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Pure Mathematics for CAPE® Unit 1 Pure Mathematics for CAPE® Kenneth Charles Sue Baisden Cadogan Chandler Mahadeo Deokinandan Unit 1 3 Great Clarendon Oxford It University furthers and Oxford © CXC® Sue This rights Press, in as British Data department objective of the UK and United of the Kingdom University excellence worldwide. Oxford Oxford in in is certain University registered by the No the a of Oxford. research, scholarship, registered other trade mark of countries Press trademarks 2014 of the Caribbean Examinations Thornes Oxford part prior of the been Ltd in asserted 2013 University this or permitted by in law, by reproduction any of licence rights outside Department, in may in writing reprographics Rights Press publication transmitted, permission appropriate to have the 2014 be reproduced, form or Oxford or by any University under terms organization. scope Oxford of the above University Press, at above. not this by system, expressly the authors Nelson published sent must impose © concerning be address You of retrieval with should in 6DP, 2013 are reserved. a Enquiries the rights without or agreed a OX2 (CXC®).. published means, Press Chandler edition stored is publishing CAPE® moral First by illustrations and Council The Press Oxford, University’s University Original All the education Text Street, circulate same Library this work condition Cataloguing on in in any any other form and you must acquirer Publication Data available 978-1-4085-2039-0 10 9 8 7 Printed 6 5 in 4 Great Britain by CPI Group (UK) Ltd., Croydon CR0 4YY Acknowledgements Cover photograph: Mark Lyndersay, Lyndersay Digital, Trinidad www.lyndersaydigital.com Page make-up Thanks for are their we copyright the the this notied, to for have holders earliest Links and If to illustrations: Kenneth contributions Although cases. due and made the Baisden, the Charles development every before effort publication publisher will to Ltd, Cadogan, of trace this has rectify Gateshead this and not any and Mahadeo third party materials websites only. contained are Oxford in any provided by disclaims contact been errors third party Oxford any all possible or in omissions in good faith responsibility website Deokinandan book. opportunity. information work. in Tech-Set referenced for in all at Contents Introduction Section 1. 1 1 5 Basic algebra Terminology and 1.20 Exponential and logarithmic functions 56 1.21 Modulus functions 58 1.22 Modulus and functions principles equations and 6 inequalities 1.2 Binary operations 8 Section 1.3 Surds 1.4 Logic 60 1 Practice questions 62 12 Section and truth tables 2 Trigonometry, geometry 14 and vectors 1.5 Direct proof 18 2. 1 1.6 1.7 Proof by induction 20 Remainder theorem and factor theorem 1.9 1. 10 1. 11 Factors of Quadratic Curve and tangent functions 64 2.2 Reciprocal trig functions 2.3 Pythagorean 2.4 Compound 2.5 Double 2.6 Factor formulae 68 a and b , n cubic 6 24 equations 1. 13 Inequalities angle formulae curves angle identities 76 expressions 36 – quadratic and expressions curves 2.8 Intersection of lines 42 1. 15 Functions 44 1. 16 Types of function 48 The a 38 and 78 32 1. 14 expression cos b sin Trigonometric 80 identities and equations 2.9 General 82 solution of trig equations 2. 10 Coordinate straight 1. 17 Inverse function Logarithms 1. 19 Exponential equations 52 and 86 geometry and lines 90 50 2. 11 1. 18 72 26 2.7 Rational 70 30 Transformation of 1. 12 identities n sketching rational cosine 22 n 1.8 Sine, logarithmic and the equation of a circle 2. 12 54 Loci 92 Equations of tangents normals to circles and 94 3 Contents 2. 13 Parametric equations 2. 14 Conic 2. 15 The parabola 100 2. 16 The ellipse 102 2. 17 Coordinates sections in 3-D 96 3. 10 Rates of change 98 3. 11 Increasing functions 146 3. 12 Stationary values 148 3. 13 Determining the stationary Unit vectors 2. 19 Scalar 2.20 Equations of 2.21 Pairs of and problems product 3. 1 a line 3. 14 Curve lines 3. 15 Tangents 3. 16 Integration 3. 17 Integration of 2 Practice questions Functions – continuity 160 sums and differences of functions 162 3. 18 Integration 164 3. 19 Calculus substitution and the area under curve 166 3.20 Deﬁnite integration 3.21 Area under 3.22 Area below the 168 124 a curve 170 126 3.4 Gradient of 130 3.5 Differentiation from ﬁrst a using and 128 curve principles x-axis area between two and curves 3.23 Volumes of revolution 3.24 More volumes of 3.25 Forming differential revolution General differentiation rule and quotient equations 134 174 176 rule 136 3.26 180 Solving differential equations chain 172 132 3.8 The 3.9 Parametric rule and differentiation 182 138 Section 4 158 1 notation Product normals 122 Limit theorems 3.7 and 118 3.3 3.6 154 116 Calculus Limit sketching 112 discontinuity 3.2 150 108 a 3 points 110 Planes Section nature of 104 2. 18 Section and decreasing and vectors 2.22 144 3 Practice questions 186 general 142 Index 188 Introduction This Study Guide has been developed exclusively with the Caribbean ® Examinations candidates, Council both in (CXC and out ) of to be used school, as an additional following the resource Caribbean by Advanced ® Proﬁciency Examination It prepared (CAPE ) programme. ® has been teaching and by a team examination. with The expertise contents are in the CAPE designed to syllabus, support learning ® by providing tools and the and requirements for full Inside features guidance this activities • On Y our answer and • of you that it is in an an Do examiner will CAPE to master to which the where your includes concepts syllabus format! electronic techniques: candidate answers short answers could understanding, be skill level questions. designed questions study to Mathematics key examination-style your speciﬁcally Pure the examination example show build refer examination sample with to and examination examination inside in you CD good provide questions, are best remember interactive answering sections for requirements developing activities your easier and guide so to provide helpful that experience feedback you can will revise areas. Answers so to course activities in achieve syllabus. type multiple-choice problem This These Y ourself you make activities from conﬁdence refer • the essay feedback T est the you Marks and help Guide assist improved. and of on Study to to included are you unique examination included can check on combination practice the your will of CD own focused provide for exercises work as you syllabus you with and practice questions, proceed. content and invaluable interactive support to help you ® reach your full potential in CAPE Pure Mathematics. 5 1 Basic 1. 1 Terminology algebra Learning outcomes To use words and and principles Language of The and functions language of mathematics mathematics is a combination of words and symbols where symbols each symbol is a shorthand form for a word or phrase. When the words and correctly symbols are properly constructed Many the of used correctly words a piece sentences used have in of mathematical the precise same way as reasoning a mathematical piece of can be read in prose. deﬁnitions. For example, You need to know the word when How to solve simultaneous a pair of ‘bearing’ used has many mathematically meanings it means when the used in direction of everyday one language, point from but another . linear equations in two Y ou need unknowns to be able cor rect to present mathematical your solutions language and using clear and symbols. n The meaning of x Symbols A used for operators mathematical Y ou are already describe operator familiar is a with rule for several combining operators or and changing the quantities. symbols used to them. means ‘plus’ or ‘and’ or ‘together with’ or ‘followed by ’, depending on context. For example, 2 a means For For ‘is not use ‘take 2 a plus 5 or together 2 and with b 5, or a followed by b away ’. means and 2 also minus have 5 or 2 familiar take away 5. meanings. comparison symbol used for comparing two quantities is and it to’. 6 ‘is means symbol x is symbols greater is than used equal are or to is to 6. which equal mean means to’. ‘not’, A for ‘is greater forward than’ slash example, , and across which a means to’. expressions, comparison between A or familiar equal Terms, T o x means comparison ‘is equal other which means used for example, Some means 5 commonest means 5 b 2 operators Symbols The ‘minus’ example, The terms, equations symbols correctly, expressions, mathematical expression you equations is a group and need and of identities to recognise the difference identities. numbers and/or variables 2 5x _______ (for example, x) and operators. For example, 2 x, 3 2y and are 3 2x expressions. The parts example, 6 of 3 an and expression 2y are separated terms in the by or expression are 3 called 2y terms. For Section An equation For is example, This a 2x 3 statement Some is equations example, example x of Therefore x an we Symbols statement 7 true are is identity can is a only true 2x saying for and x any used for for we two statement when true write that x x use that reads are ‘2 x equal 3 is in Basic algebra and functions value. equal to 7’. 5 value any quantities 1 that value the the of variable x. symbol This to can take. equation mean ‘is is For an identical to’. 2x linking statements 2 When that one is words Some statement, logically or such as connected, x for 4, is example followed x 2, by another they statement should be linked by symbols. examples of words that 2 can be used are: 2 ‘x 4 therefore x 2’, ‘x 4 implies 2 that x 2’ 2 ‘x 4 so it follows 4 hence that x 2’ ‘x 4 gives x 2’ 2 ‘x The symbols ‘therefore’ For or example, x and is ⇒ can ‘hence’ and 2x Setting out It 2’ 1 a be the 5 used x or to set out also important of following example simultaneous 3x [1] 3 ⇒ 6x [2] 2 ⇒ 6x [4] ⇒ [3] The solution Notice brieﬂy the 1 that is the what we 1 5 that’ ⇒ x or means ‘gives’. 3 3y of [1] 4y 10 [2] 9y 3 [3] 8y 20 [4] 17 y 1 y in 2 [1] and equations are way doing y gives are to 2x or to problems using cor rect words. explain the steps you take reasoning. 1 you your a 2x 17y for x where ‘implies 2x solutions 3y Substituting shows equations: 2x statements, means symbols that and The 3 your linking is link ⇒ solution important It to symbol explaining 1 and 3( 3x 1) the solution 4y 1 ⇒ of the pair 10 x 2 1 numbered. combine This them in gives order a way to of explaining eliminate one of variables. Exercise 1.1 In each and question, write down a explain correct the incorrect use of symbols 2 Find the value sin 1 Solve the equation 3x 1 1 5 3x 6 x 2 A A° 0.5 given that sin A° 0.5 A 30° 5 3x of solution. 7 1.2 Binary operations Learning outcomes Binary operations A To know perform the meaning binary of binary To of be able to identity, use the closure, commutativity, concepts inverse, example, addition, these we can rule for combine subtraction, operations, combining two members of a the real set. two members multiplication that is 4 2 6, or of division. 4 2 2, set We 4 of know 2 8 numbers the and rules 4 by for 2 2 can deﬁne other operations. For example, for a and b, where a, b are addition, members multiplication binary a associativity, We distributivity, is operations For operation and and other simple 2a of the set of real numbers, , then a and b are combined to give b operations We write Then, this for brieﬂy example, as 3 a * * 7 b 2 2a b 3 7 1 You need to know Properties of operations The meaning numbers, of the set of real An operation, *, as For example, is b commutative * addition a for on the a Multiplication on the set subtraction is of a However , any two set b real not of b b also not b by is on two example, because e.g. (2 (a 3) Addition (a e.g. b) (2 (a b) set of (b 3) 4 2 is a 4 also (3 is a 3) 4 not (2 8 set 4 because commutative because in general, 3 7 in 3 any * general, a ﬁrst a result commutative three two that (b * members members can the be second c) of real numbers a, b, c or is is associative b, is c also associative because associative a, 5 b, c because, whereas because, a, b, c in general, 2 in (3 4) ( general, 3 whereas 6 2 1 3 same 4) c), 2 e.g. c b is the set. also 7 ﬁrst, c), a, c), the 4) associative (b * b the numbers 4 7 gives of a when the c), 1 not c on (b real (b (3 subtraction a 3) Division 2 a c (2 b) e.g. the c c However , (a 4 b) b because, b either * multiplication b) on a, members (a For 3 b * b is because, associative operating a, a, 7 commutative operation a, 3 e.g. An b a combined a numbers commutative e.g. is real numbers a Division when members 2 (3 4) 2 8 4 3 1) 3 Section An operation, *, three For example, members of distributive of multiplication the is over set another a * (b ◊ distributive operation, c) over (a * b) addition ◊, ◊ when (a and * for Basic algebra and functions any c) subtraction on because a but is members 1 (b multiplication is c) not ab ac and distributive a over (b c) division ab ac because ab ___ a (b c) c b __ whereas (a b) (a c) c so a (b c) (a b) (a c) unless a 1 Example An operation Determine (a) x * 2x y * is deﬁned whether the 2x and 2y 2y 2y 2x for all real operation y * x because numbers * 2y is: the (a) x and y as x commutative * y (b) 2x 2y associative. 2x addition of real numbers is commutative. (b) the operation T aking (x x * * y) (y x, * * y and z z (2x z) x Therefore (x * * is as * * three 2y) (2y y) commutative. z * real numbers, z 2(2x 2z) 2x x * (y * z) 2y) 4y so 2z 4x 4y 2z 4z the operation is not associative. Example (a) For two real numbers, 2 x * y x Determine (b) For two Determine (a) For any * y) whether real * (b) (y (x For * * z z) y) any the x operation and the y, the operation numbers, (x * z y * (x x * real y * z ◊ (y * so ◊ x y) z) ◊ the * (y * is given by associative. operation ◊ is x(y (y * z ) z) so ◊ is given distributive by over x ◊ y xy the x z) (x * ◊ z) z) (x operation xy ◊ ◊ * y) x(y is xz * not y x, 2 2 ) x ◊ z 2 (x z 2 ) operation y and z not 2 ) xy 2 y is 2 associative. z, 2 (x z, 2 the numbers, * and 2 2 (x is 2 (x 2 2 x operation 2 ) 2 x three x, 2 2 x y, *. three * the numbers, 2 (x and y whether real operation x 2 x 2 xz 2 z z) distributive over the operation *. 9 Section 1 Basic algebra and functions Closed A set is sets closed under set, For example, any two the set integers, However , is a a not an * b of operation gives integers, and b, closed a * when another , b under is is for any member closed also an division of under two the members of the set. addition because for integer . because a b does not always 3 give an integer , for example 3 4 , which is not an integer . 4 Identity If a such is any that member under an identity For example, member 0 is of a set member the identity Also 1 there no is member b is set one b * a under members of member the a b then b of the is set, called the operation. under addition as, for any is real the no identity number identity b for a for such a 0 members that members a of b a of under b under subtraction because a multiplication, as for any a, However , is * the for 1 there there a a, However , is *, of 0 there and operation, there no is real no identity number b a for a 1 members such that a has a b a of b under division because a Inverse Any if there member is another a of set member of gives Clearly, a member can have For example, then a since Also, is a as 1 as the a is 0 is the inverse the a identity of ( any a) identity for the inverse of a only there members one members since important of of under under a 1, exception to this: 1 __ is 0 10 there is an with identity addition, a, a is if 0 a but operation combined operation. 1 __ is an when identity. member 1 __ then under which inverse the for inverse set, the an under an the meaningless. multiplication, a Section 1 Basic algebra and functions Example An operation, *, is deﬁned for all real numbers (x x and y as y) _______ x * y 2 (a) Show that the (b) Show that x (a) When set has is closed no under inverse the under operation the (x *. operation *. y) _______ x and y are real numbers, is also a real number . 2 Therefore (b) For x to the have i.e. one i.e. such set an closed inverse, member , (x is b, of under there such the operation needs that x to * be b an *. identity member , x, b) _______ that x 2 Solving Now x this is any therefore As there for gives member there is b no is no b of x , so identity identity, x b is not member has no a of single the member and set. inverse. Exercise 1.2 1 Determine set 2 of The real whether addition is distributive over multiplication on the numbers. operation * is given by 2 x for all real values Determine 3 x whether (a) commutative (b) associative (c) distributive The of operation the over * is and * y x y y. operation * is: addition. given by ___ x for 4 all positive real (a) Show that (b) W rite down (c) Determine The for the operation x, y numbers operation the identity whether ~ is by y √xy including * is 0. closed. member . each given * member x ~ y has the an inverse. difference between x and y . (a) Determine whether (b) Show that the (c) Show that each identity is closed member member is its under is this operation. 0. own inverse. 11 1.3 Surds Learning outcomes Surds The To perform operations square roots of most positive integers and fractions cannot be involving expressed exactly as either a fraction or as a terminating decimal, i.e. they surds are A not rational number exactly numbers. such when √ as left 2 is an √ as 2. In irrational this form number it is and called a can only be expressed surd. You need to know Note The meaning of a √ that 2 means the positive square root of 2. rational number Simplifying Many surds surds can be simpliﬁed. ______ ___ For example, √ 18 √ 9 2 √ 9 √ 2 3 √ 2 ______ And In √ 8 both √ 2 √ cases, 3 When 4 √ 2 a is 2 the Operations on expression i.e. (3 2 such 2 √ simplest calculation answer An √ in 2 2 possible involves the √ √ 2 surd surds, simplest 3 form. you possible should surd give your for m. surds as √ (3 2) √ (2 3) can be expanded, ______ When For (5 √ √ 2 )(2 the 3) same surd 6 3 occurs √ 3 in 2 each √ 2 √ 6 bracket ( the √ 3 )(3 2 √ 3) particular , rational (5 √ 3 can be √ 2 3) simpliﬁed. example, 2 For 2 expansion 15 6 √ 3 10 √ 3 12 (2 In √ 3 expressions 4 of √ 3 2 √ 3 4 √ 9 12) √ 3 the form √ (a b )(a √ b) simplify to a single number . example, 2 √ 3 )(5 2 2 √ 3) 5 (2 2 √ 3) ((2 25 12 √ 3) 2 4 √ 3 √ 3 4 √ 9 4 3) 13 Example Simplify (2 √ √ (2 5 )(3 5 )(3 2 √ 5) 2 √ 6 5) 3 √ 5 4 √ 5 10 ___ (2 12 4 √ 5 √ 5 √ 5 2 √ 25 10) Section 1 Basic algebra and functions Rationalising the denominator When the a fraction has a surd in the denominator , it can be transferred to numerator . When the bottom, For denominator by that surd is will a single change surd, the multiplying denominator the into fraction, a top rational and number . example, √ 2 3 _______ √ 2 3 _______ √ 5 ___ √ √ 5 √ 5 5 ___ √ √ 2 5 15 __________ 5 When top the and denominator bottom, by is of √ a b the will form a change √ b, the multiplying denominator the into fraction, a rational number . For a denominator of the form √ a b multiply top and bottom by a √ b Example √ 2 1 ___________ Rationalise the denominator and simplify √ 3( This Do fraction not start has attempt with a to single rationalising √ √ 2 3( 1 ___________ surd rationalise the and single √ 2 3( √ 2 bracket both at 2 in the the 3) denominator . same time. We will surd. 1) √ √ 6 ________________ 3 _________ √ a them √ √ 3) 3 √ 3( √ 2 3) 3( √ 2 3) ___ √ √ 6 √ 3 2 _________ √ 12 3 _______ 3( √ 2 √ 3) 2 3 3( √ √ √ 2 3 4 6 3 3 _________________ to do written some of 3 √ 6 3 √ 3 √ 2 2 9) √ √ 4 6 5 3 __________ 3(2 have 6 We √ ______________________ down these 9) every steps in 21 step in your this example, but you should be able head. Exercise 1.3 Expand and simplify __ 1 (3 2 √ Rationalise 3 )( __ √ the 3 when possible. __ √ 2) __ 2 denominator of ( __ __ 2 √ 2 √ each 3 5) surd and (1 simplify ( √ 3 when __ √ 2 2 )) possible. __ 1 2 ___ 4 √ 2 _______ __ __ 7 √ 2 1 √ 2 __ __ 2 √ √ 3 2 ____ __________ __ 5 __ 8 √ 3 2 __ √ 3 5 √ 5 __ √ 8 1 _______ __ 6 3 √ 2 ____________ 9 __ √ 2( __ √ 3 __ √ 2) 13 1.4 Logic and Learning outcomes truth Propositions A To identify simple sentence To establish compound the truth value statements To went to school today ’ is a closed sentence, but went to who school could be today ’ any is not closed because it contains the variable female. Closed p, sentences q, are called statements or propositions and are denoted etc. tables state the A converse, contrapositive a ‘Sonia of using by truth as propositions ‘she’, such and ‘She compound tables conditional and inverse proposition is either true or false. of (implication) Negation statement To determine statements whether are two The proposition ‘It is not raining’ contradicts the proposition ‘It is raining’. logically ‘It is not raining’ is called the negation of ‘It is raining’. equivalent If You need to know p is The meaning distributive binary of and proposition ‘It is raining’, the ‘It is raining’, if ~p is true. negation of p is denoted by ~ p. Truth tables For the the proposition p: p is true then ~p is false. commutative, associative for But if p is false, then operations We can We use show 1 to this logic represent in table true and form 0 to (called a represent truth table). false. p ~p 1 0 0 1 Did you know? The George a Boole system (1815–1864) using values 0 numbers in each column are called the tr uth values invented and 1 and Conjunction truth tables system is algebra. to formalise now known as logic. This Boolean The is statements raining Using We p put for the can can is symbol true or possible and across, (If it construct be all p and p: q we either in p or is raining’ cold’. a to This mean truth false, q ﬁrst can two complete q is false, is and for also of q: called ‘and’ table combinations the can ‘It p be 1 we third then p a is true or this false. and Then, column q can be of combined two as propositions. conjunction as p 0 (false) reading for must We as statements ‘It is raining’ propositions written Using 14 as the ‘It p: and is ‘It is and/or the is word raining symbol raining’ ‘it to or ‘and’ it and cold’. is mean q: This is be p q false.) q p 1 1 1 1 0 0 0 1 0 0 0 0 ‘It is is cold’ called implied so it a can be combined disjunction would of normally two be cold.’ ‘ or ’ we q p Disjunction The ‘It q (true) and cold’ conjunction write columns. the ‘It write this disjunction as p q q Section We p can can be before, p and we p q q true we can or construct in or put all the ﬁrst both ‘it is are Using The the raining’ the q table can possible two the Conditional If truth false, complete or a also third then symbol logic, example, The and in then ‘it → p p for q The is to is p the is q ‘5 q mean is one is 1 and reading p must called ‘If the q. be 0 for across (If either true.) ... a then conditional ...’ hypothesis tr ue p a except a false p q p 1 1 1 1 0 1 0 1 1 0 0 0 q we write and the statement p → q proposition q is called prime when a tr ue hypothesis inverse is not the of p → is of q is the a is prime of → ‘It conclusion. number ’ such that of p q q is of → and q is ‘6 is a of is ‘5 cold’ ~p is of → → ‘5 number ’ ‘It prime number ’ p → and q: q p is true p q p 1 1 1 1 0 0 0 1 1 0 0 1 → q p a number ’ is inverse prime inverse q converse a p → leads false. combination converse contrapositive example number ’ is the prime → ‘It ‘5 is is number ’ a prime raining’ is → ‘6 is a prime number ’. ‘It is raining’ → ‘It is cold’. ~q is → a prime ‘6 is raining’ → ‘6 is not bi-conditional statement reads ‘if p p it is is number ’ → not ‘It is a prime cold’ is ‘6 is a prime number ’ number ’. ‘It is not raining’ → ‘It is q and it is its is then as ~q of → ‘5 ~p is number ’ if q ‘It is is the then raining it is a prime → ‘5 is number ’ not conjunction converse bi-conditional simply q statement then ‘If → a → prime ‘6 is a prime number ’. statements with a p prime q raining written a → example, raining’ of contrapositive Bi-conditional ‘If of As cold’. The For is is for the ‘6 example ‘5 and A q false. false. example, The is → table converse Also For or cold’ called → p p only number ’ not if logic truth false is or Then for and functions q true column algebra statements proposition then For combinations to For be columns. true, p Basic conclusion In For for 1 q → p, that of is the (p conditional → q) (q → p). This p’. then it is cold’ and ‘If it is cold then it is statement. cold’ and raining’ if ‘If it and is cold only if then ‘It is it is raining’ can be cold’.’ 15 Section 1 Basic algebra and functions Using ‘It is We the cold’ can Start symbol and with the → q, then q → p. Lastly, the third a truth add a add fourth table can to q) mean a (q truth table for the columns. now be and p) table column of ‘if → for column conjunction and This → construct p for (p ⇔ only can for p be ⇔ if ’ we can written as write p ⇔ ‘It is raining’ ⇔ (q p) q q p q 1 1 1 1 1 1 0 0 1 0 0 1 1 0 0 0 0 1 1 1 written as a p → simpler q q truth → p table (p for → a q) → bi-conditional statement. p q 1 1 1 The 1 0 0 p 0 1 0 0 0 1 ⇔ Compound A compound combination A q p q shows are both that true p ⇔ or q is both true only when false. statements two bi-conditional compound and statement of table or combines more statement, of (p the → two or more symbols q) (q ~, → p), propositions , is , an →, using a ← example of a statement. Example Let p: p, q ‘Students Express in and the symbolic r be play the propositions: soccer ’, compound q: ‘Students statement ‘Students play soccer or basketball’ ‘Students play soccer or basketball ‘and cricket’, play r: ‘Students soccer or play basketball’. basketball but not both and students play cricket’ form. ‘Students Adding play students play cricket’ is p but to r. not this ‘Students both’ gives The way (p truth to is ( p r) table the do not r) play ~(p ~(p for a both r) and basketball’ is ~( p r). r). q compound bi-conditional soccer table statement can be constructed in a similar above. Example Construct 16 a truth table for the compound statement p q ~q 1 1 0 0 1 1 0 1 1 1 Then 0 1 0 0 0 to 0 0 1 0 0 ~q p p (~q p (~q p) p) Always start add build with p columns up statement. the and in q. stages compound Section 1 Basic algebra and functions Equivalence T wo statements same, that is are in logically the equivalent completed truth when tables their the truth ﬁnal values columns are are the identical. Example Determine whether the statements p q and ~p → q are logically equivalent. We construct a truth table for q ~p 1 1 0 1 1 The 1 0 0 0 1 ~p → 0 1 1 0 1 the statements 0 0 1 0 0 Identity This law q q truth q values are equivalent. not for the are W e p q same. and Therefore logically write p q not ~p → q law states Algebra of The → statement: p p ~p each symbols that p p and p p are both equivalent to p propositions and are called logical connectors. Example These connectors are commutative, that is p q q p and p q q p Use They are also associative, that They are also distributive over for p p (p each q) other r and p over (q the r) conditional →, example (q (q The r) → These r) (p (p properties can also q) properties compound It is q) can can (p → be also (p r) and r) proved be and using used p to p (q (q truth prove → r) r) (p (p q) q) (p → (p algebra show (p q) p p (p q) (p using the using the r) to p that (p p) distributive p (p q) distributive (p q) law q) law r) tables. the equivalence between two statements. be shown that p → q ~q → ~p Exercise 1.4 In this exercise, 1 W rite 2 (a) (b) down p, the Construct State, q a with and r are propositions. contrapositive of truth p a table reason, for ~p → whether p ~q → q and ~q p and ~q p ~q are logically equivalent. 3 p: ‘It is Using ‘The raining’, logic sun is q: ‘It symbols, shining is cold’, write and it r: ‘The down is cold in sun terms and it is is of shining’. p, not q and r the statement: raining’. 17 1.5 Direct proof Learning outcomes Direct proof Mathematics To construct simple is the Mathematicians speciﬁcally direct Proof by the use of numbers, shapes, space and change. look for patterns and formulate conjectures. They then proofs try study proofs, of to prove the truth, or otherwise, of conjectures by proof that is built counter up from axioms. The axioms are the basic rules or deﬁnitions, and all examples other true facts can be inferences conditional rules are from →). the derived those (We moves from can rules use that these are (an the by deduction, inference game allowed of for is chess each that the as same an piece, is by as using the analogy and – logic the games are basic built You need to know up The basic rules of from How to solve a n i.e. quadratic example, x a equation moves.) logic For these by factorisation or x by x x b x a x x is deﬁned …. x, to mean and from n lots this of x multiplied deﬁnition we together , can deduce that b the formula Example How to ﬁnd the area of a triangle Prove that 8 8 20 8 Using ⇒ 4x 28 Adding ⇒ x 7 Dividing 5) 8 Starting if 4( x with 4( x ⇒ 4( x This is with p (Note T opic is an example then that 1.4 know also this that converse in this of ‘ A polygon the if we the polygon four a converse s ⇒ an → 8 or a q ⇒ an each x is ⇒ side by 7, ⇒ a ⇒ p ⇒ is then that a keeps 4 keeps the equality the equality true true if i.e. true a x q. the x prove We know p ⇒ q, start equal needs from question.) contrapositive 7 7 then ⇒ has is four a 4(x 4(x implication polygon implication to another polygon four i.e. q true true of has side law 7 p say sides rhombus of to each deduction, so is also square equal distributive 20 q converse is the by q, 7 implication true p x proof can 5) but because Therefore is x ⇒ ⇒ of p that 4( x has r proof logic case, example, ‘ A a then direct ⇒ Therefore true but is p whether from true. of deduce The For 5) 4x W e 5) 5) is not 8 is → sides’ is not ~p 8 also always equal square’ ~q 5) true is true true sides. to be proved to be true. Example Prove that the sum of the interior angles of any triangle is C 180°. D ABC is DCA ECB DCA any triangle. is parallel to AB CAB CBA ACB ⇒ CAB ⇒ the 18 DE sum ECB ACB of the 180° ECB interior Alternate angles are equal Alternate angles are equal Supplementary angles 180° angles of any triangle is 180°. E Section Use of As well also counter as important A it being important for statement disproves it. to can necessary This shown is to that converse be called a prove of a to be a that statement true a is statement false. This is example, a 0 ⇒ true, is algebra and functions it is particularly implication. false counter if we can ﬁnd just one example that example 2 For Basic examples prove the 1 2 a 0 is true, but the converse a 0 ⇒ a 0 3 is false. 2 We can For example, prove odd use this a the using 9 ⇒ a 3 statement the or 3 ‘all counter as a prime example counter numbers ‘2 is a example are prime odd’ because is not number true. and 2 is W e not 0. can an number ’. Example Use a counter example to prove that the converse of the true 2 statement: The ‘n converse is an of integer ’ the given ⇒ ‘n is an statement integer ’ is false. is 2 ‘n ( is an integer ’ ⇒ ‘n is an 2 √ 2) 2 is an integer integer ’. √ but 2 is not an integer . 2 Therefore ‘n is an integer ’ ⇒ ‘n is an integer ’ is false. Exercise 1.5 2 1 Prove 2 Find 3 (a) that if x 3x 2 0 then x 1 or x ⇒ a 2 2 a counter example to show that a b 2 b is not true. 2 (b) Prove that (Start with Use in a (a) ‘n n is an odd 2k 1 counter is integer where example to ⇒ k is show n is any that an odd integer ’. integer .) the converse 2 4 5 (a) Prove that ‘x (b) Prove that the In the Prove is diagram, that twice the the of the 4c’ statement false. 2 D c 0 converse of the is area area of bx the of has triangle roots statement midpoint triangle equal of in ⇒ (a) b is also true. C AB . ABC ADC. D 19 1.6 Proof by Learning outcomes induction Proof by induction 2 Consider Establish the simple principle of proofs by these results: 1 2 1 2, 2 2 2 6, 3 3 2 12, 4 4 20 using mathematical In every case, the right-hand side is a multiple of 2. induction This suggests that the proposition 2 ‘for is any true positive but it integer does not n, n prove n is a multiple of 2’ it. You need to know The set of denoted Any positive by even integers We can W e call We start 2k number and any written using a method called mathematical induction 2 is can be the proposition odd as number 2k with the p(n) and proposition, written rephrase it as ‘n n 2m for n, m ∈ ’. 1 for k that 2 n k, ‘k k is a multiple of 2, k ∈ ’. [1] can The be it when as prove ∈ next step is to replace k by k 1 (i.e. by the next consecutive integer) A natural number is a member of 2 the set 1, 2, 3, 4, ... ⇒ (k k 1) (k 1) 2 1 k 2k 2(k 1 2 (k k ) From 1) [1] which this is a is also a multiple of multiple of 2. 2 2 Therefore of we have shown that if for any integer , n, n n is a multiple 2 2 then (n 1) (n 1) is also a multiple of 2. a multiple of 2. [2] 2 We know that p(1)is true, i.e. 1 1 is Therefore [2] shows that p(1) ⇒ p(2) so p(2) is true, then [2] shows that p(2) ⇒ p(3) so p(3) is true, [2] shows that p(3) ⇒ p(4) so p(4) is true, again again This process can be continued indeﬁnitely, i.e. for all … positive integers. 2 Therefore multiple An of you fall, one Proof to 20 by cover for on can domino have proved that, for any positive integer n, ‘n n is a 2’. analogy standing If we proof show fall, that then domino induction pushing pushing after induction any by is a row of evenly spaced dominoes end. the can positive be over over any the domino ﬁrst will domino make will the make next the whole row other . used integer to from prove a many result results proved for that a are generalised particular integer . Section The 1 proof Let has p(n) Prove three be a 2 Prove 3 Combine that that p(k Basic algebra and functions steps: proposition directly convenient distinct 1 involving 1) is n, true. then (Note assume that k is that an p(k) is arbitrary true. and integer .) p(1) is steps true. 1 and 2 to prove that p(2), p(3), p(4), ... are true. Example n Prove by induction that 10 1 is a multiple of 9, n ∈ k Assume that p(k) is 10 k 1, 1 9m, k Replace k by giving where k and m 1 are natural numbers. k 10 1 10 10 1 k 10 10 10 9 k 10(10 1) 9 10(9m) 9 which is a k k Therefore multiple when of 10 1 is a multiple of 9 then multiple is the 10 1 is also a proposition from multiple a [1] 1 10 1 is a multiple of 9, and 10 1 2 therefore 9. 9. 1 p(1) of 1 of [1] 9, 9, 3 10 1 is a multiple of 9, from [1] again 10 1 is ... n Therefore 10 1 is a multiple of 9, for all n ∈ Example 2 Prove (The by induction second 2(3) 1, ... odd the that the number nth odd is sum of the 2(2) 1, number is ﬁrst the n odd third 2n numbers odd number is n is 1) 2 Let p(n) be the proposition that 1 3 5 1) ... (2n 1) n 2 Assume Then that adding 1 the 3 5 next odd ... (2k number to both k sides gives 2 1 3 5 ... (2k 1) (2k 1) 2 k 2k 1 (k 1) 2 Therefore when 1 3 5 ... (2k 1) k then 2 1 3 5 ... (2k 1) (k 1) 2 i.e. when the sum of the ﬁrst k odd numbers is k , the sum of the ﬁrst 2 (k 1) odd numbers is (k 1) 2 Now so p(1): 1 1 2 , so the sum of the n odd ﬁrst two odd numbers is 2 , and on. 2 Therefore the sum of the ﬁrst numbers is n Exercise 1.6 Prove by induction that: 3 1 n n is a multiple of 6 for all positive integral values of n _ 1 2 the sum of the ﬁrst n natural numbers is n(n 1) 2 21 1.7 Remainder Learning outcomes theorem To apply To use the remainder general for m n factors and to theorem evaluate of a polynomial expression is theorem a the factor theorem Polynomials The and factor unknown n x a n to ﬁnd where n is a positive 1 2 x n .... a 1 x a 2 integer, a , a n , n ... a 1 , x a 1 a 1 are 0 real numbers and a 0 0 n coefﬁcients The order 5 x T wo The meaning of the notation the function 2 f(x) where a a polynomial has order of is How For x to expand polynomials power of expressions example, 0 (the x are has identical equal How (ax c)(a to factorise a power of x. For example, when they have the same order and when coefﬁcients. 2 x 5 5x order must 2 be the ax 4 2 bx same) and b cx 1, c dx e 5, if d and only 0, e if 2 of (coefﬁcients the form highest a a the polynomials 4 value is 5. f(a) each for Identical You need to know of 2 x must be equal). polynomial) quadratic The remainder theorem expression When 17 is divided 17 ___ i.e. 5 3 5 is by which called the f(x) remainder . can 7x The Substituting be written be written as 5, remainder 2. 2, as 17 5 3 In this form, 2 is divided between by these x 2, we quantities get can a quotient be and written as 2 x 2 6x relationship 7x for x 6 (quotient)(x eliminates the term 2) remainder containing the quotient, giving remainder . 