(Solution Manual) Higher Engineering Mathematics

Bird, 5

Engineering

Solution Manual

5th Edition

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Solution Manual for Higher Engineering Mathematics, The downloadable file consists of 644 pages, and the total size of the file is 5.56 MB. Purchase now only for highly discounted price.

Author(s): Bird

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HIGHER ENGINEERING MATHEMATICS 5TH EDITION JOHN BIRD SAMPLE OF WORKED SOLUTIONS TO EXERCISES ii INTRODUCTION In ‘Higher Engineering Mathematics 5th Edition’ are some 1750 further problems arranged regularly throughout the text within 250 Exercises. A sample of solutions for over 1000 of these further problems has been prepared in this document . The reader should be able to cope with the remainder by referring to similar worked problems contained in the text. CONTENTS Page Chapter 1 Algebra 1 Chapter 2 Inequalities 13 Chapter 3 Partial fractions 19 Chapter 4 Logarithms and exponential functions 25 Chapter 5 Hyperbolic functions 41 Chapter 6 Arithmetic and geometric progressions 48 Chapter 7 The binomial series 55 Chapter 8 Maclaurin’s series 65 Chapter 9 Solving equations by iterative methods 71 Chapter 10 Computer numbering systems 85 Chapter 11 Boolean algebra and logic circuits 94 Chapter 12 Introduction to trigonometry 110 Chapter 13 Cartesian and polar co -ordinates 131 Chapter 14 The circle and its properties 135 Chapter 15 Trigonometric waveforms 144 Chapter 16 Trigonometric identities and equations 155 Chapter 17 The relationship between trigonometric and hyperbolic functions 163 Chapter 18 Compound angles 168 Chapter 19 Functions and their curves 181 Chapter 20 Irregular areas, volumes and mean values of waveforms 197 iii Chapter 21 Vectors, phasors and the combination of waveforms 202 Chapter 22 Scalar and vector products 212 Chapter 23 Complex numbers 219 Chapter 24 De Moivre’s theorem 232 Chapter 25 The theory of matrices and determinants 238 Chapter 26 The solution of simultaneous equations by matrices and determinants 246 Chapter 27 Methods of differentiation 257 Chapter 28 Some applications of differentiation 266 Chapter 29 Differentiation of parametric equations 281 Chapter 30 Differentiation of implicit fun ctions 287 Chapter 31 Logarithmic differentiation 291 Chapter 32 Differentiation of hyperbolic functions 295 Chapter 33 Differentiation of inverse trigonometric and hyperbolic functions 297 Chapter 34 Partial differentiation 306 Chapter 35 Total differential, rates of change and small changes 312 Chapter 36 Maxima, minima and saddle points for functions of two variables 319 Chapter 37 Standard integration 327 Chapter 38 Some applications of integration 