Answers to
Exercises
15 (a) x −1
(b) x7
4
(e) 1/(2x )
(f ) 4x /9
(h) 25x 2/3 + 1/(4x 2/3) − 5
( j) a−1b9/2
CHAPTER 1
Exercises
1 54.62510
16 (a) xy(x − y)
(b) xyz(x − y + 2z)
(c) (a + b)(x − 2y)
(d) (x + 5)(x − 2)
(e) (x + 12 y)(x − 12 y) (f ) (9x 2 + y 2)(3x − y)(3x + y)
2 111111110000012, 37 7018
13 4558
(b) 10 101.110 101 012
(b) 27
(e) 1/6
5 (a) 1/2
(d) 32
(c) 1/212
(f ) 23
−5
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8 (a) −7 + 5Ë2
(c) 115 − 111 ÷3
9
239
169
,
577
408
,
(b) 68 + 48Ë2
(d) −1 + 23 Ë2
23 −1 ; Ë2
24 43
25 7, −2
26 10 m
10 Ë3 + Ë19 . Ë5 + Ë13
29 −2 , x , 2
11 (a) −2 < x < 10, [−2, 10]
(b) −5 , x , −1, (−5, −1)
(c) −3 < x < 4, [−3, 4]
(d) −24 , x , 0, (−24, 0)
31 (a) A = − 13 , B = 13
(c) A = 25, B = − 23
(b) A = 8, B = −5
32 A = 2, B = −1, C = 9
(b) {x: x + 3 < 1}
(d) {x: 8x − 1 < 5}
33 (a) 5
34 (a) 120
13 (b) only
(b), (c) and (d) true
(b) vb =
(b) (1 + Ë5)/2
28 (a) x , 0 and x . 5/2
(b) x , 1 and x . 2
(c) x , 0 and x . 1
(d) x , −4 and x . 23
1393
985
14 (a) va = 12 (v1 + v2)
(b) (x − 1) 2 + 2
(d) 5 − (x −2) 2
21 s = (m 2 + p 2 )t/(m 2 − p 2 ), m 2 ≠ p 2
27 (a) Ë2
60
41
− 17
÷2
(b) − 17
28
18
(d) 11 + 11 ÷5
12 (a) {x: x − 4 , 3}
(c) {x: 2x − 43 , 9}
20 (a) ( x + 12 )2 − 494
(c) 13 − 3( x − 53 )2
22 t = (u − 1)x 2/(u + 1), u ≠ −1
6 (a) 21 + ((4 × 3) ÷ 2)
(b) (17 − (6(2+3)))
3
(c) (4 × (2 )) − ((7 ÷ 6) × 2)
(d) ((2 × 3) − (6 ÷ 4)) + 3(2 )
7 (a) 1393 + 985Ë2
(c) 1 + Ë2
(b) (5 − x)/[(x − 3)(x + 1)]
(d) 3x 2 − 4y 2
17 (a) (x + 3)/(x + 4)
(c) 2/[(x − 2)(x + 12)]
3 11110.100110011001p 2
36.463 146 31p 8
Yes
4 (a) 101 110.1002
(c) x−12
(d) x 2
5/2
(g) x − 2x −1/2
(i) 2 − 1/x
(k) 1/(8b3a3/2)
2 v1v2
v1 + v2
(c) −9
(b) 0
(b)
1
4
(c) 35
(d) 11
(d) 10
(e) 84
(f ) 70
35 (a) x 4 − 12x 3 + 54x 2 − 108x + 81
(b) x 3 + 23 x 2 + 43 x + 18
(c) 32x 5 + 240x 4 + 720x 3 + 1080x 2 + 810x + 243
(d) 81x 4 + 216x 3y + 216x 2 y 2 + 96xy 3 + 16y 4
James, Glyn, and Phil Dyke. Modern Engineering Mathematics 6th Edition PDF Ebook, Pearson Education, Limited, 2020. ProQuest Ebook Central,
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AN S WE RS TO E XE RCI S E S
36 (a) y = 23 x − 2
(c) y = 25 x − 12
1
3
(e) y = x +
2
3
51 (a) 3.613 ; 0.0015, relative error bound 0.0004, 3.61
(b) 2.5351 ; 0.0176, relative error bound 0.007, 2.5
(c) 22.47 ; 0.015, relative error bound 0.0007, 22.5
(b) y = −2x − 1
(d) y = − 35 x + 3
(f ) y = −3x + 4
52 4.51
37 (x − 1) + ( y − 2) = 25
2
2
1
53 10.00 ; 0.01, 1000
−0.02 ; 0.01, 12
1
24.9999 ; 0.05, 500
1
0.996 008 ; 0.002, 500
38 4, (−2, 3)
39 x 2 + y 2 + 4x − 6y = 12
40 x 2 + y 2 − 6x − 3y + 5 = 0
54
Label
value
Absolute
error
bound
a
b
a–b
c
(a – b)/c
d
d + (a – b)/c
3.251
3.115
0.136
0.112
1.2143
9.21
10.4243
0.0005
0.0005
0.001
0.0005
0.0145
0.005
0.0195
41 2y = x + 3
42 x + y = 25
2
2
2
43 Circle through
1 1
1
1
1 12 b, 02, a c, d b , a (b + c), d b ,
2
1077
2
2
2
bd (b - c)
bd 2
bc2
bcd
(c, 0), a 2
b, a 2
b,
,
,
d + (b - c)2 (d 2 + (b - c)2
c + d 2 c2 + d 2
c
c
1
1
1
a c,
(b -c) b , a (b + c),
(b - c) b , a c, d (bc - c2 + d 2)b
2 2d
2
2d
2
is called the nine-point circle. The intersections of the
diagonals of the three rectangles in the centre of the ninepoint circle are also drawn.
Relative
error
bound
0.0074
0.0045
0.0119
Result: 10.4
y
55 0.7634 ; 0.000 72, 0.76
C (c, d)
56 (a) 0.2713 ; 0.0237
E
1
(b) 0.2715 ; 0.0072
1
57 10 (0.2709), 10 (0.2708)
The second result is more accurate since by adding the
small numbers together first their combination is given
its proper weight.
D
O
58 10 −8(0.6538), 10 −3(0.6752)
59 0.5
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A (0, 0)
F (c, o)
B (b, o)
x
44 x = 23 , x = − 23
1.7 Review exercises
45 (0, 3), (0, −3), 35 , y = 253 , y = − 253 , 10, 8
1 (a) A = ;QKD /Ë(HD 2 − Q 2K 2)
(b) −(9 ; Ë145)/8
46 (−5, 0), (5, 0), (−4, 0), (4, 0), y =
3x
3x
,y=−
4
4
47 (a) 3dp, 6sf
(d) 0dp, 3sf
(c) 0dp, 5sf
(f ) 10dp, 3sf
(b) 30dp, 3sf
(e) 0dp, 4sf
48 The answer claims unjustified accuracy: hypotenuse
= 2.236 ; 0.007 m. The angles are also subject to error.
49 (a) Absolute error bound is 121 min, relative error bound
1
is 420
(b) Absolute error bound is 1.4 min, relative error
bound is 0.04
(c) Absolute error bound is 0.005, relative error
1
bound is 116
50 0.0039, 12.9
2 (a) (x − 1)(a − 2)
(b) (a − b + c)(a + b − c)
(c) (2k + l − 3m)(2k + l + 3m)
(d) (p − q)( p − 2q)
(e) (l + n)(l + m)
3 (a) 1 cm
(b) 3.812
1

4 (a) L = 
± ÷( Z 2 − R2 ) (2π n )
 2π nC

(b) 0.1434; 0.0592; 0.4160
(L must be positive from practical considerations)
5 (a) 30 − 12Ë6
(c) 231 (14 + 11Ë2)
(b) −53 + 11Ë15
(d) 3 + 2Ë2 + 2Ë3 + Ë6
(e) 12 + 14 ÷2 + 14 ÷6
James, Glyn, and Phil Dyke. Modern Engineering Mathematics 6th Edition PDF Ebook, Pearson Education, Limited, 2020. ProQuest Ebook Central,
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1078 A NSW ERS T O EXERCISES
2 A = 2x(5 + x )
6 5, 6
(b) (−q, −5) ø (−2, 1)
8 (a) (− 12 , 23 )
(c) (−3, 1)
(d) (− 43 , 0)
a
a+c
11
1
b
b+c
x/m
Area/m2
(ii) 729 − 2916x + 4860x 2 − 4320x 3 + 2160x 4
− 576x 5 + 64x 6
13 (a) 90.5
(b) P1 = 1, P2 = 3, P3 = 5, Pr = (2r − 1),
2
3
48
4
72
5
100
3
r/m
A/m2
0.10
3.05
0.15
2.12
0.20
1.71
0.30
1.47
0.35
1.50
0.40
1.59
0.25
1.53
5 5 years
r =1
14 (a) y = 2x + 1
(b) y = (x − 7)/3 (c) y = 2x − 73
15 (y − 3) + (x − 5) = 25
2
2
16 (a) (−1, 2), 2
(b) ( 12 , − 23 ), 12 (c) (− 13 , 13 ), Ë3
17 (i) (a) (1, 2)
(b) (3, 2)
(c) x = −1 (d) y = 2
(ii) (a) (−2, 1) (b) (−2, −2) (c) y = 4
(d) x = −2
1
3
18 (2, 8), (2, 5), (2, 11), (2, 3), (2, 13), y = − , y = 493
19 ∆.477 4∆ 774∆ p 12, where ∆12 = 1010
20
Value
Copyright © 2020. Pearson Education, Limited. All rights reserved.
2
28
r* = 0.32 according to worked answer: estimated
from a graph (not drawn).
n
a
Ëa
b
Ëb
c
Ëc
d
Ëd
1
12
A(−2) = −28, area of cutting
12 (b) (i) 1 − 25 x + 25 x 2 − 45 x 3 + 165 x 4 − 321 x 5
∑ Pr = n
0
0
Absolute error
bound
7.01
0.005
2.647 6
0.000 9
52.13
0.005
7.220 111
0.000 347
0.010 11
0.000 005
0.100 548
0.000 025
5.631 × 1011 0.5 × 10 8
7.504 × 10 5 0.33 × 10 2
Correctly
rounded values
Ëa
2.65
Ëb
7.22
Relative error
bound
S 0.000 7
d 0.000 35
S 0.000 096
d 0.000 048
S 0.000 495
d 0.000 25
S 0.000 088 8
d 0.000 044 4
Ëc
0.101
Ëd
7.504 × 10 5
6 (a) 0, 2; increasing for x . 1, decreasing for x , 1,
minimum at x = 1
(b) − 25 , 2; increasing for x , −1 and x . 2,
decreasing for −1 , x , 2, maximum at x = −1,
minimum at x = 2
(c) Increasing on −1 , x , 0 and x . +1
Decreasing on x , −1 and 0 , x , +1
Maximum at (0, 0), minimum at (−1, −1) and
(1, −1)
(d) Increasing on x , 1, decreasing on x . 1,
maximum at (1, −1)
8 F(x) = (x − 1)2: f(x) shifted by 2 units in positive x
direction
G(x) = (x + 1) 2 − 2: f(x) shifted by 2 units in negative y direction
4x + 3
2−x
(c) Restriction of domain to [0, q)
Ë(x − 1), x > 1
10 (a)
1
2
(x + 3)
14 (a) odd
(d) neither
(b)
(b) even
(e) odd
16 (x3 + 3x) + (−3x2 − 1)
21 0.37 ; 0.07
17 (a) 3 − 2x
(b) 12 x + 25
(c) 0.255x + 2.478 (3dp)
22 1.714
18 (a) 3
(a) 0.0026, (b) 0.0075
23 6
(c) neither
(f) even
(b) −3
(c) 12
19 £(50 + 0.455x), £960, £(1.20x − 960), 800
20 a = 0.311
CHAPTER 2
21 m = 0.82, c = 60.9
Exercises
1 (a) [−5, 5], R, [0, 5], 0, 3, Ë(25 − x )
(b) R, R, R, 2, −1, 3Ë(3 − x)
2
24 (a) 23 x 2 + 2 x + 13
(b) 25 x 2 − x − 25
James, Glyn, and Phil Dyke. Modern Engineering Mathematics 6th Edition PDF Ebook, Pearson Education, Limited, 2020. ProQuest Ebook Central,
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AN S WE RS TO E XE RCI S E S
25 (x − 2)2 − 4(x − 2) − 2
26 (a) irreducible
(c) not irreducible
(b) not irreducible
(d) irreducible
27 (a) minimum at x = −1 of 2
3
(b) minimum at x = 2 of 0
(c) maximum at x = − 23 of 223
(d) maximum at x = 103 of − 11
20
28 (a) x , 2 and x . 4
(b) − 25 , x , 3
29 315 feet, 46 mph
30 (a) (x − 1)(x + 3)(x − 4)
(b) (x + 1)(x − 2)(x + 3)
(c) (x − 1)(x + 1)(x 2 + 2)
(d) x(x − 1)(2x + 3)(x + 2)
(e) (x − 1)2(2x − 1)(x − 2)
(f) (x 2 + 9)(x − 2)(x + 2)
31 2, 7, 139, 527, 524
32 y = (x − 5)4 + 15(x − 5)3 + 80(x − 5)2 +
165(x − 5) + 81. Since coefficients are all positive,
zeros of y must all lie to the left of x = 5, that is x , 5.
Hence the zeros of y lie between x = 0 and x = 5.
1
3
x−4
1
3
−
x −1
1
x + 23
− 23
(b)
x −1 x + x +1
1
3
(c)
5
9
x−2
4
3
+
2
−
5
9
x +1
( x + 1)
3
8
+
(d) 1 −
x−2 x−3
x−2
1
+ 2
(e) 2
x + 1 ( x + 1)2
2
3
x +1
+
−
(f) 2
x + 4 x −1 x + 5
43 (a) (Ë2, Ë2), (−Ë2, −Ë2)
(b) (Ë2, Ë2), (−Ë2, −Ë2)
(c) (− √ 25 , − √ 25 ), (√ 25 , √ 25 ), (Ë2, Ë2), (−Ë2, − Ë2)
(d) does not intersect on domain
44 (a) asymptotes: y = x − 8, x = 0,
maximum (Ë15, − 8 + 2Ë15),
minimum (−Ë15, − 8 − 2Ë15)
(b) asymptotes: y = 1, x = 1
(c) y = x, x = −5
45 y = −1 ; Ë(x + 4), ( y + 1)2 = x + 4
48 0.6, 0.8, 0.75, 36.87° = 36°52′12″;
12 5
, , 2.4, 67.38° = 67°22′48″
13 13
33 (a) x 2 − 14x + 1 = 0
(b) x 2 + 52x + 1 = 0
49 AB = 29.44, BC = 33.04 m
34 x 3 − 5x 2 + 1
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42 (a)
35 3x 2 + 22x + 378
50 30
36 (b) r = 10/(4 )1/3, h = 20/(4 )1/3
51 60
37 0.096 m3, 0.1875 m3
52 AB = 30.6 mm, AC = 26.9 mn
38 x 0 = 10.94, width of alley = 4.92 m
53 45.5 mm
39 (a) 1 + (x + 2)/[(x + 1)(x − 1)]
(b) x 3 − 2x 2 + x + 1 − 3x /(x 2 + x + 1)
55 degrees 0°
radians 0
40 (a) (−5x 2 + x − 2)/[x(x − 2)(x 2 + 1)]
(b) 2 /[(x − 1)3(x + 1)]
(c) (4x 4 − 11x 3 + 10x 2 − 5x + 4)/
[(x 2 + 1)(x − 1)2(x − 2)]
41 (a)
1
3
1
3
−
x − 2 x +1
1
1
+
(b)
x − 2 x +1
(c) 1 +
2
3
x−2
1
3
2
1079
+
+
1
3
x +1
2
9
−
2
9
x − 2 x +1
(d) ( x − 2)
x +1
1
−
(e)
x + 1 x2 + 2x + 2
1
1
− 13
+ 12 + 4
(f)
x +1 x − 2 x + 2
30° 45° 60° 90° 120° 150° 180°
5
1
1
1
1
2
4π
3π
2π
3π
6π
6π
degrees 210° 225° 240° 270° 300° 315° 330° 360°
7
5
3
5
7
4
11
2
radians 6 π 4 π 3 π 2 π 3 π 4 π
6π
57 (a) 0.3398, 2.8018, 23 π
(b) 1.8235, 4.4597,
(c) 2.6779, 5.8195, 14 π , 45 π
(d) 12 π , 23 π , 16 π , 65 π
58 12 ÷3, ÷3, 12 ÷3, 12
1
Ë(2 − Ë3), 12 Ë(2 + Ë3), 2 − Ë3
2
(a) 12 Ë3
(b) 1/Ë3
(c) 12 Ë3
(d) 12 Ë(2 + Ë3)
1
−
(e) 2 Ë(2 − Ë3)
(f ) −(2 − Ë3)
59 (a) −Ë(1 − s 2)
(c) s(3 − 4s 2)
(b) −2sË(1 − s 2)
(d) Ë{ 12 [1 + Ë(1 − s 2)]}
61 x = n (n = ;1, ;3, p )
and x = 0.9273 + 2n
(n = 0, ;1, ;2, p )
James, Glyn, and Phil Dyke. Modern Engineering Mathematics 6th Edition PDF Ebook, Pearson Education, Limited, 2020. ProQuest Ebook Central,
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1080 A NSW ERS T O EXERCISES
62
e
f
74 (a) 3
(d) 4
;Ë 12
;Ë
Ë
−12
± √23
; 12
1
2
; Ë3
75 (a) 2 ln x + ln y
(c) 5 ln x − 2 ln y
−1
;1
;Ë 13
Ë 13
76 (a) ln 4
77 (a) Ë[(1 − x)/(1 + x)]
b
c
; 12
;Ë 12
cos x
1
2
1
2
; Ë3
1
2
− Ë3
tan x
; Ë 13
;Ë 13
sin x
1
2
1
2
cosec x
2
;2
;Ë2
;Ë2
−2
;2
sec x
; 2Ë 13
−2Ë 13
;Ë2
Ë2
; 23
;2Ë13
cot x
; Ë3
;Ë3
−1
;1
;Ë3
Ë3
3
(b) 2 sin 2 θ sin 12 θ
(d) −2 cos 23 θ sin 12 θ
63 (a) 2 sin 2ucosu
(c) 2 cos 27 θ cos 23 θ
64 (a) 12 (cos 2u − cos 4u)
(c) 12 (sin 4u − sin 2u)
(b) 12 (sin 4u + sin 2u)
(d) 12 (cos 4u + cos 2u)
4
(b) x 2
b
c
d
e
f
sinh x
± 43
− 43
5
4
3
5
4
3
4
5
5
3
7
24
25
24
7
25
24
7
24
25
25
7
± 125
cosh x
8
15
17
15
8
17
15
8
15
17
17
8
13
5
12
13
5
12
5
13
13
12
5
3
4
5
3
4
3
5
3
4
12
5
13
5
12
13
5
12
5
13
13
12
tanh x
±
cosech x
±
coth x
(n = 0, ;1, ;2, p)
y
(d) ln 0.5
a
sech x
(d) Ë13 cos(u − 0. 9828), Ë13 sin(u − 5.6952)
70
(c) ln 0.75
81
(b) Ë2 cos(u − 43 ), Ë2 sin(u − 14 π )
7
(c) Ë2 cos(u − 1 π ), Ë2 sin(u − 4 π )
(b) − /6
(e) /3
(b) ln 3.2
ln(x 2 + 1) − 13 ln(x 4 + 1) − 15 ln(x 4 + 4)
80 23 π
2
67 (a) /6
(d) 2 /3
(b) 12 ln x + 12 ln y
79 ln(20 ; 6Ë10) = 3.6629, 0.025 99
65 (a) 2 cos(u − 3 ), 2 sin(u − 16 π )
66 x = 2n , 2n ; 23
(c) − 12
(f ) − 12
(b) −2
(e) 12
d
a
±
±
−
±
−
±
±
±
−
−
−
(3 + tanh 2x )tanh x
1 + 3 tanh 2 x
(b) cosh(x + y) = cosh x cosh y + sinh x sinh y
(c) cos 2x = 1 − 2 sin2x
(d) sinh x − sinh y = 2 sinh 12 (x − y) cosh 12 (x + y)
82 (a) tanh 3x =
(c) /3
(f ) − /3
84 (a) 0.7327
(b) 1.3170
(c) 0.5493
1
85 17.1383 (4dp)
86 1.0074 (4dp)
O
1
3
88 A = 250, B = −273.26
91 (a) Cusp at x = 0, maximum at x = 4, asymptote y = −x
(b) Minimum at x = 2, asymptotes y = ;Ëx − 1, x = 1
–1
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x
94
71
y
ax
2a
a
H( x) +
(l − x ) H ( x − l ) − (2l − x ) H( x − 2l )
l
l
l
95 x [1 − H(x)] − (x − 1)H(x − 1)
96 INTPT(x + 12 )
1
99 0.9401, 0.005, 0.9425
100 0.04, 0.16, 0.01, 0.006 25
O
1
/3
1
101 0.3081, 0.2829, 16.79
x
102 0.2954, 0.2688, 17.10
85
x
103 f ( x ) = − 841 x 3 + 84
–1
104
72 (a) (2e + 1)e
(d) e9
5
4x
(b) e
(e) ex/2
6
(c) e
x
y
3045 3051 3058 3064 3070 3077 3083
14.50 14.51 14.52 14.53 14.54 14.55
James, Glyn, and Phil Dyke. Modern Engineering Mathematics 6th Edition PDF Ebook, Pearson Education, Limited, 2020. ProQuest Ebook Central,
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AN S WE RS TO E XE RCI S E S
2.12 Review exercises
1 h(x) = x − 4
x [ [0, 200]
k(x) = (x 2 − 4)1/2
x [ [−20, −2] < [2, 20]
3
Price/£
Sales/000
Revenue/£000
Profit/£000
1.00
8
8
0
3x + 3
2 x

