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matemáticas puras understanding-pure-mathematics

  1. 1. IIIIIIHI-IIIIIIIIiI‘III9I I PURE ~ MATH EMATICS : I; :’I. ..J. . Sadlw D. Tiwrnireg 1 fifii 7' :0 OXFORD
  2. 2. OXFORD utdtvnitsrrv Pu/ .95 Grut Ctarendon Street. Oxford OX2 6DP Oxford University Press is 8 depantwutt ofthe University of Oxford. It funlten the University’: objective of excellence In reaareh. scholarship. and education by publishing wortdwide in Oxford New York Auckland Bangkok Buenot Aires Cape Town Chennai Dnrusalnatn Delhi Hong Kong Iatanbul Kutehi Kolhu Koala Lumpur Madrid Melbourne MuicnCtty Mttmbai Nairobi Saohnlo Shanghai Taipei Tokyo Toronto Oxford in II registered trade mark of Oxford University Press in the UK and in certain other countries 0 Oxford University Press I981 Hm pubhshtd I937 Reprinted I988. I936). l990(tI. ncc). I9~)l. I992, I993. I994, I995. I996 (tn ‘I. -c). twx. I999 (IWIICCI. mm. 2002 (cu ice I. 2003. 2t‘-‘N British Library Cataloguing in Publication Data Sadler. /t. J. Underst. -tnding pure muhematics. I. Mzthenutu: s—~Eummutiont. qoeationi. etc. I. Title II. Thoming. D. W.S. St0'.76 QA43 ISBN 0 I9-9|-1241-2 Sqhool Edition ISBN 0 I9 9|-1259-9 Trade Edition All rigtits reserved. N6 put of this publication may be reproduced. stored in a retrieval system. or tralumittod. in any form or by my means. without the prior permission in writing 0! Oxford University Press. Withtn the U. I(. . exceptions are allowed in respect of any fair dealing for the purpose of research or privnte study. or criticism or review. as permitted under the Copyright. Designs and Pucnls Act. I988. or in the ease of repronz-phi: reproduction in accordance with the terms of licences issued by the Copyright Licensing Agency. Enquiries mnoerning reproduction outside these terms and in other countries should be sent to the Rights Department. Oxford University Prest. at the address above. Printed by Bell & 8.tin LltL. GIDS'ajW$
  3. 3. Contents Introductory work IhLm. tmheLlins_L lngualities 2 Basic algebraic manigulation 5 Snr. d.L8 Quadratics I0 Trigonometric ratios of acute and obtuse angles I3 Solution of triangles I6 Three-dimensional trigonometry -2 Lfttnctlons Basic concep_ts_, ?2 Comgosite functions 34 Ilic_ittv. crsc_oLa_[unction_i: ! Ordered gairs 38 The gragh of a function 39 S E I . I . 3 2_I: ctom Basic concegu 47 E . . _ | . E2 The scalar groduct 58 Geometrical Qroofs by vector methods 63 The vector eguation ofa . straight line 65 3 Coordinate geomfly I LocL_ZI Distance between two points 72 Mid-p_qint of a line joining two goints 74 Gradient of a line joining two Qoints 77 Parallel and indicular lines 79 The guation y ; mx + c 81 Finding the Cartesian cguation of at straight ling: 8! Relationshigs shown as linear granhs 89 I , U. E 4 Trigonornetg I 1_'_rigonometric ratios of any angle 99 Trigonometrie graghs I0] _GL9ph_2kLhi. ng_I_0é Solwing trigonometric eguatione IIO Eythagorean relationshigs H4 Secant, cosecant and cotangent H5 Comgound angles I18 Qmmat mgmm pa! 5 Algebra I Indict: _1.iJ. 1? . . In Elementary theog of logarithms I36 Qunadratics I41 Polvnomials in general 152 §_Matric§ Basic concegts I59 Imnsfnrrna1it1n. t_lt£1 General relleetions and rotations I78 LP_mnututions_anil_e. ombinationa Successive ogrations 188 P_cnntttati. ons_182 Selections of any size from :2 groug 212.! Division into grougs 20-I 8_SerIes_and. tbLhinomIaLtheorern Arithmetic grogressions 208 Geometric grogreasions 212 Summation of other series and groof by . I . 2 [Z The binomial theorem for a gositive integral index 222 The binomial theorem for any rational indrx ‘$23 9 Probabilltv Simgle grobability 235 Sum and groduct laws 2343 Further conditional grobability 2-I6 Probability involving p_cn'nutations a_n<_i I . . 2 z
  4. 4. Th mdient of a curve 255 Difi'er: ntiation_ni'. .axf_258 Maximu_m, minimum and flint: of . fl . 2” Small changes 269 More about aims of inilexion 272 ll Slcetehlpg functions I Gcns: Lal_mstlmds_2Z5 Simple transformations 28:! 1 E5288 12 Calculus II: integgtion I] [HE . . Z2! K 1“ E . 1 . Ms Two applications 308 Mumm 13 Calculus III: further teehnigucs Ennction_nf_aIunctinn_3.LZ Integration of [gxfl flx)| ‘ 320 Related rates of change 322 Parameters 325 BmducLt'u1c_32r‘l gotient rule 329 Implicit functions 33] 14 Slcetehlr_rg functions [I Ingualities 3-H 15 Trlgonomfly ll Sum and product formulae 351 0 cos x 1; (1 sin 1: 353 Inverse trigonometric Ilunctions 356 C: cner: aLsolutions_361 Small angles 365 Differentiation of trigonometric functions 362 Integration of trigonometric functions 321 Differentiation of inverse trigonometric l. ‘tmetinns_3Z= ( V C antenu I6 Coordinate geometry ll Angle between two straight lines 380 Distance of a point from a line 332 Ilie. circlc_38.i The parabola 392 The ellipse and the hyperbole 396 Eolar_cooLdtnates_«1Q1 omi M matrices Differentiation and integration of vectors 11.! Eguations of a straight line in 3-D 415 Equations of a plane 423 . 'lL3_matnces_433 Sets of linear guations 439 I8 Algebra II E . I E . in Complex numbers 460 I9 Exanential and logarithmic functions Differentiating the exponential function 421 The log rithmic function 48! The exponential series 489 The lpgarithrnic series 490 20 Calculus IV: further lntggration Changg of variable 496 In tion b 502 General methods 506 Dillerential equations 509 21 Numerical methods The trapezium rule 52-! Simpson‘: rule 525 Taylor's theorem 527 Maelaurin’s theorem 528 Solution of gquations 529 c’, cos 2 and sin 2 for complex : 535 A lndex_6DQ
  5. 5. Introductory work 0.1 The number llne When we lirst leam to count. we use the numbers. I. 2. 3. 4. . . . These counting numbers fonn the set of natural numbers or positive integers. Z '. Z’ = - {l.2.3.«l. S. . ..l As our use of number progresses. we need to extend this set backwards to include new and negative numbers. This gives us the set of all integers Z. Z“ l. .. -4. -3, -2.-1.0. l.2,3.. ..| This set of integers is sufiicient for many purposes. but we soon need to introduce fractions. Any number that can be expressed in the form a/ b, where a and b can take any integer value except b = 0. is called a rational number. All the integers are rational as they can be expressed in the forth all. ll‘ we extend our set of integers Z to include all rational numbers, we have the set of rational numbers 0. There is still one group of numbers on our number line that is not included in the set 0. This is the group of numbers that cannot be expressed in the form alb for integer a and 1): these are said to be the irrational numbers. One example of an irrational number is J2. To prove that , /2 is irrational we use the method of proof by contradiction, i. e. we assume that what we are trying to prove is not true and then show that this assumption leads to a contradiction. We initially assume that . /2 can be written as a fraction a/ b in its simplest form (i. e. all cancelling carried out). where a and b are integers. "2 Suppose um, /2 -‘T: then 2 = 5, or a’ - 2h’ [I] This means that a‘ is even and so a must be even. Because a is even. we can write a = 2: Substituting for a in [I] gives (2c)’ —= 2b‘ 462 . ._ 25: 21-‘ = b’ andsobmustbeewzn. Now we have a contradiction because if both a and b are even, the fraction
  6. 6. 2 Understanding Pure Mathematics a/ b was not in its simplest form. Thus J2 cannot be written in the form a/ b and so ‘/2 is irrational. The set of rational numbers 0 together with the set of irrational numbers form the set of real numbers R and complete the number line. Note: If we express numbers as decimals, a rational number will either have a finite number of decimal places or will recur. An irrational number will have an infinite number of decimal places without recurring. 0.2 Inequalities The reader should be familiar with the notation is > b and a 2 b standing for ‘a is greater than b’ and ‘a is greater than or equal to b’ and with the notation a < b and a :1 b standing for ‘a is less than b‘ and ‘a is less than or equal to b‘. If we wish to consider a section of the number line. we can define that section by using set notation and the inequality symbols. The “cfion canbcwfittcn --3 -3 -2 -I 0 l 2 3 4 5 {x e R: - 3 4 x < 4} and this is read as ‘the set of all real x such that x is greater than or equal to — 3 and less than or equal to 4.‘ llxe R: -3 $ A’ S 4lcanalsobewrittenlx: xeR. -3 < x < 4}] If we do not wish to include the integer — 3 in our set, we would write {. 'GR: -3 < x < 4}. This would be illustrated as follows: — «J «3-2-l0 l 2 3 4 S The open circle indicates that -3 is not included in the section. If we only want to consider the integer values from -3 to +4. we would write this as (x e Z: -3 Q 2: s 4) and the set could be listed: {-3. -2. - l. 0.1. 2. 3. 4). The notation ix: x obeys some rule} can also be used for a set with one or more elements missing. For example, the set of all real numbers except the number I would be written {x E R: x at l). In a similar way lxeZ: x¢0.x-)4 l}= {. .. -5.-4,-3. -2. -l,2.3.4.5,6.. .} Nola The set lx e R: x < al is sometimes written {x2 x < a} or simply as x < a. in such cases it should be assumed that the inequalities are for real x. Inequalities can be simplified using a method similar to that used to solve equations. However. simplifying an inequality will give a range of possible values for the unknown rather than a finite number of solutions.
