3 + 6x 2 
 4
 2
3
w
( )
ie
id
ge
1
8
1
8
=  (6)4 − (6)3 + 6(6)2  −  (2)4 − (2)3 + 6(2)2 
4



3
4
3
ev
= −42 23
-R
am
br
= ( −36) − 6 23
1
Pr
op
y
es
s
-C
Hence, the total area of the shaded regions = 6 23 + 42 23 = 49 3 units2.
ev
C
U
w
x = f(y)
ie
x
-R
s
es
am
br
ev
id
g
e
a
-C
R
A
op
b
y
ni
ve
rs
ity
y
ie
w
C
Area enclosed by a curve and the y-axis
O
6 x
-R
∫
2
2
C
op
U
R
Answer
y = x(x – 2)(x – 6)
y
O
ni
ev
ie
w
C
ve
rs
ity
op
Find the total area of the shaded regions.
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
ev
ie
am
br
id
KEY POINT 9.13
w
ge
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
If x = f( y ) is a function with x ù 0, then the area, A, bounded by the curve x = f( y ), the y-axis
-R
U
id
ev
br
-R
am
s
-C
1
1
=  2(4)2 − (4)3  −  2(0)2 − (0)3 

 

3
3
y
es
= 10 23
rs
ni
op
y
ve
EXERCISE 9F
ev
ie
w
C
ity
op
Pr
Area is 10 23 units2 .
1 Find the area of each shaded region.
b
y
ie
w
ge
y
y = 2x +
ev
br
id
a
C
U
R
s
x
O
Pr
4
es
-C
op
y
c
d
1
3
w
y
op
x
O
ev
ie
id
g
es
s
-R
br
am
-C
4
w
e
5
y = 4√x – 2x
C
U
ev
ie
y = x(x – 5)
O
x
y
ni
ve
rs
ity
y
R
3 –1
x2
-R
am
y = x3 – 8x2 + 16x
O
C
x
O
4
1
=  y2 − y3 
3
 2
 0
256
x = y(4 – y)
w
R
4
ge
2
ni
∫ x dy
=
∫ ( 4y − y ) d y
4
y
4
y
C
op
Answer
0
a
ie
ev
ie
w
C
Find the area of the shaded region.
0
4
b
ve
rs
ity
op
y
WORKED EXAMPLE 9.13
Area =
∫ x d y when x ù 0.
Pr
es
s
-C
and the lines y = a and y = b is given by the formula A =
Copyright Material - Review Only - Not for Redistribution
x
ve
rs
ity
w
y
ev
ie
am
br
id
ge
R1
A
O
B
x
Pr
es
s
-C
R2
y
The diagram shows the curve y = x( x − 2)( x − 4) that crosses the x-axis at the
points O(0, 0), A(2, 0) and B(4, 0).
ve
rs
ity
op
C
w
-R
2
C
U
ni
op
y
Chapter 9: Integration
y
C
op
U
R
ni
ev
ie
Show by integration that the area of the shaded region R1 is the same as the area
of the shaded region R2.
y = x( x − 3)( x + 1)
c
(
)
y = x x2 − 9
y = x(2 x − 1)( x + 2)
d
y = ( x − 1)( x + 1)( x − 4)
ev
-R
am
ie
b
br
id
a
w
ge
3 Sketch the curve and find the total area bounded by the curve and the x-axis for
each of these functions.
es
Pr
ity
y = 3 x , the x-axis and the lines x = 1 and x = 4
e
y=
f
y = 2 x + 3, the x-axis and the line x = 3
y
ve
d
ni
op
4
, the x-axis and the lines x = 1 and x = 9
x
C
U
w
R
ev
ie
w
c
257
rs
b
y = x 4 − 6x 2 + 9, the x-axis and the lines x = 0 and x = 1
5
y = 2 x + 2 , the x-axis and the lines x = 1 and x = 2
x
8
y = 5 + 3 , the x-axis and the lines x = 2 and x = 5
x
ge
C
op
y
a
s
-C
4 Sketch the curve and find the enclosed area for each of the following.
br
y = √2x + 1
es
y
6
x
C
U
O
op
ie
1
y
ni
ve
rs
w
C
ity
Pr
op
y
3
ev
-R
s
es
am
br
ev
ie
id
g
w
e
The diagram shows the curve y = 2 x + 1. The shaded region is bounded by the
curve, the y-axis and the line y = 3. Find the area of the shaded region.
-C
R
ie
-R
s
x = y2 + 1 , the y-axis and the lines y = −1 and y = 2
-C
b
y = x 3 , the y-axis and the lines y = 8 and y = 27
am
a
ev
id
5 Sketch the curve and find the enclosed area for each of the following.
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
ge
y = 2x2 + 1
ev
ie
am
br
id
7
w
y
O
x
ve
rs
ity
C
op
y
Pr
es
s
-C
-R
9
Find the area of the region bounded by the curve y = 2 x + 1, the line y = 9 and
the y-axis.
2
y
w
ev
ie
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
12
, the x-axis and the
x2
R
ni
C
op
8 a Find the area of the region enclosed by the curve y =
ge
U
lines x = 1 and x = 4.
br
ev
id
ie
w
b The line x = p divides the region in part a into two parts of equal area. Find
the value of p.
9
y
es
Pr
y
ity
op
rs
C
258
x
2
ve
ie
w
O
op
ev
C
id
w
y
s
-C
-R
am
br
ev
y = 2√ x
Pr
es
x+y=8
op
y
O
x
ni
ve
rs
ity
Find the shaded area enclosed by the curve y = 2 x , the line x + y = 8 and the x-axis.
y
11 The tangent to the curve y = 8x − x 2
at the point (2, 12) cuts the x-axis at the
point P.
(2, 12)
C
U
op
y
y = 8x – x2
id
g
w
e
a Find the coordinates of P.
ie
8 x
s
-R
ev
O
P
es
am
br
b Find the area of the shaded region.
-C
C
w
ie
ev
ie
ge
U
R
)
ni
(
y
d
x
x2 + 5 =
.
2
dx
x +5
b Use your result from part a to evaluate the area of the shaded region.
a Show that
10
R
x
√ x2 + 5
s
-C
-R
am
y=
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
C
y
ge
12
y = √ 2x + 1
Pr
es
s
The diagram shows the curve y = 2 x + 1 that intersects the x-axis at A.
The normal to the curve at B(4, 3) meets the x-axis at C. Find the area of the
shaded region.
y
y = f(x)
U
R
ni
y
ev
br
id
ie
w
ge
Q(7, 12)
s
x
es
O
Pr
The figure shows part of the curve y = f( x ). The points P(2, 4) and Q(7, 12) lie
7
on the curve. Given that
∫ y dx = 42, find the value of ∫ x dy.
4
op
C
U
R
ni
y
y
ve
rs
2
14
-R
am
br
ev
id
ie
w
ge
A (2, 8)
s
WEB
y = g(x)
x
Pr
es
O
on the curve. Given that
∫
6
ity
The figure shows part of the curve y = g(x ). The points A(2, 8) and B(6, 1) lie
y d x = 16 , find the value of
ni
ve
rs
2
∫ x d y.
y
-R
s
-C
am
br
ev
ie
id
g
w
e
C
U
op
ev
R
8
1
es
C
op
y
-C
B (6, 1)
ie
w
259
12
ity
C
op
y
-C
-R
am
P(2, 4)
w
ie
ev
PS
C
op
13
ve
rs
ity
ev
ie
PS
C x
O
A
w
C
op
y
-C
-R
am
br
id
ev
ie
B
w
U
ni
op
y
Chapter 9: Integration
Copyright Material - Review Only - Not for Redistribution
Try the following
resources on the
Underground
Mathematics website:
• What else do you
know?
• Slippery areas.
ve
rs
ity
w
ge
9.7 Area bounded by a curve and a line or by two curves
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
-R
am
br
id
ev
ie
The following example shows a possible method for finding the area enclosed by a curve
and a straight line.
Pr
es
s
-C
WORKED EXAMPLE 9.14
op
y
The diagram shows the curve y = − x 2 + 8x − 5 and the line y = x + 1
that intersect at the points (1, 2) and (6, 7).
y
O
U
ni
C
op
C
w
ev
ie
R
ie
6
id
Area = area under curve − area of trapezium
∫ ( −x + 8x − 5 ) dx − 2 × (2 + 7) × 5
ev
am
1
1
-R
2
br
=
1
w
ge
Answer
y = –x2 + 8x – 5
y=x+1
ve
rs
ity
Find the area of the shaded region.
y
6
s
-C
1
1
=  − x 3 + 4x 2 − 5x  − 22 2
 3
 1
5
ni
op
y
ve
There is an alternative method for finding the shaded area in Worked example 9.14.
ev
y = f(x)
ie
w
ge
id
C
U
R
y
A
am
a
b
y = g(x)
ev
br
O
x
-R
ie
w
rs
ity
= 20 6 units 2
C
260
Pr
op
y
es
1
1
3
2
1
=  − ( 6 ) + 4 ( 6 ) − 5(6)  −  − (1)3 + 4(1)2 − 5(1)  − 22 2
 3
  3

Pr
op
y
es
s
-C
If two functions, f( x ) and g( x ), intersect at x = a and x = b, then the area, A, enclosed
between the two curves is given by:
A=
a
a
-R
s
es
-C
am
br
ev
ie
id
g
w
e
C
U
R
b
∫ [ f(x ) − g(x ) ] dx
y
a
or
op
b
ity
b
∫ f(x ) dx − ∫ g(x ) dx
ni
ve
rs
A=
ev
ie
w
C
KEY POINT 9.14
Copyright Material - Review Only - Not for Redistribution
6
x
ve
rs
ity
C
U
ni
op
y
Chapter 9: Integration
op
y
Pr
es
s
-C
y=x+1
C
6
x
C
op
6
y
Using f( x ) = −x 2 + 8x − 5 and g( x ) = x + 1 gives:
ni
w
ev
ie
1
ve
rs
ity
O
6
2
am
1
w
1
br
1
6
6
ie
id
2
ev
1
-R
1
6
ge
f( x ) d x −
U
∫
∫ g(x ) dx
=
∫ ( −x + 8x − 5 ) dx − ∫ (x + 1) dx
=
∫ ( − x + 7x − 6 ) d x
Area =
R
w
y = –x2 + 8x – 5
ev
ie
y
-R
am
br
id
ge
So for the area enclosed by y = −x 2 + 8x − 5 and y = x + 1:
6
es
s
-C
1
7
=  − x 3 + x 2 − 6x 
 3
 1
2
261
rs
ity
= 20 56 units2
This alternative method is the easiest method to use in the next example.
y
ge
C
U
WORKED EXAMPLE 9.15
R
op
ni
ev
ve
ie
w
C
op
Pr
y
7
1
7
1
=  − (6)3 + (6)2 − 6(6)  −  − (1)3 + (1)2 − 6(1) 
 3
  3

2
2
y
br
ev
id
ie
w
The diagram shows the curve y = x 2 − 6x − 2 and the line y = 2 x − 9,
which intersect when x = 1 and x = 7.
-R
2
2
7
ni
ve
rs
1
1
=  − x 3 + 4x 2 − 7 x 
 3

1
y = x2 – 6x – 2
y
1
s
1
7
w
e
ev
ie
id
g
es
s
-R
br
am
-C
C
U
op
1
1
=  − (7)3 + 4(7)2 − 7(7)  −  − (1)3 + 4(1)2 − 7(1) 
 3
  3

= 36 units2
7
es
7
1
O
ity
op
y
C
w
ie
ev
R
7
∫ (2x − 9) dx − ∫ ( x − 6x − 2 ) dx
=
∫ ( − x + 8x − 7 ) d x
Area =
y = 2x – 9
Pr
-C
Answer
am
Find the area of the shaded region.
Copyright Material - Review Only - Not for Redistribution
x
ve
rs
ity
Pr
es
s
y
4
O
x
Find the area of the region bounded by the curve y = 5 + 6x − x 2 , the line x = 4 and the line y = 5.
ve
rs
ity
op
C
y
y = 2x – 3
B
w
ge
U
R
ni
y
y = (x – 3)2
C
op
w
y = 5 + 6x – x2
5
2
ev
ie
-R
y
-C
1
ev
ie
am
br
id
EXERCISE 9G
w
ge
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
ie
ev
id
A
x
-R
am
br
O
2
3
Pr
y = –x2 + 11x – 18
ity
op
y
C
ev
ve
ie
w
rs
A
x
ge
C
U
R
ni
op
O
y
262
es
y
s
-C
The diagram shows the curve y = ( x − 3 ) and the line y = 2 x − 3 that intersect at points A and B. Find the
area of the shaded region.
2x + y = 12
br
ev
id
ie
w
B
-R
am
The diagram shows the curve y = −x 2 + 11x − 18 and the line 2 x + y = 12 . Find the area of the shaded region.
c
y = x 2 − 4x + 4 and 2 x + y = 12
Pr
y = −x 2 + 12 x − 20 and y = 2 x + 1
ni
ve
rs
ity
b
es
y = x 2 − 3 and y = 6
op
-R
s
es
-C
am
br
ev
ie
id
g
w
e
C
U
R
y
5 Sketch the curves y = x 2 and y = x(2 − x ) and find the area enclosed between the two curves.
ev
ie
w
C
op
y
a
s
-C
4 Sketch the following curves and lines and find the area enclosed between their graphs.
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
y
O
–4
1
x + 2 meeting at the points ( −4, 0) and (0, 2).
2
U
R
Find the area of the shaded region.
ge
y = √ 2x + 3
es
s
-C
-R
am
br
P(3, 3)
ie
id
w
y
ev
7
x
y
x + 4 and the line y =
ni
The diagram shows the curve y =
1 x +2
2
y=
C
op
ev
ie
w
C
ve
rs
ity
op
y
Pr
es
s
-C
-R
y = √x + 4
ev
ie
am
br
id
w
ge
6
C
U
ni
op
y
Chapter 9: Integration
Pr
op
y
R
x
w
rs
O
263
ity
C
Q
y
ev
ve
ie
The curve y = 2 x + 3 meets the y-axis at the point Q.
U
R
ni
op
The tangent at the point P(3, 3) to this curve meets the y-axis at the point R.
w
ge
C
a Find the equation of the tangent to the curve at P.
y
-R
am
8
O
Q
R
y
ni
ve
rs
C
ity
Pr
op
y
y = 10 + 9x – x2
es
s
-C
P(6, 28)
w
x
C
U
op
The diagram shows the curve y = 10 + 9x − x 2. Points P(6, 28) and Q(10, 0) lie on the curve. The tangent
at P intersects the x-axis at R.
w
id
g
e
a Find the equation of the tangent to the curve at P.
-R
s
es
am
br
ev
ie
b Find the area of the shaded region.
-C
ie
ev
R
ev
br
id
ie
b Find the exact value of the area of the shaded region PQR.
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
w
y
ge
y
Pr
es
s
-C
-R
am
br
id
Q
ev
ie
9
C
ve
rs
ity
op
y = 4x – x3
w
ev
ie
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
P
x
O
U
R
ni
C
op
y
The diagram shows the curve y = 4x − x 3.
The point P has coordinates (2, 0) and the point Q has coordinates ( − 4, 48).
ie
w
ge
a Find the equation of the tangent to the curve at P.
y
-R
am
10
ev
br
id
b Find the area of the shaded region.
es
s
-C
Pr
y
ity
op
O
op
y
x
ni
ev
ve
ie
w
rs
C
264
P(9, 4)
y = 5 – √ 10 – x
C
U
R
The diagram shows part of the curve y = 5 − 10 − x and the tangent to the curve at P(9, 4).
w
ge
a Find the equation of the tangent to the curve at P.
-R
am
br
ev
id
ie
b Find the area of the shaded region. Give your answer correct to 3 significant figures.
9.8 Improper integrals
Pr
op
y
es
s
-C
In this section, we will consider what happens if some part of a definite integral becomes
infinite. These are known as improper integrals, and we will look at two different types.
These are definite integrals that have either one limit infinite or both limits infinite.
∞
−2
1
1
d x.
Examples of these are
2 d x and
3
x
x
1
−∞
∫
op
C
∞
w
∫ f(x ) dx by replacing the infinite limit with a finite
a
-R
s
es
-C
am
br
value, X , and then taking the limit as X → ∞, provided the limit exists.
ie
id
g
We can evaluate integrals of the form
ev
e
U
KEY POINT 9.15
y
ni
ve
rs
∫
w
ie
ev
R
ity
C
Type 1
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
ev
ie
am
br
id
b
∫ f(x ) dx by replacing the infinite limit with a finite value, X,
We can evaluate integrals of the form
−∞
-R
KEY POINT 9.16
w
ge
C
U
ni
op
y
Chapter 9: Integration
Pr
es
s
-C
and then taking the limit as X → −∞, provided the limit exists.
∞
1
∫ x d x has a finite value and find this value.
w
X
1
R
2
1
ni
X
∫ x dx = ∫ x dx
−2
1
X
w
ge
=  − x −1 
Write the integral with an upper limit X .
U
ev
ie
Answer
y
2
1
C
op
Show that the improper integral
ve
rs
ity
C
op
y
WORKED EXAMPLE 9.16
1
∞
1
C
w
Pr
∞
Hence, the improper integral
265
ity
2
1
1
∫ x d x has a finite value of 1.
2
-C
y
w
C
am
Show that as p → −∞, the shaded area tends to a finite
value and find this value.
Pr
3
0
1
(1 – x)3
es
1
p
−3
ni
ve
rs
p
O
p
ity
op
y
0
op
C
U
w
ie
id
g
es
s
-R
br
ev
1
1
−
2 2(1 − p)2
am
=
e
0