3 f(2) divided This can 2 x 3 f(x) Now result quotient. 3 f(2) the 3 When a 3 2 __ by is when an a 2 2 7(2 x 2, 3 ) the 6 14, remainder illustration polynomial of f( x) the is so is f( x) 2 x 7x 6x 2 is 14. general divided when case: by ( ax b) then b __ f(x) (quotient)(ax b) remainder ⇒ f ( ) remainder a This result is called the remainder theorem and can be summarised as: b __ when a polynomial f( x) is divided by ( ax b), the remainder is f ( ) a Example Find the 3 2x remainder Example when 2 7x 3 is divided by 3 When 2x 2 x ax 1. and 3 Let f(x) x 3 2 2x 7x when Find 3 the b is divided by x 1, the ax values of a x and b is divided by x 3, b 1 2x 1 0 when x Using , the remainder theorem gives 2 3 therefore is when divided by 2x 2x 1 2 7x a 1 b 4 and 27 9a 3 3 1 i.e. b a and b 9a 2 [1] 1 the remainder is f( ). 2 1 f( 1 ) 2 2( ) 2 1 3 7( the 22 14 [2] 3 1 [1] [2] ⇒ 8a 16 ⇒ a in [1] ⇒ b 2 2 Substituting 2 ) remainder is 1. remainder is 4 2 x 2 for a 4 b 16, the remainder is 16. Section 1 Basic algebra and functions The factor theorem When the (x a) is remainder a is factor zero of ⇒ the f(a) This i.e. For if, for example, 4 a when 3 x polynomial x polynomial f( x), 0 is the f( x), factor f(a) 0 theorem, then x a is a factor of f( x). 3, 2 3x 3x 11x 6 81 81 4 Therefore x 3 is a factor of 27 3 x 33 6 0 2 3x 3x 11x 6 Example 3 Given 5 that when a __ 2 ax 3x divided by b x has 2, a factor ﬁnd the 2x of and a leaves and a remainder b 3 __ 1 b 8 0 ⇒ a 8b 6 [1] Using the factor theorem with x 2 4 8a 12 b 5 ⇒ 8a b 17 Using 8 1 values [1] [2] Substituting The factor ⇒ 1 for 65b b [1] theorem in can 65 be ⇒ gives used to b 1 a 2 ﬁnd [2] the remainder factors of theorem with x 2 polynomials. Example 3 Factorise 2 x x x 2. 3 If x 3 is a factor of 2 x x x 2 x x c 2 2, so )(x (x possible 3 T ry 1: (1) 1: ( 2: 3 then of c) are 1 1) (1) ( 2 0 1) ( 1) 2 0 (2) (2) 2 0 so x so x 1 is not a factor . 1 is not a factor . therefore 2 x 2. 2 (2) x and 2 2 x bx values (1) 3 T ry 2 3 T ry 2 2 x 2 (x 2)(x 3 bx c) x x 2 is a factor . 2 (b 2)x (c 2b)x 2c 2 Comparing so b 1 and 3 coefﬁcients 2 of 2c x so and c 2 x the constant gives 1 b 2 1 2 x x 2 (x 2)(x x 1) Exercise 1.7 3 1 Given that the values f(x) (x of a 3 2 5x and (x 2) are factors of x 2 ax bx 6, ﬁnd b. 2 remainder Given 1) and that is px x is a q. When f(x) is divided by x 2, the 3. (x 1) factor of f( x), ﬁnd p and q. 23 n 1.8 Factors of n a 2 extract for all factors positive integers n of a n n 6 2 a Factors of n To , 2 Learning outcomes b b 2 a b 2 is the difference between two squares, so a b 2 b (a 0 b)(a b) b 6 3 3 a Factors of b 3 From the factor theorem, when a b, 3 a 3 b 3 b You need to know 3 Therefore How to factorise quadratic (Y ou can a b verify is a this factor by of 3 a 3 b expanding ⇒ the 3 a 2 b right-hand (a 2 b)(a ab b ) side.) expressions 2 2 (a How to expand expressions 3 as (x 1) , 2 (2x 1) (3x ab b ) cannot be factorised. such 3 3 Therefore 4) 3 For example, 8 2 b (a b is 2 i.e. ab b ) 2 2 (x 2)(x 2x 2) 4 a b (a the 2 difference 2 ) (b 2 between 2 ) 2 (a b (a of the two 2 squares, 2 )(a b ) 2 using the 2 factors b)(a 2 b)(a difference b ) between two squares twice. 2 (a b ) cannot be factorised. 4 For example, Therefore a 4 4 x 16 5 a Factors of 4 2 b (a b)(a the 2 5 theorem, b can 4 (a b when is a factor (a verify 3 a a (x a this by 2 a 2)(x b b, 32 b 6 6 a 6 b 3 (a a 3 (b a 3 b the ab ) has no linear b a (a b)(a (Y ou can a , 2 which 3 (a (a b, verify b b ab 0 ) side.) 3 a 5 x 2 b a 4 (x is 3 b the 2)(x 2 b 3 3 difference 3 )(a 2x 4 b ) 2 4x 8x 16) b)(a a b b ab is a this factor by 3 of b a b 3 )(a squares. 3 b ) 3 b 3 expanding two ) 2 3 between 3 3 Therefore b factors. 4 3 when 5 b 4 right-hand 2 Now 5 2 ) 6 a b 6 2 ) 4) b 6 Therefore 5 x Factors of b 5 a 2 5 a 5 example, 4 ab 5 Therefore For 2)(x b 2 expanding 3 b ) 5 a 3 b)(a 2 b of 4 b 5 5 a (Y ou a 2 2 5 Therefore b)(a b factor 4 x 5 From 24 4 a ⇒ 2 b)(a 3 x 4 4 3 3 x Factors of a b the 0 3 ⇒ a 3 b right-hand 2 (a side.) b)(a 2 ab b ) Section 2 Neither 2 (a ab b 2 ) 6 Therefore nor 6 For example, These in (a 6 x results expressed ab 64 can one b ) can be of be 2 b)(a ab 2)(x to (x 2)(x factorise forms b 2 )(a and functions 2 ab 2 2 used the b)(a 6 x algebra factorised. 2 b Basic 2 (a 6 a 1 given any b ) 2 2x polynomial 4)(x that can 2x 4) be above. Example 3 Factorise 8x 27 completely. 3 3 8x 27 can be 3 Using written as 3 a (3) 2 b and 3 (2 x) (a replacing 2 b)(a a by 2x ab and b 3 b by ) 3 gives 2 8x 27 (2x 3){(2x) (2x 3)(4x 2 (2x)(3) (3) } 2 6x 9) Example 4 Show that (x 4 Using (a (x b)(a expanding 4 the a by x (a 1 two b b by a x bracket (x 1 x (x 1) ){(x 1) b 1) x ) 2 x(x 2 x(x gives 3 3 (x 3 ab 3 x x gives 2 b 2 1) ) 2 b)(a and 4 1) x(x 2 b)(a 3 b 4 3 1) last 4 a 2 b and (x 3 x 4 a Replacing 4 1) 2 1) 1) 1) x 2 1) x 3 (x 1) (x 1) x } 3 (x x Example 3 (a) Show that (b) Hence (a) 7x 2 7x 3 3x 3x 1 3 8x (x 3 or 3 otherwise factorise 2 3 3x 3x 1 8x 3x 3 1 8x 2 7x (x 3x 3 Using and 3 a 3x replacing 3 a (a by b)(a 2x and 1) (x 1) (x 1) 3x 1) 3 2 b ab by (x b ) 1) 3 8x 3 (x 2 b 1 2 3 3x 1 3x 3 3 (b) 3x 2 x 3 8x 3x 3 8x 2 7x gives 2 {2x (x 1)}{(4x 2 2x(x 1) } 2 3 (x 1)(7x 4x 1) 2 7x 3x 2 3x 1 (x 1)(7x 4x 1) Exercise 1.8 3 1 Factorise 2 Show 8x 3 1 (x completely. 4 that x (Hint: ( 1) 4 2) 1) 2 8(x 1)(x 2x 2) 25 1.9 Quadratic Learning outcomes and Polynomial A To investigate roots of a the nature quadratic cubic of polynomial equations equations equation equation n x a n the relationship sum and form 1 2 x n .... a 1 x a 2 x a 1 0 0 between The the the n a and has the product of roots of a polynomial equation are the values of x that satisfy the these equation. roots and the coefﬁcients of 2 ax bx c 0 The order Some, To use the relationship or of the all, of polynomial these roots gives may the not number be real. of For of the equation. example, roots the quadratic between 2 equation the sum of the roots, of the roots, the (Y ou will the wise product and 3 a the of the 2 0 has two roots, although neither of them roots coefﬁcients discover the nature of these roots if you study are Pure sum Mathematics of x the real. product x Unit 2.) pair- of 2 bx cx d 0 The nature of the roots of a quadratic equation 2 The general The values form of You need to know x of a that quadratic satisfy this equation is ax equation are given √ to expand brackets (ax of b)(a c 0. by 2 and it is the value of b 4ac that determines the 2a the nature form 2 b b 4ac _______________ How bx ________ x of these polynomial) roots. 2 Note that b 4ac is called the discriminant ________ 2 When b 4ac therefore the 0, roots are 2 √ b 4ac real and has two real and different values, different. ________ 2 When b 4ac 0, 2 √ b 4ac b ___ ∴ x 0 and x 2a so is there said is to 0 b ___ 0 2a only have one a value of repeated x that satisﬁes the equation and the equation root ________ 2 When no b real The 4ac 0, √ 2 b 4ac has no real value, so the equation has roots. relationship equation between the and the coefﬁcients of a quadratic roots 2 The If the general and form are equation of the can a quadratic roots be of this expressed equation is ax bx c 0 ) 0 0 [1] equation, as )(x (x 2 ⇒ [1] and [2] are the identical b __ 2 x equation, c __ x so we say )x that 2 ( )x coefﬁcients of x a ( x x a 2 ([1] is divided Comparing by a, so that coefﬁcients of the this identity shows b __ a c __ and a 26 are that equal.) [2] Section the sum of the roots of the equation ax Basic algebra and functions b __ 2 i.e. 1 bx c 0 is and a c __ the product of the roots is . This is tr ue whether or not the roots a are real. Example Determine (a) the nature of the roots of the equation 2 3x 2x 2 0 2 If (b) and are the roots of the equation 1 __ equation whose roots 2 2x 2 0, ﬁnd the 1 __ are and 2 3x (a) 3x 2x 2 0 so ‘b 4ac’ 4 4(6) 20 2 Therefore 3x 2x 2 0 has no 2 3x (b) real roots. 2 __ 2x 2 0 gives 2 __ and 3 3 1 __ For the equation whose roots 1 __ are and , 1 __ the sum of the roots 1 __ is ______ 2 3 1 2 3 1 __ and the product of the roots is 1 __ 1 ___ 1 __ 3 __ 2 2 3 3 2 Therefore the required equation is x x 0, 2 2 i.e. 2x 2x 3 0 Exercise 1.9a 2 1 One root Find of is the roots of the equation 3x x 0 c 0 is and the other 2 the value of c 2 2 The Find Cubic The roots the of the equation equation whose x roots 3x are 5 2 are and and 2 equations formula for Did you know? solving a general quadratic equation was known to the It ancient Greeks. However , the search for a general solution for the is thought (1501–1576) equation continued until a method was developed during the general the ﬁrst method of to publish solution for the Italy. cubic This and that method is you does difﬁcult are reasons If was Renaissance a in that Girolamo Cardano cubic not it are ‘general to real, not lead work and interested of a in formula, with. such included solution to It also but it relies numbers are is on not not at all working covered easy with in to equation. remember numbers Unit 1. For these here. ﬁnding cubic this formula, search on the internet for equations’. 27 Section 1 Basic algebra and functions The roots of a cubic equation 3 The general The order If these form of this roots are of a cubic equation , is and equation three, then By expanding form, the we can this get coefﬁcients form a of of the the the general bx it equation )(x equation relationship 2 ax therefore )(x (x is ) and between has can cx three be d 0 roots. written as 0 comparing the roots of a with cubic the general equation and form. 2 )(x (x )(x ) (x x )(x ( )x 3 Dividing the general 3 ) 2 ( form of the )x cubic ( equation by 2 ( x a ( )x b __ x )x c __ 2 x a 3 Therefore , if and are the roots of ax gives 3 )x d __ x a a 2 bx cx d 0, then b __ a c __ a d __ a 3 For is example, 2, roots the is the sum product of of the the roots roots of the pair-wise equation is 5, and x 2 the 2x product 5x of 7 0 the 7. Example 3 T wo of Find If the the is roots values the third of of the p equation and root, [2] gives therefore and from 2 px 2x q 0 are 1 and 2. q then 2 x 1 p Sum 2 2 Pair-wise 2 q Product of the product of [1] roots the of the roots [3] roots 4, from [3], [1], q p 5 8 Example 3 The Find The equation a ax 2 relationship sum of the 3 b __ cx between roots a 28 bx is ⇒ a, ( b, p) d c b ___ 3a 0 has and roots d ( p) 3 p, and [2] p Section is a root of the equation, so x 3 i.e. a b ___ ( b ) c the algebra and functions equation, ⇒ 3 ab 9a 0 3 bc 2 2ab d 2 3ab 3 ⇒ ) 3a 3 27 a b ___ ( 3a 3 by b ___ ( 3a Multiplying satisﬁes Basic 2 ) 1 9a 27a d 0 3 bc 27a d 0 0 Example 3 The roots of the equation 2 2x x 3x 1 __ Find the equation whose roots are 1 __ , the given are , and 1 __ and From 1 equation 1 __ 2 3 __ 2 1 __ 2 For the required equation, the 1 __ sum 1 __ the 1 __ of product of the roots pair-wise 1 ___ 1 ___ _____________ 3 is 1 ___ is The roots __________ 1 1 ____ The product of the roots is 2 is x 3 Therefore the required equation 2 3x x 2 0 1 Exercise 1.9b 3 1 T wo of the roots of the equation 2x 2 px qx 0 are _ 1 and 1. 2 Find the values of p and q. 3 2 The Find roots the of the equation equation whose 2 x roots 2x The Find roots a of the equation relationship 2x between p are 3 3 5x 1, 1 1 0 are and , and . 1 2 x and px q 0 are , and q. 29 1. 10 Curve sketching Learning outcomes Straight The To revise basic curve equation of any straight line can be written as y mx c where m techniques for is simple lines the gradient of the line and c is the intercept on the y-axis. sketching T o sketch points You need to know The on the graph the line. most of a straight straightforward the axes. For example, line, points to you ﬁnd need are the those coordinates where the of line two crosses 2 How to express ax bx c in 2 the form a(x p) to sketch the line 2x 3y 9 0, ﬁrst ﬁnd where the line q 1 crosses the axes: when x 0, y 3 and when y 0, x 4 , so draw 2 1 the line through (0, 3) and ( 4 , 0). 2 y 5 Curves 4 A 2x 3y 9 2 on 1 4 3 2 the the of a curve 1 1 2 3 4 coordinate curve accurate x O 5 sketch should show the shape of the curve and its position 0 such plot, as, so axes. for It should example, these also where features will in show the any signiﬁcant curve many turns. cases be A features sketch is of not an approximate. 5 1 2 Parabolas 3 2 A curve whose equation has the form y ax bx c has a characteristic 4 shape called a parabola. 5 When y has a a a minimum y value. In a 0, maximum value. both where has cases, the the curve curve turns, is as symmetrical shown in the about the line through the point diagrams. 2 T o sketch either the ﬁnd graph the of the curve coordinates of whose the equation points is y where the ax curve bx c, you can crosses 2 the axes (this symmetry to is easy ﬁnd if the ax bx coordinates c of factorises) the turning 2 or express ax curve bx of crosses the the c in the turning y-axis at form point the a(x p together point (0, use ) q with to ﬁnd the fact the that the c). y Example 5 2 Sketch then point 2 coordinates and the curve whose equation is y 2x 3x 1 4 2 y 2x 3x 1 (2x 1)(x 1) 3 1 The curve crosses the y-axis at (0, 1) and crosses the x-axis at ( 2 , 0) and (1, 0). 3 2 ( Therefore the curve is symmetrical about x 1 , 4 3 ) 8 1 4 1 (halfway between x and x 1) 2 1 1 1 1 has a minimum value where x 4 30 3 of 2( 4 3 2 ) 3( 4 2 1 ) 1 1 8 1 1 1 2 2 2 3 y 2 2 x Section 1 Basic algebra and functions y Example 2 2 Sketch the curve y x 1 (x x 1 1 1 2 y (x x) ( 3 2 curve and y has crosses a the y-axis maximum ) 4 4 value at (0, x O 2 The 3 , 2 ) 2 2 4 1) where 2 3 1 x of 2 4 4 Cubic curves 3 A curve whose equation is y 2 ax bx cx d has a characteristic shape. a 0 a 0 y 8 6 or The curve is easy or to sketch when the cubic expression factorises. 2 For example, the graph of y (x 1)(x 2)(x 3) crosses the x-axis at 1, 0), (2, 0), (3, x O 2 ( 0). 2 3 When the shows brackets that a 1 are and expanded, d 6, so and the comparing curve with crosses the ax 2 y-axis bx at (0, cx d 4 6). 1 __ The curve whose equation is y x 1 We know that is meaningless, so there is no point on the curve where y 0 x in 0. the When ﬁrst x and 0, y third 0 and when x 0, y 0 so the curve exists only 6 quadrants. 4 1 We also know that as x increases the as x approaches for positive curve gets values, closer to as x the gets larger , x-axis. x Using gets smaller , similar i.e. 2 reasoning, x For negative zero values from of x, positive as x values, approaches y 6 increases. zero, y decreases, and as 4 2 4 6 x 4 approaches , y increases. The curve gets closer and closer to the axes 6 but never Any line crosses crosses that is a them. curve called an gets closer and closer to but never asymptote 1 __ y 0 and x 0 are asymptotes to the curve y x The curve is symmetric about the line y x Exercise 1.10 1 Draw sketches of the graphs whose equations are given. Mark 2 On the graphs all signiﬁcant points on the same whose set of axes, equations draw sketches of the are 2 __ curves. y and 2y 3x 6 0 x 2 (a) y x 5x 6 (c) y x(x 1)(x 3) 2 (b) y 3x x 1 31 1. 11 Transformation Learning outcomes To understand how curves knowledge and to to use sketch the curve whose equation is y f(x) and the curve whose are equation transformed curves Translations Consider of is y f(x) 2 this curves y y f(x) 2 You need to know 3 2 The meaning of translation and y How to sketch f(x) 1 reﬂection graphs of simple x O equations Comparing of x, the Therefore points y on f(x) y value for y by For is f(x), units any the with f( x) equal 2 f(x) of in function the f, c curve x, of we y units on f(x) that than 2 equation y a particular value 2 are of 2 value f( x). units translation of above the curve y-axis. equation is y equation the is a the whose to for the f(x) is of whose cur ve parallel whose y direction cur ve the see greater points curve the 2, units positive translation consider f(x) 2 of the the is values i.e. by Now y 2 is f(x) y c f(x) y-axis. f(x 2) y y y f(x) f(x) 4 2 4 y f(x 2) x O Comparing y same the in when f(x) with value of y x in f(x 2), f(x 2) is 2 see that units the values greater than of the y are the value of x f(x). Therefore right of units curve for equal points of Using the curve similar y For f(x) any on the y f(x), 4 f(x) units function When of values y reasoning, by translation c of x-axis f, the 0, y, i.e. in the in the points the the y translation when c 0, by is the of f(x 2) f(x 2) direction of whose f(x) f(x direction y y positive cur ve y on curve curve the cur ve the and of direction 32 we 4) the is of a units in the the units is y is to the by x-axis. of the x-axis. negative x-axis. 2 translation translation parallel translation the a negative equation c are is f(x to the c) is a x-axis. direction in the positive 2 Section 1 Basic algebra and functions Reﬂections y Consider y the equation is y Comparing y f(x), given x, value a point reﬂection point y any function of Consider the curve y whose f(x) we Therefore for f(x) see of x, with that f(x) for a f(x) x O So, curve f(x) the f, in y the y the same value f(x) is the of the x-axis of f(x) f(x) the cur ve whose on for on cur ve y equation y f(x) is y in f(x) the f( is the reﬂection x-axis. x) y y f(x) y f( x) x O Comparing when the Therefore y f(x) of points symmetrical So values are with about for with x any y the the x), same we in see sign, that i.e. the f( a) y-coordinates on values f( the ( of y are equal a)) curves are y-axis. function of f( opposite the f, the cur ve cur ve y y f(x) in f( x) the is the reﬂection y-axis. y Example 6 1 ______ Sketch the curve whose equation is y x 2 1 y 1 __ Start with the curve y whose shape and x 4 position x is known. 1 y x 1 __ If f(x) 1 ______ then x is x f(x 2). 2 1 ______ So the curve y 1 __ is x a translation 2 of y by 2 units x 6 in the positive 2 2 direction of the 4 O x x-axis. 6 33 Section 1 Basic algebra and functions Example 2 Sketch the curve y 2 (x 5) y 2 y 2 (x 5) 2 x 8 6 4 2 2 2 4 2 y (x 2 5) y x 6 2 Start with y x whose shape and position is 2 Then y direction (x of the known. 2 5) is a translation negative of y x by of 5 units in 2 Therefore parallel y to the 2 (x positive the x-axis. 2 5) is a translation y (x 5) by 2 units y-axis. Exercise 1.11a 1 Sketch each of the following curves whose 2 (a) y x 4 y (c) (x y 1 ______ 2 y (d) 3 x 2 On the x same set of axes, sketch the 1 curves 3 (a) y y x (x (c) y equations are (d) y 1 (x 2) 3 2) One-way stretches Consider curve the whose 3 3 (b) are 1) 1 __ (b) equations 3 whose (x equation is y 2) af(x) y y af(x) y f(x) ay y x O Comparing points y-coordinate of the So of 34 the on y point the cur ve f(x) on y cur ve y and y f(x) y af(x) af(x) is af(x) parallel a is with times a to the the same one-way the x-coordinate, y-coordinate y-axis on stretch by a factor a y the f(x) Section Consider the curve whose equation is y 1 Basic algebra and functions f(ax) y y f(ax) y f(x) 1 x a x x O Comparing points on y f(x) and y f(ax) with the same y-coordinate, 1 __ the x-coordinate of the point on y f(ax) is times the x-coordinate a on y f(x) So the cur ve y f(ax) is a one-way stretch 1 __ of the cur ve y f(x) parallel to the x-axis by a factor a Example 2 On the same set of axes sketch the curves y x 2 , y 2x and 2 y (2x) for values of x from 3 to 3. y 2 y (2x) 30 2 y 2x 25 20 15 2 y x 10 5 x 3 2 1 O 1 2 3 2 Start with y x 2 Then double the y-coordinate of points on y x 2 to give y 2 Halve the x-coordinate of points on y x 2x 2 to give y (2x) Exercise 1.11b On of the x same from 3 set to of axes 1 __ (a) y the graphs 2 __ (b) x sketch of the curves given for values 3. y 1 ___ (c) x y 2x 35 1. 12 Rational Learning outcomes expressions Rational An To express an improper expressions expression polynomials expression as polynomial the and a sum where both the numerator and denominator are rational of proper is called a rational expression a rational 1 __ For example, 3x ______ x _____________ , , are rational expressions. 2 expression x These the (x 1)(x expressions numerator is are less 2) all x proper than the 1 rational order of expressions the because the order of denominator . You need to know When The meaning of the order of the order of the denominator , For example, How to sketch is is greater called than or equal to the order of improper . 2 x _______ x 1 _______ and 2x The factor numerator expression a polynomial the the are 1 2x improper 1 theorem use transformations to curves Expressing an polynomial There form The are two where ﬁrst improper fraction and a methods the a sum of a proper fraction we remaining method as can use to fraction involves is express an improper fraction in a proper . rearranging the numerator so that we can cancel. 2x 3 _______ For example, in the case of we x that x 1 is part the 2(x 2x 3 _______ 1) rearrange the numerator so numerator , 2 3 ________________ i.e. x We of can 1 1 can x now 2(x express 1) the 1 right-hand side as the sum of two fractions, 5 ______ ________ i.e. x 1 x 1 5 ______ We can now cancel (x 1) in the ﬁrst fraction to give 2 x 2x 3 _______ 5 ______ x The 1 2 1 x second 1 method 2 involves Start by dividing dividing x the into numerator 2x. It goes by 2 the denominator . times. _______ x 1 ) 2x 2x Multiply 3 2 2 is the x 1 by quotient 2 then and 5 is subtract the this from 2x remainder . 5 2x 3 _______ Then This 5 ______ x 2 second x method is 5 __ 12 ___ (in 1 the same way as 1 1 7 useful when the denominator ) 7 is quadratic. 3 x 2x 5 ___________ For example, to express in 2 x proper , 36 we divide by the 4x 5 denominator . a form where the fraction is 3 Section 1 Basic algebra and functions 2 x There 4 _________________ 2 x 4x ) 5 x 3 is no x term in the 2 0x 4x 3 2x numerator 5 term. 5x It 2 4x 4x so we add zero for this 2 2 x 7x Start goes x by dividing 3 x into x . times. 5 2 2 16x Multiply x subtract. Bring 4x 5 by x then 20 2 9x 25 down 5. Divide x 2 into 4x until no , and more repeat the division is process possible. 3 9x x 2x 5 ___________ x This x 4 25 2 4x second polynomial, is ___________ 2 not 5 x method we of need 4x division to ﬁnd the is 5 also other useful factor when, and given ﬁnding one it by factor of a inspection straightforward. Example 4 Given that 2x 1 is a factor of 3 2x 2 x 6x x 1, ﬁnd the cubic factor . 3 x 3x 1 ______________________ 2x ) 1 2x 4 3 x 4 2 6x x 1 3 2x x 2 0 6x x 1 2 6x 3x 2x 1 2x 1 3 Therefore cubic x 3x 1 is the factor . Example x ______ Sketch the curve whose equation is y x 1 x ______ x 1 1 __________ y x 1 ______ 1 x 1 1 x 1 4 1 ______ y 1 x 1 1 __ 2 Start with y 1 ______ , then y x x 1 1 __ is the translation of y by x x O 2 4 1 unit in positive the direction of the x-axis. 1 ______ So y 1 is x the 1 1 ______ 4 translation of y by x parallel to the 1 unit 1 positive y-axis. Exercise 1.12 1 Express each expression in a form where 3 6x _______ 2x _______ (a) the fraction is proper . 2 x x 3 ____________ (b) (c) 2 2x 1 2x 1 x 4 2x _______ 2 Sketch the curve whose equation is y 2x 4 3 Show that x 2 is a factor 4 Hence factorise x of 3 x x 3 x 1 2 2x 3x 6 2 2x 3x 6 37 1. 13 Inequalities – quadratic and rational expressions Learning outcomes Quadratic A To revise quadratic quick To solve rational sketch is the easiest way to solve an inequality such as inequalities (x inequalities inequalities 1)(x 2) 0 involving y expressions 10 8 6 You need to know 4 How to sketch a curve whose 2 equation has the form 2 y The ax c conditions for equation real bx to have a roots or to complete 4 no The roots How 2 quadratic real the x O 2 the sketch x-axis, of the i.e. (x curve y 1)(x (x 2) 1)(x 0, 2) when shows 1 x that the curve is 2 square The We of inequality know and we When 1) (x (x and x 1 x x Therefore 2, (x 1)(x solved 2) 2) is 0 algebraically. when positive x or 1 and negative x 2, depends on the signs 2). these both is 2, signs brackets (x 1) is for x are 1, negative, positive and 1 x so (x (x 2) 2 and 1)(x is x 2) is negative, 2 positive. so negative. both be 1)(x (x 2) also 1, 1)(x When can investigate When (x that whether (x So brackets 1)(x 2) are 0 positive, when 1 so (x x 1)(x 2) is positive. 2 Example 2 Find the values of a for which x ax a 0 for x 2 The curve y x ax a is a parabola with a minimum value. 2 a __ 2 Completing the square gives x ax a (x ) 2 2 a __ x ( ) 0 for all values of x, 2 2 a __ 2 so for x ax a 0, a 0 0 ⇒ 4 2 a __ Now, a 2 4 38 below 0 ⇒ 4a a a(a 4) 0 a __ 2 a 4 Section 1 Basic algebra and functions y A sketch of y a(a x O 2 4) shows that a(a 4) 0 when 0 a 4 2 Therefore x Rational An a 0 when 0 a 4 expressions expression polynomials For ax where is both called a the numerator rational and denominator are expression. example, 1 __ 3x ______ x ____________ , , are rational expressions. 2 x (x The 1)(x 2) x 1 range of values that a rational function can take 1 __ The graph of y (see T opic 1.10), shows that 0 is the only value that x y cannot take. 1 __ We can show this algebraically: y 1 __ ⇒ x x when y 0, so 1 __ there is no 1 __ i.e. 0 or of x x is undeﬁned for which y 0, 1 __ x value and y 0 but x 0 x 3x ______ Now x can take all real values when y 2 x T o ﬁnd the quadratic values that equation in y can have, we 1 rearrange the equation to give a x, 2 i.e. yx 3x y 0 2 For x to be 4y real, this equation has to have real roots, so ‘ b 4ac’ 0, 2 i.e. 9 y 0 9 __ 2 ⇒ 4 3 __ ⇒ y 3 __ and y 2 2 3 __ ⇒ 3 __ y 2 We can 2 use this information, together with the following observations, 3x ______ to sketch the graph of y 2 x 1 39 Section 1 Basic algebra and functions y 0 when x 0 y 0 when x 0 when as x approaches as x → x 0, , y y 0 very → large values (we write this as x → ), y → 0 0 y 3 2 1 20 15 10 x O 5 5 10 15 20 1 2 3 Solving inequalities It to involving rational expressions 3x ______ is easy see the values of x for which 0, but it is not so 2 x 1 x _____________ obvious for the expression (x 1)(x 2) x _____________ The values of x for which (x x 1 The x and x value 1 of and 1)(x 0 depend on the signs of x the expression 2, so we is need zero to when x investigate 0 the and sign undeﬁned of the when expression when x The easiest 0, way 0 to do x x x 1 x 2 x this is 1, to 0 1 use 0 a x 2 and x 2 table. x 1 1 x 2 x 2 x _____________ (x 1)(x 2) x _____________ Now we can see that (x 1)(x 0 when 0 x 1 and x 2) Example x _______ Solve the inequality inequality right-hand 40 is side easier is 1 __ 2x An x, 2) 2 zero. to 1 solve x if it is ﬁrst rearranged so that the 2 Section x _______ 1 Basic algebra and functions 1 __ 2x 1 x _______ x 1 __ ⇒ 2x 1 0 0 x 2 (x 1) _________ ⇒ x(2x The 1) numerator is positive for all x so the signiﬁcant values of 1 x are 0 and 2 1 We need to investigate the ranges x 0, 0 x 1 and x 2 2 1 _ x 0 0 x 1 _ x 2 2 2 (x 1) x 2x 1 2 (x 1) _________ x(2x 1) 1 __ x _______ 1 Therefore 2x for 1 0 x 2 x Example 2 x 2x k ___________ Find the values of k for which can x for all x take all real values 1 2 x 2x k ___________ Let y x Rearranging as 1 a quadratic equation in x gives 2 x x(2 y) (k y) 0 2 For x to be real, (2 y) 4(k y) (4k 4) 2 ⇒ y 8y 2 Completing the square gives (y 4) 16 4k 4 2 The minimum values provided value of that (y 16 4) 16 4k 4, is 16, i.e. k so y can take all real 3 Exercise 1.13 2 1 Find the values of x for which x 2 Find the values of k for which (kx) 3 Find the set 4 2x 1 2 (3k 2)x 4 0 for x . x 1 ________ of values of x for which x(x 0 2) 2x ______ 4 Find the range of values of y for which y 2 1 x x k ______ 5 Find the minimum value of k for which 1 for all x . 2 x 1 41 1. 14 Intersection Learning outcomes of curves investigate the intersection curve and a on the shape of a curve, a line may intersect the curve at of several a lines Intersection Depending To and points, it may touch the curve at one of these points, or it may line not intersect For example, may touch the a curve line the at may parabola any point. intersect at one a parabola point (in in two which distinct case it is points, called a or it tangent You need to know to How to linear, sketch the quadratic graphs and the parabola), or it may not intersect the parabola. of cubic functions How to solve a simultaneous one is linear pair of equations and the where other is quadratic T o The conditions for a ﬁnd curve equation roots, or to a have two repeated the points of intersection, we need to solve the equation of the quadratic and the equation of the line simultaneously. distinct root, or For no example, to ﬁnd the points of intersection of the line with equation 2 real y roots 3x 5 equations How to to ﬁnd use the the factor roots of a with the curve with equation y x 2x simultaneously. 1, we solve the y theorem A cubic rough idea equation of above which exists sketch The sketch is of of in these the this curves three case. gives cases an illustrated However , this inconclusive. nature of the solution will tell us if this x O line intersects, y x touches or misses the 1 curve. 2 2x 1 [1] y 3x 5 [2] 2 [2] in [1] There two are 3x two distinct From or ⇒ x [3], 3 5 real 2 x and 2x distinct 1 ⇒ x ⇒ (x roots so 5x 6 2)(x the 3) line 0 0 [3] intersects the curve in points. the and coordinates y 4, i.e. of (2, these 1) and points (3, are x 2 and (from [2]) y 1, 4). y Example 2 ______ Prove that the line y 3 2x and the curve y do x not intersect. 3 2 ______ Solving y 3 2x and y simultaneously x y 3 2x gives 3 [1] 2 ______ y [2] x x 3 2 ______ Substituting [2] in [1] gives x 3 2x (3 3 ⇒ 2 2x)(x 3) 2 ⇒ 2x 9x 11 0 2 ‘b 4ac’ is 81 88 which is less than zero, so there are no real The sketch shows that the line and 2 values of x Therefore 42 for the which line 2x and the 9x curve 11 do 0 not intersect. curve a do proof. not (A intersect sketch is but also this is not unreliable.) Section 1 Basic algebra and functions Example (a) Find the condition on m and c for which the line y mx c is a 2 tangent (b) Hence to the ﬁnd curve the whose equation equation of the line is y with 3x 2x gradient 2 1 that is a 2 tangent to the curve whose equation is y 3x 2x 1 2 (a) Solving y 3x 2x 1 and y mx c simultaneously gives 2 mx c 3x 1) 0 2x 1 2 ⇒ 3x For x(m the line to 2) (c touch the curve, this equation must have a repeated 2 root, i.e. ‘b 4ac’ 0, 2 so (m 2) 12(c 1) 1 (b) When m 2, 16 12(c 1) ⇒ c 3 the equation of the line with gradient 2 that is a tangent to 2 y 3x 2x 1 is 1 y 2x 3 ⇒ 3y 6x 1 0 Example Show that the 3 y x The 7x values roots line y 5x 4 of of the intersects the curve 10x x at 5 once which the and intersects 4 5x once. the curve 3 10x of f(x) 5 3 factors the it line 2 7x Possible touches are given by the equation 3 x Using 0 2 factor 2 x 7x 15x 1), 9 0 2 x 7x theorem, ⇒ when x 15x 9 1, f(x) are (x 0, (x (x 1) 9), is a (x 3) factor . 2 So f(x) (x 1)(x (x 1)(x 6x 9) 2 3) 3 the equation repeated Therefore y 7x 15x 9 0 has one single root and one root. the 3 2 x line y 5x 4 0 intersects the curve 2 x 7x 10x 5 once and touches it once. Exercise 1.14 1 Find xy the 4 value of k for which the line y kx 2 touches 3 2 Find the where 3 y nature Determine 0. of the Hence whether points sketch the on the line y the curve y curve the curve whose x 2 5x 8x 4 curve. x 5 intersects, 2 intersect the 0 equation is x touches or does not 2 2y 7 43 1. 15 Functions Learning outcomes Mappings 2 When To deﬁne mathematically the pressed, terms: function, domain, number 2 is entered in a calculator and then the x button is the the display shows the number 4. range, 2 is mapped to 4, which is denoted by 2 4 composite functions Under this rule, which is squaring the 2 x To use the fact that a function input may be deﬁned as a set 3 9, 25 625, of 0.2 ordered number , pairs 0.04, number) 2 (the 4 and square (any of real that number). 2 This is denoted This mapping by x can be x , for x represented You need to know 2 graphically against How to sketch curves by values are of ax square bx x. The of graph, x of what happens and when our we the form 2 y of values whose knowledge equations plotting a number , show that one input x O c number But the gives just mapping one that output maps a number . number to √x its square root, output only greater than numbers This x The when do or for gives 2, input to have real is (negative square written roots). as x O that output a number zero real be gives representation shows two the can x graphical mapping 4 equal not mapping √x , e.g. one of input this value values. Functions 2 For the mapping x x , for x , one input number gives one output number . The mapping The one word √x function output A x is gives used two for outputs any for mapping every where one one input input number . value gives value. function is a r ule that number for maps a each deﬁned single set of number input to another single numbers. 2 Using f for function and the symbol : to mean ‘such that’, we write f : x x 2 for x The not to mapping satisfy have function 44 they do The set x this Domain We mean and unless of is the √x function for x 0, that x maps is x to not x a for all real function values because of it range give that some real inputs for we can use particular numbers a as function any real numbers number have to as be an input excluded for a because output. is called the domain of the function. x does condition. assumed not f Section The domain is The domain does wide, or fully, If as the the also not domain is as must not the have restricted, domain numbers called be Basic algebra and functions pre-image. to we 1 contain choose all to possible make it. inputs; Hence to it can deﬁne be a as function stated. stated, we assume that it is the set of all real (). 2 The mapping domain we x x choose. 3 can Some be used examples, to deﬁne together a function with their 2 1 f : x f over graphs, any are given. 2 x 3 for x 2 f : x f(x) x 3 for x x 0 f(x) The is point on included, solid point circle. x curve, open 0 and the and For curve we the would we where denote domain not indicate be x this x part this by by 0, of 0 a the the using an circle. 3 3 x O x O 2 3 f : x This x time 3 for the x 1, 2, graphical 3, 4 representation is four discrete points. y 20 16 12 8 4 x O These 1 2 three 3 4 examples are not the same function – each is a different function. For each The is set also The domain, of there output called the notation x For the in range is A corresponding numbers is called the set of output range of the values of a numbers. function. The range f( x) represents the output function, so for 2 for x, f(x) function given a image. 2 f : x is 2, the the function deﬁned range set can x of is in numbers be 1 also above, f( x) 4, represented 7, the 3 range and 12, for is the f( x) 3, function for the deﬁned function in 3, the 19. pictorially. 2 For example, This ﬁrst f : x function number can in x also the 3 be pair is for x 1, 2, represented the value of 3, by a x, 4 can set and of be illustrated ordered the second pairs, 1 4 2 7 3 12 4 19 domain range as: where number is the the 2 value by of the f(x). set Therefore {(1, 4), (2, f : x 7), (3, x 12), (4, 3 for x 1, 2, 3, 4 can be represented 19)}. 45 Section 1 Basic algebra and functions Example The diagram B{a, b, c, d, shows Give (b) Construct (a) In two A, (b) For f 2 to doing 4 reasons a of members of the set A{1, 2, 3, 4, 5} to the 1 a 2 b 3 c 4 d 5 e set why this function, f, mapping that maps is A not to a B, function. giving your answer as a set of pairs. maps be a this maps mapping e}. (a) ordered a to different function, but to two e the (or every simplest d). f {(1, members member is to b), of of A change (2, d), B and must the (3, (d c), map two (4, e). In to A, just mappings e), (5, 4 does one in not map member A so that of 2 to B. any member There maps to are either of B. several d (or ways e), and of then a)} Example 2 The function, f, is deﬁned by and (a) (a) Find f(4) and f( 4) For x 0, f(x) x For x 0, f(x) x f( x) x f( x) x Sketch (b) f(4) f( the for for x x graph 0, 0, of x f. Give (c) the range of f. 4 f(x) 2 (b) T o sketch and that the curves part graph in of the the 2 of a 4) y ( function, xy-plane. line x So we which 4) we can can 16 use what interpret corresponds we f( x) to to f(x) x negative for positive 2 and know x about x lines 0, as values of x, 2 for x values of 0 as the part of the curve y x that corresponds x x O (c) The range of f is f( x) 0 Composite functions 1 __ 2 T wo functions f and g are given by f( x) x , x and g(x) and g(x) , x x 1 0, x They . can These be two added functions or f(x) g(x) be combined in several ways. subtracted, 1 __ 2 i.e. can x , x 0, x x 0, x x 1 __ 2 and f(x) g(x) x , x 2 They can be multiplied 2 i.e. f(x)g(x) x or divided, 1 __ x, x 0, 0, x x 2 f(x) x __ ____ and 3 g(x) x , x x 1 __ x 3 The output f g f i.e. of be made input or of g, 2 x the 1 __ 2 x can g[f(x)] g(x ), x 0, x 2 x 1 __ Therefore the function f : x , x is a 0, x is obtained by taking the 2 x function as g of the function f. This composite gf(x). 