  7. 7. hnudwnmynwd Example 1 Simplify the following inequality 6x - 3 < 2l - 2x. 6x - 3 re: 2| — 2x Roarranging gives 6:: + 2: s 2| + 3 81: —< 24 so x < 3 The answer may be left in this form or it can be expressed as the solution set {x e R: x < 3] or diagrammatically: -4 -3 -2 —l 0 l 2 3 -l 5 One important difference between the techniques used to simplify inequalities and those used when solving equations arises when we multiply or divide the inequality by a negative number. In such cases we must reverse the inequality sign. Consider the true statements 4 > 3, 3 > -2, —4 < —-2. If we multiply each statement by — I and reverse the inequality sign. we obtain the statements -4 < -3. -3 < 2. 4 > 2 which are also true. (The reader should verify that the statements obtained by multiplying by —l and not reversing the inequality signs are not true). Example 2 Simplify the following inequalities and illustrate the solutions with a diagram: (a)8-2x: <.3, (b) -5<2x+3e: 'I. (a)8-2x<3 (b) —S<lr+3<_7 ~2x$3"8 -5<lr+3 and 2x+3<7 -2x. <,-5 -5-3<2x ks‘)-3 2x25 -8<2.* 12:64 1:22) —4<x and r£2 i. e. -4<x£2 0I121J‘-£7-5~4-3-2-l0I234 The modulus sign We write Ixl to mean the magnitude or modulus of x. i. e. we are only interested in the size ofx and can disregard its sign. So I-7|:7; | -3|'—‘3: |2|=2 etc. This enables us to write inequalities of the type -l < x < I as | x| < I or -3 ft x s 3as| x| 5: 3etc. Notice that Ix] < a means that x must lie between +a and -a. i. e. x > -aandx < 0 whereas lxl > a means that 1: must either be greater than H: or less than —a. i.e. x > aorx < —a 3
  8. 8. 4 Umlersranding Pure Mathematics Examp| e'3 Express the following as inequalities of x and illustrate the solutions with a diagram: (a) ll: - 3| s 7. (b) |3x + l| > 8. (a)|2x-3|<7 (b)|3x+l| >8 2x—3;-7and2x-3<1 3x+l<-8or3x+l>8 2x2-4and 2xSl0 3x<-9or 3x>'l x3-Zand x<$ x<-3or x>2§ So -2<x<5 1’! ~r <4--4-0--r—v—-——r——. e-I-III-1? —4-3-2-lo I 2 3 4 5 6 -6-5-4-3-2-I0 l 2 3 4 S 6 Notice that we do not combine the two ranges of part (b) in the form -3 > x > 2§. Exercise A l. State whether each of the following statements are true or false. (a)4>5 (b)S<2 (c)3>l (d)2<5 (e) -1!) -2 (f) -3 < -2 (g) 2 > -2 (h) —4< -5 (i) l-'El’Ix¢2.xvf3}= (l.4,5.6,7.8.9.. ..} (j) (xeZ: x#2)= -{. ..-4.-3.-2.-l.0.3,4.. ..) (It) (xel: —I<x<2}-(—l.0.l.2} (ll iv’6z1"3<X<4}= {-3.-2,—l.0.l,2.3.4) (m)(xeZ: -2<x<5l-{-l,0.l,2,3.4.5l (N) (X691 -2<x<5l " l-1.0. 1.13.4.5} (0) (xeZ: xS2I-{xel: .r<3l (p) (xeR: x<2}-(xeR: x<3} 2. List the elements of the following sets: (n)lxel: ~2<x<4l (b)(xel: -2<x(4) (c){xeZ: —3<x<2} (d){xeZ: -3<x<2} (C) lx6Z‘: x < 5} (f) {xel’: x < 5) 3. Give the following sets of numbers in the form {x e a set of numbers: x satisfies some inequality}. (a) {-'39 -2:-'40: (*5. ""4 ‘3.-2.“"]} --r*-7--r—*-r* (c) I. .. -4. -3, -2. —l.0.I: (d) -5 -4 -3 -2 -1 o I 2 3 4 (C) —v——w—-v—- 7-- (0 : v—w —5—4-3-2—l0 1 2 3 4 -5-4-3-2-1on2 3 4 T312“ --r—~ 1-‘ —v-—'r——r— +— (3) —s -4 -3 -2 I o I 2 3 4 W -5 -4 -3 -2 —I o l 2 3 4 ‘ ‘ - W- N -s -4 -1 —2 -1 o l 2 3 4 (1) -5 -4 -3 -2 -l o I 2 3 4 4. Simplify each of the following inequalities and draw a diagram for each solutionsetforxell. (a)2x€6 (b)2x+l<5 (C)3x-I)" (d)-1-x<6 (e)4—3x>7 (f)2x-43-1 (g)Sx-3<ll-2x(h)2.s+l>3x-4(i)-5<2x—3<'I
  9. 9. (j) -7<3x+2$$(k)l<2x+'l<ll (rn)4s2-x<7 (n)-9<3—2x£l 5. Find the solution sets of the following inequalities: (a)| x-2|<5 (b)| ?.r—l| <S (c)|2x—-5|>'l (d) |4x-l| <l5 (c)|5x-l| ,24 (f) |3x-l| >5 (I) l<7x+l5<36 0.3 Basic algebraic manipulation The following examples and Exercise 3 are intended to revise basic algebraic processes. The ability to manipulate algebraic expressions correctly is an essential skill for advanced mathematics. Example 4 Simplify the following: (a) ll: + 3y - 4x (11) 2rd’ - Serf’ + 6d’c + 4c‘d (a) II! + 3y - 4x (b) 2rd’ - Sod’ + 6d’c + 4c’d =8x+3y =3cd‘+4c’d Example 5 Expand the following and simplify where possible: (:1) 5:0: - 7) (b) l4(x - 3) - 5(x - 9) (c) (3): - 2)(5x + 3) (d) (21: + 3)’ (e) 3x(x* + 4:: ~ 2) - 3(x* - 5x) (8) 5-‘(X ‘ 7) (b) “(X ' 3) ‘ 5(x - 9) (C) Introductory work (3x - 2)(Sx + 3) =5x"‘-35x = l4x—42-$x+45 = l5x’-l-9x-l0x-6 =9x+3 = l5x’-x-6 (d) (2.: + 3)-‘ (e) 3x(x’ + 4x — 2) - 3(x‘ - Sx) = -(2x+3)(Zx+3)’ ‘-= (2x+3)(4x’+ t2x+9) =8x-‘+24x'+|8.r+l? _r’+36x+27 =8x’+36x‘+S4x+27 =3x’+9x’+9x Example 6 Factorise the following: (a) l2x‘ + sxy (b) x3 + 2x — 24 (c)3a+6+xa+2x (c)2x‘-x—6 (d)4x’—9x (a) I2x’ + 8xy (b) x’ + 2.x - 24 (c) 2:’ - x - 6 = * 4x(3x + 2y) = (x - 4)(x + 6) -= (2: + 3)(x - 2) (d) 4x’-9x (e) 3a+6+xo+2x = x(4x’-9) =3(a+2)+x(a-l-2) = x(lt - 3)(2x + 3) = (a 4- 2X3 + x) —'3x-‘+l2x’-6x-3x’+l5x 5
  10. 10. 6 Understanding Pure Matln-ntaliu Example? Dividelx’ - llx + 6by(x - 2) METHOD I MBTHODZ 2.r’+4.r—3 2.r-‘-Il. r+6___2x’(x-2)+4x’-Ilx+6 x-22x’ -Ilx-I6 22: x-2 1t-1-4;! =2x‘(x-2)+4.r(x-2)-3x-i-6 -ix’ ~ H: x -2 x - 2 4x’ — 8x 3 2x‘Ix - 2) + 4x(x - 2) _ 3(x - 2) - 3x + 6 x - 2 x - 2 x - 2 -3x-I6 =2x’+4.r-3 The quotient is 2x’ + 4x - 3 The quotient is 2x’ + 4x - 3 Examples Find the remainder when x’ + 3.: is divided by x’ + x - I2 METHOD I Mtmrooz I x‘+3x _I(x’+x-I2)-I-2x+l2 . t’+x—l2)x1+3x x1+x-I2 x‘+x-I2 x’+x-I2 ___l+ 2x+I2 2x + I2 x’ + x — I2 Remainder is Zr + I2 Remainder is 2.: + I2 ExampIe9 Express as single fractions: I 2 I 2 (2x + 6) X l “-*’2*Ix+s) “’>(T: ‘37r‘(7T3-) ‘°’ 7 I779“) 5x’ . x “”Ix*= +-ar-’rs*oTTn 2 I 2 2x+6 M %+(x-+3) (b) (x+3)’—(x+3) M (7 )"(x¢| —-9) __I(x+3)+2(2) __l-2(x+3) _2(x+3)x I 20: + 3) (X + 3): — 7 (x+3)(x-3) = x+7 = I-2x—6 _ 2 2(x+ 3) (x + 3)’ 7(x- 3) _-. _x-5 ‘(£737 Sx’ , x “” = 5x’ x(x’-I) (x’+(>. r—7) x _ 5x3 x(x+I)(x-I) (x+7)(x—I) x :5.! (x+I) (x+7)
  11. 11. Improper algebraic fractions Introductory work When the highest power of x in the numerator is equal to or greater than the highest power of x in the denominator, the fraction is said to be Improper. These improper fractions can be rearranged into expressions that are not improper by long division (method l above) or by algebraic ‘juggling’ (method 2 above). The following example shows these methods. Example 10 Rearrange the following into expressions that do not involve improper algebraic fractions. x’+3 (3) }t. _“. x_fi (1,) x+3 (a) By long division I x’ - 4x + Slx’ + 3 x3-4x+$ 4x-2 x’+3 = |+_4x-2 xi-3x+5 x‘-4x-+-5 (b) By long division x-2 x+3)x’+ x-3 x‘+3x -2x-4 -2.: -6 2 x’-l-x-4 2 x+3 : x_2+x+3 E. r¢ret'uB I. Simplify the following: (:1) 2x + 6y -4- 9x (d) x‘y + Zxy’ + 3x’y (b)7x-4x-8 (c) l5a3*2ab-70‘ By algebraic ‘juggling’ x‘+3 = l(x‘-4x+$)+4.t-2 x’--$x+S x’—4x+§ 4.: -2 l(x‘-4x+5)+ x’~4x-+5 x’-4x-t-S 4:: -2 x‘-4x+5 By algebraic ‘juggling’ x’-i-x-4_x(x+3)-2(x+3)+2 x-+3 A x+§ x(x+3)_2(. :+3)+ 2 " x+3 x+3 x+3 _ 2 -x‘2+x+3 (c) 5p‘ + 8p - 3p’ - 3 2. Expand the following and simplify where possible: (3) 3-‘(X - 4) (b) 30: +- 4) - 2(x - 6) (<1) 2:0’ - 3) + )(1_ix - 4) (c) (x + 3)(x - S) (c) 5(x+3)+4(Zr-3) (0 (2x + 3)(x - 4) (g) (21: - 3)’ (h) (3.: - I)’ (i) 5x(x’ — 3:: - 4) - 2(x + 3)’ 3. Factorise the following: (a) lfix’ + 24;)’ (b) lsx’ + loxy + 20.1: (e) x‘ + 3x - I0 (d)x’-+5:-I4 (e)2.r‘-7x-l5 (f)6x’-x-2 (g) 4x‘ + 4x - I5 (h) x’ - 9 (i) l6x’ - 49y’ (j) 2x’ — l8x (| i)ax+3x+2a+6 (I) xa - Zxb + ya - Zyb ‘I
  12. 12. 8 Undnslandthg Pure Mathematics (Divide: (a)x’+2x‘-x-2by(x-I) (b)Zx~‘+9x’-4x-2lby(2x-3) (c)x‘+x’+7x-3by(x‘-x+3) (d)6x‘+l4x’-9x‘-7x+3by(1r‘- l) 5. Find the remainder when: (a) 3x’ + l3x - lis divided by x’ + 4 (b) x’ + 2x’ — 6x- Sisdividedbyx’ +4x+l (c) x‘ + 2x’ + lox’ 4: l3x + llisdividedbyx’ + x‘ + x+ I (d) 2x’ - x‘ - llx’ - lsx + Sisdividedbyzx-l 5 6. Express as single fractions: <= >2-‘+23~" <°>’s‘+"-3-' “>2-"1'—'-"-E-‘ (d’%+xil (°)2.r5+3-% ‘flu-+2y”Gi: n u»s+%: — w*, :.*°~i-, ,—L—. , (m)fx—3_—§+}. .i—“+§3; (0)%‘ (o)(xi4-%)+; ,4:—l3 1. Using the method of long division (see Example I0). rearrange the following into expressions that do not involve improper algebraic fractions: x’+6x-2 b2x*+$ )5x*+2x-ll (d)x’-5x'+9x-1 (a)x’+4x+l ()x’+l (C Jr’-l-x-2 x3-2x+3 8. Using the method of algebraic ‘juggling (see Example I0), rearrange the following into expressions that do not involve improper algebraic fractions: . r’+3x+3 x+2 2x’-4x-l-ll "’—aT4— (b)x+5 ‘°’? TW§‘ 4x+l x’+x'+3x-+5 Jr’-2x-l “": ::‘§ ‘°’"—"; ITr—' “’—: Ti— . -‘ 3.‘-5.-4 x‘+’+$’ l0—l4 (s) ‘h) 0.4surda J25 = 5. 3/8 = 2 and ‘/ l2l = ll: theaeare expressions which can be evaluated exactly. Surds are such expressions as ‘/7, 3/42. J I I0 which cannot be evaluated exactly. They are Irrational. Surds often arise in calculations. For example. using Pythagoras‘ theorem to find the hypotenuse of a right-angled triangle with its other sides 5 and 6 cm: (hypotenusc)’ - 5* + 6' -= 25 + 36 hypotenuse - /6| crn item from
  13. 13. Introductory work 9 ‘Unis is an exact answer and it is often preferable to leave an answer in this form. rather than to give a decimal approximation such as 7-8l02 . . . Expressions involving surds can often be simplified. The following rules applw (i) Ja >< J0 '-' x/ (ab) cs. J5 * J2 = t/ I0 (ii) , /a + , /1; = e. g. , /s + ‘/2 = (iii) a, /c 3; It‘/ c = (a 1; It)‘/ c e. g. 3‘/2 + 4‘/2 : 7J2 and 3,/2 — 4,/2 = -/2 It must be carefully noted that ‘/ a : t ‘/ b 3% t/ (a 3; b) for non-zero a and b e. g. / I6 + /9 # t/25. and J15 — /9 ii J7. Example I] Simplify each of the following and hence show that all three expressions are equal: (a) 1,3 (b) 3V/2 - J8 (c) 3/4’; (a) 133 (b) 3,/2 — t/ s (c) 5; = . = 333 = 3‘/2 - 2‘/2 ~ ‘/2 = /2 = /2 When fractions are involved with surds. it is normal practice to eliminate the surds from the denominator: this is called rationalising the denominator (i. e. clearing it of irrational numbers. ) For expressions of the type 7"-1, we multiply the top and bottom hy Jo. giving $3 = 7!; X :5: = Q E and for expressions of the type fit; we multiply the top and bottom by (b - ‘/ a). giving fig - i’—+'—7;’ x g . — %; .:/ E
  14. 14. I0 Umlrrstamling Pure Malhrmatirs Example 12 Rationalise the denominators of the following fractions (3) 7'1 (b) I _= l /2 I0 3 I0 ( 3 + I) ‘‘’3''2 72'72 "”73'“’-I 7“<a—I)"<.3+I> _ 13 = I0(3(3 + 1) 2 3 - I - st. /3 + 1). Exercise C l. Simplify the following: (a) 5‘/3 — 3‘/3 (b) ‘/12 + 5‘/3 (c) J200 + 1J2 - J72 (d) 6‘/3 ‘ J‘: + J48 (e) J45 + ~/3° ‘ J30 0’) (J3 + /2)(/3 ‘ ~/2) (g) (2‘/5 + , /7)(2‘/5 - , /'7) (h) (2,/3 — 3,/2)(2/3 + 3‘/2) 2. In each of the following state the ‘odd one out‘: (a) 3,/2., /12. , /is (b) 4,/3.‘/48., /I2 (c) , /2o. 2‘/3. i/12 (d) , /2o. 2‘/5. Jill (c) st/2. 2,/ s, / so (r) J6. 35. ‘/2‘/3 . I0 (3) J3. cm t/ I2. §2-.3‘/2 (I) 5»/2. 3. Express each of the following in the form a + I2‘/ c (I) (J3 + . /2)’ on (us + J3)’ re) (3/ ,3 + /2x. /3 - J2) 4. Rationnlise the denominators of the following: (a) 7'3 (b) 795 (c) 5% (d) 3—_'—J5 (c) %g (fl (3) f}: —§ an é/7%‘/7% (i) 5% 0.5 Ouadratlcs Equations of the type ax’ + bx + c - 0. (a 99 0). are called quadratic equations. There are three ways of solving such equations. 1. By fnetorlsntlon This method should only be used if ax’ + bx + c is readily factorised by inspection. Exanplel3 Solve: (a)x'+2x-24-0(b)2x’-x-6-0 (a) x3+2.x—2A=0 (b) 2x‘-x—6=0 (x+6)(x-4)-0 (2x+3)(x-2)-0 So. either 1 + 6 -= 0 So. either 2.: + 3 = = 0 or x-4-=0 or x-2-0 giving x--6 or x-4 giving x--i or x-2
  15. 15. Introductory rmrk I I 2. By the formulax = 4" 1 ~/ "” ‘ ““" 24 Example I4 Solvex’+4.r— l =0. Comparing x‘ - «ix — I = 0 with the general equation at’ + bx+ c 2 0.givesa = |. b= -4andc= -l. Substituting these values in the formula x = -bi ,4» -4ac) 241 gm x=4:, g(;5+4) 2 4- J5 or 2 - J5 (these answers are in surds) 4-24 or -0-24 (these answers are correct to 2 decimal places) ll ll 3. By completing the square This method uses the expansion (x + b)‘ = x’ + 219:: + b’. Notice that the last term. (b’). is the square of half the coefficient of 1:. (2b). Example 15 Solve x*+3x-l=0 1:’ + 3.: - l = 0 . . x‘ + 3x = I Adding the square of hull’ of the coelficient of x to each side of the equation gives . ~‘+3x+G)’= I+G)3 So (x + 3): =« '73 Ia‘-%e xu °f or as decimals x - 0-30 or -3-30 (correct to 2 d. p.) The method of wmpleting the square. used to solve ax’ + bx + c 0. can also be used to find the maximum or minimum value of the expression ax‘ + bx + c. For example, consider the expression x3 + 3x + 4: . r’+3x+4=x‘+3x+(i)‘-(i)’+4 = (x + i)’ + 1 New (x + 3)‘ cannot be negative for any value of x. i. e. (x + 3)’ 2 0. Thus x’ + 3x + 4 is always positive and will have a minimum value of 1 whenx +i = 0. i. e.whenx = -i.