332 Chapter 39 Integration using algebraic substitutions 350 Chapter 40 Integration using trigonometric and hyperbolic substitutions 356 Chapter 41 Integration using partial fractions 365 Chapter 42 The t = tan /2 substitution 372 Chapter 43 Integration by parts 376 Chapter 44 Reduction formulae 384 Chapter 45 Numerical integration 390 Chapter 46 Solution of first order differential equations by separation of variables 398 Chapter 47 Homogeneous first order differential equations 410 Chapter 48 Linear first order differential equations 417 Chapter 49 Numerical methods for first order differential equations 424 Chapter 50 Second order differential equations of the form d2y   dy   435 a dx2 b cy 0 dx Chapter 51 Second order differential equations of the form d2y   dy   441 a dx2 b cy f(x) dx Chapter 52 Power series methods of solving ordinary differential equations 458 iv Chapter 53 An introduction to partial differential equations 474 Chapter 54 Presentation of statistical data 489 Chapter 55 Measures of central tendency and dispersion 497 Chapter 56 Probability 504 Chapter 57 The binomial and Poisson distributions 508 Chapter 58 The normal distribution 513 Chapter 59 Linear correlation 523 Chapter 60 Linear regression 527 Chapter 61 Sampling and estimation theories 533 Chapter 62 Significance testing 543 Chapter 63 Chi-square and distribution -free tests 553 Chapter 64 Introduction to Laplace transforms 566 Chapter 65 Properties of Laplace transforms 569 Chapter 66 Inverse Laplace transforms 575 Chapter 67 The solution of differential equations using Laplace transforms 582 Chapter 68 The solution of simultaneous differential equations using Laplace transforms 590 Chapter 69 Fourier series for periodic functions of period 2 595 Chapter 70 Fourier series for a non-periodic functions over period 2  601 Chapter 71 Even and odd functions and half-range Fourier series 608 Chapter 72 Fourier series over any range 616 Chapter 73 A numerical method of harmonic analysis 623 Chapter 74 The complex or exponential form of a Fourier series 627 © 2006 John Bird. All rights reserved. Published by Elsevier. 1 32 4 5 2. Find the value of 5 pq2r3 when p = 2 , q = -2 and r = -1 5 5. Simplify x2 y3z x3 yz2  and evaluate when x = 1 , y = 2 and z = 3 2  8. Simplify    a3b2c 2 ab3 1  1 1 a3 b cCHAPTER 1 ALGEBRA EXERCISE 1 Page 2 5 pq2r3  5 2 22 13  5 2  4 1  -8       x2 y3z x3 yz2  = x23 y31 z12  x5 y4 z3 1  1 5 4 3 24  33 33 33 27 1 When x = , y = 2 and z = 3, x5 y4 z3 =   2 3     13 2  2  25 254 2 2 2  3 1  1  1  1  1 1  1 3  2 3   2  31 2 2  a 2 bc  a 2 b 2 c   a b c  a b2 c      When a = 3, b = 4 and c = 2, c2   9 2  ± 4 1 c2 22 4 2  1  1  1  a3b2 c 2 ab3 1  1 1 1 3 2 2 3 3  3 1  3   1  1  1     1 1  1829   1  3    a b c a b  a 3 2  b 2 3 2  c  2   a6 b c  a3 b c         3 2 3 1 a 2 b2 c 11 1  3 = a 6 b3 c 2 or 6. Evaluate  a 2 bc 2  a 2 b 2 c  when a = 3, b = 4 and c = 2   1 3 1  1     a2 b a2 b 6 a11 3 b c 3 5 © 2006 John Bird. All rights reserved. Published by Elsevier. 