4 g( x ) = 3x + 1
x + 1
3x – 1

1.05
7
7.35
0.35
1.10
6
6.60
0.60
÷7
(b) 12 ; j 2
÷
(d) 1, − 12 ; j 211
3 (a) −3 ; j2
(c) − 12 ; j
(e) ;Ë3, ;jË2
4 (a) 24 + j18
(c) −1 − j5
(b) 2.35 m2
2 (a) 25.6 cm
1.15 1.20 1.25 1.30
5
4
3
2
5.75 4.80 3.75 2.60
0.75 0.80 0.75 0.60
x −3
−3 x −1
−1 x 0
0 x 1
x 1
(b) 17 − j19
(d) −26 − j13
5 (a) −1 − j5
(c) (1 − j7)/25
(b) (9 + j19)/13
(d) −1 − j2
6 (a) 10
(c) 251 (47 − j4)
(e) j
(g) j 173
(b) −3 − j4
(d) −j
(f ) 5 − j12
1
(h) − 178
(5 + j8)
7 (a) 2 − j7
1081
(b) −3 + j
8 (a) −1 + j, −1 − j
(c) j6
(d) 23 + j 23
(b) −2, 1 + jË3, 1 − jË3
9 ;3 + j2
6 0.37 ; 0.005
8 (x − 1)4 + 7(x − 1)3 + 14(x − 1)2 + 13(x − 1) + 4
9 (a)
2
1
−
x − 4 x −1
(b) 1 −
5
4
+
13
4
x +1 x −3
2
7
11
+
−
(c)
9( x − 1) 9( x + 2) 3( x + 2)2
(d)
1
13
2
21
13
(5x − 7)
+
x − x +1 x +3
(b) 2 + j3
(d) 2 − j3
11 (a) Ë2, 14
(c) 5, − tan −1 43
(e) 2, 23
(b) 2, − 16
(d) 2, − 13 π
(f ) 2, − 23
12 w = 5 − j4, z = 2 + j3
13 x = 12 , y = − 23
1
14 2 + j2, 2
15 15 (7 + j4)
10 (a) 2 cos 23 θ sin 12 θ
(b) 2 cos 25 θ cos 12 θ
3θ
11θ
cos
(c) −2 sin
2
2
Copyright © 2020. Pearson Education, Limited. All rights reserved.
10 (a) 3 + j2
1
(c) 13 (2 + j3)
1
16 130
(451 + j878)
17 x = 14 , y = − 43
−1 1
2
−1
11 (a) 2Ë5 sin(u − a), a = tan
(b) Ë65 sin(u − a), a = −tan 8
(c) 2 sin(u + 16 π )
13 2
15 0.0025, 0.300
16 ÷ 13
17 (b) 12 DË3
18
11
4
− j 134
19 2Ë2, 1/Ë2, 5 /12, /12
20 (a) 16[cos(11 /12) + j sin(11 /12)],
1
[cos(7 /12) + j sin(7 /12)],
4
4[cos(−7 /12) + j sin(−7 /12)]
(b) 15[cos(−5 /6) + j sin(−5 /6)],
3
[cos(− /2) + j sin( /2)],
5
5
[cos( /2) + j(sin /2)]
3
21 z = 0.0024, argz = −1.9728, z = −0.0009 − j0.0022
18 1, 2, 0, 1, 3, 7, 5, 5, p
0.833 89, 0.551 94
22 (a) e1.61+j0.927
21 r = 3/(2 sin u − cosu)
23 (a) 14.2026 + j14.2026
CHAPTER 3
Exercises
2 4 + j, −2 + j3, 2 + j4, −9 + j3, −1 + j12, 5 + j3
(b) e0.693+j2p/3
(b) 0.1839 + j0.3186
(b) 1∠0
24 (a) 1∠ /2
(d)
Ë2∠− /4
(c) 1∠
(f
)
Ë5∠( − tan−1 12 )
(e) Ë6∠− /4
−1 5
(g) Ë13∠(tan−1 23 − ) (h) Ë74∠(−tan 7 )
−1 28
(j) 53∠(tan 45 − )
(i) 5∠0
James, Glyn, and Phil Dyke. Modern Engineering Mathematics 6th Edition PDF Ebook, Pearson Education, Limited, 2020. ProQuest Ebook Central,
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Created from monash on 2022-06-03 01:16:25.
1082 A NSW ERS T O EXERCISES
25
4
5
43 (a) x = 5, a straight line
(b) circle centre (1, 0), radius 3
(c) circle centre (− 45 , 0), radius 43
(d) half-line, y = x − 2, x . 2
+ j 75 , 15 √65∠tan −1 47
(b) 2 ÷3 + j 2
1
11
26 (a) − 125 π , − 12
π
27 (a) 128, − 13
3
(c) 161 , 23π
(b) 1024, 0
29 (a) 12 cosh 1 − j 12 ÷3 sinh 1
(b) cosh 43
π
÷3
π
(c) 12 sinh + j cosh
3
3
2
1
(d)
÷2
30 (a) 12 (4n + 1) + j cosh−12
(b) 12 (2n + 1) + j(−1)n+1 sinh−1 43
(c) 12 (4n + 1) + j cosh−13
(d) cosh−12 + j(2n + 1)
tanh u sec2 v
1 + tanh 2u tan 2 v
tan v sech 2u
y=
1 + tanh 2u tan 2 v
sech 2 2
2 tanh 2
+ j
2
1 + tanh 2
1 + tanh 2 2
33 0.1645 − j0.1214
48 x 2 + y 2 − 4x − 2y + 1 = 0, z − 2 − j = 2,
34 (a) −2 + j2, −4
(c) 117 + j44, −527 + j336
(e) 8, −8 + j8Ë3
(b) −j8, −8 − j8Ë3
(d) −8, −8 + j8Ë3
(f ) 8, −8 − j8Ë3
35 (a) 18 cos 4u + 12 cos 2u + 83
(b) 43 sin u − 14 sin 3u
37 2
Copyright © 2020. Pearson Education, Limited. All rights reserved.
2
3
+ kπ ), k = 0, 1, 2
38 (a) 2 ∠(− π + kπ ), k = 0, 1, 2, 3
(b) 2∠( 16 π + 23 kπ ), k = 0, 1, 2
(c) 18−1/3∠( 16 π − 43 kπ ), k = 0, 1, 2
(d) 1∠( 14 π + 12 kπ ), k = 0, 1, 2, 3
(e) 4∠( 13 π + 83 kπ ), k = 0, 1, 2
(f ) 34 −1/ 4∠( 12 tan −1 35 − kπ ), k = 0, 1
1/ 4
1
24
1
2
39 1.455 − j0.344, 0.344 + j1.455, −1.455 + j0.344,
−0.344 − j1.455
40 2.529 + j2.743, 0.471 + j2.257
41 1, − 12 ± j 12 √3
2kπ
2kπ
+ j sin
, k = 1, 2, … , n
n
n
(a) j2 cot 15 k , k = 1, p , 4
(b) 23 (1 + j cot 16 kπ ), k = 1, p , 5
cos
42 5, 13
(b) z + 2 = 2
(d) Re(z 2) = 1
47 (a) Circle, centre (1, 0), radius 2
(b) Circle, centre ( 12 , 0), radius 23 π
(c) Circle, centre (2, 3), radius 4
(d) Half-line, y = 0, x . 0
(e) Circle, centre (− 138 , 0), radius 158
(f ) Semicircle, centre ( 12 , − 12 ), radius 12 Ë2, through (0, 0)
= 0.9994 + j0.0366
∠( 12π
45 (a) Straight line, y = 1
(b) Circle, centre (0, 2), radius 1
(c) Circle, centre (0, 45 ), radius 43
(d) Circle, centre (÷ 13 , 0), radius 2 ÷ 13
(e) Rectangular hyperbola, xy = 1
(f ) Ellipse, foci at (1, 0), (0, −1), through (0, 0)
(g) Hyperbola, foci at (1, 0), (0, −1)
(h) Half-line, y = x − 2, x . 2
(i) Half-line, y = ÷3x − 23 ÷3, x 23
( j) Circle, centre (0, 2), radius 1
46 (a) Re[(3 + j)z] = 2
(c) z + 1 − j2 = 3
32 x =
7/ 6
44 circle is z + 2 = 2
line is Re((3 + j) + z) = −2
 z−j 
π
arg 
 =±
2
 z − 4 − j
49 Part of x 2 + (y − 1)2 = 2
50 (x − 3)2 + y 2 = 4
51 (a) u = x + y, v = y − x
(b) u = (x − 1)2 − y 2, v = 2(x − 1)y
(c) u = x(x 2 + y2 + 1), v = y(x 2 + y2 − 1)
52 a = ( j − 2)/5, b = 3(1 + 2j)/5
56 u2 + v2 = 1
57 100 + j100.12
58 83 + j 83
3.7 Review exercises
1 (a) 12 + j9
(b) 2 + j
(c) 11 + j2
(d) 7 + j24
(e) 5
(f ) (1 − j2)/5
(g) (18 + j14)/5
(h) tan−1(3/4) = 0.6435
(i) 5Ë5 [cos(0.9653 + 3k ) + jsin(0.9653 + 3k )],
k = 0, 1
2 x = ; 23 , y = ;2
3 101 (7 + j9)
James, Glyn, and Phil Dyke. Modern Engineering Mathematics 6th Edition PDF Ebook, Pearson Education, Limited, 2020. ProQuest Ebook Central,
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Created from monash on 2022-06-03 01:16:25.
AN S WE RS TO E XE RCI S E S
4 (a) Circle centre (− 13 , 43 ), radius 23 Ë2
(b) Re  1  = − 1


2
 z − 2
6 Centre (R2, 12 vL), radius 12 vL
7 (a) 32 cos6u − 48 cos4u + 18 cos2u − 1
13 419.8 − j238.8, 0.5928 × 10 −3 + j1.0518 × 10 −3
14 1 + j3, 15 (3 + j11), 15 (7 + j11)
25
15 Mod 13
, arg = −154°17′ = −2.6927 rad
16 (a) 0.22 ; j0.49 (b) 1.44 + j1.57 (c) 10.48 + j19.74
(d) 0.80 + j0.46 (e) 1.09 + j0.83
 2 R 0 X − 2 RX0 
19 θ = tan −1  2
2
2
2 
 R + X − R0 − X 0 
20 4.46 − j2.06
(b) ;(0.35 + j0.40)
(d) −1.26 + j1.71
24 1∠18°26′, 1∠108°26′, 1∠198°26′, 1∠288°26′
25 21/6e j(1/9+k/3)p, k = 0, p , 5
27 −(vu + v2v), −(v2u + vv); 14 π
r 2 < − 271 q 3
28 1 − j2, 2Ë5
31 Circle u2 + v 2 − 12u + 16v = 0
Centre (6, −8), radius 10
32 v + 3u = 5
33 Circle u2 + v2 − 25 u + 1 = 0; Centre ( 45 , 0), radius 43 ;
Maps to region outside circle
Copyright © 2020. Pearson Education, Limited. All rights reserved.
(c) (0, 0, 1)
(f ) Ë3
12 PQ = (4, −5, 11), PQ = 9Ë2
direction cosines 4/(9Ë2), −5/(9Ë2), 11/(9Ë2)
13 Ë134 N, (7, 2, 9)/Ë134
14 a = 4 b = 1
15 Ë21
Ë17
g =2
Ë38
16 (1, 1, −1.414)
17 (a) AB = (6, 0, −1), AC = (1, 2, −1) (b) 7
(c) ( 37 , −76 , 17 ) (d) Ë37, Ë6 (e) 12 (7, 2, −2)
18 (1, 4), (4, 0)
19 PQ = QR = (1, 5, −3) and PQ : QR = 1 : 1
20 distance = 13/5, t = 1/5
21 (a) 0.7974 + j0.3685
(b) r = 0.8784
u = 24°49′ = 0.4329 rad, 1.098
23 (a) −0.04 + j0.28
(c) 0.92 + j0.27
(e) −0.04 + j0.28
(b) (2, 4, 25 )
(e) 3
(h) ( 23 , 23 , 13 )
11 (a) (3, 3, 1)
(d) Ë2
(g) (√ 12 , √ 12 , 0)
1083
22 1 − j2 length = Ë20
23 12 − j 12
 W 2 + W 22 − W 23 
25 θ = sin −1  1