  16. 16. I2 Understanding Pure Mathematics Example 16 Find the maximum value of 5 - 2.: - 41:‘ for x e R. 5-2.x-4x’ -4(x’+}x)+5 -4(x= +;x+, I,)+,4,+s -4(x+ l)’ +4 =5‘ -4(x+i)’ Now (x + 1)’ 2 0. Thus 5 - 2x - 4:‘ hasamaximum value of‘). or 5}. whenx = -5. tllll Exercise D l. Solve the following equations by factorisation: (a). r’—x—l2-0 (b)x’+Sx-24*0 (c)x‘+36‘-'l3x (d)2.r’-x-6=0 (e)1x’-7x-4=0 (f)3x‘+5x-l2=0 6 (g)x(x+5)= (h)4x’-9x= -3-x (i)x(x-I)-2(x+$)=0 . 2 5 - 3 " -l mx-f+2=o (lt)3x-5-= "x (l)1%-I2 2. Solve the following equations by using the formula x ’— —b if $1 _ 4“), giving your answers correct to two decimal places: (a). r’-3x+l-0 (b)x’-x-4=0 (c)x‘+7x+5==0 (d)2.“+3x-I“-0 (e)3x’-6x+2=0 (l')4x‘-3:: -2=0 3. Solve the following equations by using the formula x -- ‘b it (flag. leaving your answers in surd form: (a). "*3.'+l=0 (b)2.r’-6x+I=0 (c)3x’-6x+l=0 4. Solve the following equations by ‘completing the square‘ (leave your answers in surd form): (a)x‘+6.'-l=0 (b)x’-l-4:: -3==0 (c)x’+x-l'=0 (d)x’-x-330 (e)x’-3.x-5:0 (l')2x’-6x+l=0 (g)3x*—4.r—2=~0 (h)4x’-6x+l=0 (i)3x’-6x-l~0 Each of the expressions given in questions 5-13 has a maximum or minimum value for x 5 R. Find (a) which it is. maximum or minimum. (b) its value. (c) the value ofx for which it occurs. 5.. "+4.r-3 6.x’-6x+l 7.3-2x-x’ 8.x’-5.r+I 9.5+-Zr-x’ 10.6-x‘+8x Il.2.r’+3.r+l l2.3x*+2x+2 13.! -5x-2.1:’ I4. Use the method of ‘completing the square‘ to show that the solutions of ax’ + bx + c = Oare given by x = —b 1 (b, _ 4“)
  17. 17. Introductory work I3 0.6 Trlgonometrlc ratios of acute and obtuse angles Acute angles Trigonometric ratios of acute angles are usually defined by using a right- 90_ _ A c angled triangle. Denoting the sides of the triangle by a. b and r: sinzl= % cosA= -3 tnnA‘—‘;5_ ; . “ ltfollowsthat Sm’! = 9 + £= = 9 = tanA cos A b b c A . B (‘. .- b These are general results and should be remembered: Also sin A = g »-— cos (90‘ - A) and cos A = - sin (90‘ -- A) sin x e cos (90' - x) (i. e. sin x " cosine of complement) cos x -= sin (90' - 1:) (cos x = sine of complement) Particular angles 30‘ suit! 60‘ Suppose APQR is equilateral. with sides 2 units and that PM is the perpendicular biscctor of QR. " QM = I unit Using Pythagoras‘ theorem MP1 + MQ’ - PQ1 or MP = - V/ (2' — I’) So MP = ‘/3 Since APQR is equilateral. PQM -. 50' and QPM 30'. From APQM sin 30' 1%: cos 30’ 3?; tan 30' = vigor 3/3-3: and sin 60' 2 3?; cos 60' -— %; tan 60' —— 31‘; — /3. 45‘ Consider a right-angled triangle which is isosceles and in which the equal sides A are I unit in length. The equal angles will each be 45'. I I Using Pythagoras‘ theorem BC’ = - I’ + I’ or BC - , _/2. . . _ I 2_ . _ 2_ . _ l _ B c Hence sm45 -35 or 3%. cos45 -54-. tan45 —T—l. / a Although the trigonometric ratios of 30'. 45‘ and 60' have all been obtained separately. some of these could have been deduced from others: to find cos 60‘. knowing sin 30‘. we could say cos 60‘ -* sin (90' — 60') - sin 30' - 5. to find tan 30'. knowing sin 30' and cos 30' we could say tan30‘ = -—' ~“ : ';3— andsoon.
  18. 18. I4 Undr-mandinx Pure Mulhmrallrs 0’ and 90‘ In the AABC. as the angle 1: decreases so does the length of the side BC. As x approaches 0'. BC approaches zero and AC approaches AB in length. i. e. as x tends to zero (written x -+ 0). then ac - o and AC -o AB, ’ B but sinx= B—(: andasx-0.BC-00. Thus sin0'= XoE--'0. AC alsocosx= %gandasx-o0.AC-oAB. 'l'huseos0‘= Qg - I. and un. r=: —gandasx~0.BC~0. Thustan0"—‘-A25 =0. Similarly. as . r -o 0‘. so AC3 -o 90‘. Thus. considering the trigonometric ratios of ACE as x —o 0': sin90' = £% - l. cos90‘= Xo(E = Oandtan90' = These results should be rnernorised: sin 90‘ E l . . ms 90. 6 = 00 (infinity). o 1 2 J- «Q 12 2 I Example 17 Show that cos’ 30' + cos 60' sin 30' = I. The left-hand side is cos’ 30‘ + cos 60' sin 30‘ = - cos 30'(oos 30') + cos 60‘ sin 30’ - sfl :12 11 —.3.1 2"2 " 2": ’-1+4 -= lasrequired. Obtuse angles Trigonometric ratios of obtuse angles cannot be defined by means of a right- anglod triangle. The sine, cosine or tangent of an obtuse angle is the sine. cosine or tangent of the supplement of the angle. with the appropriate sign. If 9 is an obtuse angle: sin 0 = +sin (l80‘ - 9) i. e. sine of the supplementary angle. cos 0 = -eos (l80' - 0) i. e. -cosine of the supplementary angle. tan 0 -= -tan (l80' - 0) i. e. —tangent of the supplementary angle. A full definition of the trigonometric ratios of angles of any size is to be found in section 4.].
  19. 19. Inlroclucmry work I5 Example 18 Write each of the following as a trigonometric ratio of an acute angle (a) sin I55‘. (b) cos I40‘. (c) tan I30‘. (a) sin I55‘ = +sin (l80‘ - I55‘) A + sin2$' (la) cos I40‘ = -oos(l80' — 140') = -cos 40' (c) tan I30’ = -tan (l80' - I30’) = -tan 50' Example 19 If sin 35' = 05736 write down the values of (a) sin I45‘, (b) cos I25‘ (3) sin I45’ = +sin (l80' - I45’) (b) cos I25‘ = -cos (l80‘ - I25’) - +sin 35' - -cos 55' = +&5736 = -sin 35' -0-S736 Example 20 Given that sin 9 = ,7; and that 9 is an acute angle. find: (n) cos 9. (b) tan 9. First sketch a right-angled triangle containing an angle 9 and with two sides 15 of length 7 and 25 units. such that sin 9 = ,7, . ‘I Using Pythagoras‘ theorem. the third side of the triangle n . /(25’ - 7’) -- 24 Since 9 is acute. all trigonometric ratios of 9 will be positive, hence (a) cos9-it (b)tan9= ,9. Example 2| Given that sin 9 = H and that 9is an obtuse angle. find (a) cos 9. (b) tan 9. As 9 is obtuse, sketch a right-angled triangle containing an angle (l80’ - 9) and with two sides of length 24 and 25 units. 25 2.: As sin 9and sin U80‘ - 9) are numerically equal. sin (l80‘ - 9) ‘— 33. Using Pythagoras‘ theorem. the third side of the triangle = ‘/ (25‘ — 24’) = = 7 (a) cos 9 = -cos (l80' - 9) (b) tan 9 = -tan (l80‘ - 9) = -,7’ = --If. " 0
  20. 20. I6 Undrutundmg Pure Mrarhmratirs Exercise E All the questions in this exercise should be answered without the use of a calculator or tables 1. Write each of the following as trigonometrical ratios of acute angles: (a) sin I30‘ (ls) cos 130‘ (c) tan I30‘ (d) sin I40‘ (e) cos I70’ (l) cos I60‘ (g) tan I00‘ (h) sin 95' 2. Write down the values of the following. leaving surds in your answers: (a) sin 30' (13) cos 30‘ (c) tan 45' (d) sin 45‘ (e) tan 60' (f) sin 90‘ (g) cos 90' (h) sin I50‘ (i) sin 135' 0) cos I20’ (It) sin ISO‘ (1) cos I80‘ (in) ten I35’ (n) cos 150‘ (0) tan I20‘ (p) sin I20’ 3. Show that: (s) sin’ 30‘ + sin’ 45‘ + sin’ 60‘ = i (b) sin 60‘ cos 30' + cos 60' sin 30' = l (c) sin’ 45' + cos’ 45‘ - I 4. If sin 20' = 0-342 write down the values of: (a) sin 160' (1)) cos 70‘ (c) cos H0‘ 5. If sin 40’ = 0-643 and cos 40‘ = 0-766 write down the values of: (a) sin 50' (b) cos 50' (c) sin I40’ (d) sin l30' (e) cos I40’ (f) cos l30' 6. lfsin A = 098 and cos A = 02, find the value oftan A. ‘I. ll‘ sin 8 = 0-954 and cos B = 0-3. find the value of tan 8. 8. If sin 9 - i and Ois acute. find the value of (a) cos 9 (b) tan 9. 9. ll‘ sin 9 = § and Ois obtuse. find the value of (it) cos 9 (b) tan 0. 10. ll‘ sin 9 = 1-‘; and Ois obtuse. find the value ol'(a) cos 0 (1)) tan 0. II. Find the value of . r in each of the following. given that x is acute: (a) sin 50' = cosx (b) cos 30' = - sin 2: (c) cos (40' + x) - sin 30' (d) sin (20' + x) -= cos 50' (e) cos (3: - I0’) - sin IO’ (U cos (2.: + 40') - sin 40' 0.7 Solution of triangles Cosine rule Consider AABC in which CN is the perpendicular from C to BA (produced if necessary——see Case 2). bet the sides ofthe triangle be a. b and c. the angle CAB be A and AN = x, CN = h. (3 Case I Using Pythagoras‘ theorem ACBNa‘ = Ir* +(c-X)’ a 1’ ACAN b’ - It’ + x’ Eliminating It‘ a‘-b‘-x’+(c-x)’ So a’ - b‘ + c‘ - Zcx 3 From AANC x = b cos A '—‘_"""‘——" Hgncc g3 -u I)’ + c’ —— 25¢-gag‘ C
  21. 21. Introductory work I7 Case 2 Using Pythagoras‘ theorem ACBN a‘ = Ir’ + (e + x)’ ACAN b’ '4 Ir’ + x’ Eliminatingh’ a‘= =b’-x’+(c+x)’ So a’ '= b’ + c’ + Zcx From AANCx = boos(l80' — A) = -bcoszl Hence a’ = b’ + c‘ - 2bccosA This is the cosine rule and it is frequently used to determine a side or angle of a given triangle. Example 22 Find the length of the side BC in each of the following triangles: (8) C (11) C 6’ ‘*0 so” if 3 I2 cm A B I2 cm A By the cosine rule By the cosine rule BC‘ = I2’ + 8‘ — 2(l2)(8)cos32' BC‘ =7 12* + 8’ - 2(l2)(8) cos I4 = I44 -41- 64 - l92(0-8480) = I44 + 64 - I92(-0-7660) = 208 — I62-8 = = 208 + I47-l giving BC = 6~72 cm giving BC = I8-8 cm 0 Example 23 55° 3% Find the angle Hin the given triangle. 7 cm By the cosine rule 7‘ = 5‘ + 4’ - 2(4)(5) cos 9 wsa_ 25+ I6-49 40 = -0-2 We now need to find the angle whose cosine is -0-2. Since the cosine of the angle is negative, the angle must be obtuse. Using a calculator, press the inverse cosine button (marked cos" or are cos) to give cos" (-0-2) = l0l-$4‘ (correct to 2 d. p.) Alternatively using tables of cosines. search for 0-2 and read off the appropriate angle (78‘ 28'). As Ornust be obtuse 0 -= I80‘ - 78‘28‘ = l0l ~54‘.