2   3. Remove the brackets and simplify: 24 p  235 p  q  2  p  2q 3qEXERCISE 2 Page 3 24 p  235 p  q  2  p  2q 3q= 24 p  215 p  3q  2 p  4q  3q = 24 p 30 p  6q  4 p  8q  3q = 24 p 26 p 11q = 24p – 26p + 11q = 11q – 2p 2 y  4  6 y  3 4  5 y = 2 y  4 6 y  3 4  5 y  2 y  2 3y 12  5 y = 2  3 y  12 3 y a2  3ab  2a  6b  ab = a2  3ab  2a  ab  a2 6b 6a2b ab a 6b 2  a2  ab = ab 8. Simplify a2  3ab  2a  6b  ab 6. Simplify 2 y  4  6 y  3 4  5 y © 2006 John Bird. All rights reserved. Published by Elsevier. 3 t t t t l g 2 23. Solve the equation: 1 3a  2 5a  3 1  0 4. Solve the equation: 3 t 1t  6 EXERCISE 3 Page 4 1  1 from which, (5a + 3) = -(3a – 2) 3a  2 5a  3 i.e. 5a + 3 = -3a + 2 and 5a + 3a = 2 – 3 Thus, 8a = -1 and a = - 1 8 If 3 1 6 then 3  6 1 t   i.e. 3  6  6 from which, 6  6  3  3 Hence, if 6 = 3 then  6  2 3 and t = 22 = 4 t  t 2 l If t  2then  and 2       t 2 gt 2 from which, l  g    or l  4 2 If m  L L  rCR then m  L  rCR  L i.e. mL  mrCR  L 7. Transpose m  L for L L  rCR 6. Make l the subject of t  2l g t t t t t t l g g © 2006 John Bird. All rights reserved. Published by Elsevier. 4 from which, mrCR  L  mL  L   mand L  mrCR   m If x y  1 r 2 1 r 2 then x 1 r 2   y 1 r 2   from which, and x  xr 2  y  yr 2 x  y  yr 2  xr 2  r 2  y  x  Thus, r 2  x  y x  y and r    8. Make r the subject of the formula x 1 r 2 y 1 r 2  x  y   x  y © 2006 John Bird. All rights reserved. Published by Elsevier. 5 2. Solve the simultaneous equations 5a  1 3b 2b  a  4  0 3. Solve the simultaneous equations x  2 y  49 5 3 15 (1) 3x  y  5  0 7 2 7 (2) EXERCISE 4 Page 5 5a + 3b = 1 (1) a + 2b = -4 (2) 5 × (2) gives: 5a + 10b = -20 (3) (1) – (3) gives: - 7b = 21 from which, b = 21 7 = -3 Substituting in (1) gives: 5a + 3(-3) = 1 from which, 5a = 1 + 9 = 10 and a = 10 = 2 2 8 = 4 27 Substituting in (3) gives: 3x + 40 = 49 and 3x = 49 – 40 = 9 from which, x = 3 If 8x2  2x 15  0 then (4x – 5)(2x + 3) = 0 3 2 4.(b) Solve the quadratic equation by factorisation: 8x2  2x 15  0 15 × (1) gives: 3x + 10y = 49 (3) 14 × (2) gives: 6x – 7y = -10 (4) 2 × (3) gives: (5) – (4) gives: 6x + 20y = 98 27y = 108 (5) from which, y = 10 hence, 4x – 5 = 0 i.e. 4x = 5 i.e. x = 5 4 and 2x + 3 = 0 i.e. 2x = -3 i.e. x =  © 2006 John Bird. All rights reserved. Published by Elsevier. 6 5  52  4(2)(4)5  (25  32) 6.