1
=
2 
 2(1 − x )  p
y


1
=
(1 − x )−2 
 ( −2)( −1)
p
-C
C
w
ie
ev
R
0
∫ (1 − x ) dx
=
∫ (1 − x ) dx
Area =
y=
s
Answer
y
ie
1
.
(1 − x )3
ev
br
id
The diagram shows part of the curve y =
-R
ge
U
WORKED EXAMPLE 9.17
R
op
ni
ev
ve
ie
1
rs
y
∫ x dx = 1 − 0 = 1
op
∴
1
→0
X
s
As X → ∞,
1
X
es
-C
= 1−
-R
am
br
ev
id
ie
1
1
=−  −− 
 X   1
Copyright Material - Review Only - Not for Redistribution
x
ve
rs
ity
1
2 → 0.
2 (1 − p )
am
br
id
As p → −∞ ,
Pr
es
s
-C
y
Type 2
1
.
2
-R
Hence, as p → −∞, the shaded area tends to a finite value of
ev
ie
w
ge
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
1
−1
2
1
1
2
∫ x d x is an improper integral because x tends to infinity as x → 0 and it is
2
0
1
C
op
ni
However,
1
∫ x dx is an invalid integral because x is not defined when x = 0.
2
U
R
ev
ie
w
1
For example,
y
C
ve
rs
ity
op
These are integrals where the function to be integrated approaches an infinite value
(or approaches ± infinity) at either or both end points in the interval (of integration).
w
ge
well-defined everywhere else in the interval of integration.
am
br
ev
id
ie
For this section we will consider only those improper integrals where the function is not
defined at one end of the interval.
-C
-R
KEY POINT 9.17
b
es
s
∫ f(x ) dx where f(x ) is not defined when x = a can be
We can evaluate integrals of the form
a
rs
op
y
ve
KEY POINT 9.18
ni
ev
ie
w
C
266
ity
op
Pr
y
evaluated by replacing the limit a with an X and then taking the limit as X → a, provided the limit
exists.
b
∫ f(x ) dx where f(x ) is not defined when x = b by
C
U
R
We can evaluate integrals of the form
a
br
ev
id
ie
w
ge
replacing the limit b with an X and then taking the limit as X → b, provided the limit exists.
0
2
Pr
op
y
-R
5
∫ x dx.
s
-C
2
Find the value, if it exists, of
es
am
WORKED EXAMPLE 9.18
Answer
5
is not defined when x = 0.
x2
2
X
ev
ie
id
g
5 5
−
X 2
es
s
-R
br
=
w
e
5
5
=−  −− 
 2  X 
C
=  −5x −1 
Write the integral with a lower limit X .
y
−2
X
op
∫ 5x d x
am
∫
2
ni
ve
rs
5
2 dx =
x
X
U
2
-C
R
ev
ie
w
C
ity
The function f( x ) =
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
w
ge
C
U
ni
op
y
Chapter 9: Integration
am
br
id
ev
ie
5
tends to infinity.
X
As X → 0 ,
2
5
-R
∫ x dx is undefined.
Hence,
2
Pr
es
s
-C
0
ve
rs
ity
p
y
ge
Answer
3
ie
id
1
2
am
−
0
p
) (
s
es
y
ve
p
267
3
dx → 6 2 .
2−x
ni
U
ie
ev
∫
As p → 2,
Pr
)
=6 2 −6 2− p
ge
0
br
ev
id
ie
w
Hence, as p → 2 the shaded area tends to a finite value of 6 2.
-R
am
EXERCISE 9H
x
op
w
= −6 2 − p − −6 2
2
ity
(
p
0
p
rs
C
op
y
-C
1 

3
=
(2 − x ) 2 
1
   ( −1)


 2
0
=  −6 2 − x 
O
-R
br
p
ev
0
3
√2 – x
w
∫ 2 − x dx
=
∫ 3(2 − x ) dx
Area =
y=
C
op
ni
U
R
find this value.
R
y
3
.
2−x
Show that as p → 2 the shaded area tends to a finite value and
The diagram shows part of the curve y =
C
ev
ie
w
C
op
y
WORKED EXAMPLE 9.19
∫
3
0
3
dx
3−x
e
s
∫
25
∫
∞
5
dx
x
f
1
dx
(1 − x )2
i
0
2
-R
s
es
-C
am
br
ev
ie
id
g
w
e
C
U
h
y
4
g
∫
op
Pr
∫ xx
dx
ni
ve
rs
ie
4
ity
∞
d
R
ev
∫
es
∫
w
C
op
y
-C
1 Show that each of the following improper integrals has a finite value and, in each case, find this value.
−2
∞
∞
2
4
10
dx
dx
dx
a
b
c
2
5
3
x
x
−∞ x
1
4
Copyright Material - Review Only - Not for Redistribution
8
4
4
dx
x−4
∞
2
∫

4
∫  x + ( x + 2)  d x
1
2
3
ve
rs
ity
20
(2x + 5)2
p
x
20
.
(2 x + 5)2
Show that as p → ∞, the shaded area tends to the value 2.
w
ev
ie
The diagram shows part of the curve y =
ve
rs
ity
C
op
y
O
Pr
es
s
-C
y=
-R
am
br
id
ev
ie
w
ge
y
2
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
y
C
op
U
4
∫ xx
dx
e
∫
2
5
dx
1 (2 x − 1)2
2
9
c
25
s
es
y
y = x2
ity
rs
op
ni
O
ni
ve
rs
C
ity
Pr
op
y
es
s
-C
-R
am
br
ev
id
ie
w
ge
y = x2
y
es
s
-R
br
ev
ie
id
g
w
e
C
U
R
op
w
We can approximate the volume, V , of the solid by a series of cylindrical discs of thickness
δx (corresponding to a small increase in x) and radius y (corresponding to the height of the
function).
am
x
5
y
C
U
When this area is rotated about the x-axis through 360° a solid of revolution is formed.
-C
2
y
ve
O
The volume of this solid is called a volume of revolution.
ie
2
Pr
y
op
C
w
ie
ev
R
1 

∫  x + x  dx
0
Consider the area bounded by the curve y = x 2, the x-axis, and the lines x = 2 and x = 5.
ev
2
-R
-C
9.9 Volumes of revolution
268
12
∫ x x dx
0
f
ie
2
dx
x+4
w
0
am
5
∞
ev
d
b
ge
∫
∞
4
6
dx
x
id
∫
∞
br
R
a
ni
3 Show that none of the following improper integrals exists.
Copyright Material - Review Only - Not for Redistribution
2
5
x
ve
rs
ity
C
U
ni
op
y
Chapter 9: Integration
y
5
w
ev
ie
As δx → 0, then V →
y = x2
∫ πy dx.
2
-R
2
am
br
id
∑ πy δx.
ge
The volume of each cylindrical disc is πy2δ x. An approximation for V is then
2
op
y
KEY POINT 9.19
x
O
Pr
es
s
-C
This leads to a general formula:
the boundary values x = a and x = b is given by the formula V =
2
C
op
y
a
w
ge
U
WORKED EXAMPLE 9.20
R
b
∫ πy dx.
ni
ev
ie
w
C
ve
rs
ity
The volume, V, obtained when the function y = f( x ) is rotated through 360° about the x-axis between
br
op
y
1
 9  dx
1  3x + 2 
2
C
O
ity
−2
1
2
y
ev
ve
ie
w
rs

 81
=π
(3x + 2)−1 
1
 3( −1)
2
ge
C
U
R
ni
op
 −27 
=π

 (3x + 2)  1
-R
s
-C
am
br
ev
id
ie
w
 −27   −27  
= π 
−
  8   5  
81π
units3
=
40
Pr
op
y
es
Sometimes a curve is rotated about the y-axis. In this case the general rule is:
ity
2
a
-R
s
es
-C
am
br
ev
ie
id
g
w
e
C
U
R
b
∫ πx dy.
y
the boundary values y = a and y = b is given by the formula V =
op
ni
ve
rs
The volume, V, obtained when the function x = f( y ) is rotated through 360° about the y-axis between
ev
ie
w
C
KEY POINT 9.20
9
3x + 2
s
2
2
∫
= π 81(3x + 2) d x
∫
y2 d x = π
es
∫
2
y=
Pr
Volume = π
-R
am
-C
Answer
y
ev
id
ie
Find the volume obtained when the shaded region is rotated through 360° about
the x-axis.
Copyright Material - Review Only - Not for Redistribution
1
2
x
269
ve
rs
ity
ev
ie
w
ge
am
br
id
WORKED EXAMPLE 9.21
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
y
Find the volume obtained when the shaded region is rotated through 360°
about the y-axis.
-R
5
ve
rs
ity
y
Using the given y = x 2.
id
ie
w
ge
 y2 
= π

 2 2
ity
op
Pr
y
es
s
-C
-R
am
br
ev
 25   4  
= π 
−

 2   2  
21π
units3
=
2
rs
Find the volume of the solid obtained when the shaded region is rotated through
360° about the x-axis.
op
2x2 + y2 = 9
P (2, 1)
ie
w
ge
O
ev
id
br
x
-R
am
Answer
y
C
U
R
ni
ev
ve
ie
w
C
WORKED EXAMPLE 9.22
y
270
x
C
op
2
O
U
2
R
5
∫ y dy
x 2 dy = π
2
ni
∫
5
y = x2
Pr
es
s
-C
y
op
C
w
ev
ie
Answer
Volume = π
5
1
es
s
-C
When the shaded region is rotated about the x-axis, a solid with a cylindrical hole
is formed.
2
2
Pr
op
y
The radius of the cylindrical hole is 1 unit and the length of the hole is 2 units.
∫ y dx − volume of cylinder
= π ( 9 − 2x ) d x − π × r × h
∫
ity
2
2
0
2
y
ie
w
0
2
ni
ve
rs
C
Volume of solid = π
ev
2
op
C
U
R
2
= π  9x − x 3  − π × 12 × 2
3

0
-R
s
es
-C
am
br
ev
ie
id
g
w
e
16 


= π   18 −
− (0 − 0)  − 2 π

3 


32 π
units3
=
3
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
ev
ie
am
br
id
EXERCISE 9I
w
ge
C
U
ni
op
y
Chapter 9: Integration
1 Find the volume obtained when the shaded region is rotated through 360° about the x-axis.
-R
y
y=
ve
rs
ity
c
w
2
x √x
ev
1
-R
br
am
y
5
3–x
y=
x
4
O
–1
x
1
s
-C
O
x
2
ie
ge
U
R
id
y=
1
d
ni
y
O
x
2
4
2x – 1
C
op
1
y
y
y
op
C
ev
ie
w
O
b
Pr
es
s
2
y = x2 + x
-C
a
es
a
271
3
ity
11
y
b
Pr
y
w
rs
C
op
y
2 Find the volume obtained when the shaded region is rotated through 360° about the y-axis.
y = √2x + 1
y
C
O
x
br
ev
ie
x
id
O
w
ge
2
op
1
U
R
ni
ev
ve
ie
y = x2 + 2
y
a
, where a . 0. The volume
x
obtained when the shaded region is rotated through 360° about the x-axis
is 18 π. Find the value of a.
-R
ity
x
2
y
op
C
U
w
e
y = √x3 + 4x2 + 3x + 2
ev
ie
id
g
es
s
-R
br
am
-C
1
y
ni
ve
rs
C
w
ie
O
4 The diagram shows part of the curve y = x 3 + 4x 2 + 3x + 2 . Find
the volume obtained when the shaded region is rotated through 360°
about the x-axis.
R
ev
a
y= x
Pr
op
y
es
s
-C
am
3 The diagram shows part of the curve y =
Copyright Material - Review Only - Not for Redistribution
–2
O
1
x
ve
rs
ity
y
am
br
id
ev
ie
w
ge
5 The diagram shows part of the line 3x + 8 y = 24 . Rotating the
shaded region through 360° about the x-axis would give a cone of
base radius 3 and perpendicular height 8.
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
-R
Find the volume of the cone using:
3x + 8y = 24
x
8
O
Pr
es
s
-C
a integration
3
y
b the formula for the volume of a cone.
b Find the volume of the solid formed when the enclosed region bounded by the curve, the x-axis and the
y-axis is rotated through 360° about the x-axis.
ve
rs
ity
ev
ie
w
C
op
6 a Sketch the graph of y = ( x − 2)2 .
y
ni
C
op
y
7 The diagram shows part of the curve y = 5 x − x.
U
R
The curve meets the x-axis at O and P.
br
9
− 1 that intercepts the
y2
y-axis at the point P. The shaded region is bounded by the curve,
the y-axis and the line y = 1 .
-R
P
es
y
b Find the volume obtained when the shaded region is rotated
through 360° about the y-axis.
rs
w
y
2
.
x
The line y = 7 intersects the curve at the points P and Q.
ni
op
y
ve
ie
ev
ie
w
ge
2
. The shaded area is
2x + 1
rotated through 360° about the x-axis between x = 0 and x = p.
y
br
ev
id
O
es
y=
Pr
p
x
y
y = √ 25 – x2
y
ity
ni
ve
rs
25 − x 2 . The point
op
C
w
ie
ev
O
es
s
-R
br
P(4, 3)
3
U
id
g
e
b Find the volume obtained when the shaded region is rotated
through 360° about the x-axis.
am
2
2x + 1
O
a Find the volume obtained when the shaded region is rotated
through 360° about the y-axis.
-C
w
C
op
y
Show that as p → ∞, the volume approaches the value 2 π.
11 The diagram shows part of the curve y =
P (4, 3) lies on the curve.
x
-R
s
-C
am
10 The diagram shows part of the curve y =
ie
y=7
Q
C
U
R
2
y = 3x + x
P
a Find the coordinates of P and Q.
ev
x
O
9 The diagram shows part of the curve y = 3x +
b Find the volume obtained when the shaded region is rotated
through 360° about the x-axis.
R
9
–1
y2
1
Pr
a Find the coordinates of P.
x=
ity
op
C
272
y
s
-C
am
8 The diagram shows part of the curve x =
x
P
O
ev
id
ie
b Find the volume obtained when the shaded region is rotated
through 360° about the x-axis.
w
ge
a Find the coordinates of P.
y = 5√x – x
Copyright Material - Review Only - Not for Redistribution
4
x
ve
rs
ity
C
U
ni
op
y
Chapter 9: Integration
am
br
id
ev
ie
w
ge
12 The diagram shows the curve y = 4 − x and the line x + 2 y = 4 that
intersect at the points (4, 0) and (0, 2).
y
2
-R
a Find the volume obtained when the shaded region is rotated through
360° about the x-axis.
op
y
Pr
es
s
-C
b Find the volume obtained when the shaded region is rotated through
360° about the y-axis.
x + 2y = 4
ve
rs
ity
C
w
ev
ie
y
x2 + y2 = 100
w
ge
b Find the volume of water in the bowl.
–8
op
Pr
y
es
s
-C
-R
am
br
ev
14 Use integration to prove that the volume, V cm , of a sphere with radius
4
r cm is given by the formula V = πr 3.
3
273
ity
rs
C
w
y
ve
∫ f(x ) dx = F(x ) + c.
op
d
[ F( x ) ] = f( x ), then
dx
If
∫ x dx = n + 1 x
●
∫ (ax + b) dx = a( n + 1) (ax + b)
+ c (where c is a constant and n ≠ −1)
ev
n +1
am
1
n +1
+ c ( n ≠ −1 and a ≠ 0)
s
-C
n
-R
br
n
id
●
ie
w
ge
1
C
U
R
●
ni
ev
ie
Integration as the reverse of differentiation
Integration formulae
y
w
∫ f(x ) dx, where k is a constant.
ev
a
es
s
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br
a
C
b
a
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k f( x ) d x = k
op
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b
a
e
U
∫ f(x ) dx = [ F(x ) ] = F(b) − F(a ).
id
g
∫
b
b
f( x ) dx = F( x ) + c, then
am
●
∫
If
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Rules for definite integration
●
Pr
∫ k f(x ) dx = k ∫ f(x ) dx, where k is a constant
∫ [ f(x ) ± g(x ) ] dx = ∫ f(x ) dx ± ∫ g(x ) dx
op
y
●
w
C
●
es
Rules for indefinite integration
ev
x
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id
3
Checklist of learning and understanding
R
x
y
C
op
ni
U
R
The bowl is filled with water to a depth of 3 cm.
P
4
O
13 A mathematical model for the inside of a bowl is obtained by rotating
the curve x 2 + y2 = 100 through 360° about the y-axis between y = −8
and y = 0. Each unit of x and y represents 1cm.
a Find the volume of the bowl.
y =√ 4 – x
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ve
rs
ity
b
a
a
a
Pr
es
s
b
y
ve
rs
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y
C
op
Area under a curve
ev
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w
y = f(x)
w
x
id
b
ie
a
O
ev
The area, A, bounded by the curve y = f( x ), the x-axis and the lines x = a and x = b is given
by the formula:
∫ y dx when y ù 0
A=
b
∫ f(x ) dx when f(x ) ù 0 ).
a
es
s
-C
a
(or A =
-R
b
am
br
●
ge
U
R
ni
A
y
a
-C
∫ f(x ) dx = − ∫ f(x ) dx
C
op
b
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∫ [ f(x ) ± g(x ) ] dx = ∫ f(x ) dx ± ∫ g(x ) dx
a
●
b
ev
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am
br
id
b
●
w
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C
U
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Cambridge International AS & A Level Mathematics: Pure Mathematics 1
Pr
b
C
274
ity
op
y
y
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w
A
ve
a
y
C
U
x
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b
∫ f( y) dy when f( y) ù 0).
a
es
s
-C
Pr
ity
ni
ve
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A
a
b
y = g(x)
x
y
O
-R
s
es
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am
br
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id
g
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C
U
R
ev
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w
C
op
y
y = f(x)
op
a
y
(or A =
ev
id
b
∫ x dy when x ù 0
-R
A=
w
ge
The area, A, bounded by the curve x = f( y ), the y-axis and the lines y = a and y = b is given
by the formula:
br
●
am
R
O
op
ni
ev
x = f(y)
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ve
rs
ity
ev
ie
am
br
id
The area, A, enclosed between y = f( x ) and y = g( x ) is given by the formula:
A=
b
-R
●
w
ge
C
U
ni
op
y
Chapter 9: Integration
∫ [ f(x ) − g(x ) ] dx
Pr
es
s
-C
a
op
y
where a and b are the x-coordinates of the points of intersection of the
functions f and g.
ve
rs
ity
∞
●
w
C
Improper integrals
Integrals of the form
∫ f(x ) dx can be evaluated by replacing the infinite limit with a finite
a
Integrals of the form
y
∫ f(x ) dx can be evaluated by replacing the infinite limit with a finite
U
R
b
●
C
op
ni
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value, X , and then taking the limit as X → ∞, provided the limit exists.
ie
ev
id
b
∫ f(x ) dx where f(x ) is not defined when x = a can be evaluated by
-R
br
Integrals of the form
am
●
w
ge
−∞
value, X , and then taking the limit as X → −∞, provided the limit exists.
a
Pr
∫ f(x ) dx where f(x ) is not defined when x = b can be evaluated by
op
a
rs
ity
replacing the limit b with an X and then taking the limit as X → b, provided the limit exists.
U
R
between the boundary values x = a and x = b is given by the formula V =
∫ πy dx.
2
w
ge
a
ev
id
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The volume, V , obtained when the function x = f( y ) is rotated through 360° about the y-axis
br
●
b
op
y
The volume, V , obtained when the function y = f( x ) is rotated through 360° about the x-axis
ni
ev
●
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Volume of revolution
C
C
w
es
b
Integrals of the form
y
●
s
-C
replacing the limit a with an X and then taking the limit as X → a , provided the limit exists.
b
∫ πx d y.
2
a
y
op
-R
s
es
-C
am
br
ev
ie
id
g
w
e
C
U
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ev
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w
ni
ve
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C
ity
Pr
op
y
es
s
-C
-R
am
between the boundary values y = a and y = b is given by the formula V =
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275
ve
rs
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ge
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
am
br
id
ev
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w
END-OF-CHAPTER REVIEW EXERCISE 9
The function f is such that f ′( x ) = 12 x 3 + 10 x and f ( −1) = 1.
1
2
∫  5x − x2  dx.
4
dy
6
=
− 5x and the point (3, 5.5) lies on the curve. Find the equation of the curve.
dx x 2
8
3
− 3 and that f(2) = 3. Find f( x ).
A curve has equation y = f( x ). It is given that f ′( x ) =
x+2 x
A curve is such that
ve
rs
ity
5
y
w
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1
ie
O
br
id
[5]
C
op
x = 62 + 1
y
3
U
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ni
y
[4]
x
ev
ev
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w
C
op
3
[3]
Pr
es
s
-C
Find
y
2
[3]
-R
Find f( x ).
6
+ 1. The shaded region is bounded by the curve, the y-axis, and
y2
the lines y = 1 and y = 3. Find the volume, in terms of π, when this shaded region is rotated through 360°
about the y-axis.
[5]
y
es
s
-C
-R
am
The diagram shows part of the curve x =
Pr
ity
a
State the value of x for which f( x ) has a stationary value.
[1]
b
Find an expression for f( x ) in terms of x.
[4]
y
C
op
y = 6x – x2
U
R
ni
7
y
rs
C
w
ie
ev
A function is defined for x ∈  and is such that f ′( x ) = 6x − 6. The range of the function is given
by f( x ) ù 5.
ve
op
6
276
ev
br
id
ie
w
ge
y=5
x
-R
am
O
s
a
Sketch the curve y = ( x − 3)2 + 2.
b
The region enclosed by the curve, the x-axis, the y-axis, the line x = 3 is rotated through 360° about the
x-axis. Find the volume obtained, giving your answer in terms of π.
[6]
op
y
ni
ve
rs
ity
Pr
[1]
-R
s
es
-C
am
br
ev
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id
g
w
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C
U
R
ev
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w
C
8
[6]
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q4 June 2010
es
op
y
-C
The diagram shows the curve y = 6x − x 2 and the line y = 5. Find the area of the shaded region.
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
y
-R
O
Pr
es
s
U
C
ev
id
s
es
O
A
x
Pr
y
277
ity
rs
op
C
w
Find the coordinates of B and C.
[2]
ve
ie
-R
br
am
-C
B
y
op
ni
w
U
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ie
ev
2
√x + 1
y=1
Pr
op
y
es
s
-C
am
y=
-R
br
id
y
x
ity
C
ni
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rs
2
.
x +1
2
 4
op