  22. 22. I8 Umlt-rxmmling Pure Mrrlfirnraliu Sine rule Consider the triangle ABC in which the sides are a. b and c and the angles are A. B and C‘. The sine rule states that Proof: Let CD be the perpendicular from C to AB (produced if nccessary—see Case 2) and let CD = Ix. Casel C C1581 l I l III I l I ' __. ... D a A 0 From ABCD sinB= g From ABCD sinB—£ or Ir= asinB . ..(l] or Ir-asinB . ..[l] From AACD sinA —g From AACD sin(l80‘ - 0-; or h= bsinA . ..[2] or Ir= bsin(l80'-A) = bsinA . ..[21 a b Thus. eliminating h from equations [I] and [2] gives a sin 8 = b sin A or . = —. — _ . . u 1 c a = b = c It can similarly be proved that 3% - BEE. hence 33 33-? 81.“ C. Example M Find the length of the side BC in the given triangle. . BC _ 8-3 By “I6 SIDC I'l. IlC - H 8-3 sin 55' so BC Slll 77'- BC = 7-IS cm.
  23. 23. Introductory Mwlt I9 Example 25 Find the angle 1: in the given triangle. . 8 6 By the 9118 rule 3 So sin x - 6 70. I 07048 By calculator or tables sin" 0-7048 = 44-8|’ but we must remember that (l80‘ - 44-81') would also have a sine of 04048. and so x - 44-8|‘ or I80‘ — 44-8l' - 44~8l’ or l3S~l9' In this example the obtuse value of the angle can be discarded as it is not possible when one angle of the triangle is already known to be 70‘. Thus x = 44-81’ In some cases both answers will be possible. Example 26 Find the angle at in the given triangle. 3 4 gm ‘ By the sine rule ‘in 38. - --— A -tun sin .1: So sin 1: - ‘ “g 33 - 08209 hence . r '- 55-17‘ or (l80' — 5S~l7') i. e. I24-83' In this case both answers are possible and there are two triangles that could be drawn from the original information. C, If x. - 55-17' and x, - I24-83', then these are triangles ABC, and ABC, as shown on the right. Note an the solution of triangles Suppose we had to solve the AABC shown (i. e. find all unknown sides and anglu). We would first find c. using the cosine rule, and then have a choice as to which angle. A or B. we would find next. If we use the sine rule. we should then have to decide whether or not the obtuse angle answer is applicable. It is therefore best to find the angle opposite the smaller side first (i. e. angle 3) as this must be smaller than the angle opposite the larger side and will therefore be acute. Alternatively. angle A or 8 could be found by using the cosine rule.
  24. 24. 20 Umlc-muudinx Pure Matllrmaricr Area of a triangle In addition to the rule: area = for finding the area of a triangle. there are two other useful formulae: I Area - {ah sin C (or fhc sin A or lac sin 8). Proq/2 Consider AABC in which in is the length of the perpendicular from H to AC. Area - MC :4 h a = lb X (BC sin C) = lab sin C‘ The other two variations of the formulae can then be obtained from this by use of the sine rule. A 2 Area = , ’'[. r(. ' — a)(. r -- b)(. r ~ c)] + h + . . where s = " 2-——r. the semi-perimeter. This second formula was lint stated by the Greek Mathernatician Hero and is called Hero's formula. (The proof is not given here). Example 27 Find the areas of the given triangles: (3) (b) § on 7 cm fié‘ 6 cm 4 cm Area= §><5X6xsinS0’ _4+-7+9_ = Il-Scm’ 3 : §_—_-Io Area = / [l0(lO - 4)(l0 - 7)(l0 - 9)] = J I80 - I3-4 cm’ Exercise F I. Find the length x in each of the following: (3) (c) r V V D 3 (en , ,% (f)
  25. 25. 2. Find the angle 0 in each of the following: (3) lb) 3 3. Find the areas of the following triangles: (3) A (d) «bf 30., I0 em 4. Solve the triangle ABC given that A = 66'. C‘ = 44'anda = 7cm. 5. Solve the triangle ABC given that A r 45'. c —- Scrnandb = 6cm. 6. Solve the triangle ABC given that C - 50', c - Semanda - l0cm. 7. From a ship A, two other ships 8 and C. lie on bearings of 320‘ and 060‘ respectively. If B is 5 km from A and C is 3 km from A. find the distance from B to C. 8. Three points A. B and C all lie on level ground with B due south of A. The point C lies 250 m from A on a bearing N551? and C is 400 in from B. Find the batting of C from B to the nearest degree. 9. A ship A is 7 km away from a lighthouse L on a bearing 080‘ and a ship B is 5 km away from the lighthouse on a bearing 2|O'. Find the distance and bearing of A from B. 10. Three points A. B and C lie in a straight 7cm 2% V 3 9b < 6° dar- 9 I 4 cm — 1 — (b) fl". .. 5m - -tcm Introductory work 2| 9cm line on level ground with 8 between A and C. A vertical must stands at A and is supported by wires attached to its top and to points on the ground. One such wire is fastened at B and another at C. The wire to C makes an angle of-$0‘ with CA and the wire to B makes an angle of 60' with BA. If BC — 22 m. find the length of the wires and the height of the mast. II. A gardener encloses a triangular plot of land by using an existing hedge of length I6 m for one side and fencing. of total length 20 m. for the other two sides. II‘ the area of the plot is 24‘/3 m’ find the lengths of the sides of the triangle. I2. From a harbour H two ships A and B are situated Zx km due north and x kin on a bearing N a B respectively (a being an acute angle). Use the cosine rule to show that the distance between the two ships is given by x‘/ (5 - 4 cos a).
  26. 26. 22 Untlnnlundrng Pure Malhemaliat 0.8 Three-dimensional trigonometry Angle between a line and a plane in Suppose a line OP meets a plane ABCD at the | point 0. The angle between the line OP and D C the plane is then defined as the angle between OP and the line which is the projection of OP on the plane ABCD. The projection of OP on the plane ABCD is OQ where PQ is perpendicular to the plane ABCD. Hence the required angle is POO. Angle between two planes Suppose two planes ABCD and ABRS intersect in the line AB. as shown. D The angle between the planes is the angle between two lines. both of which are at right- angles to AB. which intersect on AB, and lie 3 one in each plane. The required angle is then R MEN. Line of greatest slope All lines drawn on an inclined plane are not equally steep. In the diagram ABC is horizontal and parallel to OX. The vertical distances of A. B and C above 0 are all equal to It. Hence. since OA < 08 < OC then I’ > I. > A 6K 6? oc thus sinv. > sin 9, > sin 0, i. C. 0' > 0; > 9) The steepest line drawn on the inclined plane will be drawn from O at right-angles to the line ABC. Since ABC is parallel to the line OX, this line of greatest slope will also be at right-angles to OX. When we say a plane is inclined at an angle 0 to the horizontal. we mean that the line of greatest slope on the plane makes an angle 9 with the horizontal plane.
  27. 27. Introductory work 23 Angle between skew lines When considering a two-dimensional situation. two lines that are not parallel will intersect. However. in three dimensions two lines which are not parallel to each other will only intersect if the lines themselves are in the same plane. i. e. coplanar. Two lines which are not coplanar are said to be skew lines. Q Consider the skew lines LM and P0: AB is a line drawn parallel to PQ so that AB and LM intersect (i. e. AB is parallel to PO and is coplanar with LM). The angle between the skew lines PQ and LM is then defined as the angle 0 between the lines AB and LM. Exunple 28 Tltecuboid shown has PQ - 24cm. PS -18cm and QL = 7 cm. Calculate. to the nearest degree: (it) the angle between the line SL and the plane PQLM (b) the angle between the planes LMSR and PQRS (c) the angle between the skew lines PK and ML. (a) PL is the projection of SL on the plane PQLM. Hence the angle between Sl. and the plane PQLM is the angle between SL and LP i. e. st? tan SLP - -§TP‘ I3 " ]7(7= + 243) '. SCP = 36‘ (to nearest degree) (la) Planes LMSR and PQRS meet in the line RS. R0 is in the plane PQRS and is perpendicular to RS. LR is in the plane LMSR and is perpendicular to RS. so the angle between the planes is QRL. tan QRL = % QRL = 2|‘ (to nearest degree).
  28. 28. 2-l Under. sramI: 'niz Pure Marliemurlai (e) MI. is parallel to PQ. so the angle K between the skew lines PK and ML equals / the angle between PK and PQ i. e. KPQ. _ K0 tan ‘ . . :47’ + '3') 18 cm L 24 / KPQ = 39' (to nearest degree) / ‘Ian P 24 cm Q Example 29 A particular hillside may be considered to be a plane inclined at I5‘ to C the horizontal. A straight path up the hillside makes an angle of 30' with the line of greatest slope. Find. to the nearest degree. the angle the path makes with the horizontal. Let ABC D be a portion of the hillside and the path be represented by AV. AK is the projection of AV in the horizontal plane. AH is horizontal and AG is a line of greatest slope on the hillside. Let AG = a and VAK > 9. From AAGH GH U a sin I5‘ hence VK = a sin I5‘ . __ _a_ : a From AAGV cos 30 — Av or AV ms 30. In AAKV sin 9 = }’§ = asin I5‘ x °°‘ 3°‘ 8 a . . sin0-sinl5'oos30' giving 0 I3’ (to nearest degree) A Example 30 A rectangular-based pyramid PQRST has its vertex T vertically above the mid-point M of the side PQ. If P0 24 cm. QR S cm and TQ H 20 cm calculate. to the nearest degree: (a) the angle between the planes RTS and PQRS. (b) the angle between the skew lines RT and QP. (a) Let K be the mid-point of SR. Planes RTS and PQRS meet in RS. TK lies in the plane RTS and is perpendicular to RS. MK lies in the plane PQRS and is perpendicular to RS. The required angle is TR“. which is marked a on the diagram. M _ 5/9°’ _' ll‘) . _ '3! MK 5 5 giving a 73' (to nearest degree) tuna‘
  29. 29. Introductory work 25 (b) RS is parallel to QP. so the angle between the skew lines RT and QP equals the angle between RT and RS i. e. the angle TRS. marked fl in the diagram. From ATRK tan - -1% but TK - / (s= + TM’) and TM’ - 20* -- i2= - not , /(52 + I6’) tan fl - -~——-- I2 giving TRS - S4‘ (to nearest degree) Example 31 A particular hillside may be considered to be a plane inclined at 22‘ to the horizontal. A man walks 60 m due north up a path which follows a line of greatest slope on the hillside. He then walks due west across the hillside to a point F. From P he follows a straight path leading back to his starting point 0 and the length of this path is 85 m. Calculate the inclination of the path OP to the horizontal and the total distance the man has walked. Let OQ be the path due north and let N and M be the feet of the perpendicular: from Q and P to the horizontal plane through the point 0. Let POM = 0. From AOQN QN - 60 sin 22' hence PM = 60 sin 22' From Aom sin a - %l-; ,'- — 5° or 9 = 15-3‘ Total distance walked = or) + or + P0 ° = 60 + / (853 — 60*) + 85 = 205-2 m The inclination of the path OP to the horizontal is l5-3' and the man walks 205-2 m. Exercise 6' I. The diagram shows a cuboid ABCDEFGH with CD = 9 cm. DC - |2cmand FB = Scm. Find (a) the angle between the line EC and the plane ABCD. (b) the angle between the planes EFCD and ABCD, (e) the angle between the skew lines EC and AD.
  30. 30. 26 Undrrstandmg Pure Mathematics 2. The diagram shows a wedge ABCDEF E with ABCD a horizontal rectangle and ABFE a vertical rectangle. BC - I2 cm. F co = 5 cm and BF = 3 cm. Find f D (1) the angle between the line FD and the 3 cm 5 / base ABCD. i °" (b) the angle between the planes El-‘CD B C / and ABCD. '2 °"' (c) the angle between the lines BE and EC. (d) the angle between the skew lines EC and F8. 3. ABCDE is a pyramid with ABCD the square horizontal base. AB - 6 cm andAE = BE = CE = DE = 8cm. Find (a) the height of the pyramid. (b) the angle between the line AE and the base. (c) the angle between the plane EAB and the base. 4. ABC is I horizontal triangle, right-angled at B, and BCDE is a vertical rectangle. BE - 5 cm. ac - l0crn and EAR - 30'. Find (a) BAC. (b) DAC. (c) DAB. 5. The diagram shows a right triangular prism of length l0 cm. The rectangular base BCFE is horizontal. ABC and DEF are vertioul isosceles triangles with A8 = AC = $cmandBC = 6cm. Find (a) the height of AD above the base BCFE. (b) the angle between the plane DBC and the base. (c) the angle between the line DC and the base. 6. The figure ABCDEF shown in the F diagram has equilateral triangles ABC and DEF. of side 8 an. slanting in towards the base such that each triangle makes an C E angle of 60' with the rectangular base ABED. Find (it) how much shorter CF is than BE, 4 (b) the angle between CB and the base to the nearest degree. A 3 7. From the base of at vertical tower of height 30 m point A lies due south and point 8 lies in 3 direction S60'W with A and 8 both on the nine horizontal level as the base of the tower. The top of the tower has angles of elevation of l2’ from A and I5‘ from 8. Find the distance from A to B to the nearest metre.