2W1W2


 W 2 + W 23 − W 12 
φ = sin −1  2

2W2 W3


26 F = (−940, 124, −31) N
(7.93 m, −1.04 m, 0 m), T = 1342 N
27 (a) 14
(d) (12, 0, −6)/Ë5
(b) 6
(e) −24
28 (a) 98.0° = 1.711 rad
(c) 145
(c) (2, 1, 6)/Ë41
(f ) (12, 4, 8)
(b) 64.8° = 1.130 rad
(d) 3 or −4
29 1, −1, 2
30 Ë45, Ë55, 27.8°
31 4 units
CHAPTER 4
32 145
Exercises
33 Ë5/2
2 183.3 km, 270 km
38 r 2 − (r · â)2 = R2
4 60° or −60° to the positive z axis
39 Ë3, 70.5° or 1.23 rad
5 13 a + 23 b
40 X < 2.98 m
7 OC = 2a + b, OD = 2a + 2b, OE = a + 2b
41 (a) (3, −2, −1)
(d) −1
8 20.62 m s−1, 14.05°
9 8Ë2 kilometres per hour from the NW
10 70.71 N
(b) (−1, 1, 0)
(e) 1
42 (a) (−5, −3, 1), (−10, −6, 2)
(b) (5, 3, −1), (10, 6, −2)
(c) a and c are parallel
43 −8i − 6k, 2i − j
James, Glyn, and Phil Dyke. Modern Engineering Mathematics 6th Edition PDF Ebook, Pearson Education, Limited, 2020. ProQuest Ebook Central,
http://ebookcentral.proquest.com/lib/monash/detail.action?docID=6036640.
Created from monash on 2022-06-03 01:16:25.
(c) (5, −4, −2)
(f ) (2, 2, 1)
1084 A NSW ERS T O EXERCISES
4.6 Review exercises
44 Ë75, (−25, 5, 35)
45 (1, −13, −7), (−6, 21, −12)
3
5
48 (a) (8, 1, 6), (4, 1, 3)
4
5
(b) (− , 0, )
(c)
5
2
51 (48, 72, 0)/Ë14
(b) Ë35
52 (−8, −32, −4)/Ë21
2 (a) (3, 4, 5)
53 (a) (0, 1, −1) (−2, 1, −1) (−2, 1, 0)
(b) (0, 0, 0) (0, 0, −2) (−3, 3, −1)
(c) (−3, 3, −3)
3 (a) (1, 2, 0), (2, 1, 1)
(c) 1
54 ;(−3, 5, 11)/Ë155
0.9968
(b) Ë5
(d) 112.2° = 1.96 rad
(b) 1 or −4
8 (1, 1, 1), (−5, −11, 1)
56 mv = eB
9 E = e(0.550, 0.282, 0.282)
57 15
10 (a) 2x + 3y + 6z + 28 = 0
59 8
61 (a) (−5, 3, −7)/Ë83
62 73 (1, 1, 1)
1
3
u1
u2
u3
v1
v2
v3
w1
w2
w3
63
4 (a) −4
(c) 34/3
5 ( 23 , 13 , 23 ), (− 35 , 0, 45 ); (1, 5, 22)
55 Distance = 1.92
(b) (0, 1, −4)/Ë17
(b) 5
(b) Ë3/2
11 (a) 2x + 3y − z = 10; 10/Ë14
12 P(2, 4, 4), Q(1, 2, 3)
(−2, −4, 0)
(2, −13, 11)
13 (a) 0
(b) 15(1, 1, −2)
15 (−90, −36, 12), 85.3° or 1.49 rad
64 a = −1/F2
65 (a) (c · a)(b· d) − (c · b)(d · a)
(c) −(a · b)(a × c)
16 (11, −12, 5); 76.8°; (−11, 12, −5)/Ë290
(a) −11x + 12y − 5z = 8
(b) −11x + 12y − 5z = −4
(c) 12/Ë290
66 (a) (3, 3, 3)
(b) (1 + s, 2 + s, 3 + s)
(c) x − 1 = y − 2 = z − 3
17 r = (−3, 0, 1) + l(8, −8, −8) + m(5, 1, −3)
r · (−1, 2, −1) = −6
67 yes, no, no, yes r = (2 + s, 1 − s, 1 − s)
18 x + 2y − 2z = −1
69 (3, 4, 0), 43.5° = 0.759 rad
19 (a) (0, 0, 1), (1, −1, 0), (0, 1, −1)
(b) 1
(c) 3, −3, 2
(d) 3, 2, 1
71 r = (2 − t, 2t, −1 + 4t)
2 − x = 12 y = 14 (1 + z), no intersection
Copyright © 2020. Pearson Education, Limited. All rights reserved.
1 (a) Ë93
(b) (17, −3, −10)/Ë398
(c) 85.8° = 1.50 rad, 47.0° = 0.820 rad
(d) (2, 13, −13)/6
72 Ë35
73 r · (0, −1, 1) = 1
−y + z = 1
74 r · (1, −1, 1) = 2
75 r · [b × (c − a)] = a · (b × c)
20 r = (2 − t, 3 − 3t, 2t)
(a) Ë(61/14)
(b) (0, −3, 4)
(c) (19, 15, 18)/14
21 a = r · a′ b = r · b′ g = r · c′
22 Taking i along OA and j along OB then
F = v2(1.4, 1.65) and OC = (−1.4, −1.65) m.
76 r = (0, −5, 10) + l(1, 2, −3)
77 r = (1, 2, 4) + t(1, 1, 2) (− 25 , − 23 , −3)
78 79.0° = 1.38 rad
79 (a) r · (2, 3, 6) = −28
(b) 5
CHAPTER 5
Exercises
80 r = (1 + 2t, −1 + 4t, 3 − 4t), 41.8° = 0.729 rad
81 r · (1, −5, 3) = 28
82 r = (−1 + 14t, t, 1 − 8t)
2
3 Ë29, r · (−18, 36, −27) = −9
1 (a) not possible
1
(b) 3
1
(c) not possible
James, Glyn, and Phil Dyke. Modern Engineering Mathematics 6th Edition PDF Ebook, Pearson Education, Limited, 2020. ProQuest Ebook Central,
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Created from monash on 2022-06-03 01:16:25.
AN S WE RS TO E XE RCI S E S
(d) not possible
8 9 10
(e) 

7 10 9
0 1 1
2 (a)  3 2 −1


0 0 3
4
− 23 
−2
3


(b)  0 − 43 − 43 
 0 −2
0
 1 − 12

(c)  12
1
 0
1
1
2
1
2
1
2




9 7 13 18  9 5
(b) 



7 12  8 5 18 2
37 33
(c) 26 36


29 28
 1 2 0


(b)  1 1 −1
10 4 5
2 −2
14 (a) 
1
2
 2
2 0
, X =  5
15 AB = BA = 

− 2 
0 2
1
 3 −2

4 (b) 0
3 −1
 1 6 3
18 x 2 + y 2 + z 2
x 2 + 4y 2 + 7z 2 + 5xy + 8xz + 11yz
x + 2 y + 3z = 2

3x + 4 y + 5z = 3
5x + 6 y + 7z = 4
5 1, 1, 2, 2
b = −1
g=2
8 a = 12 ( p + q − r), b = 12 ( p − q + r),
g = 12 (−p + q + r)
9 (a) l = 1
m = −1
30.5
27.5


10 Average = 19.5


11.5
11.0


Copyright © 2020. Pearson Education, Limited. All rights reserved.
13 (a) No, yes, yes, yes, no, no
 4 7 −1
3 

18 10 14
6 a=1
31000
 9 000


11 16 900


 340

18

14 750
14 600


 270


 122

9

2 2 −1 0 2
12 


2 2 −1 0 2
2
2
1 3  2



1
3
2
2
2


1 1  −2 −2 −2
 2 2


 2 2
−1 −1
1085
n=3
32.57
26.43


Weighted average = 19.14


11.43
10.43


2
BA = 1

2

CB not defined, CA = 4
3

19 AB = 0 2,
1 4


3 4 
20 (a) 
,
2 3
4 4 
2 2


1 2
−1 2

, BC = 
0 1
 2 1
−1 2
1 2
1 4, AC not defined
2 3
−1 2 0 2
(b) 
, 

 0 1 0 0
9 0 
both 

0 9
−3 12
both 

30 3
First set does not commute, second set commutes
Bricks – type C and sand
b + d b 
23 
D 
 0
1
1 1 1

24 (a) A = 
 B = 1
1 1 1
1
1

(b) A = [1 1 1] B = 1
1
1 0
0 0
(c) A = 
 B = 1 1
0
0




James, Glyn, and Phil Dyke. Modern Engineering Mathematics 6th Edition PDF Ebook, Pearson Education, Limited, 2020. ProQuest Ebook Central,
http://ebookcentral.proquest.com/lib/monash/detail.action?docID=6036640.
Created from monash on 2022-06-03 01:16:25.
1086 A NSW ERS T O EXERCISES
1
 1 25 23   0 12
2
  1
5

25 A =  2 −1 2 + − 2 0 −2
 23
0
2 1 − 21 2

 1 0 1  1 0  1 0
40 2, 
, 
, 2 

−3 2 −3 2 0 1
44 (a) −1.6569, 9.6569
(b) 4.6667 ; j0.623 61
(c) 2, 3 ; j
7
 41 15
26  −9 63 − 40


−13 38
41
47 x (2x + 1) (x − 1)
27 £2273.88
51 Non-singular, singular, non-singular, singular
45 (a) −0.1884
2
 1 3 2  3 6 5



28 0
5 2  6 7 8
2 −2 1 − 4 1 −3
799.8 800 
30 800  800 



800.2 800 
1

 0

−  0

0
2
3
1
3
2 − j
 j 1 ,


−√ 

√ 
1
2
1
2
 16
a  16
b = −10 + A −10 and B = A



  
1
3
 5 −7 −1


5 2
− 4
 2 −1 −1
98 −92
−100

1 
53 169
 − 47 −79 −50
6 −87 −8

33 n = 3
 1 0 0

55 −1
1 0 ,

 2 −2 1
Copyright © 2020. Pearson Education, Limited. All rights reserved.
−1 0 1
34 Minors = −1 −2 −1
2 −2 −2
−1 0 1
Cofactors = 1 −2 1
2 2 −2
A = 2
35 (a) −19
(d) 1
(b) 130
(e) −3
 1 1 0
56 0 1 2 ,


0 1 3
 1 0 0


0 1 0 ,
0 0 1
(c) −65
36 −4, 16, 16, −32
37 3
d − b
38 
−
a
 c
 2 −1 0
39 − 4 3 −1


 1 −1 1
2
0 0 −1
1 0 −1
−

0 1 −1


0 0
1
1 0 0 0 
 1 −1 −1 
0 12 0 0 

52 0
,
1 0 , 
0 0 13 0 
0 0

1 
0 0 0 14 
1
3
2
3
31 h = 13 , k = 23 , l = 13 , m = 16
 1
32 A = √ 2
√ 12
(b) 100
2
1
4
4 −8 6


0 2 −1 ,
0 0 2
 1 1 0


0 1 2 ,
0 0 1
 24 −20 3


4 −1
− 4
 4 − 4 2
 1 1 0


0 1 0 ,
0 0 1
 −1 −3 2 


3 −2 
 2
−1 −1
1
−3 2 2
1
57  2 −3 2 ,
5
 2 2 −3
58
1
4
4
8
− 4
1 

6
11
12
−

68 
 36 −2 − 4
 0.68 − 0.32 − 0.32


0.68 − 0.32
− 0.32
− 0.32 − 0.32
0.68
8
−5 22
1 

11
−
19
2

49 
 13 −18 −11
James, Glyn, and Phil Dyke. Modern Engineering Mathematics 6th Edition PDF Ebook, Pearson Education, Limited, 2020. ProQuest Ebook Central,
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Created from monash on 2022-06-03 01:16:25.
AN S WE RS TO E XE RCI S E S
1
0
59 A = A−1, B −1 = 
0

0
1
0
( AB ) −1 = 
0

0
1
60 (a )  
2
68 a = u1 b = (−u1 + u2)/p c = (−u1 + u3)/q
d = (u1 − u2 − u3 + u4)/pq
0 0 0
0 0 1
1 0 0

0 1 0
69 x = 0.5889u
−1
( c)  
 2
−3
72 (a)  0
 
 2
4 
3
(d )  
2
 
1
73 x = 1
y = 5l
y = −2m
Copyright © 2020. Pearson Education, Limited. All rights reserved.
y = 0.7083
z = 1.9167
2.6844
(b) 0.0234


2.3569
t = 2.9583
−1.3424
 1.2860

(c) 
 2.4458


 0.5511
−142.53


 −50.52, det = 0.001453
 262.77
79 Solution: 1, 2, 2, 3
After 5 iterations: 0.989, 1.99, 1.98, 3.00
z = 7l
z=0
80 Solution: −0.083, 0.708, 1.917, 2.958
After 3 iterations: −0.189, 0.634, 1.868, 2.920
65 −6, −3, −2
 0.1875
− 0.0625



 0
 0

− 0.0625


 0.1875
z=2 t=3
78 4.5, 8, 10.5, 12, 12.5
x = 2, y = 1, z = 2
(b)
− 253 
1 
(c)  150

 2
 75 
−74.17


76 −25.54 , det = 0.002 725
 140.11
−7 − 6 5
1 

63
2
0 2
12 
 1 − 6 1
 0.1896
− 0.0604

− 0.0167
67 
 0.0167
− 0.0021

− 0.0021
y=2
74 x = −0.0833
− j
62  1
 
 j
66 (a) a = 0
c = −0.5825
−3
(b)  2
 
 4
 1.1602
75 (a) − 0.0515


 0.0065
x=l
x=m
z = 0.2222u
71 y1 = 1.8936 y2 = 4.6809 y3 = 8.1489 y4 = 10.7660
 cos (− π8 )
61 
π 
−sin (− 8 )
64 a = 1
a = −6
y = 0.4222u
70 a = −0.4011 b = 1
f(1) = 1.4345
0 0 0
0 0 1
0 1 0

1 0 0
 1
 
( b) − 4 
 0
1087
1
1 
7
9 
6
λ 
(c) λ 
 
λ 
−2
(d)  1
 
 3
81 (a) 0.8
− 0.0021
− 0.0604
− 0.0167
0.0167
− 0.0021
0.2088
− 0.0551
− 0.0218
0.0171
− 0.0021
− 0.0551
0.2103
− 0.0564
− 0.0218
0.0167
− 0.0218
− 0.0564
0.2103
− 0.0551
− 0.0167
0.0171
− 0.0218
− 0.0551
0.2088
− 0.0604
− 0.0021
0.0167
− 0.0167
− 0.0604



0.1896
(b) 1.1
(c) no convergence
 10 
82  20 
−30 
There is no convergence in 50 iterations, even from a
 10.1


starting value of  19.9
−29.9
83 I1 = 0.5172, I2 = 0.4914, I3 = 0.8017
84 0.1685, 0.3258, 0.5282, 0.7188, 0.9563, 1.0063,
0.8063, 0.6064, 0.4059, 0.2079
85 (a) 2, 2, [2/3, −1/3]
(c) 2, 2, [1, −t, t]
(e) 2, 3, inconsistent
(b) 1, 2, inconsistent
(d) 2, 2, [2 − t, 1, t]
(f ) 4, 4, [0, 1/3, 0, 1]
James, Glyn, and Phil Dyke. Modern Engineering Mathematics 6th Edition PDF Ebook, Pearson Education, Limited, 2020. ProQuest Ebook Central,
http://ebookcentral.proquest.com/lib/monash/detail.action?docID=6036640.
Created from monash on 2022-06-03 01:16:25.
1088 A NSW ERS T O EXERCISES
86 a = 2 gives 1, 2 and inconsistent equations; a = −1
gives 1, 1 solution [1 − 2t, t]; otherwise solution is
 −α (α − 1)α 
 α − 2 , α − 2 
87 (a) 2
(b) 3
96 Eigenvalues 3, 3
eigenvectors [1, 0], [0, 1]
Eigenvalues 3, 3
eigenvector [0, 1]
Eigenvalues 5/2, 5/2 eigenvector [1, −2]
Eigenvalues 0, 0
eigenvector [1, −2]
88 (a) Rank = 2, (−2 + t, 5 − 2t, t)
(b) Rank = 2, no solution
89 Rank = 3, rank = 3; ( m, −1, 1, −m)
90 (a) x = −1, y = 13 (2 − 4m), z = m
(b) No solution
(c) x = 111 (−9 − 45l + 13m), y = 111 (5 − 8l + 5m)
z = l, t = m
(d) Unique solution x = −1, y = −1, z = 1
92 Rank = 4 implies points not coplanar; rank = 3
implies the points lie on a plane; rank = 2 implies the
points lie on a line; rank = 1 implies the four points
are identical
94 (a) l2 − 4l + 3, eigenvalues 3, 1
(b) l2 − 3l + 1, eigenvalues 2.618, 0.382
(c) l3 − 6l2 + 11l − 6, eigenvalues 3, 2, 1
(d) l3 − 6l2 + 9l − 4, eigenvalues 4, 1, 1
(e) l3 − 12l2 + 40l − 35, eigenvalues 7, 3.618, 1.382
(f ) l2 − (2 + a)l + 1 + 2a, eigenvalues
1 + 12 a ; 12 Ë(a2 − 4a)
1  1
95 (a) 2, 0;   ,  
1 −1
−1  2  2
     
 2 ,  2 , −1
 2 −1  2
Copyright © 2020. Pearson Education, Limited. All rights reserved.
(c) 9, 3, −3;
(d) 3, 2, 1;
(b) 4, −1;
2 1  0
     
2 , 1 ,  2
1 0 −1
(h) 4, 3, 1;
 2  2  4
     
−1 , −1 ,  1
−1  0 −2
2 −1
3 ,  1
   
 1 −2  3
     
97  2 ,  1 , 0
−1  0  1
98 (a) 5, 1, 1;
1 −2 −1
    
1 ,  1 ,  0
1  0  1
(b) 2, 2, −1;
−1 8
   
 1 , 1
 0 3
(c) 2, 2, 1;
 3  4
   
 1 ,  1
−2 −3
(d) 2, 1, 1;
2  0 1
     
1 , −2 , 3
2  1 0
−3
99 One eigenvector  1
 
 1
100 2, 1, 1;
−1 −1 1
     
 1 ,  1 , 0
 1  0 1
(e) 14, 7, −7;
2  6  3
     
6 , −3 ,  2
3  2 − 6
101 3, [1, 0, 0, 1]; 2, [0, 1, 0, 0] and [0, 0, 1, 0];
−1, [−1, 0, 0, 1]
(f ) 2, 1, −1;
 1  1  1
     
−1 ,  0 ,  2
−1 −1 −7
−1  0 2
104 3, 2, −6;  1,  1, 1
     
 1 −1 1
(g) 5, 3, 1;
−2  1  0
     
 −3 , −1 , −1
 1  0  1
 1
105 −1
 0
James, Glyn, and Phil Dyke. Modern Engineering Mathematics 6th Edition PDF Ebook, Pearson Education, Limited, 2020. ProQuest Ebook Central,
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Created from monash on 2022-06-03 01:16:25.
AN S WE RS TO E XE RCI S E S
5.10 Review exercises
3
1
2
−2 −3 −1
1 