2
-R
br
ev
iii The shaded region is rotated through 360° about the y-axis. Find the exact value of the volume of
revolution obtained.
[5]
[5]
s
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q11 June 2012
es
am
[1]
ie
1
w
e
∫  y − 1 d y. Hence find the area of the shaded region.
id
g
ii Find
2
4
can be written in the form x = 2 − 1.
x +1
y
C
U
Show that the equation y =
y
The diagram shows the line y = 1 and part of the curve y =
-C
[5]
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q10 November 2011
O
i
[4]
C
iii Find the volume obtained when the shaded region is rotated through 360° about the y-axis.
w
ie
y = √ 1 + 2x
ie
ge
y
ii Find the equation of the normal to the curve at C.
ev
[6]
The diagram shows the curve y = 1 + 2 x meeting the x-axis at A and the y-axis at B.
The y -coordinate of the point C on the curve is 3.
i
R
[2]
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q8 November 2012
w
R
ni
C
op
ii Find, showing all necessary working, the area of the shaded region.
y
ve
rs
ity
The diagram shows the curve y2 = 2 x − 1 and the straight line 3 y = 2 x − 1.
1
The curve and straight line intersect at x =
and x = a , where a is a constant.
2
i Show that a = 5.
11
ev
x
2
10
R
a
1
–
y
op
C
w
ev
ie
3y = 2x – 1
y2 = 2x – 1
-C
9
ev
ie
am
br
id
w
ge
C
U
ni
op
y
Chapter 9: Integration
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ve
rs
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ev
ie
am
br
id
w
ge
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
1
−
1
12 A curve has equation y = f( x ) and is such that f ′( x ) = 3x 2 + 3x 2 − 10.
Pr
es
s
[3]
iii It is given that the curve y = f( x ) passes through the point (4, −7). Find f( x ).
[4]
ve
rs
ity
y
op
[4]
ii Find f ′′( x ) and hence, or otherwise, determine the nature of each stationary point.
C
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q9 June 2013
y
x=4
y=
ev
br
id
P
x
8
. The curve intersects the y-axis at A(0, 4) . The normal to
3x + 4
s
-C
-R
am
O
The diagram shows part of the curve y =
8
√ 3x + 4
w
ge
Q
C
op
y
B
A
(0, 4)
U
R
ni
ev
ie
13
ie
w
-R
1
By using the substitution u = x 2 , or otherwise, find the values of x for which the curve y = f( x ) has
stationary points.
-C
i
C
es
i
Find the coordinates of B.
[5]
Pr
ii Show, with all necessary working, that the areas of the regions P and Q are equal.
[6]
y
U
C
w
ge
y = (3 – 2x)3
O
x
es
s
-C
ie
-R
am
br
id
( 12 , 8)
ev
R
ni
ev
14
y
ve
ie
w
rs
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q10 June 2015
op
278
ity
op
y
the curve at A intersects the line x = 4 at the point B.
Pr
ity
ii Find the area of the shaded region.
[5]
[6]
ni
ve
rs
w
y
op
-R
s
-C
am
br
ev
ie
id
g
w
e
C
U
R
ev
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Cambridge International AS & A Level Mathematics 9709 Paper 11 Q10 November 2013
es
C
op
y
1 
The diagram shows the curve y = (3 − 2 x )3 and the tangent to the curve at the point  , 8  .
2

i Find the equation of this tangent, giving your answer in the form y = mx + c.
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ve
rs
ity
-R
y
y
Pr
es
s
-C
15
ev
ie
am
br
id
w
ge
C
U
ni
op
y
Chapter 9: Integration
Q (2, 3)
1
2
y = 1 x2 + 1
2
C
op
y
P (0, 1)
U
and y =
br
ev
id
The diagram shows parts of the curves
x
1
y = (4x + 1) 2
w
ge
O
1 2
x + 1 intersecting at points P (0, 1) and
2
ie
R
ni
ev
ie
w
C
ve
rs
ity
op
y = (4x + 1)
α
-R
Find α , giving your answer in degrees correct to 3 significant figures.
s
-C
i
am
Q(2, 3) . The angle between the tangents to the curves at Q is α .
ii Find by integration the area of the shaded region.
[6]
es
[6]
279
y
op
y
op
-R
s
es
-C
am
br
ev
ie
id
g
w
e
C
U
R
ev
ie
w
ni
ve
rs
C
ity
Pr
op
y
es
s
-C
-R
am
br
ev
id
ie
w
ge
C
U
R
ni
ev
ve
ie
w
rs
C
ity
op
Pr
y
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q11 November 2014
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
ge
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
ev
ie
am
br
id
1
dy
= 2 x 2 − 3. Given that the curve passes through the point ( −3, −2) , find the equation
dx
of the curve.
[4]
1
−
dy
= 2 − 8(3x + 4) 2 .
A curve is such that
dx
i A point P moves along the curve in such a way that the x-coordinate is increasing at a constant rate of
0.3 units per second. Find the rate of change of the y-coordinate as P crosses the y-axis.
[2]
4
The curve intersects the y-axis where y = .
3
ii Find the equation of the curve.
[4]
Pr
es
s
-C
-R
A curve is such that
y
ve
rs
ity
ev
ie
w
C
op
y
2
w
CROSS-TOPIC REVIEW EXERCISE 3
3
1
dy
= 3x 2 − 6 and the point (9, 2) lies on the curve.
dx
i Find the equation of the curve.
[4]
ii Find the x-coordinate of the stationary point on the curve and determine the nature of the stationary
point.
[3]
U
R
ni
C
op
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q4 June 2016
s
-C
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q6 June 2010
es
i Find the equation of the curve.
[4]
ity
op
1
ii Find the set of values of x for which the gradient of the curve is less than .
3
w
ie
y
y
C
U
w
ge
y = (2x – 1)2
y2 = 1 – 2x
ity
Pr
op
y
ni
ve
rs
C
U
i State the coordinates of A.
w
ie
ev
-R
s
es
am
[1]
[6]
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q7 November 2016
br
id
g
e
C
ii Find, showing all necessary working, the area of the shaded region.
op
y
The diagram shows parts of the curves y = (2 x − 1)2 and y2 = 1 − 2 x , intersecting at points A and B.
-C
w
ie
x
A
es
O
s
-C
-R
am
br
ev
id
B
ie
5
op
ni
ev
ve
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q7 June 2011
R
ev
R
[3]
rs
C
280
dy
3
1
=
and the point  1,  lies on the curve.
 2
dx (1 + 2 x )2
Pr
A curve is such that
y
4
-R
am
br
ev
id
ie
w
ge
A curve is such that
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
ev
ie
am
br
id
1
Pr
es
s
U
id
-R
am
br
ev
2r cm
The diagram shows a metal plate. The plate has a perimeter of 50 cm and consists of a rectangle of width
2 r cm and height x cm, and a semicircle of radius r cm.
1
[4]
a Show that the area, A cm 2, of the plate is given by A = 50 r − 2 r 2 − πr 2.
2
Given that x and r can vary:
50
[4]
b show that A has a stationary value when r =
4+π
c find this stationary value of A and determine the nature of this stationary value.
[2]
y
ve
op
A line has equation y = 2 x + c and a curve has equation y = 8 − 2 x − x 2.
i For the case where the line is a tangent to the curve, find the value of the constant c.
C
U
R
8
ni
ev
ie
w
rs
ity
Pr
es
s
-C
y
op
y
w
ge
x cm
ie
R
ni
r cm
C
op
7
C
[3]
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q10 November 2016
ve
rs
ity
w
1
ii Given that the curve also passes through the point (4, 10), find the y-coordinate of A, giving your answer
as a fraction.
[6]
C
op
y
-C
i Find the x-coordinate of A.
ev
ie
−
A curve has equation y = f( x ) and it is given that f ′( x ) = 3x 2 − 2 x 2 .
The point A is the only point on the curve at which the gradient is −1.
-R
6
w
ge
C
U
ni
op
y
Cross-topic review exercise 3
[3]
br
ev
id
ie
w
ge
ii For the case where c = 11, find the x-coordinates of the points of intersection of the line and the curve.
Find also, by integration, the area of the region between the line and the curve.
[7]
s
9
.
2−x
es
dy
and determine, with a reason, whether the curve has any stationary points. [3]
dx
ii Find the volume obtained when the region bounded by the curve, the coordinate axes and the line x = 1 is
rotated through 360° about the x-axis.
[4]
Pr
ity
iii Find the set of values of k for which the line y = x + k intersects the curve at two distinct points.
ie
ni
ve
rs
[4]
y
op
-R
s
-C
am
br
ev
ie
id
g
w
e
C
U
R
ev
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q11 November 2010
es
C
op
y
i Find an expression for
w
-R
am
The equation of a curve is y =
-C
9
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q11 June 2014
Copyright Material - Review Only - Not for Redistribution
281
ve
rs
ity
4
for x > 0 .
2x + 1
ev
ie
10 A function f is defined as f( x ) =
-R
am
br
id
w
ge
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
Pr
es
s
-C
a Find an expression, in terms of x, for f ′( x ) and explain how your answer shows that f is a decreasing
function.
[3]
[4]
c On a diagram, sketch the graph of y = f( x ) and the graph of y = f −1( x ) , making clear the relationship
between the two graphs.
[4]
ve
rs
ity
C
op
y
b Find an expression, in terms of x, for f −1( x ) and find the domain of f −1.
1
1
y
[4]
[2]
[5]
id
ie
w
ge
U
R
ni
ev
ie
C
op
w
−
dy
2
= x 2 − x 2 . The curve passes through the point  4,  .
 3
dx
i Find the equation of the curve.
d2 y
ii Find 2 .
dx
iii Find the coordinates of the stationary point and determine its nature.
11 A curve is such that
br
ev
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q12 June 2014
es
s
-C
-R
am
2
12 The function f is defined for x . 0 and is such that f ′( x ) = 2 x − 2 . The curve y = f( x ) passes through the
x
point P (2, 6).
i Find the equation of the normal to the curve at P.
[3]
Pr
iii Find the x-coordinate of the stationary point and state with a reason whether this point is a maximum or
a minimum.
[4]
w
rs
C
282
[4]
ity
op
y
ii Find the equation of the curve.
ve
ie
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q9 November 2014
y
1
9
−
.
x −1 x −5
i Find the x-coordinate of the point where the normal to the curve at P intersects the x-axis.
op
ge
C
U
R
ni
ev
13 The point P (3, 5) lies on the curve y =
am
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q11 November 2016
s
es
y = (x – 2)4
A (1, 1)
O
x
B
y
ni
ve
rs
C
ity
Pr
op
y
C
w
C
U
op
The diagram shows part of the curve y = ( x − 2)4 and the point A(1, 1) on the curve.
The tangent at A cuts the x-axis at B and the normal at A cuts the y-axis at C .
br
w
-R
iii Find the area of the shaded region.
[6]
[2]
[4]
s
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q10 June 2013
es
am
a
, where a and b are integers.
b
ev
ii Find the distance AC, giving your answer in the form
ie
id
g
e
i Find the coordinates of B and C.
-C
ie
ev
R
-R
y
-C
14
[6]
ev
br
id
ie
w
ii Find the x-coordinate of each of the stationary points on the curve and determine the nature of each
stationary point, justifying your answers.
[5]
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
P (6, 5)
y
op
-R
ve
rs
ity
O
1
2
Pr
es
s
y = (1 + 4x)
C
w
ev
ie
am
br
id
y
-C
15
w
ge
C
U
ni
op
y
Cross-topic review exercise 3
Q (8, 0)
x
1
y
C
op
ni
ev
ie
The diagram shows part of the curve y = (1 + 4x ) 2 and a point P (6, 5) lying on the curve.
The line PQ intersects the x-axis at Q(8, 0).
U
R
i Show that PQ is a normal to the curve.
[7]
id
ie
w
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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]
br
ev
[In part ii you may find it useful to apply the fact that the volume, V , of a cone of base radius r and vertical
-R
am
1 2
πr h.]
3
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q11 November 2015
283
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height h, is given by V =
Copyright Material - Review Only - Not for Redistribution
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Cambridge International AS & A Level Mathematics: Pure Mathematics 1
am
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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
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[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
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am
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id
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a
-C
C
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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]
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id
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am
6 cm
y
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w
C
284
[3]
C
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ni
B
ge
5
[2]
[3]
1 − tan x
≡ 2 cos2 x − 1.
1 + tan2 x
2
U
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w
C
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y
2
-R
It is given that f( x ) = 2 x −
1
Copyright Material - Review Only - Not for Redistribution
[3]
[2]
[3]
ve
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ev
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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
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d
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c
[1]
Pr
es
s
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b
Find the range of f .
-R
9
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Practice exam-style paper
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-R
am
-C
[4]
es
Pr
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ev
id
br
-R
am
es
s
-C
ity
Pr
op
y
-R
s
es
am
br
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id
g
w
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C
U
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C
w
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[5]
285
ity
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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
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ity
am
br
id
a −4, 3
a x.2
a x = 2, y = 3
a 2 5
Pr
es
s
-C
c (5x + 4)2 − 20
d (3x − 7)2 + 12
a −9, 1
b −6, 2
c −5, 7
d 2, 7
e −6, 3
f
a − 2 ± 11
b 5 ± 23
c − 4 ± 17
−10, 1
id
y
w
ie
c −4.19, 1.19
d −3.39, 0.89
e − 1.39, − 0.36
f
x=
y
5
op
4.93
3.19
−0.217, 9.22
b ± b2 − 4ac
b
; the solutions each increase by .
2a
a
ev
-R
a ( − 3, 9), (2, 4)
7
b  −8,  , (2, 1)

2
c ( − 10, 0), (8, 6)
d ( − 2, − 7), (1, 2)
e (2, − 2), (10, 2)
f
g (2, 4)
h ( − 3, 1), (9, 7)
2

7
45
g  x +  −
2
4

2

3
7
h  x −  +
2
4

ni
ve
rs
( )
), (4, −9)
2
a 9 and 17
3
2 21 cm and 5 25 cm
4
ev
U
e
3 21 cm and 9 cm
s
-R
2
7
9
d 2 x +  −

4
8
(
−3, −4 13
( − 1, − 3), (2, 1)
j
( − 5, − 24), (5, 1)
l
(4, − 6), (12, 10)
n ( − 1, 3), (3, 1)
o (6, − 2), (18, − 1)
es
br
am
m
b 3( x − 2)2 − 13
id
g
2
5
33
c 2 x +  −

4
8
k ( −6, −2),
( x − 2)2 − 12
8, 1 21
y
f
(2, 2), (10, − 2)
op
e ( x + 2)2 + 4
i
C
2

15 
225
d  x +  −
2 
4

w
2

3
9
c  x −  −
2
4

ity
b ( x + 4)2 − 16
ie
op
y
Pr
es
s
-C
1
a ( x − 3)2 − 9
a 2( x − 3)2 + 1
−1.64, 0.24
Exercise 1D
am
br
b − 5.24, − 0.76
2
3
4
b 20 cm, 21cm, 29 cm
Exercise 1B
-C
a −0.29, 10.29
C
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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
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Pr
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ve
1
2
−3 ± 3
2
Exercise 1C
1
ity
d 4
−5,
e
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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
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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
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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
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Cambridge International AS & A Level Mathematics: Pure Mathematics 1
Copyright Material - Review Only - Not for Redistribution
b 13 − 19 and 13 + 19
ve
rs
ity
x = 4 21 and y = 16 or x = 16 and y = 4 21
7
r = 5, h = 13
8
a ( − 3, 5) and (2, 0)
b 5 2
-C
1
b  , 0
2 
14
a 2, 8
w
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-R
w
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-R
s
es
8
A: y = ( x − 4)2 + 2 or x 2 − 8x + 18
y
op
a
y = x 2 − 6x + 13
B
y = x 2 − 6x + 5
C
y = − x 2 + 6x − 5
D
y = − x 2 + 6x − 13
E
y = x 2 + 6x + 13
F
y = x 2 + 6x + 5
G
y = − x 2 − 6x − 5
H
y = − x 2 − 6x − 13
s
es
Pr
op
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ni
ve
rs
a ∪-shaped curve, minimum point: (3, − 1), axes
crossing points: (2, 0), (4, 0), (0, 8)
br
(
y = 5 + 3x −
12
Proof
w
),
1 2
x
2
)
-R
s
am
11
ie
−1 43 , −2181
axes crossing points: ( −5, 0), 1 21 , 0 , (0, −15)
-C
y = 3x 2 − 6x − 24
ev
id
g
e
( − 7, 0), (2, 0), (0, − 14)
(
10
C
U
( −2 21 , −20 14 ), axes crossing points:
op
y
b Student’s own answers
b ∪-shaped curve, minimum point:
c ∪-shaped curve, minimum point:
1
1
( x − 2)2 or 6 + 2 x − x 2
2
2
A
-R
a = 2, b = − 40, c = 128
C
w
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ev
Proof
ve
5
Exercise 1F
R
b ∩ -shaped curve, maximum point:
 1 , 1 1  , axes crossing points:
 4 8
7
ni
a = 2, b = − 9, c = 7
2
 − 1 , 0 , (1, 0), (0, 1)
 2 
9
-C
4
1
9
1
− 2 x − 

8
4
C: y = 8 −
am
c 4 10
a
ie
a
6
B: y = 4( x + 2)2 − 6 or 4x 2 + 16x + 10
b 4
4
d
25
1
f
, 25
9
b (4, 4), (16, 8)
es
3
−4 14 when x = 3 21
ev
e
5
U
R
c
−1, 2
ge
a
l
id
2
3
±
2
4, 6 14
1 1
,6
9 4
1 9
,1
4 16
x−6 x +8 = 0
br
C
w
ie
k
±2
( −2 14 , −6 81 ) , minimum
C
op
y
g ± 3
i
b
ge
-C
e ±1
( 2 21 , 13 14 ) , maximum
2
N
D N
D
+
,
−
2 2N 2 2N
b −1, 2
2
d ±
,± 5
2
1
f 1,
2
h No solutions
1
j − ,1
2
am
c ± 5, ±1
id
a ±2, ±3
br
1
b
9
49
a 2 x +  −