(a) Solve the quadratic equation, correct to 3 decimal places: 2x2  5x  4  0 If roots are x = 2 and x = -5 then (x – 2)(x + 5) = 0 i.e. x2  2x  5x 10  0 i.e. x2  3 x  10  0 If 2x2  5x  4  0 then x    2(2) 4 4 Hence, x = 5  57 4 = 0.637 or x = 5  57 4 = -3.137 5. Determine the quadratic equation in x whose roots are 2 and -5 5  57 © 2006 John Bird. All rights reserved. Published by Elsevier. 7 3. Determine 10x2 11x  6 2x  3EXERCISE 5 Page 8 2x  3 10x2  15x - 4x - 6 - 4x - 6 Hence, 10x2 11x  6 2x  3 = 5x - 2 x  y x2  2xy  y2 x3  x2 y 2x2 y  3xy2 2x2 y  2xy2 xy2  y3 xy2  y3 Hence, x3  3x2 y  3xy2  y3 x  y = x2  2 xy  y2 x 1 5x2  5x 4x + 4 4x - 4 8 Hence, 5x2  x  4 x 1 = 5x + 4 + 8 x  1 6. Find 5x2  x  4  (x 1) x3  3x2 y  3xy2  y3 5. Divide x3  3x2 y  3xy2  y3  by (x + y) 5x - 2 10x2 11x  6 5x + 4 5x2  x  4 © 2006 John Bird. All rights reserved. Published by Elsevier. 8 x  3 5x4  15x3 18x3 18x3  54x2 54x2  2x 54x2 162x 160x + 1 160x - 480 481 5x4  3x3  2x 1 3 2 481 Hence, x  3 = 5 x  18 x  54 x  160 x  3 8. Determine 5x4  3x3  2x 1  (x  3) 5x3 18x2  54x 160 5x4  3x3  2x  1 © 2006 John Bird. All rights reserved. Published by Elsevier. 9 2. Use the factor theorem to factorise x3  x2  4x  4 EXERCISE 6 Page 9 Let f(x) = x3  x2  4x  4 If x = 1, f(x) = 1 + 1 – 4 – 4 = -6 x = 2, f(x) = 8 + 4 – 8 – 4 = 0 hence, (x – 2) is a factor x = 3, f(x) = 27 + 9 – 12 – 4 = 20 x = -1, f(x) = -1 + 1 + 4 – 4 = 0 hence, (x + 1) is a factor x = -2, f(x) = -8 + 4 + 8 – 4 = 0 hence, (x + 2) is a factor Thus, x3  x2  4 x  4 = (x + 1)(x + 2)(x – 2) Let f(x) = 2x3  x2 16x 15 If x = 1, f(x) = 2 – 1 – 16 + 15 = 0 hence, (x – 1) is a factor x = 2, f(x) = 16 – 4 – 32 +15 = -5 x = 3, f(x) = 54 – 9 – 48 + 15 = 12 x = -1, f(x) = – 1 – 1 + 16 + 15 = 29 x = -2, f(x) = -16 – 4 + 32 + 15 = 27 x = -3, f(x) = -54 – 9 + 48 + 15 = 0 hence, (x + 3) is a factor 2x3  x2 16x 15  2x3  x2 16x 15 (x 1)(x  3) x2  2x  3 x2  2x  3 2x3  4x2  6x 5x2 10x  15 5x2 10x  15 Hence, 2 x3  x2  16 x  15 = (x – 1)(x + 3)(2x – 5) 4. Use the factor theorem to factorise 2x3  x2 16x  15 2x - 5 2x3  x2 16x 15 © 2006 John Bird. All rights reserved. Published by Elsevier. 10 Let f(x) = x3  2x2  x  2 If x = 1, f(x) = 1 – 2 – 1 + 2 = 0 hence, (x – 1) is a factor x = 2, f(x) = 8 – 8 – 2 + 2 = 0 hence, (x – 2) is a factor x = 3, f(x) = 27 – 18 – 3 + 2 = 8 x = -1, f(x) = -1 – 2 + 1 + 2 = 0 hence, (x + 1) is a factor Hence, x3  2x2  x  2 = (x – 1)(x – 2)(x + 1) If x3  2x2  x  2  0 then (x – 1)(x – 2)(x + 1) = 0 from which, x = 1, x = 2, or x = -1 6. Solve the equation x3  2x2  x  2  0 © 2006 John Bird. All rights reserved. Published by Elsevier. 