  31. 31. 8. 9. 10. A vertical mast PQ has 4 equal wires attached to its top P. The other ends of the wires are attached to points A. 3. C and D on the ground. level with Q. If PA makes an angle of 60‘ with the ground. find the height of the mast and the angle between the plane PM! and the ground if (a) ABCD is a square of side 6 rn, (b) ABCD is a rectangle with AB = 6 m and BC = 8 m. A rectangular plot of land ABCD lies in a horizontal plane. A pole of length 3 m is held vertically at C and the top of the pole has an angle of elevation of I6’ from B and 25’ from D. Find the area of the plot and the elevation of the top of the pole from A (to the nearest degree). VABCD is a pyramid with ABCD as the square base of side I0 m. V is situated vertically above a point P inside ABCD thatis5mfromthelineADand3m from the line AB. ll‘ VA = V8 = I3 m find (a) the angle between the plane VAB and the base, (b) the height of the pyramid. (c) the angle between VA and the base. (d) the angle between the plane VDC and the base. Points A. B and C all lie in the same horizontal plane with B on a bearing N30'W from A and C on a bearing N20‘E from A. A vertical tower of height 30 m stands at B and the angle of elevation of its top. from A. is I8‘. A vertical tower of height 20 m stands at C and the angle of elevation of the top of the tower from A is 20’. Find the distance from B to C to the nearest metre. Introductory work 27 12. From an observation point at sea level, an aircraft is observed on a bearing N30‘E, elevation 20‘ and at height 600 m. The aircraft then flies due cast for I000 in. without altering its height. What will be its bearing and angle of elevation from the observation point then? (Give answers to the nearest degree. ) I3. A hillside forms a plane surface inclined at 25‘ to the horizontal. A straight path up the hillside makes an angle of 60' with the line of greatest slope. Find. to the nearest degree. the inclination of the path to the horizontal. 14. A road is to be constructed up a hillside which may be eonsidered as a plane surface making an angle of 40' with the horizontal. At what angle to the line of greatest slope must the road be «instructed if it is to make an angle of 20‘ with the horizontal? (Give your answer to the nearest degree. ) 15. VABCD is a pyramid with the vertex V situated perpendicularly above the centre of the square base ABCD. ll‘ 0 is the angle between the edge VA and the base, and 4: is the angle between the plane VAB and the base. show that tan ¢ = ' , /2 tan 0. 16. The diagram below shows a right triangular prism ABCDEF with the rectangular base ABCD horizontal and triangles ABE and DCF vertical. AE = EB. AB = xand BC = X‘/2. The angle between EFCB and the horizontal is 9 and the angle between EC and the base is 4:. Show that 3 tan d = tan 9. F
  32. 32. 28 Undemandmg Pure Matlmnatiu I7. ‘I11: diagram shows a cuboid ABCDBFGH with A3 = 4:. AB = 3x and BC = y. I! the angle between the skew lines BH and AD is 9 show that ysin9- $xcos0= 0. 18. Two vertical masts BD and CE. each of height Ii. have their bases 3 and C on level ground with C to the east of B. A point A lies on the same level as B and C and is due south of B. The angle of elevation of D from A is 0 and angle DAB = = ql. If the angle of elevation of E from A is a show that sin a = -—- sin Boos 4:.
  33. 33. 1 Funcfions 1.1 Basic concepts Four children. Ann. Bob. Carol and David. are given a spelling test which is marked out of live; their marks for the test are shown in the arrow diagram on the right. By choosing any one name from the set of names. we can find the mark that relates to it. Any relationship which takes one element of one set and assigns to it one and only one element of a second set is said to he a function. The first set is said to be the domain of the function and the second set is the eo-domain. We say that each element of the first set is mapped onto its Image in the second set. The set of all images will be a subset of the eo—domain. and is called the range. Thus in the above example. the domain is {Ann. Bob. Carol. David} the co-domain is {l. 2. 3. 4. S} the range is B. 4. 5} Notice that a function can map more than one element of the domain onto the same element of the range. e. g. Ann -o 4 and Carol - 4. Such functions are said to he many-to-one. Functions for which each element of the domain is mapped onto a different element of the range are said to be oneto-one. Relationships which are one-to-many can occur. but from our definition above. they are not functions. The following diagrams illustrate these facts. oni. ~to-one function many-to-one function the domain is (A. B, C} the co-domain is {V. X. Y. Z} the range is (W. Z} the domain is (A. B. C} the co-domain is (W. X. Y. Z} the range is (W. X. Y} one-to-many relationship iill § Ow)- Because this is a one-to-many relationship, it is not a function. If every element of the co-domain is the image of at least one element of the domain. then the function maps the domain onto the co-domain. Otherwise the function maps the domain into the co-domain.
  34. 34. 30 UIt¢lrr. tIamIi'n; : Pure MuIh¢'maIir. t: Chap! r'r I Figure A shows a one-to-one function mapping (A. B. C) into (W. X. Y. Zl Figure B shows a one-to-one function mapping (A. B. C} onto (W. X. Y) Figure C shows 1! many-to-one function mapping (A. B. C) into (W. X. Y. Z} Figure D shows a many-to-one function mapping {A. B. C) onto (W, X) For most functions which concern us. the domain will be a set of numbers. Suppose a function has the set X as the domain and is such that it doubles any element x of the domain to give the corresponding element of the range. This function would be written f: x -o 2:: or f (x) = ‘ . .x. either form being acceptable. As this function would map the number 2 onto its image. the number 4. we could write this fact as f: 2 -v 4 or asf(2) - 4. Other letters such as g or /1 may be used in place offif we wish to distinguish between functions. Example I Dmw arrow diagrams for the functions (at) f: .t -o Zr (b) g: x —o 3x + l (c) ft: x - . t'’ for the domain l~ I. 0. I} and state the range of each function. (. i) / : . r ~ 2x (5) 3: . r -o 3.: + I (c) It: x - x’ — — 1 _ 7 range is {- 2. 0. 2) range is (- 2. l. 4) range is (0. I} It is sometimes helpful to think of functions as ‘machines’. A box of numbers (the domain) is put into the machine. The machine then alters each number according to some rule and outputs the new numbers into a second box (the range). 0 - 2 This machine subtracts 3 from This machine changes each each number. number to the number 5. The domain is {l. 2. 3. 4) The domain is (I. 2. 3. 4} The range is {-2. - I. 0.1} The range is {5}
  35. 35. Furirlioru 3 I Example 2 State the range of each of the function machines for the domains shown. (1!) function (h) 3 2 -o machine -o ? -2 0 - l -o l 0 fzx -o Zr + 3 I 2 tn) to. I. 2. 3: " ” 2‘ ‘ 3 :3. 5. 7. 9:; the range is :3. 5. 7. 9: (b) i~2. -1.0. 1.2: " ""3 :4. 1.0:; the range is :0. 1.4: Note that it is usual to state the elements of the range in ascending order of magnitude. Therefore the first element of the domain will not necessarily be mapped onto the first element of the range. Example 3 The functions [and g are given asf(x) ~'= x + 3 for x > 0 and g(x) = x’ for -2 :2 . r :2 3. State the range of each of these functions. lfx ,2 0. then . r + 3 3 3. Thus the range offwill bcf(x) 2 3. If -2 < x s 3. thcn0 < x’ S 9. Thus the rangcofgwillbco < gtx) Q 9. Example 4 lff(. ') = 4x - 3 and g(. ') = x’ for the domain of all real x. find: (u) / (2). (b) It ~ 2). (c) g( - 3). (d) the possible values of a if [(0) - g(a). (alf(2)*4(2)-3-5 (b)f(-2)**4(-2)-3= -| | (C) g(-3) = (-3)’ = 9 (d)f(a) = 4a — 3andg(a) = a‘ Thus. if f(0l ' 8(0) wehave 4a - 3 * a‘ = a’-4a-l-3 giving ¢r—' lor3 lfthe domain of a function is not stated. it should be assumed to be the set of all real numbers for which the function is defined. Example 5 The following functions map an element x of the domain onto its image y. i. e. f: x -o _| '. For each of the three functions below. state (i) the domain for which the function is defined. (ii) the corresponding range of the function, (iii) whether the function is one-to-one or many-to-one. (a)f: .t'-». t'+3 (b)f: .'-o‘/ x (c)f: .r-oi xi (a)f: .'-v. r 4- 3 (i) The function is defined for all real x. so the domain is R. (ii) For this domain. the range will contain all elements of R. so the range is R. (iii) Each element of the range is obtained from only one element of the domain. so the function is one-to-one.
  36. 36. 32 Understanding Pure Mathematics: Chapter I (h) f: x -v ‘_/ x. (i) The function is not defined for negative at. so the domain is {x e R: x 2 0:. (ii) For this domain. the range will contain all positive numbers in R. Remember that the symbol J is defined to mean the positive square root. The range is therefore {y e R: y 2 0}. (iii) Each element of the range is obtained from only one element of the domain. so the function is one-to-one. (c) f: .' -. 3‘'—, (i) The function is defined for all real it except 3: = 0. We write the domain as {x e R: x 4 0}. (ii) For this domain. the range will contain neither zero nor any negative numbers because x’ (and hence l/ x’) will be positive. The range is therefore {y e R: y > 0}. (iii) Here the elements of the range can be obtained from more than one element of the domain. e. g.f(3) = l andf(- 3) -— 5, so the function is many-to-one. Exercise 1/! l. State which of the following arrow diagrams show functions. (3) (1?) (C) (d) j =3 D; 1 P’ A 2. State which of the following arrow diagrams show: (i) a one-to-one function mapping into the co—domain, (ii) a one-to-one function mapping onto the co-domain. (iii) a many-to-one function mapping into the co-domain. (n) (5) (0 (<1) V07 3. State the range of each of the following ‘function machines‘ for the domains shown. (2!) swim (bl rwn~—m
  37. 37. 4. Draw arrow diagrams for the functions]: x -o x + 2. g: x -o x’ + I and Ir: x -o (. r +" l)‘ for the domain (—'2. - I. 0, I. 2) and state the range of ench function for this domain. 5. Dmw arrow diagrams for the functions]: x -- (xi. 3: I -« | .r| i nnd ft: x -4 Ix — II for the domain (~ 2. - I. 0. I. 2) and state the range of each function for this domain. 6. if/ (.r) = it + 3 find (:1) [(2). (b)/ (— I). (C) [(6). (d) the value ofa if/ (a) - o. 7. if x(. r) - . r‘ ~ 6 find (it) 5(4). (h) ; ;(—4), (c) (‘(2). (d) the pouibic values ofa if 3(0) e a. 3. lff(. ') - Zr‘ and g(x) = 3 - x find (it) [(3). (h) f(— 3). (c) x(— 3). (ti) the possible values ofa iff(a) - 5-(u). 9. The functionfis given by / (x) - ax -i b. If/ (3) - Jandf(»l) —- 5. find a and Ir. Ii). The mnctlon g is given by g(x) = ax’ - b. lfx(2) -- 5 and g(-— I) - 2. find the value: ofa and b and hence find g, -(- -1). ii. Bitch of the following functions mnps itll element x of the domnln onto its imnrte y. i. e. f(. r) - )2 Find the range of each function for the given domains and state whether the function is one-to-one or muny-to-one. (at) f: x -o . r -4- 3 with domnln (. r: 0 <_ x «T 4). (h) f: .r —- . r - 2 with domain ix: 0 4 x s 4). (c) f: .r —o It with domain (. r: 0 1-: .r 4 3). (d) / : .1 -- Zr with domain ix: -3 < . r < 3). (c) [2 . r -o . r' with domain ix: -3 C . r <_ 3). (f) f: .r —o , /.r with domain (. r: 0 < x < 25). (gift x —o | .r| with domain ix: »3 . < x < 3). (h) f: .r -o . r' with domain R. (i) f: .r -o | .r| with domain R. (j) I: .r —~ % with domain (at: x > ll. (it) f: .r - Ir‘ 4- -I with domain R. (I) f: .r -9 with domain (. r e R: x :4 ll. I2. The following functions map an element x of the domain onto its image y. i. e. [: .r - _t'. For each function state (i) the domain for which the function is defined. (ii) the corresponding rnnizc of the function. (11) [2 . ' -9 2x. ([1) /2 , r —v lxl. (c) / :1 -o ; . (d) f: .r - :3. f - . 3 I3. fund 3 are defined as follows: /: x -v ‘E: (J, 2 : 6 __ _‘ Jx for 0<. r<.3 "J x’ for J<. '<6 Explain why 1; is it function but f is not. . ' + 2 I’ 0 <_ . .-: 2 I4. fund g are defined in follows: I: x -v ‘.3: 2 Q ; 4 4 ’_ . x’ for 0 A? .r 4: 2 ‘J _ 3.: for 2 < .1 < 4 Explain whyfis #1 function whereas g is not. Functions 33
  38. 38. 34 Unalcrsrrrndzng Pun» Matlirmurirsr Chapter I 1 .2 Composite functions Consider the functions Ax) = .r - I and g(x) = it’. Using the ‘machine’ analogy. suppose we input the numbers ll. 3. 5. 7} into machinefand then take the output from this machine and put it into machine g. -o 25 I69 l 8l This combined or composite function is written gf(x) or simply gf. Notice that the function f is performed first and so is written nearer to the variable x. The set ll. 3. 5. 7: is the domain for the composite function and : l, 25. 8|. I69: is the range. Example 6 lfftx) = = Zr and g(. r) —- x’ - I. find the range of each of the following functions for the domain {-2. - I. 0. I. 2} ta) / (x). (b) xtx). (c) fztx). td) it/ (X) (u) : ~2. -L0. 1.2; 14 4-4. -10. 2.4:; the rangeisl-4. -2.0. 2. (b) 1-2. -I. o. I. 2: -‘~ 0.0. -lktherangeisl-l.0.3) (c) I -2. -l.0. l.2l 5 {3.0. -I}-146.0. -2}: the rangeist-2.0.6} (d) : —2. -1.0. 1.2; 1 {-4. -2. o. 2. 4: Is us. 3. —u; the range isl—l.3. tsp Example 7 lff(. r) : 2.: and gut) *2 3.r + I. find (at) [(2). (b) g(3). (c) fg(2). and express g/ ‘(. r) as ii single function Ii(x). (in) IQ) * 3(2) (b) 8(3) = 3(3) + 1 (c) 8(2) = 3(2) + | =- = I0 = 7; 5°/ ’x(2) " f(7) = I4 fix) = 1r. sogftx) = g(2x) = 3(2.tr) + I '_*6X+l h(x)=6x+l 1.3 The Inverse of a function Consider a function f which maps each element . r of the domain X onto its image y in the range Y. i. e. f: x —o)- where . re)( and yeY. Can we find the inverse function. written 1" '. that has domain Y and range X. which maps y back to 1.’? i. e. f": y - x
  39. 39. Function: 35 First recall the definition of a function. Any one-to~one or many-to-one relationship is a function. However. if we attempted to find the inverse of a many-to-one function. we would obtain a one-to-many relationship which is not a function. Thus. only a one-to-one function can have an inverse function. The following diagrams illustrate these points. domain X Lvrange Y domain Y Lerangc X I is a one-to-one function f" is a one-to-one function domain U -L°t’ange V domain V Ltmngc U " ""' x is a many-to-one function g" is one-to-many and so it is not .1 function To find the inverse of one-to-one function. we write the separate operations of the function as a flow chart. We then reverse the how chart. writing the inverse of each operation. Example 8 Find the inverses of these functions. (a)jI. ') = Zr + 3 (b) g(. r) = 2 - x (c) Itfx) = 3-‘ - 3 (a) First write the function as a flow chart with input x and output 2x + 3 ~E-«- Now reverse the flow chart. and write the inverse of each operation With input x. this will now output the inverse function. - 3 So the inverse function r' is x -. " 2 Check: [(2) = = 7. / "'('l) = 2.