4
8
4
U
b
53 
5
− x− 
4 
2
y
y = −2x − 3
2
a
ie
13
3
Pr
(2, 3)
ity
12
rs
7x + y = 0
a 2( x − 2)2 − 3
b x=2
ni
11
ve
rs
ity
op
2 53
( − 3, 0), (4, 0), (0, 12)
w
10
2
Pr
es
s
a ( − 2, 1) and (3, − 1)
y
am
br
id
9
 1 , 12 1  , axes crossing points:
4
2
C
op
6
ge
7 cm and 11cm
C
w
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R
d ∩ -shaped curve, maximum point:
5
Exercise 1E
ev
C
U
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y
Answers
Copyright Material - Review Only - Not for Redistribution
287
ve
rs
ity
Pr
es
s
U
w
-R
id
8
k,6
9
−3 , m , 1
10
k.6
y
op
Proof
y
 1 , 0
2 
C
op
5
25
1
a  3x −  −
b − ,x,2


3
2
4
3
x = ±2, x = ±
2
x , − 9 − 2 3 or x . − 9 + 2 3
e
3
4
k , 1 or k . 2
es
s
-R
5
w
57
8
1
d k,
2
25
f k,
16
1
2
Proof
s
13
2
id
g
am
3
2
k , − 4 3 or k . 4 3
2
k = − 10 or k = 14
f
b k,
br
c k,2
e k.
es
Pr
8
9
7
1
d k = 0 or k = 2
a k . −13
-C
R
5
ni
ve
rs
b k = 4 or k = 1
e k = 0 or k = −
k , − 2 or k . 6
End-of-chapter review exercise 1
ity
b = − 2, c = − 35
a k = ±4
1
c k=
4
6
U
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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
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w
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ity
C
7
Pr
y
x , −5 or x . 8
3
a 1, x <
b −1 , x , 0
2
c − 1 < x , 1 or x > 5
op
6
− 5, − 9
2
s
b −7 < x , 1
c x , − 2 or x > 3
288
k < −2 2
ie
br
5
11
1
am
-C
4
Proof
Exercise 1I
id
g x < − 9 or x > 1
7
5
i − ,x,
2
3
5
−3 , x ,
2
a 5<x,7
ge
R
e x , − 4 or x . 1
10
y
d −3 , x , 2
1
3
f − ,x,
2
5
h x , − 2 or x . 5
ni
c − 12 < x < 1
9
8
k=
ev
3
4
5
,x,
3
2
a −9 , x , 4
e
7 − 2 10 , k , 7 + 2 10
p2
20
25
k<
8
Proof
7
ve
rs
ity
w
C
op
y
c x < − 7 or x . 1
f
C
op
a x < − 5 or x > 5
2
ev
-C
e −6 < x < 5
-R
am
br
id
c 4<x<6
ev
ie
ev
ie
b x , − 2 or x . 3
3
d − ,x,2
2
1
1
f x , − or x .
2
3
b −5 < x < −2
3
2
d − <x<
2
7
1
f x , − 4 or x .
2
b x , 7 or x . 8
1
13
b k.
2
12
39
26
c k.
d k.−
8
5
e 5 − 21 , k , 5 + 21
a k.
ev
a 0<x<3
1
6
w
ge
Exercise 1G
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
b (2, 1), (5, 7),
b (5, 25)
1
3 5 ,  − , 5
 2 
ii k = 3 or 11
11
i
( 2 21 , 2 21 )
ii m = −8, ( − 2, 16)
12
i
1
2( x − 1)2 − 1, (1, − 1) ii  − , 3 21
 2

w
y
C
op
ni
b domain: x ∈ R, − 3 < x < 2
range: f( x ) ∈ R, − 7 < f( x ) < 20
b − 13 < f( x ) < − 3
ie
ge
a f( x ) . 12
c − 1 < f( x ) < 9
1
e
< f( x ) < 16
32
a f( x ) > − 2
d 2 < f( x ) < 32
3
f
< f( x ) < 12
2
b 3 < f( x ) < 28
c f( x ) < 3
d − 5 < f( x ) < 7
a f( x ) > 5
b f( x ) > − 7
c −17 < f( x ) < 8
d f( x ) > 1
8
a f( x ) > − 20
b f( x ) > −6 13
9
a f( x ) < 23
b f( x ) < 5
10
a
s
Pr
rs
ve
2( x − 3)2 − 13
ni
4
ity
x−4
5
f −1( x ) =
U
6
br
w
4
function, one-one
y
s
es
Pr
a 2( x − 2 ) − 3
y
15
2
-R
c x ∈ R, −3 < x < 5
s
es
am
b Many-one
-C
a = 1 or a = − 5
ev
4 x
br
2
14
12
e
id
g
O
–2
13
f( x ) > k − 9
a2
g( x ) <
+5
8
a=2
11
U
2
R
b − 1 < f( x ) < 5
op
w
ie
ev
4
4
2
–2
ni
ve
rs
6
–4
O
ity
C
8
C
op
y
10
2
w
-C
-R
am
g function, one-one h not a function
a
289
ie
e function, one-one f
ev
id
c function, one-one d function, one-one
2
y
a function, one-one b function, many-one
ge
R
Exercise 2A
7
ie
C
w
ie
ev
es
y
op
3 − 2x
3
1
6
-C
2
w
-R
am
Prerequisite knowledge
y
br
ev
id
5
2 Functions
10
x
4
a domain: x ∈ R, − 1 < x < 5
range: f( x ) ∈ R, − 8 < f( x ) < 8
4
1
iii y − 3 = − ( x − 2)
5
1
2
b each input does not have a unique output
U
w
ev
ie
ve
rs
ity
i
C
10
O
op
op
c x < 1 or x > 9
R
2
Pr
es
s
-C
2
y
a 25 − ( x − 5)
-R
4
a Proof
c 2,x,5
9
y
6
b (6, 29)
c k = 1, C = (2, 5)
8
a
ev
ie
a Proof
3
ge
7
b k = − 4 or k = − 20
am
br
id
a
C
( 121 , −2 )
6
C
U
ni
op
y
Answers
Copyright Material - Review Only - Not for Redistribution
b k=4
x
ve
rs
ity
C
c RR( x ), domain is x ∈ R, x ≠ 0,
range is f( x ) ∈ R, f( x ) ≠ 0
ev
ie
w
ge
d QPR( x ), domain is x ∈ R, x ≠ 0,
range is f( x ) ∈ R, f( x ) . 1
-R
b domain: x ∈ R
range: f( x ) ∈ R, f( x ) > 2
Pr
es
s
1
2
C
c f( x ) > 3
w
5
a g is one-one for x > 3, since vertex = ( 2, 2 )
y
op
a −3
7
a f( x ) > − 9
b f −1( x ) = − 3 +
w
x + 32
2
a k=3
10
11
a f −1( x ) =
s
es
Pr
ity
f −1( x ) = 3 + 9 − x
ii Domain is x < 9, range is 3 < f −1( x ) < 7
1
b Domain is x < 4 23
a f( x ) =
5−x
a = 5, b = 12
9
12
4x + 3
x + 1 −1
, g (x) =
3
2x
1
(1 + 3 x + 3 )
2
b Domain is − 2 < x < 122
a f −1( x ) =
C
U
e
13
id
g
a PQ( x ), domain is x ∈ R,
range is f( x ) ∈ R, f( x ) > − 1
a f( x ) = ( x − 5)2 − 25
ev
ie
b f −1( x ) = 5 + x + 25 , domain is x > − 25
es
s
-R
br
am
x−2
2
b Proof
2( x + 1)
a ff ( x ) =
for x ∈ R, x ≠ − 3
x+3
b Proof
c −2 or 1
-C
b x > −3
C
6
b fg( x ) > −6 14
b QP( x ), domain is x ∈ R,
range is f( x ) ∈ R, f( x ) > 1
y
a f −1( x ) = − 1 + 3 x + 4
b x <1
b No inverse since it is not one-one
ni
ve
rs
op
y
w
R
a 4x 2 + 2 x − 6
5−x
2x
y
b ( x − 1)2 + 3
19
b f −1( x ) = − 2 + x + 4
-R
a x < − 1 or x > 3
18
ie
Pr
ity
16
f −1( x ) = 2 + 3 x + 1
4
rs
b −1
f
a f −1( x ) =
ve
ni
U
a 2( x + 1) − 10
-C
15
2
7 − 2x
x −1
a Domain is x > − 4, range is f −1( x ) > − 2
b i
14
17
3x + 8
x
3
8
19
k>−
2
Proof
13
c f −1( x ) = 5 + x − 3 d f −1( x ) =
op
±4
ge
12
c gg
f fgf
id
Proof
b gf
e gfg
br
11
x−3
b g −1( x ) = 2 +
am
9
b f −1( x ) =
ev
2
y
1
or 3 21
2
4
− or 0
3
−9
x+2
4x + 9
a fg
d ff
x+8
5
ie
id
br
am
b −4
op
C
8
10
ie
b
a f −1( x ) =
e f −1( x ) =
5 79
b −4 21 or −
-C
a (2 x + 5)2 − 2
w
ie
ev
c hh
g SP( x ), domain is x ∈ R, x > − 1,
range is f( x ) ∈ R, f( x ) > −1
w
a a = 3, b = −12
6
a −
x +1
b kh
PS( x ), domain is x ∈ R, x > − 1,
range is f( x ) ∈ R, f( x ) > − 1
-R
3
c 231
s
a hk
b 3
es
2
7
ev
U
a 7
f
Exercise 2C
1
ge
1
6
ve
rs
ity
w
ev
ie
R
Exercise 2B
5
R
domain: x ∈ R, x > 3,
range: f( x ) ∈ R, f( x ) > − 2
f
4
290
e domain: x ∈ R, x ≠ 2,
range: f( x ) ∈ R, f( x ) ≠ 0
ni
C
op
y
d domain: x ∈ R, x ≠ 0,
range: f( x ) ∈ R, f( x ) ≠ 0
e RQQ( x ), domain is x ∈ R, x ≠ − 4,
range is f( x ) ∈ R, f( x ) ≠ 0
C
op
-C
c domain: x ∈ R
range: f( x ) ∈ R, f( x ) . 0
ev
a domain: x ∈ R
range: f( x ) ∈ R
am
br
id
16
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
y
O
2
4
6
x
291
w
ie
ity
rs
y
4 − 2x
x
−1
c Domain is 0 , x < 2, range is f ( x ) > 0
b f −1( x ) =
op
C
y
f –1
y=x
ie
d
w
ve
ev
s
x
a Symmetrical about y = x
4
b Not symmetrical about y = x
ni
ve
rs
c Symmetrical about y = x
es
s
-R
br
ev
ie
id
g
w
e
C
op
a Proof
5
y
d Symmetrical about y = x
U
R
f
O
x
ity
f –1
6
es
4
Pr
2
–6
am
–2
-R
br
am
-C
O
C
w
ie
f –1
a 0 , f( x ) < 2
3
4
–4
ev
f
–4
2
–2
y=x
4
–4
y=x
–2
-C
6
–2
ni
id
ge
6
f
x
y
ev
x
U
y
b
6
2
Pr
y
op
C
w
ev
ie
6
5
4
C
op
c
-R
4
3
s
2
–4
op
y
C
w
ev
ie
-R
ge
id
br
am
-C
O
–6
R
2
1
d f −1 does not exist since f is not one-one
x +1
a f −1( x ) =
2
b Domain is − 3 < x < 5, range is − 1 < x < 3
2
f
2
–2
–4
f
O
4
–2
–6
1
ni
U
R
–4
2
y=x
f –1
f–1
3
es
ev
ie
y
6
–6
4
Pr
es
s
op
c (fg)−1( x ) = g−1 f −1( x )
w
C
ve
rs
ity
7−x
ii
6
a
y=x
6
-C
7−x
a
6
14 − x
b i
6
Exercise 2D
1
y
5
y
16
c
b Proof
1± 5
2
b and c
c
15
x +1
x
ge
a f −1( x ) =
am
br
id
14
U
ni
op
y
Answers
Copyright Material - Review Only - Not for Redistribution
b d = −a
ve
rs
ity
 −1
c Translation  
 0
 2
d Translation  
 0
ni
U
7
a = 2, b = − 3, c = 1
ev
-R
a
s
x
3
2
ity
1
rs
–4 –3 –2 –1 O
–1
4
s
3
w
y
op
d y = 3x 2 − 2 x − 5
w
a Reflection in the x-axis
b Reflection in the y-axis
ie
ev
c Reflection in the x-axis
s
-R
d Reflection in the x-axis
es
am
-C
y = 2 x 2 + 3x + 1
b y = 2x 4
C
3
a y = − 5x 2
c
4 x
U
2
e
1
2
br
–4
–4
id
g
–3
–3
ni
ve
rs
w
ie
1
1
–2
ity
C
4
–2
1
y
y
–4 –3 –2 –1 O
–1
2
–4 –3 –2 –1 O
–1
es
–4
2
4 x
3
Pr
-C
op
y
–3
3
3
C
x
–2
c
2
4
ev
3
4 x
y
ie
b
-R
2
3
op
ni
ge
1
2
–4
id
br
am
–4 –3 –2 –1 O
–1
–3
U
R
4
1
–2
ve
w
ie
ev
y
1
y
4
Pr
op
C
y = x 2 − 6x + 8
es
4
–4
3
ev
6
ie
ge
3
–3
2
R
y = ( x + 1)( x − 4)( x − 7)
br
am
-C
y
2
–2
b
c b = −1
5
1
1
4 x
2
–3
1
–4 –3 –2 –1 O
–1
O
–1
–2
Exercise 2F
2
292
–2
b a=2
id
4
3
ve
rs
ity
op
C
w
ev
ie
R
3
–4
–4
 2
Translation  
 4
f
1
y
 0
b Translation  
 −5 
y
 0
a Translation  
 4
 1
e Translation  
 0
y
a
2
C
op
-C
g y = ( x + 1)2 + x + 1 h y = 3( x − 2)2 + 1
2
w
x−3
y=
x−2
f
4
3
-R
e
2
y=
x+5
d y = x2 + 1
Pr
es
s
y = 7x2 − 2x + 1
y
w
am
br
id
c
a
ev
ie
b y =5 x −2
a y = 2x2 + 4
1
4
ge
Exercise 2E
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
O
2
6
–4 –3 –2 –1 O
–1
w
ie
s
es
–3
Pr
–4
293
c
ity
4
6 x
y
ve
op
2
C
U
1
ge
–4 –3 –2 –1 O
–1
w
ie
ev
d
y
4
x
1
2
3
4
x
-R
4
3
ity
2
ni
ve
rs
1
y
–4 –3 –2 –1 O
–1
op
c Stretch parallel to the y-axis with stretch
factor 2
C
U
–2
–3
–4
es
s
-R
br
ev
ie
id
g
w
e
d Stretch parallel to the x-axis with stretch
1
factor
3
am
3
–4
Pr
a Stretch parallel to the x-axis with stretch
1
factor
2
b Stretch parallel to the y-axis with stretch
factor 3
-C
R
ev
ie
w
C
3
2
–2
–3
es
y = 162 x 3 − 108x
1
s
br
id
-C
+2
op
y
e
y=2
x −1
y
4
3
b y = 3x 3 − 3
1
d y = x 2 − 4x + 10
2
am
a y = 6x 2
x
–2
–6
c
1
ni
ie
ev
R
–4
2
2
rs
w
2
–2
4
3
–4 –3 –2 –1 O
–1
-R
am
-C
y
op
C
O
3
y
U
y
6
–2
2
4
4
–4
1
y
b
2
–6
4 x
–4
ev
id
br
b
3
–3
ge
–6
2
–2
x
ni
–4
1
C
op
ev
ie
w
–2
R
4
1
ve
rs
ity
–2
3
-R
2
–4
y
4
2
Pr
es
s
am
br
id
4
y
op
C
–6
a
w
1
6
ev
ie
a
y
-C
1
Exercise 2H
ge
Exercise 2G
C
U
ni
op
y
Answers
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
3
x
–2
b y = − x 2 + 4x − 5
6
a y = 2g( − x )
b y = 3 − f( x − 2)
7
a Stretch parallel to the y-axis with stretch
 0
1
factor followed by a translation  
2
 3
3
w
4
 0
translation  
 2
x
ev
2
br
1
b Reflection in the x-axis followed by a
ie
id
ge
1
-R
am
–2
s
ity
4
rs
ve
1
3
4
x
parallel to the y-axis with stretch factor
-R
s
Pr
2
1
2
3
 0
x-axis, translation  
 4
U
e
b y = 2 − f( x )
id
g
1
x −1 + 3
2
a y=−
10
a y = 3[( − x + 4)2 + 2] = 3(4 − x )2 + 6
ev
b y = 3( x − 1)2
b y=−
1
( x − 1) − 3
2
9
-R
b y = 3[( −( x + 4))2 ] + 2 = 3( x + 4)2 + 2
s
am
a y = 3( x − 1)2
es
br
y = 2f( x − 1) + 1
-C
3
y-axis with stretch factor 2, reflection in the
y
ev
R
–4
c
x
op
ie
–2
a y = 2f( − x )
4
ni
ve
rs
w
–4 –3 –2 –1 O
–1
 3
c Translation   , stretch parallel to the
 0
ity
1
w
op
y
C
3
es
-C
y
4
–3
2
 −1
b Translation   , stretch parallel to the
 0
1
y-axis with stretch factor , reflection in the
2
 0
x-axis, translation  
 −2 
ie
br
am
h
1
2
ev
id
–3
w
ge
–2
–4
2
 −5 
a Translation   followed by a stretch
 0
U
R
–4 –3 –2 –1 O
–1
8
ni
ev
ie
1
C
w
2
op
C
3
 6
c Translation   followed by a stretch parallel
 0
1
to the x-axis with stretch factor
2
d Stretch parallel to the y-axis with stretch
 0
factor 2 followed by a translation  
 −8 
y
op
Pr
y
g
294
–4
es
y
-C
–3
C
–4 –3 –2 –1 O
–1
a y = 2x2 − 8
U
2
x
5
ni
R
3
10
C
op
ve
rs
ity
4
5
O
b
ie
op
y
a
y = x2
y
4
y
Pr
es
s
-C
y
3
–4
C
w
2
–3
f
ev
ie
1
-R
1
2
 1
b y =  x − 5 