2 For f(x) gf(x) x means , x the and function 2 i.e. 46 gf(x) g(x g(x) g of 3x the 3x 1, x x function 2 ) 1, f( x), function and is written Section fg(x) means the function f of the function 1 Basic algebra and functions g( x), 2 i.e. fg(x) This f(3x shows composite For any the input 1) that the function composite domain values of (3x 1) composite x function fg( x) is not the same as the gf( x). function of , g. gf( x), Therefore f(x) the is the range range of f of must f and be this range included in gives the g. Example f, g and h are functions given by 2 f(x) x , x as (a) Find a (b) Calculate , g(x) function the 2x of value x: of: 1, x fg (i) , 1 x, x ghf. (ii) gf(3) (i) h(x) hfg(3) (ii) gg(3). (iii) 2 (a) fg(x) (i) f(2x 1) (2x 2 ghf(x) (ii) gh(x 1) , x 2 ) g(1 x 2(1 ) 2 x ) 1 2 3 2 (b) gf(x) (i) g(x 2x , x 2 ) 2 2x 1, gf(3) 2(3) 1 19 2 hfg(x) (ii) hf(2x 1) h((2x 1) 2 ) 1 (2x 1) , 2 hfg(3) gg(x) (iii) 1 g(2x gg(3) (7) 1) 4(3) 3 48 2(2x 1) 1 4x 3 15 Example 1 _______ f and g are functions of x such that f( x) and 2x Find gf(x) 1 _______ T o x 1 g(x). change 1 _______ to 2x x, we need to ﬁrst take the reciprocal of 1 , 2x 1 1 __ so let h(x) , then hf(x) 2x 1 x 1 T o change 2x 1 to x, we need to halve 2x 1 and then add , 2 1 so let j(x) 1 x , 2 1 ___ jh(x) 1 then jhf(x) 2 1 (2x 1) 2 x 2 1 __ 2x g(x) 2 Exercise 1.15 1 The and function f(x) (a) Find (b) Sketch x the f is for given x value the 0, of by x f(5), graph of f( x) f( 3) the x for x 0, x . and f(0). function. 2 2 The (a) functions (i) (ii) (b) (i) (ii) Find the Sketch Find f g are function the the Sketch and curve curve given whose function the given given whose by by f( x) , x and g(x) 2 x fg( x). equation by x is y fg(x) is y gf(x) gf( x). equation 3 _______ 3 f and g are functions such that f( x) and 2x gf(x) x. Find g(x). 1 47 1. 16 Types of function Learning outcomes Codomain The To deﬁne mathematically codomain of a function is the possible values that can come out of a the function. terms: one-to-one function (injective function), function onto This (surjective function), many-to-one, onto function one-to-one will include and (bijective function) know include other all the function or codomain To prove whether simple function is or not a the values values one the we actual as that The will have values values well. come not that that seen might come codomain out of a before). come (i.e. useful function We out out is of can a the range) when (e.g. then a we but do may not complicated choose as a function. given one-to-one or 2 For the function f : x x for x 1, 2, 3, we can onto 1 choose the from to codomain to be the set of integers 2 1 10 1 inclusive. 3 4 5 You need to know 2 6 7 The meaning and range of function, domain 8 3 9 10 domain codomain One-to-one functions A function different to - one A is one-to -one member of the when each codomain. member The of the function f domain deﬁned maps above to is a a one- function. one-to - one function is also called an injective function Onto functions A function by a is onto member without a of when the every domain, matching member i.e. member of no the of the members codomain of the is mapped codomain are to left domain. 2 For x example, 2, 1, when 0, 1, g 2 is given and the as g( x) codomain x for is 2 0 the 1 set {0, 1, member image 4}, of in the the the diagram codomain shows has at that every least 0 one 1 domain. 1 Therefore An onto g is an onto function is function. also called a 2 4 domain codomain surjective function. 2 The function some f : x members of x the for x 1, codomain 2, 3 given do not onto the above have an is not image surjective in the because domain. 2 However , because number , not one 48 h : x every and real for of x number , every one-to - one member x positive because the when real more it positive squared, number than codomain, is e.g. one has a real real member both 2 and numbers maps of 2 to a square the map is root. domain to surjective positive 4. real But h maps is to Section 1 Basic algebra and functions Bijective functions A function This is a member comes For of the from just to one is both one-to - one where each codomain one example, maps T o that function member f : x one 2x for member member of and of ). of where the x and of is the each called a domain member of bijective maps the to function. one codomain domain. ) onto member onto and onto (each Therefore f is a is one-to - one member bijective (each of member comes of from function. summarise: injective but Every domain not surjective member maps member of to the of a surjective the Some different domain codomain. not map member and but injective members of the every codomain to of the Every same different of mapped member domain codomain member is bijective the and member codomain is of the to member codomain to. of maps a of the every the mapped to. Example 2 Determine the whether codomain is the function injective, given surjective by or f( x) x 2, x onto neither . 2 x 2 are not When the 2 for mapped x 3 domain Therefore f all to. and maps is x x to not , therefore Therefore 3, one f(x) f is some not 7, member members therefore of of the codomain surjective. the more than one member of codomain. injective. 2 So f(x) x 2, x onto the codomain is neither injective nor surjective. Exercise 1.16 1 2 Let A f {(1, {1, 2), Show that Let A f {(1, f is {1, (a) Show (b) Is f 2, (2, 1), 4} and (3, 1), the (4, one-to - one 0, (0, that an 3, 4), f onto 1, 0), is 2} not and and (1, function 1), f : A A be given by 3)} onto. the (2, function f : A A be given by 4)} one-to - one. function? Give a reason for your answer . 49 1. 17 Inverse function Learning outcomes Inverse functions f To deﬁne term mathematically is the function where f( x) 2x for x 2, 3, 4 the 2 4 3 6 4 8 2 4 3 6 4 8 inverse function The domain The mapping of the range {2, 3, can can 4} be be maps to the reversed, mapped i.e. back range each to {4, 6, 8}. member the domain. You need to know We can express this reverse mapping as 1 x The meaning What The a of shape of is a x a function curve when for x 4, 6, 8 2 one-to-one function equation is its y a quadratic of x or a cubic function This It is is a function called the in its inverse own right. function of f where f( x) 2x 1 Denoting function the inverse we write f function of f by 6, 8 f , 1 1 (x) x for x 4, 2 of x Notice that the range of Not every function has an the of the inverse function is the domain function. inverse. 2 Consider for the example, function both f( x) map to 4. x for x When this , which mapping is is such that reversed, 2 and each 2, value of 2 x maps this is not domain i.e. to two a function. maps only values to a x, for Hence different one-to - one A of only f has Y ou can When tell any whether line a have an and function parallel to the of an the onto f( x) is x-axis to both where 2 each codomain and 2, member have an of and the inverse, inverse. inverse an maps functions member functions function example 4 only if f is a one-to - one function. one-to - one will cut the from the graph graph only of once, y f f(x) is one-to - one. y y x O one-to - one The The line The graph of diagram y A 50 shows of a coordinates gives the curve and that is its one-to - one inverse obtained by reﬂecting y f(x) in the x reﬂection whose not a function x O the point are A( a, ( b, coordinates of a), A b) i.e. on the curve y interchanging f(x) the x- is the and point A y-coordinates of Section Therefore we can interchanging Now the x obtain and y coordinates the in of equation the A of equation on y f(x) y the can reﬂected curve 1 Basic algebra by y f(x) be written as [ a, f(a)]. Therefore A(a, the coordinates the reﬂected of A' curve on is the such reﬂected that the curve output and functions are of f [f( a), is a], i.e. mapped the to equation the input of b) of f. x O A(b, Hence if the equation of the reﬂected cur ve can be a) written 1 in the for m y g(x), then g is the inverse of f, i.e. g f y Any curve whose equation can be 2 y written be in the reﬂected However , form in the this y line f(x) y reﬂected x can x curve may not 2 y have an equation that can be x written 1 in the form y f (x) 2 The diagram shows the curve y the line x x O and The its reﬂection equation of in the image y x curve is √x is 2 x y ⇒ y a function. √x and x not y 2 y (We on x can the y this reﬂected maps case see to two cannot function of from the curve, values be diagram one of value y. written So as in x , x of this a x.) y However , 0 as, if we change the domain 0 to 2 give x so the , function then does f have is f( x) a an x , x one-to - one 0, function inverse. x O Example 1 Find y f (4) 5x when f(x) 5x 1, x 1 1 For the reﬂected curve x 5y 1 ⇒ y 1 (x 1) and (x 5 1 1 function, so f (x) (x 1), f 1) is a 5 1 1 (4) 5 (4 1) 1 5 Exercise 1.17 2 1 A function f is deﬁned by f : x (3 x) , x 3, x 1 Deﬁne 2 The x f (x) fully. functions f 1 Find: 3 (a) and g are given by f( x) 2x, x and g(x) 2 x, (a) Show g 1 f that 1 (x) f( x) (b) (x – (gf) 1)(x (x) – 2)(x – 3), x does not have an inverse. 1 (b) Redeﬁne f(x) with a different domain, so that f (x) exists. 51 1. 18 Logarithms Learning outcomes Indices Logarithms To use the laws of logarithms depend on the laws of indices, so here is a reminder of these to laws. simplify expressions p a a (a q a a p p a a q p p q q 3 For example, x For example, x For example, (x q x x 3 pq ) 4 x x 4 3 a 3 3 4 3 ) 4 7 x x 4 1 4 x 12 x You need to know 1 0 1 __ n a 1, n n a , a √ a n a The 2, value 3, of simple powers of 5 Logarithms 2 We can read the This this The statement base 10 relationship 2 In the raised can be the power to form the power is relationship 2 or the which called can is 2 to 100 then the to the a as power rearranged is whole 10 gives give base the 10 100. same must information, be raised to i.e. give 100. logarithm be abbreviated logarithm log 2 to the to base read 10 of 100 100 10 3 In the same way, 2 8 ⇒ 3 log 81 ⇒ 4 log 8 2 4 and 3 81 3 2 Similarly, log 25 2 ⇒ 25 5 ⇒ 3 9 5 1 1 and log 3 9 The base of a 2 2 logarithm can be any positive number , so c b a ⇔ log b c, a 0 a The symbol ⇔ means that each of these facts implies the other . 0 Also, as a 1 this means log 1 0, a i.e. the The logarithm power of a 4 16, to 4 any base number is zero always gives a positive result, __ 1 2 1 positive 2 e.g. of , … 16 c This means that, if log b c, i.e. b a , then b must be positive. So logs a of positive but the logarithm Natural There numbers is of a negative number does not exist. logarithms an irrational mathematics. This exist, constant It is was number denoted ﬁrst by that e named appears and e by is in equal Euler several to who different areas 2.71828… showed that as x 1 __ x → , (x ) → e x 1 __ Newton discovered that the sum 1 1 ______ 1 1 __________ 1 2 1 2 3 1 ______________ 1 52 2 3 4 … → e as more and more terms are added. of Section When e is used logarithms as and the are base for denoted logarithms by ln they are called 1 Basic algebra and functions natural x y ln x means log x so ln x y ⇔ e x e Logarithms by lg x or with log x, a base i.e. if of the 10 are base is called not common given, it is logarithms taken to be and 10. denoted So log Did you know? x y means log x and log x y ⇔ 10 x Natural 10 logarithms Napierian invented Evaluating scientiﬁc base e or the calculator can be used to ﬁnd the values of logarithms with 10. ‘ln’ – in called Napier 1614 he logarithms tables Use button to evaluate natural logarithms and the ‘log’ explanatory related an to internet ‘Napier’s Use also logarithms. John logarithms published A are text natural search and logarithms. to look up bones’. button x to evaluate button) is Laws of common used to logarithms. evaluate The powers of e button (usually above the ‘ln’ e. logarithms x Given x log b and y log a x Now bc Therefore log (a bc c then a y b and a c a y )(a x ) x ⇒ y bc i.e. y a log a bc log a b log a Example c a 2 Express This is the ﬁrst law of logarithms and, as a can represent any base, the law is applies used for to all the the logarithm of logarithms any product the formula in provided that the same √ log pq r in terms of this simplest possible logarithms. base 2 √ log pq r 2 Using x and y again, a law for the log of a fraction can be log p log q log √ r found. 1 x b __ a __ b __ ⇒ x log p 2 log q y log r 2 a y c a c b __ Therefore log a ( b __ ) x y i.e. log a ( c ) log b log a c a Express n A third law allows Example c us to deal with an expression of the type log b a 3 log p n log q 4 log r x n Using x log b x ⇒ n a n b i.e. a b a as x __ Therefore a single logarithm. n log b ⇒ x n log a b i.e. log a b n log a b a 3 log p n log q 4 log r n 3 These are the most important laws of logarithms. Because they are log p log n any base we do not include a base, but in each of these laws 4 log r true 3 for log q p every n q ____ 4 logarithm must be to the same r base. b __ log bc log b log c, log n log b log c, log b n log b c Exercise 1.18 1 Find (a) the log value of: 16 (b) log 2 2 2 (c) log 4 Express in terms of the 8 4 simplest possible logarithms: p __ (a) x ______ 2 log (b) ln 5x (c) log p√q (d) ln q 3 Express as (a) log p (b) ln 3 x a single log q 1 logarithm: (c) 2 log p 5 log q 1 ln x (d) 2 ln x ln (x 1) 4 53 1. 19 Exponential Learning outcomes and logarithmic Exponential equations equations x An To solve logarithmic exponential exponential equations the base of has x as part of the index, for example, 2 3 8 including When changing equation and a logarithm you solution need is to solve 2 For an exponential equation, ﬁrst look to see if the obvious. example, for x 3 5 2 Therefore 125, 5 x notice that 125 5 3 5 so 2 x 3 ⇒ x 1 You need to know When The laws How to of logarithms simplify the change solution the index is to logarithms x For example, for not a obvious, taking logarithms of both sides can factor . 3 3 8, taking logs of both sides gives log 8 _____ (x 3) log 3 log 8 so x 3 log 3 Therefore x 3 1.892... ⇒ x 4.89 log 8 that is NOT equal to log . log 3 well have used s.f.) 8 __ _____ Note (3 Note also that we could equally 3 natural logarithms. Example x Solve The the equation left-hand x 2 side of 2(2 this ) 3 equation cannot be simpliﬁed so taking logs x will not help, but using x Let y 2 y x , then 2 will. 2 __ x 2 2(2 ) 3 ⇒ y 2 3 ⇒ y 3y 2 0 y (y 2)(y 1) 0 ⇒ y x So 2 1 or y 2 x 1 ⇒ x 0 or 2 2 ⇒ x 1 Example x Solve the x 4(3 equation 4(3 x )(5 x ) 7 ⇒ (3 x )(5 logs gives 7 x )(5 ) x T aking ) ln (3 1.75 x )(5 ) ln 1.75 0.207 (3 s.f.) ⇒ x ln 3 x ln 5 ln 1.75 ln 1.75 __________ x ln 3 ln 5 Logarithmic A logarithmic x, T o for equation example, solve a equations ln ( x contains 2) logarithmic 1 the logarithms of expressions containing ln x equation, again look to see if the solution is obvious. 2 For example, for log (2x 1) 2 log 2 x, we can write 2 log 2 then log (2x 1) log 2 x x as log 2 2 x 2 2 ⇒ 2x ⇒ x 1 x 2 2 When single 54 the solution logarithm is and not obvious, then remove express the the 2x 1 logarithmic logarithm. 0 ⇒ terms x as 1 a , Section For example, for 3 log x log 2 16 1, collecting the logarithmic 1 Basic algebra and functions terms 2 3 x ___ on one side and expressing as a single term gives log 1, then 2 16 3 x ___ removing the log gives 1 __ 16 2 3 Therefore x 8 ⇒ x 2 Example Solve the ln x 2 equation ln (x ln x 1) 2 ⇒ ln x ⇒ ln ln (x 1) ln (x 1) 2 x ______ x x ______ 2 ⇒ x 2 1 e 1 2 2 So x(1 e e ______ 2 ) e ⇒ x 1.16 (3 s.f.) 2 1 e Changing the base of When logarithmic the bases simpliﬁed change If x to the log a single base c of of and logarithm terms logarithmic the we a are term. different, T o do they that, we cannot need to be be able to logarithm. want to change the base of the logarithm to b, then a x x log c ⇒ c a a log c b _____ T aking logarithms to the base b gives log c x log b a ⇒ x b log a b log c ln c ____ b ______ i.e. c log and in particular log a c a log a ln a b The base of an exponential expression can x T o express change x a as a power of e, then using x in a similar way. p a e gives x ln a p, therefore x ln a a e Example Solve the equation 3 log x 1 2 log 3 First change the base of x 9 log x to 3. 9 log x log 3 _____ log x x 3 _____ , 9 log 9 2 3 3 log x 1 2 log 3 x ⇒ 3 log 9 x 1 log 3 x 3 1 1 2 log x 1 so log 3 x ⇒ 3 x 3 2 1.73 (3 s.f.) 2 Exercise 1.19 2x 1 Solve the 2 Solve the ln(x 3 Solve 1) the equation (4 ln 2 equation x 2 )(5 simultaneous 1 1 ) x 6 equations ln y and ln(x log x log 2 2 2y 1) 0 2 x 3 4 Given that ln y 3, ﬁnd the value of x given that ln x 4 log 5 8 y 55 1.20 Exponential Learning outcomes and logarithmic functions Exponential functions x The To deﬁne exponential mapping x 2 2 is such that 2 1 2 2 4, 2 2 , and and any 4 real number maps to a single real number . logarithmic functions x Therefore x 2 , x is a function. x But is x not a ( 2) x has a real value only when x is an integer , so ( 2) , x function. You need to know x However , The the deﬁnition meaning of of a function a for any value of a 0, x a , x is a function. and x one-to-one The function f( x) a , x is called an exponential function. function x How to ﬁnd an For inverse function is The the meaning laws of of logarithms all f(x) values of a 0, x a 0, therefore the range of f( x) a , x 0 and logarithms X The meaning of The natural y curve a logarithms x The family of curves whose equations are y a go through a point that 0 is common point (0, to all of them: when x 0, a 1, i.e. they all go through the 1). x y x y When a 1, y 1 When a 1, and x 1 3 10 x y x 2 0, a increases as x increases x 2 1 y ( 2 (e.g. ) 2 3 , 2 10 , ..., 2 , ...) 8 x and when x 0, a decreases 2 (e.g. 2 as x decreases 3 , 2 10 , ..., 2 6 , ...), but never reaches 0, x i.e. as x → , a → 0 x When a 1, the opposite happens: y ( (e.g. x (1.5) as x → , 2 1 4 ) 4, ... → 0 ( 2 a increases 5 1 ) 32...) 2 x as x → , 2 x This 4 graph shows the curve y a for some different x O 2 a 2 values of x X 4 The f(x) inverse of the function a x The 2 function f( x) a is x If y a a one-to - one function, so it has x where f(x) a x an inverse. 1 , we obtain the graph of y f (x) by reﬂecting x y We a in can the line obtain y the equation of this reﬂected curve 1 y, y so the equation of y f by interchanging x and y (x) is given by x a . T aking logarithms to x y a base a, we get log x y, i.e. y log a Therefore when x a x f( x) a 1 , f (x) log x a The function log x has domain x 0. a y log x a (The range of a function is the domain of the inverse function.) 1 The function f : x log x, x 0, x is called a logarithmic a O 1 x x function and it is the inverse of f : x a , x x The graph shows the curves with equations y a and y log x a 56 Section 1 Basic algebra and functions X e The functions When a ln x and e, x f(x) f(x) The e , ln x, x x is 0, logarithmic x called the is function exponential called is function the and the inverse vice function logarithmic of the and function. exponential versa. x The graph shows the curves y e and y ln x y 10 x y e 8 6 4 y lnx 2 4 x O 2 2 4 6 8 10 2 4 x Note and that the the x-axis y-axis is an is an asymptote asymptote to to the the curve curve y y ln x variations of the e x These sketches show some simple graph of y e Example y y y 3x Given O O x x ﬁnd (0, f(x) e 1, x ∈ , 1 f (x). 1) (0, 1) 3x When x y e y x y y e e e interchanging (0, 1, x x and y gives 3y 1) x e 1 3y O ⇒ e x ⇒ 3y ln (x 1 x 1) 1 1 Therefore f (x) ln (x 1), 3 x 1, x Exercise 1.20 1 On the same set of axes, x y 2 On y 1 the 1 e sketch the x , y same ln x, 1 set y e of , y axes, ln x 1 Given f(x) 1 e 4 Given f(x) 1 2 of graphs of e sketch 1, y the ln (x x 3 graphs x 1) 1 , x , ﬁnd f (4). 1 ln x, ﬁnd f (2). 57 1.21 Modulus functions Learning outcomes The x modulus of y The To deﬁne the modulus its x is written as | x| and it modulus function means and of the positive value of x whether or properties not x itself e.g. |2| Hence is 2 the positive and | graph or 2| of y negative, 2 |x| can be found y O x O x x You need to know from part How to sketch Algebraic simple the of graph the of graph y for x by changing which y is the negative curves to methods for the equivalent positive values, i.e. by y solving reﬂecting the part of the graph where y is inequalities negative Hence in we the x-axis. deﬁne the function f : x |x|, ⎫ |x| x for |x| x y 0 ⎬ as The |x| for x 0 always positive Now the for positive any |x| x |y| y for two or zero, x for y square real 0 so 0 then x root of numbers and we |x| x x x can , x |x| y |y| so |x| y so x i.e. and for 2 If write |x| __ 2 √ x y, x 0, and |y| y y conversely if x | x| |y| ⇔ x 2 y follows 2 ⇒ x ⇒ (x that | x| |y| ⇔ x y)(x y) 2 that x The can , then y y illustrate 2 y when y | x| last this y y x y x y y) y) 2 i.e. We y 0: (x shows 2 0 (x table and 2 y x The 0 2 2 |y|, It 2 y 2 and 2 x for 2 i.e. Now |x| | x| 2 |x| properties of is x on property a is |x x |y| y| y, ⇔ y |x| x y |y| diagram. x The y modulus representing x y x x x a number number is equal from to zero, the distance shown here of as the the point vertical line. y y The diagram when y of that but x and when shows y one are is that both positive positive and or the both other negative, |x y| |x| |y| is negative, | x y| |x| |y|, so | x y| |x| |y| 0 58 Section The The the C modulus of graph curve for For of C any which f( x) example, curve W e y then to (x sketch in y negative. 1)(x reﬂect whose equation is y equation f(x), The by is y remaining |(x |f(x)| reﬂecting 1)(x in can the sections 2)| the and functions we are start be found x-axis by not the from parts of changed. sketching the x-axis the part of this curve which is below the x-axis. y y y (x 1)(x any function f, the mapping x 1 |f(x)| 1)(x 2)| is 2 also a function. y the graph of 1 |(x x O Example Sketch 2) x O y algebra 2) y For Basic a function curve with 1 |1 1 x| and y write 1 x 2 2 1 the equations y in (1 x) 2 1 non-modulus each part on form the of sketch. x O Start with a sketch 2 of 1 y 1 x, then reﬂect the 2 part the below the x-axis in x-axis. Example 2 Sketch the graph of y 2 |x 4| 2 Start with positive y |x direction 4|, of the then translate the curve by 2 units in the y-axis. y y 10 10 8 8 6 4 2 2 O x 4 2 4 4 2 2 x O 2 4 2 4 Exercise 1.21 Sketch the graphs of: 1 ______ 1 y 1 |x 1| 2 y | x 2 | 3 y |(x 1)(x 1)(x 2)| 59 1.22 Modulus Learning outcomes equations To solve modulus inequalities Intersection T o and equations ﬁnd the points of intersection between two graphs, we need to solve and the equations of the graphs simultaneously. When those equations inequalities involve a modulus, modulus For a sketch helps to identify example, x equations in non- form. to ﬁnd the y values y You need to know of those where the graph 1 2x (1 2x) y of y y How to sketch the graph |x intersects (x the y |1 2x|, we draw |f(x)| a 2| of graph y How to solve linear sketch and on it y the (x 2) and equations quadratic write equations of each section in 0.5 2 non-modulus form. O How The to solve x inequalities There properties of modulus are two intersection, functions (1 2x) points one x of where 2 ⇒ x 3 1 and one where 1 2x x 2 ⇒ x 3 2 Alternatively, using the property that | x| |y| ⇔ 2 |x 2| |1 2x| ⇒ (x 2 2) 2 x y : 2 (1 2x) 2 x 4x 4 4x 3 0 4x 1 2 ⇒ 3x 8x ⇒ (3x 1)(x 3) 0 1 x or x 3 3 Check: 1 when x 2 , |x 2| when It is x 3, We |x can 2| that the can 5 an 2x| 1 or g(x) by and |1 of 2x| x found give such f(x) as the 5, so x this of |2 x 1| and is a solution 3 x x 3x 3 is also method that modulus following g(x) so using values involving using then 1 , 3 values equation above, |f(x)| |1 sometimes equations solve illustrated when squaring Solving and 3 essential because 2 1 3 are a are not solution. checked solutions. signs by sketching graphs as fact: f(x) g(x) Example Solve the equation |2 x 1| 3x 1 2x 1 3x gives x 1 and (2x 1) 3x gives x 5 Check: when x when x 1, |2x 1| 3 and 3x 3 1 , |2x 1| 5 3x 5 x is 5 60 the 3, so only solution. x 1 is not a 1 , 5 1 Therefore 3 and so x is 5 a solution. solution 2) Section Example 1 Basic algebra y and functions y x 2 1 ______ Solve the equation x 2 | x 3 | 1 ______ From the sketch, x 2 | x 3 | where 1 ______ x 2 2 ⇒ x x x 7 0 3 1 ⇒ x 3.19 (3 y s.f.) 1 y x x and sketch shows that we only want the positive 3 x 3 O (the 3 root) where 1 ______ x 2 2 ⇒ x ⇒ Solving Simple inequalities inequalities can be involving solved x x 5 0 3 from a x 2.79 or modulus sketch of 1.79 (3 s.f.) signs the graphs. y y y (x x a a) b a a For a example, b x Otherwise the a the sketch b shows O that b the x a inequality | x a| b is true for b method used is the same as for equations. y y Example 3 x y y Solve the inequality From the sketch, |3 x| x| (3 x) x |x| y |3 |x| where 3 x x x, 3 i.e. where x 2 3 |3 x| |x| for x x O 2 3 2 Exercise 1.22 1 Solve (a) these |3x equations: 2| 5 2 Solve (a) the |2x following 1| |1 inequalities: x| 1 ______ (b) |2 x| |x| (b) | x | x 1 |x 1| x (c) |e 2| 1 (c) |ln x| 1 61 Section 1 Practice questions 3 1 f(n) x n 10 2n Solve (a) the equation e x 4e 3 0 2 (a) Show that (b) Hence f( k 1) f(k) 3(k k 1) Find (b) log by 3 prove for all by induction positive that integer f( n) values is of the x values 2 log 2 2 divisible of x for which 1 x n. 11 (x 1) 4 x and (x 3 2) are factors of 2 px qx 16x 12 n 2 f(n) 9 (a) Show (b) Hence 1 that f(k prove by 1) 9f(k) induction (a) Find the (b) Hence that f( n) is solve 4 8 for all positive integer of p and q. values of the equation divisible x by values 8 3 px 2 qx 16x 12 0 n. 2 x ______ 12 Given that y for x 3 3 Prove by induction that n n is a multiple of the for all positive integer values of all real values of x, ﬁnd 1 6 range of values of y. n. ___ √ 18 1 ________ 13 4 p and q are Construct __ Simplify (a) propositions. a truth table √ 2 to show the truth 3 3 values Given (b) that x 5 , ﬁnd the value of log 5. x of ~p → Hence q and p determine equivalent q whether ~p → q and p q are 14 Find the range of values of x for y which x 1 ___________ statements. 0. 2 x 5 p and q are 5x 6 2x ______ propositions. 15 Sketch (a) the graph of . x (a) W rite (b) Use down the contrapositive of ~p → On (b) the algebra of propositions to show 1 ~q a separate diagram sketch the graph of that 2x ______ ~p → (p ~q) q → y p | x 1 | . 2x ______ Solve (c) 6 Sketch (a) the graphs the equation x | of x 1 | . 2 __ y x 1 and y | x 16 | The the same diagram. Show the * y the Find (b) points the where range of the graphs values of x 7 Prove 1 by 2 x x for binary … operation 2 x * y n (n 1)(2n all State y . whether the operation is: closed (b) associative (c) distributive (d) such that x over has multiplication an inverse. 2 17 (a) * is deﬁned Sketch Solve the the graph of inequality y |ln 2 x| |1 x | 2 by 2 x x x, 1) y and y 18 for for reasons (a) (b) The y which 6 8 with n __ 2 3 by that 2 2 deﬁned | induction 2 1 | is intersect. 2 __ x * 2 coordinates Explain of operation 2 x on binary with Given f(x) 2 ln x, sketch the graph of: . reasons whether the operation * (a) y f(x) (b) y f is: 1 (a) closed (x) commutative. (b) x 19 3 9 f(x) When 2x f(x) is 5x px divided the 1 6 2x 5(3 1 ) by (x 2) the remainder 2 f(x) (x 1) for x and 10. 1 _______ _ 1 One root of the equation f( x) 0 is g(x) . 2 (a) Find (b) Factorise real 62 equation q. 20 is Solve 2 the values roots f( x) of of and the p and hence state f( x) the for (x q. equation number 0 of (a) Find (b) Explain not. x 0, x . 1) fg(x) and why gf(x). g( x) has an inverse but f( x) does Section 1 Practice questions 1 Find (c) g f(x). Deﬁne (d) a 28 domain for the function 2 h(x) 21 The f (x function {(2, 3), f (3, (a) Show (b) Suggest is f that given 2), that so (4, is h where f(x) h function f is deﬁned by ⎧ 1 1) The x 2, x 2 4 x, x 4 ⎨ exists. ⎩ for (a) Find (b) Explain x . ff(2). by 4), onto (1, why f does not have an inverse. 2)} but not one-to - one. ln x 29 a change to one of the ordered Show that Hence of f to give a function g such that g is e ﬁnd, in terms (ln 6 and of e, the value of both 2 ln 3 ln 13) e onto x. pairs . one-to - one. 1 Using (c) your ordered deﬁnition of g, give g as a set of pairs. 30 Find the range of f where f is deﬁned by 2 x ______ f(x) 22 Find the the line relationship x ay b between 0 is a a and tangent b such to the for that x curve Hence all x . 1 sketch the curve whose equation is 2 y 2x Hence x ﬁnd tangent 4 the with coordinates gradient 1 of the touches point the where y f(x) p and the curve. 31 Draw 23 Find the maximum value x ______ of k for q are a propositions. truth table for ~p q. which 2 k (x 1) 2 x 1 32 for all real values of Give (a) x. a counter example following statement The of sum any is two to show that the false. prime numbers is an is an 2 24 Given that 4 x 3 x 1 is a factor of even 2 3x 3x 3x 4, 2 Prove (b) ﬁnd the number . other factor and hence even 4 3 x that if n is any the x 3x 2 same diagram, sketch the x 2 , y x (x 1) 33 2 __ √y x __ Simplify √y _______ __ √y x √y 2 , y x 1, 34 for _______ following x 2 y n completely. curves: (a) 2 3x __ On n integer . x 25 integer , factorise Find the (x 2) conditions satisﬁed by a and b such 2 y (b) for ln x, 0 y x that 2 1 ln x, y 1 is a factor of (x a)(x 2 3b 2b ) ln x, 2 4 x y (c) 26 The e roots 3 x , y 3e of the 6x 3x , y e , for 2 x 2 35 Solve 36 The the inequality |x 16| 1. equation function f is deﬁned by 2 2x 5x 3 0 2 x are , and f(x) . and 27 the equation whose 1. Factorise roots are 1, 2, x 1 x 1 { x Find 2, 1 (a) Sketch (b) Find the graph f(0) (i) of y f(x). f(2) (ii) completely: 2 (c) g(x) x 3x 10 4 (a) 81x 16 Find 3 (b) (a b) the points of intersection of the curves 3 b y f(x) and y g(x). 63 2 Trigonometry, 2. 1 Sine, cosine Learning outcomes and To revise circular To revise the line rotates an from angle its initial position OP sine, cosine about the ﬁxed point O 0 measure to a and vectors tangent functions The deﬁnition of When geometry any other position OP , the amount of rotation is measured by the angle and between OP and OP . 0 tangent functions When the rotation rotation is is anticlockwise, clockwise, i.e. a the negative angle angle is the angle is positive, and when the negative, represents a clockwise rotation You need to know The The of sine, an cosine angle in a and rotation of OP can be more than one revolution. tangent right-angled P P triangle The exact cosine values and of tangent the of sine, 30°, 45° 45° and O 60° P O P 0 How of to use curves to O P 0 0 transformations help with curve sketching P Q The r radian Degrees r and (sometimes revolutions called are circular two units measure) used is to measure another unit angle. used to The radian measure angles. 1 rad O r P One radian by The an is arc circumference the angle equal of a in circle subtended length is 2 r to so at the the the centre radius of number the of of a circle circle. radians in one 2 r ____ revolution is 2 Therefore 2 rad 360°. r The diagrams show some other angles measured in radians. 3 π π 2π π 2 6 π π 2 4 π 3 y The P(x, sine, cosine and tangent functions y) When OP is coordinates deﬁned 1 drawn of P on are ( x, x- and y), y-axes, then the where sine, OP cosine 1 and as: y y , 1 θ O 64 x x y x __ __ sin cos __ and 1 tan for x unit and tangent the functions are Section The 2 Trigonometry, geometry and vectors sine function __ Measuring in radians, for 0 , OP is in the ﬁrst quadrant, y is 2 __ positive and increases in value from 0 to 1 as increases from 0 to 2 sin Therefore increases from 0 to 1 __ For , OP is in the second quadrant, y is positive and decreases 2 __ in value from 1 to 0 as increases from . to 2 sin Therefore decreases from 1 to 0. 3 ___ For , OP is in the third quadrant, y is negative and decreases 2 3 ___ in value from 0 to 1 as increases from to 2 sin Therefore decreases from 0 to 1. 3 ___ For 2 , OP is in the fourth quadrant, y is negative and 2 3 ___ increases in value from 1 to 0 as increases from to 2 . 2 sin Therefore increases from –1 to 0. y P(x, y) y y y P(x, y) θ θ 1 1 O y x x O y y y θ θ 1 x O O P(x, This and OP As shows sin moves OP from in round rotates 1 sin that varies to 0, is positive value the quadrants clockwise, and so for between sin on. The 0 < 1 and < 1. 1 x and The negative pattern for repeats y) P(x, itself y) 2 as again. decreases graph sin of from f( ) 0 to sin 1, then shows increases these properties: θ 1 π O π π 3 π θ 1 This The graph also shows that, for any angle , sin ( ) sin cosine function x __ For any value of , cos 1 P(x, y) P(x, y) 1 1 x θ θ x θ θ x x 1 P(x, y) 1 P(x, y) 65 Section 2 Trigonometry, geometry and vectors As OP moves decreases increases The round again to again graph of 1, to y the 1. then So cos quadrants, cos cos increases , looks like to sin , like cos decreases 0 and varies in from the in 1 to fourth value 0, and then quadrant between 1 and 1. this: θ 1 α 3π 2π α α O π π 2π 3π θ 4π 1 The curve ) cos ( is symmetrical about the vertical axis, showing that cos Comparing the curves for cos sin , and we see that when we translate __ the cosine curve by in the positive direction of the horizontal axis, 2 __ we get the sine curve, i.e. sin cos ( ) 2 The tangent function y __ For P(x, any value , of tan x y) As OP rotates through the ﬁrst quadrant, x decreases from 1 to 0 and 1 y y __ y increases from 0 to 1. Therefore the fraction increases from 0 to very x __ large O values, and as → x , tan → 2 y __ The behaviour of in the other quadrants shows that in the second x quadrant third fourth The tan is quadrant quadrant cycle Therefore then the negative tan is tan is repeats graph of and increases positive and negative and f( ) 2π tan looks The cosine and sin , sine, 1 For f( ) is continuous repeats every cos , is repeats in the , and to in the 0 this: (i.e. 1 continuous every 2π θ and tangent functions functions show that: has pattern 1 sin has no breaks) and a that 2 rad cos ( cos like tangent sin ) sin ( sin to π cosine f( ) 0, 0 from π the to θ sine, For 66 increases Properties of the of from itself. tan graphs from increases cos ) (i.e. 2 rad 1 cos has no breaks) and has a pattern that Section For f( ) tan , the range ) tan ( is is undeﬁned when has a pattern that geometry and vectors tan ... repeats 3 ___ __ , , 2 and Trigonometry, unlimited __ tan 2 , 2 … 2 rad. every Example 5 ___ Find exact values for: 5 ___ sin (a) (a) From the sin From (b) sin 3 3 √ 3 ___ __ the sketch, 5 ___ 2 the tan 3 __ 2 tan 3 √ 3 3 f(θ) f(θ) 1 4 0.5 2 π π 3 3 π π sketch, 5 ___ 1 __ cos 3 f(θ) From (c) __ cos 3 tan (c) 3 sketch, 5 ___ 5 ___ cos (b) 3 π 3π θ π π π θ 2π 3 2 2 2 2 1 4 π 3π θ π 2 2 0.5 1 π π 3 2 2 π π O π 3 2 3 π π O π 2 Example Sketch the maximum graphs value of of the the following functions for values of in __ (a) f() sin 2 (b) the range 0 2 . In each case give the function. f() 3 cos 2 f() (c) __ ( 2 sin ) 2 2 __ (a) Comparing y sin 2 with (b) Start with y cos , then (c) Start with y sin (this is sin 2 y f(ax) shows y sin 2 that reﬂect the curve in the stretched is compressed by horizontal axis, stretch to a factor of 2 horizontal sin 2 parallel axis. goes to the curve Therefore through two to cycles by the a factor vertical translate the of axis 3 the every one cycle sin of up the vertical by translate by direction of 2 of axis) 2 parallel then in the the positive horizontal axis, units lastly stretch by a factor to the vertical 1 sin 2θ f(θ) 1 2 3 f(θ) cos θ 2π 4π 5π 3 3 3 θ π ( 1 θ π) 2 1 2 3 2 sin 2 0.5 π 2 4 O 2 axis. f(θ) f(θ) f(θ) of axis. parallel f(θ) factor then and for a horizontal parallel and curve by this O O π π 3 0.5 3 π 2π 3 3 π 4π 5π 3 3 2π θ π θ 3 1 2 1 2 4 The maximum value is 1. The maximum value is The 5. maximum value is 2. Exercise 2.1 7 ___ 5 ___ 1 Find exact values for: (a) cos ( ) (b) tan ( 2 Sketch the graphs of the following 9 ___ ) 4 sin (c) ( ) 4 functions for 2 values of in the range 0 2 __ (a) f() tan 2 (b) f() 2 cos (c) f() sin ( ) 2 3 W rite down the maximum and minimum values of each function. _ 1 (a) 5 cos (2 ) (b) 5 cos 2 (c) 2 sin 2 67 2.2 Reciprocal Learning outcomes To deﬁne and use trig functions The the reciprocal trigonometric functions reciprocal The reciprocals of the three main trigonometric (trig) functions have their trigonometric functions own names: 1 _____ 1 _____ cosec You need to know where cosec secant The the properties sine, and cosine graphs and and is is sec abbreviation the of abbreviation tan cosecant, for cot cos the cot 1 _____ sin sec is the abbreviation for cotangent. of tangent The cosecant function functions The graph of f( ) cosec is given below. f(θ) 1 2π π θ 3 f(θ) sin θ 1 The graph the shows cosec multiple that: function of (this is not is to continuous; be expected it is undeﬁned because these when are the is any values of 1 __ where sin 0 and is undeﬁned) 0 the and The The cosec function takes all real values except for similar to values between 1 1. secant function graph of f( ) sec , shown below, is the graph for cosec θ) f(θ) cos θ 1 O 2π 2π θ 3π 1 The properties function: 68 of the secant function are similar to those of the cosecant Section 2 Trigonometry, geometry and vectors __ it is not it takes continuous; it is undeﬁned when is any odd multiple of 2 The The all real values except for values between 1 and 1. cotangent function graph of f( ) cot is given below. f(θ) f(θ) tan θ 1 2π The properties tangent of π the cotangent function θ π are similar to those of the function: it is undeﬁned it takes From the curve for all real graph when is any multiple of values. we can see that the curve for cot is the reﬂection of the __ tan in the x-axis, and it is translated by in the direction of 2 the negative x-axis. __ cot Therefore tan ( __ ) ) tan ( 2 2 cos Example θ 1 Find, in radians, the smallest value of for which sec 2 1 2 1 _____ cos 1 __ sec 2 O π 2π 3 3 the sketch, the required value of 4π 5π 3 3 π 2 ___ From 2π θ is 1 3 2 1 cos θ Example 7 ___ Find the exact value of 1 cosec 6 1 2 7 ___ cosec 1 ______ 7 6 7 __ 6 sin O 6 π 7 ___ From the sketch, sin 6 θ 1 1 __ 2 2 7 ___ cosec 2 1 6 Exercise 2.2 7 ___ __ 1 Find the exact value of: (a) cot 2 Find, in radians, the smallest value 5 ___ sec (b) 4 (c) cosec 4 of x for 3 which: 2 ___ (a) cot x 1 (b) sec x __ (c) cosec x 1 √ 3 69 2.3 Pythagorean Learning outcomes The of To derive and Pythagorean use identities relationship any between the sine, cosine and tangent angle the y identities P(x, y) You need to know x O The deﬁnitions of and tangent functions The deﬁnitions of the the sine, cosine reciprocal trig functions y Pythagoras’ For any , angle we know that sin , cos , OP How to solve a quadratic y sin _____ But equation cos __ tan OP x y x ____ ____ y x ____ ____ theorem __ OP OP x sin _____ i.e. for all values , of tan [1] cos Using the reciprocal functions, this identity can be written as cos _____ cot sin The Pythagorean identities y x O P(x, y) 2 For any angle , a right-angled triangle y 2 x ____ 2 Dividing by OP gives ( can be ( ) 2 cos 2 y 2 OP 2 sin identities 1 [2] 2 (cos these x 1 2 use which OP i.e. can in ____ ) OP We drawn 2 to means (cos ) produce , etc.) further identities. 