  40. 40. 36 Understanding Pure Mathematics. ’ Chapter I (b) First write the function as a flow chart. Now reverse the flow chart. writing the inverse of each operation. _| 0- 0- 4- X So the inverse function g". is: -o 2 - x, i. e. g is its own inverse. Check: g(3) = -l. g"(-I) -—' 2 — (-I) ‘* 3. (c) First write the function as a flow chart Now reverse the flow chart (notice that the operation ‘invert' is its own inverse) I o- 0- +3 0- x x + 3 So the inverse function Ii" is x —. 1711 Check: 5(2) = -2}, h"(-2§) z .1. = 2, Example 9 lffzx-ox + landg: x-oéfiodinsirnilarformz (a)f". (b) 3“. Mix. (d) :3. (e) Us)"- (it) 1': .t-—v-ox+l (b) g: x-o -o -9% S0f"Z. Y-93 ‘I 3 Check: /: 3 —. 4. / ~': 4 -. 3. Sax": x -» ; (i. c.xis its own inverse) Check: gt 6 4 }. 3'': f —o 6.
  41. 41. (cl sax-_% (d) 3"-x-% f: .%-v%+I g: %-M? ‘-= x 3 s°fr"""§+‘ Sogg: x-f (c) In So(flg)": x-x_| Check: fg:3 -0 2. (fg)":2 -9 3. The reader may notice similarities between this work on inverse functions and the process of ‘changing the subject of a formula‘ which may have been met in earlier ymrs. For example. in part (a) of Example 8. we found that for f(x) = = 2:: + 3 then 1" (xi - " ‘ 3 as follows: . This is similar to making 1: the subject of y - 2x + 3 y= -2:-l-3 rearranging y-3=2x then x~= L? inverse functions can be obtained using this technique. Example 10 Find the inverse: of the functions. (3) f(x) = 3x - I. (b) g(. ') = 3 x-I" (a) lff(x) = _v. we require f"(y) = x Ify= =3x-I. lfy- _ 2- + I thcnx 3 "Rafi: So. given y. we can return to x using x-I‘ x-I 3 the expression y 3' I. . 3 the expression - + I. ' + I 7 T'‘“‘ f''‘'‘‘’ “' 3 Thus g"(. ') r J3‘ +1 orx= §+l . V Functions (b) If 30:) - y. we require g"(y) = x So, given y, we can return to x using 37
  42. 42. 38 Umlrt-srundirrg Pure Matltemutit-1: C haptrr I Restricting domains Although we have said that only one-to-one functions have inverses. we can consider the inverse of a many-to-one function if we restrict the function to a domain for which it is one-to-one. For example. I: x —- Jr’ is a many-to-one function for 1| domain R. but if we restrict the domain to Ix s R: x ,2 0:. the function is then one-to-one and will have an inverse. E. rerct'. se I3 I. If flx) = 5x and g(. r) = x’ + 3 find (1!) fl3l. (bl 8(2). (C) 8/0). Id) I30)» (6) 88(3). (7) [(3). 2. If jlx) ? Zr + I and g(x) = x’ find the range of each of the following functions for the domain {-3. -2. — I. 0. I. 2. 3) (:1) fix). (b) xix). (C) fzlxl. (d) x/ (x). 3. lfflx) = —. r. g(x) = I - x’ and h(x) = |x| find the range ofeach of the following functions for the domain (*2. - I. 0. l. 2): (a) ft. lb) xi. (C) sh. (d) ft/ I. (c) hef- 4. lff: x —o 3x. g: x - 2x - I and In x -o x‘ express the following as single functions: (3) fit. (bl Kfl (C) 8'! » (<1) /18. (C) /7!. (7) / '8". (8) 85/- 5. If f : x -o It and g: x -o x -+ 4 state which of the functions fl‘. fg. gf or gg corresponds to (a)x—o2.t+-1,(b). r-ox+8,(c)x-o2x+8. 6. lff: x -o x’ and g: x -o x + 2 state which of the functionsfiifg. gfor gg corresponds to (a)x—ox+4. (b)x-ox’-I-2. (c)x—ox’+4x+4. 7. lff: it --~ x + I g: x -t x’ and h(x) -o écxpress the following in similar form (:2) /3. (b) / h. (cl sh. id) kg. (c) hfx. ll‘) haf- 8. Find the inverses of the following functions: (a)f: .r->3.r—2 (b)fIX-0; (c)f: .r-o5-; r (d)f: x—o(. r+l)’ (e)/ zx-. J1r+2 tr)/ zx-.2-% (h)f= -r-°2x'_6 _ I (8) f- X " _; T'§ 9. lff: x -. 1: + 3. g: x -. :r - 2 and 1.: x -. 3 find in similar form I («'1)f)t- (19) UK)". (0) hit. (dl lbs)”. (0)5813 (T) (ht<f)". 1.4 Ordered palrs Consider the sets A = II. 2} and 8 = {2. 3. 4}. The diagram on the right represents functionf: x -o x + 2 where x 6 A and y e 8. Alternatively we could show this relationship by listing the possible values of x and y as ordered pairs. For the function above. the ordered pairs would be (I. 3) and (2. 4). . v—~. '+2
  43. 43. Functions 39 The eartesiatt product of the two sets A and B is written A X B and is the set of all possible ordered pairs (x. y) where . r e A and y e B. For the sets A -I ll. 2: and B ‘-7 (2. 3. 4). the cartesian product would be: {(1. 2). (I. 3). (I. 4). (2. 2). (2. 3). (2. 4)! Example II If A = l2. 3: and B - (4. 5. 6. ‘I. 8}. write down all the ordered pairs (x. y) such that x e A. _l‘ e B and . r is a factor of y. The required ordered pairs would be (2. 4). (2. 6). (2. 8). (3. 6). Example 12 Four members of the cartesian product of two sets A and B are (2. l). (2. 3). (4. I) and (4. 5). If there are a total of six such ordered pairs in the cartesian product A X B. find: (a) the two missing members of A K B. (b) the elements of A and B. (c) the ordered pairs (x. y) such that . r e A. _v E B and x is less than y. (a) From the ordered pairs (2. I). (2. 3). (4. l) and (4. 5). set A must contain the elements (2. 4) and set B must contain ll. 3. 5). Thus the two missing members of». X B are (2. S) and (4. 3) (b) A - (2.4: and B - ll. 3. 5) (c) The required ordered pairs are (2. 3). (2. 5). (4, 5). 1.5 The graph of a lunctlon Consider the function f: .r —o x + 3 which maps the domain X onto the range Y. lfx e X and y e Y such that f: x -o y. we can write dovm the ordered pairs (x. y). If X = R. then Y = R and there would be an infinite number of such ordered pairs (x. y). However. if we restrict the domain X to certain values. say X = (-2. - I. 0. I, 2}. we can then list the ordered pairs: (-2. I). (- I. 2). (0. 3). (I. 4), and (2. 5). This set of ordered pairs can be plotted as points on a graph. If these points are joined. we obtain the graph of the function f: x -o x + 3 for the domain ix 6 R: lxl 6 2}. We say that the graph of this function has the equation y - x + 3; the ordered pair fixing each point gives the x- and )'-coordinates of that point on the graph. As the value of . r varies over the whole domain. so each corresponding value of y can be obtained; x and y are called variables with x the independent variable and y the dependent variable. -2 -l (l The equation need not always be given with y as the ‘subject’. For example. y = 3 + Scanbewrittenasy - x = 3.)’ - x - 3 = 0.etc.
  44. 44. 40 Undersrandirrg Pure Marlxentaricr: Chapter I Clearly. if the domain were the set of all real numbers R. we could not plot the entire graph of the function. Usually we are only interested in certain values of x or in a particular section of R for which a graph is required. The ordered pairs are usually written as a table of values. For the ordered pairs obtained above. the table of values would be Example 13 For each of the following functions (i) write down the equation of the function. (ii) construct a table of values for the given domain. (iii) plot the graph of the function for that domain. (a)f: x-o2x+I for | x|<2. (b)j’: x-+3 for Osxsd, (c)/ ':x-ox’+2x-2 for —4<x: s2. (a) (i) The aquation will be y -= 2.: + I (b) (i) The equation will be y -I 3. (ii) (ii) In this case y will equal 3 for every value of x.
  45. 45. l J i Fwrcliolu 4} (c) (i) The equation will he y = .r‘ + 2x - 2 (ii) In this case the equation is more complicated so we will build up each y value gradually. Iflflfllflflfl (iii) HHHHHHI EIEHHII In some cases. the chosen points may be insulficient to draw the graph with certainty. so it will then be necessary to take extra values for x to obtain more points. In Example I3 part (c). we could have included 1: = = -} and x = - I} to give two more points at the part of the graph which is difficult to draw. Example 14 State which of the following points lie on the line y = 2x - 3: (2. I). (-2. — I). (-I. -5). For a point to lie on a particular line the coordinates of that point must satisfy the equation of that line. Substituting . r = 2 into y = 2x — 3 gives y -= I. (2. I) does lie on the line. Substituting x = -2 into y = 2x - 3 gives y = -7. (-2. — I) does not lie on the line. Substituting x = (- I. -5) does lie on the line. Thus the points (2. I) and (- l. - 5) lie on the line y = Z: — 3. -lintoya 2x - Jgivesy-'~' -5. Example 15 Find where the line y = 2x - 6 cuts (a) the x-axis (b) the y-axis. (:1) Any point on the x-axis will have a y-coordinate of zero. Substituting y = Olnto y = Z: - égives 0 = 2x - 6 i. e. x -‘ 3 y - 2x - 6 cuts the x-axis at (3. 0). (b) Any point on the yaxis will have an x-coordinate of zero. Substituting 3: I-' 0 intoy H 2.: - 6 givcsy '- -6 ' y = 2x - 6 cuts the y-axis at (0. -6).