11 2. Determine the remainder when x3  6x2  x  5 is divided by (a) (x + 2) (b) (x – 3) 4. Determine the factors of x3  7x2 14x  8 and hence solve the cubic equation x3  7x2 14x  8  0 EXERCISE 7 Page 11 (a) Remainder is ap3  bp2  cp  d where a = 1, b = -6, c = 1, d = -5 and p = -2 Hence, remainder = 1(2)3  6(2)2 1(2)  5 = -8 – 24 – 2 – 5 = -39 (b) When p = 3, remainder = 1(3)3  6(3)2 1(3)  5 = 27 – 54 + 3 – 5 = -29 Remainder is ap3  bp2  cp  d where a = 1, b = 7, c = 14, d = 8 Let p = 1, then remainder = 1(1)3  7(1)2 14(1)  8 = 30 Let p = -1, then remainder = 1(1)3  7(1)2 14(1)  8 = -1 + 7 – 14 + 8 = 0, hence (x + 1) is a factor Let p = -2, then remainder = 1(2)3  7(2)2 14(2)  8 = -8 + 28 – 28 + 8 = 0, hence (x + 2) is a factor Let p = -3, then remainder = 1(3)3  7(3)2 14(3)  8 = -27 + 63 – 42 + 8 = 2 Let p = -4, then remainder = 1(4)3  7(4)2 14(4)  8 = -64 + 112 – 56 + 8 = 0, hence (x + 4) is a factor Hence, x3  7x2 14x  8 = (x + 1)(x + 2)(x + 4) If x3  7x2 14x  8  0 then (x + 1)(x + 2)(x + 4) = 0 from which, x = -1, x = -2 or x = -4 © 2006 John Bird. All rights reserved. Published by Elsevier. 12   Remainder is ap3  bp2  cp  d where a = 2, b = -1, c = -7, d = 6 Let p = 1, then remainder = 2(1)3  (1)(1)2  (7)(1)  6 = 2 - 1 – 7 + 6 = 0, hence (x - 1) is a factor Let p = 2, then remainder = 2(2)3  (1)(2)2  (7)(2)  6 = 16 - 4 – 14 + 6 = 4 Let p = -1, then remainder = 2(1)3  (1)(1)2  (7)(1)  6 = -2 - 1 + 7 + 6 = 10 Let p = -2, then remainder = 2(2)3  (1)(2)2  (7)(2)  6 = -16 - 4 + 14 + 6 = 0, hence (x + 2) is a factor Let p = -3, then remainder = 2(3)3  (1)(3)2  (7)(3)  6 = -54 - 9 + 21 + 6 = -36 2x3  x2  7x  6 2x3  x2  7x  6 The third root can be found by division, i.e.  x 1)(x  2 x2  x  2 x2  x  2 2x3  2x2  4x 3x2  3x  6 3x2  3x  6 Hence, 2x3  x2  7x  6 = (x – 1)(x + 2)(2x – 3) If 2x3  x2  7x  6  0 then (x – 1)(x + 2)(2x – 3) = 0 from which, x = 1, x = -2 or x = 1.5 6. Using the remainder theorem, solve the equation 2x3  x2  7x  6  0 2x - 3 2x3  x2  7x  6 © 2006 John Bird. All rights reserved. Published by Elsevier. 13 2. Solve the following inequalities: (a) x  1.5 (b) x + 2  5 2 4. Solve the following inequalities: (a) 7  2k  1 4 (b) 3z + 2  z + 3 5. Solve the following inequalities: (a) 5 – 2y  9 + y (b) 1 - 6x  5 + 2x CHAPTER 2 INEQUALITIES EXERCISE 8 Page 13 (a) x  1.5 i.e. x  2(1.5) i.e. x  3 2 (b) x + 2  5 i.e. x  5 – 2 i.e. x  3 (a) 7  2k  1 i.e. 7 – 2k  4 i.e. 7 – 4  2k i.e. 3  2k and k  3 4 2 (b) 3z + 2  z + 3 i.e. 3z – z  3 – 2 i.e. 2z 1 and z  1 2  4  y or y   4 3 3  4  x or x  - 1 8 2 (a) 5 – 2y  9 + y i.e. 5 – 9  y + 2y i.e. -4  3y i.e. (b) 1 - 6x  5 + 2x i.e. 1 – 5  2x + 6x i.e. -4  8x i.e. © 2006 John Bird. All rights reserved. Published by Elsevier. 14 1. Solve the inequality: t  1  4 3. Solve the inequality: 2x 1  4 EXERCISE 9 Page 14 If t  1  4 then -4  t + 1  4 -4  t + 1 becomes -5  t i.e. t  -5 t + 1  4 becomes t  3 Hence, -5  t  3 If 2x 1  4 then -4  2x – 1  4  3  x 2 Hence,  3  x  5 2 2 x  5 2 1 k  3 means 1 – k  3 and 1 – k  -3 i.e. 1 – 3  k and 1 + 3  k i.e. k  -2 and k  4 5. Solve the inequality: 1 k  3 -4  2x – 1 becomes -3  2x and 2x – 1  4 becomes 2x  5 and © 2006 John Bird. All rights reserved. Published by Elsevier. 