 2
1
( x − 5)2
4
c
2
–4 –3 –2 –1 O
–1
a y=
ev
ie
4
am
br
id
4
w
ge
y
e
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
-R
1
–2
i
( x − 2)2 − 4 + k
ie
ii f( x ) > k − 4
ev
-R
iv f −1( x ) = 2 + x + 4 − k , domain is x > k − 4
− 5 < f( x ) < 4
ii
y
es
s
i
Pr
295
ity
f
C
U
op
x
6
w
ge
ie
-R
s
es
4( x − 3)2 − 25, vertex is (3, − 25)
Pr
ii g( x ) > − 9
i
1
2
x + 25 , domain is x > − 9
2( x − 3)2 − 5
iii f( x ) > 27
ii 3
y
9
op
x+5
, domain is x > 27
2
( x − 1)2 − 16
ii −16
id
g
i
C
e
10
w
U
iv f −1( x ) = 3 +
ev
ie
iii p = 6, q = 10
es
s
-R
br
am
i
ity
x
O
b y = 3x 2 + 6x
-C
8
iii g−1( x ) = 3 −
ni
ve
rs
C
w
–2
1
 ( x + 2 ) for − 5 < x < 1
3
iii f −1( x ) = 
5− 4
for 1 , x < 4
x

ev
id
br
am
-C
op
y
y
8
x
y
ve
ni
ev
R
ie
y=x
f–1
O
 3
by translation  
 0
ev
ii 3
iii p = 2
the y-axis or reflection in the y-axis followed
a
x
w
6
rs
y
4
iii f ( x ) = 3 + 4 − x , domain is x < 0
U
-C
y
op
C
2
ie
w
−( x − 3)2 + 4
−1
 −3 
b Translation   followed by a reflection in
 0
R
y
i
–2
–4
2
C
op
ni
5
7
O
3
O
–1
–3
1
2
End-of-chapter review exercise 2
f–1
2
–2
ge
id
br
am
 −5 
followed by translation  
 0
2
3
Pr
es
s
ve
rs
ity
y
op
C
parallel to the x-axis with stretch factor
f
4
x
 −10 
Translation 
followed by a stretch parallel
 0 
1
to the x-axis with stretch factor or stretch
2
25
7
− 9 x − 

4
6
a
y=x
y
y = f(x)
O
–2
w
ev
ie
w
ev
ie
ge
-C
y
y = g(x)
1
x + 2 for x > − 2
b
 −2 
translation  
 0
R
a f −1 : x ֏
4
y-axis or reflection in the y-axis followed by
12
C
 2
Translation   followed by reflection in the
 0
am
br
id
11
U
ni
op
y
Answers
Copyright Material - Review Only - Not for Redistribution
iv f −1( x ) = 1 + x + 16
ve
rs
ity
br
Prerequisite knowledge
am
ity
rs
ev
ie
C
b x + 2 y = −8
c x + 2y = 8
d 3x + 2 y = 18
a y = 2x + 2
b 5x + 3 y = 9
c 7x + 3 y = −6
6
a y=
3
x+8
2
b (0, 8)
a y=
y
8
op
a (6, 3)
4
x + 10
3
2
b y=− x+7
3
b
( −7 21 , 0 ), (0, 10)
w
C
U
7
a 2 y = 5x + 33
-R
s
( − 2, 6)
a y = 3x + 4
(8, 2)
9
es
10
b 9x + 5 y = 2
a 2 y = 3x − 3
c 12 21
br
k=2
-C
9
b y = − 3x − 1
c 39
id
g
units
a y = 2x + 1
5
e
2
am
8
k=4
38 21
d 100
op
s
es
Pr
b ( − 1, 9)
ni
ve
rs
a ( − 2, − 1)
ity
a = 2, b = − 1
c 2 41, 2 101
7
c 4 145
w
4
op
y
C
R
ev
ie
w
6
b a = − 4, b = 16, c = 11
ie
b = 3 or b = −5 54
a (6, 6)
-R
4
b −2
ie
id
3
br
a = 3 or a = − 9
1
2
c a = 6 or a = − 4
c 2x − 3 y = 9
am
3
a
y
ve
ni
ge
2
-C
17 units2
a = 10, b = 4
c 2x + 3 y = 1
b PQ = 197, QR = 146, PR = 3 5 ,
not right angled
5
11
1
a PQ = 5 5, QR = 4 5, PR = 3 5 ,
right-angled triangle
2
b 5
Exercise 3C
b 4 − 21, 4 + 21
Exercise 3A
1
a 1
-R
Pr
b −5
a ( x − 4)2 − 21
4
es
b 6
U
R
ev
ie
w
C
3
8
s
-C
1
a −
6
2
a
3
c 7 21
op
296
(0, − 26)
10
y
2
7
9
( −4 21 , −2 ), 13
1
6
k=
C
op
ge
id
3 Coordinate geometry
(7, − 1)
5
7
k = 2 or k = 3
5
ni
1
( x − 2)
2
4
w
iv f( x ): half parabola from (0, 10) to (2, 2);
g( x ): line through O at 45°;
f −1( x ): reflection of f( x ) in g( x )
b Not collinear
ie
ve
rs
ity
3
1 1
,
5 6
Proof
2 5
− ,
5 2
a
y
2
U
C
w
ev
ie
1
ev
-C
ii 2 < f( x ) < 10
y
2( x − 2)2 + 2
v f −1 ( x ) = 2 −
R
Exercise 3B
v h −1( x ) = − x + 2
iii 2 < x < 10
w
A( − 5, 5), B(7, 3), C( − 3, − 3)
2
iii b = 2
op
i
12
fg( x ) = 2 x − 3, gf( x ) = 4x + 4x − 1
ii a = − 1
1
iv ( x 2 − 3)
2
13
iv k = 22
2
b 8 2
ev
i
a (5, 2)
-R
12
11
am
br
id
iii − 1 , x , 7
ii f > − 11
ev
2( x − 3)2 − 11
Pr
es
s
i
ge
11
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
Copyright Material - Review Only - Not for Redistribution
b 33
ve
rs
ity
E (4, 6), F (10, 3)
11
10
12
(14, −2)
13
a y = −3x + 2
b ( −1, 5)
c 5 10, 4 10
d 100
ge
a Proof
14
( x − 5)2 + ( y + 3)2 = 40
15
( x − 9)2 + ( y − 2)2 = 85
16
( x + 3)2 + ( y + 10)2 = 100 ,
( x − 13)2 + ( y + 10)2 = 100
17
a i
1+ 2
b i
3+2 2
3 2
b (0, 0),
2
d (5, − 3), 2
Exercise 3E
( 4 25 , 154 )
15
a y = 2x − 7
16
x + y = 8, 3x + y = 3. Other solutions possible.
U
a x + y = 64
b ( x − 5) + ( y + 2)2 = 16
)
Pr
2
2
5
y
2
6
(4, 3)
7
a ( x − 12)2 + ( y − 5)2 = 25 and
w
ge
ie
ev
-R
End-of-chapter review exercise 3
1
s
es
Pr
ity
9
( x − 4) + ( y − 2) = 20
5
6
a a = 5, b = −2
2
c y = − x − 3 25
5
2
i 16t
7
(13, −7)
8
a ( −2, 2), (4, 5)
e
id
g
es
ev
s
-R
br
2
am
b (4, −5)
op
( x − 5)2 + y2 = 8 and ( x − 5)2 + ( y − 4)2 = 8
10
C
8
4
i
ii Proof
w
Proof
3
4
1
49
and
ii
24
9
4
a = −4, b = −1 or a = 12, b = 7
2
ie
7
U
( x − 6)2 + ( y + 5)2 = 25
-C
ie
ev
R
6
2 , a , 26
y
ni
ve
rs
–6
w
C
op
y
–4
2
( 5 − 5, 6 + 2 5 ), ( 5 + 5, 6 − 2 5 )
b Proof
6 x
4
b y = −2 x + 16
( x − 2)2 + ( y − 10)2 = 100
-C
am
2
br
O
–2
ie
ity
rs
( x + 2)2 + ( y − 2)2 = 52
2
,m,2
29
a (0, 6), (8, 10)
d 20 5
ve
4
−
c
ni
( x − 2)2 + ( y − 5)2 = 25
4
5
U
3
Proof
id
R
ev
ie
w
C
op
y
c ( x + 1)2 + ( y − 3)2 = 7
1
3
25
d x −  +  y +  =


2
2
4
3
s
2
2 5
y
(
2
es
2
2
-R
am
-C
2
w
g (4, − 10), 6
e ( − 7, 0), 3 2
( −1, −4), (5, 2)
1
3 10
f (3, − 4),
2
1
1
h 3 2 , 2 2 , 10
C
br
id
c (0, 2), 5
ii Student’s own answer
ev
a (0, 0), 4
ge
1
ii Student’s own answer
op
b
b ( x + 1)2 + ( y − 4)2 = 20
y
ve
rs
ity
)
y=
C
op
-R
Pr
es
s
ii x + y = 7
Exercise 3D
R
13
ev
ie
am
br
id
-C
y
(
12
ni
ev
ie
w
C
b
2 21 , 4 21
( x − 3)2 + ( y + 1)2 = 16, (3, − 1), 4
3
21
x−
4
2
( x − 5)2 + ( y − 2)2 = 29
11
y = 4 21
a i
op
14
10
w
10
C
U
ni
op
y
Answers
Copyright Material - Review Only - Not for Redistribution
b y = −2 x + 5 21
297
ve
rs
ity
iii ( −1, 8), 2 10
2
b p = −1
a y=− x+3
3
c ( x − 6)2 + ( y + 1)2 = 26
5
U
ni
C
op
y
4
a (19, 13)
b 104

 10
, 10 
b k , −12, k . 12
a 
 3

4
a y=− x+2
b Proof
3
c ( x − 15)2 + ( y − 7)2 = 325
w
ie
-R
ity
rs
ve
a 1< x < 5
8
( k , −2 k )
9
6 5
10
a k = 14
11
a ( −1, −11), (6, 3)
12
a k=−
13
i
C
1
x+4
3
25
b k,−
12
b y=
w
1
b x.5
2
fg( x ) = 5x, range is fg( x ) > 0
ii g −1 ( x ) = 4 − 2 x , domain is 0 , x < 2
5x
a b = −5, c = −14
ity
a 2 y = 3x + 25
b ( −3, 8)
16
a 36 − ( x − 6)
b 36
2
a 3( x + 2)2 − 13
a a = 12, b = 2
C
18
op
c 6 , x , 18
x
b ( −2, −13)
y
17
b −3
26 − x
2
2
2
a ( x − 8) + ( y − 3) = 29
id
g
w
c g−1( x ) = −3 +
19
ie
U
ii −3 , x , 8
15
e
4
(2.5, −20.25)
b i
c x < 36, g ( x ) > 6 d g−1( x ) = 6 + 36 − x
es
s
-R
br
am
2
-R
s
es
Pr
ie
(–2, –42)
-C
3
−1
ni
ve
rs
(2, 22)
O
–4
2
ie
ge
id
br
am
-C
y
w
C
(2, –6)
b
1
b −13, 3
7
14
4 x
O
op
y
–4
O
ev
30
–1
–2
ev
(–2, 26)
ni
2
2
2
x = ± ,x = ±
3
2
y
a
R
ev
ie
w
1
U
C
Cross-topic review exercise 1
–3
op
op
298
1
s
d Proof
es
c Proof
Pr
b 4 − 2 3, 4 + 2 3
3
2
ev
id
br
a 4, (4, −2)
y
17
-C
am
16
ev
y
ge
14
15
R
C
6
ve
rs
ity
C
ii Proof
y = −x 2 + 6 x − 8
y
y
y = −2 x + 6, (3, 0)
i
w
ev
ie
R
5
Translation   , vertical stretch with stretch
0
factor 2
Pr
es
s
8 6
ii (0, −2),  , 
 5 5
y = 2x − 2
12
13
4
5
op
i
-C
iii (5, 12)
11
a = 5, b = −2
w
10
19
19
c
− 113,
+ 113
2
2
ii ( −1, 6)
i 2, m = 1
3
ev
ie
a ( −2, −3)
am
br
id
b y=−
1
x + 4 34
2
9
-R
ge
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
Copyright Material - Review Only - Not for Redistribution
b 5x + 2 y = 75
x
ve
rs
ity
w
-R
Degrees 0 45 90 135 180 225 270 315 360
w
ev
-R
7
8
12.79°
Pr
es
s
C
w
a 13 cm
b 2.275 cm
a 0.5 rad
b 0.8 rad
ie
d 28 π cm
4
15.6 m
5
a 19.2 cm
b 20.5 cm
c 50.4 cm
a 0.927 rad
ity
6
b 4 cm
ni
ve
rs
c 17.6 cm
a 14 cm
7
b 11.8 cm
c 25.8 cm
a 13 cm
b 2.35 rad
C
U
8
f
80°
g 54°
h 105°
10
Proof
j
810°
k 252°
l
n 420°
o 202.5°
ie
ev
-R
es
s
48°
w
e 240°
id
g
c 56.6 cm
a Proof
br
0.727
c 6 π cm
ev
-R
3
f
b 3π cm
9
am
5π
6
a 2 π cm
d 15°
-C
7π
4
2π
π
7π
6
5 π 11π
2π
3
6
d 0
c 30°
m 225°
3π
2
c 0.622
3
e
2
7.79 cm
b 60°
81°
2π
3
b 14.1
2
n 7π
36
e
Radians 4 π
3
a 90°
i
π
2
a 0.644
1
ge
id
-C
op
y
C
w
ie
ev
R
2
3π
π
3
y
s
es
ni
b
l
3π
2
Degrees 240 270 300 330 360
Exercise 4B
2π
9
5π
d
18
5π
f
6
h 7π
6
j 5π
3
U
π
9
5π
c
36
π
e
36
3π
g
4
5π
i
4
k 13π
36
π
m
20
o 10 π
3
a
am
1
br
R
Exercise 4A
π
6
6
Pr
rs
5.14, 15.4 cm 2
ve
w
ie
ev
ity
op
3
C
13, 67.4
5π
4
π
ie
id
br
y
Prerequisite knowledge
2
3π
4
y
U
Radians 0
am
-C
4 Circular measure
(12 + π ) cm, 3π cm 2
π
2
b. Degrees 0 30 60 90 120 150 180 210
b k = ±4 10
1
π
4
Radians 0
ii 4x − 2 y = 21
(4, 5), (10, 2)
a i
d 87.1°
a.
ge
R
5
range is (fg)−1 ( x ) < −3
24
c 76.8°
e 45.3°
x + 26
, domain is x > 28,
6
ii (fg)−1( x ) = −
b 45.8°
y
fg( x ) > 28
b i
a 68.8°
C
op
a k = −2
d 3.49
op
c ( −2, 0), (18, 0)
3
d y=− x+6
4
c 0.820
op
b 10
ve
rs
ity
a (8, 0)
4
ni
23
ev
ie
w
C
22
b 0.559
e 5.59
Pr
es
s
op
y
c
a 0.489
ev
ie
ge
-C
c
a
3
x + 7 −1
18
, g (x) = 5 −
f (x) =
x
3
Proof
2
3 17
17 
3
− x + 
b − , 

2 4 
4 
2
−5 and −1
d (1, −2), ( −1, 4)
−1
b
21
1
2
a x=
am
br
id
20
C
U
ni
op
y
Answers
Copyright Material - Review Only - Not for Redistribution
b 43.4 cm
299
ve
rs
ity
a 1.25 rad
b 40 cm 2
5
a 1.75 rad
y
b 4.79 cm
am
br
id
U
11
 2 πr 2
 3 −
12
π
100  1 + − 3  cm 2


3
d 13.9 cm 2
2
b Proof
c 1
d
6
5
f
b Proof
15
4
15
c
16
a
i
id
-R
s
es
Pr
ity
U
ii
e −1
f
θ =
y
π
4
π
3
cos θ
3r 2α
ii
+ πr 2
2
-R
1
sin θ
π
3
θ =
3
1
3
1
2
1
2
3
2
2
2
3
2
1
s
2
π
5
id
g
e
iii α =
br
2 πr + rα + 2 r
am
i
-C
10
4α cos α + 4α + 8 − 8 cos α
c
C
i
tan θ
w
9
ii 6.31
ie
i
1
4
b
θ =
ev
8
Proof
ii r 2


π
r  cos θ + 1 − sin θ + − θ 


2
es
i
3
f 1
6
ni
ve
rs
7
5
f
24
25
12
19
5
3
5
2
15(3 − 5 )
4
15
15
15
16
75 − 4 15
15
1
2
d
ev
br
iii A = 36, θ = 2
b
3+ 2
2
2− 3
e
2
1
a
2
2 −2 6
d
2
c
am
-C
op
y
i
R
ev
ie
w
C
6
1
4
C
5
ii 8 + 5 π
4
ii
+ 4 tan α + 2α
cos α
r(1 + θ + cos θ + sin θ )
ii 55.2
7
π
AC = r − r cos θ
ii
+2 3−2
3
Proof
ii 36 − ( r − 6)2
a
f
c
op
i
d
w
U
4
4
25 3 25 π
b
−
4
8
ge
3
2
b
e 4 + 15
End-of-chapter review exercise 4
1
e
y
ity
rs
ve
a Proof
3
3
2
3
4
4
5
b
b
e
ni
ie
ev
R
3
5
20
d
3
2
a
3
a
op
es
Pr
y
14
w
13
1 
a  tan x +
cm b 0.219 rad

tan x 
a 15 − 5 3 + 5 3 π
6
π
i α =
8
i 8 tan α − 2α
b −5,
a 0, 5
s
-C
3r 2 
cm 2
2 
op
C
300
ii 630°
ie
id
a Proof
am
10
iii
w
ge
b 36.5 cm 2
br
c 51.7 cm 2
1
5π
6
iii 195°
ii 4 π
ie
a 29.1cm 2
d r
b i 30°
Exercise 5A
25
( 2 3 − π ) cm2
6
b
1
1 + r2
π
a i
4
-R
9
a
3
ni
8
-R
ve
rs
ity
 32 3 − 32 π  cm 2