2 2 Dividing 2 cos sin ______ 2 sin 1 by cos gives 1 1 ______ 2 cos 2 sin _____ then using tan cos 2 gives 1 tan 2 sec [3] cos 2 2 Dividing cos 2 sin cos ______ 2 1 by sin gives 1 _____ 1 2 sin 2 cos _____ then using cot sin 2 gives cot 2 1 cosec sin These 70 identities can also be used to solve some trig equations. [4] Section 2 Trigonometry, geometry and vectors Example 2 Solve the equation 2 cos x sin x 2 Using [2] gives 1 for 2 cos x 1 sin 0 x 2 2 x so 2 cos x sin x 1 becomes 2 2(1 sin x) sin x 1 f(x) 2 ⇒ 2 sin x This a is sin x 1 quadratic (2 sin x 0 1 equation 1)(sin x 1 in sin x 0 x O π 1 ⇒ sin x or sin x 1 2 1 5 ___ __ ⇒ x , , 6 These identities 3 ___ 6 can 2 be used to prove the validity of some other identities. Example Prove This that (1 identity cos A)(1 has yet to be sec A) proved working. Work separately on the LHS cos A)(1 (1 so do sin A tan A not left-hand sec A) use the side complete (LHS) identity and the 1 cos A sec A 1 cos A sec A in your right-hand side (RHS). cos A sec A cos A _____ cos A sec A cos A 2 2 1 cos A __________ sin A ______ (1 These cos A)(1 identities can sec A) also be sin A tan A sin A tan A to sin A cos A used sin A _____ cos A eliminate cos A RHS trig ratios from a set of equations. Example Eliminate from the equations x 3 sin and y 2 sec x __ x 3 sin sin ⇒ 2 __ and y 2 sec ⇒ cos 3 2 x __ ( ) 2 __ ( 3 y y 2 2 ) 1 Using cos 2 sin 1 Exercise 2.3 2 1 Solve 2 Simplify the equation sin x _____ sec 2 x tan x 3 for 0 x 2 3cos x ______ cos x 3 Eliminate 4 Prove the from sin x the identity equations tan A cot A x 4 tan and y 2 cos sec A cosec A 71 2.4 Compound Learning outcomes angle formulae The identity T rigonometric To derive sin (A tan (A and B), use cos (A B) You need to know Pythagoras’ The are not cos A cos B sin A sin B distributive, i.e. cos (A We can B) is NOT equal to cos A cos B and below, The functions identities for B) cos (A B) derive which We ﬁnd the two the the correct shows length a of identity circle PQ of for radius using two cos ( A 1 B) unit using centre different the diagram O. methods and then equate results. cosine formula y theorem P properties of the trigonometric functions Q S A B N Using the cosine 2 PQ From formula 2 1 2 the in OPQ x M O gives 2 1 2 cos (A 2 cos (A B) B) diagram OM cos B and ON cos A so QS ( QM sin B and PN sin A so PS (sin A Using Pythagoras’ theorem in 2 PQ PQS ( cos A cos A 2 2(cos A cos B (sin A cos sin B) sin B) 2 cos B) 2 cos B) 2 gives 2 cos A 2 B 2 cos A cos B sin 2 A sin B 2 sin A sin B sin A sin B) 2 Equating the two 2 ⇒ The angles identity 72 is expressions 2cos (A cos (A A and true B) B for B) can all for PQ 2 2(cos A cos B cos A cos B be any angles. size gives sin A sin B) sin A sin B and the proof is similar . Therefore the Section Compound The we identity can use Replacing derived it B angle to by B in Trigonometry, geometry and vectors identities above derive 2 is one of the compound angle identities and others. [1], cos ( A B) cos A cos B cos ( A B) cos A cos (B) cos A cos B sin A sin B [1] sin A sin (B) sin A sin B [2] __ Replacing A by A in [1] gives 2 __ cos __ ( (A B) ) cos ⇒ Replacing B by B in [3], [3] cos B sin B) sin A cos B sin ( A B) sin A cos (B) B) gives ( A ) sin B 2 sin (A by ) sin ( A sin A cos B cos A sin B [3] sin A cos (B) cos A sin B [4] sin A cos B cos A sin B ______________________ __________ [1] A 2 Dividing __ ( 2 cos (A B) cos A cos B sin A cos B __________ sin A sin B cos A sin B __________ cos A cos B cos A cos B ______________________ tan ( A B) cos A cos B __________ sin A sin B __________ cos A cos B cos A cos B tan A tan B ______________ ⇒ [5] 1 Replacing B by B in [5] tan A tan B gives tan A tan B ______________ tan ( A B) [6] 1 Y ou need to lear n these recognise Collecting the identities together identities either and be tan A tan B able to side. gives sin ( A B) sin A cos B cos A sin B sin ( A B) sin A cos B cos A sin B cos ( A B) cos A cos B sin A sin B cos ( A B) cos A cos B sin A sin B tan ( A B) tan A tan B _______________ 1 tan A tan B tan A tan B _______________ tan ( A B) 1 tan A tan B 73 Section 2 Trigonometry, geometry and vectors Using These compound identities angle can be identities used to prove further identities. Example Prove that LHS cos ) sin A cos 2 sin A cos ) sin (A cos A sin 2 sin A sin A cos cos A sin 1 Therefore These sin ( A LHS identities 2 sin A can be used to RHS solve equations. Exam tip The range is given in radians so the Example __ answer must be given in radians. Solve the equation cos sin ( for ) values of in the range 3 0 __ cos sin ( ) 3 __ sin __ cos cos sin 3 3 √ 3 ___ 1 sin cos 2 2 √ 3 ___ (1 1 ) cos ⇒ (2 ⇒ tan 2 √ 3) cos 2 identities can sin √ 3 1.31 rad The sin 2 (3 be s.f.) used to ﬁnd exact values of some trig ratios. Example Find the sin 15° exact value sin (45° of sin 15°. 30°) sin 45° cos 30° √ 3 ___ 1 ___ √ 2 2 √ 3 ____ 1 ____ √ 2 2 √ 2 √ √ 6 2 ________ 4 They can be used to simplify expressions. Example Simplify This is cos 74 the cos RHS cos 2 cos 2 of [2] sin with A sin sin 2 sin 2 and B 2 cos ( cos ( cos 1 __ 2 √ 2 cos 45° sin 30° 1 ___ 2 ) ) 2 Section These identities can also be used to eliminate an angle from two 2 Trigonometry, geometry and vectors equations. Example __ from Eliminate the equations x sin ( ) and y cos 4 __ x sin ( ) 4 __ sin __ cos sin cos 4 1 ___ cos √ √ 1 ___ 1 ___ 2 2 sin √ y √ 2 x( 4 1 ___ sin 2 √ 2 y) sin 2 Then, using (x 2 2y gives 2 y) 2 2x 1 cos 2 √ 2 ⇒ 2 sin 2xy y √ 2 1 1 Exercise 2.4 1 2 Find the exact value (a) sin 75° (b) sin 50° cos 40° Prove of: cos 50° sin 40° that: cot A cot B 1 ______________ (a) cot (A B) cot A (b) 3 sin (60° ) cot B sin (120° ) Simplify: (a) sin cos 3 cos sin 3 tan P tan 3P _______________ (b) 1 4 Solve tan P tan 3P the equations for 0 __ (a) cos ( ) sin sin 3 __ (b) cos ( ) 4 5 Eliminate from x the equations __ 4 tan ( ) and y 2 tan 4 3 _ 6 Given that sin , ﬁnd the exact value of: 5 __ (a) sin ( ) 4 __ (b) tan ( ) 4 75 2.5 Double Learning outcomes angle Double The To derive and use ratios of angle compound identities angle identities are true for any two angles, A and B, so identities for they trig identities multiple can be used for two equal angles, i.e. when B A angles Replacing B by A in the compound sin 2A angle formulae gives 2 sin A cos A You need to know 2 The compound angle cos 2A tan 2A 2 cos A sin A identities 2tan A __________ 2 1 The identity for cos 2 A 2 cos can be 2 A sin A sin tan expressed A in 2 A cos A (1 other forms 2 A (1 because 2 cos A) 2 cos A 1 and 2 cos 2 2 2 sin A) 2 sin Therefore A 2 cos 1 2 sin A 2 A sin A 2 cos 2A 2cos { A 1 } 2 1 The last two identities above can be 2sin A rearranged as 1 _ 2 cos A (1 cos 2A) (1 cos 2A) 2 1 _ 2 sin A 2 Y ou also need alter native some As with The of the the to lear n identities most the useful previous following all double involving for they illustrate can some identities The simplifying identities, examples angle cos 2 A. in trigonometric be of including identities used their in a this all set the are expressions. variety of problems. uses. Example 3 Given that is an acute angle and that sin , ﬁnd the value of : 5 (a) cos 3 (b) tan 2 3 Given is acute sin and (a) cos 3 cos (2 ) 3 4 , then cos 5 and cos 2 cos 2 (2 cos sin 2 [2( 1) cos 16 ___ 2 sin 1 ] 2 5 2 2 tan _________ tan 9 __ 1 16 76 24 ___ 4 ______ 2 1 sin compound cos 4 __ 25 3 tan 2 9 ___ 4 __ ) 25 (b) a 2 7 44 ____ 5 4 Using tan 5 125 angle identity Section 2 Trigonometry, geometry and vectors Example Solve the equation cos 2 x cos 2 x 3 sin x 3 sin x 2 for Using the 0 x 2 2 2 ⇒ (1 2 sin ⇒ 2 sin ⇒ (2 sin x x) 3 sin x so 2 that terms identity the in cos 2x equation 1 contains 2 sin x only sin x 2 x 3 sin x 1 1)(sin x 1) sin x or 1, so x 0 5 ___ __ 1 ⇒ 0 , 2 , 6 6 Example 3 Prove LHS the identity sin 3x sin 3 x 3 sin x sin (2x sin 2x cos x 2 sin x 4 sin x x) cos 2x sin x 2 cos x cos x (1 2 sin x) sin x 2 2 sin x(1 sin 3 x) sin x 2 sin x 3 3 sin x 4 sin x RHS 3 Therefore sin 3 x 3 sin x The Note as that well as we a identity used a 4 sin for sin 3 x compound double angle the x is angle identity in worth remembering. identity this and a Pythagorean identity proof. Example Eliminate from equations x 1 _____ y cos 2 and y cosec 2 and x 1 2 sin sin 2 __ x 1 2 ⇒ y (1 x) 2 2 y Exercise 2.5 12 ___ 1 Given that cos and that is acute, ﬁnd the exact value of: 13 (a) sin 2 (b) cos __ __ (Hint: use a double angle identity with A 2 ) 2 2 1 tan A __________ 2 Prove the identity cos 2 A 2 1 3 Solve 4 Eliminate the equation t from tan the tan x tan 2x 2 equations x A for 0 cos 2t x and y sec 4t 77 2.6 Factor formulae Learning outcomes The factor formulae The To derive and use identities convert formulae in this set are called the factor formulae because they the factor expressions such as sin A sin B into a product. identities Starting with the sin A cos B compound angle formulae cos A sin B sin (A cos A sin B sin (A B) You need to know sin A cos B The properties of the B) trig adding gives 2 sin A cos B sin (A B) sin (A B) [1] 2 cos A sin B sin (A B) sin (A B) [2] functions The compound angle subtracting gives Now, the identities using other cos A cos B cos A cos B adding cos (A sin A sin B cos (A 2 cos A cos B gives right-hand following compound sin A sin B gives subtracting The two 2 sin A sin B side of each of angle formulae B) B) cos (A these cos (A B) B) formulae cos (A can cos (A be B) [3] B) simpliﬁed [4] by the substitutions. 1 P A B ⎫ A ⎧ (P Q) 2 ⇒ ⎬ ⎨ ⎩ 1 Q A B B (P Q) 2 1 _ Then sin P sin P sin Q sin Q 2 sin 1 _ (P Q) cos 2 2 1 _ 1 _ 2 cos (P Q) sin 2 cos Q Q) [5] (P Q) [6] 2 1 _ cos P (P 2 cos 1 _ (P Q) cos 2 (P 1 _ cos P cos Q 2 sin For may ﬁnd it easier to remember Q) [7] 1 _ (P Q) sin 2 Y ou 2 (P Q) [8] 2 the last group using words. example, sum The ﬁrst of sines group, twice [1]–[4], is sine used (half sum) when we cosine want to (half express difference) a product as a Exam tip sum Identities [4] and [8] involve a or difference. For example, using [1] 1 minus sin 5x cos 3x (sin 4x sin x) 2 sign, so take care when using these. The second difference sin 5x As well as identities, calculus 78 group, as a sin 3x being the [5]–[8], product. problems used when example, we using want to express a sum or [5] 2 sin 4x cos x useful factor is For for solving formulae which we are will some trig equations particularly consider in useful Section and for 3. proving certain some types of Section 2 Trigonometry, geometry and vectors Example cos 2x cos 2y ______________ Prove that sin 2x cos 2x 2sin (x cos 2y ______________ LHS tan(y x) sin 2y y) sin (x y) ______________________ sin 2x Using sin 2y 2sin (x y) cos (x sin (x [8] and [5] y) y) ___________ cos (x y) sin (y x) Using __________ sin A sin ( A) cos 2x cos (y x) tan (y x) and cos A cos ( A) RHS cos 2y ______________ sin 2x tan(y x) sin 2y Example Solve the equation cos 2x ⇒ 2 cos 3x ⇒ cos 3x either cos 4x cos x cos 2 x 0 or cos 4x Using cos x i.e. in 3 of x times ⇒ 3x between the 0 3 ___ , x , x __ x This ax The , , 2 6 example cos ax of 5 ___ __ , 6 in b. a T o is true 2 3x between 0 and 6, 6 3 ___ , 11 ____ , 6 2 6 3 ___ or 2 6 a special values that of 11 ____ 2 includes ﬁnd 11 ____ , 2 , 6 times same 3 ___ , 7 ___ , 2 2 2 9 ___ , 2 need values 5 ___ __ __ ⇒ , 2 we , 6 7 ___ x x __ cos x 0 7 ___ 5 ___ , 2 2, and range for ⇒ or 0 0 2 values for [7] __ For 0 0 0 cos 3 x 0 of x case in a of an given equation range, we of the need form to ﬁnd values range. for equations of the form sin ax b and tan ax b Exercise 2.6 __ 1 Simplify sin x ( __ ) sin x ( 3 2 Prove 3 Solve the the identity ) 3 sin equation A sin 3 x sin 2A sin 4x sin 3A sin 2x sin 2A (1 for 0 x 2 cos A) 79 2.7 The expression Learning outcomes The The To express a cos b sin You need to know b sin can be reduced to a single term such as ) can ﬁnd i.e. of values angle for r and by expanding r cos ( ) using a identity, r cos ( if r (cos then properties b sin ) compound The a cos expression sin b as We a cos expression r cos ( r sin/cos ( cos a cos sin ) sin ) a cos b sin a cos b sin the Comparing the coefﬁcients of cos sin and gives trigonometric functions r cos The compound a [1] r sin and b [2] angle formulae b __ [2] [1] tan ⇒ a 2 and [1] 2 2 [2] ⇒ r 2 2 sin (cos 2 ) a 2 b _______ r ⇒ However , always For it use is not the example, expansion of r (cos Comparing the sensible to express r cos ( cos 2 a method to 2 √ 4 cos sin to Using rely work ), 2 b sin ) of the the sin formulae for 1 r and . Y ou should result. 3 sin which coefﬁcients on out 2 cos as r cos ( ), start with the gives 4 cos cos 3 sin and sin gives r cos 4 and r sin 3 r sin 4 _______ 3 Then tan and r 2 √ 2 3 4 5 4 3 4 cos Therefore 3 sin 5 cos ( ) where tan 4 We can This also time express start r (sin Comparing with cos the the cos 4 cos 3 sin expansion sin ) coefﬁcients of as of r sin ( r sin ( 4 cos cos ) ), which gives 3 sin and sin gives r cos 3 and _______ 4 Then tan and r 2 √ 4 2 3 5 3 4 Therefore 4 cos 3 sin 5 sin ( ) where tan 3 Exercise 2.7a 1 Express 2 Express Finding a cos ﬁnd a cos the 80 12 sin the the form: (b) 24 sin in the ) and r sin ( ) form: (b) maximum in ) 7 cos r cos ( (a) T o 5 cos r cos ( (a) r sin ( ) minimum values of b sin maximum b sin maximum we and and express minimum it as minimum a values single values. sine of or an expression cosine. We of can the then form ‘read’ Section For example, 3 sin x to ﬁnd 2 cos x, the we maximum can express r (sin x cos Comparing coefﬁcients r of 2 √ sin x and minimum x values 2 cos x as 3 sin x cos x Trigonometry, geometry and vectors of ): r sin (x 2 cos x gives cos r 3 and r sin 2 ___ 2 2 2 3 sin cos x sin ) ______ Therefore and 2 3 √ 13 tan and 3 ___ So 3 sin x The 2 cos x maximum √ value 13 of ) sin (x the sine of We an do angle not is 1 need and to evaluate the minimum value ___ is 1, therefore the maximum value of 3 sin x 2 cos x √ is 13 and the ___ minimum T o ﬁnd occur , value the we value do √ is 13 . of need x to at which the maximum and minimum values : evaluate 2 tan ⇒ 0.588 rad Make sure your calculator is in radian mode 3 __ ) sin (x is maximum when ) (x and minimum when 2 3 ___ ) (x , i.e. when x 2 The equations For we a cos of the example, can a cos to express b sin solve cos x cos r as x √ 3 b sin a single x 0.588 rad √ 3 sin x sin x as c or cosine makes solving straightforward. 1, for 0 x 2 , ), r cos (x √ cos x c sine b sin r sin x sin 3 sin x __ 2 ⇒ and 2 a cos form r cos x cos i.e. 0.588 rad 2 equation Expressing 3 ___ __ 4 tan and √ 3 so 3 __ cos x √ 3 sin x 1 becomes 2 cos x ( ) 1 3 __ ⇒ cos x ( 1 __ ) 3 2 f(x) 1 2 π O π 2π 4π 5π 3 3 π 3 x 2π 3 1 2 1 __ T o ﬁnd values of x in the range 0 to 2 we need to ﬁnd values of x 3 __ in the range __ to 2 , 3 __ cos x ( 3 __ 1 __ ) ⇒ 3 x 2 __ 3 5 ___ __ , 3 , 3 3 2 ___ x 0, , 2 3 Exercise 2.7b 1 Express Hence and 3 cos x ﬁnd the the values 4 sin x in maximum of x at the and which form r cos (x minimum they occur ). values in the of 3 cos range 0 x x 4 sin x 2 __ 2 Solve the equation cos x sin x √ 2 for 0 x 2 81 2.8 Trigonometric Learning outcomes The In To prove To solve trigonometric important previous in a given equations identities we have introduced some trigonometric at a time. equations We range pages and identities group trigonometric identities now collect them together: sin _____ tan cos 2 cos 2 sin 1 sec 1 cosec You need to know 2 1 The properties of the tan 2 2 cot trigonometric 2 ratios sin (A sin (A cos (A cos (A B) sin A cos B B) sin A cos B cos A sin B B) cos A cos B sin A sin B B) cos A cos B cos A sin B sin A sin B tan A tan B ______________ tan(A B) 1 tan A tan B tan A tan B ______________ tan(A B) 1 sin 2A tan 2A tan A tan B 2 sin A cos A 2 tan A __________ 2 1 tan A 2 cos 2 A sin A 2 cos 2A { 2 cos A } 1 2 1 A 1 2 cos 2 sin A (1 cos 2A) 2 1 2 sin A (1 cos 2A) 2 2 sin A cos B sin (A B) 2 cos A sin B sin (A B) 2 cos A cos B cos (A B) 2 sin A sin B cos (A B) sin (A B) sin (A B) cos (A B) cos (A B) 1 sin P sin Q 2 sin 1 (P Q) cos 2 sin P sin Q 2 cos (P Q) sin cos Q 2 cos (P Q) 2 1 Q) 1 2 cos P (P 2 1 1 (P Q) cos 2 (P 1 cos P cos Q 2 sin 1 (P Q) sin 2 82 Q) 2 (P 2 Q) identities one Section Proving trigonometric T o prove The an identity, examples with a mixed and speciﬁc set identities. work with convert when it is the in to the identities. Some In work on previous Exercise guidelines more Trigonometry, geometry and vectors identities sensible exercises of 2 on complicated one side pages 2.8a where to at have you start a all are time. been asked associated to prove are: side Exam tip all ratios to sine and cosine Remember a multiple angle is involved, start with that and break it guidelines, to ratios of a single angle using the compound angle formulae multiples multiples use the vice The factor sin 3 A), and the double angle formulae for these points practice will are only help strategies that work for you you. even sin 2 A) formulae to express a product of ratios as a sum, or versa. following both (e.g. (e.g. and for develop odd that down sides, example but illustrates that it may be necessary to work on separately. Example Prove that There the are multiple single RHS: sin 3 A angle A (sin 2A (sin 2A angles and sin A)(1 on both sides, 2 cos A) so we will start with the RHS and express that in terms of ratios of simplify. sin A)(1 2 cos A) (2 sin A cos A sin A(2 cos A sin A(4 cos sin A)(1 1)(1 2 cos A) 2 cos A) 2 We now LHS: turn sin 3A to the LHS and sin 2A cos A 2 sin A cos express that in A terms 1) of ratios of the This form for sin A (2 cos 2 sin A (4 cos A cos 2A is chosen because we have the factor 2 A A 1) sin A sin A (2 cos angle cos 2A sin A 2 single and we want the other factor to involve cos A 2 A 2 cos A 1) 2 Therefore sin 3 A A (sin 2A 1) RHS sin A)(1 2 cos A) Exercise 2.8a Prove the following identities. 1 cos A _________ 2 1 (cot A cosec A) 1 2 cot 2A cosec 2A cosec ______________ 2 8 sec cosec cos A sin cot A cot A cot B 1 ______________ 9 3 1 cot(A B) sin A _________ cos A _________ cot A 1 tan A sin A sin( ) 10 2 ___ sin( 3 sin 4 ) sin 5 sin 6 3 3 1 cos 2A __________ 5 cot B sin 4 ______ sin 3 sin 5 ______________ __ 4 cos A cot A 11 cos 3A 12 tan(A 4 cos A 3 cos A tan A sin 2A sin B _______________ cos 2A cos 2B _______________ 6 cos 2B cot(A B)cot(A B) tan A cos A cos(A cos x cos y B) _ 1 ____________ cos 4x cos 2A 4 7 B) 8 cos 13 2 x 8 cos x 1 tan (x y) 2 sin x sin y 83 Section 2 Trigonometry, geometry and vectors Solving trigonometric Any of a cos x the trig identities, b sin x equations together r sin/cos (x with ), the can be in a given range transformation used to help solve a trig equation. There are whether an two the identity. equation The tables together equation If to general a this form that with exhaustive, Equation an will not that follow nor Equations is approaches factorise, possible involves list most appropriate are they to solving either (and only of the method it containing one its often one trig for given is try of them. or to one categories solving First form not), ratio common see by using reduce the angle. of The trig lists equations, are not angle only Method 1 a cos x b sin x 0 Divide 2 a cos x b sin x c T ransform 3 a cos 2 in equations. infallible. category x trig b sin x Use c the by the cos x, the provided LHS to Pythagorean LHS in terms of that cos x identities ratio 0 ) r cos (x one to express only 2 a sin x b cos x c b sec x c 2 a tan 4 x a cos x a sin x Equations Equation 1 a cos x b tan x b tan x Multiply 0 multiples of one category 2 cos ax b cos 2x Use c b cos 2x The methods equation, may of a be nor able trig Practice that cos x angle double angle formulae containing trig to ratios reduce of to only Solve for then ax in a divide times by x the required range a c listed do to do they be not give always simpliﬁed Sometimes the lead only to quickly you the way when may of solving quickest part need to of a particular solution. it is classify An equation recognised each side as of help you recognise the best way to tackle any part an independently. will 0 c c identity. equation 84 the equation and sin ax provided Method an a sin x cos x, 0 containing by equation. Section 2 Trigonometry, geometry and vectors Example Solve the sin 2 cos equation cos 2 sin tan 3 for values of in __ the range 0 2 sin 2 cos cos 2 sin ⇒ sin 3 ⇒ sin 3 ⇒ sin 3 ⇒ sin 3 (1 ⇒ sin 3 sin 3 The LHS is the expansion of sin (2 ) tan 3 sin 3 ______ cos 3 sin 3 sec 3 sec 3 ) 0 sec 3 or 0 0 1 __ For values of in the range 0 , we need to ﬁnd values of 3 2 3 ___ in the range 0 , 2 sin 3 0 3 ⇒ 0 or and sec 3 1 ⇒ 3 0 __ 0, 3 Example Solve the cos 4 Using This cos 4 equation cos 2 the factor cos 3 formula cos 3 gives ⇒ 2 cos 3 cos ⇒ cos 3 (2 cos ⇒ cos 3 as a on factor cos 3 cos 2 1) 0 cos 0 for 0 0 the of cos 3 ﬁrst the two terms gives a factor cos 3 LHS. 0 1 0 or 2 3 ___ __ 3 , 5 ___ __ , 2 so 2 2 __ , 6 5 ___ , 2 6 2 ___ or 3 __ i.e. __ , 2 5 ___ 2 ___ , 6 , 3 6 Exercise 2.8b Give answers Solve range the 0 that are following not exact equations correct for values to of three in signiﬁcant the ﬁgures. Solve range the 0 following equations for values of in the 2 2 1 cos 2 sin 2 sin sin 2 cos 1 7 2 sec 8 4 sin 2 tan 2 cos 2 2 cos 1 5 sin 12 cos 9 13 sin 3 5 cos 6 sin 5 sin 2 0 2 4 cos cos 10 tan 2 11 sin 3 12 4 sin 1 1 _________ 1 4 3 2 1 _________ 3 1 cot 2 4 cos 0 2 2 sin sin 2 cos 3 0 sin cos cos 3 sec 85 2.9 General Learning outcomes To ﬁnd the general solution solution General trig equations solutions of trigonometric equations of The trig of equation cos x 1 has a ﬁnite number of solutions in a given range equations of values All of x possible has one values solution of in cos this x occur range, in the namely range x x and cos x 1 0 You need to know However , In The properties of fact cos x cos x 1 1 for for an every inﬁnite number multiple (both of values positive of and x. negative) of 2 the trigonometric functions cos The main x trigonometric identities 1 6π 4π O 2π 2π 4π 6π x 1 Therefore the x when solutions 2n This The is for of n called the range 1 of values can be of x is written unrestricted, as the general Within the cos x general solution solution of range x cos x , of the the c, equation. |c| equation 1 cos x c has, in general, two solutions. cos x 1 c 6π 4π O 2π 2π 4π 6π x 1 If one The 2 . solution graph 86 f( x) Therefore multiples We of can of is we 2 write x , then to this cos x can get and has the to solution as the a other pattern general solution that is x repeats solution of every the interval equation by x 2n , where n is an of adding integer . Section Therefore the general x where is a solution 2n solution of in the for the n equation cos x c 2 Trigonometry, geometry and vectors is range x Example 1 __ Find the general solution of the equation cos x 2 1 __ cos x 2 __ ⇒ x in the range x 3 Therefore the general solution is __ x 2n 3 The All general possible the solution of values equation sin x of sin c x has sin x occur two in c, |c| the range solutions sin in 1 0 this x 2 and, in general, range. x 1 c π O π π x π 1 If the The smaller graph Therefore of 2 We is to can an of solution f( x) we can and write sin x get the to this is x , then has a the pattern general other that solution solution repeats of sin is every x c by interval adding of 2 multiples solution as x 2n and x 2n where n integer . Therefore x where the 2n is general the solution and x smallest of (2n the solution equation 1) in the for range sin n 0 x c is 2 x Example √ 2 ___ Find the general solution of the equation sin x 2 √ 2 ___ sin x 2 __ ⇒ x as the smallest solution for values of x in the range 4 0 x 2 __ Therefore the general solution is x 2n __ or 4 x (2n 1) 4 87 Section 2 Trigonometry, geometry and vectors The solution of tan x general c __ All possible values of tan x occur in the range __ x and 2 equation tan x c has one solution tan in this the 2 range. x 10 5 3π 2π π π O 2π 3π x 5 10 The graph of f( x) tan x has a pattern that repeats every interval __ if x is the solution of tan x c in the range x 2 solution x can be written n , n Therefore , the , so general 2 as the general x solution of n , the n equation tan x is the solution in the c is __ where of __ range __ x 2 2 Example Find the general solution of the equation tan x 1 __ tan x 1 ⇒ x __ for values of x in the range __ 4 x 2 2 __ Therefore the general solution is x n 4 When a multiple multiple angle. angle Then is use involved, that to ﬁrst ﬁnd ﬁnd the the general general solution solution for the for angle. Example 1 __ Find the general solution of the equation sin 2 x 2 1 __ When sin 2x , the solution for 2x in 2 5 ___ __ 2x and 2x 6 6 Therefore the general solution for __ 2x 2n and 6 5 ___ 2x 2n 6 giving the general solution x 12 88 for x as 5 ___ ___ n and x 12 n 2x is the range 0 2x 2 the single is Section 2 Trigonometry, geometry and vectors Example Find the general solution of the sin x sin x ⇒ ∴ 2 sin x cos 2x sin x (1 Either 2 cos 2x) sin x Then 0 0 0 ⇒ the x general equation 2 sin x cos 2x 0, , solution cos 2 x in the 0 interval 0 x 2 x n is 2 ___ 1 __ Or 2 or ⇒ 2x in 2 the interval x 3 __ 2 ___ Then the general solution is 2x 2n ⇒ x n 3 3 __ i.e. x n or x n 3 Example Find the general solution of the equation x __ x __ cos sin 2 x __ x __ cos ) where r √ 2 and tan 1 2 sin 2 1 2 1 ___ __ x __ cos ( x __ cos ⇒ r cos 2 x __ 1 x __ sin 2 2 ( 2 ) 4 √ 2 For values of x in the range need values of ( 2 __ x __ ( __ ) 2 4 ⇒ x Therefore x , __ ) from 4 __ 2 __ to 4 __ 2 4 __ or 4 __ x __ we or the 4 0 general solution is is x n Exercise 2.9 Find the general solutions of the equations. _ 1 1 sin x 1 5 cos 3x 2 2 2 cos x 1 6 sin 3 tan x 1 7 tan 2x 8 sin 3x x sin x 0 __ √ 3 __ 4 √ 2 cos x 1 sin x 0 89 2. 10 Coordinate Learning outcomes geometry The gradient of a and straight straight lines line y The To revise the gradient of gradient of a straight line is given a A(x by straight ﬁnding the increase in y divided by , y 1 ) 1 line the increase in x when moving from y To ﬁnd the angle between two B(x one straight point on the line to another point , y 2 the line. x x 1 To ﬁnd from a 2 lines on y 1 ) 2 the distance straight of a 2 point line x O You need to know y y 1 2 _______ In the diagram, the gradient of the line is . x The Pythagorean The compound The exterior of This is also the value x 1 identities 2 tan angle formulae Therefore angle property of the gradient of a straight line is equal to the tangent of the a angle that the line makes with the positive direction of the x-axis. triangle When two Therefore lines if are the perpendicular , equation of a the product of their is y mx c, form y line L gradients any line is 1. perpendicular 1 ___ to L will have an equation of the x k m The The angle lines L between two and L 1 and m have lines gradients y m 2 L 1 respectively, 1 where 2 L 2 m tan 1 and m 1 The angle tan 2 2 between the lines is 1 tan Therefore tan( 1 α 2 ) 2 θ tan tan 1 2 1 2 _______________ x O 1 tan tan 1 Therefore the angle, , 2 between lines with gradients m and 1 m is given by tan 1 m m 1 2 Example L is the line y 3x 5 and L 1 is m 3 the angle and m 1 between 2, so ( 2) the line 2x y 1 0. the lines? angle between the lines is given 1 is the 3 ___ 4 2) obtuse __ 4 (3)( 3 ___ 5 __ ___________ tan This the 2 3 90 is 2 What angle ⇒ 5 between 4 the lines; the acute angle is m 2 m 1 2 __________ by Section The distance of The In is distance the of a diagram, the length point from point the of a the from distance line a line of the is a Trigonometry, geometry and vectors line the point 2 perpendicular A( a, b) from distance. the line y mx c y AN. A(a, b) B((b B is a point on the line such that AB is parallel to the x-axis, so b is c)/m, θ the d y-coordinate of B. N b c ______ Therefore from the equation of the line the x-coordinate of B is m b c ______ This gives the length of AB as b c ma ___________ | a | | θ | m m x O AN d AB sin y m ________ tan But m, 2 _______ sin therefore Using 1 cot mx c 2 cosec 2 √ m 1 b c ma ___________ _______ Hence d | | 2 √ m Note the that right AB. The the also 1 diagram line, is point then we needed x-coordinates that sin ( But the the modulus between Note in of if is is on A and the the the write because of obtuse, ) sin , so A would the left a of the (b length line. c)/m of AB is If for A were the the on length of difference B. angle result is ABN valid when the line has a negative gradient. The distance of the point ( a, b) from the line y mx c b c ma ___________ _______ is given | by | 2 √ m 1 Example Find the distance of the point (2, 3) from the line 2 Rearranging the equation of the line as y x ⇒ a 2, b 3, m the distance and of 4 0 3 4 c 3 Therefore 3y 3 2 2x 4 3 (2, 3) from 2x 3y 4 0 is given by ___ 4 4 3 √ 17 13 ______ 3 3 ___________ ________ | 2 √( | 2 ) 1 13 3 Exercise 2.10 1 Find y 2 A is the 3 4 A acute 2 the line Find y the 3x point that the 3 Find 2x (2, 5) bisects distance between of 3y and 1 B is the lines whose equations are 0 the point ( 5, the angle AOB, where the point (1, from 5) 2). O the is Find the line the equation of origin. whose equation is x point point angle and P( a, (1, a b) is equidistant from the line 2y 5x 1 and from the 1). relationship between a and b 91 b) 2. 11 Loci and Learning outcomes the equation To deﬁne the meaning of loci context of the To ﬁnd a given a condition restricted the Cartesian equation of To ﬁnd on the possible positions of a point P , then P to a particular set of points. This set of points is called the of When locus P is the Cartesian equation the this point ( x, y), condition is the relationship called between x and y that the y of Cartesian a placed P . satisﬁes is xy-plane locus circle in is the a Loci When of equation of the locus of P . circle P(x, For example, condition on is the if A is the P( x, y) is point that (1, the 4) and gradient y y) the of AP 4 ______ You need to know 2, then gradient of AP is x How to ﬁnd the the Cartesian equation of the 1 A(1, locus 4) of distance x O y 4 ______ between two points P is x How to ﬁnd the equation of How 2 2x 1 a ⇒ straight y 2 line to form a perfect square Example A point the P(x, point Find the y) (1, is 2) the as it equation same is of distance from the the locus line of y from x 3. P . P(x, The x distance 3 The is 3 of P from the line P from the point y) x distance of (1, 2) is _________________ 2 √(x the x 2 1) (y 2) equation of the locus of P is given _________________ 2 by 3 x √(x 1) 2 ⇒ (3 ⇒ 9 ⇒ y x) (y 2) 2 (x 1) 2 6x 2 x 2 (y 2) 2 2 x 2x 1 of the y 4y 4 2 4y 4x 4 0 Exercise 2.11a Find the Cartesian following equation of 1 P is the same distance from the point 2 P is the same distance from the points 3 P is twice distance from the line the The Cartesian A circle point. 92 locus is P( x, y) when P satisﬁes the conditions. the locus equation of of points that a are y (0, 4) (1, and 2) 5 the and as it ( is line 2, x 6 4). from the point (2, circle a constant distance from a ﬁxed 0). Section If P(x, y) is any point on the circle of radius r and centre C( p, q), 2 then Trigonometry, geometry and vectors y _________________ 2 CP r and CP √(x 2 p) (y q) P(x, 2 (x 2 p) (y q) 2 r C(p, i.e. the Cartesian equation of a circle, centre 2 radius r is (x 2 p) (y ( p, q) the equation of 2 (x a circle, centre 2 2) (y ( 1)) (2, 2 q) 2 9 ⇒ 1) r and radius centre at the ( 1, circle 3) whose and its equation radius equation where f, g of and the c form are 2 2gx y can y 4x is (x 2y 4 2gx be (y 3) 2fy rearranged c 8 has its [1] 0 2fy g g 2 f c 2 g) and [2] [1] as 2 Comparing 2 y (x x 0 2 1) 2 g ⇒ 2 by 8. 2 given 2 x constants, x is √ is 2 An 3, x 2 x 2 Conversely q) and O Therefore y) r f f 2 (y f) shows 2 2 c [2] that 2 y 2gx 2fy c 0 is the equation of a circle ___________ 2 with centre ( g, f) and 2 √g radius f c ___________ 2 provided 2 √g that f 2 Notice xy that the coefﬁcients of c is a real number. 2 x and y are equal and that there is no term. Example Find the 2 x centre and radius of the y 2x 4y 2 Rearranging (x 1) ( 1, 2) 4 whose equation is 0 2 x y 2 2x 4y (y and 2) the 1 radius as 4 4 4 comparing 1 0 as gives the centre as the point 1. 2 2 2 Alternatively x circle 2 2 x 0 y 2x 4y 4 0 with 2 y 2gx 2fy c gives g 1, f 2 and c 4 __________ 2 Then using ( g, f) and √g 2 f c, gives the centre as the point __________ ( 1, 2) and the radius as √ 1 4 4 1 Exercise 2.11b 1 Find the radius 2 Find the 2 x 3 is of the circle whose centre y 6x the 2 is (3, 2) and whose and radius centre 2y 6 and of the circle whose equation is the circle whose equation is 0 radius of 2 (Hint: 2y 6x divide the 4y 1 equation 0 by 2.) 2 4 centre 5. 2 Find 2x equation Explain why equation of the a equation x 2 y 2x y 6 0 cannot be the circle. 93 2. 12 Equations of tangents Learning outcomes The equation of When To ﬁnd the equations of normals to know To ﬁnd be a the condition for tangent a tangent to coordinates of the a circle centre and at a the circles given radius of a point circle, can use the fact that the tangent at a given point on the circle is circles perpendicular the normals to tangents we and we and to a a line to the radius through the point of contact. to circle Example Find You need to know the point equation A(3, 11) of the on the 10y tangent at the circle C(2, 2 How to ﬁnd the centre x and 5) 2 y 4x 8 0 A(3, radius from First equation of a rearrange the equation 2 The basic facts geometry How of to ﬁnd a about of Therefore points a line 2) 2 (y 5) 8 4 25 the circle the intersection as circle (x 11) the Cartesian of and The a the gradient centre of AC of the circle is the point C(2, 5) 6 curve 1 the gradient of the tangent at A is 6 How to ﬁnd the distance of a 1 So point from a the equation of the tangent at A is y ( 11) (x 3) 6 line ⇒ x The The normal to normal through the to a a of 69 0 curve curve point 6y is the line perpendicular to the tangent to a curve contact. normal tangent In the case through The of the a circle, point of the condition for There are two normal is the line containing the radius contact. methods a line to for be determining a tangent to whether a line is a a circle tangent to a circle. The ﬁrst distance the 94 the uses the centre fact of the that a circle line to is the a tangent line is to equal a circle to the if the radius of circle. The the method from second method equations of the uses line the and fact the that there circle are will be solved a repeated root simultaneously. when Section 2 Trigonometry, geometry and vectors Example (a) Find the values of c 2 to (b) the Find the circle the y equation which of 2x the the 8 diameter 2 x ∴ the x of y the 2 y First For line c 0 is a tangent 0 circle that is parallel to tangent. 