15 2. Solve the inequality: 2t  4  1 t  5 3. Solve the inequality: 3z  4  2 z  5 EXERCISE 10 Page 15 If 2t  4  1 then 2t  4 - 1  0 i.e. 2t  4  t  5  0 and 2t  4  t  5  0 t  5 t  5 t  5 t  5 t  5 i.e. t  9 t  5  0 Hence, either (i) t + 9  0 and t – 5  0 or (ii) t + 9  0 and t – 5  0 (i) t  -9 and t  5 and both inequalities are true when t  5 (ii) t  -9 and t  5 and both inequalities are true when t  -9 Hence, 2t  4 t  5  1 is true when t  5 or t  -9 If 3z  4  2 then 3z  4 - 2  0 i.e. 3z  4  2(z  5)  0 and (3z  4)  2(z  5)  0 z  5 z  5 z  5 (z  5) z  5 i.e. 3z  4  2z 10 z  5  0 i.e. z 14  0 z  5 Hence, either (i) z - 14  0 and z + 5  0 or (ii) z – 14  0 and z + 5  0 (i) z  14 and z  -5 i.e. -5  z  14 (ii) z  14 and z  -5 Both of these inequalities are not possible to satisfy. Hence, 3z  4  2 is true when -5  z  14 z  5 © 2006 John Bird. All rights reserved. Published by Elsevier. 16 3. Solve the inequality: 2x2  6 8. Solve the inequality: 4k  52  9 EXERCISE 11 Page 16 2x2  6 i.e. x2  3 hence, x or x  - 3k 2  2  10 i.e. 3k 2  12 and k 2  4 Hence, -  k  i.e. -2  k  2 t 12  36 then (t – 1) or (t – 1)  - i.e. (t – 1)  6 or (t – 1)  -6 i.e. t  7 or t  -5 4k  52  9 then 4k + 5 or 4k + 5  - i.e. 4k + 5  3 or 4k + 5  -3 i.e. 4k  -2 or 4k  -8 and k   1 2 or k  -2 6. Solve the inequality: t 12  36 4. Solve the inequality: 3k 2  2  10 3 3 4 4 36 36 9 9 © 2006 John Bird. All rights reserved. Published by Elsevier. 17 1. Solve the inequality: x2  x  6  0 3. Solve the inequality: 2x2  3x  2  0 7. Solve the inequality: t2  4t  7  0 EXERCISE 12 Page 17 x2  x  6  0 thus (x – 3)(x + 2)  0 Either (i) x – 3  0 and x + 2  0 or (ii) x – 3  0 and x + 2  0 (i) x  3 and x  -2 i.e. x  3 (ii) x  3 and x  -2 i.e. x  -2 2x2  3x  2  0 thus (2x – 1)(x + 2)  0 Either (i) 2x – 1  0 and x + 2  0 or (ii) 2x – 1  0 and x + 2  0 (i) 2x  1 i.e. x  1 2 (ii) 2x  1 i.e. x  1 2 and x  -2 both of which are not possible and x  -2 thus -2  x  1 2 z2  4z  4  4 i.e. z2  4z  0 i.e. z(z + 4)  0 Either (i) z  0 and z  -4 i.e. -4  z  0 or (ii) z  0 and z  -4 both of which are not possible Since t2  4t  7  0 then t  22 - 7 – 4  0 5. Solve the inequality: z2  4z  4  4 © 2006 John Bird. All rights reserved. Published by Elsevier. 18 i.e. t  22  11 and t – 2  thus, t  2   11 or t – 2  - or t  2   11    11 11 © 2006 John Bird. All rights reserved. Published by Elsevier. 19 2. Resolve 4(x  4) x2  2x  3 into partial fractions. 4. Resolve 3(2x2  8x 1) (x  4)(x 1)(2x 1) into partial fractions. 