8
1
29
22
−
−
−
−
,
,
8
1
29
9 (a) 
,

22 
6 −2 −8
−6 2 8
12 18 − 40  12 0 0



1 (a)  0 0
8  18 0 0
 0 0
4 − 40 8 4
z = 2, y = 1, x = 2
1
9
 8

(b) Y = 18
3
35
−


 8 − 4 − 6
 8 9 −7 12 12 − 6



0
(b) 0 2 0  0 4
0 0
5  0 0 16
3
 6 11
(c) Z = 15 2 −59


16 −7 −16
4 14 −3


5
0 0
0 0 4
10 (a) 3, 0, 2, 1; det = 12
(b) 1, 2, 3, 4
2 l = −1, m = 2
l = 2, m = −1
11 1, 2, 3
 3
3 Normal strain =  −3 


3÷2 
Shear strain = 0
12 If c ≠ 0 then rank = 2
if c = 0 then rank = 1
 0.0051
 0.9712


13 a = − 0.3931


− 0.0760
 0.0283


4 (a − b)(b − g)(g − a)(a + b + g)
5 u = 1: (1 + 2a, −3a, a)
u = 2: (2a, 1 − 3a, a)
 3 6 5


A =  6 7 8
− 4 1 −3
 1 3 2

6 A = 0
5 2

2 −2 1
2
3
 3 −2 −1


A−1 =  2 −1 0
− 4
3 1
Copyright © 2020. Pearson Education, Limited. All rights reserved.
1089
−11
 
X =  −1
 15
14 a = 0.4424, b = −1.5037, c = 1.5023, d = −0.0611
max at x = 0.74, f = 0.4065
15 rank B = 2, AAT = I, A−1 = AT
x1 = 2.444, x 2 = −2.556, x3 = −1.222
16 (b) x1 = 44, x 2 = − 48, x3 = −39, x4 = 33
2 −2 −1
1

T
7 (a) P =  1 2 −2 , the solution X= P −1b
3
2
1 2
 0
17 (a) −1
 
 2
−8
 
(b) −5
 12 
18 (a) 4, 3, 2;

 0.5774  0.1961  0

 
 

 0.5774 ,  0.5883 ,  0.7071
− 0.5774 − 0.7845 − 0.7071
exists
Q

 Ix − A
(b) E = 

Qx Qy
 Ixy −
A


0

2
x
Qx Qy
A
Q2y
Iy −
A
Ixy −
0
 8 −10 3
8 (a) B =  3 − 4 1


−1
0 0
(b) k = 8.2316, k = −1.9316
f(2) = 0.2200 f(3.5) = −0.4228

0


0

A
0.2033 0.1374 − 0.4472
 

 

0.6505 , 0.8242 , − 0.8944

0.7318 0.5494  0
(b) 5, 3, −1;
(c) 9, 6, 3;
1
3
−1   2   2 
  1  1 
 2  , 3 −1  , 3  2 
 2   2  −1 
19 l = 9 a = 1 b = 6
James, Glyn, and Phil Dyke. Modern Engineering Mathematics 6th Edition PDF Ebook, Pearson Education, Limited, 2020. ProQuest Ebook Central,
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Created from monash on 2022-06-03 01:16:25.
1090 A NSW ERS T O EXERCISES
1  0  −1 
20 0  , 1  ,  0 
     
1  0   1 
21 After 100 iterations rounded down to the nearest integer
70 98 136
56 78 109
and the largest eigenvalues are
42 59
81
0.9963, 0.9996, 1.0029
21 29
40
22 (a) 3, 1;
0.7071  0.7071
0.7071 , − 0.7071


 
6 (a) True
(b) False
(c) False
7 (a) {n [ N: 11 < n < 32}
(b) {11, 13, 15, 17, 19, 21, 22, 23, 24, 25, 26, 27, 28,
29, 30, 31, 32}
(c) Ā = {n [ N: 11 < n < 32}
B = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 20, 21, 22, 23,
24, 25, 26, 27, 28, 29, 30, 31, 32}
(d) A > B = {1, 3, 5, 7, 9, 11, 12, 13, 14, 15, 16, 17,
18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28,
29, 30, 31, 32}
(e) Ā ø B = A > B (see (d))
(b) 0.8794, −1.3473, −2.5321;
0.4491  0.8440  0.2931

 
 

0.8440 , − 0.2931 , − 0.4491
0.2931 − 0.4491  0.8440
26 E1 = 4E2 + 3I2; I1 = 3E2 + 25 I2
CHAPTER 6
11 (a) A > B
(b) ∅
(d) U
(e) A
(g) A > (B > C)
Exercises
1 A = {1, 2, 3, 4, 5, 6, 7, 8, 9}
B = {−4, 4}
C = {5, 6, 7, 8, 9, 10}
D = {4, 8, 12, 16, 20, 24}
2 A < B = {−4, 1, 2, 3, 4, 5, 6, 7, 8, 9}
A > B = {4}
A < C = {n [ N: 1 < n < 10}
A > C = {n [ N: 5 < n < 9}
B < D = {−4, 4, 8, 12, 16, 20, 24}
B > D = {4}
B>C=∅
Copyright © 2020. Pearson Education, Limited. All rights reserved.
5 (a) A < B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 16,
18, 20}
(b) A > B = {2, 4, 6, 8, 10}
(c) A < C = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 16, 32}
(d) A > C = {2, 4, 8}
14 (a) 5
(b) 25
15 (a) 20
(b) 27
(c) A
(f ) A ø (B > C )
16 4
17 (a) C = {a, b, d, i}, B ø C = {a, b, i}
B > C = {a, b, i}, A > B > D = ∅
A ø F = {a, b, c, f, i}, D ø (E > F) = {b, c, e, h, i},
(D ø E) > F = {b, c, i}
(b) B ø C = {c, d, e, f, g, h}, C ø E = {c, e, f, g, h, i}
D ø E ø F = {b, c, e, f, h, i}
(c) L1: {b, c, d, e, f, g, h, i}
L 2: {b, c, d, e, f, g, h, i}
L3: all elements
3 A < B = {n [ N: 1 < n < 10}
A > C = {1, 5, 9}
A>B=∅
B < C = {1, 2, 4, 5, 6, 8, 9, 10}
B > C = {4, 8}
18 (a) 1 if p = 1, q = 1; 0 otherwise
(b) 0 if p = 0, q = 0; 1 otherwise
(c) 0
(d) 1
4
A>B
A<B
AøB
A>B
A>B
19 (a) p · q + pˉ · qˉ
(b) (p + pˉ) · (q + qˉ )
(c) p + q + pˉ + qˉ
(d) p · q + r · s
(e) pˉ · q · s + pˉ · qˉ · r · s + p · q · r · s + p · qˉ · s + p · q · s
(f ) p · q · r + p · q · t + p · q · u + p · s · uˉ + p · v
20 pˉ · qˉ + r + ˉs + qˉ · t
James, Glyn, and Phil Dyke. Modern Engineering Mathematics 6th Edition PDF Ebook, Pearson Education, Limited, 2020. ProQuest Ebook Central,
http://ebookcentral.proquest.com/lib/monash/detail.action?docID=6036640.
Created from monash on 2022-06-03 01:16:25.
AN S WE RS TO E XE RCI S E S
21
p
q
0
0
0
0
1
1
1
1
0
0
1
1
0
0
1
1
r
0
1
0
1
0
1
0
1
pˉ qˉ rˉ pˉ · q · rˉ pˉ · q · r p · qˉ · rˉ p · q · r f
1
1
1
1
0
0
0
0
1
1
0
0
1
1
0
0
1
0
1
0
1
0
1
0
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
1
27
s2
s3
s2
s3
s2
s3
s1
0
0
1
1
1
0
0
1
Lamp
s1
s3
s2
(s1, s2 and s3 are the three switches)
28
f =p
ˉ · q · rˉ + pˉ · q · r + p · qˉ · rˉ + p · q · r
22 (a) p · q
(c) p · q + pˉ · qˉ
(e) 1
1091
c2
c3
c2
c3
c1
(b) pˉ · r
(d) p + q + r
(f ) q + r
Machine
c1
c2
c3
(c1, c2 and c3 are the three contacts)
23 (a) pˉ · q + p · qˉ + p · q
(b) (p + q) · ( p + qˉ ) + p · (rˉ + qˉ )
(c) p · (q + pˉ) + (q + r) · pˉ
29
A
Machine
24
p
r
p
r
q
p
A
B
C
q
(a)
s
t
(b)
30 (a) p + q
p
and
q
p
q
p
p
(b) ( p + qˉ ) · r
p
r
p
q
p
q
(c)
(d)
or
q
Copyright © 2020. Pearson Education, Limited. All rights reserved.
and
q
not
25
D
r
J
F
F
H
J
H
A
B
A
C
(c) 0
(d) p · r + s · pˉ
p
q
and
26
or
p
B
C
(A, B and C are the panel of three)
not
and
s
James, Glyn, and Phil Dyke. Modern Engineering Mathematics 6th Edition PDF Ebook, Pearson Education, Limited, 2020. ProQuest Ebook Central,
http://ebookcentral.proquest.com/lib/monash/detail.action?docID=6036640.
Created from monash on 2022-06-03 01:16:25.
f
1092 A NSW ERS T O EXERCISES
31
p
not
q
not
or
and
p·q
p·q+p·q
q
p
p
or
p
p
not
p
not
and
or
and
not
not
not
and
q
q
and
q
q
or
or
q
p
not
and
not
p
or
q
and
q
p
(e)
not
p
or
q
(a)
r
not
p
not
q
not
r
not
r
not
p
not
p
not
and
32 (a) Fred is not my brother
(b) 12 is an odd number
(c) There will be no gales next winter
(d) Bridges do not collapse when design loads are
exceeded
r·p
and
and
33 (a) F
or
(b) T
(c) T
(d) F
q
34 (a) T
(b) F
(c)–(e) are not propositions
(f ) Truth value is not known
r
p
not
q
not
and
35 (a) A ` B
~
(c) Ã S (B ` C )
(b)
p
not
q
not
and
p · (q + r · s)
r
p
not
s
not
p
not
q
not
and
or
p
not
q
not
and
or
r
s
and
and
(b) A S C
~
(d) C S B
36 (a) It is raining and the Sun is shining therefore there
are clouds in the sky
(b) It is raining therefore there are clouds in the sky
and hence the Sun is shining
(c) If it is not raining then the Sun is shining or there
are clouds in the sky
(d) It is not the case that it rains and the Sun shines,
and there are clouds in the sky
Copyright © 2020. Pearson Education, Limited. All rights reserved.
s
37 (a) x 2 = y 2 S x = y for positive numbers x and y
(b) x 2 = y 2 S x = y for x = 1 and y = −1 (one of
many possible answers)
(c)
p
or
q
38 (a) n = 4
and
p
p+q·r
(b) n = 3
(c) n = 7
or
r
p
or
r
or
q
and
p
or
q
and
r
(d)
and
r
39 (a) B ` C S A
B ` C S Ã
(b) If x 2 + y 2 > 1 then x + y = 1; if x 2 + y 2 , 1
then x + y ≠ 1
(c) If 3 + 3 = 9 then 2 + 2 = 4; if 3 + 3 ≠ 9
then 2 + 2 ≠ 4
James, Glyn, and Phil Dyke. Modern Engineering Mathematics 6th Edition PDF Ebook, Pearson Education, Limited, 2020. ProQuest Ebook Central,
http://ebookcentral.proquest.com/lib/monash/detail.action?docID=6036640.
Created from monash on 2022-06-03 01:16:25.
1093
AN S WE RS TO E XE RCI S E S
40 (a)
(c)
and
(d)
~
A
B
A`A
T
F
F
T
~
~
~
~
A
B
A~ B
F
F
F
F
F
F
T
T
T
F
T
T
T
T
(b)
7
p
(a)
=
p
B
C
F
F
F
F
T
F
F
F
T
F
T
F
q
p
(b)
=
A`B A`BSC A`BSC
A
(c)
p
q
p
q
T
F
F
T
F
F
T
T
F
T
F
T
F
F
F
T
F
T
F
T
F
T
F
T
T
F
T
F
T
T
T
T
T
T
F
p
(d)
q
p
r
r
p
p
p
=
q
r
q
r
47 1, 4, 9, 16
8 (a) ( A B D) ( A C D) ( A B D)
(b) (B > C ) ø (C > B)
6.7 Review exercises
9 (a) x · y · z · u + x · y · z · u + x · uˉ
1 (a) A ø B = {8, 9}
(b) C − A = {3, 7, 8} C > B = {6, 9}
2 (a) A > B = {2, 4, 6, 8, 10}
(b) A > B > C = {10}
(c) A ø (B > C) = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12,
14, 20}
Copyright © 2020. Pearson Education, Limited. All rights reserved.
p
q
=
F
p
3 (a) Ā = {n [ N: 11 < n < 20}
(b) Ā ø B = {1, 3, 5, 7, 9, 11, 12, 13, 14, 15, 16, 17,
18, 19, 20}
(c) A ø B = {13, 15, 17, 19}
(d) A > (B ø C ) = {1, 3, 5, 7, 9}
B
A>B
R
z
u
x
y
z
u
u
(b) x · y + z· uˉ + x · y · z
z
y
A>B
Only equals A ø B if region R does not exist, that is
A>B=∅
u
z
x
g = U (the universal set)
A
y
x
4 Statement (a) is true.
5 (a) f = A
(b)
x
y
z
(c) (i) 0
(ii) 0
x
y
z
u
x
y
z
u
x
y
z
u
x
y
z
u
x
y
z
u
x
y
z
u
x
y
z
u
James, Glyn, and Phil Dyke. Modern Engineering Mathematics 6th Edition PDF Ebook, Pearson Education, Limited, 2020. ProQuest Ebook Central,
http://ebookcentral.proquest.com/lib/monash/detail.action?docID=6036640.
Created from monash on 2022-06-03 01:16:25.
1094 A NSW ERS T O EXERCISES
10 (a) p
T
T
F
F
(e) p
T
T
F
F
q p ` q (b) p q p ~ q
T
T
T T
T
F
F
T F
T
T
F
F T
T
F
F
F F
F
A
B
q
T
F
T
F
pˉ pˉ` q p q p ~ q
F
F
T
T
F
F
T
T
T
T
F
T
T
F
T
F
(c) p
T
T
F
F
C
q pSq
T
T
F
F
T
T
F
T
A`B
T
T
F
F
C Sp
T
T
T
T
(d) To modify the circuit we introduce the chairman’s
vote E. If N denotes No and Y denotes Yes, the new
circuit must have the output
Nnew = (Nold + T) · E
Ynew = (Yold + T ) · E
where T = Tie. Hence the modified circuit will be
No (old)
Hence (p ~ q) ` (p q) S p is a tautology.
(b) p + q + rˉ
11 (a) qˉ · p + pˉ · q
(c) p · q · r + q · rˉ· s
Yes (old)
Yes (new)
T
E
12 (a) (i) p · r + q · rˉ· s + qˉ · r · ˉs
(ii) p · qˉ · rˉ + pˉ· rˉ· s
Copyright © 2020. Pearson Education, Limited. All rights reserved.
No (new)
T
E
A tie is now impossible.
13 C1 · C2 · F1 · F2 · F3 + C1 · C3 · F1 · F2 · F3
+ C1 · C2 · C3 · F1 · F2 · F3
where Ci = call button on floor i
and Fi = 1 if lift is on floor, 0 otherwise
15 N · (V + R· M). That is, no dope smoking occurs if Neil
is absent and either vivian is absent or Mike is present
and Rick is absent.
14 (a) Let the four people be labelled A, B, C, D.
The truth table is then as given below:
16 p · q · p · r + q · rˉ = q · rˉ
A
0
0
0
0
0
0
0
0
1
1
1
1
1
1
1
1
B
0
0
0
0
1
1
1
1
0
0
0
0
1
1
1
1
C
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
D
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
Yes
0
0
0
0
0
0
0
1
0
0
0
1
0
1
1
1
No
1
1
1
0
1
0
0
0
1
0
0
0
0
0
0
0
Extracting from this table those inputs that cause a
Yes, No or Tie (Y, N or T) we have
(b) Y = Ā· B · C · D + A · B · C · D + A · B · C · D
+ A·B·C·D + A·B·C·D
N = Ā· B · C · D + Ā· B · C · D
+ Ā· B · C · D + Ā· B · C · D + A · B · C · D
T = Ā· B · C · D + Ā· B · C · D + Ā· B · C · D
+ A·B·C ·D + A·B·C·D + A·B·C ·D
(c) Y = A · B · D + A · B · C + A · C · D + B · C · D
N = Ā· B · C + Ā· B · D + Ā· C · D + B · C · D
T does not simplify.
Tie
0
0
0
1
0
1
1
0
0
1
1
0
1
0
0
0
q
q·r
r
q
q + r (= q · r)
r
17 (a) F
(d) F
(g) No
(b) No
(e) F
(h) F
(c) No
(f ) F
(i) T
18 (a) (i) p q p S q q S p p 4 q (ii) q p p S q
T T
T
T
T
T T
T
T F
F
T
F
T F
F
F T
T
F
F
T F
F
F F
T
T
T
F F
T
(b) (i) False (ii) False
19 PO · T + PO · PF · T + PO · PF · T
= PO · T + PF · T + PO · PF · T is minimal
James, Glyn, and Phil Dyke. Modern Engineering Mathematics 6th Edition PDF Ebook, Pearson Education, Limited, 2020. ProQuest Ebook Central,
http://ebookcentral.proquest.com/lib/monash/detail.action?docID=6036640.
Created from monash on 2022-06-03 01:16:25.
AN S WE RS TO E XE RCI S E S
PO = 1 when pressure in oxidizer tank > required
minimum
PF = 1 when pressure in fuel tank > required minimum
T = 1 when time < 15 min to lift-off
L = 1 when panel light is on
21 p · q · r · s, p · q · r · ˉs,
p · qˉ · r · ˉs, p · qˉ · rˉ· ˉs,
pˉ · q · rˉ· s, pˉ · q · rˉ· ˉs,
pˉ · qˉ · rˉ· ˉs
p · q · rˉ· s, p · q · rˉ· ˉs,
p · qˉ · rˉ· ˉs, pˉ· q · r · s,
pˉ · qˉ · r · s, pˉ · qˉ · r · ˉs,
Converse
22 (a) If I do not go, the
train is late
(b) If you retire, you
will have enough
money
(c) You cannot do it
unless I am there
(d) If I go, so will you
p · qˉ · r · s,
pˉ · q · r · ˉs,
pˉ · qˉ · rˉ· s,
Contrapositive
If the train is early,
I will go
If you do not have
enough money, you
will not retire
I can do it if you are
there
If you do not go nor
will I
23 ‘If you were a member of the other tribe, what would
you answer if I asked you if your God was male?’
The answer is then definitely false!
1095
15 £2700k, 11
16
1 + (n − 2) ÷ x
1− x
17 3.1365 (4dp)
18 25 , 27 , 29 , 112 , 132 , 152
19 £66 116, £128 841, after 7.3 years
20 (a) 11 781
(d) 3154 − 1
(c) 1 − ( 12 )153
153
(f ) 154
(b) 1 205 589
(e) 1 217 370
21 9
23 x =
10r
, 117.46
1
1 − (1 + 100
r)− n
24 2 −
2+n
2n
25
a(1 − r n )
dr
+
[1 + nr n − 1 − (n − 1) r n ]
(1 − r )2
1− r
26 (a) A2n + 3
(c) A(−1)n + 23 ( 12 )n
(b) A3n − 5(n + 12 )
(d) A2n + 23 n2n
27 £1770, {10 000, 9430, 8792, 8077, 7276, 6379, 5375,
4250, 2989, 1578, −3}
CHAPTER 7
28 (A + n)/n2
Exercises
1 (a) 13, 1, 95
(b) 6, 10, 14
(c) − 64, 16, −4
2 xn+1 = 83 xn, x0 = 5
3 p = −3, q = 13, x0 = 13, x1 = 10, x2 = 7, x3 = 4
xn+1 = xn − 3
Copyright © 2020. Pearson Education, Limited. All rights reserved.
5 45, 57.5, 63.75, 66.875
6 (a) 40
 v
7  
V
8
(b) 16 736
(c) 35
n+1
× 100
31 (a) A5n + B2n
(b) A3n + B(−2)n
nπ
nπ 