  46. 46. 42 Undenlanding Pure Murhmmtlcr: Chapter I It is important to realise that it is possible to write equations and to draw graphs for relationships between x and y that are not functions. The equation x‘ + y’ = 25 is a relationship between x and y and the graph of all possible ordered pairs (x. y) gives a circle. However this relationship is not :1 function because every 1: value does not have one and only one associated y value. e. g. for x -» 0. y could equal 5 or -5. Exercise I C I. If A — I3. 4. 5. 6} and B '— ll, 2. 3. 4), write down all the ordered pairs (x. y) such that x e A. y e B and x is twice y. 2. If A A ' I2. 3. 4}. write down all the ordered pairs (x. y) such that x e A. y e A and x is greater than y. 3. HA - {-2. - I. 0. I. 2) and B = {0. I. 2. 3. 4}. write down all the ordered pairs (x. y) such that x e A. y e B and y = x’. 4. Three elements of the eartesian product A X B are (2. 3). (2. 4) and (3. 5). If there are six such ordered pairs in the eartesian product. find (a) the sets A and B. (b) the other thrx elements of A X B, (c) set C-. a subset ofA X B. such that C - {(x. y): x E A. y e B and x - y}. 5. IIA ~ (I. 2. 3| and 3 ‘ll. 2. 3. 4. 5. 6}. find the ordered pairs ofset C given that C = {(x. y): are A. ye B and y = 2:}. 6. State which of the following points lie on the line y - 8 - 3x. (2. 2): (‘ L s)- (| . 5)! (49 -4)‘ ‘I. If all of the following points lie on the line y = 2x - 6. find the values of a. b. c. d and e: (S. a). (2. b). (- 2. c). (d. 2). (e. 8). 8. If the point (2, 2) lies on the line y '— ax - 4, find the value of a. 9. If the points (2. I) and (-2. - ll) lie on the line y '= ax + b. find the values of a and b. 10. Find where the following lines cut (i) the y-axis (ii) the x-axis. (n)y-Jr-4 (b)y—‘1x-4 (c)y= -I2-2x (d)y= lx+3 (e)y+2x~8 (i')y+5x=3 (g)2)'-x—|2 (lt)y—x’-3x+2(i)y= x‘+x-6 For each of the functions in questions I] to 20, (a) write down the equation of the function. (1!) construct a table of values for the given domain. (c) plot the graph of the function for that domain. ll. f:x—~x+l for | x|£3 l2.f: x-ox-2 for | x|$4 l3.f: x—o2x+3 for 12:13.3 I-tfzx-ox for -2.<, x<4
  47. 47. Ftuu-Ilmu -l3 l5.f: x«->5 for | x|€3 I6./ ';x-o -2 for | x|<3 I7.f: .v—~. ~‘+3.r-2 for —S§x= <2 l8.f: .t-ox’-2.»: -4 for —3{; x<5 l9.f: .r-Zr’-4x—5 for -Zsx-<.4 20.f: .v: -l0+x—. r‘ for -3<. r.<,4 1.6 Some further considerations Odd and even functions Any function for which f( -x) = [(3) is atlled an even function. Two examples of even functions are fzx-ox‘ and gzx-olxl c-2-ft-3)-r9 = f(3) cert-3)-* 3 = 3(3) Any function for whichfl -x) * —f(x) is culled an odd function. Two examples of odd functions are f: x—~x and gzx-x’ e. g.f(-3) —o -3 = -f(3) c. g. g(-3) - -27 = -3(3) (The words even and odd are used because the functions f: x -o x‘ have the property that f(—x) = /'(. r) for even values of n and the property that f(-xl = -f(x) for odd values of n). it should be noted that most functions are neither odd nor even. Example 16 Show that the function f(x) -—- 4x-‘ - x is an odd function. f(-x) f 4(-x)’ - (ex) —— -4x’ + x - -(4x’ - x) = -f(x) Thus f(. r) = 4x’ - x is an odd function. Consider the graph of an even function that maps x —o y. i. e. f(x) = y. Because the function is even. we know that / (-x) - y. Thus for every point (x. y) on the graph of the function. there will also be a point (—x. y). The graph of an even function will therefore be symmetrical about the y-axis. Consider the graph of an odd function that maps x -o y, i. e. f(x) = y. Because the function'is odd. we know that / '(—. r) = -y. Thus for every point (x. y) on the graph of the function. there will also be a point (-x. -y). The graph of an odd function will therefore be unchanged under 2: I80‘ rotation about the origin.
  48. 48. 44 Understanding Pun Mathematics: Chapter I Example I7 For each of the graphs shown below, state whether they are graphs of odd functions. even functions or neither of these. (a) . r (b) -" i (C) . r (d) > ‘ : .' Ar ‘- (a) The graph is symmetrical about the y-axis. Therefore it is the graph of an even function. (h) The graph is unchanged under a I80‘ rotation about the origin. Therefore it is the graph of an odd function. (e) The graph is unchanged under a I80’ rotation about the origin. Therefore it is the graph of an odd function. (d) The graph is neither symmetrical about the y-axis nor is it unchanged under a I80‘ rotation about the Therefore the graph is neither an odd nor an even function. Graphs ofinverse functions If the functionfmaps x —o y thenf" maps y -o 1:. Thus for every point (x, y) on the graph of the function f, there will exist a point (y. x) on the graph of the function 1'". For example. a point (2. I) on the graph of I will mean there is a point (I. 2) on the graph of f". A point (I. 0) on the graph offwill mean there is a point (0. I) on- the graph off " and so on. So we could obtain the graph off" by finding the ordered pairs (1. y) for f: x —o y and piotting them as (y. x). This will give the same line that we would obtain by reflecting the graph of I in the line y = x as the following graphs illustrate. (a) fzx-ox+ 3 for xeR (h) g. x-+2x+ l for rel! (c) hzx-ox‘ for l. reR: x>0} f": x-ox - 3 for xek x-I h": x-o‘/ xfor{xeR: x)0} g": x-o for xef! 2 ‘I’ _r = It 4- I
  49. 49. Fumlahn 45 Exatelse ID I. Show that each of the following functions are odd functions: (a) _I'(. ‘) I 7.'. (b) j(. ‘) = ." + . '. (c) _](. ') = Zr-‘ - fix. 2. Show that each of the following functions are even functions: (u) f(. ') - 4.1’. (b) f(. r) - 2 + x’. (c) f(. r) = 3x’ + Zlxl. 3. For each of the following functions. state whether they are even. odd or neither of these: I (I) f(. ') I 4 - 3.1". (b) _I(. ') = 3x‘ + x. (c) f(. ') -4 . t‘ -- ; . (d) f(. ') - . '‘ + | .'| . (e) fix) - . r’ + | .r| . 4. For each of the following graphs. state whether they are graphs of odd functions. even functions or neither of these. ta) _o~ (b) _. - (C) y. - (d) v. . ' ' ‘ 5. Find the equations of the lines obtained if each of the following lines is reflected in the line _i- - x. . - E - = - . - - - - ' (a) _l - 3. (b) .1 4 x. (c) _| 2x 4. (d) _i J‘ + 2. Exercise IE Exausfuarlon questions I. Find the range of the function f: x - ]x - l| corresponding to a domain -3 < x £ 3. (Cambridge) 2. The functionf: .t- 4% 4- bis such thatf(- I) - I) andf(2) - 9. (i) State the value of x for which I is not defined. (ii) Find the value of a and of b. (iii) Evaluate [(4) and I "(4). (Cambridge) ‘ “’—*’ ll‘) I] I3 3. The arrow diagram on the right represents part of the mapping f: x -o “.7: b. .2 ,2 (i) Find the value ofa and of I». " H (ii) Find the element that under this mapping has an image of 4. '0 '0 (Cambridge) ‘8’ 3 4. Given the functions f: x -o 2x - 3 and g: .r -o find in similar form 7 7 . _ . . . . - ' 6 6 (I) f '. (II) .17. (In) xx. (Iv) fx- (V) (fr) '. 5 5 (Cambridge) 4 4
  50. 50. 46 Undemuntling Pure MalIu'nIaIr'r: .' Chapter I 5. Express in terms of the functions f : x -o _/ x and g: x —o . r + 5. (i) x -o (x + 5). (ii) 3: -o x - 5. (iii) x -s x + I0. (iv) . ' - ‘. ' + I0. (V) x -o x’ + 5. (Cambridge) 6. Given the functions f: x - ls — S and g: x —o express in similar form: (i) 13:. (ii) )9. (iii) g". ts‘ = xx. ete. ). Express in terms of one or both of] and g. (iv) x -» §(x + s). (v) x -. %. (vi) x -. 4x - I5. (Cambridge)
  51. 51. 2 Vectors 2.1 Basic concepts Suppose an insect walks directly from a point A to a point B. 30 cm from A. 3 and then to point C. 20 cm from 3 (see Figure I). The insect has walked / ,5 50 cm altogether but it is clearly not S0 cm from its original position. i. e. si All + BC 15 AC. However. it is true to say that. in travelling from A to B iv and then from B to C. the insect arrives at the same point C as it would '/ have done had it travelled directly from A to C. The position of C. relative / ’ to A. is not changed by the route taken by the insect. We can write " Tfi + E - KB A Figure! where A-B. indicates that both the length and direction of AB are being considered. Quantities involving both magnitude and direction are called vector quantities. Quantities involving only magnitude are called aealar quantities. Thus #75 is the vector whose length is equal to that of the line All and whose direction is that from A to B. The vector a. would be the same length as Iii but will be in the opposite direction. i. e. EA’ will be in the direction from B to A. We write M1] or simply A3 for the length of the vector K3. Thus. from Figure I. Iii) = - 30cm and )tTc'| - 20cm or simply AB - 30cm and BC - 20cm. If we wish to find the vector A? we could. given the necessary bearings. N find its length and direction by the use of trigonometry. Taking the directions of Ti and 53 as N601’. and SIM: respectively. we first make a rough sltetch: By the cosine rule AC’-30’+20’-2¥30X20cos70' AC = 29-83cm By the sine rule 20 _ AC 33 SI! ) 75' _ sin0_ 20sin‘I0‘ " '2—9-s3‘ 9 = - 39-05' Thus A-6 is of length 298 cm and direction sso-95': Alternatively. a solution could be obtained by matting an accurate scale drawing.
  52. 52. 48 Undrn-landing Pun’ Malllevuulir-. r: Chapter 2 Vectors may also be written using single letters written in heavy type: 5 I : £ 6 The arrows on the diagram indicate the directions of the vectors. Thus c - a + h. The length of vector a is written | a| or simply as a. Nam Whilst text books can use heavy type to indicate vectors. this is not easy when writing by hand. The reader is advised to show these single letter vectors underlined. Thusc = a + biswrittenasc = a_ + b Equal vectors For two vectors to be equal. they must have the same magnitude and the same direction. Thus if we represent vector a by a line segment of a certain magnitude and direction. then any other line segment of the same magnitude and direction will also equal a. Parallel vectors Two vectors a and II are parallel if one is a scalar multiple of the other. . i. e. ifa - lb. /1/I If 1. is positive. the vectors are parallel and in the same direction: i. e. they are like parallel vectors. If A is negative. the vectors are parallel and in opposite directions: / i. e. they are unlike parallel vectors / Example I In the parallelogram OABC. 67¢ = a and (T5 = e. The point D lies on AB and is such that AD: DB - l:2. Express the following vectors in terms of a and c: lalfffi (bit? lclfi ldlfi’) (elcTt3 (r)l‘)‘C. First draw a diagram: C O a A (a) 55 is the same length as (‘)3 and is in the same direction. - EB’ - O_A° or CB - a
  53. 53. V¢'(‘I0f3 49 (b) E? is the same length as (73. but is in the opposite direction. . . tTc‘= -E5 or 3-6-= -a (c) H is the same length as (TC and is in the same direction. : T§= e (d) D is one third of the way along AB. - 7tT)'= §7t§ zic (c) (T)-(T5:-+3!-5 -: t+§e (f) ITC. ‘=5§+§? .‘ or I7c‘= D—A’+H3-ta = §c+(-st) = —§c-a+e = ic-a =1:-a Example 2 OABC is a trapezium with 52 = a. i = e and 53 pamllel to and twice as long as The points D and E are the mid-points of AB and CB tespectively. Find the following vectors in terms of a and e: (a) 63‘ (b) Ti __gc) I-T)’. _ Hence show that CA is parallel to and twice as long as ED. First draw a diagram: 7'6'+6E‘+fi3’ -a+e+2a : t+c (a) EX~"C_5+5X (b) K3 = -c-+-a llllll R + [T3 :33’ — gifts’ :1 - }(a + e) from (b) - lta - c) ‘Now E—D° = {(a - e) = -gt-e+a)= gcTt' or2fi)'= <":7° Thus EX is patallel to and twice as long as i-E3. (c) F53
  54. 54. 50 Untlrr. tmmlm_c Pure Mulht-man’c. r.' C huptrr .7’ Non-parallel vectors For two non-parallel vectors a and b. if Aa + [.15 = era 4- fib then A = a and p ‘- fl i. e. we muy equate the coeliicients of vector at appearing on one side of the equation with those of vector at appearing on the other side. and similarly for vector b. Proof": If / Ia + pl! = Ga + fill then la-ca -flh-uh all “ al " hm" ll) ~-ill But (A — a) and (3 — ; r) are scalars and so statement [I] means that a and b are either parallel vectors. which we know they are not. or /1—a~'0 and fl-u=0 i. e. J. ‘ a and [3 -- )1. Example 3 The diagram shows a parallelogram OABC C B E with (i. =—- a and i = e. D is a point on A8 such that AD: DB = 2:l. OD produced e meets CB produced at E. fl — It 5?) and D fie’ ——- ta. Find (:2) E’ in terms ofa and k. (1)) DE in terms of h. it and e. 0 " A (c) the values of It and k. (a) t? €=kc'§ (b) 17:’-—-110.15 - ka - h(: t + jc) (c) To determine 1: and It we need a vector equation containing It and It. Now 53 = 55 + I-53 i. e. ka = -fie + h(a + fie) It'a- -lc+lnt+ihe equating coellicients ofa gives I: w It equating coellicients ofe gives 0 '- -l + {It A -tandk-t Exercise 2/! l. A ship is initially at a position A and (to the nearest degree) of the tanker from travels 6 km on a bearing 055‘ followed its original position. by 8 km on a bearing ISO‘ to reach a final 3. A hilt-er leaves a position A and walks position B. Find. by scale drawing. the 550 in in a direction S60'W followed by distance and bearing of B from A. 700 rn in a direction Nl0’W. Find. by 2. An oil tanker travels 200 lm on a bearing calculation. the distance that the hiker is I60’ followed by 300 km on a bearing then from A and the direction in which 200'. Find. by calculation. the final she must wall: if she is to return directly distance (to the nearest km) and bearing to A.