2 (a) for 2 x 2x point 8 ( 1, 0 0) ⇒ is (x the 2 1) centre of y the 9 circle and the radius is 3. method the circle, line the x y distance c of 0 the (i.e. line y x c) from ( 1, to be 0) is a tangent to the 3. b c ma ___________ _______ ∴ using d | |, 2 √ m m 1 where d 0 c (1)( | Second method Solving x ⇒ 3 ⇒ 1 y c 0 and (x (x 0 and to be 1) y (x c) a 2x(1 tangent c) this c 8 0’ 9 9 0 equation 2 1 3 √ 2 simultaneously gives 2 2 4ac 2 2x line c 2 1) 2 ⇒ ‘b 2 2 i.e. b √ 2 the 1, 1 c ______ | √ 1 For 1) ______ a gives _______________ 3 3, 1 ⇒ 4(1 must have equal roots, 2 c) 8c 64 0 0 1 2 ⇒ c 2c 17 ___ √ 2 72 ________ ⇒ c 3 √ 2 2 (b) The i.e. diameter through The i.e. 1, diameter Therefore y x the goes ( is through the centre of the circle, 0). parallel equation to of the the tangents diameter so is its gradient (y 0) is 1. 1(x ( 1)), 1 Exercise 2.12 2 1 Find at 2 the the equations points Determine 2 x on of the whether the tangents circle the where line 3x to y the circle (a) y 4x Explain 2 x y 6y 11 0 8y 4y 5 0 is a tangent to the circle 0 why the circles y 4x 12y (b) Find the coordinates (c) Find the equation Find (x 1) 2 (y 2) 9 and 2 the 4 2 2 3 2 x 1 point the of of 36 of the the 0 touch. point of common contact normal to condition to the the the two circles. circles through contact. that m and c satisfy if the line 2 tangent of circle whose equation is x y mx c is a 2 y 6x 5 0 95 2. 13 Parametric Learning outcomes equations The deﬁnition of When To deﬁne To ﬁnd a a the Cartesian equation relationship between x and y is difﬁcult to work with, it is easier given in to express each of x and x y in terms of a third variable. This of variable curve parameter parameter often a direct a is called a parameter parametric form 2 For example, equations A point in are P(x, the equations called y) is the on t , parametric the curve given y t 1, equations by these t is of the the parameter . The curve. equations if and only if the You need to know 2 coordinates How to ﬁnd the centre of a circle from P are (t , t 1). and By radius of giving t any value we choose, we get a pair of corresponding values of its x and y. For example, when t 2, x 4 and y 1. Therefore (4, 1) is a equation point The Pythagorean and the trig identities angle identities By double on this giving t curve. several other values we can plot points and draw the curve. t 3 2 1 0 1 2 3 x 9 4 1 0 1 4 9 y 4 3 2 1 0 1 2 y 4 2 x O 2 4 6 8 10 2 4 6 The relationship Cartesian The between parametric equations and equations Cartesian equation of a curve can be found by eliminating the parameter . 2 In the case of the equations x t , y t 1, eliminating t gives 2 x (y When , the For x 96 1) the trig parametric identities example, 2 sin a [1] curve and equations are useful has y these involve to help trigonometric eliminate parametric 3 cos 2 [2] equations: ratios of an angle Section 2 Trigonometry, geometry and vectors 2 The cos 2 identity equation of the 1 2 sin can be sin ⇒ ﬁnd the Cartesian y __ and [2] cos 2 ⇒ 2 3 y 2 x __ __ Therefore to curve. x __ [1] used 1 2 ( 3 ) 2 2 ⇒ It 2y is not always equations. to 6 3x easy to However , parametric convert the a Cartesian Cartesian equation equation of a to circle parametric can be 2 For example, radius 3 and converted equations. the circle centre whose (2, equation is (x 2 2) (y 1) 9 has 1). y 4 P(x, y) 3 2 θ C(2, N 1) x O 2 6 2 From the diagram, CN 3 cos x 2 PN 3 sin y 1 therefore therefore Hence, of the x 2 circle, 3 cos where and is y the 3 cos 3 sin 1 3 sin are the parametric equations parameter . Exercise 2.13 1 Find the Cartesian are given (a) x equation of the following curves whose equations parametrically. 2 t 1, y (b) x t t 1 _____ 2 1 _____ , y 2 1 (c) 2 x Show t sec , that t y the 3 tan curve x represents Give the a whose 3(1 parametric sin ) equations and y are 3 cos circle. coordinates of the centre and the radius of the circle. 97 2. 14 Conic sections Learning outcomes Conic There To deﬁne the conic is sections a set of curves called conic sections that come from the sections intersection of a plane and a right circular cone. You need to know The deﬁnition of a right circular When the plane is perpendicular to the axis of cone the The meaning of the axis of not a cone, the always curve is a considered circle. to be (The one of circle the is conic sections.) cone Did you know? This group among the of curves earliest is believed whose to be properties When were the plane inclination They were studied extensively ancient Greeks. Apollonius wrote conic was an eight-volume sections translated Khayyam last is at an angle less than the investigated. near around by the the curve is the an slant of the curved surface, ellipse Perga treatise 300 into Arabic the of of on BC. This by Omar beginning of the millennium. When the surface, plane the is curve parallel is a to the slant When plane of the the is the at cone an curved formed 98 of curved parabola called is double-ended angle greater surface, a a and than two -part hyperbola the the slant curve is Section Like many purely many mathematical academic such activity activities, with mathematical no conic interest activities, they sections in their end up were investigated applications. having wide And as 2 Trigonometry, geometry and vectors a like scientiﬁc uses. It was in planets round the ﬂight One of the of a in in a same Sun most time ball path from the a about property axis is to that shape another scientiﬁc to its fact, found whose thrown important In discovered our that moon the and orbits of satellites the move orbits. Galileo a Kepler ellipses. telescopes parabola to are that elliptical follows giant comes parallel century in cricket the rotating the Earth vertical mirrors This 17th round the Around to the focus an is object a player is a applications light. (A projected parabola. For at an angle example, the parabola. is the parabolic use of surface parabolic is made by axis.) of the reﬂected parabola onto one that point means (called all the light coming focus). Focus The Hubble mirror In the space which next parabola topics and telescope collects we currently orbiting the Earth has a parabolic starlight. use coordinate geometry to ﬁnd equations for the ellipse. 99 2. 15 The parabola Learning outcomes The We To deﬁne a parabola as a parabola are already familiar with the shape of a parabola – the graph of locus 2 y To ﬁnd and the the Cartesian the parametric ax bx c is a parabola. equation equations One of of that parabola focus) We the any and can properties point a now standard on a ﬁxed use of a parabola parabola straight this is line property discovered equidistant (called to the derive by from the a ancient ﬁxed point Greeks (called is the directrix). the Cartesian equation of the parabola. You need to know The The meaning of a by How to ﬁnd simplest equation is y obtained parameter taking and the Cartesian the the ﬁxed ﬁxed line point as x as A( a, 0) a, P(x, equation of a then locus P(x, y) is such that PA y) PN N How to ﬁnd between The the two distance points condition for tangent to a a line to be a curve a 2 2 PA (x 2 a) y (x A(a, 2 and PN 2 O a) (x a) (x a) 4ax 2 y ⇒ PN x 0) 2 (x a) 2 2 ⇒ y 2 y whose 4ax is vertex the is at Cartesian the and origin, whose equation whose focus is of line the the of standard symmetry point ( a, parabola is the x-axis 0) 2 For example, point (2, the equation y 8x gives a parabola whose focus is at the 0). Example y 2 Find the focus and vertex of the parabola (y 1) 8(x 2) and 10 hence sketch the curve. 2 Comparing (y 1) 8(x 2) 5 2 with Y 4aX whose focus is at X a, Y 0 and vertex is at focus X 0, gives Y y when Y 0 O 1, Y X 0, y x 2, 1, a when 2 X 0, x 2 5 Therefore The line When 100 X the of vertex is symmetry 2, x 4, at is so the Y the point 0, i.e. focus is (2, y 1). the 10 1, point (4, 1). x 6 8 10 Section Parametric The equations for the parametric equations of standard the standard 2 Trigonometry, geometry and vectors parabola parabola are 2 x Therefore the coordinates of at any and point y on 2at the parabola can be written as 2 (at , 2at). Using that parametric apply to coordinates any point on means the that we can ﬁnd general properties parabola. Example Find the equation of a chord that passes through the focus of the 2 parabola x at and y 2at from any point on the parabola. y 2 P(at O A chord of a curve is (a, a line , 2at) x 0) joining any two points on the curve. 2 P(at , 2at) is any The equation point on the parabola. 2 of the line through ( a, 0) and (at , 2at) 2at _______ is y (x a) 2 at a 2 ⇒ y(t 1) 2t(x a) Exam tip Example Find the value of a for which the line y 2x 1 is a tangent to the This problem highlights the 2 parabola y 4ax importance Solving the equation of 2 (2x For line and the parabola simultaneously solutions gives are checking valid in that the context of 2 1) the the of line 4ax to ⇒ be a 4x 4x(1 tangent to the a) 1 curve, 0 this the equation must problem. have 2 equal roots, i.e. 16(1 a) 16 ⇒ 1 a 1 ⇒ a 2 2 (a 0 which is is not not a a valid solution to the problem because it gives y 0, parabola.) Exercise 2.15 2 1 Find the focus and vertex of the parabola 2 (a) (y 2) given by: 16x (b) x 3t 3 The y 2 , y parametric 16t. Find the coordinates of the points of equations in terms of of a p, the curve are x equation of 8t , the 6t chord 2 Find, intersection joining the points on the curve where t 2 of and t p 2 the line y x 6 and the parabola x 4t , y 2t 101 2. 16 The ellipse Learning outcomes The One To deﬁne To ﬁnd the ellipse as a of properties of an ellipse discovered by the ancient Greeks is when a point is constrained so that its distance from a ﬁxed point and and parametric the locus that the Cartesian ellipse equations of a ﬁxed straight line are in a constant ratio which is less than 1, the an locus is an ellipse. ellipse The the of You need to know The meaning How to ﬁnd of a position ﬁxed the of line. ellipse the ellipse These are depends eccentricity of the (Notice that when We ﬁnd the also on value and 1, on called the ellipse e depends the is the the of position focus the the constant denoted deﬁnition and by of the ﬁxed directrix. ratio; this point The is and shape called the e gives a parabola.) parameter the Cartesian will simplest Cartesian equation for an ellipse. This is when a __ equation of a locus the point (ae, 0) is the focus and the line x is the directrix. e How to solve quadratic y 2 a __ 2 PN inequalities 2 x ( and ) 2 PS (ae 2 x) y e 2 Now ePN PS so e 2 2 PN PS P(x, e y) 2 a __ 2 2 x ( (x a by b ) 2 ae) N y e 2 ⇒ x 2 (1 e 2 ) 2 y 2 (1 e ) 2 2 y x 2 Replacing a 2 (1 e __ 2 ) gives 2 shape of the curve can be x 0) 1 2 a The S(ae, O __ a b x deduced from the equation as e follows: 2 2 2 y x __ 2 __ 2 1 ⇒ x 2 2 (b 2 a y a __ y 2 ) and x 0 2 b b b 2 so 2 b y Therefore 0 b ⇒ y (b y)(b b, and y) 0 as _______ a __ x 2 a a O x 2 √b y , b the curve is symmetrical about the b y-axis 2 2 y x __ __ Solving for 1 for y , gives similar 2 a results 2 2 b x, i.e. a x y b a and the curve is symmetrical about the x-axis. Did you know? Also A property of the ellipse is that The sum of the distances between and point on between the each focus ellipse is x 0, and when y 0, curve is symmetrical and symmetrical foci and use this to draw an a about both axes and so we can see y it has a x e e ellipse. B The the line focus focus A ( ae, 0) O is is axis called and its 2a A x (ae The is 2b directrix line called axis B directrix AA' major length 102 that directrices. a can a constant. x You x the two foci when the and through the its BB' minor length is Section 2 Trigonometry, geometry and vectors 2 2 y x ___ ___ 2 1 is the Cartesian equation of an ellipse 2 a b 2 where a major b axis of with b foci length 2 at 2a a 2 (1 ( ae, and 0) e ), and minor (ae, axis of 0), length 2b Example 2 2 y x __ Find the eccentricity of the ellipse __ 9 Comparing and b the e 2 b a the standard equation gives a 3 e 2 ) gives so e 9(1 e ) 9 3 Parametric equations of The parametric x Therefore 4 √ 5 ___ (a cos , with 2 (1 5 2 ⇒ equation 2 2 Using given 1 4 the coordinates an equations a cos of ellipse any and point y of on an ellipse are b sin the ellipse can be written as b sin ). Example 3 ___ __ Find the length of the chord joining the points A where and B where 3 x a cos and y __ The coordinates of A where __ are a cos of B where 2 ellipse , i.e. ( cos ) , 2 2 √ a 2 ____ 3 ___ , ), b sin 4 i.e. ( 4 √ b 2 ____ ) , 2 2 2 a __ 2 the √ 3 b ____ a __ ) 3 3 ___ (a are 4 AB b sin 3 3 ___ coordinates __ , ( 3 The on 4 b sin ( b ___ 2 √ 2 1) 4 ( √ 2 √ 2 3) 4 ________________________ 1 ⇒ AB 2 √a (3 2 2 √ 2) b (5 2 √ 6) 2 Exercise 2.16 1 Find the lengths of the major and minor axes of the ellipse 2 2 y x ___ __ 25 Hence 2 (a) 1 9 sketch W rite the down ellipse. the coordinates of F and 1 x (b) 4 cos , Find, in y terms F , the of , the lengths of F P and F 1 (4 cos , (c) Hence two foci of the ellipse 2 3 sin P where P is the point 2 3 sin ). show that the sum of the lengths F P , 1 F P 2 and F F 1 is 2 constant. 103 2. 17 Coordinates Learning outcomes A, To deﬁne a To deﬁne three-dimensional B AB, and BC are and introduce the unit three points in A space. the line AC (the are displacements. magnitude segment AB, of i.e. AB 2 m) Each is the and has a length 2 m a vectors deﬁnite j, C magnitude of i, and vectors vector coordinates To 3-D Vectors in direction in space. k ___ › C The To ﬁnd the To ﬁnd a magnitude of a displacement from A to B (written as AB) vector followed by the displacement from B to C 2 m is B vector sum equivalent to the displacement ___ ___ write There this are as AB many to C. › BC other A ___ › › We from AC quantities that are deﬁned by magnitude and You need to know direction Pythagoras’ and can be represented by vectors. theorem A vector is a quantity speciﬁc A scalar be as A of quantity represented the length vector the can line represents is by of be a a one real piece is of the fully string direction by does a the both in deﬁned Length, for not straight magnitude of has direction number . represented represents the that which magnitude by magnitude example, depend line and and a space. the on is its segment a alone scalar can direction. where direction and quantity, of the the line length segment vector . B B B → a → BA A A A ___ › The line vector and can the be arrow denoted shows by the AB, where direction, ___ A i.e. and B from are A the to B. end A points vector in of the the › opposite for direction example, is denoted by BA. The vector can also be denoted by, a Properties of vectors The the T wo are If magnitude line of a vector representing vectors are a is written as | a| or a, so |a| is the length of a equal if their magnitudes are equal and their directions equal. two vectors magnitude then b a but and b have opposite the same directions, a ta a If t is a positive real number , then ta is b in the same magnitude 104 direction t|a|. as a but of a Section 2 Trigonometry, geometry and vectors Addition of vectors If the sides represents AB the and BC vector represent sum a the vectors a and b, the third side, AC, b b C C B B b a a a a b D A D A b Notice that diagram) and b follow each whereas a a in b is other the round opposite the sense triangle (clockwise (anticlockwise in in the this case). The order above in show, which i.e. a a and b b b are added does not matter , as the a This rule can be extended diagrams a to cover the b c d A E addition of as many vectors as we wish a to add. In the the diagram vector notice the sum that a, a b, side c b AE and c d represents d. B d Again follow each b other round the pentagon in the D same c sense, but opposite Note a that, equally b c although represent vector called a d is usually this in the C some in appears three to be two -dimensional, it can dimensions. and displacement vectors has no displacement However , diagram vectors Position vectors A sense. particular position in space. Such a vector is vector vectors ___ represent the speciﬁc position of a point, for › example of the the point vector A OA, relative where to O is a ﬁxed origin, represents the position O. ___ › OA is called represented the by any Coordinates T o locate we start a Any by giving other its other vector line of of A. the It is same unique length and and cannot be direction. in three dimensions point from O. position a in three ﬁxed point can distances y dimensions origin, be the point located from O in y each of three directions. mutually Therefore perpendicular we need three x coordinates to locate a point in 3-D. P(x, y, z) O We use the familiar x- and x y-axes, z together Then (x, y, with any z) a point relative third has to axis O z. coordinates the origin O. z 105 Section 2 Trigonometry, geometry and vectors y Cartesian unit vectors A (0, 1, unit vector has a magnitude of one unit. 0) j (1, i 0, i is a unit vector in the direction of Ox j is a unit vector in the direction of Oy 0) k x (0, 0, 1) k is a unit vector Therefore the in the position direction vector , of Oz relative to O, 4j of any point P can be given in z terms y For of i, j and example, k the point P distant 3 units from O in the direction of Ox 4 units from O in the direction of Oy 5 units from O in the direction of Oz 5k 4j 3i ___ P(3, 4, 5) › x O has coordinates (3, 4, 5) and OP 3i ___ › This can also be written as 5k 3 OP ( 4 ) 5 ___ › z When P is position y Then r any point vector xi of yj with coordinates ( x, y, z), then r OP is the P . zk x or r ( y ) z Displacement P(x, y, vectors are also given in the same way. For example, z) r the vector 2i 3j 2k can represent the position vector of the point x O P(2, and we 3, 2) but direction can it as assume can OP . that equally Unless it is a represent we are any told vector that displacement a of the vector is a same magnitude position vector . z Addition y 3i and subtraction of vectors in i, j, k form k V ectors 2j in i, subtracting j, k the form can be coefﬁcients added of i, j, and and subtracted k by adding or separately. 5j For example, when r 2i 5j k and r 1 3i 2j 3)k 3k 2 3k then r r 1 r (2 3)i 5i 3j (2 i (5 2)j ( 1 2 r 1 2 2k x 2i and r r 1 z y The The P is 7j magnitude of magnitude the Using P 3)i (5 ( 2))j ( 1 3)k 2 point of (4, a 3, Pythagoras’ 4k a vector 4i 3j in i, 2k j, k form is the length of OP where 2). theorem twice gives 4 2 OB 2 OA OB 2 2 AB 4 2 2 x O 3 2 2 OP 2 2 BP 2 (4 2 ____________ B OP √ 2 4 2 3 2 2 ) 3 2 4 2 3 2 2 ___ 2 2 √ 29 z ____________ 2 For 106 any vector r xi yj zk, |r| √x 2 y 2 z vector , Section 2 Trigonometry, geometry and vectors Parallel vectors T wo vectors if v and v 1 For example and 3i 3i 2j k are parallel when v 2 2j is k also tv 1 is parallel parallel to to 6i 3i 4j 2j where t 2 2k k (t (t 2) 1) Equal vectors V ectors v a 1 only if a i b 1 a 1 j c 1 and 2 k and v 1 b b 1 a 2 and c 2 i b 2 j c 2 k are equal if and 2 c 1 2 Example Determine whether the vector 2i i 6j k is parallel to: 1 i (a) 2j k (b) 3j k 2 2i (a) 6j not k is not a multiple of i 2j k , so these vectors are parallel. 1 2i (b) 6j k 2( i 3j k) so these vectors are parallel. 2 Example ___ › A is the point ( 1, 3, 2) and ___ B the point (3, 0, 1). Find | AB|. ___ › › |OA| ___ i ___ › AB is 3j A( AO 1, and | OB| 3i k ___ › 2k › 3, OB 2) When in drawing three axes, as they However, B(3, 0, 1) this a diagram dimensions, gives draw the include reference showing not complicate always a do the points the diagram. origin as point. O Remember , the same and sense vectors be in to the be added opposite must sense go to round their the diagram in the sum. ___ › AB ( 4i ___ i 3j 3j 2k) (3i k) k ____________ ___ › 2 √ |AB| 4 2 2 3 1 √ 26 Exercise 2.17 1 P is the point (1, 4, 2). ___ › Give 2 a | OP| 3i (a) 3a (b) a in 5j i, j, k 2k. W rite 1 3 a 2 ( and and ﬁnd down the the length of OP . vectors: 2 ) and b 0 A form 1 ( ) are the position vectors of the points 4 B. ___ › Show the vectors a and b on a diagram, and ﬁnd the magnitude of BA. 107 2. 18 Unit vectors Learning outcomes To ﬁnd a unit vector problems Unit vector A and parallel to unit vector parallel to has a a magnitude given vector of one unit. a ___ › given vector The vector a 2i – 6j ____________ To solve problems in three |a| 2 √ 2 6 3k is represented by OA ___ 2 2 3 √ 49 7 dimensions O 7 units You need to know a How in to three add and subtract vectors dimensions a A How to ﬁnd the magnitude of a 1 unit vector 1 The properties of cubes and right Therefore the unit vector parallel to a is the magnitude of a, 7 prisms 1 i.e. the unit vector parallel to a is (2i – 6j 3k) and is denoted by â 7 A unit vector in the direction of v is denoted by v̂ v ____ and is given by |v| Example 8 Find a unit vector in the direction of v ( 1 ) 4 ____________ 8 ___ 1 __ |v| √ 64 1 16 √ 81 9 v̂ ( 9 1 ) 4 Exercise 2.18a 1 Find a unit vector in the direction of the vector i 2j 2k 3 2 The position vectors of the points A and B are ( 2 1 ) and 4 ( 2 respectively. ) 0 ___ › Find a unit Solving T o solve Mark the to problem origin diagram. you length can Mark add and in but what displacement 108 parallel to AB. problems represent that in a the vector three do all not the need what you direction to to information you vector . dimensions, attempt to ﬁnd. need to another it on When ﬁnd. line helps draw the the a to be as diagram diagram Remember can draw axes is a clear these and draw given, that any represented diagram. clutter by lines copy line the it so equal same Section 2 Trigonometry, geometry and vectors Example OABCDEFG is a cube of side 4 units. O is the origin and the unit M D G vectors M is i, the j, and k are midpoint parallel of the to edge OA, OE and OC respectively. DG. E F Find, in i, j, k form, ___ vectors ___ › ____ › OA (a) the › OG (b) OM (c) 4 ___ › OA (a) 4i ____ Each ___ › OG (b) edge of the is 4 units long › › OA cube ___ ___ › AB BG 4i 4k 4j C B k ____ ____ 4j j 4k ____ › › OM (c) 4i › OG GM O ____ ____ › ____ › A i ___ › › 1 GM GD and GD OA –4i 2 ____ › 1 OM (4i 4j 4k) – (4i) 2 2i 4j 4k Example The and position 4i – 3j – vectors 2k of the points A and B are 2i 5j – 3k respectively. ___ › Find the vector The vector ___ of of magnitude magnitude 5 5 units units in in the the direction of AB. B direction › of AB is ﬁve ___ times the unit vector in the › direction ___ of AB. ___ › AB A ___ › › OB OA (4i 2i ___ – – 3j 8j – 2k) – (2i 5j – 3k) k ___ › O √ |AB| 69 the ___ › unit vector in the direction of AB is 5 ____ 1 ____ ___ (2i – 8j k) so the required vector is √ ___ (2i – 8j k) √ 69 69 Exercise 2.18b 1 The ___ position vectors of ___ points › OA A and B 2 are OABCDE –i 2j k and OB i – bj k OA OD Find, in The respectively. (a) is a right triangular prism with › terms ___of b, the unit vector in 2 units, unit and OD vectors OC i, j, 2 units and k and are AB parallel 4 to units. OA, respectively. the › E direction of Find AB. the unit ___ › (b) Given that | AB| 2, ﬁnd the value of b vector ___ parallel D › to DB. 2 j C k B 4 O 2 A i 109 2. 19 Scalar product Learning outcomes The angle between two vectors π Y ou To deﬁne the scalar product can obtuse two either the acute angle or α α the angle for the angle between two lines, vectors i.e. use of To ﬁnd the angle between either or two However , vectors their the angle directions between when they two both vectors is converge deﬁned or both as the angle between diverge. a f You need to know c θ The Cartesian form of a vector How product θ e two to expand the θ d b of brackets (a) The The the (b) scalar scalar angle product product between a . (c) b of a two and vectors b ab cos and is a where and b denoted is the is deﬁned by a . angle b, as ab cos where is i.e. between a and b Parallel vectors When a and either b a are . b parallel, then ab cos 0 or a . b a ab cos a b a b π Now cos 0 1 for and for and cos 1, therefore parallel vectors in the same parallel vectors in opposite direction a directions a . b . b ab ab 2 In the For special the unit case when vectors, i . i i, j j . a and j b, a . b a . a a k, k . k 1 Perpendicular vectors __ When a and b are perpendicular , then a . b ab cos 2 __ but cos 0, a therefore 2 for In perpendicular particular , for the vectors unit a and vectors i, j b, a and . b k, 0 π 2 i 110 . j i . k j . k 0 b Section The scalar When a product of vectors x i y 1 a . b (x i y 1 involve Now i the . i z a j . . j b i k x . example x j 2j y . and vectors i y j z 2 z k, 2 k) 2 j, i . involving k and perpendicular z 2 j . i k . i, j that . j and are all k . k, zero as they vectors. z z 1 k) . k) 2 (x 1 j terms i geometry 1, y 1 y y 2 of 1 (3 i gives 2 i 2 k x 1 For b 2 product (x and (x involving 1 i.e. k brackets terms scalar z k) Trigonometry, in Cartesian form 1 1 these with therefore j 1 Expanding together j 1 2 i y 2 (2i j z 2 5j k) x 2 2k) x 1 (3)(2) y 2 ( y 1 2)(5) z 2 z 1 (1)( 2 2) 6 Example 2 Find the value of a for which the vectors 1 ( 2 1 ( a ) are 2 2 a ( ) ) 4 a 2 2 1 The and 1 perpendicular . ( 2 ) vectors a are perpendicular when 4 a 2 0 2 Example ___ › The position vectors of points A and B are OA i 2j 3k ___ › and OB i 3j 2k, respectively. ___ ___ › Find the angle ___ between ___ ___ › √ 14 and ___ | OB| √ 14 , › OA OB 1 6 ___ 6 1 ___ › › |OA| |OB| × cos AOB 1 cos AOB ___ ___ › › 1 1 ___ __________ so OB. ___ › ∴ and › |OA| ___ › OA 14 |OA|| OB| ⇒ AOB 1.50 rad Exercise 2.19 1 a 4i Find 2 a Show 2i j . 3j b and that i 5k and the 7j b angle 3k 2i 2j between is a 4k and perpendicular b. to both i j 2k and 3k 111 2.20 Equations Learning outcomes of To deﬁne and the line Straight A a straight lines line is in three dimensions uniquely located in space if: vector, Cartesian parametric equations of a it line is parallel through it a passes to a ﬁxed given point, through two vector , i.e. it has a known direction, and passes or ﬁxed points. You need to know The vector The difference position vector and line line L passes through the point A whose position vector is a and is a parallel displacement a between The a equation of to the vector b vector y How to add and subtract The Cartesian form of a vectors vector in L three dimensions b The scalar product of two vectors A a O x P(x, r y, z) z P(x, y, z) is any point on the line. ___ ___ › If r is any the real position ___ › i.e. r and OP a for any OA is different For is 112 › OP , then AP tb, where t can take b value a the of a , b is This although example, line. to the 3i the line k this is b equation called the position values. parallel the r AP , and unique i.e. › where a P , ___ › r Now of value. ___ Now vector position is a vector means line is i vector vector a of parallel of point any the on equation vector that point a point to on vector the the the of line. the on line the line line. line so equation it of can a have line is many not unique. whose and the gives vector 2j k equation is the is r position i 2j vector of k a (3i point on k) Section 2 Trigonometry, geometry and vectors Example W rite (a) down vector r 5i j Determine (b) the position equation vector of two points on the line whose is 2k (3i whether the 4j 6k) vector 6i 8j 12k is parallel to the line. (c) Show (a) Comparing gives that a Giving gives r the 8i shows 6i (c) 8j The vector to 3i of 3i 4j 2i Parametric y, z) is then line the P(x, 2j many (like The is perpendicular one 4j 8i point on on 3j 4k with the the are (3i 4j parallel to the 2k is 2(3i 4j 6k 6k z) the is 3j yj give and 5k 4j is 6k), parallel and 3j 3k on zk is the of 3 2i 3k) to 3j the with if point xi yj line, 3k is 12 the so zero, 3i 2j 5k k parametric the line example, and the 1 vectors of r a b 8j 12k is the scalar then the vectors any This of vector means any the line not line r zk x i x x of z y j a, of a equations 4k is bj ck is (8i 4j can i any 4j 6k) 6k) k on j line of in z k terms so it can equations line same (ai of line. line the the in the the although 1 z the parametric found y 6 on point the be equations unique x 5 is way, bj of a unique. i.e. if ck) 1 (ai bj ck) 1 y y b, z z 1 equations ai 0 line. 5k point that 1 the (8i parametric are line a c 1 where (x y vector x 1, parallel to ) is a point on 1 the line. equations x 7j gives of the to 2j coordinates 18 1, the taking line 4 , y position on equations any 6 1 b line. the r and 8 , equation) Therefore 2i line. a position 6i 2 values. vector line j called the a 1 are so therefore 1 For r the line. line, 6k) perpendicular 3i i, the are is different then are 6k) point to line. coefﬁcients parametric y, 6k point equations 3i is another and j 4j any parameter have 2k (2i x Now 3k (3i this equations of xi Equating the line. 4j 6k) Therefore These the 2k so gives 5i 3j perpendicular . (3i P(x, 4k the 12k parallel are If r 3i also product j on that 2k 3j 5i points j value 2i 5i j any Comparing (b) 5i Therefore two r vector of a 1 line parallel to 2 , y where the 3 ( 1, 3, 7 , 1) z is a 1 point 4 on the line and line. 113 Section 2 Trigonometry, geometry and vectors Example Find and the parametric B(0, 1, equations of the line through the points A(1, 1, 4) 2). A B O ___ ___ › AB ___ › is parallel to the line and AB ___ › › OB OA ___ › Therefore Using x AB A(1, x 1, a, j 4) as y a 1, b the (i point b, on z y 2, the the c z 4k) line and c i 2k comparing gives x 1, with y 1 equations 2 , 1 2j 1, z 1 4 1 2 z equations of with j 1 parametric , 1 Cartesian Starting 2k 1 Therefore x a y 1 and parametric of line are 2 4 a the line equations of a line, x x a, 1 y y b, z z 1 c and solving each equations of a line, x y x y the Cartesian (x Using the of the equation on ai y a the bj x 1, ck forms line line and and ) a a described in a each of vector ) c is a point on the line 1 is equations of three of z ( b 1, and 1 ______ a where z 1 ______ point gives i.e. 1 ______ Any for 1 vector parallel to the line. line above them, that is can you be used can parallel to to ‘read’ the describe the the coordinates of a line. Example State whether r 2i 3j r (3 )i T o direction ﬁrst r line 2i r (3 i 114 this j but the (i of j the and 4)k lines )j (2 i 4k) We j to equations (2 line. parallel ( equation (3 j )j 2k is with (i each the )i 4k whether of 3j line 2k lines (3 determine the so the are can the 4k) line parallel vector ⇒ ( we this second is parallel. parallel ‘read’ 4)k 4k) are to r 3i i therefore j the need from needs the to the ﬁnd a rearranging vector (i j 3j 2k ( 4k) lines vector equation are parallel. of in the ﬁrst. 4k) i j 4k), Section 2 Trigonometry, geometry and vectors Example The point A(3, 2, C 2) is the and midpoint B(2, 1, 5). D of the line is the point segment (1, joining 4, 1). A D Find the Cartesian equations of the line through C and D. C T o ﬁnd the Cartesian equations of the line through C and D, B we need a vector parallel to CD. ___ ___ › C is the point ___ with ___ › ___ › AC vector › OA AC O ___ › 1 and position › 1 AB (OB 2 OA) 2 1 {(2i j 5k) (3i 2j 2k)} 2 1 ( i j 7k) 2 ___ › 1 OC (3i 2j 2k) ( i j 7k) 2 1 (5i 3j 3k) 2 ___ ____ › ___ › CD OD (i › OC 1 1 4j k) (5i 3j 3k) Using D(1, equations 1) as a 3i 11j k) point on the line and 3i 11j , z k as a vector parallel to CD gives the Cartesian as y x 1 ______ 4, ( 2 2 4 z 1 ______ ______ 3 11 1 Example The line L has equations x 3 , 2 y 1 2 5 and the 1 line L has equations x 1 4, y 3, 4 z 1 . 2 Find the angle between L and L 1 L is parallel to the 2 vector 3i j 2k and L 1 4i If is parallel to the vector 2 3j is the k angle between L and L 1 ( 3i j 2k) (4i 3j , k) | 12 3 2 ( √ 9 the scalar j 2k| product, 1 3i |4i 3j k| cos ___________ __________ ⇒ using 2 4 )( √ 16 9 1) cos 17 ________ cos ⇒ √ 14 2.67 rad This the ___ ___ is √ 26 the acute obtuse angle is angle between the lines; 0.471 rad Exercise 2.20 1 A is the point parametric (1, 2, 2) equations and for B the is the line point 2 2 The line l has equation r 1 1 ( 5 r ( 1 that l and 1 0, 5). and Find vector and B. ) 2 ( ) and the line l has 2 1 3 ) ( a ) 8 1 Given A 4 5 equation (2, through l are perpendicular , ﬁnd the value of a. 2 115 2.21 Learning Pairs of outcomes lines Pairs of T wo To in determine three whether dimensions intersecting or two are lines lines in in space space may be parallel or not parallel, in which case they may lines intersect or they may A non-parallel not. parallel, skew pair of lines that do not intersect are called skew You need to know The different forms for the O equations of a line in three dimensions The condition for vectors to be Parallel parallel lines Intersecting lines Skew lines separated in space How to solve simultaneous a pair of equations Parallel It is easy vectors For r lines to tell that are example, 2i k parallel i.e. 2i j parallel the are whether k j and Non-parallel lines each are line parallel from because their you can ‘read’ the equations. lines (2i because two to the k) and vectors 4i 2j r j parallel 2k ( 2k to (4i the 2(2i j 2j 2k) lines, k)), are parallel. lines r 2 a μb 2 2 P r a 1 r λb 1 1 1 r r 1 2 O r 2 T wo lines whose vector equations are r a 1 intersect if values of and can be b 1 found for and 1 which r 2 r 1 If no The such values parametric determining 116 can be found equations whether two of then lines lines the are lines easiest intersect or are to are a 2 r 2 skew. work skew with when b 2 Section 2 Trigonometry, geometry and vectors Example Show x of If that 2 their the the 2, y point lines lines of 4 x 1 , z , 6 y Solving When When then lines values equating of equations 3, Therefore the the these x 2, 3 and ﬁnd and the coordinates intersection. intersect equating z intersect the 1 and , 1 3 x these y 2, values intersect at gives gives 2, y y of the 2 1 3 and 2 z and and point ( values z give 2, 2 [1] 4 [2] 2 (ﬁrst 0 line) (second the 2, gives 2 0 x and of same line) point on each line, so 0). Example Determine whether the lines r i k (i 3j k) and 1 r 2i 3j k (4i j 5k) are parallel, intersect or are skew. 2 The are lines not W riting x are not parallel the 1 equations , in y 1 i parametric 2 4, 3 , z , 3 z 2 Equating x and x 1 Equating and 4i j 4k y and [1] 1 1 5 2 gives 1 gives 3 2 4 [1] y 3 [2] 2 and these 2 1 [2] values gives z 0 and 1 values are and z 1 These k 1 y 2 Solving 3j form: 1 x With because parallel. 0 1 2 not equal, therefore the lines do not intersect and are skew. Exercise 2.21 Show that the lines r ( 1 ) 2 2 T wo r lines which intersect 2i 9j 13k ai 7j 2k (i 5 2 2 1 and ) r 2j ( 1 0 2 have 3 ( 1 ) 0 ( ) are skew. 2 equations 3k) and 1 r (i 2j 3k) 2 Find: (a) the value (b) the position of (c) the angle a vector between of the the point of intersection lines. 