5. Resolve x2  9x  8 x2  x  6 into partial fractions. CHAPTER 3 PARTIAL FRACTIONS EXERCISE 13 Page 20 Let 4(x  4)  4x 16  A  B  A(x  3)  B(x 1) x2  2x  3 (x 1)(x  3) (x 1) (x  3) (x 1)(x  3) Hence, 4x – 16 = A(x – 3) + B(x + 1) If x = -1, -20 = -4A from which, A = 5 If x = 3, 12 – 16 = 4B from which, B = -1 Hence, 4( x  4)  5  1 x2  2 x  3 ( x  1) ( x  3) Let 3(2x2  8x 1)  A  B  C  A(x 1)(2x 1)  B(x  4)(2x 1)  C(x  4)(x 1) (x  4)(x 1)(2x 1) (x  4) (x 1) (2x 1) (x  4)(x 1)(2x 1) Hence, 6x2  24x  3 = A(x + 1)(2x – 1) + B(x + 4)(2x – 1) + C(x + 4)(x + 1) If x = -4, 96 + 96 -3 = A(-3)(-9) from which, 189 = 27A and A = 7 If x = -1, 6 + 24 -3 = B(3)(-3) from which, 27 = -9B and B = -3 If x = 0.5, 1.5 - 12 -3 = C(4.5)(1.5) from which, -13.5 = 6.75C and C = -2 Hence, 3(2 x2  8 x  1)  7  3  2 ( x  4)( x  1)(2 x  1) ( x  4) ( x  1) (2 x  1) Since the numerator is of the same degree as the denominator, division is firstly required. © 2006 John Bird. All rights reserved. Published by Elsevier. 20 3x - 2 3x3  2x2 16x  20 3x  2  3 x  2   x2  9x  8 8x 14 x2  x  6 x2  x  6 8x + 14 Hence, x2  x  6  1 x2  x  6 Let 8x 14  8x 14  A  B  A(x  2)  B(x  3) x2  x  6 (x  3)(x  2) (x  3) (x  2) (x  3)(x  2) Hence, 8x + 14 = A(x – 2) + B(x + 3) If x = -3, -24 + 14 = -5A form which, -10 = -5A and A = 2 If x = 2, 16 + 14 = 5B from which, 30 = 5B and B = 6 x2  9 x  8 2 6 Hence, x2  x  6  1  ( x  3) ( x  2) x2  4 3x3 12x 2x2  4x  20 2x2  8 - 4x + 12 Hence, 3x3  2x2 16x  20  12  4x (x  2)(x  2) x2  4 Let 12  4x  12  4x  A  B  A(x  2)  B(x  2) x2  4 (x  2)(x  2) (x  2) (x  2) (x  2)(x  2) Hence, 12 – 4x = A(x + 2) + B(x - 2) If x = 2, 4 = 4A from which, A = 1 If x = -2 20 = -4B from which, B = -5 Hence, 3 x3  2 x2  16 x  20  1 5 ( x  2)( x  2) ( x  2) ( x  2) 7. Resolve 3x3  2x2 16x  20 (x  2)(x  2) into partial fractions. 1 x2  9x  8 © 2006 John Bird. All rights reserved. Published by Elsevier. 21 2. Resolve x2  7x  3 x2 (x  3) into partial fractions. EXERCISE 14 PAGE 22 Let x2  7x  3  A  B  C  A(x)(x  3)  B(x  3)  Cx2 x2 (x  3) x x2 (x  3) x2 (x  3) Hence, x2 + 7x + 3 = A(x)(x + 3) + B (x + 3) + C x2 If x = 0 3 = 3B from which, B = 1 If x = -3 9 – 21 + 3 = 9C i.e. -9 = 9C f rom which, C = -1 Equating x2 coefficients: 1 = A + C from which, A = 2 x2  7 x  3 2 1 1 Hence, x2 ( x  3)  x  x2  ( x  3) 4. Resolve 18  21x  x2 (x  5)(x  2)2 into partial fractions. Let 18  21x  x2  A  B  C  A(x  2)2  B(x  5)(x  2)  C(x  5) (x  5)(x  2)2 (x  5) (x  2) (x  2)2 (x  5)(x  2)2 Hence, 18  21x  x2  A(x  2)2  B(x  5)(x  2)  C(x  5) If x = 5 18 + 105 – 25 = 49A i.e. 98 = 49A from which, A = 2 If x = -2 18 – 42 – 4 = -7C i.e. -28 = -7C from which, C = 4 Equating x2 coefficients: -1 = A + B from which, B = -3 18  21x  x2 2 3 4 Hence, ( x  5)( x  2)2  ( x  5) ( x  2) ( x  2)2