3 
5 3
cm
3
1.86 cm 2
7
2
r
1 + r2
b
c
Pr
es
s
-C
op
C
w
ev
ie
b 3.042 cm 2
c 5.16 cm 2
1 + r2
a
y
4
1
C
op
b 1.5 rad
w
a 1.125 rad
Prerequisite knowledge
d 54 π cm 2
ev
ie
3
6
R
b 20 π cm 2
ev
2
a 12 π cm 2
9π
c
cm 2
4
a 867 cm 2
1
5 Trigonometry
ge
Exercise 4C
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
Copyright Material - Review Only - Not for Redistribution
π
6
1
2
2+ 3
3
ve
rs
ity
d 40°
Pr
es
s
ve
rs
ity
−2
− 2
−
2
3
y
1
2
ev
-R
f 180°
a 1
b 5
c 7
d 3
e 4
f
301
y
2
y
2
C
op
s
es
Pr
ity
e 180°
w
ie
ev
-R
O
90
180
270
360 x
180
270
360 x
s
–1
–2
y
2
b
a
1 + a2
a
d −
1 + a2
b
b
1 − b2
y
1
O
C
w
90
ie
–1
1 − b2
ev
d
−
es
s
-R
id
g
br
am
c − 1 − b2
1
2
d 120°
U
1− b
3
2
c 360°
a
e
a
2
1
3
w
ie
12
13
−1
b 180°
es
-C
a
1 + a2
ve
b
b
-C
8
2
− 3
1
2
21
b
θ = 210°
a 360°
1
2
b −
a a
c
3
Pr
21
5
a − 2
3
5
a −
13
op
y
C
w
ie
R
ev
7
3
ity
id
br
a −
2
1
h −
2
am
4
1
rs
ge
f
θ = 135°
Exercise 5D
U
d
θ = 120°
C
op
ni
U
y
op
C
w
ie
ev
R
4th quadrant
6
1
cos θ
id
d − cos 65°
π
f − sin
8
2π
h tan
9
3
b −
3
-C
c − tan 55°
π
e − cos
5
3π
g − cos
10
1
a −
2
2
c −
2
3
e −
2
3
g −
3
3
5
sin θ
b cos 55°
am
a − sin 10°
br
Exercise 5C
2
2
13
c − 7
4
a −
11
ge
8π
3
3
5
tan θ
ni
C
w
ev
ie
R
c 688°
5
12
3
d −
4
3
b
13
d − 7
3
b −
op
-C
y
op
π
6
4π
i 3rd quadrant,
9
a 125°
g 3rd quadrant,
1
10
d 3rd quadrant, 30°
π
f 2nd quadrant,
3
π
h 1st quadrant,
3
π
j 4th quadrant,
8
b −160°
5π
d
4
13π
f −
6
e 1st quadrant, 40°
e
-R
a 2nd quadrant, 80° b 3rd quadrant, 80°
c 4th quadrant, 50°
3
c
ni
ve
rs
2
12
13
w
b 40°
c 20°
a −
9
ev
ie
a 70°
am
br
id
1
ge
Exercise 5B
C
U
ni
op
y
Answers
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
br
–1
180
-R
y
op
C
-R
s
O
es
Pr
2π x
π
3π
2
2π x
y
2
y
ii
π
2
ie
id
g
w
e
O
op
1
ev
–1
–2
es
s
-R
br
am
3π
2
–3
360 x
U
R
–4
π
–2
ni
ve
rs
–3
π
2
–1
ity
270
180
–2
-C
1
C
-C
op
y
O
C
y
3
2
1
–1
a i
ev
id
4
y
2
90
360 x
270
w
360 x
270
am
–3
180
ie
180
br
–2
w
C
s
90
ge
U
1
O
ie
360 x
es
ve
ni
ev
2
R
90
rs
w
ie
3
–1
ev
300
y
ity
y
5
4
f
240
Pr
y
op
C
e
180
–3
i
–4
302
120
–2
360 x
270
60
ie
90
C
op
O
-C
–3
1
am
–2
2
ev
O
–1
360 x
270
y
3
h
U
1
180
–2
ge
2
id
R
3
90
–1
ni
y
4
O
y
360 x
300
ve
rs
ity
op
C
w
ev
ie
d
w
ev
ie
240
w
180
-R
120
y
60
y
2
1
Pr
es
s
-C
O
g
ge
y
am
br
id
c
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
1
3π
2
C
w
ve
rs
ity
1
a a = 3, b = 5
ge
y
w
90
14
a = 5, b = 4, c = 3
15
a A = 2, B = 6
y
op
2π x
ev
s
es
-C
am
–4
-R
br
–2
ie
id
g
y = cos 2x
3π
2
C
e
π
w
U
ev
π
2
180
–4
s
es
Pr
O
–2
ni
ve
rs
ity
2
y = 3 sinx
O
f
ie
-R
ev
id
br
2π x
y = 2 sin x
2
b 2
y
10
8
b 1 < f( x ) < 5
op
ve
b
3π
2
op
y
C
4
y
13
303
4
π
w
7
C
op
ie
-R
a a = 3, b = 2
–3
b 2
a y
s
Pr
ity
12
rs
a = 3, b = 1, c = 5
6
-C
am
π
2
4
π
11
U
ie
R
O
–2
b k=
y = 2 + cos 3x
3
1
–2
–3
y = sin 2x
ni
y
4
2
–1
π
c (0, 0),  − , −2 
 2

es
y
op
C
w
–2
–1
2π x
3π
2
π
–1
a
π x
ev
br
π
2
b 4
6
π
2
w
ge
id
y = 1 + cos 2x
am
-C
O
O
–π
2
–π
U
y
3
y
3
2
ni
a
1
ie
a
C
y
op
–2
2π x
-R
-C
π
–1
C
w
ev
ie
R
a = 3, b = 2, c = 3
 π   5π
  9 π   13π

, −1 
, 1 , 
, −1  , 
b  , 1 , 
8   8
  8
  8

5
ev
9
10
π
2
2
R
a = 4, b = 2, c = 5
Pr
es
s
O
8
ev
ie
ge
y
2
am
br
id
iii
U
ni
op
y
Answers
Copyright Material - Review Only - Not for Redistribution
b 5
270
360 x
ve
rs
ity
d 197.5°, 342.5°
e 126.9°, 233.1°
f 116.6°, 296.6°
g 60°, 300°
h 216.9°, 323.1°
π 5π
b
,
3 3
d 3.92, 5.51
2 π 4π
f
,
3 3
h 2.19, 5.33
2
c 1.25, 4.39
ni
e 1.89, 5.03
ge
3
id
b 17.7°, 42.3°, 137.7°, 162.3°
-R
es
y
op
ni
C
U
x + 4
3 
0.298, 1.87
7
a 0°, 150°, 180°, 330°, 360°
-R
b 0°, 36.9°, 143.1°, 180°, 360°
s
es
e 0°, 60°, 180°, 300°, 360°
Pr
8
–1
b 56.3°, 123.7°, 236.3°, 303.7°
a 30°, 150°, 270°
y
9
2π x
b 45°, 108.4°, 225°, 288.4°
c 0°, 109.5°, 250.5°, 360°
d 60°, 180°, 300°
id
g
w
e
3π
2
a 60°, 120°, 240°, 300°
op
π
0°, 76.0°, 180°, 256.0°, 360°
C
U
π
2
d 0°, 116.6°, 180°, 296.6°, 360°
f
ity
w
ev
ie
e 0°, 180°, 199.5°, 340.5°, 360°
br
f
70.5°, 120°, 240°, 289.5°
s
-R
g 19.5°, 160.5°, 270°
es
am
a 26.6°, 206.6°
6
f
4−x
b f is one-one, f −1 ( x ) = cos −1
 2 
3π
a
2
5−x
b f −1 ( x ) = sin −1
,3<x<7
 2 
-C
a 90°, 210°
c 139.1°, 175.9°
ge
br
am
-C
op
y
C
2
ie
h 5.77°, 84.2°
π 7π
b
,
2 6
d 0.0643, 2.36, 3.21, 5.51
3π
f 0,
2
b 56.3°, 236.3°
d 18.4°, 108.4°, 198.4°, 288.4°
f
4
ev
R
g 58.3°, 148.3°
c 0°, 78.7°, 180°, 258.7°, 360°
O
6
f 116.6°, 153.4°
e 278.2°
y
6
a 2 < f( x ) < 6
5
e 24.1°, 155.9°
c 119.7°, 299.7°
b f −1 ( x ) = sin −1

a −7 < f( x ) < −1
4
4
5
id
ie
ev
R
3
d 105°, 165°
w
π
4
2π
e
3
16
a
25
c
c 38.0°, 128.0°
ie
a 0
ni
ve
rs
w
C
2
Pr
f 135°
π
b
4
π
d −
6
π
f −
3
16
b
9
ity
e −60°
op
304
rs
d −90°
ve
c 60°
y
1
s
b 30°
-C
a 0°
am
Exercise 5E
a 26.6°, 153.4°
ev
y = 6 + cos x
g 0.848, 2.29
br
17
a 0.305, 2.84
C
op
90
y
120 x
U
R
–4
y = 2 + sin x
C
c 45.6°, 314.4°
ev
w
b 23.6°, 156.4°
w
60
–2
ev
ie
a 56.3°, 236.3°
ve
rs
ity
C
O
1
ie
op
y
2
16
w
ev
ie
-R
f
30
x +5
, 0 < f −1( x ) < 2 π
4 
Exercise 5F
-C
4
a −9 < f( x ) < −1
b f −1 ( x ) = 2 cos −1

am
br
id
6
7
ge
y
8
Pr
es
s
c
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
h 30°, 150°, 270°
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
op
π 5π
,
6 6
End-of-chapter review exercise 5
a 0.565, 2.58
11
3π
7π
2.03,
, 5.18,
4
4
am
br
id
b
C
1
2
a Proof
e Proof
f
Proof
5
a Proof
b Proof
6
30° or 150°
c Proof
d Proof
a Proof
b Proof
7
30° or 150°
8
i
e Proof
f
a 4 + 3 sin2 x
b 4 < f( x ) < 7
9
a (sin θ + 2) − 5
b 4, −4
10
a Proof
ii 45°, 135°, 225°, 315°
10
i
60° or 300°
ii 120°
11
Proof
π 5π
a
,
6 6
i f( x ) < 3
rs
ve
ni
13
w
s
es
b 30°, 150°, 210°, 330°
5
a Proof
b 72.4°, 287.6°
6
a Proof
b 65.2°, 245.2°
7
a Proof
b 41.8°, 138.2°, 270°
8
a Proof
b 30°, 150°
9
a Proof
b 66.4°, 293.6°
10
a Proof
b 70.5°, 289.5°
11
a Proof
3−x
iv f −1( x ) = 2 tan−1 

 2 
i
30° or 150°
15
i
Proof
ii n = 3, θ = 290°
i
17
i 1.68
C
op
ni
ve
rs
ii 194.5° or 345.5°
Proof
16
w
e
ie
b 30°, 150°, 210°, 330°
ev
id
g
π x
ii 54.7°, 125.3°, 234.7°, 305.3°
es
s
-R
br
am
-C
π
2
14
ity
Pr
a Proof
U
op
y
4
C
w
ie
ev
f
ie
id
-C
b 18.4°, 116.6°
b 60°, 131.8°, 228.2°, 300°
R
ii 3 − 2 3
ev
a Proof
y
3
b −2.21, 0.927
O
-R
3
ii 109.5° or 250.5°
i
iii
ge
a Proof
br
2
iii 20
Proof
12
b 76.0°, 256.0°
am
a Proof
ii 4
i
U
R
1
2π θ
9
1 − 4a 2
4a
, cos θ =
b sin θ =
2
1 + 4a
1 + 4a 2
Exercise 5H
y
w
ie
ev
8
ity
4
2
y= 1
2
–1
y
op
7
C
w
ie
ev
Proof
π
y
d Proof
y = cos 2θ
op
c Proof
O
C
-C
b Proof
-R
a Proof
s
h Proof
es
g Proof
y
6
Proof
Pr
f
y
1
C
op
ni
ge
br
d Proof
am
e Proof
x=±
4
b Proof
id
a Proof
c Proof
ve
rs
ity
y
d Proof
1 − k2
k
c −k
d Proof
c Proof
5
b
c Proof
U
R
4
1 − k2
a
3
2
39.3° or 129.3°
b Proof
op
C
w
ev
ie
3
Pr
es
s
-C
9 sin2 x − 3
1.95
3
Exercise 5G
1
a = 1, b = 2
-R
2
w
10
ev
ie
ge
U
ni
y
Answers
Copyright Material - Review Only - Not for Redistribution
305
ve
rs
ity
b −32 x15
3
a 2n + 3
14
a x5 + 5x 4 y + 10 x 3 y 2 + 10 x 2 y3 + 5xy 4 + y5
16
a y − 3y
Pr
ity
3
rs
ie
a 1 + 16x + 112 x
b 1 − 30 x + 405x 2
7
21 2
c 1+ x +
x
d 1 + 12 x 2 + 66x 4
2
4
5103
5103 2
e 2187 +
x+
x
2
4
f 8192 − 53248x + 159 744x 2
2
C
ev
-R
s
es
a −84
35
c
4
b 5940
d −9720
7
−224 000
8
−41184
9
40 095
10
a 128 + 320 x + 224x 2 b 1 − 28x + 345x 2
op
y
7920
C
Pr
ity
ni
ve
rs
w
ii 16 − 8 5
11
a 1024 + 5120 x + 11520 x 2
s
-R
b 1024 + 10 240 y + 30 720 y 2
es
-C
am
c 32
d 792
b 56
c 1 − 3x + 3x 2
br
16 + 8 5
b i
3
id
g
a 1 + 3x + 3x + x
2
c 364
b n
6
e
b 97 + 56 3
6
5
U
a 16 + 32 x + 24x 2 + 8x 3 + x 4
d 5005
w
id
br
am
-C
op
y
C
5
c 495
n( n − 1)
a
2
n( n − 1)( n − 2)
c
6
a 45
h 512 + 1152 x 2 + 1152 x 4
f
±2
b 84
g 256 + 1024x 2 + 1792 x 4
d 16
−32
5
g 768
h −
2
A = 486, B = 540, C = 30
a 35
ie
g 16x 4 − 96x 3 + 216x 2 − 216x + 81
9x 27
27
+
+
h x6 +
2 4x 4 8x 9
a 12
b 10
4
R
ev
ie
w
3
b y5 − 5 y 3 + 5 y
op
ni
U
ge
R
8x 3 + 36x 2 y + 54xy2 + 27 y3
e 40
b 36 2
y
ve
4
e x 4 − 4x 3 y + 6x 2 y2 − 4xy3 + y 4
c −90
3
ie
b 1 − 4x + 6x 2 − 4x 3 + x 4
c x 3 + 3x 2 y + 3xy2 + y3
2
y
a p = 8, q = 8
C
op
15
s
y
op
a x 3 + 6x 2 + 12 x + 8
f
C
x11
es
-C
b 11 − 3n
d 8 − 12 x + 6x 2 − x 3
w
13
2
Exercise 6A
C
5
w
a 125x 6
w
12
b 142
-R
2
ie
a 1 + 4 y + 6 y2
ev
id
br
b 9x 3 − 9x 2 − x + 1
am
a 4x 2 + 12 x + 9
ev
11
1
1
1
54
Exercise 6B
Prerequisite knowledge
306
10
Pr
es
s
U
ge
6 Series
−216
b 113 100
ni
iii f is one-one
R
2π x
3π
2
π
x−5
iv f −1( x ) = 2 cos−1 

 3 
a x8 − 4x 6 + 6x 4 − 4x 2 + 1 b −16
ev
C
ev
ie
w
π
2
ve
rs
ity
op
y
2
O
8
9
f
-C
4
16 + 112 x + 312 x 2 + 432 x 3 + 297 x 4 + 81x5
-R
6
7
ev
ie
ge
y
8
am
br
id
ii
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
13
16 + 224x + 1176x
14
a = −2, b = 1, p = −364
15
n = 8, p = 256, q = −144
ge
b 1
am
br
id
2
-R
Pr
es
s
-C
13
a $17 715.61
14
Proof
15
Proof
16
Proof
-R
s
es
ity
Proof
19
a 4 − 3 sin2 x
20
a Proof
C
0.25, 199.21875
7
0.5, 9
ie
w
6
8
b 900
-R
9
s
es
a −0.25, 256
b 204.8
11
a 90
b 405
12
a 36
b 192
13
93.75
14
a = 2, r =
e
id
g
3
5
π
π
,x,
3
2
a 5π
17
a Proof
-R
ev
16
c Proof
es
s
br
am
b 40.5
10
15
U
ar5 , ar14
−1, −1
ni
ve
rs
f
ity
d No
Pr
b 3, 15 309
52
165
2
a
, 13.5
3
ev
U
ge
a No
1
1
c − ,−
3 27
e No
-C
y
57
ɺ ɺ = 57 + 57 +
+…
a 0.57
100 10 000 1000 000
b Proof
b Proof
-C
am
br
id
b 9a
op
y
C
w
ie
ev
R
2
, 810
3
5
Exercise 6D
6
4
307
y
18
5
32
y
a a = 8d
c 26 23
4
3
op
17
b 1 91
d −36 74
a 3
3
ni
16
1
( 5n − 11)
2
9°
2
3
−10.8
3
,8
2
64
2
rs
10, −4
ve
14
Pr
b 20
op
7, 8
C
w
ie
ev
R
ie
id
br
am
-C
a 17, −4
13
C
op
$360
1
op
11
Exercise 6E
C
25
b $94 871.71
w
10
b 2,
ie
5586
1
3
a
ev
9
ve
rs
ity
31
ni
8
U
1817
ge
7
b 2059
11
w
y
d −3160 x
1442
4
b 48.8125 m
12
6
3
n
b −1957
a 7, 29
2
3
a 8 
 4
a 1037
c 38 13
5
1
10
21
b 35, 3535
7
15
b 255
d 700 59
a 22, 1210
4
12
a 765
48
x +1
40, −20
y
R
ev
ie
w
3
8
9
a + 6d , a + 18d
op
C
2
−8, 2
c −85
Exercise 6C
1
7
w
a 1 − 4x + 7 x 2
ev
ie
12
C
U
ni
op
y
Answers
Copyright Material - Review Only - Not for Redistribution
11π
8
b Proof
b
ve
rs
ity
-R
es
Pr
ity
rs
ve
c 15
d 225, maximum
a Proof
b 625 − ( r − 25)2
c 25
d 625, maximum
2
40 000
200 
b
−πr−

π
π 
40 000
, maximum
d
π
ii 0.9273
a Proof
i
Proof
ni
U
i
r 2 (tan θ − θ )
ii 12 + 12 3 + 4 π
14
i
2 − 5 cos2 x
ii −3 and 2
16
i
-R
b 2187
i
Proof
ii 26.6°, 153.4°
−5 < f( x ) < 3
ii (0.253, 0), (0, −1)
y
4
es
s
-C
iii
a d = 2a
op
y
b 99a
15
a a = 60, d = −10.5
b 42 23
16
a 17
b r = − 5, S = 7
4
7
17
a
18
i
19
a a = 10, b = 45
π
b i 0 ,θ ,
ii 1.125
3
i x = −2 or 6, 3rd term = 16 or 48
Pr
b n = 22
ity
2
–π
2
ni
ve
rs
–π
4
π
4
op
U
ii 22100
w
e
C
–4
-R
16
27
s
ii
–6
ev
ie
id
g
br
O
–2
b 16
am
y = f(x)
y
1
5
41000
-C
y
b 225 − ( r − 15)2
13
15
a a = 44, d = −3
C
a Proof
iii 0.685, 2.46
es
ie
Proof
iii 5.90 cm 2
b − 17 , 3
7
ii 243 − 405x + 270 x 2
14
w
ie
ev
11
12
id
am
2
a −
3
c 1312.2
br
23
i
b 112
c Proof
ge
1 + 10 x + 40 x 2
a i
b 5940
R
a 14
625
8
s
-C
y
C
w
135
2
a 6561x16 − 17 496x15 + 20 412 x14
a 1 + 8 px + 28 p2 x 2
20
6
10
b −37 908
13
a −25.6
y
7
12
5
b 12
b 27 79
-R
40
11
a 1 − 10 x + 40 x 2
9
br
6
10
4
ii 35.3°, 144.7°, 215.3°, 324.7°
am
5
864
25
16800
−
9
a 729x 6 − 2916x 3 + 4860 b −5832
op
2
8
ev
ve
rs
ity
ni
3
3
U
5
3840
8
ge
2
2
7
id
240
x180
C
op
op
1
4
R
b
End-of-chapter review exercise 6
op
R
ev
ie
w
7
308
3
,n=6
2
a x = −3 or 5, 3rd term = 24 or 40
4
b −
5
a 4
C
6
1
Pr
es
s
-C
b 12.96, 68
y
5
Cross-topic review exercise 2
C
−2.5, 22.5
3
a
5
w
4
b 384, 32
ie
a 2
a d = 6, a = 13
w
3
22
b 16
ie
a 100
ev
2
b 115.2°
b a = 12 , r = 5
7
7
w
b 788.125
a 2 14
ev
a 352
am
br
id
1
21
ev
ie
ge
Exercise 6F
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
Copyright Material - Review Only - Not for Redistribution
π x
2
ve
rs
ity
7
3
4
− 3
2
x
x
1
3
i 6 x+
+
2 x
x3
a 3
5
c −
4
15
8
−3
9
−8
10
( −2, −10), (2, − 6)
g 14x +
17
−2 , x , 3
18
x < −1 and x >
19
Proof
op
b 9x 8
1
d − 2
x
2
f
3
3 x
e 10(5x − 2)7
C
U
ie
-R
ev
g −56(4 − 7 x )3
1
( x + 2)2
16
c
(3 − 2 x )2
72
e −
(3x + 1)7
50(2 x − 1)4
21
(3x − 1)6
h
5
j −16x(2 − x 2 )7
f
l
d
y
f
16( x + 1)
x 2 ( x + 2)2
1
a
2 x −5
2x
c
2x2 − 1
h
b
es
s
-R
ev
ie
3
b 16(2 x + 3)7
8
91
d
x + 1