+ B sin
(c) ( 15 )n  A cos


2
2 
(d) A5n + Bn5n
(e) A(− 12 )n + B
1
13
1
32 (a) 4 + 12 (5n ) − 12 (2 n )
(b)
(c) − 16 n − 361 + A3n + B(−2)n
33 (b) (1 − n)a n
4
9
+ 149 (− 12 )n + 13 n
(c) (3 + (2a−10 − 0.3)n)a n
34 T2 = 2x 2 − 1, T3 = 4x 3 − 3x, T4 = 8x 4 − 8x 2 + 1
1 n
∑ ( xl − 10)2
n l =1
35 (a) Nt = 2t
9 2.618
81 256 625
10 {0, 13 , 16
25 , 109 , 321 , 751}
11 {1, 1.5, 1.4, 1.417, 1.414, 1.414, 1.414}
12 1.222
13 5
14 (a) 16, 31
29 (a) 0 (b) −3 × 2n (c) 0 (d) 75 × (−2)n
(a) and (c) satisfy recurrence relation
(b) 10, 20, 40, 80
 1 + ÷5 
(b) Nt = 

 2 
t +1
t +1
 1 − ÷5  
−
  ÷5
 2  
36 (a) 0.5, 0.4, 0.3, 0.2353, 0.1923, 0.1632; S0
(b) 0.4615, 0.4722, 0.4789, 0.4831, 0.4859, 0.4879;
S0.5
(c) 2, 2, 1.817, 1.682, 1.585, 1.513; S1
(d) 1.5, 1.5625, 1.5880, 1.6018, 1.6105, 1.6165;
Se1/2 = 1.6487
James, Glyn, and Phil Dyke. Modern Engineering Mathematics 6th Edition PDF Ebook, Pearson Education, Limited, 2020. ProQuest Ebook Central,
http://ebookcentral.proquest.com/lib/monash/detail.action?docID=6036640.
Created from monash on 2022-06-03 01:16:25.
1096 A NSW ERS T O EXERCISES
(e) 1.4142, 1.5538, 1.5981, 1.6119, 1.6161, 1.6174;
S 12 (1 + Ë5)
(f ) 0, 0, 1.2990, 2, 2.3776, 2.5981; S
37 (a) 1, 1, 53 , 2.5, 3.4, 4.3333; diverges to infinity
(b) 1, 0, −1, 0, 1, 0; oscillates between 1, 0, −1
(c) 1, 3, 1, 3, 1, 3; oscillates between 1 and 3
38 (a) 10
(d) 25
(b) 19
(e) 18
(c) 1 000 002
41 (a) convergent
(c) divergent
59 (a) −1
(d) 0
(c) n − 1
(b) 1
(d) n
62 (a) Undefined at x = 0, continuous for x ≠ 0
(b) Infinite discontinuity at x = 2, continuous for x ≠ 2
(c) Finite discontinuity at x = 0, continuous for x ≠ 0
(d) Finite discontinuities at x = ;Ën, n = 0, 1, 2, p
66 a = −1.879, b = 0.347, g = 1.532
(b) convergent
(d) divergent
67 (a) is convergent, (b) convergent and (c) divergent
Root is 0.771
68 5.4267, 5.3949, e1 ° 0.05
1
2+N
(b) 4 − N − 1 , 4
,1
2N + 1
2
1
1
1
(c)
−
,
4 2( N + 1)( N + 2) 4
69 a0 ø 1.9, ak ø 12 (2k + 1)
un+1 = cos−1(−sech un), a0 = 1.8751
43 (a) 1 −
44 (a) divergent
413
(a) 999
46 19
33 ;
(b) divergent
(b) 10
99
7.12 Review exercises
(c) convergent
300
(d) 1729999
(c) 1
50 (a) x < 1
(b) x [ R
51 (a) (−x 2 ) r ( x < 1)
(c) x < 1 (d) x < 1
(b)
x 2r +1
2r + 1
(| x | 1)
(c) (−1)r(r + 1)x r ( x < 1)
(2r − 3)!
x r r 1 (| x | 1)
(d) − 2 r − 2
2
r!(r − 2)!
1  r + 2 (−1)r 
2 +
(| x | 12 )
(e)
2 r 
10 
(−2)(−3)(− 4)
1⋅ 2 ⋅ 3
(− 21 )(− 23 )(− 25 )(− 27 )
(d)
1⋅ 2 ⋅ 3 ⋅ 4
(b)
1.1905, six multiplications
55 (a) (1 + 2x 2)−1
1− x
ln(1 − x )
(c) 1 +
x
3 1 − 0.2(− 12 )t,
n
(b) (1 − x)−1/2
x2
(d)
ln(1 − x 2)
1 − x2
1
n
5 (a) A2 + B3
(c) A2n + B3n + 12 4n
(b) (A + Bn)2n
(d) A2n + B3n + 3n−1n
8 200 + 20(− 25 )t
9 2 + A cos nu + B sin nu, tan u = √7
3
10 g ø 0.577 235; compare the true value 0.577 216
11 (a) divergent
(c) divergent
12 (a) 410
333
(f ) (1 + x)x 4r ( x < 1)
5⋅ 4
1⋅ 2
( 1 )(− 12 )(− 23 )
(c) 2
1⋅ 2 ⋅ 3
1 1000, 850, 700, 550, 400, 250, 100
1000, 681, 464, 316, 215, 147, 100
2 £361, £243, £141, £53 for r < 23.375
48 1.082 322 1; summation from right allows full account
to be taken of the accumulative effect of small terms
Copyright © 2020. Pearson Education, Limited. All rights reserved.
(b)
(c) 2
65 0.830
42 94
56 3.1415
58 (a) 3
1
2
64 1.75
40 2.2, 2.324, 2.418 996, 2.450 262
Estimate = 2.465 5011
Limit = 2.465 571
54 n = 8,
(b) − 23
63 (a) Upper bound is 7, lower bound 5
(b) Upper bound is 3, lower bound −1
39 70%
52 (a)
57 (a) 13
(b) 143
333
13 (a) convergent
(c) convergent
14
(b) convergent
(d) convergent
(c) 101
999
(d) 1724
3333
(b) divergent
(d) divergent
x (1 + x )
(1 − x )3
15 a = 1, b = − 13 , c = 451
x < 0.2954
tan 0.29 is given to 4dp; tan 0.295 has an error of 12 unit
in 1dp, but when rounded to 4dp gives an error of 1 unit
16 x − 16 x 3
5
35
x − 16 x 3 + 403 x 5 − 112
x 7 + 1152
x9
James, Glyn, and Phil Dyke. Modern Engineering Mathematics 6th Edition PDF Ebook, Pearson Education, Limited, 2020. ProQuest Ebook Central,
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Created from monash on 2022-06-03 01:16:25.
AN S WE RS TO E XE RCI S E S
17 0.095, 31
17 r = l/4
19 2.718 586 07
2.718 357 88
2.718 281 81
18 £1.25, 7500
20
r
Mr
19 (a) 9x 8,
(d) 16x 3 + 10x 4,
(g) 1 + 1/(2Ëx),
0 1
2
3
4
5
6
7
8
0 2.64 1.93 2.13 2.06 2.13 1.93 2.64 0
2
22 F = W (l − a) (l + 2a) − WH( x − a)
l3
W (l − a)2a
W (l − a)3 (l + 2a) x
M=
−
2
l
l3
+ W(x − a)H(x − a)
23 (a) positive values of Ë(1 − sin2u)
(b) A = −5/128
25 a = − 13 , −4/45
(b) 1
(c) 2x
1
(e)
2÷x
(f )
−1
(1 + x )2
2 (a) 4x − 5
(b) −1
(c) (1, −15), ( 12 m + 23 , 12 m 2 + 12 m − 15)
(d) −1
(e) y = −x − 14
(b) 1
3 (a) 6x 2 − 6x + 1
(c) (1, 3), ( 14 [1 ; Ë(1 + 8m)], 14 [12 − 3m ; mË(1 + 8m)])
(d) m = 1, −1/8 (e) y = x + 2, y = −0.125x + 3.125
Copyright © 2020. Pearson Education, Limited. All rights reserved.
21 (a) (−3x 4 − 2x 3 − 3x 2 + 6x +1)/(x3 + 1)2,
(b) (4 − 3x 2)/[(x 2 + 4)2Ë(2x)],
(c) (1 − 2x − x 2)/(x 2 + 1)2,
(d) (x 1/3 + 2)/[3x 1/3(1 + x1/3)2],
(e) (x 2 + 2x − 1)/(1 + x 2)2,
(f ) 3x(2 − x)/(x 2 − 2x +2)2
24 331, 465, 7, −31
Exercises
4 (a) minimum at x = −1/2
(c) maximum at x = 3/2
5 3ax 2 + 2bx + c
 1,
2t − 1,

6 v(t ) = 
 3,
 −1,
0 t 1
1 t 2
2t3
3t 9
8 13 m2 min−1
10 3W(2x − l)2/4l 2
12 mT
16
(d) 6 − 8x + 27/x 2,
(f ) 8x 3 + 9x 2 + 4x
23 103 Ë2
CHAPTER 8
(d) 3x 2
20 (a) 18x 5 + 75x4 − 2x − 5,
(b) 20x 3 + 3x 2 + 30x − 27,
(c) (21x 2 + 10x − 3)/(x 3/2),
(e) (3x 3 − x 2 + x − 3)/(2x 5/2),
(c) −8x,
(f ) −1/x 3,
(i) −1/x 4
22 2ac + ad + bc, (ad − bc)/(cx + d)2, 6ax + 5b,
(bax 2 + 2acx)/(bx + c)2
24 1.233 577
1 (a) 0
(b) 23 Ëx,
(e) 12x 2 + 1,
(h) 7x 5/2,
1097
dx
∝ (a − x )(b − x )
dt
(b) minimum at x = 1/3
(d) maximum at x = 1/2
25 (a) 10x − 2
(b) 12x 2 + 1
(c) 24x 23
(d) 12x 3 − 6x 2 − 20x + 11
(e) 36x 5 + 20x 3 − 54x 2 + 12x − 15
(f) 1/(x + 1)2
(g) 1/(x − 2)2
2
(h) 2(2 − x)/(x − 4x + 1)2
(i) (6 − x 2)/(x 2 + 5x + 6)2
26 (a) 45(5x + 3)8
(b) 28(4x − 2)6
5
(c) −18(1 − 3x)
(d) 3(6x − 1)(3x 2 − x + 1)2
(e) 6(12x 2 − 2)(4x 3 − 2x + 1)5
(f) −5(4x 3 − 1)(1 + x − x 4)4
27 (a) 512(x + 2)6(3x − 2)4(9x + 4)
(b) (5x + 1)2(3 − 2x)3(37 − 70x)
(c) ( 12 x + 2)(x + 3)3(3x + 11)
(d) 2(x 2 + x + 1)(x 3 + 2x 2 + 1)3 ×
(8x 4 + 19x 3 + 16x 2 + 10x + 1)
(e) (x 5 + 2x + 1)2(2x 2 + 3x − 1)3 ×
(46x 6 + 57x 5 − 15x 4 + 44x 2 + 58x + 6)
(f) (2x + 1)2(7 − x)4(37 − 16x)
(g) (3x + 1)4(21x 2 + 74x + 19)
30 b = 56, h = 144, w = 90
31 (a) 1/Ë(1 + 2x)
(b) (3x + 4)/[2Ë(x + 2)]
(c) (3x + 2)/(2Ëx)
32 (a) (2x 2 + 4)/Ë(4 + x 2)
(b) (9 − 2x 2)/Ë(9 − x 2)
(c) (2x 2 + 4x + 4)/Ë(x 2 + 2x + 3)
(d) 23 x −1/3 − 14 x −3/ 4 (e) 23x ( x 2 + 1)−2/ 3
(f) (8x − 3)/[3(2x − 1)2/3]
James, Glyn, and Phil Dyke. Modern Engineering Mathematics 6th Edition PDF Ebook, Pearson Education, Limited, 2020. ProQuest Ebook Central,
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Created from monash on 2022-06-03 01:16:25.
1098 A NSW ERS T O EXERCISES
33 (a) −2/(x + 3)3 (b) 1 − 1/x 2 (c) −1/(x 2 − 1)3/2
(d) 2(2x + 1)(2 − 9x − 12x 2)/(3x 2 + 1)4
53 (a) 10 x ln 10
(b) −2−x ln 2
3
2
(c) (2x + 3x + 6x + 6)(x − 1)5/2[(x + 1)1/2/(x 2 + 2)2]
34 (a) 3 cos(3x − 2)
(b) −4 cos3x sin x
(c) −6 cos 3x sin 3x = −3 sin 6x
(d) 25 cos 5x − 12 cos x
(e) sin x + x cos x
(f) −sin 2x/Ë(2 + cos 2 x)
(g) −a sin(x + u)
(h) 4 sec24x
54 x 2e−2x((3 − 2x) sin x + x cos x)
35 (a) 1/Ë(4 − x 2)
(c) (x tan−1x + 1)/Ë(1 + x 2)
57 −
(b) −5/Ë(1 − 25x 2)
(d) 1/Ë(3 + 2x − x 2)
x sin −1x
(f) 1 −
÷(1 − x 2)
(e) 3/(1 + 9x 2)
37 (a) −9x2cos2(x3)sin(x3)
3
1
sin 2 x cos x
(c) 2
(b)
21
1 + 3cos 2 x
÷(1 + sin 3 x)
(d) −
sin÷ x
2÷x
(b) − 12 e−x/2
(d) xe5x(5x + 2)
(f ) −ex/(1 + e x )2
(h) aeax+b
39 (a) 2/(2x + 3)
(b) 2(x + 1)/(x 2 + 2x + 3)
(c) 1/(x − 2) − 1/(x − 3 (d) (1 − ln x)/x 2
(e) 5/[(2x + 1)(1 − 3x)] (f ) (2x + 1)/[x(x + 1)]
40 (a) 3 cosh 3x
(c) 3x 2 cosh 2x + 2x 3 sinh 2x
(e) cos x sinh x − sin x cosh x
(b) 4 sech24x
(d) 12 tanh 12 x
(f ) −sinh x/cosh2x
41 (a) 2/Ë(4 + x 2)
(b) 2/Ë(x 2 − 1)
(c) 1/(1 − x 2)
−1
x sinh x
(e) Ë(4 − x 2)/x
(d) 1 +
÷(1 + x 2)
(f ) [(1 + x 2) − 2x tanh−1x(1 − x 2)]/[(1 − x 2)(1 + x 2)2]
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56 cot 12 u
x (2 x 2 + 2 y 2 − a 2 )
y (2 x 2 + 2 y 2 + a 2 )
58 (a) (ln x)x−1[1 + (ln x) ln ln x]
(b) (2 ln x)x ln x−1
(c) − 53 x(5 − 2x 2)(1 − x 2)−1/2(2x 2 + 3)−7/3
59 (a) [(3 − 2x) ln x + 1]x 2e−2x
36 278 (volume of sphere)
38 (a) 2e2x
(c) (2x + 1) exp(x 2 + x)
(e) e−x(1 − 3x)
(g) 12 e x/Ë(1 + e x)
55 y = x + 1; (0, 1), (−1, 0)
ex
(b) [(x − 1) sin 2x + 2x cos 2x] 2
x
60 (a) (6 + 19x 2 + 12x 4)x/(1 + x2)3/2
(b) (1 − 2x − 2x 2)/(1 + x + x 2)2
(c) −(84x 2 + 6y 2 y′ + 6xyy′2)/(1 + 3xy 2) where
y′ = −(28x 3 + y 3)/(1 + 3xy 2)
(d) (6x − 2y′ − 6yy′2)/(3y 2 + x) where
y′ = (1 + y − 3x 2)/(3y 2 − x)
61 (a) −(2 + t 2)/(sin t + t cos t)3
3t
t
(b) 18 cosec sec3
2
2
64 (a) 2 x −
2
,
x3
2+
6
x4
θ
a
65 cot , − 2
2 y
2
3
 1 + 2t  2 (1 + t )
66 
 ,
 1 + t  3(1 + 2t )3
42 e−2pa/v
67 2, 0
43 a = 26, b = 39
68 (a) 34e3x, 3ne3x
6
, (−1)n−1(n − 1)!/(2 + x)n
(b) −
(2 + x ) 4
12
12
+
,
(c)
5
(1 + x )
(1 − x )5
44 horizontal side 1/Ë2
46 u, α1 ln 3, 3α u / 2
47 (1 − t tan t)/(tan t + t)
48 y = 1 − (Ë2 − 1)x
2+x
,y≠1
1− y
( x − 1) y
(b)
, x ≠ 0, y ≠ 1
x(1 − y)
49 (a)
1
2
1
(−1)n 