117 2.22 Planes Learning outcomes Deﬁning There To determine Cartesian the vector equation of a are several one one (b) You need to know and add and subtract one only only three to deﬁne and only plane three one therefore a unique plane, for example: plane two one can be given can given plane drawn points be through specify drawn to intersecting can be drawn a three contain lines non-collinear unique plane two specify a perpendicular intersecting unique to a plane given vectors direction in one therefore and lines, (c) to ways plane points, How plane and (a) a at a given distance from the origin, therefore the normal to dimensions a The scalar product The equations of two plane and the distance of the plane from be drawn the origin specify a unique vectors plane of a line in three one (d) and only one plane can through a given point and dimensions perpendicular The meaning of a unit a vector normal to to the a given plane direction, specify a therefore unique a point on the plane and plane. normal (A nor mal therefore to a The vector We of use a plane is perpendicular the any to line any equation of deﬁnition of a perpendicular line a plane in the to the plane. A normal plane given in (d) to derive the vector plane. N A P r a n r a O ___ › A is a The point vector on n is the plane and perpendicular OA to a the plane. ___ › P is AP any is a point line on in perpendicular the the to plane plane, and and OP is r therefore n ___ › AP r a, therefore the scalar i.e. This 118 is called the vector is plane.) product (r a) equation of a . of n r plane. a 0 and n is zero, equation Section The vector equation of a plane can be written in another 2 Trigonometry, geometry and vectors form. N P a A r n a r θ O In If the diagram, ON But r . ON d, then in n̂ OP cos Therefore r perpendicular This Therefore an is the distance triangle . n̂ to OPN, of d the plane from the origin. OP cos d d the is the unit vector vector equation can equation of be n̂ equation and multiplied the for m r . of distant n by a d any D plane from that the is origin. scalar. represents a plane D __ perpendicular to n and distant from the origin. n̂ Example The vector equation Find the distance 2i j 2k Dividing form r . is a both n̂ of of the vector sides a plane plane is r . from (2i the perpendicular by |2 i j 2k) 12. origin. to 2k| j the plane. converts the equation to the d 1 |2i j 2k| 3, so the equation becomes r . (2i j 2k) 4 3 Therefore the plane is 4 units from the origin. Example (a) Find the equation of the plane that is 4 vector ( 4 and ) contains the point 5 ( 1 (b) Find (a) Using r . ( the form 4 2 4 5 ) 1 ( of the (r a) . plane n . ( 4 the ) from 0 ⇒ the r . origin. n a . n 0 gives 4 ( ) 3 4 ) 0 1 4 r to 3 distance the perpendicular 2 4 ) 25 0, i.e. r 1 . ( 4 ) 25 1 25 ________________ 25 ____ _______________ (b) The distance of the plane from the origin 2 √4 2 4 2 ( 1) ___ √ 33 119 Section 2 Trigonometry, geometry and vectors Example 3 Show that the line whose equation is r ( 1 2 ) 2 ( is ) parallel to 5 2 8 the plane r . ( 2 ) 5 4 If the line normal is to parallel the to the plane, then it is perpendicular to the plane. 8 2 The line is parallel to 2 ( and ) the plane is perpendicular to ( 2 ( ( ) 2 5 ) The point unit vector 0, therefore the line is parallel to the plane. 4 The Cartesian n̂ 4 8 2 Now ) 2 5 li P( x, y, equation of z) is any point on a plane the plane that is perpendicular to the where mj nk N P(x, y, z) a r n li mj nk O Using r . n̂ (xi ⇒ lx d gives yj my zk) (li nz mj nk) d d Therefore lx my where and li Multiplying the d mj this Cartesian is the nk ai the bj plane d is the Cartesian distance is equation ax Then a unit by a of the vector constant equation plane from of the a plane origin perpendicular to gives general the more the plane. form of equation, i.e. of nz ck from is a the by vector origin is cz D perpendicular given to the plane and the distance by D _____________ ____________ √ 2 a For example, 3i from 2j the 6k the is origin 2 equation b 2 3x 2y perpendicular is given c to the 6z 21 ___ ____________ 3 120 21 and by 21 _____________ √ plane 2 2 2 2 6 7 3 represents the a distance plane of where the plane Section 2 Trigonometry, geometry and vectors Example (a) Find the A(1, (b) Hence (a) The (1, Cartesian 5, 2), B(1, write down Cartesian 5, 2) equation 1, 1) a equation satisﬁes the 1, 1) satisﬁes therefore These This also three is all the [1] using b b are the gives gives 6b [4] gives b c ax the contains by the points D of to cz the plane. D plane, [1] the plane, D 4c any is [2] enough D to [3] ﬁnd multiple a, of b and ax c in terms by cz of D D. is plane. 4b [3] of 2c that 4). perpendicular plane because of is equation 5a need a plane 1, that 5b the equations we equation [2] 5[2] C of a similarly, the equation a also of C(5, vector therefore (1, and 3c c 0 [4] 4D [5] 6D ___ 3[5] 11 3D ___ 8D ___ Then [4] gives c and [2] gives a 11 Therefore the equation of 11 the plane is ai 3D _____ 6D ___ x y 11 bj ck 8D ___ z 11 D 11 A ⇒ 3x 6y 8z 11 B C (b) ai bj ck Therefore is 3i perpendicular 6j 8k is to the plane ax perpendicular to by the cz D plane. Exercise 2.22 1 Find, the in point Hence 2 3 the the (1, 1) (a) 5, and that . 3 (3i Show in the that Hence is is j the ﬁnd and the D, is the equation line L 2k) point equation perpendicular distance of with parallel the plane n perpendicular the 2 . 1) Cartesian z r down Show r (b) (2, write Find 0 form to of the the to the plane P x plane vector from that vector equations the to the plane plane the of i the that 3i 2 2j whose , j k. origin. contains contains the point k. y vector 1 , equation is 6 (0, 8, equation 1) of is the contained line in parallel the to L plane. that is contained P . 121 Section 1 Give exact 2 Practice values questions 13 for: The equation 2 7 ___ 4 ___ ( cos (a) x ) sin (c) ( y 7 ___ ( ( cosec (d) the a circle 6y 3 is 0 equation gradient is of the diameter of this circle 1. ) 3 14 Give the whose 5 ___ ) 4 2 2x 4 Find tan ) 3 (b) of 2 minimum value of 3 The focus of a parabola directrix is Find equation the the line y of is the point (2, 4) and the 8 the parabola. __________ __ 2sin( ) 3 15 The (a) equations of two circles are 5 __ 3 Given sin A and that A is 2 acute, (x 13 ﬁnd the value the 2 cot x y 4x ﬁnd the 2 c 0 value of c such Find the coordinates of that the the circles point of touch. the contact circles. Find the solution of the equation the of the of contact. point common tangent equation 2 cot 4 16 values of between Prove (a) . and that tangent to the line the circle 2 (x Prove 6y the through 5 9 (c) for 4 cot 2 _______ 2 sin 2 tan 1) identity tan Hence Find of (b) (y 2 cot A (b) (b) Prove (a) 2 and 4 1) of: sin 2A (a) 2 3x 2y 12 0 is a 2 1) (y 2) 13 that Find (b) cos A cos 5A sin A sin 5A the coordinates of the point of contact cos 6A of the line and the circle. __ 6 Express cos x √ 3 sin x in the form r cos (x ). 17 Find the equations of 2 to Hence ﬁnd the maximum and minimum the circle the tangents from the origin 2 x y 8x 6y 16 0 values __ of cos x √ 3 sin x and the values of x at which 18 they occur in the range 0 x Show (a) that the line y 2x 2 is a tangent to 2 the parabola whose parametric equations are 2 x 7 Show 5 cos ) Find (b) 2 (2 sin , y 8t all values of the coordinates of the point of the line and parabola. . 2 19 8 Prove the 9 2 2x cot Solve the The equation 2 x 2x tan x (a) Sketch (b) Find the sin 2 sin (a) Find the 0 y x of between 0 and cos 4x 2 9y general cos 3 solution of the of the foci. points and of intersection the of the y 2y 25 the circle The of the locus of the point P( x, at equations each of of the these tangents points. line l has vector equation 2 y) r distance of P from the line y 4 is twice ( 2 ) ( 4 the distance between P and the point (2, 0 ) 1 1). and the line 1 has vector equation 2 3 2 12 Find the Cartesian parametric x 122 2 equation equations cos 2 and y of the are 5 curve whose r ( 1 ) ( 1 line 0 cos 2 equation the the circle 1 when ellipse. coordinates 5 1 Find 36 2 Find equation 21 11 2 . (b) the the 2 Find is equation x 10 ellipse 4cos 2x 20 values an 2 sin sin 3 for of identity 2 sin contact 29 of for 4t that sin where 1 ) 2 and are scalar parameters. to the 0 Section Show (a) that position the lines vector of intersect their and point of give the 28 Show that the 2 the acute angle between whose Practice vector questions equation is intersection. r Find (b) line 2 the two ( lines. 1 1 ) ( 1 5 4 whose equation is ) parallel to the plane 1 22 The (a) position r 2i 3j r 5i k, r i vectors of three points are j the Find (a) the equation of the plane these ﬁnd ) 8 the equations three the point (2, 1, of 5) the that is which to the plane whose equation is points. 3x Hence 5 parametric through perpendicular (b) ( k vector contains . 1 line Find r 2k, 29 is distance of the plane from 3y z 9 the Find (b) the coordinates of the point of origin. intersection 23 The equations of two lines 5 , y 2 3 , z 1 4 2, y 3 6, Show z 1 that 2) Show (b) Show that the lines are the vector 2i j k is these as if P( x, it Find P(a cos is y) is from the twice the centre , sin b 1 perpendicular two to the plane. as far origin from then P the lies point on and radius of the a circle. parallel. 31 that and 2 circle. (a) line and (4, x this are: 30 x of ) and cos Q(a 1 points on the , sin b 2 ) are 2 ellipse lines. 2 2 y x __ ___ 2 The Cartesian equation of a line l 1 2 a 24 b is 1 y x 1 ______ 2 5 the Find the (b) Deduce equation of the chord PQ. 2 and (a) z 3 ______ ______ 2 Cartesian equation of a line l y x 1 ______ 3 the z 1 ______ ______ that the equation of the tangent to is 2 ay ellipse sin at the point cos bx cos , (a b sin ) is ab 2 3 a 2 Given that l and l 1 value 25 The 3x of Cartesian y are perpendicular , ﬁnd the 32 2 P(5t , 10t) 2z equation of a plane is (a) Find (b) M 7 any point Find the distance of this plane from the on Cartesian is the midpoint Find the Cartesian as (a) is a curve. a. t equation of the of line equation the curve. OP . of the locus of M varies. the origin. 33 Show (b) that the line whose vector (2i 2j equation (a) Prove that is 2 cos r lies i in the Hence from 2j (b) down the The position distance of the Find the 2j relative to k) a the hence vectors and ﬁxed acute ﬁnd general line cos of points A and B x 1 solution of the equation 2 x 5 sin x 2 Show that the (3i origin 6j 2k) angle respectively r O. the area of 4 OA triangle and whose vector equation is ( 1 ) t 8 between line are is Find sin origin. 0 (2i 2 34 26 2 x 3k) plane. write the 4k OB ( 0 ) 2 contained in the plane whose and 2 OAB. equation is r . 0 ( ) 8 1 27 Show r . (i and that r . Hence 5j the (2i ﬁnd planes 2k) the 10j whose equations are 10 4k) distance 8 are between parallel. the two planes. 123 3 Calculus 3. 1 Functions Learning outcomes 1 – continuity To investigate the meaning of the and discontinuity using that functions . we For have a drawn function so to far be in this unit continuous, have there involved must be no of breaks functions graphs of continuous continuity discontinuity Continuous functions Most and in its graph and no points at which it is undeﬁned. graphs 2 For example, graph x in of its y f( x) f(x) x has for no x breaks is a and continuous f( x) has a function real value because for every the value of domain. You need to know y The meaning of a function The meaning of the domain 10 of a function The shapes 5 of graphs of simple functions O 3 2 x 1 1 2 3 Discontinuity The graph below illustrates the graph of a function over the domain x . y 10 5 discontinuity x O 4 4 6 8 5 10 There point is a clear missing break from in the the graph graph where where x x 2. 2. There These is also breaks a are called y discontinuities . The function Some is functions continuous are for all continuous other even values though of the x graphs have breaks in them. O x 1 __ For example the function f( x) , x 0, x is continuous because x 1 __ although the graph of y has x is 124 not in the domain of f( x). a break in it where x 0, this value of x Section 3 Calculus 1 Example (a) Sketch the x f(x) graph State, its the x 4 4, x 4 { , x (b) of 4, with a reason, function x given by whether the function is continuous over all of domain. f(x) (a) 10 5 4 x O 2 2 4 6 8 5 (b) There is so curve the a break is where not x 4 and continuous x over 4 all is the included in the domain, domain. Example (a) Sketch the graph of the function given by 2 x f(x) , State 0 x 0 { , x, (b) x whether x the function is continuous. f(x) (a) 5 x O (b) There are in nature the missing is no 5 breaks of the because in this graph f( x) is graph where deﬁned and x when although 0, x there 0. is there not is a a change point Therefore the function continuous. Exercise 3.1 Sketch each of the following functions and state whether the function is continuous. x 1 f(x) x, x {2, 4, 4 6} f(x) x x 1 ______ 2 f(x) , x x f(x) 1, x 0 x 0 , 1, 5 f(x) x 9, x 4 x 4 1, x 4 x 5 { x { x x 1 x, 3 1, { 9, , x , x 125 3.2 Limit notation Learning outcomes Limits 1 __ In To introduce basic concepts T opic 1.10 we looked at the behaviour of as the value of x gets large. of x 1 __ limits This table of values shows that approaches zero as x approaches x To of investigate f(x) above as x and the behaviour approaches inﬁnity. a from below x 5 10 100 1000 10 000... 0.2 0.1 0.01 0.001 0.000 1... 1 __ x You need to know 1 __ We The meaning The shape of write this as → 0 as x → , and we say that the limiting 1 __ 1 __ of the curve y value x continuity of 1 __ as x → is 0. The notation for this statement is lim x→ x 0 x x 1 __ How to sketch simple functions If we now look at as x → 0, there are two cases to consider because x x can approach x zero from approaches 0 positive from 20 As x → 0 from x → 10 from ( approaches 0 5 100), 10 0 from 1 __ ( 1000)… ), → ( 100), 1 __ ( 1000)… ( ), 0.001 → x 1 __ x→0 x 0.01 lim above 1 _______ 10), 0.1 Therefore values. 20 0.001 1 ______ ( negative below 1 _____ (i.e. from 1 ______ 10), 0.01 0 x 5 1 _____ ( 0.1 As below or above 1 ____ (i.e. values 1 __ ) does not have a unique value so lim x ( For lim [f(x)] to exist and equal k ) does not exist. x x→0 then x→ →a as x → a and as x from above } The limit of f( x) as x → a → from a from above f(x) → k below is written as lim f(x) and the x→a of f(x) as x → a from below is written as lim f(x) x→a Limits and discontinuity We can now For f(x) to be deﬁne a discontinuity continuous where x in terms a, lim x→a 126 of f(x) limits. lim x→a f(x) f(a) limit Section 3 Calculus 1 2 x For example the function f( x) , sketched Using the in T opic 0 0 , x, is x x { x , 3.1. condition for continuity at x 0, 2 lim x 0, lim x 0 and x→0 f(0) x→0 0, therefore However applying x f(x) 4, x 4 4, x 4 , 3.1, gives is this { x T opic f( x) condition x lim (x continuous , 4) to where 8 at x the x 0 function and 4, which lim (x is 4) at x also sketched in 0 x→4 x→4 They are not equal so there is a discontinuity 4 Example 2x The function f is deﬁned as f( x) { Find lim x 1 x 1 , x (a) 1, x 2 , f(x) x→1 (b) Find lim f(x) x→1 (c) Hence First show sketch that the f( x) is graph continuous y at x 1 f(x) y 5 x O 2 (a) From 2 the graph, 4 lim f(x) lim (2x 1) 1 x→1 x→1 2 (b) From the graph, lim f(x) x→1 lim x 1 x→1 2 (c) f(1) lim 1 f(x) 1 lim f(x) f(1) x→1 x→1 Therefore f(x) is continuous at x 1 Exercise 3.2 (a) Sketch the graph of the function given by 2 f(x) (b) Find x , 2 x { 1 , x for values of 3 x 3 2 lim x , x 1 f(x) x→1 (c) Find lim f(x) x→1 (d) Hence show that f( x) is continuous at x 1 127 3.3 Limit theorems Learning outcomes To list and use the The limit theorems 1 lim If f(x) F then lim kf(x) kF where k is a constant. limit x→a x→a theorems 2 For example, lim 2 x 4, therefore lim x→2 sin 3x 3 4 12 x→2 _____ To ﬁnd and use lim →0 2 If lim f(x) F and if lim x→a g(x) G then lim [f(x) x→a g(x)] FG g(x)] F x→a 2 For example, lim You need to know x 4 and x→2 lim (x 1) 3, x→2 2 therefore The notation for limits The trigonometric double How to factorise x (x If lim f(x) F and if lim x→a 4 3 g(x) G The factor 12 then lim [f(x) G x→a 2 For y example, lim x 4 and x→2 x→a 3 1)] angle and factor formulae 3 [x x→2 3 formulae lim theorem lim (x 1) 3, x→2 2 therefore lim [x x 1] 4 3 7 x→2 f(x) F __ ____ 4 If lim f(x) F and if x→a lim g(x) G then, provided G 0, lim x→a x→a ( g(x) ) G 2 For example, lim x 4 and x→2 lim (x 1) 3, x→2 2 x ______ therefore 4 __ lim x x→2 1 3 0 __ Now is meaningless, but these theorems can be used to ﬁnd the limit 0 0 __ of a function which appears to be . The example below illustrates this. 0 Example 2 x 9 ____________ Find ( lim ) 2 x x→3 7x 2 (x x 9 ____________ 12 3)(x 3) x 3 ______ _____________ provided that x 3 2 x 7x The as x limit gets 12 as (x x We 4) approaches closer towards. 3)(x and do closer not want 3 x means to 3 the to 4 we see value want what when the values value x 3, they so of are we the function tending can say that 2 x 9 ____________ lim x→3 ( x 3 ______ ) 2 x 7x 12 lim x→3 lim ( ) x (x 4 3) 6 ___ x→3 __________ lim (x 4) 6 1 x→3 sin x _____ The limit of as x → 0 x This is an directly, origin. 128 so important we start limit with and the the limit graphs of y theorems sin x do and y not help x to close ﬁnd to it the Section 3 Calculus 1 y x 0.6 x 0.4 0.2 π π 6 6 x 0.2 0.4 0.6 sin x _____ When x 0, 0 __ 0 __ x __ and is 0 meaningless. 0 __ For x , 6 sin x and x are nearly equal, and as x approaches 0 from 6 sin x _____ above, 1 x sin x _____ Also as x approaches 0 from below, 1 x sin x _____ Therefore lim Note that this result is valid only 1 x x→0 when x is measured in radians. Example sin 2 ______ Find lim →0 sin 2 ______ 2 sin cos ___________ sin _____ sin 2 ______ lim sin _____ →0 lim →0 lim lim 1 2 sin 2 _______ 2 2 sin 2 ______ lim →0 (2 cos ) →0 sin 2 ______ Alternatively, 2cos 2lim 2 →0 →0 2 2 Exercise 3.3 2 x 5x 6 ___________ 1 Find the limit of as x x → 2 2 2 1 x _______________ 2 Find the limit of 3 x as x → → 0 1 2 x x 1 sin 3 sin _____________ 3 Find the limit of as 2 sin x _____ 4 Use the results above to show that the function f( x) has a x discontinuity where x 0 129 3.4 Gradient Learning outcomes of a Tangents, The Find the point on gradient the of curve a curve at line chords joining two and points normals on a curve is a chord a curve A line that touches a curve point at a point is a tangent to the curve. of contact You need to know A The a deﬁnition straight The of the gradient chor d of B line concept of a limit The gradient of The a curve gradient of gradient We can curve ﬁnd at another a the point point B by on a of gradient A at a cur ve the of given at a point point tangent to A the is deﬁned cur ve at as the A. a taking the curve B close As B to A A. moves gradient closer of to tangent closer the the at to chord gradient A, the gets of the A. limit (gradient of chord AB) gradient of tangent at A B→A 2 We can use this to ﬁnd the gradient of the curve y x at the point 1 where x 2 1 A is the point on the curve where x , 2 1 B is a point on the curve whose x-coordinate is a bit larger than 2 h We denote the ‘bit larger ’ by is sometimes used as an alternative x notation for the ‘bit larger’ 1 So B is the point on the curve where x x 2 y is not a variable – it is a preﬁx 8 that means ‘a small increase in’ 6 the 1 B( 4 1 δx 2 1 A( 1 , 2 2 δx) ) value follows the y variable means that a small 2 2 increase in the means small value of y, t ) 2 2 value x O 3 130 of it. So 2 1 1 2 3 a of t, and increase so on. in the Section 1 The coordinates of A are ( of B are ) 1 ( x, 2 ( 2 x) ) 2 1 1 2 ( 1 4 1 coordinates Calculus 1 , 2 The 3 x) 2 4 ____________ The gradient of AB is 1 1 x 2 2 1 1 2 x (x) 4 4 _________________ 1 x x gradient of tangent at A limit (gradient of chord AB) B→A limit (1 x) 1 x→0 1 at the point where x 2 , the gradient of y x is 1. 2 2 We can apply the same process to a variable point on y x 2 A B is is any the point point x-coordinate on on of the the curve curve so its coordinates whose can x-coordinate is be a denoted little by larger ( x, than x ). the A, 2 i.e. x x. Hence B has coordinates 2 (x (x x, (x x) ) 2 x) x _____________ The gradient of AB y is (x x) x 2 x 2 2xx 2 (x) x _____________________ x 2 B(x δx, (x δx) 2 2xx (x) x(2x ____________ x) ___________ 2 A(x, x 2x x ) x x O The gradient of the tangent at A lim (gradient lim (2x of x) x AB) 2x x→0 We can now use this result to ﬁnd the gradient at any particular point 2 on y For 2 x example, 3 6, at and the point where x where 4 x the 3, the gradient gradient is 2 of 4 the curve is 8 Exercise 3.4 1 Use the method above to ﬁnd the gradient at any point on the on the 3 curve Use (a) 2 your x Use y the x result to 1 ﬁnd (b) method above the x to gradient on the curve where: 5 ﬁnd the gradient at any point curve 2 y x Use your 2x result to ﬁnd the gradient at the points where: 1 (a) x 0 (b) x 2 131 ) 3.5 Differentiation from ﬁrst Learning outcomes The In Find the gradient function of principles gradient function the previous topic, we found that the gradient of any point on the a 2 curve curve by y x is given by 2x differentiation from ﬁrst 2 principles Now 2x is a function and it is called the gradient function of x . 2 Because the 2x is derived from x , 2x is often called the derived function or derivative You need to know dy ___ The gradient function of a general curve y f(x) is denoted by f(x) or by dx The concept How to of a limit manipulate algebraic f(x) fractions limit of sin The as The factor formulae → 0 x f(x x, δx, f(x δx) δx) f(x) f(x)) δx x O dy ___ 2 So for y x , we write f(x) 2x or 2x dx f(x x) f(x) ______________ For any curve y f(x), gradient of AB x f(x x) f(x) ______________ and the gradient at A lim x x→0 f(x x) f(x) ______________ i.e. f (x) lim x x→0 We it can is use called this general formula differentiation from to ﬁnd ﬁrst the gradient of any function, principles Example 1 __ Differentiate from ﬁrst principles. 2 x 1 __ f(x) , 1 ________ so f(x x) 2 2 x (x x) 2 1 ________ f(x x) − f(x) 2 x 1 __ (x 2 (x 2xx 2 x (x) _____________ 2 x) 2 x) _____________ x 2 (x 2 x) x 2 (x 2 f(x x) f(x) 2xx ______________ f(x) lim x→0 2x x __________ lim when y (x x) dy 1 __ x 2 __ ___ , 2 x 132 3 dx x __ 2 4 x x(x 2 2x ____ 2 x→0 lim 2 x x→0 (x) _____________ 3 x x) 2 x x) and Section 3 Calculus 1 Example Differentiate f(x) f(x sin x, x) − sin x so f(x f(x) from ﬁrst x) sin (x principles. sin (x x) − x) sin x Using 1 2 cos (x x 2 1 2 cos (x 1 x) sin 1 x cos (x 2 2 ____________________ f(x) lim 1 x) sin x 2 2 __________________ x x→0 a factor formula 1 x) sin 2 lim 1 x→0 x 2 1 Now as x → 0, cos (x x) → cos x and, provided that x is in 2 1 sin x 2 _______ radians, → 1 1 x 2 f(x) cos x dy ___ i.e. if y sin x, then cos x dx Example __ √ Differentiate the curve f(x) at xl the from point ﬁrst principles where x and hence ﬁnd the gradient of 4 __ 1 √ xl x 2 1 1 f(x x) f(x) (x x) 2 x 2 1 1 ((x x) 2 x 2 1 1 )((x x) 2 x 2 ) _____________________________ 1 1 ((x x x x) 2 x 2 ) x _______________ 2 Using ((x x) 1 1 2 2 x (a b)(a b) 2 a b ) x _______________ 1 1 ((x x) 2 x 2 ) x 1 _________________ f(x) lim x→0 x((x dy 1 if y x 2 , dy 4, x 2 ) x→0 (x x) 2 x x 2 1 2 2x 2 2 x 2 2 1 1 __ ___ x 1 1 __ ___ 1 1 1 dx When 2 lim 1 __ ___ i.e. x) 1 _____________ 1 1 dx 1 __ (4) 2 2 4 Exercise 3.5 Differentiate the following functions from ﬁrst principles. 1 1 f(x) kx 2 f(x) cos x 3 f(x) 5x where k 4 f(x) 5 (x 6 f(x) x 2 2x 1) 3 sin 2x 133 3.6 General Learning outcomes differentiation Differentiation of a constant y The To derive rules for equation y c, where c is a constant, differentiating represents a straight line parallel to the simple functions y c c x-axis. Therefore the gradient of the line is zero, dy i.e. when y x O ___ You need to know c, 0 dx dy The equations of straight ___ lines For example, when y 5, 0 dx dy ___ The meaning of dx y Differentiation of ax y The equation represents a y line ax, where through a the is a constant, origin y with gradient a a dy ___ Therefore when y ax, a dx x O dy ___ For example, when y 4x, 4 dx n y Differentiation of The table shows some of the x results from T opic 3.5: 1 3 2 x y x x 1 2 2 x x f(x) 1 dy 1 __ 2 ___ 3x 2x 2 2 x 3 x 2x 2 dx These that results power suggest and that reduce to the differentiate power by a power of x, we multiply by 1, dy ___ n i.e. when y x , n 1 nx for all values of n dx dy example, when y x dy ___ 10 For , 9 10x ___ 4 and when y x , dx 5 4x dx n Differentiation of y ax 3 The result from Exercise 3.5, question 3, shows that when y dy ___ 2 15x 2 5 × 3x dx This is a particular example of the general result, dy ___ n i.e. when y ax n , 1 anx where a is a constant. dx dy 1 example, when y 4x 2 , 134 1 dx 3 1 ___ For 4 × 1 x 2 2 2x 2 5x , Section y Differentiation of The result from Exercise f(x ) 3.5, Calculus 1 g( x ) question 3 5, shows that when dy ___ 2 y x 2x 1, 2x 2 dx This is dy the d ___ ___ same 2 (x dx as differentiating d ___ ) (2x) dx each term separately, i.e. d ___ dx (1) dx d d ___ The notation ___ f(x) means the differential of f(x) with respect to x, i.e. dx This result is f(x) f(x) dx true for the differential of the sum or difference of any functions, dy d ___ ___ i.e when y f(x) g(x), dx T wo other results from T opic 3.5 are d ___ f(x) dx g(x) dx important: dy dy ___ when y sin x, ___ cos x and when y cos x, dx All these We have results used are the important letters sin x dx y and and x for you the need to remember variables, but any them. letters can be ds ___ used, for example when s 2t cos t, 2 sin t dt (Letters s and concerned t are with often used for displacement and time in problems movement.) Example 3 dy 6x 5x 2 _____________ ___ Find when y Exam tip dx 3x d ___ 3 f(x) 3 6x 5x 2 _____________ 6x ____ 5x ___ 2 ___ 3x 3x f(x) d ___ 3x 3x dx dx ____ ( g(x) ______ ) d ___ g(x) 5 2 2x 2 3 dx 1 x 3 dy 5 2 y 2x 2 ___ 1 x 3 3 so 2 dx 4x 2 x 3 Exercise 3.6 dy ___ Find when y is: Exam tip dx 4 1 5 5x 4 3 cos x d d ___ f(x) dx d ___ g(x) f(x) dx ___ g(x) dx 3 2x 3 ________ 1 __ 2 6 2 x x 3 3 4x 4 (3x 2x 4)(2x 5 1) 7 x(2 8 5 sin x √ x) 4 cos x 135 3.7 Product Learning outcomes rule The If To for derive and use y differentiating derive uv, where a product and use if x f(x) and v product of functions g(x) is a small increase increases in in y, the u value and v, of x, and y, u and v are the then a formula for y a u a of corresponding differentiating rule rule for differentiating and To quotient a formula functions and quotient y (u uv u)(v v) of u(v) v(u) (u)(v) functions As y uv, this y simpliﬁes u(v) to v(u) (u)(v) You need to know Dividing How to differentiate sums by x gives y and How to of powers differentiate of sin x u x x and y lim The limit x→0 dy u x dv ___ v ___ , x v ___ x ___ lim dx x→0 x du ___ u ___ , lim dx x→0 and x lim u dx 0 x→0 theorems dy y ___ Trigonometric lim dx identities dv ___ ___ Therefore v x ___ Now cos x u ___ v ___ ___ differences ( ) u x x→0 (uv) u dx Y ou may T o ﬁnd it easier the to a ﬁrst , dx du ___ v dx differentiate multiply v dv ___ d ___ i.e. du ___ dx dx remember product of function this r ule in words: functions, by the differential of the second function then of add the the second function multiplied 2 For by the differential ﬁrst. example, if y x 2 cos x, then using u x and v cos x dy ___ 2 −x 2 sin x (cos x)(2x) 2x cos x − x sin x dx Exercise 3.7a dy ___ Use the product rule to ﬁnd when y is: dx 1 3 (a) x (b) sin x cos x sin x 2 3 (d) x (e) (x (3x 4) 2 1 __ (c) sin x x 136 1) cos x gives Section The rule for differentiating 3 Calculus 1 a quotient of functions u __ If y , where u f(x) and v g(x) v and if x is a small corresponding increase increases in in y, the u value and of x, and y, u and v are the v, u u _______ then y y v u __ y v u u _______ , y u __ v v v v vu uv __________ 2 v Dividing by x vv gives u ___ v ___ v y u x x __________ ___ 2 x v y , x dv ___ v ___ ___ lim x→0 vv dy ___ Now lim dx x x→0 lim x→0 du ___ y ___ Therefore lim dx ( ) 0 2 dy dv ___ u dx dx ___________ ___ v du ___ y v u v if lim x→0 u __ i.e. and dx dx dx __________ x x→0 x dv ___ v dy ___ du ___ u ___ , dx then 2 v Y ou need to important. computer dx remember One way monitor – to this v r ule. remember ‘visual The it display is order by unit’ in the using or numerator the VDU. old So, word VDU sin x _____ For example, if y is for a comes ﬁrst. 2 , then using u sin x and v x gives 2 x 2 dy x cos x 2x sin x ________________ ___ x cos x 2 sin x _______________ 4 dx 3 x x 2 Alternatively, writing y as a quotient, i.e. y x sin x, dy ___ product rule gives 3 2x x using the x cos x 2 sin x _______________ 2 sin x and cos x 3 dx The disadvantage simpliﬁcation of of the x writing result a is quotient often as a product is that the complicated. Exercise 3.7b dy ___ Use the quotient rule to ﬁnd when y is: dx sin x _____ (a) tan x ( x ______ ) (d) 2 cos x x 1 2 x ______ (b) (e) x cot x 1 cos x _____ (c) __ √ x 137 3.8 The chain Learning outcomes rule Differentiating When To derive the chain rule and it to differentiate is a composite If write y x is a small increase y y The meaning of a x → 0,y and composite u the in also dy y y of lim ( where u f(x), we x and y and u are the then zero. y x→0 lim x ( x y u ___ ___ ( lim ) dy product du dx dy dy ___ du ___ ___ i.e dx is known composite as the function chain gf( x) by ) x u→0 du ___ ___ of functions This ( lim u u→0 a ) u x→0 simple differentiate quotient u ___ ___ ) functions and of u, approach to value and ___ dx differentials How gf(x) x ___ u Therefore The y theorems function when u ___ ___ x As in increases ___ You need to know limit i.e. g(u). corresponding The function, composite functions composite function to can use y a du r ule and making dx can the be used to differentiate substitution u 6 For example, when y dy ___ (2x 3) 6 , using u 2x 1 gives y u du ___ 5 then a f(x) 6u and du 2 dx dy dy ___ dy du ___ ___ ___ dx du 5 gives dx 5 6u 2 12u dx dy ___ Substituting 2x 1 for u gives 5 12(2x 1) dx Y ou f(x) will not when y usually be given the substitution, so you need gf(x) Example _______ Find f(x) when f(x) 2 √ 2x 5 _______ 1 Let y 2 √ 2 2x 5 then if dy dy 4x dx du dy u 2 dx 2 2 4x √ u 2x ___ u f(x) 2x _________ _______ √ 2x 138 2 − 5 1 ____ √ y 2 1 ____ ___ dx 5, 1 du gives − 1 __ and du ___ ___ 2x ___ dx ___ dy du ___ Therefore u √ u √ u u 2 to recognise Section 3 Calculus 1 Example __ Differentiate sin 3 ( with ) respect to 4 When is the variable, we replace x __ Let y sin 3 ( with __ and ) u 3 4 dy with d y sin u dy du ___ d then 4 d ___ du ___ ___ Using ___ 3 and d du cos u gives du dy ___ (cos u)(3) 3 cos d __ 3 ( ) 4 After a bit of practice the substitution For example, with mentally you could simpler and go functions, write straight down you the should differential be able to make directly. from dy ___ 3 y 2(3x 1) to 2 (2)(3)(3)(3x 18(3x 1) dx 2 1) Example dy 2 _______ ___ Find when y dx We (x could use the 3) quotient rule for this, but by writing the equation as 1 y 2(x 3) y 2(x 3) , we can use the chain rule: i.e. dy ___ 1 ⇒ 2 (2)( 1)(x 3) dx dy 2 ________ ___ ⇒ 2 dx (x 3) Exercise 3.8a 1 Differentiate each function with respect to x. 4 (a) (3x 3 4) (d) (2 (c) sin 2x x 1 _____ 3 (b) cos x) (e) sin x 4 2 Differentiate 3 Find cos dy respect to . 1 ________ ___ ______ when dx with y √ x 2 1 139 Section 3 Calculus 1 T o differentiate the more complicated functions, it is sensible to write down substitutions. Example dy ___ __ 2 Find given y (3x 2) sin 2x ( ) dx This 3 is the product ﬁnding the differential product of two of composite each functions. composite t (3x − and by apply the u 3x 2 so t u du ___ 2u and du 3 dx dt ___ dt ___ then du ___ dx du dx 6u 6(3x − 2) __ s start then 2 2) dt ___ ⇒ Let will and rule. 2 Let We function sin 2x ( __ and ) u 2x ( du ___ so ) 3 s sin u 3 ds ___ ⇒ 2 and dx cos u du ds ___ then ds ___ dx du ___ du dx (cos u)(2) 2 cos __ 2x ( ) 3 Now using dy the ds ___ ___ t dx product rule gives dt ___ s dx dx __ 2 (3x 2) × 2 cos 2x ( __ ) sin 2x ( ) 3 × 6(3x __ 2(3x − 2) (3x − [ 2) cos 2x ( We of can extend three where the functions, u dy f(x) and dy ___ chain i.e. 3 sin 2x ( )] then dv 3 rule rule where dv ___ ___ dx chain to y using cover functions hgf(x), the by extended du ___ du dx __________ 2 For example, when y √cos(x 1) , 1 2 then 140 u x 1 and v 2) __ ) 3 Extending the − 3 cos u so y v 2 that using y version are of a h(v) the composite and v chain g(u) rule, i.e. Section dy 3 Calculus 1 1 ___ 1 then 2 v ( sin u) 2x 2 dx 1 1 2 cos u ( sin u) 2x 2 2 x sin(x 1) ____________ __________ 2 √cos(x 1) Example 1 ____________ Differentiate 2 sin (3x 1) 1 ____________ First we can write 2 as sin (3x 1) 2 sin (3x 1) 2 Now sin where (3x f(x) 1) (3x − hgf(x) 1), 2 g(x) so we sin x need and two h(x) x substitutions. 2 Let u y sin 3x and v (3x 1), 1 sin u 2 then y v dy dy ___ then dv ___ ___ dx dv du ___ du dx dy ___ gives 3 2v × cos u 6(sin u) × 3 dx 3 cos u 6 cos (3x 1) ____________ 3 sin (3x 1) Exercise 3.8b ______ 4 1 Differentiate 2 Find (2 x 1) √ 2 x dy 1 with __ ___ when y cos 2 ( to x. 2 ) d respect sin 4 5 3 Find 4 Find f(x) when f(x) sin [(x 4) ] _____________ dy ___ 4 when y √1 (3x 1) dx 141 3.9 Parametric and general differentiation dy d x ___ Learning outcomes ___ The formula 1 d x dy This is a formula we dy need to be able to differentiate a parametric ___ To ﬁnd when the equation of equation dx a curve is given parametrically dy y ___ When y f(x), x ___ ___ then dx lim x→0 dy y → 0 as x → 0 1 lim dx ( y→0 meaning of parametric y ) y dx ___ ) 1 dy dx ___ ___ The x ___ so dy 1 ( x→0 ___ But You need to know lim x i.e. 1 dy dx equations How to differentiate simple Parametric differentiation functions dy ___ When The product rule, quotient the chain The limit of a curve is given parametrically, we ﬁnd in dx of the parameter . rule dy dy ___ can rule terms and equation theorems If x f(t) and y g(t), then the chain rule dx dt ___ Using the formula above, dy dy ___ 1 becomes dx dt , dt ____ dx f(t) 1 _____ 2 For dx ___ ___ dx g(t) dy ___ then dy ___ dt dx dt dy dt ___ ___ dx dt dx ___ dx so dt ___ ___ gives example, when x t and y then t dy dx ___ 2t 1 1 _______ ___ and Using the quotient rule 2 dt dt dy dy ___ 1) dx ___ ___ (t dx dt dt 1 _______ 2t 2 (t 1) 1 _________ 2 2t(t 1) Exercise 3.9a dy ___ Find in terms of the parameter when: dx 2 (a) y t 1 __ 3 , x 1 (c) t x 2 , y (1 t) t (b) y cos , x 2 sin General differentiation In the previous functions are 142 and topics derived summarised in we have rules the for table. given the differentials combinations of of various functions. types These of results Section 3 Calculus 1 dy ___ y General results dx d ___ c (a constant) 0 (f(x) g(x)) f(x) g(x) dx n d ___ n1 x nx (af(x)) af(x) dx dy n dy ___ n1 ax du ___ ___ anx dx du dx dy dx ___ ___ sin x cos x 1 dx dy d ___ cos x dv ___ (uv) sin x du ___ u dx v dx dx du ___ v d ___ 2 tan x sec x ( dx Any of asked The these can be used u ) 2 v v directly unless their derivation is for . next applying some results dv ___ dx dx __________ u __ exercise the gives most functions practice direct may fall in method into recognising to more ﬁnd its than the type of differential. one category, function Remember so two rules and that may 3x ________ be needed. For _______ example, √ 4x a composite some function. expressions is Remember before a quotient and the denominator is 1 also that differentiating you may be able to simplify them. Exercise 3.9b dy ___ 1 Find when y is equal dx to: ______ 2 √ 1 x ________ 2 (a) sin (x 1) 1 x ______ 3 (b) (1 (c) x ) sin x (d) 2 x 2 Differentiate each of the 1 following 3x ________ 3 _______ (a) √ 4x functions (e) (4 x with respect 5 x to x. 2 ) (i) x ( j) ( cos 2x 1 2 cos x _____ x 2 ___________ (b) x 1 ______ (f ) 2 sin x x 1 _____ 4x 4 x ) 1 x 2 ___________ (c) (g) 2 cos x x 4x 4 sin x _____ (d) (1 2x) tan x (h) x dy ___ 3 Find in terms of the parameter when: dx 2 (a) x (b) x t 1, y 2 (c) x cos , y 3 tan t 1 _____ 1 _____ , t y 2 1 t t 143 3. 10 Rates Learning outcomes To solve rates of of change Rate of increase change The problems gradient increase in of a straight line, y mx c, is calculated from y ____________ from increase in Therefore increase one point on the line to another point on the line. x the in x, gradient i.e. the measures rate of the rate increase of at y which with y increases respect to per unit x You need to know dy ___ gives The chain the gradient of the tangent at a point on the curve y f(x), rule dx How to differentiate dy simple ___ so is a measure of the rate at which y is increasing with respect dx functions to x For at that point example, at on the the curve. point y where dy ___ 2 x 2 on the curve y x 8 , 2x 4 dx So where x 2, y is increasing at the 6 rate in of 4 units for every unit increase x. 4 Note that because varies this the as x is rate only at true which where y x 2 changes varies. 2 dy ___ At the point where x 2, 4 dx The negative decreasing in sign at 4 shows units per that y is unit increase 4 x O 2 2 4 x Connected The chain variables variable rule y u rates of is and and useful x, we we change when know want to we the ﬁnd know rate the of the relationship change rate of of y change with of x between respect with the to a respect to u Example 1 __ The equation of a curve is y 4 x A point P is increasing Find the moving at rate the at along constant which the the rate curve of so 0.01 that units x-coordinate is dy rate at which y is increasing is the 4 , dy dt 1 __ ___ therefore 2 x dx x dx ___ The rate of change of x with respect to t is dt 144 is second. when x 1 ___ so time. 1 __ y y-coordinate dy , dt is per increasing ___ The the 0.01 where t seconds Section dy dy ___ dt ___ ___ Using dt ___ dt dx dx Calculus 1 100 ____ 1 __ gives 0.01 2 dx 3 2 x x 2 dx ___ Now dt ___ using dx ___ 1/ dt x ____ gives dx dt 100 dx ___ When x 1, 0.01 dt Therefore when x 1, x is increasing at the rate of 0.01 units per second. Example 3 Air is per second. Find leaking the rate out of a spherical of change of a of the balloon radius at the when constant the volume rate of of the 0.3 cm balloon 3 is 36 cm 4 (The volume sphere is V 3 r ) Give your answer correct to one 3 signiﬁcant ﬁgure. 