22
b
w
U
e
id
g
br
am
k 6x 2 ( x + 2)( x + 4)2
g −
b 6x 2 + 8
e 16x 3 − 24x
10 x( x 2 + 3)4
a −
2
0
d 2x + 1
10
2
f
− 2
3
x
x
i
op
Pr
es
s
2
1
h − 5
x
c 10 x − 3
309
C
f
w
ge
id
c 3x5
1
2
y
a 6( x + 4)5
1
c −20(3 − 4x )4
b 15x 4
3
d − 2
x
a 10 x − 1
y
a = −10.5, b = 18
ev
−2
a 8x 3
4
-C
16
ni
ve
rs
C
w
ie
ev
R
a = 4, b = −6
−3
d
10
e − 3
3x
2
g
x
5
15
b −8, 2
Exercise 7B
ve
b
op
y
4
2.95 2.99
h 3x
g 5x
a = −5, b = 2
-R
s
2.8
-C
e 0
14
ni
4
1
c
2
a 5x 4
4
c − 5
x
a
br
R
3
Pr
es
2.5
EF
ity
2
DF
rs
Gradient
CF
b 3
2
ev
AF BF
Chord
am
ie
w
C
op
a
a = 2, b = −7
ity
y
Exercise 7A
1
13
b 0.5
C
op
ni
ge
id
br
am
-C
4
a 4( x − 2)−3
3
−
2
y = 2x + 1
12
5
4
a ( −2, 7), (3, −8)
11
U
R
1 −1
d
x
2
5
2
f − x3
5
b 2(3x + 1)−5
e 3x −2
3
w
b 5x 3
1 2
x
2
c
ev
ie
2
3
a 3x 2
1
2
ve
rs
ity
1
ev
ie
w
C
Prerequisite knowledge
5
1
+
2
x
4 x3
w
op
y
7 Differentiation
h 3−
ie
-C
ii 70
-R
Proof
b i
6
Pr
es
s
17
am
br
id
ge
 x +1
iv f −1( x ) = sin−1 
 , domain is −5 < x < 3,
 4 
1
1
range is − π < f −1( x ) < π
2
2
a 250
C
U
ni
op
y
Answers
Copyright Material - Review Only - Not for Redistribution
d
5( x 3 − 5)4 (2 x 3 + 5)
x6
3
−
( x − 5)2
32 x
− 2
( x + 2)2
45
−
6
2 ( 3x + 1 )
49(4x − 5)
−
(2 x − 5)8 x8
1
2x + 3
3x 2 − 5
2 x 3 − 5x
ve
rs
ity
d 2y = x − 1
ni
a 4y = x + 4
b y = x −1
7
( −2.5, 8.5)
7
9
a y = 4x − 68
20
a 2+ 3
x
( −7.5, 2.5)
10
( −6.2, 6.6)
11
a ( −32.6, 28.4)
12
a (2 3, 8 3 ), ( −2 3, −8 3 )
C
w
−
+
+
+
+
ev
11
a Proof
12
a = −4, b = 4
Pr
ity
rs
ve
ni
U
b −8, 8
C
1
3x 3 +
2
−3 59
7
4x 2
-R
Proof
4
−15(3 − 5x )2 − 2, 150(3 − 5x )
es
27
2 (3x + 1)5
10
a y = −4x + 18
h 24x 2 − 36x + 20
11
a Proof
ni
ve
rs
7
ii Proof
w
y=−
( −0.8, 15.5)
1
x
2
es
s
12
b y=
1
x +1
4
ie
ev
b i
-R
id
g
ii (16.5, 0)
op
ity
6
−
C
Pr
(4.25, −7.5)
5
y
s
3
9
f
br
am
−
Proof
U
4
(4x − 9)3
4x − 30
g
x4
5x + 12
i −
4 x5
−
10
d 48(2 x − 3)2
e −
+
e
36
x4
-C
w
ie
ev
R
c −
+
Proof
b 30 x − 14
a 2
0
8
b (0.4, −5.48)
Exercise 7D
1
−
9
ge
a −7
−
8
15
i 4x + 5 y = 66
24
i 5− 3
x
37.4
C
16
0
x.5
id
-C
73
op
y
15
+
8
br
y = 0.6x + 1.6, 30.96°
dy
dx
End-of-chapter review exercise 7
am
14
7
−
es
-C
y
op
C
R
ev
ie
w
(4, −1)
6
d2 y
dx 2
b 317.2 units2
13
5
+
-R
b (0.6, 2.48)
b Proof
4
s
am
b y = 4x − 7.5
b (17, 0)
3
op
6
2
ie
(0, 7.5)
1
w
5
0
ie
a Proof
b x + 4y = 0
2
81
x
id
4
br
a x + 4y = 4
8
c 10
48
−
(2 x − 1)9
d 5x − 6 y = 3
3
b 4
y
y = 3x + 9
6
U
b 8x + y = 17
c x − 6y = 9
310
a −1
a y = 3x − 7
c
2
4
5
ge
1
R
ev
ie
Exercise 7C
4 − 8(2 x − 1)3 , −48(2 x − 1)2
y
ve
rs
ity
a = 5, b = 3
w
8
f
C
op
y
op
C
7
9
4 (1 − 3x )3
80
3
3 (2 x + 1)7
d −
3
w
4
3
(5, 1)
4,
6
3
x3
ev
12
b −
15 x
−6
4
e
Pr
es
s
5
30 45
−
x 4 x7
4
45
−
3
x
4 x7
a
c
-C
10
2
ev
ie
h
f
-R
1
(2 x − 5)3
g −
4
3
3x + 1
6
3 (2 − 3x ) 4
ge
2
3 3 (5 − 2 x )2
am
br
id
e −
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
Copyright Material - Review Only - Not for Redistribution
ii (2.1, 8.25)
ve
rs
ity
ii 10 34
Proof
14
i
Proof
15
1
iii  , 4 34 
2

1
i y − 2 = ( x − 1), y − 2 = −2( x − 1)
2
ii 2 21
c ( −3, −12) maximum; (3, 0) minimum
b (2, − 2), maximum
-R
a a = −15, b = 36
Proof
d x , 0 and x . 8
e x , 1 and x . 4
f
a x , 1 13
b x . 4.5
c 2,x,5
d −1 , x , 3
ity
rs
ve
1
3 x
2
a a = −6, b = 5
b minimum
c (0, 5), maximum
d (2, − 11), −12
a a = 4, b = 16
b minimum
y
op
ni
a a = 54, b = −22
b minimum
C
b ( −1, 14)
ie
w
ge
a a = −3, b = −12
br
ev
id
c (2, −13) = minimum, ( −1, 14) = maximum
d (0.5, 0.5), −13.5
-R
Exercise 8C
s
ii 97.2
Q = 5x − 36x + 162
40 − 2 r
b Proof
θ=
r
r = 10
d A = 100, maximum
50 − x
y=
b Proof
2
A = 312.5, x = 25
c
3
a
c
y
a
U
b ( −0.5, 6.25) maximum
2
ii 108
2
op
a (2, 4) minimum
P = 9x 2 − x 3
C
ity
Exercise 8B
b i
c i
ni
ve
rs
7 , x , 20
a y = 9−x
1
es
Proof
op
y
8
a 3x 2 − 10 x + 160
d ( −3, −17) minimum; (1, 15) maximum
b x = 1 23 , 151 23 cm 2
5
a Proof
b A = 37.5, maximum
ie
-R
a QR = 9 − p
c
2
p= 3
s
am
(1, − 7) maximum; (2, −11) minimum
-C
f
6
ev
id
g
br
e ( −1, − 4) minimum
w
4
e
c ( −2, 22) maximum; (2, −10) minimum
es
C
O
c x , 0 and 0 , x , 3
13
1.5 , x , 3.5
8
, increasing
(1 − 2 x )2
2x − 6
, neither
( x + 2)3
Proof
8x + 20, 8x + 20 ù 0, if x ù 0
1
12
x , − 4 and x . 2
7
9
1
c x , 0 and x . 2
Pr
6
11
am
5
10
−2 23 , x , 2
-C
4
f
2
–1
U
R
3
y
w
ie
ev
c x , −1.75
x , 2 and x . 6 23
3
-R
b x .1
Pr
a x.4
e
4
s
18
(1 − 3x )3
es
-C
b
y
2
C
op
ve
rs
ity
ni
id
am
a 10(2 x − 1)4
(2, 6) minimum
y
br
9
9 x
,
4 x
2
f
3+ k
3−k
, minimum; x =
, maximum
2
2
(0, 1), minimum; (1, 2), maximum;
(2, 1), minimum
x=
b −2 , x , 3
3 9
b − 3 , 4
x x
op
C
w
6
7
U
a 6x − 1, 6
ev
ie
b −3 , x , 1
ge
2
1
w
a a=3
9
Exercise 8A
ie
5
8
3
ev
b −44, 81
8 Further differentiation
c
R
4
e (4, 4) maximum
18 18
,
≠0
x3 x3
a −2, 3
3
Pr
es
s
y
op
C
w
ev
ie
R
ev
ie
d ( −2, −28) maximum; (2, 36) minimum
-C
am
br
id
9 9
ii  − , 
 2 4
6 12 
iii  ,
, E not midpoint of OA
 11 11 
a x , −1 and x . 3
b (1, 12) minimum
w
ge
i
1
a (9, 6) minimum
2
13
Prerequisite knowledge
C
U
ni
op
y
Answers
Copyright Material - Review Only - Not for Redistribution
b Proof
d A = 12 3, maximum
311
ve
rs
ity
b V = 486, x = 3
3
0.09 units per second, increasing
7
1, 6
8
0.016 units per second
1
,3
3
w
ev
ie
y
C
op
w
ie
b
1.024 cm s −1
5
4π
π
cm s −1
10
ii 8 π cm2 s −1
3
Maximum
4
k = 0.0032 kg cm 3 , 0.096 kg day -1
5
ev
s
ity
8
op
C
w
ie
i
A = 2 p2 + p3
ii 0.4
i
Proof
ii 20 m by 24 m
i
Proof
ii 120, minimum
4(6 − x )
ii Proof
3
iii A = 72
d 2 y 16
dy
8
i
= − 2 + 2, 2 = 3
dx
x
dx
x
d2 y
ii (2, 8), minimum since 2 . 0 when x = 2
dx
d2 y
( −2, −8), maximum since 2 , 0 when x = −2
dx
πx
i y = 30 − x −
ii Proof
4
iv Maximum
iii x = 15
i
y=
id
g
w
e
C
U
op
y
ni
ve
rs
9
y
2
2
cm s −1
45 π
300 π m 2 hr −1
10
es
s
-R
br
am
-C
C
10
cm
π
a 40 π cm3 s −1
a 2
ie
6
ie
32 π cm2 s −1
-R
1.25 units per second
es
5
Pr
0.08 units per second
1
cm s −1
3
-R
s
es
Pr
ve
ni
U
4
ev
R
id
w
op
y
3
9
13
7
1
units per second
300
−0.04 units per second
2
a Proof
6
−0.315 units per second, decreasing
1
10
1
d 1241, maximum
-C
Exercise 8D
a
3
cm s −1
120
1
b
cm s −1
7
9
b −
cm s −1
100 π
b
End-of-chapter review exercise 8
ge
c 13 13
9
b i
b Proof
br
a r = 20 h − h2
am
15
a Proof
ev
w
ie
ev
R
14
8
12
rs
13
7
0.125 cm s −1
1
cm s −1
320
9 π cm2 s −1
11
ity
op
C
312
5
ev
id
br
am
-C
y
12
π cm3 s −1
0.003 cm s −1
Pr
es
s
ge
U
R
11
4
π cm 2 s −1
5
18 cm3 s −1
4
6
ni
10
2
ve
rs
ity
w
C
op
y
9
1
-R
am
br
id
c Maximum
288
b Proof
a y= 2
x
c 432, 12 cm by 6 cm by 8 cm
1
1
a y = 1 − x − πx b Proof
2
4
dA
1
d2 A
1
c
= 1 − x − πx ,
= −1 − π
2
dx
4
4
dx
2
4
d
e A=
, maximum
4+π
4+π
5 − 2 r − πr
a h=
b Proof
2
dA
d2 A
c
= 5 − 4r − πr, 2 = −4 − π
dr
dr
5
25
e
, maximum
d
4+π
8 + 2π
50 − 2 x
a r=
b Proof
π
100
2500
= 14.0,
= 350, minimum
c
( π + 4)
(4 + π)
432
a h= 2
b Proof
r
d A = 216 π, minimum
c r=6
2
500 − 24x
b Proof
a y=
10 x
5 10
d Proof
c
6
160 3
a h=
− r
b Proof
2
r
c r=8
-C
8
ev
ie
Exercise 8E
ge
a Proof
7
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
C
5 364 
b Maximum at  − ,
, minimum at (1, 4)
 3 27 
Pr
es
s
y
f
5
3
y = 2x 2 + 2x 2 + 2
f( x ) = 4 + 8x − x 2
y
C
w
ie
ev
-R
s
4
x
1 2
x − 10 x + 3
2
b x , − 2 and x . 1 23
a y = 2 x 3 + 6x 2 + 10 x + 4
10
y = 2 + 4x − 2 x − x
11
a −3.5
y
9
2
w
-R
13
Proof
b P = minimum, Q = maximum
1
a −6
b y = x 2 − 6x + 2
2
2
2
f ′( x ) = 2 x − 2, f( x ) = x 2 + − 4
x
x
ev
12
b
3
s
b 5x 4 + c
2
d − 2 +c
x
b y = 42 x − 97
a y = x3 +
x +c
id
g
br
a 2x6 + c
3
c − +c
x
313
(4, 20)
O
e
5
w
y
s
7
es
ni
ve
rs
3
+ 6x 2 + c
y=
7
−
5
12 x 2
e
am
4
7
2x 2
-C
R
ev
ie
w
d
y = 2 x + 3x − 7
8
U
C
c
es
ity
rs
ve
ni
U
ge
op
y
b
es
a
Pr
3
6
2
y
ity
d
2 2
x −1
3
5
4
id
c
y = 2 x − 2x +
3
+2
x
y = 2 x 3 − 6x 2 + 5x − 4
3
y = 5x 3 + 2 − 2
x
a y = 2x2 x + 2x − 1
br
R
b
3
f
y=
2
3
am
a
1
y=− 2 +c
4x
x4
f( x ) = x5 −
+ 2x + c
2
x6 x3
f( x ) =
+
− x2 + c
2
3
1
8
f( x ) = 3x 2 − 2 − + c
x
x
3
3
f( x ) = − 6 + − 4 x + c
x
2x
x 3 5x 2
y=
+
+c
3
2
3x 4 2 x 3
y=
+
+c
2
3
x4
− 2 x 3 − 8x 2 + c
y=
4
x2
1
5
y=
−
+ +c
2
x
4
4x
Pr
-C
e
y = 3x 4 + c
-C
2
ev
ie
w
C
op
c
w
ie
5
x
b y = 2x7 + c
3
d y=− +c
x
f y =8 x +c
y
a y = 5x 3 + c
b y = 2x3 − x2 + 5
6
d y = x2 + − 4
x
a y = x3 + x + 2
4
c y = 10 −
x
e y = 4 x −x+2
ev
b 10 x − 4 +
Exercise 9A
1
4
C
op
ni
U
ge
1
b (0, 9)
a 24x 7 − 13
10
Exercise 9B
-R
3
3x 3 3x 3
+
+c
d
10
4
x
3
−
+c
2 2 x2
3
2x 2
20
x2 4 x
g
+
+c
h
+
+c
7
x
6
3
12
9
−
+c
i 2 x2 +
x
4x4
b −1
id
2
a −19
2
a − , 5

 3
br
1
am
R
Prerequisite knowledge
8x 2
c −
+ 2x2 + x + c
3
x
1
+
+c
e
f
2 2x
7
ve
rs
ity
ev
ie
w
C
op
 2 p 4 p3 
ii (0, 0) minimum,  − ,
maximum
 3 27 
iii 0 , p , 3
9 Integration
x3
− 3x 2 + 9x + c
3
op
 2 p 4 p3 
 − 3 , 27 
b
op
i
−
C
13
ii Maximum
10
+c
x
f
ie
Proof
5
-R
i
-C
12
4 x
+c
3
x 3 5x 2
a
+
+ 4x + c
3
2
e
ev
ie
4
3
ge
a −2 , x ,
am
br
id
11
U
ni
op
y
Answers
Copyright Material - Review Only - Not for Redistribution
ve
rs
ity
a y = 9 + 3x − x 2
16
a y = 2 x x − 6x + 10 b (4, 2), minimum
17
(1, 7), maximum
18
a y = x 2 − 5x + 2
1
y
w
ev
b 84 25
ie
-R
es
s
b 8
20 56
d 5 13
op
a 11 56
C
3
b 40 21
d 14
e 16
f
5
b 6
6
a 48 3
4
3 13
7
10 23
8
a 9
b 1.6
9
a Proof
b 3− 5
10
18 23
11
a ( −1, 0)
12
10 21
13
34
op
C
w
s
es
9
y
ev
26
-R
14
ie
s
es
Pr
U
1
( x + 3)8 + c
4
1
b
(2 x x − 1)5 + c
5
b
ie
-R
w
Proof
y
a 21 13
c 15.84
ity
a 15 x (2 x x − 1)4
( x + 1)4
4 x
a
b 5 21
id
g
br
7
am
6
b 2 151
f
a 7 51
e
b
d 4
a 15x 2 ( x 3 − 2)4
Pr
ity
U
( x − 3x + 5)
+c
3
4( x + 3)7
a
x
6
26
3
d 21121
ni
ve
rs
a 6(2 x − 3)( x 2 − 3x + 5)5
-C
ie
6x
(4 − 3x 2 )2
2
R
ev
5
1 2
( x + 2)4 + c
8
1
b
(2 x 2 − 1)5 + c
20
2
b − 2
+c
x −5
1
+c
b
8 − 6x 2
b
3
c 4 32
id
-C
op
y
C
a
w
4
a k = −2
f 18
8
4
b
45
2
4
b
a 20 x(2 x 2 − 1)4
2
3
br
am
a 8x( x 2 + 2)3
1
4
15
4x
a − 2
( x + 5)2
c
ge
y = 4 2x − 5 − 2
Exercise 9D
6
1
b y = 5 2x − 3 − 4
b y = 8 3x + 1 − 2 x 2 − 2 x + 5
6
d
ev
a Maximum
b 3
Exercise 9F
rs
5
5
ve
a x + 5y = 7
4
ni
C
y = 3( x − 5)4 − 1
4
R
ev
ie
w
3
ve
rs
ity
ni
U
y = 2 x−2 +5
op
c
314
2
5
e 2
3
-C
a
a 10
c
1
b y = (2 x + 5) 2 − 7
3
2
d y=
+6
3 − 2x
y
2
am
i
3
ge
br
g
1
(3x + 1)6 + c
18
1
d − (1 − 2 x )6 + c
4
5
1
(2 x + 1) 2 + c
f
5
2
h −
+c
(2 x + 1)2
b
id
e
1
(2 x − 7)9 + c
18
2
(5x − 2)9 + c
45
4
3
− (5 − 4x ) 3 + c
16
4
3x − 2 + c
3
5
+c
32(7 − 2 x )4
1
y = (2 x − 1)4 + 2
8
11
2
107
c
6
37
e
8
a
C
op
Pr
es
s
y
op
C
w
ev
ie
R
c
c −6
2
Exercise 9C
a
16
9
d 21
5
f
2
b
e 9
b x + y = −1
c (1, −2)
1
a 7
-R
am
br
id
-C
b 5 y = x + 21
ev
ie
15
ge
( −11, 408 13 )
w
Exercise 9E
14
C
U
ni
op
y
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
Copyright Material - Review Only - Not for Redistribution
b 90
ve
rs
ity
3
57 61
4
y
a y=
9
a y = −8x + 16
b 108
1
10
a 2y = x − 1
b 8.83
2
ve
br
-C
16 π
3
25 π
d
4
124 π
b
15
y
s
es
Pr
ity
ni
ve
rs
op
3
C
i
w
 20 