n! 
+
n +1
(1 + x )n + 1 
 (1 − x )
69 a4 sin(ax + b)
50 y = x + 2, y = 4 − x
72 (a) (x 2 − 20) cos x + 10x sin x
(b) (x − 4)e−x
(c) 216(273x 2 + 39x + 1)(3x + 1)9
51 y = 4 − 64x
73 121 (145)3/2
52 43
75 Ë2, (1, 2)
James, Glyn, and Phil Dyke. Modern Engineering Mathematics 6th Edition PDF Ebook, Pearson Education, Limited, 2020. ProQuest Ebook Central,
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1099
AN S WE RS TO E XE RCI S E S
76 133/2/16
(b) 23 Ë2x 3/2 + c
104 (a) 95 x 5/3 + c
1 4
2 3
(c) 2 x − 3 x + ln x − 2x + c
77 (2 − 2x + x ) /[2x(1 − x )], | x | < 1
2
4 3/2
2
1
+c
x
(f) 18 (2x + 1)4 + c
(g) − 83 (1 − 2x)4/3 + c
3
8 7
12 5
(h) 7 x + 5 x + 2x + x + c
2x
+c
( j)
(i) 43 sin(2x + 1) + c
ln 2
(d) 2ex + 23 sin 2x + c
78 16 (13)3/2
79 (a) Minimum (1, 0), maximum ( 23 , 271 ),
inflection ( 65 , 541 )
(b) Minimum (1, 43), maximum (−5, 151),
inflection (−2, 97)
(c) Minimum at x = −2, inflexion at x = 1
80 (a) Minimum (−2, − 13 ), maximum (2, −3), inflection
(−(4 − 3Ë4)/(3Ë4 − 1), 3(3Ë4 + 1)/(3Ë4 − 1))
(b) Maximum (4, 54, e−4)
(c) Minimum (0, 0), maximum (2, 4e−2),
inflections (2 ; Ë2, (4 ; 2Ë2)e−2 ; Ë2)
(d) Minimum at x = 13
1
(b) − 156
105 (a) 43
(c) 27/2/5 − 109
106 (a) −
(e) 13 x 3 + 3e x +
(e) 13
(d) 1
1
+c
x
(b) 23 (x + 1)2/3 + c
1
+c
x
(d) sin x − cos x + c
(f) sin−1(x − 1) + c
83 (0, 0) minimum, (1/Ëe, K/(2e)) maximum.
dv
dv
S 0 as x S 0+)
not defined at x = 0 but
(Note:
dx
dx
3 + 4x
+c
3 − 4x
(g) 13 sin−13x + c
(2 x + 1)
+c
(i) sin −1
÷5
84 S = 8 a2
(k) sin−1(x − 2)/3 + c
(c) 2x 2 − 7x −
81 d = 10 Ë(2/ ), h = 10 Ë(2/ )
3
3
82 d = 8.0 (1dp),
(e)
h = 9.9
85 b [ [80, 82.2]
86 (a) x = 43
1
24
107 (a) 25
(b) x = 2
ln
(b) 29
(h) sin−1 12 x + c
(j) sin−1(2x − 1) + c
(l)
1
2
tan−1 12 (x + 3) + c
(d) 23
(c) 3
(e) 132
108
f(x)
87 Distribute wash water equally.
1
88 In year k a volume (1 − a)/(1 − a11−k ) of standing
timber should be felled, a growth factor
–4
O
1
4 x
Copyright © 2020. Pearson Education, Limited. All rights reserved.
89 1.035, 0.92, 0.88
91 (a) 5.436 = 2e, 8.155 = 3e
(b) 5.440 (h = 0.01), error depends on h2
(c) 8.00 (h = 0.01)
g(x)
92 1.5432
1
2
94 Brian by (6Ë3 − 4Ë6) s
97 −6
–4
7W/8 − 4Wx / l 0 x l / 4
101 F = 
l/4 x l
−W/8
109
O
f(x)
p2
4
102 Depth h satisfies 1000 h2(3 − 2h) = 6t
103 (a) 17 x7 + c
(c) − 15 cos 5x + c
(e) 13 tan 3x + c
3
(g) − + c
x
(i) 14 sec 4x + c
4 x
(b) 13 e3x + c
(d) 18 (2x + 1)4 + c
(f ) 2 ln x + c
(h) 12 sin 2x + c
p2
8
–2p
p
O
( j) (4x − 1) + c
1
6
3/2
James, Glyn, and Phil Dyke. Modern Engineering Mathematics 6th Edition PDF Ebook, Pearson Education, Limited, 2020. ProQuest Ebook Central,
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p
2p x
1100 A NSW ERS T O EXERCISES
110 (a) −x cos x + sin x + c
(b) 19 (3x − 1)e3x + c
(c) 161 xx 4(4 ln x − 1) + c
(d) − 131 e−2x(3 cos 3x + 2 sin 3x) + c
(e) 12 [(x 2 + 1) tan−1x] − 12 x + c
(f) 14 [(2x sin 2x + cos 2x) + c
−2
111 (a)
(b) 9 ln 3 − 269
112 (a) 13 (1 + x 2)3/2 + c
1
+c
(c) −
2 (1 + x 2 )
(c) 13 (2e3 + 1)
(d) Ë(x 2 − 1) + c
(h) −Ë(4 − x 2) + c
114 (a) 12 ln(x 2 + 4x + 5) − tan−1(x + 2) + c
(b) −2Ë(5 + 4x − x 2) + 7 sin−1[(x − 2)/3] + c
(c) 12 x − 12 ln sin x + cos x + c
(b) 181
2
(c) ln 4
(d) 2e(e − 1)
(b) x ln x − x + c
116 (a) x sin−1x + Ë(1 − x 2) + c
(c) x cosh−1x − Ë(x 2 − 1) + c
(d) x tan−1x − 12 ln(x 2 + 1) + c
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117 (a) 15 [ln x + 1 + 4 ln x − 4 ] + c
2
(b) ln | x − 2 | −
+c
( x − 2)
x
(c) ln
+c
x +1
x −1
(e) 12 ln
+c
x +1
(g) 12 ln
x ( x − 2)
+c
( x − 1)2
(d)
1
2
x + sin 2x + c
(e)
x + 14 sinh 2x + c
(f)
1
5
cosh(5x + 1) + c
(b) 12
(3x 2 − 2)(1 + x 2)3/2 + c
 3
(b) − sinh −1   + c
 x
(c) 2Ëx − 6 ln(3 + Ëx) + c
1
15
122 (a) 2Ë(1 + x) − 2 ln[1 + Ë(1 + x)] + c
(b) 151 sin3x(5 − 3 sin2x) + c
(c) 2 sinËx − 2Ëx cos Ëx + c
123 (a) ln tan 12 x 113 a = 23 , b = −1
1
3
2
−1 1
2 (x + 1) + c
2 ln(x + 2x + 5) − 2 tan
7
115 (a) 6912
x − 14 sin 2x + c
1
2
121 (a)
(b) 13 sin4x + c
1 x

+ tan −1 x + c


2 1 + x2
1
2
120 (a) 0
(e) ln x 2 + 3x + 2 + c
1
(f ) 24 (6 − 8 sin2x + 3 sin4x)sin4x + c
(g)
(c)
(b) ln
1
4 − ÷7 + 3 tan 12 x
ln
+c
÷7
4 + ÷7 + 3 tan 12 x
(c)
2 + 3 tan 12 x
+c
3 − 2 tan 12 x
 x − 1
 + (x − 1)Ë(3 + 2x − x2) + c
126 (a) 2 sin−1
 2 
(d) 131 ln
 x − 3
 +c
 2 
(b) cosh−1
 x − 2
 +c
2 
(c) sinh−1
 x + 2
 +c
 3 
(d) 2Ë(x2 + 4x + 13) + sinh−1
(e) 16 (2x2 − x − 9)Ë(3 + 2x − x2) − 2sin−1
(d) ln | x + 1 | + 1 + c
x +1
x −1
1
(f ) ln
+ +c
x
x
127 197
10 π
(h) 13 ln 1 + 2 x + c
1− x
128 2
(i) 2x + 23 ln x − 1 − 13 ln x 2 + 2 + 1 +
10
 2 x + 1
tan −1 
 +c
 ÷3 
3÷3
1 + tan 12 x
+c
1 − tan 12 x
 x − 2