3 ___ When V 36, r 3 √ dr ___ We require when the volume is decreasing at the rate of dt dV ___ 3 0.3 cm per second, i.e. when 0.3 dt 4 From V dV ___ 3 r , dV ___ dt ___ dr 0.3 × dt ___ ⇒ dr dt ___ dt ___ 2 4 r gives dt dr ___ Then 4 r dr dV ___ Using 2 3 40 __ dr 2 r 3 dr 3 ______ 1/ 2 dt 3 ___ When r 40r dr 1 ______ dr ___ , 1 3 √ The radius dt is 0.006 3 120 decreasing at the rate of 0.006 cm per second. Exercise 3.10 1 The area of a circular oil slick on a lake is increasing at the rate of 2 2 m 2 per second. Find the rate 3 m. Give The of your equation change answer of of the radius correct to of three y 2 sin . increasing at the a curve is the slick when signiﬁcant A point is the radius is ﬁgures. moving along the _ 1 curve so that is constant rate of radians per 4 7 _ second. Find the rate of change of y when 6 3 A right circular contains grain, cone has which is its axis pouring vertical out of and a its hole in vertex the downwards. vertex at 50 cm It rate _ 1 3 of the per second. The semi-vertical angle of the cone . is Find 6 the of rate the of change circular of the surface of height the of grain grain is in the cone when the radius 2 m. 145 3. 11 Increasing Learning outcomes and decreasing functions Increasing and decreasing functions dy To determine whether ___ a function The value of at any point on a curve whose equation is y f(x) dx is increasing or decreasing measures the rate at which y measures the rate at which the to is increasing function as f( x) x is increases, increasing to ﬁnd the rate of change How to of The y a function differentiate respect 3 Consider , How f (x) with x You need to know i.e. for graph f(x) example, shows and y the the function relationship given by between f( x) the x 3x 2 curves f(x) y simple 2 f(x) 3x 3 functions How to differentiate products, 5 quotients and composite functions 3 f(x) How to rational solve quadratic x 3x 2 and inequalities x O 3 2 5 This the graph x-axis above the shows (i.e. that f(x) x-axis (i.e. f( x) 0), f(x) Therefore and So, to to determine determine decreasing f(x) a f (x) is as x increases increasing as x 0 0 when when function is is positive f(x) f(x) is the range of values is of x f(x) increasing or or decreasing, negative. for which the function 3x 2x 4 decreasing. 2 f(x) ⇒ 3x 2x f(x) 6x f(x) 0 4 2 1 when 6x 2, i.e. when x 3 1 Therefore f(x) is decreasing for values of x less than 3 146 is below f (x) is increasing 2 is f (x) where decreasing. Example Find where increases 0). f(x) f(x) whether whether is and we need Section 3 Calculus 1 Example 2x ___________ Determine the range of values of x for which f( x) is increasing. 2 x 5x 4 2x ___________ f(x) 2 x 5x 4 2 2(x 5x 4) 2x(2x 5) __________________________ ⇒ f(x) 2 (x 2 5x 4) 2 2x 8 _____________ 2 2 (x 5x 2(2 4) x)(2 x) _______________ 2 [(x f(x) is zero x 4 x 4, 1)(x when and x 4 x 1 x 4)] 2 so and we 2, x 2 need to investigate 2 x and f(x) 1, 1 is undeﬁned the x sign 2 of when f (x) and x when 2 2 Note The that table [(x 1)(x shows the x 2 2 4)] sign of 0 for f (x) 4 all for 4 values the x of x different 2 ranges 2 x of x 1 1 x 2 x x x 2 2 [(x 1)(x 2(2 x)(2 4)] x) _______________ 2 [(x 1)(x 4)] Therefore f(x) hence is f(x) 0 for increasing 2 for x 2 1 x and for 1 1 and for x 1 2, x 2 Exercise 3.11 1 Find the range of values of x for which the 3 f(x) is 2 x function given by given by 2 4x 4x increasing. Find the range of values of x for which the function 2x ________ f(x) 2 (1 is x) decreasing. 147 3. 12 Stationary values Learning outcomes To deﬁne and ﬁnd Stationary values stationary We have seen in T opic 3.11 that when f (x) is positive, f( x) is increasing values as x increases, and that when f (x) is negative, then f( x) is decreasing as x increases. You need to know There may f(x) neither The How to differentiate is value be of How to f( x) at nor such Therefore differentiate and f( x) is neither negative a point is but is called increasing equal a to nor decreasing so zero. stationar y value f(x) 0 ⇒ f(x) has a stationary value. products, Consider quotients where positive simple functions points the graph of y f(x). y composite At the points A, B y, is and C, f( x), functions and therefore increasing Therefore nor the neither A decreasing. values of y at C these points are stationary values, B dy ___ i.e. 0 ⇒ y has a dx stationary The value. points stationar y At on the the graph where y has a stationary value are called points stationary i.e. x O points, tangents are the gradients parallel to of the the tangents to the curve are zero, x-axis. y or f(x) has a stationary value dy ___ Therefore at a stationary point , { or f(x), 0 dx the tangent parallel to to the the cur ve is axis Example Find the stationary 3 f(x) x f(x) x 3 At values of the function given by 2 2x x 1 x 1 2 2x stationary 2 values, ⇒ f (x) f(x) 3x 4x 1 0, 2 ⇒ ⇒ 3x 4x (3x 1)(x 1 0 1) 0 x 1 1 ⇒ x or 3 1 When x , f(x) 3 __ 1 2 27 9 23 __ 1 1 3 and when x 27 23 __ Therefore the stationary values of f( x) are 1 and 27 148 1, f(x) 1 Section 3 Calculus 1 Example 1 ___________ Show that the curve whose equation is y has no 2 x stationary 1 ________ 1 2 ________ ___ ⇒ 2 2 x dy 1 ___________ y 2x points. 2x (x 1 3 1) dx (x 1) there are dy ___ At stationary values, 0, but no values of x for dx 2 ________ which 0, 3 (x 1) 1 ___________ therefore the curve y has no stationary values. 2 x 2x 1 Example x ________ Show that the curve whose equation is y has only one 2 (x stationary point and x 1 ________ ___ ⇒ 2 (x 1) it. dy x ________ y ﬁnd 3 1) dx (x 1) dy ___ The curve has a stationary point where 0, dx x 1 ________ i.e. where 0 3 (x 1) x 1 ________ There is only one value of x for which 0, i.e. x 1 3 (x Therefore the curve has only one 1) stationary point. 1 When x 1, y 1 , so the stationary point is the point ( 1, 4 ). 4 Example 2 The curve Find the y ax values of a bx and 8 has a stationary value of 5 when x 1. b dy ___ 2 y ax bx 8 ⇒ 2ax b dx y has a stationary 5 and 2a Solving a [1] b b and value 8 of 5 ⇒ a when b x 1, 3 [1] 0 [2] [2] simultaneously gives a 3 and b 6 Exercise 3.12 2 1 Find the stationary value 2 Find the stationary value 3 by f(x) x of of the function given the function given by f( x) x 5x 1 2 6x 12x 2 2 x ______ 3 Find the stationary points on the curve y x 4 Show that there are no stationary points on the 1 curve 2x ___________ y 2 x 5x 4 149 3. 13 Determining the Learning outcomes Turning A To deﬁne maximum curve points stationary points points f(x) can have several stationary points and the shape of the and curve minimum y nature of and points close to one of these points belongs to three different types. of inﬂexion y To distinguish stationary between points A To of deﬁne the second differential a function C You need to know B How to differentiate simple x O functions How to ﬁnd How to stationary determine a function is values whether increasing Moving along the curve in the positive direction of the x-axis: or 1 Near the point A, the gradient of the curve goes from positive through decreasing zero to negative. A The 2 Near zero value the to of point y is at B, 3 At a C value the point here, A the is a maximum called gradient the of tur ning maximum the curve point. value goes of from y (or of negative f(x)). through positive. B The called of y is at gradient on but the the called A is zero curve sense is a called but moves of minimum the the minimum gradient through curvature tur ning C. does does The point. value not curve change of (or change does from y of sign not f(x)). as turn clockwise to anticlockwise. A point to There are changes, the Note and at that least a cur ve a other at a A point turning the terms which (and in and of the vice the B is is diagram and one is where between not changes called a the B from point sense and necessarily of C. zero. clockwise inﬂexion. of curvature Therefore However the zero. maximum refer cur vature versa) inﬂexion point They values. at points between value. stationary 150 two one gradient gradient on anticlockwise only and to minimum the do behaviour not of a mean greatest function close value to its Section Distinguishing There are three between ways of 3 Calculus 1 stationary values distinguishing between stationary values. y A A A 1 2 C C 2 B B 2 1 C 1 B x O First Look For method at A the (a points on maximum either side of, value): y and at close A to, the y at A y at A stationary values. 1 y at A 2 For B (a minimum value): y at B y at B y at B 1 y at B 2 For C (a point of inﬂexion): y at C y at C y at C 1 y at C 2 We can summarise these observations in Maximum V alues either the of Both y side a table: Minimum smaller Both Inﬂexion larger One one of smaller and larger stationary value Second For this method method stationary we look at the gradient on either dy A (a of, and close to, maximum value): ___ at A 0, ___ at A 0, at dy (a minimum value): dx dy dy ___ at B 0, ___ at B 0, at B 1 dx (a point of inﬂexion): dy ___ at C 0, ___ at C 0, at 1 results are summarised in Maximum the C 0 2 dx These 0 dx dy dy ___ C 2 dx For 0 2 dx ___ B A 1 dx For the dy dy ___ For side values. dx dx table: Minimum Inﬂexion dy ___ Sign of 0 0 0 or 0 dx either and side, at, stationary a value 151 Section 3 Calculus 1 Third For A method (a maximum value): As a point moves along the curve dy ___ through A, then as x goes increases dx from positive to negative dy ___ so decreases as x increases. dx dy ___ Therefore is a decreasing function dx at For B (a minimum value): A. As a point moves along the curve dy ___ through B, then as x increases goes dx from negative to positive dy ___ so increases as x increases. dx dy ___ Therefore is an increasing function dx at For C (a point of inﬂexion): A. As a point moves along the curve dy ___ through C, then as x increases goes dx from again positive to to zero positive then increases values. dy ___ Therefore itself has a stationary dx value at C. Second derivative dy dy ___ The rate of change ___ of is the derivative of with dx respect to x. This is dx 2 d y ____ called the second derivative of y and is denoted by or when y f(x), 2 dx by f(x). 3 For example, when y x 2 2x , dy ___ 2 3x 4x and dx 2 d y d ___ ____ 2 (3x 4x) 2 dx dx 6x 4 Returning points, to we the can third now method express for the distinguishing observations between above in derivative. 2 dy d ___ At A, y ____ is a decreasing function therefore 0 2 dx dx 2 dy d ___ At B, y ____ is an increasing function therefore 2 dx dx 2 dy d ___ At C, y ____ has a stationary value, therefore 2 dx 152 dx 0 0 stationary terms of the second Section 3 Calculus 1 2 d y ____ This is the easiest method to use. However can also be zero at a 2 dx 2 d y ____ maximum or minimum value, so when 0 either the ﬁrst or second 2 dx method These has to results be used are to distinguish summarised in the between stationary values of y table: Maximum Minimum 2 d y ____ Sign of at Negative a Positive 2 dx (or stationary zero) (or zero) value Example 4 Find the stationary distinguish points between on the curve y 3 3x 4x 2 and them. dy 4 y ___ 3 3x 4x 2 ⇒ 3 2 12x 12x dx dy ___ At stationary points, 3 0 ⇒ 12x ⇒ x ⇒ x 2 12x 0 dx 2 When x Therefore 0, y (0, 2) 2 and and (1, when 1) are x (x 1, 1) 0 y stationary or 0 x 1 1 points. 2 d y ____ 2 36x − 24x 2 dx 2 d y ____ When x 1, 36 24 0 2 dx (1, 1) is a minimum point. 2 y d ____ When x 0, 0 which is inconclusive, so we look at the signs 2 dx dy ___ of each side of x 0 dx dy ___ 1 When x 1 , 2 dy x , Therefore (0, 12 ___ 8 4 point of dx 1) is 12( a 2 ) 0 2 12 ___ 2 1 3 ) 2 ___ 1 When 12( dx 0 inﬂexion. Exercise 3.13 Find the them stationary when 3 (a) x f( x) points on the 2 3x curve 3 y f(x) and distinguish between is: 3x 1 (b) x 2 2x 4 x 2 (c) (x 2) 153 3. 14 Curve sketching Learning outcomes Features to T o To use a variety of techniques sketch curve whose when shape is sketching unknown, we curves can often ﬁnd several to features sketch a look for by observation and by calculation from the equation of the curve. curves The main features where the stationary vertical the range the behaviour where to curve look for crosses are: the axes You need to know How to ﬁnd stationary and distinguish The meaning How to ﬁnd How to points between of an points and horizontal asymptotes them of values of y asymptote of y as x → ∞ limits solve rational the function is increasing or decreasing. inequalities Not How to convert an rational function improper to a proper A all of picture diagram these of as features the you curve ﬁnd need can be to be built considered up by for marking a particular these curve. features on them. rational function How that to ﬁnd a the range of rational function values can take Example x 3 ______ How to determine whether Sketch the curve whose equation is y x a function is increasing 2 or decreasing First ﬁnd where the curve crosses the axes: 1 when x 0, y 1 and when y 0, x 3 2 1 Therefore the curve goes through the points (0, 1 ) and (3, 0). 2 Now look for asymptotes: 1 ______ as x → ∞, 1 → x The value of y is 1 so the line y 1 is an asymptote. 2 undeﬁned when x 2 so the line x 2 asymptote 1 ______ and lim x→2 These (1 1 ______ ) x ﬁndings are ∞ and lim 2 shown (1 x→2 on the ) x ∞ 2 diagram. y 6 4 2 x O 4 2 4 2 4 6 154 is an a Section Next investigate stationary 3 Calculus 1 points: x 3 ______ y x 2 1 ______ 1 x 2 1 1 (x dy 2) 1 ________ ___ ⇒ 2 dx (x 2) no values dy ___ There are of x for which 0 so there are no stationary dx values. dy ___ Also 0 for all values of x except x 2, therefore y is increasing as dx x increases. Check the range of 2y x 3 ______ y ⇒ We x now y: 3 so 2 therefore of _______ x values y the curve have x has real values except when y 1, 1 does enough not cross the information to line y sketch 1 the curve. y 6 4 2 O 4 x 2 2 4 6 Example 2 (x 1) _________ Sketch the curve y x(2x Check when for x intercepts 0, 0, y is on 1) the undeﬁned. axes: Therefore the curve does not cross therefore the curve the y-axis. when y touches (This the must x 1. x-axis be a This at (1, is a repeated root, 0). stationary point.) 155 Section 3 Calculus 1 Check for asymptotes: 1 y is undeﬁned when x 0 and x , 2 1 therefore the lines x 0 and x are vertical asymptotes. 2 2 (x 1) ________ y 2 2x x 2 x 2x 1 ___________ 2 2x x 1 __ 3x 2 _________ 2 4x 1 __ 3x 2 __________ 2 2x(2x 2 2x 1) 3x 2 __________ 1 __ As x → ∞, ( 2 2x(2x as x → ∞, 1 __ ) → from ( 2 2x(2x below 2 3x 2 __________ 1 __ and 1) 1 __ → ) 1) from above. 2 1 __ Therefore y is a horizontal asymptote. 2 1 __ When y , 2 1 __ 1 __ 3x 2 __________ 2 2x(2x 2 3x 2 __________ 2 __ ⇒ 1) 2x(2x 0 ⇒ x 1) 3 1 __ Therefore the curve crosses the line y 2 __ where x 2 3 y 6 4 2 4 x O 2 2 4 2 4 6 This diagram shows the features found crosses the asymptote so far . 1 As the curve y only once, and does not 2 cross the vertical minimum asymptotes, it is clear that the point (1, 0) is point. 1 We can also deduce that y when x 0 2 1 and y 0 when x 2 We now check for other stationary points. 2 (x 2 1) 2(x dy _________ y ___ ⇒ 2 1)(2x x) (x 2 x(2x 1) dx x (x 1)(3x 2 (2x 1) ______________ 2 x dy ___ 1 dx 156 0 when x 1) 1 (4x 1) _________________________________ and 3 2 (2x 1) 1) a Section 3 Calculus 1 1 Therefore y has stationary points where x 1 and where x 3 2 2 ( ) 3 ________ 1 When x 1, y 0 and when x , y 3 1 ( We will use determine (The the values point and values their ) x either side of 3 the stationary points to nature. chosen the of 4 1 )( 3 must curve must be close be to the value continuous of between 1 1 5 __ 3 4 3 12 4 x at the those x stationary values.) 1 y 4 4 4 2 0 0 0 1 ( Therefore , 4 ) is a maximum point and (1, 0) is a minimum point. 3 We of x could for also check which information y to is the range increasing sketch the of values and that decreasing, y can but take we and now the have values enough curve. y 6 2 O 4 2 x 2 4 2 4 6 Exercise 3.14 Sketch the graphs of the following curves. 1 ______ 1 y 2 x x ______ 2 y 2 x 2 x ________ 3 y 2 (2 x) 157 3. 15 Tangents Learning outcomes and normals Equations of tangents and normals dy ___ To ﬁnd the equations of tangents We know that represents the gradient function of a curve whose dx and normals to curves including equation curves given whose equations is y f(x) are We parametrically can the therefore tangent When x to a, ﬁnd the y the curve, f(a) gradient at and any the of the given curve, point gradient at on that and the hence the gradient of curve. point is f (a). You need to know Therefore (a, How to differentiate How to y use quotient to the rule product and the is equation given of the tangent to the curve y f(x) at the point by simple functions f(a)) the rule, chain the rule f(a) The the f(a)(x nor mal point of to a) a curve contact of is a the line perpendicular tangent. differentiate functions Therefore to at the the tangent point and ( a, through f(a)) the 1 ____ gradient of the normal is , and the equation of the normal to the f(a) The relationship between the curve gradients How of to ﬁnd perpendicular the equation at the point y f(x) at the point ( a, f(a)) is given by lines of a 1 ____ y f(a) (x a) f(a) straight line ______ dy 1 ________ ___ For example, when y √ x 1, dy x 3, 2 √ x 1 1 __ ___ When ______ dx , dx so y the y 14 4x 4 5 gradient of the curve at the 4y x 5 4 1 __ point where x 3 is y 3 4 x 1 2 x O 2 1 1 When x 3, y 2, therefore the line through (3, 2) with gradient is a 4 tangent to the curve at the point (3, 2). 1 The equation of this tangent is y 2 (x 3) 4 ⇒ The is equation y ⇒ of the 2 4(x y 14 normal When the the curve dy 158 of a and curve expressed dt 5 point normals for curves whose parametric g(t) dx ___ ___ dx be x 4x equation can dy ___ by are this 3) Equations of tangents equations at 4y ____ dt f(t) as is given (f( t), as g(t)) x and f(t) the and y gradient g(t), a point function is on given Section Therefore the equation of a tangent to the curve can be given 3 Calculus 1 as g(t) ____ y g(t) (x f(t)) f(t) Then any particular value of t gives the equation of the tangent at that point. Similarly, the equation of a normal to the curve can be given as f(t) ____ y g(t) (x f(t)) g(t) Example Find the equation of the tangent and the normal to the curve __ x 3 cos , y 4 sin , at the point where 6 f( ) f() ⇒ so the 3 cos 3 sin equation g( ) and of a 4 sin g() and tangent to 4 cos this curve is 4 cos ______ 4 sin y 3 cos ) (x 3 sin (This is curve at a general any equation; point on the it gives the equation of a tangent to the curve.) __ Therefore the equation of the tangent at the point where is 6 given by √ 3 __ 4( ) 2 4( y √ ) 3( (x 2 √ 3 __ _____ 1 √ 4 3 ___ )) ⇒ y 2 3 3 ___ (x 2 1 3( ) 3 2 ) 2 The equation of a normal to this curve is given by 3 sin ______ 4 sin y 3 cos ) (x 4 cos (Again any this point is on a general the equation giving the equation of a normal at curve.) __ So the equation of the normal at the point where is given by 6 1 3( ) 2 y 4( √ 3 __ _____ 1 ) 3( (x 2 4( √ 3 3 ___ 3 ___ )) ⇒ y 2 2 √ 3 __ (x √ 4 ) 2 3 ) 2 Exercise 3.15 1 Find the equation of the tangent and normal to the curve whose 2 _______ equation is y at (x 2 Find the equations the point where x 7 3) of the tangent and y 1 normal to the curve whose 1 _____ equations are x , 1 t t 159 3. 16 Integration Learning outcomes Reversing differentiation 2 When To deﬁne reverse integration of as x is differentiation apply basic principles with respect to x the derivative is 2x and Therefore to differentiated the of when the derivative of an unknown function is 2 x then the 2 unknown function could be x integration This process operation of of ﬁnding a function differentiating, is from called its derivative, which reverses the integration You need to know The The differentials of constant of integration simple 2 functions We know 2 x that 2x is the derivative of x , but it is also the derivative of 2 5, x 3 2 In fact, 2x Therefore is the the derivative result of of x integrating c, where 2x is not c is any constant. a unique function but is of 2 the c form is x called c the where c constant is of any constant. integration 2 x ∫2x c is called the integral of 2x with respect to x and is where x ∫… function means any f(x) ‘the function we integral reverses of the … with process respect of to x’. differentiating, f(x) dx f(x) ∫3x differentiating 2 x x any 2 with respect 3 dx for c 3 example, so have ∫ so as c dx Integrating For written 2 dx c and it follows ∫x that to x gives 1 2 dx 3x 3 x c 3 1 (We do not need to write c in the second form, as c represents any 3 constant in either expression.) n Now we know 1 ______ n ∫ x that dx n the Therefore For rule can example, dx 1 n is (n 1)x to integrate a power be used ∫6x to integrate 6 __ 3 dx by x ∫3x of the any 3 __ 4 c 3x increase that power by 1 and power. power of x except 1. 4 x c, 2 0 dx x, new c and ∫4x 4 ___ 3 dx x 2 160 so c 4 ∫3 x 1 divide This of 1 x n derivative 2 2 c 2x c Section 3 Calculus 1 Exercise 3.16a Find the (a) ∫5x (b) ∫4x following integrals. dx (c) ∫6 (d) ∫4x 7 dx Families of dx ∫x (f ) ∫5x 3 dx 2 (e) dx 6 dx curves dy ___ When 2x, then y ∫2x 2 dx x c dx 2 Therefore Each The the value graph of equation c gives shows a y x different some c represents member members of this of a the family of curves. family. family. y 15 10 5 T o ﬁnd the equation the curve. For example, if of a y ∫3x y x x O 4 particular member , we need to know a point on 2 dx 3 then If c we also and know hence the that (2, c 5) equation is of a point the on the particular curve we can ﬁnd the value of curve. 3 When x 2 and y 3, y x c ⇒ 3 8 c ⇒ c 5 3 Therefore the equation of the curve is y x 5 Exercise 3.16b 1 Find the which 2 Find y the which y equation ∫6x ∫9x the curve that goes through (1, 5) the curve that goes through ( 1, and for dx equation of 2 of 2) and for 4 dx 161 3. 17 Integration of sums and differences of functions Learning outcomes Integration of a sum or difference of functions dy To deﬁne the integral of ___ sums When y f(x) g(x) we know that f(x) g(x) dx and differences of functions Therefore For You need to know it follows example, ∫(cos ∫(f(x) that x sin x) dx g(x) ) dx sin x f(x) cos x g(x) c c c dy The differentials of ___ simple When y f(x) g(x) we know that f(x) g(x) dx functions The the The meaning reverse of of meaning integration as Therefore of the constant of How to solve simultaneous For follows example, The effect curve by a on g(x) ) dx f(x) g(x) a pair 2 ∫(2x sin x) dx 2x) dx x x x cos x c and c of equations Integration of ∫(f(x) that 2 ∫(1 integration it differentiation the equation of translation parallel a multiple of a function a to dy ___ When y af(x) we know that the y-axis af(x) dx ∫af(x) Therefore For example, ∫6x af(x) 2 dx c ∫3(2x 2 3 ) dx T o ∫ (f(x) Note g(x) ) dx ∫(f(x) that / ∫ f(x) dx g(x)) dx 2x c summarise, is ∫ g(x) dx not equal and to ∫ (af(x) ) dx ∫f(x) dx Example ∫(x Find ∫(x 3 4 cos x) dx 3 4 cos x) dx ∫x 1 3 4∫ cos x dx dx 4 x 4 sin x c 4 Exercise 3.17a Find 162 the following integrals: 2 (a) ∫9x (b) ∫x(3x (c) ∫(5 dx 3 sin x 4) dx (Hint: 6 cos x) dx Multiply out the bracket.) / a∫ f(x) dx ∫g(x) dx Section 3 Calculus 1 Example 2 d y ___ The equation of a curve is such that 6x 14. T wo points on the curve have coordinates (1, 2) and 2 dx ( 1, 12). Find the equation of the curve. 2 d y ___ is the second derivative of y with respect to x, so we need to integrate twice. 2 dx dy ___ The ﬁrst integral gives dx 2 d y dy ___ 6x 14 y ∫(6x 14) dx ⇒ ∫(3x can to integrate now When x When x Solving again to ﬁnd an expression 3 [1] use 1, y 1, y and c) dx the 2, [2] y y 14x c of ⇒ 7x the points 2 6 12 8 simultaneously for y so we need to introduce another unknown constant. 2 x ⇒ 12 3 Therefore ⇒ coordinates 3x dx 2 14x 2 dx need We ___ ⇒ dx We dy ___ 2 gives c cx on d the curve c d, i.e. c d, i.e. 4 and d to c ﬁnd d c the d values 4 4 of c and d [1] [2] 0 2 x 7x 4x Example dy ___ The equations of a family of curves is given by 4x 3 dx Sketch the graphs of two members of this y family. 10 dy 8 ___ 4x 3 ⇒ y 2 ∫(4x 3) dx 2x 3x c dx 6 We can The use any simplest value is c of c to get one member of the family. 4 0 2 This gives y 2x 3x x(2x 3) which is a parabola that passes 2 3 through O and ( , 0 ) 2 Then any other value of c translates the curve 2 The sketch shows y a curve by c units up the y-axis. 2 1 O x 2 2x 3x and y 2x 3x 2 Exercise 3.17b dy ___ 1 The equation of is such that 3 cos x and the curve dx __ passes through the point , ( 4 ) . 2 Find 2 A the curve equation passes of the through curve. the points (0, 1) and (1, 6). The equation of 2 d y ___ the curve is such that 6x 2 dx Find the equation of the curve. 163 3. 18 Integration Learning outcomes using Integration When To use substitution to substitution we use using the substitution chain rule to differentiate a composite function the integrate 3 result is often a product of functions. For example, when y 4 (x 4x) , functions 3 then with u 4 x 4x so that y u , the chain rule gives dy ___ 3 4u 2 2 (3x 4) 3 4(3x 3 4)(x 4x) dx You need to know In The chain rule general, dy dy ___ The differentials of simple u dx f(x) and du ___ ___ if g(u) then using the chain rule gives du ___ du y g(u) dx dx functions Therefore, The integrals of using integration as the reverse of differentiation simple functions du ___ ∫g(u) dx g(u) c [1] du g(u) c [2] dx ∫g(u) Now Comparing [1] and [2] gives ∫g(u) Replacing g(u) by f(u) du g(u) dx ∫f(u) dx … du c gives du ___ ∫f(u) du dx du ___ Therefore … dx du ___ This means that ((a integrating function of ) u) with respect to x is dx equivalent to integrating (the same function of u) with respect to u. du ___ This means that the relationship ... dx … du is not an equation dx an identity For – example, it is to a pair of ∫3x ﬁnd equivalent 2 3 (x operations. 4 4) dx we can use the substitution 3 u x This 4 gives ∫3x 2 3 (x du ___ Then as 4 4) 3x , ... dx ∴ ∫3x 2 dx ∫3x ∫u 4 u dx 2 dx … du becomes 4 1 ... 3 x dx 2 du Since 5 u c 5 1 3 (x 5 164 2 dx 4 u du ___ 2 dx 5 4) c ... 3x dx ... du ... du or Section 3 Calculus 1 Example Use the substitution u sin x u sin x to 2 ∫cos ﬁnd x sin x dx du ___ ⇒ cos x, dx du ___ ∴ ... dx … du ⇒ ... cos x dx ... du dx Therefore ∫cos 2 x sin x dx ∫u 2 du 1 3 u c 3 1 3 sin x c 3 Example ______ Use the substitution u √ 1 ______ x to ( √ x 1 ∫ ﬁnd ) x dx ______ 1 du ___ u √ 1 x ⇒ 1 (1 2 2 x) and u 1 x 2 dx 1 du ___ ∴ ... 1 dx … du ⇒ … (1 2 x) dx ... du 2 dx ______ ⇒ ...dx ...2 √ 1 x du …2u du ______ Hence ( √ x 1 ∫ x ) dx ∫(u 2 2 1)(u)(2u) du 2 3 u 4 2 2u ) du c 3 5 3 2 ∫(2u 3 u 5 2 (1 x) 2 (1 5 x) 2 c 3 3 __ 2 (1 x) 2 (3(1 x) 5) c 15 ________ __ 2 3 2)√(1 (3x x) c 15 Exercise 3.18 1 Use the substitution u 2x 2 Use the substitution u x 3 Use the substitution u x 4 Use the substitution u tan to ﬁnd ∫cos 2 1 4 to to to ﬁnd ﬁnd ﬁnd 2x dx ∫6x ∫(x ∫sec 2 (x 3 1) dx 6 1)(x 2 4) du 2 tan d d ___ (Hint: 2 (tan ) sec ) d 165 3. 19 Calculus Learning outcomes and examine the area under a curve Calculus The To the process topics in Section 3 of this book all come under the umbrella name of of calculus ﬁnding the area under a curve Calculus studies You need to know Our The The meaning of deﬁnition limits of from differentiation in The We many mathematics, development over many of limits, change would that place. electronic look involve It exploration has of derivatives when the very rate different change and applications space devices. to the Apart and of in to integrals; change without predict almost the of obvious basically calculus. what will everything development from it varies. the It we tiny is used happen do to when today, microchips scientiﬁc uses, calculus principles Did you know? with world takes the covers As study things situations change found from ﬁrst the modern model is how branches of the up of build calculus centuries, but to the full range area will x- disciplines look the and of under now Consider the occurred a at area y-axes a A in as economics and graphic design. curve how and such the the the fundamental diagram vertical below line theorem enclosed through a of by value calculus the of is curve derived. y f(x), x the y deﬁning step Newton in was made by Sir Isaac y about Gottfried Wilhelm independently but about 8 the after contribution that to symbols introduce f(x) Leibniz made years 1665–1667 . same step, Newton. The Leibniz made that was are dy ___ still used today (including and ∫ ). dx A Leibniz is considered the founder of by modern many to be mathematics. x O We can strips ﬁnd as an shown approximate in the next value for this area by dividing it diagram. y y (x, y) y δx x O 166 x f(x) into vertical Section The area of The area is a typical then strip of height approximately y the and sum xa We write this as A This of y x ∑ between approximation x of all x the is Calculus 1 y x strips. xa y x where ∑ x0 values width 3 y x means the sum of all x0 0 and improves x as x a gets smaller , so we can write xa A lim x→0 y x (∑ ) x0 Considered putting We If now A is x, x, look the vertical this way, together at area line gives (i.e. a area different enclosed through a the under integrating) approach by the a the point corresponding curve different to curve ( x, small y) y is the ﬁnding on process that involves elements. f(x), the increase A. the x- curve, in A, and then y-axes a small and the increase in A y y (x, f(x) y) δA A y δx x x O A is approximately width i.e. equal to the area of the rectangle of height y and x, A y x A ___ ∴ y and this approximation gets better as x → 0 x dA ___ A ___ Now lim x→0 x dx dA ___ hence y dx This the gives the connection differential of A with between respect to ﬁnding x, i.e. A A as is a the summation reverse of a process and differential, hence A This is called the fundamental ∫ y dx theorem of calculus . 167 3.20 Deﬁnite Learning outcomes integration Deﬁnite We To deﬁne and compute know integration that the area between a curve y f(x), the x- and y-axes and the deﬁnite line integration through a value of x is given ∫ by y dx, i.e. by ∫ f(x) dx T o ﬁnd this area up to the line x b, we ﬁnd ∫ f(x) dx and substitute b T o ﬁnd this area up to the line x a, we ﬁnd ∫ f(x) dx and substitute a for x for x You need to know How to integrate simple Then the x is b area the between difference the curve, between the these x-axis two and the lines x a and calculations. functions y How to use substitution to y f(x) integrate functions A a O 2 For example, when y x x b ∫ 2, y dx ∫(x 1 2 2) dx 3 x 2x c 3 1 Then the area up to x 1 is 1 3 (1) 2(1) c 2 3 1 and the area up to x 2 is c c 3 2 3 (2) 2(2) c 6 3 3 2 Therefore and x the 2 between 1 c the curve y x 2, the x-axis and x 1 is 2 (6 area ) 1 (2 3 c ) 4 3 . Notice that the unknown constant disappears. 3 b This process is called deﬁnite integration . It is written ∫ as f(x) dx and it a means the integrand deﬁnite We can value when of x the integrand a. The when values a x and b b minus are called the value the of limits the of the integral write the example above more precisely as 2 (x 2 1 2 ∫ 2) dx [ 8 3 x 2x 3 ] ( 1 4 ) ( 3 1 1 2 ) 4 3 3 1 The unknown because it substituted The constant cancels and second out. the value is of integration The order then square in which does brackets they subtracted are from not need show to the the to be substituted ﬁrst included values of x (top to one be ﬁrst). value. b ∫ f(x) dx can be found only when f( x) is continuous from x a to x b x a 1 1 __ For example ∫ ( 1 168 1 __ ) x dx cannot be found because is x undeﬁned when 0 Section 3 Calculus 1 Example 2 ∫ Evaluate sin x dx 0 2 2 ∫ sin x dx [ cos x] 0 0 __ cos ( cos 0) 2 0 ( 1) 1 Exercise 3.20a Evaluate the following deﬁnite integrals: 2 1 3 ∫ (a) 2 4x dx ∫ (c) (x 1 4 ∫ (b) 3) dx ∫ (d) general need x (x 5) dx 1 Deﬁnite not dx 3 (2x 2 In 1) 1 to when we substitute need given integration to as do the use a back this for limits to substitution substitution to a using give the deﬁnite ﬁnd the u answer integral g(x) in to ﬁnd terms because corresponding of we an x. can values of integral, we However , we use the do values of u Example 2 3 Use the substitution u 2 x 2 to evaluate ∫ x 3 (x 4 2) dx 0 3 u x 2 { and 2 ⇒ … du x 0 ⇒ u 2 x 2 ⇒ u 10 2 ∫ dx 10 2 ∴ … 3x x 3 (x 1 4 2) dx ∫ 0 4 u du 3 2 10 __ 1 [ 5 ] u 15 2 100 000 ______ 32 __ 15 15 99 968 _____ 15 Exercise 2.20b 3 6 1 Use the substitution u x 1 to evaluate ∫ 7(x 1) dx 1 __ 4 2 2 Use the substitution u sin x to evaluate ∫ cos x(1 sin x) dx 0 ______ 1 2 3 Use the substitution u 2 1 x to evaluate ∫ x √ 1 2 x dx 0 169 3.21 Area under Learning outcomes a curve Calculating the From To calculate the area under T opics 3.19 area and 3.20 under we a curve y now a know that the area between the curve curve 6 y and f(x), x the b x-axis shown and in the the lines x diagram y a f(x) is 5 b the You need to know value ∫ of f(x) dx 4 a 2 How to integrate simple functions For example, the x-axis the and x area 1 between and x y 2 x 3 , is 2 2 How to evaluate a deﬁnite ∫ by 2 1 2 given x dx [ 8 3 x ] 3 7 1 3 1 3 3 b 1 integral 1 f(x)dx a How to sketch curves x O a b 1 Example Find the area The curve is The sketch enclosed a by parabola shows the the curve which area cuts y the required. (It (1 x)(2 x-axis is at x always x) and 2 the and sensible to x-axis. x y 1 draw a sketch.) 1 A ∫ (1 x)(2 (2 x x) dx 2 1 1 2 = ∫ x ) dx [2x 1 1 2 x 3 ] x 2 3 2 2 1 1 2 3 (2 8 ( ) 4 2 ) 3 1 4 2 3 2 1 x O 1 The area is 4 square units. 2 Exercise 3.21a Find the area lines given. 1 y 4x 2 y x 3 y √ enclosed by each of the following curves, the following curves the x-axis and 3 , x 0, x 2 2 Find x, the 1, x area 170 y 1 5 y x(1 1, 1, x enclosed 2 4 x x x) x 1 4 by each of and the x-axis. the Section The area between a curve and the 3 Calculus 1 y-axis 2 The the We diagram line y can First This shows the area between two different the curve y x 1, the y-axis and 10. ﬁnd this area in ways. method uses the bounding fact that rectangle the minus area the required area is equal between the to the curve area and of the the y x-axis. 10 When y 10, x 3 so the area shown is equal to 8 (area of the rectangle bounded by x 0, y 0, x 3 and y 10) 6 (area between the curve, the x-axis, x 0 and x 3) 3 4 2 Therefore the area required ∫ 30 (x 1) dx 0 2 3 1 [ 30 3 x x ] 3 0 x O 1 2 3 4 3 4 27 __ {( 30 3 ) (0) } 30 12 18 3 The area is Second This square units. method uses Consider The 18 area a a of direct approach. horizontal this strip strip is y of length x and approximately x y width y 10 x 8 δy y10 The area required lim y→0 ( x y ∑ y 6 ) y1 4 10 We know that this limit is equal to ∫ x dy 1 2 10 Now ∫ x dy means integrate x with respect to y so we need to ﬁnd x in terms x O 1 of 1 2 y ______ 2 From y x 1, x √y 1 ______ 10 9 3 1 9 2 so the required area is ∫ (√y 1 ) dy ∫ u 2 du [ 3 u 2 ] 18 square units Using the substitution u y 1 0 0 1 Exercise 3.21b 2 1 Find the y-axis area and enclosed the line y by the curve 3 y √ x, the 2 Find the y-axis y area and enclosed the line y by the curve y x , the 4 y 10 4 8 6 2 4 2 O x 2 4 6 8 10 O x 2 171 3.22 Area two below the x-axis The area between below the To calculate below To the areas of The areas between a curve and the x-axis that is axis curves x-axis calculate area curves Learning outcomes and deﬁnite y integral 2 between 2 1 3 ∫ curves x dx 4 [ ] x 4 1 1 3 1 4 3 4 This can be 4 interpreted as the area You need to know 3 between the curve y x , the x-axis How to integrate How The to ﬁnd the lines x area below deﬁnite and above 1 and x 1 2 1 2 2 a integral 1 the ∫ x-axis 1 1 3 curve simple functions the x O 2 and x dx 4 [ x ] 4 2 2 3 1 How to sketch curves 4 3 4 How to ﬁnd the points If intersection of two 4 of we look at the diagram, the y area curves 3 between and is the the curve lines equal to x the y 2 area x , and the x found x-axis 1 above, 3 i.e. 3 square units. 4 (The curve about has rotational symmetry O.) x O 2 1 3 But the ∫ integral x dx is negative. 2 This is gives because the the length of value a of y vertical that strip is b negative, so ∫ y dx will be negative a when This and y is negative means the you x-axis for need when a to x be part b careful of the when area is ﬁnding below the the area between a curve x-axis. Example Find the First draw is below area a the Therefore between sketch: x-axis we need the The and to curve sketch the ﬁnd y 3x(x shows area each that between area 2), x the 2 the x-axis area and and between x 4 is the x lines 0 above and the x 0 x 2 and x 4 y x-axis. separately. 