  55. 55. 4. Four towns A. B. C and D are such that B is 50 lm from A on a bearing 080'. C is 70 km from B on a bearing I20‘ and D is 40 km from C on :1 bearing 2l0‘. Find. by sale drawing. the distance and bearing of (a) D from A. (b) A from D. 5. HA1 -= st and E5 - 3a which ofthe following statements are true? (a) A? is parallel to C-0‘, (b) A1 is equal to 6-D’. (c) A3 is three times as long as CD. (d) C-3 is three times as long as A3. 6. State which of the following vectors are parallel to the vector at + 2b. (a) 3:: + 6b (b) 3: (c) {ti-+5 (d)a-2|: (cl J98 + U25 7. State which of the following vectors are parallel to the vector 2:: - 3b. (a) 2 + 6b (b) a -}h (c) 4:: - 6h (d) 2:: + 3b (e) 321 - 2h 8. Find the values of the scalars It and k in the following vector statements given that a and h are not parallel. (a)Ira+3h=2a+kl) (h)Ita+kb=3b-2a (c) ha + 2: = kh - h (d) a(It + 3) = h(k - l) (e) ha-hh=5a-ka+h-kh (f)ha+ka+3hlc= :t+h+3hl> (g)ka+ha+kh= a-Zhh (h) 2Int -lt+ l2kh = 2:: - Jkn +6Ith 9. The diagram shows a parallelogram OABC with('T' = aand(_)C -» c. Disapoint on CB such that CD: DB = l:3. C D B 3 t Express the following vectors in terms of ti and e. taiifa‘. (I917. tcifiii. anti)’. (e) KB. Vectors 5| It). The diagram shows a triangle OAB with OK = aand5B = h. CisapointonAB such that AC: CB = 3:l. Express the following vectors in tenns of at and b. (a) R’. (b) R. (c) 53’. (d) 6'6. ll. OABC is a trapezium with (Tl = it, E 3 c and C3 = 3:. D is the mid-point of AB. Express the following vectors in terms of a and e. (a) as’. (b) Ha‘. (ci§_o'. id) 63. I2. OAB is a triangle with OA = a and OB - b. C is a point on A8 such that AC: CB = l:2 and D is a point on OB such that 0D: DB = 221. Express the following vectors in terms of a and h. (a) A_B’. (h) A7‘, (c) ti‘. :3. oaac is a parallelogram with (Ti = at and(_)—(3 — c. DisapointonOCsuch that 0D: DC = l:2 and E is a point on AC such that AE: EC = 2: I. Show that OB is parallel to and three times as long as DE. 14. OABC is a parallelogram with OK = :t and 65 = e. D is the mid-point of CB and on meets AC at E. ir6i«. ' - 1.66 and Ali = 1:33 find: (n) W in terms of Ir, it and c. (b) R in temts ofk, a and e. (c) the values of It and k. 15. oaac is :1 trapezium with M = 3:. 63 - eandCB - 2a. Disapointon AB such that AD: DB = 3:l and OD crosses CA at E. trA—r~. ' = IrA_C and i «- 5:53 find the values of It and k.
  56. 56. 52 Understanding Pure Mathematics. ’ C hapter 2 2.2 Posltlon vectors and unit vectors As we saw in section l.5. the position of any point in a plane can be stated in terms of an ordered pair (. r. y). the coordinates of the point. This ordered pair gives the perpendicular distance from the point to each of the coordinate axes. The point P in the diagram has coordinates (3. 2) with respect to the origin 0. We could also define the point P by giving the distance and direction of P from the origin i. e. by stating the vector 5!‘. This vector is called the position vector of P. In addition to writing this position vector as W or as a single letter. say p. it can also be expressed in terms of its horizontal and vertical components. in one of the following ways: (i) As a column matrix. 6P -« (3) (ii) By using unit vectors. A unit vector is a vector of length one unit in a given direction. If we denote a horizontal unit vector by l and a vertical unit vector by j then W = 3| + 2]. (If a third dimension is required, we use It to represent a unit Vector at right angles to the plane containing l and 1. Chapter I7 covers three-dimensional vectors. ) If we represent the vector Jl + 2] by a line segment of the required length and direction. then any other line segment of the same length and direction will also represent the vector 3| + 2]. Thus in the diagram. at. b. e. d and p all equal 3l + 2]. However. with 0 as the origin. only p is the position vector 3l + 2] as it is the vector from O to the point P(3. 2). Note also (from Pythagoras’ theorem) that a vector at + bj has magnitude , /(a* + b’).
  57. 57. Vectors 53 Example 4 For each of the vectors shown below: (i) express the vector in terms ofl and i. where i is a unit vector in the Ox direction and j is a unit vector in the Oy direction, (ii) find the length of each vector. (iii) find the angle 0. (i) a= 3l+4j (i) h= -2l+ 3] (i) c= -3l- 2] (ii) lll - J13’ 4' 4’) (ii) lbl ' Jl('2l’ + 3'] (ii) | ¢| ' J[(“3)’ 4' (-2): ] - 5 units - , /I3 units - ‘/13 units (iii) tan 0 - 1 (iii) tan 0 - Q (iii) tan 0 = i ' 0 - S3-l‘ 0 - 56-3’ 0 = 33-‘K Example 5 Find the unit vector in the same direction as the vector 2l - 1. 2| - lis a vector of length J5 units. Thus a vector parallel to 2i - j and of length l unit will be ‘-32: — l) or= §3<2i -1) The required unit vector is 3é§(2l - j). Note: Whilst we use I and j for the unit vectors in the direction of the x- and y-axes respxtively. a unit vector in the direction of some vector a is usually written I. so_g = _'_ I-I Example 6 Find a vector that is of magnitude 6 units and is parallel to the vector i + j. The vector! 4» jhas magnitude . /(I! + I’) = J2 units 6 Thus a vector parallel to I + 1 but of length 6 units will be -750 + j) - 1/2: + 3./21
  58. 58. 54 Ullderxlandinx Pure Mathematics. ‘ Chapter 2 Example 7 if A has position vector a. B has position vector II and C has position vector c. find the following vectors in terms of a. b and e: (a) K3 (b) Ex’ (c) ETC‘ (d) A-6. First draw a sketch showing the position vectors: (a)K§-. A—(5+67I§--a+borb-a ib)B_A. =B—0°+(_)_A°= —b+aora-b (c)B—(‘-B—0°+0_C‘= --h+e ore-It (d)A_(‘-I-3+(-)-C: --a+eore-a Example 8 Point A has position vector a. point B has position vector 21) and C is a point such that ATC‘ = - so - 3.. Show out A. B and C are collinear. First make ll rough sketch with O as this origin: /Ti - A-0’ + 01 = -a + 2]) but A7.‘ = as - 3. = 3(-I + 2|! ) = 375 Thus H and A_(f are parallel and, as the point A is common to both these vectors. the points A. B and C are collinear. Addition and subtraction of vectors in component vector form When vectors are expressed in terms of I and 1, it is easy to add two or more vectors to find their combined or resultant vector. From the vector diagram shown. we know that 61-’ + F0‘ - 075 . ..m Taking i and] to be the unit vectors in the directions Ox and Oy respectively. we can see from the diagram that (-J? - 2! + 4], W =4: —jandcTQ’ = 6! + 31. Substitution into [I] gives (Ii 4- 4]) 1- (4l --j) "' 65 + 3] Thus to add vectors expressed in terms of i and j. we add the i components together, 2 + 4 = - 6. and the j components together. 6 + —l - 3. Applying a similar method for subtraction sivcs 61‘-Po-t2i+4n—t4i~i> = -21 + 5]
  59. 59. Vector: 55 To interpret 6:5 - P_Q. we can think of it as 6-5 + (--P3). The vector (-PQ) is equal in magnitude to %' but opposite in direction. 50. (Tl; - 175 means that we must add to 53 a vector equal in length. but opposite in direction. to that or F6. Similarly. multiplying a vector expressed in terms of I and j by a scalar is straightlbrward. Ira = 31+ -ljtherI2a - 6i + 81. 3:1 = 9] + I21. etc. Example 9 lfa = '- 3i + 4) and! ) = 2i 4» 8] find: (:1) n + b. (b) la + In], (c) a - 2h, (d) is - lb]. (a)n+h = (3i+4i)+(2i+8i) (b)| a+b| -|5i+I2j| = st + I2] = / (52 + I2’) = I3 units (c) a - 2b = (3! + 41) - 2(2i + 81) (d) la — 2:»; = | -I - I2}! = -I - I2] — , /[(-l)‘ + (—|2’)] — , /I45 units Example 10 It‘ a point A has position vector i + 2j and B has position vector Si + 5, find the position vector of the point which divides AB in the ratio I: -3. Note first that the ratio involves two numbers of dillerent sign. This indicates that the required point divides AB externally rather than internally. The dilference between these two cases is illustrated below: Point P dividing AB internally in the ratio l:3 Point I’ dividing AB externally in the ratio l:3. (By writing this as l: -3 we need not A state that the division is external). X, .- In this case the vectors A_l5 and W are in the In this case the vectors K7’ and F3 are in same direction. opposite directioms. Hence the use of the minus sign in the ratio.
  60. 60. 56 Um! cr. rIamIin_g Pure Math: -marirx: C haprcr 2 To solve the given question we first draw a diagram showing the points A, B and P. position vectors a. b and p respectively. (Note that the diagram is an aid to calculation and need not be drawn accurately). Now 0-}; —‘ p = (TA. + Xi‘ = at + u n + l(-ls + a) in - lb Buta = I + Zjandb ~ Si-I-j 9 ‘- ill + 21) — ltsr + I) R -I + gj The point dividing AB in the ratio l: —3 has position vector -l + £1. The general formula for the position vector of a point dividing a line in a given ratio is given by the section theorem: The point P which divides the line AB in the ratio Am has position vector pa + Ah I + p p where p = and a and b are the position vectors of A and B respectively. m 2! 5' D -, + ) + | ( + I + (-3) = -i + 1} usbcfore. Applying this formula to example I0 gives p = Other base vectors in two dimensions Any vector lying in the plane of the unit vectors I and I can be expressed in the form xi + yj. The unit vectors I and j are base vectors from which other coplanar vectors can be built up. Although it is usually convenient to use I and j as the base vectors. any pair of nomparallel coplanar vectors :1 and It could be used instead. Thus for three coplanar vectors a. b and c. it is possible to express any one of the vectors in terms of the other two. i. e. e = 1.2: + p. h b «T 1,1: + We a '— Jt, b + me where 7.. . p. . 1,. 11,. 1, and it, are suitable scalars. Example I I With a = and b = as base vectors. express 1: -*~ and 2 If e=1l. a+pb If d= /'la-l-pb (iv) - *6) * "(-3 d = (3) in the form 2.2 + pl).
  61. 61. Vrrlors 57 Thus 5=3A+2p Thus 3=3A+2p and |7=-‘SJ. -u and 2=5/1-, u Solving simultaneously gives .1 -= 3 and 11 = = - 2 Solving simultaneously gives 11 = .7, and ; i = ,9, ' c=3a—2h d-~.73a+,9,h Exercise 28 1. Taking I and j to be unit vectors in the directions Or and 0y respectively. give the position vectors of the points A to H shown in the diagram. Each square in the grid is a unit square and O is the origin. 2. For each of the vectors shown below: (i) express each vector in terms of I and j where I is a unit vector in the 0: direction and j is a unit vector in the 0y direction. (ii) find the length of each vector. (iii) find the angle 0. la) (b) (C) -4 -3 -2 -I 3. lfal + 8j is parallel to 2i + 4). find the value of a. 4. ll‘ :1 = I + 2] find i. a unit vector in the direction of xi. 5. lfb = Ill - j find 5. a unit vector in the direction of b. 6. Find a vector that is of magnitude 39 units and is parallel to St + I2). 7. Find a vector that is of magnitude 3J5 units and is parallel to 2i - J. 8. Find a vector that is of magnitude 2 units and is parallel to 4i - 3). 9. If the point P has position vector 2i -4- 3] and point Q has position vector ‘It + 4;, find: (a) :76. (b) ('23. 10. If the point P has position vector 7i - 3] and point Q has position :0 vector 5| + 5). find: (a) PQ. (b)
  62. 62. S8 b'ndersIandin; ; Pure Malhcmam-rs C Irupler .7 II. The point P has position vector - Si -4» 3] and Q is a point such that F6 - 7i — j. Find the position vector of Q. I2. The point P has position vector 3! - 2] and Q is a point such that or’ -» 2i ~ 31. Find the position vector ol'Q. 13. Using :1 - and b = ( A as base vectors. express the vectors e : and d -- in the form ). a + pb. 14. Given that a 3i — jand b 2i '4' i. find: (.1) la]. (b) | b|. (c) a + I), (d) | a + b| . 15. Given that SI — and b — (_ mid: (at) as + 2!). (b) la + Zhl. (c) 2:: + 319. (d) |2n + 3b]. I6. The point A has position vector 3i + jand point B has position vector lol 4- 3. Find the position vector of the point which divides AB in the ratio 3:4. 1 17. The point C has position vector (3) and point D has position vector 9 Find the position vector of the point which divides CD in the ratio 4: -3. I8. The point K has position vector 3i +~ Zj and point L has position vector I + 3]. Find the position vector of the point which divides KL in the ratio (a) 4:3. (b) 42-3. 19. The three points A, B and C have position vectors a. II and e respectively. W: =— 3b — 2:. show that A. B and C are collinear. 20. The three points A, B and C have position vectors l - j. Si - 3] and Ill - 6] respectively. Show that A. B and C are collinear. 2|. Points A. B. C. D. E and F have position vectors (-31) and (_ respectively. Find which of the points C. D. E and F are collinear with A and B. 2.3 The scalar product The scalar product of two vectors a and In is defined as the product of the magnitudes of the two vectors multiplied by the cosine of the angle between the two vectors. Writing this scalar product as a . I) it follows from the definition that :1 . h = |a[| b[cos 9 The term scalar product is used because | a|. |h| and cos Gare all scalar quantities and so their product will be at scalar quantity. We read a . It as ‘a dot b‘ and for this reason the scalar product may also be referred to as the dot product.

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