 2  +c
/2
# Ë(1 + cos x)dx,
2
48
130 54
35 π , 5 π
( j) 2 ln x + ln(x − 1) − tan x + c
132 (a) 23 , ( 118 , 25 )
x −1
3
+
+c
x+2
x+2
134 20, 103 , 130
(l) −2 ln | x − 1 | + ( 12 + 23 ÷5) ln | x − 12 − 12 ÷5 |
− ( 23 ÷5 − 12 ) ln | x − 12 + 12 ÷5 | + c
119 (a) 161 (4 cos 2 x − cos 8x) + c
(b) 241 (sin 12 x + 6 sin 2x) + c
2
131 41.8 Ω
−1
(k) ln
1
2
0
21
(b) 158 π , ( 16
, 0)
135 ( 25 , 1), ( 12 , 0)
137 0.6109, 0.6463, 0.6549, 0.6569; 0.6577
138 0.1526
139 5.869 849
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AN S WE RS TO E XE RCI S E S
(y) −e−x(x 2 + 2x + 2) + c
(z) 2 Ë 13 tan−1[Ë 13 (2x + 1)] + c
140 246 A h
141 76.09
3 y = 12x − 8, y = 121 (49 − x)
142 1.1114 (4dp)
4 y = 15 (4x + 6), y = 1 − x, 89 Ë2
8.13 Review exercises
x 2 +x
1 (a) (2x + 1)e
6 Maximum ( 23 , 271 ), minimum (1, 0), inflection at x = 65
x 2 (9 − x )
(b)
(3 − x )3
(c) 5 cos(5x − 1)
(d) (ln tan x + 2x cosec 2x)(tan x)x
1
(e)
(f) − 12 (1 + x)−3/2
÷(1 − x 2 )
(2 x + 1)
1
(g) −
(h) −
( x − 1)2 ( x + 2)2
1 + x2
(i) 3 cos(3x + 1)
(j) x 2(1 + 3 ln x)
2
2
x (9 − x )
(k)
(l) −sech 2x
(3 − x 2)2
1
(n) 2x sin 2x + 2(x 2 + 1)cos 2x
(m) ÷ 2 sinh 12 x
e ÷x
4−x
(o)
(p)
3
2 ÷x
( x + 2)
1/ 2
(8 − 7 x )(2 x − 1)
( x + 1)6
(q) 2 cosec 2x
(r)
(s) sin x + x cos x
1
(v) −
1 + sin x
(t) 2xe x
(u) 2x ln 2
2
1
(w) −
x ÷( x 2 − 1)
(x) x 2(3 cos 2x − 2x sin 2x)
(y)
3x 2 + 1
2 ÷( x 3 + x + 3)
(z) −
e − x (2 + x )
(1 + x )2
2 (a) 29 x 3/2(3 ln x − 2) + c
(b) ln(x 2 + 2x + 2) + tan−1(x + 1) + c
(c) ln 29
(d) 12 + 14 √ 13 π
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(e) ln
(g)
(i)
1
2
2
3
1
2
1
2
1101
( x − 2) 2
+c
x −1
+ 23 √ 13 π
Ë(x 3 − 1) + c
(f ) 13 tan3x − tan x + x + c
(h) 13 sin−1 23 x + c
( j) x 2 − x + 2 ln x + 1 + c
(k) tan−1 12 (x + 3) + c
(l) 2(2 − x) cosËx + 4Ëx sin Ëx + c
(m) 1
(n) 14 (sinh 2 − 2)
(o) − 301 (1 − 3x)10 + c
(p) 12 sin x − 101 sin 5x + c
(q) x(ln 2x − 1) + c
(r) −e−x /2 + c
(s) 12 cosh −1 23 x + c
(t) 7 /2
(u) π8 − 14 ln 2
(v) − 151 (4 − 3x)5 + c
(w) 101 sin 5x + 12 sin x + c
(x) x sin−1x + Ë(1 − x 2) + c
2
7 Maximum 10.55 when u = 4.42 and 1.28, minimum
1.45 when u = 2.85 and 5.99
8 (a)
wL4
16 EI
L
(1 ± ÷ 13 )
2
(b)
9 L = 100 m, W = (200/ ) m
10 Local minimum (0, 0), local maximum (3, −3)
asymptotes x = 2, x = 6, y = 1
11 0.4446 cf. 0.4425
13 0.782 80, 14 π , error = −0.002 60
(b) ln 67
15 (a) 1076
15
16 − 14 π
sec4t, − 14 πsec t cosec3t
17 −3, 18, 5Ë10/9
18 158
19 43 π ab
20 23 π
22 12 (sinh−12 + 2Ë5)
23 158
24 (a) 3 a2
(b) 4a
(c) cycloid has cusps at these values
25 (a)
13
15
1
4
− π
(b)
5
12
−
1
2
(d) 8a
ln 2
29 0.785, 0.626, 0.624; 2.62
30
/2, 16 π (53/2 − 1)
48
33 (a) 54
35 π , 5 π
(b) ( ÷23 − π6 , 65 π − 25 ÷3)
35 (a) 5.21 × 106
(b) 7.76 × 106
38 (d) 1.910
(i) 0.000, 0.191, 0.375, 0.541, 0.682, 0.798, 0.888,
0.953, 0.995, 1.008
39 (a) u
x
y
0
1.0
0.00
0.23
0.9
0.21
0.33
0.8
0.27
0.42
0.7
0.31
u
x
y
0.74
0.3
0.27
0.84
0.2
0.22
0.97
0.1
0.15
1
2
0.49
0.6
0.32
π
0.0
0.00
(f ) 0, , 12 π , cos−1(Ë(5/7)), cos−1(−Ë(5/7))
, ;5Ë10/49)
(h) ( 25
49
James, Glyn, and Phil Dyke. Modern Engineering Mathematics 6th Edition PDF Ebook, Pearson Education, Limited, 2020. ProQuest Ebook Central,
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Created from monash on 2022-06-03 01:16:25.
0.57
0.5
0.32
0.65
0.4
0.30
1102 A NSW ERS T O EXERCISES
CHAPTER 9
37 2, 3
Exercises
38 cos y, −x sin y
1 (a) 1/4
(e) 2/3
(i) /4
(b) 1/2
(f ) − 4/3
(c) 1/4
(g) 2
(d) 1
(h) 3/2
2 (b) is convergent, 0.860 334
40 (a) exy(y cos x − sin x), exyx cos x
2 xy
y2 − x 2
,− 2
(b) 2
( x + y 2 )2
( x + y 2 )2
3 f ′(x) . 1 near x = a
4 1.618 034
5 xn+1 = xn − 101 (x 3n − 2x − 1)
6 (a) f ′(x) , 1 (x , 1)
39 (a) 3x 2y + 4x + y, x 3 + 18y + x
(b) 3(x + y 2)2, 6y(x + y 2)2
3x + y
y+x
,
(c)
2
2
1/ 2
2
(3x + y + 2 xy)
(3x + y 2 + 2 xy)1/2
(c)
(b) f ′(x) . 1 (x . 1)
7 1.5, 1.49, 1.48, 1.48, 1.47, 1.47, 1.46, 1.46, 1.45, 1.45;
Ë2 = 1.41
− x 2 − 2 xy + 2 y 2 + 6 x 2 − 4 xy − 2 y 2 + 6
,
( x 2 + 2 y 2 + 6) 2
( x 2 + 2 y 2 + 6) 2
41 (a) −x/z, −y/z
9 1 + 2 x + 12 x 2 + 16 x 3 + … = x + e x
46 −1 + 12 Ë3, −tan−12
10 y3 = 1 + 2 x + 12 x 2 + 61 x 3 − 241 x 4
1
y4 = 1 + 2 x + 12 x 2 + 61 x 3 + 241 x 4 − 120
x5
47 2 tan−1(r tan u) +
11 x − 13 x 3 + 15 x 5 − 17 x 7 + … (−1 x 1)
17 7
x +…
12 x + 13 x 3 + 152 x 5 + 315
17
1 2
ln cos x = −{ 6 x + 121 x 4 + 451 x 6 + 2520
x8 + . . . }
18 l . 18
(c) − 23
(d) −1
(e) − 13
(f ) −1
20 b0 = 82.82, b1 = − 245 , b2 = − 0.0018
21 X = ln 4 ø 1.386
2t 3 + 3t − 1
÷(t + 3t 2 − 2t + 1)
4
(b) 4xt(x 2 − 2t 2)/(2x + 3t)
53 (a) 10.5
(b) 19Ë 341
54 195 cm s−1
22 0.0006
55 Ë(1 + 4t 2 + 9t 4)
23 1.175 201 21
56 −6e−2s + 2e−s−t, −6e−2t + 2e−s−t
24 (a) 1st order
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2r tan θ + πr 2 sec2θ
1 + r 2 tan 2θ
48 2sexcos y − 2texsin y
−2texcos y − 2sexsin y
49 (a)
(b) 14
1 − yz 1 + xz
,
xy − 1 1 − xy
43 (a) 2xy + 3yz − 4z 3xy, x 2 + 3xz − 2x 2z3,
3yx − 6z 2x 2y
(b) −ye2z sin xy, −xe2z sin xy, 2e2z cos xy
31 6
x + …)
8 e(1 − 12 x 2 + 16 x 4 − 720
19 (a) 43
(b)
(b) 2nd order
(c) 3rd order
25 −0.1038
26 2.732 051, 4.872 977
58 fxx = fxy = fyy = fyx = 0, fxz = −3 sin 3z, fyz = −6 sin 3z
fzz = −9(x + 2y) cos 3z, fzx= −3 sin 3z, fzy = −6 sin 3z
60 −3
27 0.576 368 88
66 a = 3, b = 23 π
28 n = 2, 0.0203
29 (a) 0.643 283
57 fxx = y(2 + xy)exy, fyy = x3exy, fxy = fyx = x(2 + xy)exy
(b) 0.6875
1
30 0.4627, h = 256
32 (1, 2t, 3t 2), (0, 2, 6t)
dU
= (1 + 2t 2)T̂ (t), where
33
dt
1
2t
2t 2
r(t ) =
k
i+
j+
2
2
1 + 2t
1 + 2t
1 + 2t 2
68 0.018 702, 0.02
69 0.029 65 m3, 0.0295 m3
70 173 ; 4 m
71 − 23
72 3%
74 0.5%
75 35% increase
James, Glyn, and Phil Dyke. Modern Engineering Mathematics 6th Edition PDF Ebook, Pearson Education, Limited, 2020. ProQuest Ebook Central,
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AN S WE RS TO E XE RCI S E S
76 (a) xy 2 + x 2y + x + c
(b) x 2y 2 + y sin 3x + c
(c) Not exact
(d) z 3x − 3xy + 4y 3
11 8.155 299, 8.154 959, 8.154 845
77 −1, y sin x − x cos y + 12 (y 2 − 1)
16 −0.21, 0.01
78 m = 2
8x 5 + 36x 4y + 62x 3y 2 + 63x 2y 3 + 54xy 4 + 27y 5 + c
17 0.61%
1103
12 4 m s−1, 4 m s−2
13 ( 12 t 2 + 16 t 3 + t ) i + ( 121 t 4 − t ) j + t 2 k
79 (a) (0, 0), maximum; (10, 0), saddle
(b) (0, 0), maximum
(c) (−1, 3), saddle
(d) (−1, 23 ), saddle; (1, 23 ), minimum
(e) (0, −1), saddle; (0, 3), saddle; (−1, 1), maximum
(f ) Minimum at ( 12 , 13 ); degenerate and stationary sets
x = 0 and y = 0
(g) (1, 1), minimum
18 −0.2%
19 −3.33%
20 (b) 2u
22 (a) 2
25 f ″(z)/(4tËt)
81 Maximum at (0, 0); saddle at ( 13 , 13 )
27 −y/(x 2 + y 2) + const
82 N = 2000, n = 2000, P = 250
28 Maximum at (0, 0), saddle points at (3, 3), (−3, −3),
(1, −1), (−1, 1)
2
3
4
3
83 Minimum at ( , )
29 Minimum at (0, 0), saddle at ( 12 , 23 )
20 (π 2 − 16)
12 (20 − π 2)
84 a =
,b=
5
π
π4
30 x = ( 23 )2/3 , y = ( 23 )2/3 , z = 22/3 · 3−1/6
85 x = 2, y = 2
31 Saddle at (0, 0) and (0, 4), maximum at (2, 2),
minimum at (−2, 2)
86 Minimum T = − 14 at ( 12 , 0)
32 x = 0, y = ;3 (max); y = 0, x = ;3 (min)
maximum T = 94 at (− 12 , ± 12 ÷3)
200
3
87 x =
2
3
1
3
33 7.4163/a
,θ = π
4
3
35 y = −x cot cx
88 ( , )
89 ( 13 , 12 , 61 )
90 1, 2
1
3
CHAPTER 10
2
3
2
3
91 (− , − , − )
Exercises
Copyright © 2020. Pearson Education, Limited. All rights reserved.
92 (− 414 , 47 )
1 (a) First-order, dependent variable x, independent
variable t, linear, homogeneous, ordinary
differential equation
(b) Second-order, dependent variable x, independent
variable t, linear, homogeneous, ordinary
differential equation
(c) First-order, dependent variable x, independent
variable t, nonlinear, ordinary differential equation
(d) First-order, dependent variable x, independent
variable t, linear, nonhomogeneous, ordinary
differential equation
(e) Second-order, dependent variable x, independent
variable t, linear, nonhomogeneous, ordinary
differential equation
93 285
92
9.10 Review exercises
1 0.2575
4 12
5 18 (sin 2k − 2k cos 2k)
6 2.09
7 (a) For these series see Section 6.3.5.
8
/2
9 (a) 12
(b)
10 (a) 6, x = 0
/4 (c) 161
(d) 14
(b) 3, x = 23 π
(e) 25
(f ) 23
(c) −1, x = 0
2 (a) Second-order nonlinear ordinary differential
equation, dependent variable p, independent
variable z
James, Glyn, and Phil Dyke. Modern Engineering Mathematics 6th Edition PDF Ebook, Pearson Education, Limited, 2020. ProQuest Ebook Central,
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1104 A NSW ERS T O EXERCISES
(b) Second-order linear nonhomogeneous ordinary
differential equation, dependent variable s,
independent variable t
(c) Third-order nonlinear ordinary differential
equation, dependent variable p, independent
variable y
(d) First-order linear nonhomogeneous ordinary
differential equation, dependent variable r,
independent variable z
(e) First-order linear homogeneous ordinary
differential equation, dependent variable x,
independent variable t
(f) First-order linear nonhomogeneous ordinary
differential equation, dependent variable x,
independent variable t
(g) Third-order nonlinear ordinary differential equation,
dependent variable p, independent variable q
(h) Second-order nonlinear ordinary differential
equation, dependent variable x, independent
variable y
(i) First-order linear homogeneous ordinary
differential equation, dependent variable y,
independent variable z
3 (a) x(t ) = 43 t 3 + C
(b) x(t ) = 201 t 5 − 13 t 3 + Ct + D
1 4t
(c) x(t) = 16 e + Ct + D
(d) x(t) = Ae−6t
1
(e) x(t) = ln t + 125
cos 5t + Ct 2 + Dt + E
2Ë2t
(f ) x(t) = Ae + Be−2Ë2t
4 (a) x(t) = 23 t 3 + Ct + 2 (b) x(t ) = − 14 sin 2t −
t
+ 25
π
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(c) x(t) = 4t + D
(d) x(t) = 2 − t 2
(e) x(t) = 12 e−2t + Ct + a − 12
(f ) x(t) = C − cos 2t
e
(et − e−t )
(h) x(t ) = 2
(g) x(t) = e2t
e −1
5 (a) Under-determined
(b) Fully determined, boundary-value problem
(c) Fully determined, initial-value problem
(d) Under-determined
(e) Fully determined, boundary-value problem
(f ) Fully determined, initial-value problem
(g) Under-determined
(h) Under-determined
(i) Fully determined, boundary-value problem
( j) Fully determined, initial-value problem
(k) Fully determined, boundary-value problem
(l) Fully determined, initial-value problem
1
6 y( x ) =
[w(a − x)4 − 4R(a − x)3
24 EI
+ 4a2(aw − 3R)x − a3(aw − 4R)]
At A the boundary condition is y(a) = 0 so R = 3aw/8
Maximum displacement is y = 0.005 42 wa4/EI
11 (a) x(t) = Cekt
(c) x(t) = Ct b
(b) x(t) = Ce2t
(d) x(t) = (2a ln t + C)1/2
3
1/ 2
12 (a) x(t) = (67 − 3 cos t)
 163 2 
− 
(b) x(t ) = 
2
t
13 (a) x(t) = (t1/2 + C)2
(c) x(t) = Cexp( 12 et )
(e) x(t) = (1 − Ceat)−1
(b) x(t) = cos−1(Cecos t−t)
(d) x(t) = (3et + C )1/3
(f ) x(t) = (C − 2 cos t)1/2
1/3
2
14 (a) x(t ) = −2 ± ( 23 t 3 + 2t )1/2
4 (t − 1)
3 + e 2 sin t
(b) x(t ) =
(c) x(t ) =
4−t
3 − e 2 sin t
−a
t
(d) x(t) = −ln(1 + e − e )
(e) x(t) = [12(t ln t − t + 1)]1/3
15 K = 2/75, x(10) = 20/7, x(50) = 100/23, x S 5 as t S q
16 t = Ë(m/Kg)tanh−1 12
17 A(t ) =
1
[1 − (1 + 6α Kt )− 1/6 ]
α
18 (a) x(t) = ;tË[2 ln(Ct)]
−t
(c) x(t ) =
ln Ct
(b) x(t) = t(ln Ct 3)1/3
19 x(t) = ; t(4 ln t + 256)1/4
1/ 2
t C

 − 1
(b) x(t) = t cot−1(ln(1/Ct))

÷3  t 3
1/ 2
t 
C
 
(c) x(t ) = 1 ±  − 11 
(d) x(t) = t sin−1Ct
t
 
2 
20 (a) x(t ) =
(e) x(t) = t ; (2t 2 + D)1/2
2t
÷(2 − t 4 )
(c) x(t) = t ln(ln 12 t + e2)
4t 2
(e) x(t ) =
5 − 4t
21 (a) x(t ) =
(f ) x(t) = −t ln(−ln Ct)
(b) x(t) = ; 14 (9t 2 − 32)1/2
(d) x(t) = ; t[ln(ln t 2 + e4)]1/2
22 (a) x(t) = t + 3 ; (2t + C)1/2
(b) x(t) = 12 [;(2t + C )1/2 − t − 1]
(c) x(t) = 12 [;(2t + C )1/2 − t]
(d) x(t) = t − 1 ; (2t + C)1/2
(e) x(t) = Aet − 2t − 4
(f ) x(t) = 12 t − 2 + Aet
(g) x(t ) = −
1
− 2t
t+C
23 (a) x(t) = ;Ë(C − t 2)
(b) x(t) = ;Ë(C + t 2)
2
(c) x(t) = −t ; Ë(C + 2t ) (d) x(t) = t 2 ; Ë(C + t 4)
(e) 12 x 2 − xt + 12 t 2 − t = C
(f ) x 2 + xt + t 2 = C
24 (a) x(t) = 1 ; Ë(1 − 2t − t 2)
(b) x(t) = 12 [−t ; Ë(3t 2 + 4)]
James, Glyn, and Phil Dyke. Modern Engineering Mathematics 6th Edition PDF Ebook, Pearson Education, Limited, 2020. ProQuest Ebook Central,
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AN S WE RS TO E XE RCI S E S
(c) x(t ) =
2
(1 ± (1 − t 2 )1/ 2 )
t2
25 (a) Not exact
(c) x sin(x + t) = C
(e) x + ext = C
(g) Not exact
(d) x(t ) =
2−t
cos t
(b) Not exact
(d) Not exact
(f) (x + Ët)2 = C
(h) t ln(x + t) = C
26 (a) x(t) = sin−1(1 − t) − t
(b) x(t) = [ 12 (15 − t)]2/3 − 2t
(c) x(t) = ;[t 2 ; (1 − 4t)1/2]1/2
1 1
(d) x(t ) = 4 exp  −  − t
 2 t
13
Ë(6t 4 + C )
2t
4
1 + 2t + Ce 2t
(c) x(t) = ;Ë(Ce2t − 2et )
1
(d) x(t) =
t(1 + Ct )
(b) x(t) =
1
t ÷(3 − 2t )
6
(b) x(t) =
÷(12 − 11e 6t )
38 (a) x(t) =
27 Must have b = e; then ax 2 + 2bxt + ft 2 + C = 0
28 Must have h(t) = dg/dt; then x = −C/g(t)


10
(c) x(t) = 3 
3t 
 3 cos (t ) + 9 sin(t ) − 13e 
29 Must have k = −1; then x ln(x + t) + C = 0
(d) x(t) = − 12 Ë(5t 9 + 3t)
30 Must have k = 2; then x = [sin−1(C/t 3)]/t
39 X(0.3) = 1.269 000
31 (a) x(t) = 23 + Ce−3t
(b) x(t ) = − 14 t − 161 + Ce 4t
(c) x(t) = − 12 e−4t + Ce−2t
(d) x(t) = Ce−t /2 − 2
40 X(0.25) = 2.050 439
41 X(1) = 1.2029
42 X(0.5) = 2.1250
2
43 Xa(2) = 2.811 489, Xb(2) = 2.819 944
32 (a) x(t ) = − 23 + 27 e 2t
(b) x(t ) = 13 t − 19 + 109 e −3t
(c) x(t ) = 12 t 3 − 3t ln t − 23 t
44 Xa(2) = 1.573 065, Xb(2) = 1.558 541
45 Xa(1.5) = 2.241 257, Xb(1.5) = 2.206 232
33 (a) x(t) = Cet − 2t 2 − 5t − 5
(b) x(t ) = − 14 t 2 − 18 + Ce 2t
d
+ t2
dt
d
−k
(c) L =
dt
2
C
2
2
(c) x(t ) = 1 −  sin t + cos t +

t2 
t
t2
C
6
6
1 3
(d) x(t ) =  − 2 + 3 − 4  e t + 4
t t

t
t
t
Copyright © 2020. Pearson Education, Limited. All rights reserved.
37 (a) x(t) =
(e) x(t) = 12 sin 2t ln(tan 12 t) + C sin 2t
(f ) x(t) = 13 t 2 + Ce−2t
(g) x(t) = Ce−1/t − 4
3
1
2
34 (a) x(t) = (1 − e )
1
5
(c) x(t ) = t −
2t 2
1
25
(d) x(t) = 1 + e
1 −2 t
3
+ e
t−1 −t
(b) x(t) = 2e t
18 − 25e 2 −5t
+
e
75e 5
1/t−1/2
2
2
1
(e) x(t ) = 12 + t 2 + 1 − +  e t − ( 23 + e)
2

t
t
t2
1+cos t
(f ) x(t) = U(1 + e
)
35 T(t) = Tin + Ce−AUat/V
αρgh e −pt
2αρgh (1 − e −pt )
+
(1 + αβ + αγ 0)
(2 + 2αβ + αγ 0)
(2 + 2αβ + αγ 0) A
where p =
2αρd
36 Q(t ) =
1105
d
− 6t 2
dt
46 (a) L =
(b) L =
47 (a) independent
(b) dependent
48 (a) k1 = 2, k2 = −2, k3 = −1
(b) k1 = 1, k2 = −1, k3 = −1, k4 = 1
d
d3
d2
− f (t )
(b) L = 3 + sin t 2 + 4t 2
dt
dt
dt
d2
d
(c) L = 2 + sin t − t − cos t
dt
dt
d
d
b
cos t
−
(e) L =
(d) L = sin t −
dt
t
dt
t
2
d
d
2 d
t
− te
(g) L = t 2 + (2t − t 2 ) − t
(f ) L =
dt
dt
dt
d2
d
(h) L = t 2 + 3 − t
dt
dt
49 (a) L =
2
50 (a) dependent
(c) independent
(e) dependent
(g) independent
(b) independent
(d) independent
(f) dependent
(h) dependent
James, Glyn, and Phil Dyke. Modern Engineering Mathematics 6th Edition PDF Ebook, Pearson Education, Limited, 2020. ProQuest Ebook Central,
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1106 A NSW ERS T O EXERCISES
(i) dependent
(k) independent
( j) independent
51 (a) 2, −1, 1, 1
(c) 0, 2, −1
(e) 0, −1, 0, 1, 6
(b) −3, 3, 2, 1
(d) 1, −1, 0, 1
52 (a) x(t) = A + Bt + Ct 2 + Dt 3
(b) x(t) = Ae pt + Be−pt
(c) x(t) = A cos pt + B sin pt + C cosh pt + D sinh pt
(d) x(t) = A + Be−2t
(e) x(t) = A + B cos 2t + C sin 2t
(f ) x(t) = Ae−t + Bte−t
(g) x(t) = Aet + Btet + Ce−t
53 LM =
2
1 d3
2
 d
−  2 + 4 + e t 2
3
t
 dt
t dt
2
4