2 2 A ∫ (3x 6x) dx [ 3 2 2 x ] 3x (8 12) 0 4 0 0 4 2 B ∫ (3x 6x) dx [ 3 2 x 3x B 4 ] (64 48) (8 12) 20 2 2 O x 1 Therefore the area required 20 4 24 square 2 units. A 4 2 Note that ∫ 0 172 (3x 6x) dx gives the value of the area of B minus the area of A 3 4 5 Section 3 Calculus 1 Exercise 3.22a 1 (a) Sketch (b) Find the the curve area y x(x enclosed 1)(x between 2). this curve and the x-axis. 2 2 (a) Sketch (b) Find Area When a curve area y you need an which idea are to This ﬁnd will where shown x the area any points the 1. by this curve and the x-axis. curves show the in enclosed between two diagram. give the the next between areas of two that curves, may intersection be are. you should below the There are draw x-axis two and methods, example. Example 2 Find the area between the curves y x and y 3x(2 x) y 4 2 The curves intersect where x 3x(2 6x 0 3) 0 x) 3 2 ⇒ 4x 2 1 ⇒ x(2x ⇒ x 0 or x 1 1 2 x 1 First 1 2 3 method 1 The area between y 3x(2 x), the x-axis and x 1 2 1 1 1 1 2 2 2 is given by ∫ (6x 3x ) dx [ 2 3 3x x ] 27 ___ 27 ___ 4 8 0 area between y x 8 1 2 The 27 ___ 0 0 , the x-axis and x 1 is given by 2 1 1 1 1 2 1 2 ∫ x dx [ 9 __ 2 3 x 3 ] Therefore Second the 9 __ 0 0 8 0 area 8 between the curves 27 ___ 9 __ 8 8 is 9 __ square units 4 method 2 A vertical strip in the area has length (y 6x 3x 2 ) (y x 2 ) 6x 4x 1 1 1 1 2 2 Therefore the area between the curves is ∫ (6x 4x 2 ) dx 4 [3x 0 2 3 x 3 ] 0 27 ___ 9 __ 4 2 9 __ 0 square units 4 Exercise 3.22b 2 1 Find the area between 2 Find the area enclosed by the y-axis enclosed by the curve the curves y 2 x and y 2 y x x 3 and the curves and 3 y 3 16 Find y the x area y (x 2)(x 2) and the line 3x 173 3.23 Volumes Learning outcomes To calculate when round part the the of a of revolution Volume of revolution volume formed curve is rotated x-axis When an object formed area is rotated about a solid of straight line, the three-dimensional You need to know How to integrate simple The volume is called formed is a called a revolution volume of revolution functions The How to evaluate a line about which rotation takes place is an axis of symmetry of deﬁnite the solid of revolution. integral All The formula for the volume of cross-sections of the solid that are perpendicular to the axis of a rotation are circular. cylinder The is diagram rotated shows the completely solid about of the revolution formed y shaded area f(x) x O calculate the y y T o when x-axis. the perpendicular volume to the of axis this of x O solid we divide it into ‘slices’ rotation. y x O When the When one to this ﬁrst, The 174 cut the cuts cut is are is V, of together , through and volume volume close of this then this slice the point another slice is a each is slice P( x, cut is y) is and made approximately small approximately increase, the at a V, a volume distance of a cylinder . the x cylinder . of the volume solid from V up the Section This ‘cylinder ’ has radius y and depth 3 Calculus 1 x y P(x, y) P(x, y) y y x O δx δx 2 Therefore V y y V ___ x 2 ⇒ x This approximation V ___ i.e. x → ⇒ 0, 2 y dx V the rotated y x Therefore as dV ___ 2 lim x→0 When improves = area ∫ 2 y dx between completely a cur ve about the y x-axis f(x), the the x-axis, volume of x the a and solid x b for med is is b given by the deﬁnite integral V 2 ∫ y dx a Example ___ Find the the volume x-axis and generated the line x when 2 is the area rotated between completely the curve about the y y √ 3x , x-axis. 4 3 When we integrate with respect to x, the limits of the integration must 2 be values of x 1 2 ___ 2 2 2 V ∫ y dx ∫ 0 (√ ) 3x dx O 1 0 ∫ 2 3 2 3 x 1 1 2 3x dx [ 2 x ] 2 (6 0) 6 2 0 0 3 Therefore the volume Note values required is 6 cubic units. 4 that for volumes of revolution are usually given in terms of Exercise 3.23 2 1 The area enclosed completely about by the the curve x-axis. y Find 4 the x and volume the of x-axis the solid is rotated generated. 2 2 The x area 1 volume enclosed and of x the 1 by is solid the curve rotated y x , the completely x-axis about and the the x-axis. lines Find the generated. 175 3.24 More volumes Learning outcomes Rotation When To ﬁnd the volume a section rotated about of a the an revolution about the area is rotated y-axis about the y-axis, the volume formed is calculated generated in when of curve a similar way to rotation about the x-axis. is y-axis y To ﬁnd when curves or the the the is volume area generated between rotated about two the x-axis y-axis x P(x, δy y) You need to know O How to integrate x simple functions How to sketch How to ﬁnd curves T wo intersection the of points two of slices cylinder parallel of radius to x the and x-axis, depth a distance y apart, form an approximate y curves 2 The The formula for the volume of volume of this cylinder is x y. a V ___ cone ∴ dV ___ 2 x ⇒ 2 y x ⇒ V ∫ 2 x y dy Therefore between the y volume a and y generated b is when rotated the part of completely the cur ve about the y y-axis f(x) is b 2 given by V ∫ x dy a Note to y, that so must ∫…dy the be means function values of to that be the integration integrated must has be in to be done terms of y with and respect the limits y Example 2 The region Find the deﬁned volume of by the the inequalities solid y x 2, 2 y 3 is rotated about the generated. y-axis. y y 2 The equation of the curve is y x 2 3 3 3 2 2 V ∫ x y 2 dy with y x x 2 2, 2 2 2 i.e. x y 2, gives 3 O 3 1 V ∫ (y 2) dy 2 [ y 2y 2 ] 2 2 9 {( 1 6 ) (2 4) } 2 = 2 1 Therefore the volume generated is 2 176 cubic units. x O x Section 3 Calculus 1 Exercise 3.24a 2 1 Find the the volume x-axis is generated rotated when completely the area about enclosed the by y 9 y 1 x and y-axis. 3 2 The region enclosed completely about Rotation of When an formed area has subtracting volume i.e. if a between the f(x) is two , the Find curves We formed rotation the x y-axis the and volume between two section. volume by y y-axis. area hollow formed y an by the of by the equation of is rotated can ﬁnd outer the line about an axis, volume of the of inner the this solid solid curve by from the outer curve and y g(x) is the 2 of the inner b curve, then the volume between them is given by b 2 ∫ rotated curve, 1 equation is curves the rotation the generated. y 2 ∫ dx 1 a y dx where a and b are the values of x where the 2 a curves intersect. y y O O An x x alternative method may simplify the working. This involves just one integral. A slice (An If, through annulus for a is value the the of x, solid area y is gives a shape between the two whose cross-section concentric y-coordinate of a is an annulus. circles.) point on one of the curves 1 and y the corresponding point on the other curve, where 2 > y 1 2 the y area of the cross-section is , y then 2 1 y 2 (y y 1 ) 2 2 x 2 The volume of a slice of thickness x is then 2 (y y 1 2 Therefore V = 2 (y V ___ 2 ) x y 1 ⇒ 2 ) y 1 ⇒ V This 2 (y ) y 1 2 dx = is 2 2 x ∴ δx dV ___ 2 (y ) x 2 ∫ (y 2 2 ) x y 1 useful 2 when the equations of the curves are similar . For example, 2 1 __ when y 1 __ 1 and 1 y x Each , 2 then problem A sketch y 1 2 x should of the be ( 1 __ 1 ) ( x assessed cur ves 2 1 __ 2 y 2 to deter mine involved will help 2 __ ) x the you best do 1 x method. this. 177 Example Find the volume of the solid formed when the area enclosed by the ______ curve y about First the ﬁnd √ x 1 and the line 2y x 1 is rotated completely x-axis. where the curve and line intersect. ______ 1 √ x 1 (x 1) 2 1 ⇒ x 1 2 (x 1) 4 2 ⇒ (x 1) ⇒ (x 1)(x 1 ⇒ (x 1)(x 5) ⇒ x 5 When 1 x Therefore Next 4(x or 1, x y the sketch 1) 4) 0 0 0 0 and graphs the when intersect x at 5, (1, y 0) 2 and (5, 2) y graphs. 3 The hollow rotating about the the the line x-axis volume found section of by completely is the without formed a cone, cone and can 2 be integration. 1 x O 1 1 2 3 4 5 1 This cone has base radius 2 units and height 4 units. 16 __ Therefore the volume of the hollow cone is 3 1 (using the formula V 2 r h ) 3 ______ The volume between given x generated 1 and x when 5 is the section rotated of the completely curve y about the by 5 5 1 V ∫ (x 1) dx [ 2 x x ] 2 1 1 25 __ 1 {( 5 ) 2 8 )} 8 8 3 1 2 16 __ Now ( = 3 8 Therefore the required volume is 3 178 cubic units. √ x 1 x-axis is 6 Section 3 Calculus 1 Example The by diagram the line y shows x the 1, area the y enclosed curve 6 ______ y √ x 1, the x- and y-axes and the 5 line x 4 4 Find the volume generated when this 3 area is rotated completely about the 2 x-axis. 1 This volume needs to be calculated in x O 1 two are by separate calculations different the for rotation volume the of volume the generated as by line the the limits 1 3 1 generated and the rotation of the curve. The x volume 0 and x generated 4 is when rotated 4 section the of the x-axis line y x 1 between is 4 2 ∫ the about (x 1) 2 dx ∫ 0 (x 2x 1) dx 0 4 1 __ 3 [ x 2 x x ] 3 0 1 __ 3 ( (4) 124 _____ 2 (4) 4 0 ) 3 3 ______ The volume between x generated 1 and x when 4 ______ 4 is the section rotated of about the the curve x-axis y √ x 1 is 4 4 1 __ 2 ∫ (√ x 1 ) dx ∫ (x 1) dx [ 1 __ [( the 124 _____ 9 ___ 3 2 volume of ] 1 1 __ 2 (4) 4 ) ( 2 Therefore x 2 1 1 2 x 9 ___ 2 (1) 1 )] 2 the solid formed 2 is 221 _____ = cubic units. 6 Exercise 3.24b 2 1 Find and the the volume line y generated 3x is when rotated the area completely between about 2 2 3 The area between and x 2 (a) Draw is a the rotated sketch curves about y the showing x line the curve y x x-axis. 2 , y x-axis. the the y x Find x the 1 1 and the volume and the lines x 0 generated. curve 3 y (b) The x (x area 1 is 1) between enclosed rotated by x the 1 and line and completely about x the the 2 curve, x-axis. between Find the x 0 and volume generated. 179 3.25 Forming Learning outcomes differential Differential equations equations 2 dy d ___ Any To use differential equation with terms involving , model situations involving and so on, is called a 2 dx to y ____ , equations differential dx equation change dy ___ An equation involving only terms in is called a ﬁrst order differential dx equation. dy ___ You need to know For example x y 2 is a ﬁrst order differential equation. dx 2 The meaning of rates of d change y ____ If an equation involves , it is called a second order differential 2 dx The relationship between equation. quantities that are proportional Rates of increase dy ___ W e know that represents the rate at which y increases with respect to x dx When the varying value of a quantity P depends on the change in another dP ___ quantity Q, then the rate of increase of P with respect to Q is dQ Such changes expands and the occur when it is frequently heated. temperature is T, in everyday When then the the life, volume rate at for of a which example, quantity the liquid of volume liquid of the is V liquid dV ___ increases with respect to changing temperature can be modelled by dT Another the example number , n, is of the proﬁt, books sold P, made (among by a other bookseller . factors). So This the depends rate at on which dP ___ proﬁt increases as n changes can be modelled as dn Formation of differential The If motion you are velocity of and If velocity an particle with with object is to through a to modelled you where respect respect falls often mechanics acceleration, displacement of a studying equations will be velocity time and by a differential familiar is the with rate of acceleration is equation. displacement, change the of rate of change time. medium that causes its velocity v to decrease dv ___ with respect to time at a rate that is proportional to its velocity, then dt dv ___ measures the rate of increase with respect to time, so is negative. dt dv ___ As is proportional to v, we can model this movement with the dt dv ___ differential equation dt 180 kv where k is a constant of proportionality. Section 3 Calculus 1 ds ___ As v is the rate of change of the displacement, s, v, we can also dt 2 d s ___ model this movement with the equation kv 2 dt Note we that when assume we that are the told change the is rate with at which respect to a quantity time is unless changing, we are told otherwise. Example Form a differential W ater is depth of in the leaking water equation from is a to model cylindrical decreasing is the tank following such proportional that to information. the the rate at which volume of water the left tank. dh ___ The rate at which the depth, h, is changing is dt 2 The volume of water in the tank r is h dh ___ is negative as h is decreasing, dt dh ___ 2 ∴ r ∝ h dt and r are both constants so we can write this equation as dh ___ kh dt Exercise 3.25 Form 1 a differential When bacteria number that 2 3 A of its to s. moving C grown is in a displacement, rate at increases Grain is volume, volume which at difference 4 are cells to model in a the culture, proportional to following the the rate data. of number increase of cells of the present at time. body The of equation a rate being of s, cereal its drained grain grain from with between V, of a straight in line a ﬁxed crop respect ﬁnal a remaining in the is time H, hopper . hopper so point grows to height, from the moves is that is such The rate inversely which and the its that is change proportional its height, proportional present rate inversely of of h cm, to the height. change of proportional the to the hopper . 181 3.26 Solving Learning outcomes differential Solving differential Solving To solve differential equations a differential equations equation means ﬁnding a direct relationship equations between the variables. dy ___ For example, the general solution of the differential equation 2x dx You need to know 2 is How to integrate simple T o functions How of to ﬁnd x the constant integration from given information How to use y x get and a unique solution, we need a pair of corresponding values of y Differential solution equations will additional substitution c involve pieces of often two involve unknown information to a constant constants. get a of In unique proportionality, this case we so need the two solution. to ds ___ For example, given the differential equation integrate kt, and that s 1 when dt t 0 and that s ds ___ 6 when 1 kt ⇒ s t 10, then 2 kt c 2 dt s 1 when t 0 gives s 6 when t 10 1 c __ 1 gives 6 50k 1 ⇒ k 10 __ 1 ∴ s 2 t 1 10 We also second need two additional derivative integrations are involved needed, pieces in each the of of information differential which will when equation. introduce a there In this is a case two constant. 2 d y ____ For example, if 2 3x then integrating once gives 2 dx dy ___ 3 x c dx Integrating 1 y again gives 4 x cx k that y 4 If we this If know gives we also 5 know 4 5 that when 2c y 1 c k x 2, k [1] when x 1, 1 we have 1 [2] 4 1 Solving [1] and [2] simultaneously gives c 1 and k 4 2 2 d y ____ Therefore the solution of the equation 2 3x is 2 dx 1 y 1 4 x 4 1 x 4 2 4 ⇒ T o solve do not to 182 4y the a x x 2 differential need to equation, understand differential you every equation. need detail to know about the how to integrate situation that it. gives Y ou rise Section 3 Calculus 1 Example A body moves so that at time t seconds its displacement, s metres 2 d s ___ from a ﬁxed point O is modelled by ds ___ √ t. When t 0, 5 and 2 dt s 2. value Find of 2 s the direct predicted relationship by ds ___ t model between when t s and t. Hence ﬁnd the 4 3 1 d s ___ this dt 2 ⇒ 2 t 2 dt 2 c 3 dt ds ___ When t 0, 5, ∴ c 5 dt 3 ds ___ t t 0, s 2 5 ⇒ s 3 dt When 5 __ 4 2 t 2 5t k 15 2, ∴ k 2 5 __ 4 Hence s t 2 5t 2 15 128 ___ When t 4, s 8 __ 22 30 15 15 8 __ The model predicts that s 30 when t 4 15 Exercise 3.26a 1 The rate of change of a quantity dr ___ with respect 2. Find r to is given by __ 3 sin . When , d 2 r r in terms of . 3 The variation of a quantity P with respect to r is modelled by the 2 d P ____ differential 2 equation 12r 6r. It is known that when model predict that the r 1, 2 dr dP ___ P 6 and 1. What does this value of dr P will be when Integration Many r by differential 3? separating the variables equations used to model situations are of the form dy ___ f(y). We cannot integrate f( y) with respect to x, so we need to change dx the form of the differential equation. du ___ We know from T opic 3.18 that ∫ f(u) dx = ∫ f(u) du, where u is a dx function of x. dy ___ Therefore ∫ f(y) dx = ∫ f(y) dy dx dy ___ This means that integrating ((a function of ) y) with respect to x, is dx equivalent to integrating (the same dy ___ Now we can write = f(y) dx then ∫( function ___ f(y) dx y) with respect to y. dy 1 ____ ___ f(y) dx as = dy 1 ____ of 1 1 ____ ) dx = ∫ 1dx becomes ∫( f(y) ) dy = ∫ 1 dx 183 This is called integration have effectively by separating the variables dy in going from f(y) ∫( to dx is to gather containing all x the on terms the containing other side, i.e. y on we what we 1 ____ ___ done because one have side f(y) and ) all ‘separated’ dy = the the ∫ 1 dx terms numerator dy ___ and denominator of dx dr ___ For example, = , dt ∫ r dr ∫ = 2 dt then multiplying by r gives r r 1 so dr ___ 2 __ given ⇒ = 2 dt 2 r 2t c 2 Example The is atoms in modelled at any a as given radioactive inversely time, t, material are proportional measured in to days. disintegrating the number Initially of there at a rate atoms are N that present atoms present. (a) Form and solve a differential equation to represent this information. (b) Half the model mass disintegrates predicts that it will in 200 take days. for Find how three-quarters long of the the mass to disintegrate. (a) If n is rate the of number change dn ___ of n of atoms is = ⇒ dt n n ∫ Therefore at any given time, then the dn ___ k __ ∴ present negative, k where k is a constant. dt n dn 1 ⇒ ∫ k dt kt 2 n c 2 1 Initially, i.e. when t 0, n N, ∴ 2 N = c 2 1 hence 1 2 n kt 2 N 2 2 1 (b) When t 200, n N 2 1 2 1 ( ∴ 2 N 200k 1 200k 2 N 2 3 ⇒ ) 2 3 ____ 2 N so k 8 2 N 1600 1 i.e. 2 n 3 ____ 1 2 N t 2 2 N 2 1600 3 When 1 of the mass has disintegrated, n 4 1 2 2 1 ( ⇒ N 4 N ) 3 ____ 4 1 2 N t __ 1 2 N ⇒ 2 1600 3 ____ 32 ⇒ t 1 t 2 1600 250 3 The model predicts that it will take 250 days for of 4 disintegrate. 184 the mass to Section 3 Calculus 1 Example The rate of modelled people being already person (a) increase as was Form in the number , proportional infected. Nine to n, the people of people square were infected root of infected the 5 by a virus number days after is of the ﬁrst infected. and solve a differential equation to represent this information. (b) How will many take days, for to 100 the nearest people to be day, does the model predict that it infected? 1 dn ___ (a) When n people are infected, kn 2 where k is a constant. dt 1 ∫ ∴ When (This 1 2 n t dn is ﬁrst ∴ c 0, not the ∫ n k dt 2n 2 kt c 1 given person ⇒ explicitly is but the days are counted from when infected.) 2 4 When t 5, n 9 ⇒ 6 5k 2 ⇒ k 5 1 4 ∴ 2n 2 t 2 5 4 (b) When n 100, 20 t 2 5 45 __ ⇒ t 2 The model 100 people predicts to be it will take approximately 23 days for infected. Exercise 3.26b 1 1 The velocity, v m s , of a ball rolling along the ground is such that, _ 1 t seconds t a 2 0 and direct W ater it that v is dripping a dripping, 2 at when a damp radius which v t of the tap dv 3, between from circular the rate ∫ started, relationship forming The after 2 v on patch. the k dt. solve and to a the that v differential concrete hours patch r cm, Given 5 when equation to give t. T wo damp radius, ∫ of was the surface after the where tap it is started 20 cm. damp patch is increasing is 1 __ modelled as being proportional to . r (a) Form and number (b) How for 3 Grain long, the is conical solve of differential equation elapsed after to nearest hour , the radius pouring pile a hours of the from whose a damp h is on tap does patch hopper height the to to giving starts the model reach a barn increasing at r to a in terms of t, the drip. predict it will take 1 m? ﬂoor rate where that is it forms a inversely 3 proportional doubles height after has to a h . The time grown to T. initial Find, height in of terms the of T, pile the is 2 m time and after the height which its 6 m. 185 Section 1 3 Practice questions 9 Find: A spherical balloon is losing air at the rate of 3 h ___________ (a) _____ lim √ h→0 0.5 cm per second. __ 2 Find √ 2 the balloon x 2 ___________ rate of when change the of radius the is radius of the 20 cm. lim (b) 2 x x→2 2x _ 4 8 (The volume of a sphere of radius r 3 r is . ) 3 2 (a) The function f is given by 10 Find the range of values of x for which the 4x ________ 2 x f(x) { 1, x function 2, 2x, x ∈ lim f(x) and by f( x) Show increasing. x) that 2 f(x). x _________ x→2 x→2 2, lim is (2 11 Find given 2 x y 2 (3x Hence explain continuous (b) Repeat at whether x or not f( x) has 2 one stationary value Find the stationary 3x { 1, x 3 x 3 y x 1, ∈ 3x the curve 2 4x 6x 12x 5 and from ﬁrst principles: 13 distinguish The curve value sin 2x __ (b) it. 2 x Differentiate (a) ﬁnd on 3 between 3 3 and points 4 3) for (a) 12 f(x) is of when x 4 y 2 ax when them. x b has x 0 and a of a, b and c. a maximum minimum value c 2. √ x Find the values dy ___ 4 Find when: dx 14 (a) y (2x 1)(3x (b) y 4 sin x (c) y Sketch the curve whose 2) equation is 4x ________ y 2 (2 2 x 2x 1 ___________ x Given that y (Y ou can Find the use your results equation of x sin x through d ﬁnd the point the , ( 10.) curve which goes 1 ) and for which 2 y and . y 2 ∫(5 cos ) d dx 16 Find question ____ dx 6 from __ 2 dy ___ x) 1 15 5 3 cos x f(x) when f(x) A curve passes through the points (0, 1) and is: (1, 1). The equation of the curve is such that x ______ (a) 2 d 2 x y ____ 1 ______ 4 6x 2 dx (b) x √ x 1 Find the equation of the curve. x _____ (c) sin x ______ (d) √ 17 2 x (a) Use ∫ __ (e) cos 2x ( 2x ( sin 2x u x to ﬁnd cos 2x sin 2x dx 2 3 (b) Use ∫ ___ in u 2 ) dy Find substitution __ sin ) 3 7 the 1 terms of t the substitution 2 6x(x 1 to ﬁnd 4 1) dx when: dx 18 2 (a) y 2t, x (b) y 3 cos t, t Evaluate: 2t 4 x 4 (a) 5 sin t ∫ (3x 4) dx (b) ∫ 2 8 The equation of a curve 2 cos d __ 2 is 2 19 (a) Use the substitution u x 1 to evaluate 3 y 3(x 5) ( ∫ A point is increasing moving along the at the constant Find the rate curve rate ______ 2 of so that 0.2 cm x is per x √ 2 x 1 ) dx 1 (b) Use the substitution u __ second. of change of y when x is 4 3 ∫ 1.5 cm. 186 0 cos (sin 1) d sin to evaluate Section 20 Find (a) the area enclosed by the The (b) curve initial radius of 3 the Practice balloon questions was 10 cm. 2 y 4 x and the T en x-axis. radius, 3 Find (b) the the area y-axis enclosed and the by lines the y curve 1 and y y x seconds after r cm, of air the started balloon to escape, was , How 2 long balloon will is it be before the radius Sketch (a) the curve (The of the 2 cm? 4 21 the 5 cm. volume of a sphere 3 r is .) 3 y Find (b) the area (x 1)(x enclosed 1)(x by this 2) curve and the 28 Given that x-axis. y 3 cos 2x, 2 d y ____ 22 Find the area enclosed by the ﬁnd curves in terms of y 2 dx 2 y 2 x 1 and y 5 x __ 44 2 23 Find the volume generated when the 29 area The ( point ) , 3 is a point of inﬂexion on the 9 3 enclosed x-axis by and the the curve line x y 1 x is 1, the rotated y-axis, curve the completely 3 y about the Find Give your answer in terms Find (a) the equation of the tangent to the x 2 at the point on the curve a sketch x- the and to show curve the and the equation area tangent point ∫ 4 where x of the 1 tangent to the curve at dx 10 where a 0. Find the value of a 2 x a between the bx 1 __ 2 Draw (b) 1 where 30 x 6x curve 2 y 2 ax of the 24 x-axis. enclosed in part (a) and 31 Solve the differential equation y-axes. 2 d y ____ Find (c) the volume generated when the area 6x 4 3 2 dx described in part is (b) rotated completely dy ___ about: given that, when x 0, and y 9 dx (i) the x-axis (ii) the y-axis. 32 Find the between 25 Solve the differential volume the generated when the area curves equation 2 y 2 x and y 8 x dy ___ 2 6y is dx given that y 3 when x rotated completely about the x-axis. 1 y 33 26 Solve the differential equation dy x __ ___ dx given that y 3 y when x 2 x O 27 Air is escaping from a spherical balloon 2 that is proportional to V at a rate 3 , where V cm is the The volume of the diagram shows the area enclosed by the balloon. 2 curve (a) Use the information above to form y where has t elapsed escape. equation seconds is from in the terms time when the of in air V x 2x 2, the x- and y-axes and the a tangent differential and seconds started to the curve at the point where x 2 t (a) Find this (b) Find the area. that to rotated volume generated completely about when the this area is x-axis. 187 Index 0 128, circumference 129 64 0 clockwise rotation codomain A acceleration a cos acute angles addition algebra 9, 110, 115 propositions a lines triangle between annulus 110, of a curve between two under curves below x-axis and curves compound angle identities compound statements associative operations 31, of axes, and 57, axioms (z) 154, minor symmetry third 8, 98, sections proportionality 10 base of 52 a logarithm changing bearings 55 6 bi-conditional bijective binary Boole, 52 statements functions George 14 algebra 14 168 128, 124, 180, 182, 15 geometry in three graph 70, 90–1 dimensions 105–7, 68 68 82 cosine formula cosine function cosine graphs 72 64, 66, cotangent function cotangent graphs 65–6 67, 86 68, 69 69 examples (solids) 19 109 cubic curves cubic equations 31 27–9 curves area C c (constant of calculators calculus integration) 53, Cardano, families 81 of a circle 92–3, of a curve 96–7 of an of a line of a parabola of a plane 102, 114, form Cartesian unit rule circles 64, equation 188 103, deduction 92–3, of 106 144, 130 94, 93, 98 97 30–1, 174, 37, of 175, 133 51 38, 39, 154–7 32–5 176 terminating decreasing 111 138–41, 101, decimals, 100 vectors 132, D 103 115 120–1 Cartesian chords cylinders 97 96–7 163 reﬂected transformation 92 173 33, 130–1, of sketching 31, 161, of equations 167 27 equations ellipse of two of of gradient theorem Girolamo Cartesian between equations 168 166 fundamental chain 160, 164 functions integrals deﬁnite integration denominator 169, 13, rationalising function second 12 146 18 deﬁnite derived 184 125 15 cosecant cubes 8 160, (k) functions function counter 49 operations Boolean 15–16 (c) cosecant cosine 15 17 integration coordinates B 80 134 of coordinate 174 105 18 base logical of converse 76–7, 98–9 contrapositive 156 78 14 120, continuous 102 72–5, 178 constants 9 72–5, 15 conjunction 17 164 16 statements connectors, 170–1 138–41, formulae 172–3 connectors major 171 173 166–8, associative axes, y-axis 9 46–7 angle conic between asymptotes functions 17 8, compound cones 98 area axes operations conditional Perga 53 commutative conclusion 111 177 Apollonius logarithms differentiating 18 50 connectors 17 90 vectors 64 49, commutative composite 10 64 between of 80–1 90, 8, of angles common 180 +b sin 48, 172, 175 168–9 36, 39 13 (derivative) 152–3 132 108–9 Index differential equations differentiation of composite of a functions constant parametric from ﬁrst 180–5 132–5 138–41, 164 of quotients 54 functions 56, expressions 6, of 132–3 functions of 136 functions factor formulae factor theorem factorising 137 78–9 23 24–5 n reversing table factors 160 142–3 of y ax of y ax of y f(x) of y x 57 36–7 F principles products equations exponential 134 142 of exponential of families 134 ﬁrst n a of order b 24–5 curves 161, differential 163 equations 180 n focus 134 g(x) 135 point fractions, 99, 100, improper 102 36 n direct 134 proof directrix 18–19 100, discriminant of a 181, vectors point from connectors operations domain 44–5, dominoes double 9, 10, 47, 183 105, distributive 8, a derived 106 line 91 17 48, of exponential 56, 124 modulus onto types e (eccentricity e (irrational of an ellipse) number) 52, 102 56 58–9 48, 49, 50 48, trigonometric, E 126–7 56–7 subjective 76–7 124–5, 57 146–7 50–1, logarithmic 50, identities 132 discontinuity inverse 36–7 125 146 increasing 9 20 angle 124, decreasing distributive division 46–7 continuous 180, displacement 49 composite 14–15 displacement 44–7 bijective 102 26 disjunction distance functions of 49, 50 reciprocal 68–9 48–51 fundamental theorem of calculus 167 53 x e G 57 ellipses 98, 99, Cartesian equal a 42–3, a diameter 102, logarithmic normals of a 158, parabola of a quadratic of a of equivalent Euler 52 130–1, a straight of a tangent of a 132, 90, 130–1, 133 144 144 68 sine 158–9 67, 86 69 function secant 101 142, line 66, 50–1 68 65, straight curve 66, 67, line tangent 66, 87 30 67, 88 H 51 h 7, 94, 112–13, equivalence 182 132–3 of cosine 116–17 159 26–7 tangents vector 112–15, 26 simultaneous of equations 90–1 71 reﬂected roots 90, 120–1 polynomial curve cotangent 100, 96–7, plane a cosecant 42–3, differential coordinate graphs 103 54–5 parametric solution, function 163 54 30, of 161, 95 ellipse line 96–7, 180–5 exponential 134–5 142–3 gradient 97 of of a general 92–3, curve of differentiation table 27–9 differential 99 general geometry, a an 103 60–1 of of 102, 92 circle cubic of 107 7, Cartesian of equations vectors equations Galileo 102 42–3, 158, 60 159 118–20 (small Hubble increase telescope hyperbola in) 130 99 98 hypotheses 15 17 operations 164 I identity 7, 26 189 Index compound double law angle angle 72–5, image 47, implications 82–3, by improper fractions 36 130, 131, functions 136, 137 inﬁnity, 61, injective points of integers 150, 48, 151, 49, 50, 152, deﬁnite laws of long 172, a difference of a multiple of a sum of of functions 160 using substitution 162 162 164–5 163 power variables of angles x of 183–4 160 a triangle 18 intersection curves between and two lines lines 116, of planes and inverse inverse 42–3 graphs of connectors division functions irrational numbers 44, induction mathematical operators maximum turning maximum values turning minimum values models 183, modulus functions modulus signs modulus of of indices laws of logarithms e 52, Gottfried length 104 (surds) Newton, 12–13 128, 180, 182, non-parallel 190 150, 151, 152, 80–1 58 8, 9, 10 to 184 a a 52, 53 Sir Isaac 52, lines 166 116, 117 158 curve 94 of plane 158, 118, 13, 159 120 36, 39 6 O obtuse 53 Wilhelm Omar 166 angles onto 128–9 30, 113–14, 144 116, 116, 117 117 115, 116–17 90, Khayyam one-way 126–7 90, logarithms 14 functions functions operators, orbits origin of 110, 8, 48, 49, 50 164 mathematical 120 49, 34–5 48, 9, planets 119, 115 98 stretches operations intersecting 152, 53 equations lines of 151, 99 153, 157 80–1 58–9 negation one-to - one of 31, 60–1 53 52 Leibniz, gradient 8 150, 185 John 99 laws equations 184, x numerator theorems 30, (Napierian) mathematics limit 56 20–1 6, points natural L notation 30, minimum to limit 51, points Napier , proportionality) Johannes of 50, mathematical normals language 46, 56 K of 45, 50 non-parallel (constant 17 N 98 50–1, number Kepler , 53 52 36–7 multiplication 60 15 irrational k 52, numbers 117 cones 10–11, (Napierian) negative reverse 162 functions functions separating 53 53 M rule twice 56 55 53 of mapping of 54–5 56–7 175 168–9 of 52–3, 52, natural 153 56 160–5 deﬁnite of function logical 169, 102 logarithmic 167 160 integration interior 154, 10 integrals 92, evaluating 126, 112–13 180 point common 112 of equations 176 function equations of 120 117 logarithmic 38–41 119, 104 dimensions logarithms 146–7 approaching inﬂexion, a 120 118, of heating of base quadratic a loci 36–7 52 inequalities 116 planes 116, three liquid 18–19 expressions indices in vector improper increasing skew 96–7 20–1 48 small to segments 70–1 proof 45, increase, 116–17 114, perpendicular 10–11 trigonometric induction, of parallel parallel Pythagorean of pairs 80 17 members by 76–7, 76–7 6, 8 50, 56 153, 156, 157 Index reciprocal P pairs of lines parabolas 30, Cartesian tangent parabolic 98, 99, to 42, parallel lines parallel vectors parameter 100–1, equations of 114, 108, of an of a line ellipse of a parabola perpendicular vectors vector point to points of 120 by of the y-axis 176–7 of product scalar quantities function order graphs 150, 151, 152, 153 sines, 26 36, 106, 109, 112 48, solid 50 equations of propositions curves 20–1 72, identities line straight lines stretches, 105 subjective 70–1 using quadratic quotient u 71 inequalities 62–3, sum 38–41 122–3, of 154–7 131, 136, 137 of 15 in three 148–53, 30, 42–3, dimensions 155, 90, 156–7 112–15, 116–17 112 34–5 functions 48, 49, 50 7 139, 8, sines 164–5 140, 9, 164–5, 169, 171 10 78 12–13 6–9, symmetry 142 15–16 integration symbols 137, 39, 145 one-way 138, surds 186–7 21 rule 38, 174–5 equations subtraction 26–7 questions quotient 130, values/points straight Q of 37, 16 substitution roots in volume stationary equations 87 117 revolution conditional 20–1 58 theorem 67, 30–1, 116, compound 17 Pythagorean 60 statements 140 14–17, of Pythagoras’ quadratic 66, bi-conditional x 42–3, 78 increase of 7, 64–5 65, lines small 20 47, skew 180 82 sum sketching 39 105, 70, equations 10–11 simultaneous 118–20 152–3 differential closed 115 68–9 second 18–19 of 177–9 104 derivative sets, 120–1 110–11, second function induction algebra about sphere, direct properties 178 scalar 180 proof, curves x-axis sine 136, 174–9 the sine 44–5, rule between 91 22, of 64 area sine vectors 160 174–5 about secant 111 109 proﬁt an 99 integers product proof 158–9 120 110, equations pre-image prism 118, inﬂexion polynomials positive 91, equations line polynomial position 142, 120 equation a volume of 175 orbits of S lines planets’ differentiation revolution, 142 101 perpendicular Cartesian 22 116–17 distance 118, 21, 26 solid 103 113–14, 68–9 rotation 110 96–7, perpendicular 64, theorem roots clockwise equations functions 50–1 revolution, 96 parametric planes repeated 116 107, 33, remainder reversing differentiation () 170 99 parametric pi 102, 100 101 mirrors trigonometric reﬂection 116–17 axis 14, 30, of 16, 31, 52, 167 102 174 R radians 64, radius range 93, 45, T 129 radioactivity tangents 184 94 47, 48 rate of change 144–5, rate of decrease 180 rate of increase 144, rational expressions rational numbers rationalising real numbers the 180, circles to curves to 180 36–7, 10–11, graph 13 44–5, 49 of 148 94, 158, 130–1, 42, telescopes 64, 66, 159 144 101 (trigonometric) function 13 denominator 130, of parabolas tangents 39–41 158 94–5 equations 181 gradient 12, 8–9, (lines) to 70, 82, 90 66 67, 88 99 191 Index terminating terms theorem three decimals 12 equal of calculus 167 straight lines transformation translations triangles, parallel in 105–7, in of 32, interior 34, 18 functions, trigonometric identities tables values turning see of unit reciprocal 82–3, 68–9 vertex virus 31, 150–3 minimum, stationary points 48–51 U u (substitution) unit vectors 110, 106, 111 109, 112 104 of 106 108–9 180, 181 100 infection 185 volume maximum, function 110 121 of 106, velocity 96–7 14–15 30, 108, 105, subtraction 14–17 points also position 84–9 106 106 plane properties angles trigonometric a of 107, perpendicular 32–5 59 equations truth to 112 curves trigonometric truth 108–9 105, 107 magnitude dimensions coordinates types displacement 6 138, 139, 140, 108–9 164–5, 169, of a cone of a cylinder 178 of revolution of a solid of a sphere 175, 176 174–9 175, 176, 178, 179 145 171 X x-axis, rotation about 178 about 176–7 n V x variables VDU 7, 182, 18 183–4 137 Y x vector equations of a line of a plane vectors 192 of a y-axis, 56 rotation 118–20 104–9 addition angle y 112–13 Z 105, between 106, two 107 110, 111 z (third axis) 105 Pure Mathematics for CAPE® Achieve your potential Developed guide in will CAPE® Written exclusively provide Pure by an Mathematics essential begins range ● the ● ● with additional experienced syllabus the key features Engaging the Caribbean Examinations support to Council®, maximise your this study performance Mathematics. information with of you with and in team designed that teachers examination, an easy-to-use learning activities of outcomes to enhance help you and this experts study guide double-page from your develop the of the the CAPE® covers format . syllabus study the in all Each and the topic contains subject , analytical Pure skills such a as: required for examination Examination tips with essential advice on succeeding in your assessments Did You Know? boxes to expand your knowledge and encourage further study This study choice The also questions examiner Pure guide and feedback, Mathematics Caribbean Nelson at sample to build a fully interactive examination skills and answers confidence to CSEC® produce and a Council series (CXC®) of of O x f o rd incorporating with in has Study multiple- accompanying preparation for the CAPE® worked Guides exclusively across a wide with range of CAPE®. How Pa r t CD, examination. Examinations Thornes subjects includes University P re s s to get in touch: web www.oup.com/caribbean email schools.enquiries.uk@oup.com tel +44 (0)1536 452620 fax +44 (0)1865 313472 Unit 1