 4,

3 
13
ii P and Q are both
14
i
y = −24x + 20
-R
e
b 24 π
a ∪-shaped curve, y-intercept = 4,
vertex = (2, 0)
32 π
b
5
1
iii f( x ) = 2 x 2 + 6x 2 − 10 x + 5
U
39 π
4
a 24 π
i
29.7°
9
8
ii 1
ev
id
g
s
15
es
6
11
12
br
5
C
op
C
-R
am
10
b
am
4
-C
3
w
w
9
Proof
ie
f
a ∪-shaped curve, y-intercept (0, 11),
vertex at (3, 2)
483π
b
5
9
i Proof
ii
4
i B (0, 1), C (4, 3)
ii y = −3x + 15
2π
iii
15
5π
iii
i Proof
ii 1
3
1
i
or 9
9
1
3
3 −
3 −
1
ii f ′′( x ) = x 2 − x 2 at x = max, at
2
2
9
x = 9 min
ev
id
d Proof
op
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C
2
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b Proof
71π
5
15 π
c
8
81π
a
2
6
a
ev
7
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Exercise 9I
1
5
y
a Proof
4
op
3
25 3
4
x − 20 x − + c
x
3
6 5 2
y = 30 − − x
x 2
4
f( x ) = 6 x + 2 + 2 − 10
x
194 π
9
a 1
b f( x ) = 3x 2 − 6x + 8
10 23
ie
h 1
2
g 6 3
20
i
9
Proof
es
f 16
Pr
e 50
ity
d 4
rs
5
4
s
1
b
256
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a 2
3
f( x ) = 3x 4 + 5x 2 − 7
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End-of-chapter review exercise 9
e Proof
w
b 171π cm 3
8
c Proof
ie
C
13
128 π
3
32 π
b
5
b
U
C
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R
ev
12
Proof
52 π
a
3
8π
a
3
1888 π
cm 3
a
3
Proof
250 π
9
14
c −
ev
11
b
1
(2 3 − 3)
2
b 64
1
R
3125 π
6
b 16 π
b
1
x+2
3
a y = −3x + 46
Exercise 9H
R
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7
a (0, 3)
1 
a  , 7  , (2, 7)
3 
-R
Pr
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b
10 23
op
6
8
10
a 36
1
3
1 13
a (25, 0)
9
c 36
5
7
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2
10 23
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26 23
am
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1
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Exercise 9G
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Answers
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15
7
a Proof
b Proof
1250
= 175 (3 s.f.), maximum
c
( π + 4)
i 12
ii x = −1 or x = −3, 1 13
3
Proof
4
a 2187 − 10 206x + 20 412 x 2
3
a
(4 π − 3 3 ) cm 2
2
b (2 π + 3 + 3 3 ) cm
b −30 618
a ( x − 3)2 + ( y + 2)2 = 20
b x − 2 y = 17
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5
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6
7
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dy
9
dy
i
=
, no turning points since
≠0
2
dx (2 − x )
dx
81π
iii k , −8, k . 4
ii
2
8
8
<0
a f ′( x ) = −
2,
(2 x + 1) (2 x + 1)2
4−x
b f −1( x ) =
,0,x<4
2x
c y
8
rs
9
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6
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1
3
1 −2 1 −2
x + x
2
2
y=
i
2
2
y − 6 = − ( x − 2) ii y = x 2 + + 1
7
x
10
a 3x + 4 y = 17
11
a −
ii x = −1 (max), x = 2 (min)
16
32
− 2 x, 3 − 2
x2
x
431π
5
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3π
2
b
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i 13
π
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iii x = 1, minimum since f ′′(1) . 0
13
π
2
2π x
c 0.927 rad, 5.36 rad
6−x
d g−1( x ) = cos−1 

 5 
ity
ii
O
op
1
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3
2
s
x
4
i
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f
Pr
-C
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4
2 2
2
x − 2x 2 −
3
3
iii (1, −2), minimum
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12
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C
10
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12
y=x
4
27
8
b (3, 21)
ii
a 1 < f( x ) < 11
b
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f–1
a 7, −8
1
b i
3
a 21 − 2( x − 3)2
c x < 3 − 13, x > 3 + 13
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10
y
i
2
f ′( x ) = 2 +
y
6
316
284 π
3
C
op
i
9
ii
15
. 0 for all x
x4
y = − x 3 + 6x 2 − 12 x + 11
1
ni
5
8
Proof
Practice exam-style paper
ve
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3
y=−
+ 1 ii x . 1, x , − 2
2(1 + 2 x )

1
1
ii
 , 0
2

6
4
2
ii −
9
27
3
40
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ii x = 4, minimum
y
4
C
i
i
17
4
op
3
y = 2 x 2 − 6x + 2
3
ii
w
16
(3x + 4) 2 + 12
3
Pr
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ii y = 2 x −
5

 3
B  , 0  , C  0, 
4

 4
w
2
i
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14
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2
y = x 3 − 3x + 7
3
i −0.6
w
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Cross-topic review exercise 3
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1
units per second
240
b ( −2, −12), maximum
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General form of a circle: the equation
x 2 + y2 + 2 gx + 2 fy + c = 0 , where ( − g, − f ) is the centre
ni
B
and g 2 + f 2 − c is the radius of the circle
Geometric progression: each term in the progression is a
constant multiple of the preceding term
Gradient function: the derivative f ′( x ) is also known as the
gradient function of the curve y = f( x )
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Chain rule: the rule for computing the derivative of the
composition of two functions
Common difference: the difference between successive
terms in an arithmetic progression
Common ratio: the constant ratio of successive terms in a
geometric progression
Completed square form: the equation
( x − a )2 + ( y − b )2 = r 2, where ( a, b ) is the centre and r is
the radius of the circle
Completing the square: writing the expression ax 2 + bx + c
in the form d ( x + e )2 + f
Composite function: a function obtained from two given
functions by applying first one function and then applying
the second function to the result
Convergent series: a sequence that tends to a finite number
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Decreasing function: a function whose value decreases as
x increases
Definite integral: an integral between limits whose result
does not contain a constant of integration
dy
Derivative: denoted by
of f ( x ); gives the gradient of a
dx
curve
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Differentiation: the process of finding the gradient of a
curve
Differentiation from first principles: the process of finding
the gradient of a curve using small increments
Discriminant: the part of the quadratic formula
underneath the square root sign
Domain: the set of input values for a function
N
ev
Normal: the line perpendicular to the tangent at a point on
a curve
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Many-one: a function which has one output value for
each input value but each output value can have more
than one input value
Mapping: a diagram to show how the numbers in the
domain and range are paired
Maximum point: a point, P, on a curve where the value
of y at this point is greater than the value of y at other
points close to P
Minimum point: a point, Q, on a curve where the value of
y at this point is less than the value of y at other points
close to Q
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Identity: a mathematical relationship equating one
quantity to another
Improper integrals: a definite integral that has either one
limit or both limits are infinite, or a definite integral where
the function to be integrated approaches an infinite value
at either or both endpoints in the interval (of integration)
Increasing function: a function whose value increases as x
increases
Indefinite integral: an integral without limits whose result
contains a constant of integration
Integration: the reverse process of differentiation
Inverse function: the inverse of a function, f −1( x ) , is the
function that undoes what f( x ) has done
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Basic angle: the acute angle made with the x-axis
Binomial: a polynomial with two terms
Binomial coefficients: the coefficients in a binomial expansion
Binomial theorem: the rule for expanding (1 + x ) n or ( a + b ) n
C
Factorial: 6 × 5 × 4 × 3 × 2 × 1 = 6 ! (read as ‘6 factorial’)
First derivative: see Derivative
Function: a rule that maps each x value to just one y value
for a defined set of input (x) values
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Amplitude: the distance between a maximum (or minimum)
point and the principal axis of a sinusoidal function
Arithmetic progression: each term in the progression differs
from the term before by a constant
Asymptote: a straight line such that the distance between a
curve and the line approaches zero as they tend to infinity
ie
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Glossary
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Range: the set of output values for a function
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Cambridge International AS & A Level Mathematics: Pure Mathematics 1
Reference angle: the acute angle made with the x-axis
d2 y
or f ″ (x ) and is used to
dx 2
determine the nature of stationary points on a curve
Second derivative: denoted by
Series: the sum of the terms in a sequence
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Self-inverse function: a function f where f −1( x ) = f( x ) for
all x
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Solid of revolution: the solid formed when an area is
rotated through 360° about an axis
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Stationary point: a point on a curve where the gradient is
zero, also known as a turning point
ev
Tangent: a straight line that touches a curve at a point
Term: a number in a sequence
Turning point: a point on a curve where the gradient
is zero, also known as a stationary point
op
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Radian: one radian is the angle subtended at the centre of
a circle by an arc that is equal in length to the radius of a
circle
ev
V
Vertex: the vertex of a parabola is the maximum or
minimum point
Volume of revolution: the volume of the solid formed when
an area is rotated through 360° about an axis
ity
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am
Quadrant: the Cartesian plane is divided into four
quadrants
−b ± b2 − 4ac
Quadratic formula: the formula x =
,
2a
which is used to solve the equation ax 2 + bx + c = 0
ie
id
T
br
Q
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Pr
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Parabola: the graph of a quadratic function
Pascal’s triangle: a triangular array of the binomial
coefficients, where each number is the sum of the two
numbers above
Period: the length of one repetition or cycle of a periodic
function
Periodic functions: a function that repeats its values in
regular intervals or periods
Point of inflexion: a point on a curve at which the direction
of curvature changes
Principal angle: the angle that the calculator gives is called
the principal angle
ev
ie
P
Roots: if f( x ) is a function, then the solutions to the
equation f( x ) = 0 are called the roots of the equation
-R
am
br
id
One-one: a function where exactly one input value gives
rise to each value in the range
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completing the square 6–8
graph sketching 19
proof of the quadratic formula 10
composite functions 39–41
conic sections 71
connected rates of change 228–9
practical applications 230–2
constant of integration 244–5
convergent series 176
cosine
angles between 0° and 90° 118–20
general angles 123–6
graph of 128–9
inverse of 137, 138
transformations of 132
trigonometric equations 140–4
trigonometric identities 145–7, 149
basic angle (reference angle) 121–2
binomial coefficients 160–4
binomial expansions 156–8
binomial theorem 161
decreasing functions 213–15
definite integration 250–2
area bounded by a curve and a line
260–1
area bounded by two curves 260–1
area enclosed by a curve and the
y-axis 255–6
area under a curve 253–5
improper integrals 264–7
volumes of revolution 268–70
degrees 100
converting to and from radians
101–2
derivatives 193
first and second 205–6
see also differentiation
Descartes, René 83
differentiation 191–6
addition/subtraction rule 195
chain rule 198–200
constants 195
increasing and decreasing functions
213–15
notation 193
power rule 194–5
practical applications of connected
rates of change 230–2
practical maximum and minimum
problems 221–3
rates of change 227–9
real life uses of 212
Pr
es
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Galileo Galilei 2
Gauss, Carl 167
geometric progressions 171–2, 180
infinite geometric series 175–8
sum of 173–4
gradients 75, 192–3
and equation of a straight line 78–80
of tangents and normals 201–3
see also differentiation
s
-R
ev
factorial notation 163–4
factorisation, quadratic equations 3–5
first derivative test 217
first derivatives 205
functions
composite 39–41
definitions 34
domain and range 35–7
graph of a function and its inverse
48–9
increasing and decreasing 213–15
inverse of 43–5, 48–9
many-one 35
one-one 34–5
quadratic see quadratic functions
self-inverse 44
transformations of 51
combined transformations 59–63
reflections 55–6
stretches 57–8
translations 52–3
trigonometric see cosine; sine;
tangent; trigonometry
use in modelling 34
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calculus 191
see also differentiation; integration
Cantor, Georg 181
Cartesian coordinate system 83
quadrants of 122
chain rule, differentiation 198–200
connected rates of change 228–9
circles
area of a sector 107–8
equation of 82–7
intersection with a line 88–90
length of an arc 104–5
right-angle facts 85
codomain (range) of a function 35–7
inverse functions 43, 45
combined transformations of functions
59–60
trigonometric functions 132
two horizontal transformations 61–3
two vertical transformations 60–1
common difference, arithmetic
progressions 166
common ratio, geometric progressions 171
communication vi
completed square form, equation of a
circle 83
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equation of a circle 82–3, 86–7
completed square form 83–4
expanded general form 84–5
equation of a straight line 78–80
tangents and normals 202
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scalar multiple rule 195
second derivatives 205–6
stationary points 216–19
tangents and normals 201–3
differentiation from first principles 193
discriminant 24
domain of a function 35–7
inverse functions 43, 45
-R
addition/subtraction rule,
differentiation 195
amplitude of a periodic function 129
angles
degrees 100
general definition of 121–2
radians 101–2
arcs, length of 104–5
area bounded by a curve and a line
260–1
area bounded by two curves 260–1
area enclosed by a curve and the y-axis
255–6
area under a curve 253–5
arithmetic progressions 166–7, 180
sum of 167–9
asymptotes 129
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Index
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Index
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scalar multiple rule, differentiation
195
second derivatives 205–6
and stationary points 218–19
sectors, area of 107–8
self-inverse functions 44
graphs of 48
sequences
arithmetic progressions 166–9
geometric progressions 171–4
infinite geometric series 175–8
series 156, 167
binomial coefficients 160–4
binomial expansion of (a + b)n 156–8
simultaneous equations 11–13
sine
angles between 0° and 90° 118–20
general angles 123–6
graph of 128–9
inverse of 136, 138–9
transformations of 130–1
trigonometric equations 140–4
trigonometric identities 145–7, 149
sketching a graph 17–19
quadratic inequalities 22
solids of revolution 268
sphere, surface area of 14
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ity
quadrants of the Cartesian plane 121
quadratic curves, intersection with a
line 27–8
quadratic equations
completing the square 6–8
functions of x 15–16
number of roots 24–5
simultaneous equations 11–13
solution by factorisation 3–5
es
s
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radians 101
area of a sector 107–8
converting to and from degrees
101–2
length of an arc 104–5
simple multiples of π 102
range (codomain) of a function 35–7
inverse functions 43, 45
rates of change 227–8
connected 228–9
practical applications 230–2
recurring decimals 176
reference angle (basic angle) 121–2
reflections 55–6
revolution, volumes of 268–70
right-angle facts for circles 85
roots of a quadratic equation 24–5
op
parabolas 17–19
intersection with a line 27–8
lines of symmetry 17–19
maximum and minimum values
17–19
paraboloids 18
parallel lines 75
Pascal’s triangle 157–8
periodic functions 129
perpendicular lines 75
points of inflexion 216
power rule, differentiation 194–5
principal angle 137, 138
problem solving vi
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Lagrange’s notation 193
latitude 104
Leibnitz, Gottfried 193, 239
length of a line segment 72–4
limits of integration 250
quadratic formula 10
quadratic functions 2
graph sketching 17–19
lines of symmetry 17–19
maximum and minimum values
17–19
quadratic inequalities 21–3
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one-one functions 34–5
oscillations 117
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Newton, Isaac 193, 239
normals, gradient 201–3
nth term of a geometric progression
171–2
nth term of an arithmetic progression
166–7
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320
many-one functions 35
mappings see functions
maximum points 17–19, 216–17
practical problems 221–3
and second derivatives 218–19
mid-point of a line segment 72–4
minimum points 17–19, 216–17
practical problems 221–3
and second derivatives 218–19
modelling vi–vii
ni
identities, trigonometric 145–7, 149
image set of a function see range
(codomain) of a function
improper integrals
type 1 264–6
type 2 266–7
increasing functions 213–15
indefinite integrals 241
inequalities, quadratic 21–3
infinite geometric series 175–8
infinity 181
improper integrals 264–7
inflexion, points of 216
integration 239, 249
area bounded by a curve and a line
260–1
area bounded by two curves 260–1
area enclosed by a curve and the
y-axis 255–6
area under a curve 253–5
definite 250–2
of expressions of the form (ax + b)n
247–8
finding the constant of 244–5
formulae for 241–3
improper integrals 264–7
notation 240
as the reverse of differentiation
239–40
volumes of revolution 268–70
intersection of a line and a parabola
27–8
inverse functions 43–5
graphs of 48–9
trigonometric functions 136–9
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horizontal transformations,
combination of 61–3
line segments
gradient 75
length and mid-point 72–4
parallel and perpendicular 75–7
see also straight lines
lines of symmetry, quadratic functions
17–19
longitude 104
Pr
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gradients of parallel and perpendicular
lines 75–7
graph of a function and its inverse 48–9
graph sketching 17–19
quadratic inequalities 22
graphs of trigonometric functions
transformations 130–3
sin x and cos x 128–9
tan x 129
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Index
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Pr
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Underground Mathematics vii
vertex of a parabola 17–19
vertical transformations, combination
of 60–1
volumes of revolution 268–9
rotation around the x-axis 269–70
rotation around the y-axis 268–9
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waves 117
321
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trigonometric identities 145–7, 149
trigonometry 117
angles between 0° and 90° 118–20
general definition of an angle 121–2
graphs of functions 127–33
inverse functions 136–9
ratios of general angles 123–6
transformations of functions 130–3
turning points 17–19, 216
see also stationary points
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tangent (trigonometric ratio)
angles between 0° and 90° 118–20
general angles 123–6
graph of 129
inverse of 137
trigonometric equations 141–3
trigonometric identities 145–7, 149
tangents to a curve 27
gradient 201–3
terms of a sequence 166
trajectories 2
transformations of functions 51
combined transformations 59–63
reflections 55–6
stretches 57–8
translations 52–3
trigonometric functions 130–3
translations 52–3
trigonometric functions 131
trigonometric equations 140–4, 149
w
ge
stationary points 17–19, 216–17
practical maximum and minimum
problems 221–3
and second derivatives 218–19
Stevin, Simon 176
straight lines
equation of 78–80
intersection with a circle 88–90
intersection with a quadratic curve
27–8
see also line segments
stretch factors 57–8
stretches 57–8
trigonometric functions 130
sum of an arithmetic progression 167–9
sum of a geometric progression 173–4
sum of an infinite geometric series
175–8
sum to infinity of a series 176–8
Copyright Material - Review Only - Not for Redistribution