 d
+  3 + + 6t + (4t − 2) e t
t
 dt
t
+ (4t − 6t 2 − 1)et
d2
1 d3
4

 d
t
ML =
e
−
(
+
4
)
+  6t − + 4tet
 dt
t dt 3
t
dt 2 
− 6t 2et + 12
d2
 df
 d
+  f1 2 + f1g2 + f2 g1
 dt
d t 2  dt
dg 2
+ f1
+ g1 g2
dt
2
d
 df
 d
ML = f1 f2 2 +  f2 1 + f1 g2 + f2 g1 
 dt
 dt
dt
dg1
+ f2
+ g1 g2
dt
54 LM = f1 f2
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55 (a) x(t) = Aet + Be3t/2
(b) x(t) = e−t(A cos 2t + B sin 2t)
(c) x(t) = Aet + Be−4t
(d) x(t) = e2t(A cos 3t + B sin 3t)
56 (a) x(t) = 17 (3et − 10e−2t/5)
(b) x(t) = e3t(2 cos t − 6 sin t)
(c) x(t) = 12 (e3t − et )
57 (a) x(t) = et/4[A cos( 14 Ë27t) + B sin( 14 Ë27t)]
(b) x(t) = Ae(Ë13−3)t + Be−(Ë13+3)t
(c) x(t) = e−t/2[A cos( 12 Ë3t) + B sin( 12 Ë3t)]
(d) x(t) = Ae4t + Bte4t
(e) Aet + Be2t/3 + Ce−2t/3
−t
t
(f) x(t) = Ae + e [B cos(2Ë2t) + C sin(2Ë2t)]
(g) x(t) = A + et[B cos(Ë2t) + C sin(Ë2t)]
58 x(t) = et(A cos t + B sin t + C cos 2t + D sin 2t)
59 (a) x(t) = e t /2 [cos( 12 ÷5t ) − ÷ 15 sin( 12 ÷5t )]
(b) x(t) = 2(t − 1)e2(t−1)
(c) x(t) = e −5t /2 [cos( 12 ÷7t ) + ÷ 17 sin( 12 ÷7t )]
(d) x(t) = 16 (7t + 33)e−(t+3)/3
(e) x(t) = 27 e t − 4e 2t + 23 e 3t
(f ) x(t) = (2 − 5t + 4t 2)e−2(t−1)
60 x(t) = et/2[A cos( 12 Ë3t) + B sin( 12 Ë3t) + Ct cos( 12 Ë3t)
+ Dt sin( 12 Ë3t)]
61 x(t) = Ae4t + Be−t + Cte−t + Dt 2e−t
62 (a) x(t ) = 29 − 13 t + Ae−t + Be3t
38
(b) x(t) = − 15 t 2 + 14
+ Ae(1+Ë6)t + Be(1−Ë6)t
25 t − 125
t
(Ë5+1)t/2
(c) x(t) = −5e + Ae
+ Be−(Ë5−1)t/2
63 (a) x(t) = − 18 cos 4t + 241 sin 4t
+ e 3t /2 [ A cos( 12 ÷7t ) + B sin( 12 ÷7t )]
1
e−3t + Ae2t/3 + Bte2t/3
(b) x(t) = 121
3
÷2
− (1− ) t
56
(c) x(t ) = − 105
289 cos 2t + 289 sin 2t + Ae
3
÷2
+ Be −(1+ )t
33
(d) x(t) = 45 t − 16
+ e − t /2 [ A cos( 12 ÷15t ) + B sin( 12 ÷15t )]
(e) x(t) = t − 2 + Ae−t/4 + Bte−t/4
72
21
cos 3t − 625
sin 3t + Ae4t + Bte4t
(f ) x(t) = − 625
1 −5t
2t
(g) x(t) = 52 e + e [A cos(Ë3t) + B sin(Ë3t)]
(h) x(t) = −t 2 − 6t − 24 + 15 e−2t + Ae(Ë7/Ë3−1)t/2
+ Be−(Ë7/Ë3+1)t/2
(i) x(t) = − 45 te −3t − 654 cos 2t − 657 sin 2t + Ae t + Be −3t
( j) x(t) = 161 − 14 t cos 4t + A cos 4t + B sin 4t
(k) x(t) = − 47 t − 43 te 4t + A + Be4t
17
21
cos 2t + 1460
sin 2t + Ae t + B e −2t
64 (a) x(t) = 1460
+ Ce(2+Ë3)t + De(2−Ë3)t
(b) x(t) = − 121 e 2t − 391 te −2t + Aet + Be−2t + Ce(2+Ë3)t
+ De(2−Ë3)t
(c) x(t) = − 12 t 2 − 29 t − 694 − 121 e − t + Aet + Be−2t
+ Ce(2+Ë3)t + De(2−Ë3)t
1
11
65 (a) x(t) = 125
cos t + 250
sin t − 271 (t + 1)
+ (A + Bt + Ct 2)e3t
(b) x(t) = − 18 et + (A + Bt + Ct 2)e3t
(c) x(t) = 16 t 3 e 3t − 271 (t + 1) + (A + Bt + Ct 2)e3t
(b) v = Ë7, z = 2 Ë 17
66 (a) v = 3, z = 1
67 (a) a = 1, B = 4
(b) p = 1.4, q = 0.25
(c) b = 2.2, g = 1.21
68 (a) v = 4p, ζ =
a
4p
(c) v = 1.78, z = 0.12
(e) v = 0.51, z = 2.48
69 (a) a = , b =
(c) q = 8, r = 4
2
1
, ζ = 7÷( 12 α )
÷(2α )
(d) v = 5h, z = 4
(b) ω =
(b) a = 0.4 , b = 4 2
7
28
,b=
(d) a =
π
2π 2
James, Glyn, and Phil Dyke. Modern Engineering Mathematics 6th Edition PDF Ebook, Pearson Education, Limited, 2020. ProQuest Ebook Central,
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AN S WE RS TO E XE RCI S E S
70 Ωmax = vË(1 − 2z2), only exists if z2 < 12
1
A(Ω max ) =
2ζω 2 ÷(1 − ζ 2)
71 2.52 m s−1 (approximately 5 knots)
72 m > 621 N m−1 s
73 73 pF > C > 7 pF
dx
x(0) = 1
= v,
dt
dv
= 4xt − 6(x 2 − t)v, v(0) = 2
dt
dx
x(0) = 0
(b)
= v,
dt
dv
= sin v − 4x, v(0) = 0
dt
74 (a)
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75 X(0.3) = 0.299 90
dx
76 (a)
x(1) = 2
= v,
dt
dv
= −4Ë(x 2 − t 2), v(1) = 0.5
dt
dx
x(0) = 1
(b)
= v,
dt
dv
= w,
v(0) = 2
dt
dw
= e2t + x 2t − 6etv − tw, w(0) = 0
dt
dx
x(1) = 1
(c)
= v,
dt
dv
= w,
v(1) = 0
dt
dw
= sin t − x 2 − tw, w(1) = −2
dt
dx
x(2) = 0
(d)
= v,
dt
dv
= w,
v(2) = 0
dt
dw
= (x 2t 2 + tw)2, w(2) = 2
dt
dx
x(0) = 0
(e)
= v,
dt
dv
= w,
v(0) = 0
dt
dw
w(0) = 4
= u,
dt
du
= ln t − x 2 − xw, u(0) = −3
dt
dx
x(0) = a
(f )
= v,
dt
1107
dv
= w,
v(0) = 0
dt
dw
w(0) = b
= u,
dt
du
= t 2 + 4t − 5 + Ë(xt) − v − (v − 1)u, u(0) = 0
dt
77 X(0.65) = −0.834 63
78 X0.01(0.4) = 0.398 022
X0.005(0.4) = 0.397 919
step size required is <0.0024
X0.002(0.4) = 0.397 856
79 s tends to around 6.3%. With double the inflow s tends
to about 11.1%.
10.13 Review exercises
1 (a) Second-order nonlinear ordinary differential
equation, dependent variable x, independent
variable t
(b) First-order nonlinear ordinary differential equation,
dependent variable z, independent variable x
(c) Third-order linear nonhomogeneous ordinary
differential equation, dependent variable p,
independent variable s
2 (a) Under-determined, x = 16 t 3 + At + 1
(b) Fully determined, x(t ) = 241 t 4 − 247 t 2 + 14 t
(c) Over-determined, no solution exists
(d) Fully determined, x = 161 e 4t − 14 te 4 − 161
3 x(t) =
C
÷(C + e −2 at )
4 (a) x(t) = cos−1(sin t − 1)
(b) x(t) = ln(ln t + e2)
(t −8)/3
(c) x(t) = e
(d) x(t) = t cos−1(cos 1 − ln t)
1
(e) x(t) = − 2 [t ; Ë(17t 2 + 16)]
(f ) x(t) = t2t
(g) x(t) = t(3 − ln t)
(h) x(t) = t ; Ë(4 − 6t 2)
3
÷[4 + a(t − 1)]
t
sin −1 (π − t )
(c) x(t) =
t
ln(2 + e 8 − t )
(e) x(t) =
t
5 (a) x(t) =
6 (a) x(t) = 94 e 2t − 12 t − 14
(c) x(t) = 15 (e2t + 9e−3t )
(b) not exact
(d) not exact
(b) x(t) = 12 (e−t + e−t )
(d) x(t) = 1 + (e − 1)ecost+1
2
7 X0.1(0.4) = 1.125 583, X0.05(0.4) = 1.142 763
Richardson extrapolation estimates the error as
0.017 180, so, to obtain an error less than 5 × 10 −3,
a step less than 0.0146 should be used.
James, Glyn, and Phil Dyke. Modern Engineering Mathematics 6th Edition PDF Ebook, Pearson Education, Limited, 2020. ProQuest Ebook Central,
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1108 A NSW ERS T O EXERCISES
8 X0.05(0.25) = 2.003 749, X0.025(0.25) = 2.004 452
Richardson extrapolation estimates the error as
0.000 703, so, to obtain an error less than 5 × 10 −4,
a step less than 0.0179 should be used.
9 x(t ) = (20 − t ) −
(20 − t )
400
(f ) x(t ) = 131 + 85 e − t −
22 (a) ω = ÷2, ζ =
3
e −2t
(44 cos 3t − 27 sin 3t )
65
7
2 ÷2
(b) v = p1/4, z = 12 p3/4
11 a = 4kT 30
(c) ω =
13 y(t) = y0 + CË(x − x0 )
÷q
, ζ = a ÷(2q)
÷2
(d) ω = ÷(2α ), ζ =
14 Half life is ln 2/k
7
÷(2α )
15 Time to 95% of final value is ln(20)L/R
23 (a) a = 2, B = 4
(c) a = 2, c = 8
16 Tyre life is approximately 29 500 miles
d2
d
+ sin t − 9, f (t ) = − cos t
2
dt
dt
d3
d2
d
(b) L = 3 + t 2 + t(t − 4) + 1, f (t ) = − e t
dt
dt
dt
d
−t
t
(c) L =
− e , f (t ) = e
dt
d2
(d) L = 2 + 4, f (t ) = cos Ωt
dt
d3
1
d
, f(t) = −ln(t 2 + 4)
(e) L = t 2 3 − 2
t + 2t + 4 dt
dt
17 (a) L =
18 (a) sin t − cos t
Copyright © 2020. Pearson Education, Limited. All rights reserved.
19 (a)
df
dg
=
dt
dt
(b)
df
dg
=
dt
dt
2t ( k 3 − 2 )
t3
t2
− 2 + 3 +
(1 − e − kt )
3k k
k
k4
U
 U k − U kt 
ln  +
e 
(b) x(t) =

k −U k
k
8
26
2 3
t
+
t
−
(c) x(t) = 15
5
15
(d) x(t) = − 174 cos t − 171 sin t − 174 e − 4(t −π )
25 (a) x(t) =
d
dt
d2 f
d2g
= 2
2
dt
dt
 (1 − p)t + 4  1− p
27 (a) x(t) = 

4p

(b) x(t) = 1
(c) x(t) = ( 14 − 23 t ) −1 / 2
(d) x(t) = 12 t 2 + 1
(d) 2
1
and
20 (a) x(t) = Aet + Be2t + 101 sin t + 103 cos t
(b) x(t) = Aet + Be2t + Ce−3t + 16 t + 367
(c) x(t) = Aet + Be2t + Ce−3t + 15 te2t
(d) x(t) = Ae4t + te4t
96
736
t + 2197
(e) x(t) = e−3t/2 (A sin t + B cos t) + 134 t 2 − 169
4
(f ) x(t) = e−3t/2 (A sin t + B cos t) − 16
75 cos t + 25 sin t
(g) x(t) = Ae2t + Be4t + Ce−t + 18 t 2 − 163 t + 13
64
(h) x(t) = et(A cos 2t + B sin 2t) + 18 e−t
−t
(i) x(t) = Ae + Be + Ce − te
2t
t
24 x(t) = t − Ce−t + D
(a) x(t) = 121 e−2t + Ce4t + D
(b) x(t) = −ln(cos(t + C )) + D
(c) x(t) = Ct 3 + D
26 x(t) = C tan( 12 Ct + D)
(a) x(t) = Ae pt + B
(b) x(t) = −ln(t + C ) + D
(c) x(t) = ;Ë(C − ln(D − t))
(b) 0 (commutative)
(c) 0 (commutative)
(b) a = 4 , B = 2
(d) p = 150, q = 6Ë2
4t
1
6
2t
1
5
1 t
6
2
25
+ e
( j) x(t) = e (A cos 2t + B sin 2t) + t +
+ 14 te t sin 2t
21 (a) x(t) = 15 (1 − e−t cos 2t − 12 e−t sin 2t)
(b) x(t ) = −2t + 5 + 27 e t − 23 e− t /3
(c) x(t) = (12e−t + 30te−t − 12 cos 2t + 16 sin 2t)/25
(d) x(t) = 3et − 2 − e2t
(e) x(t ) = − 75 e t + 43 e 2t + 151 e − 4t
 µα − β 2

m
ln 
V 2 + 1
2( µα − β )  T − µmg

m
Time to take off is
÷((µα − β )(T − µmg))
28 Length of runway is

× arctan 



 µα − β  
V

 2
 T − µmg  

29 X0.025(2) = 0.847 035, X0.0125(2) = 0.844 066
Richardson extrapolation estimates the error as
0.002 969, so we have X(2) = 0.84
C
32 R = 2
L
James, Glyn, and Phil Dyke. Modern Engineering Mathematics 6th Edition PDF Ebook, Pearson Education, Limited, 2020. ProQuest Ebook Central,
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AN S WE RS TO E XE RCI S E S
CHAPTER 11
(m) e−t(cos 2t − sin 2t)
Exercises
(o) − e t + 23 e 2t − 12 e −2t
2
(b) 3 , Re(s) 0
s
1
(d)
, Re(s) −1
(s + 1) 2
(c) 0
(d) 3
(e) 2
(h) 0
(i) 2
( j) 3
s
1 (a) 2
, Re(s) 2
s −4
3s + 1
(c)
, Re(s) 0
s2
(b) −3
(g) 0
2 (a) 5
(f ) 0
3 (a)
(b)
(c)
(d)
(e)
(f )
(g)
(h)
(i)
( j)
(k)
Copyright © 2020. Pearson Education, Limited. All rights reserved.
(l)
5s − 3
, Re(s) . 0
s2
42
6
− 2
, Re(s) . 0
4
s
s +9
3s − 2
4s
, Re(s) . 0
+ 2
2
s
s +4
s
, Re(s) . 3
s2 − 9
2
, Re(s) . 2
s2 − 4
5
3
2s
, Re(s) . 0
+ −
s + 2 s s2 + 4
4
, Re(s) . −2
( s + 2) 2
4
, Re(s) . −3
2
s + 6s + 13
2
, Re(s) . −4
( s + 4) 3
36 − 6s + 4s 2 − 2s 3
, Re(s) . 0
s4
2s + 15
, Re(s) . 0
s2 + 9
s2 − 4
, Re(s) . 0
( s 2 + 4) 2
18s 2 − 54
, Re(s) . 0
(s 2 + 9) 3
2
3s
, Re(s) . 0
(n) 3 − 2
s
s + 16
s +1
2
3
+ 2
+ , Re(s) . 0
(o)
( s + 2) 3
s + 2s + 5 s
(m)
4 (a)
(c)
(e)
(g)
(i)
−3t
−7t
1
4
4
1
4 −3t
9
3
9
1
64
−2t
1
8
−t
−e )
(e
− t− e
(d) 2 cos 2t + 3 sin 2t
(4t − sin 4t)
(f ) e−2t(cos t + 6 sin t)
(1 − e cos2t + 3e−2tsin 2t)
e (cos 2t + 3 sin 2t)
−3t
(k) −2e
(l)
(b) −e−t + 2e3t
1
5
t
(h) et − e−t − 2te−t
(j) 12 e t − 3e 2t + 112 e 3t
+ 2 cos(Ë2t) − √ sin(Ë2t)
1
2
(q) 9e
(r)
1
9
−2t
−e
(n) 12 e 2t − 2e 3t + 23 e− 4t
(p) 4 − 29 cos t + 12 cos 3t
[7 cos( Ë3t) − Ë3 sin( 12 Ë3t)]
−3t/2
1
2
e − t − 101 e −2t − 901 e − t (cos 3t + 3 sin 3t)
5 (a) x(t) = e−2t + e−3t
4 t /3
(b) x(t ) = 35
− 263 (cos 2t + 23 sin 2t )
78 e
(c) x(t) = 15 (1 − e−t cos 2t − 12 e−t sin 2t)
(d) y(t) = 251 (12e−t + 30te−t − 12 cos 2t + 16 sin 2t)
(e) x(t ) = − 75 e t + 43 e 2t + 151 e − 4t
(f) x(t) = e−2t(cos t + sin t + 3)
13 t
e − 13 e −2t + 14 e − t (cos 2t − 3 sin 2t)
(g) x(t ) = 12
(h) y (t ) = − 23 + t + 23 e − t [cos(Ë2t) + Ë 12 sin(Ë2t)]
(i) x(t ) = ( 18 + 43 t ) e −2t + 12 t 2 e −2t + 83 − 12 t + 14 t 2
(j) x(t ) = 15 − 15 e −2t /3 (cos 13 t + 2 sin 13 t )
(k) x(t) = te−4t − 12 cos 4t
(l) y(t) = e−t + 2te−2t/3
(m) x(t ) = 45 + 12 t − et + 125 e 2t − 23 e − t
25
(n) x(t ) = 209 e − t − 167 cos t + 16
sin t − 801 cos 3t
3
− 80 sin 3t
6 (a) x(t ) = 14 ( 154 e 3t − 114 e t − e −2t ), y (t ) = 18 (3e 3t − e t )
(b) x(t) = 5 sin t + 5 cos t − et − e2t − 3
y(t) = 2et − 5 sin t + e2t − 3
(c) x(t) = 3 sin t − 2 cos t + e−2t
y(t ) = − 27 sin t + 29 cos t − 12 e −3t
(d) x(t ) = 23 e t /3 − 12 e t , y(t ) = −1 + 12 e t + 23 e t /3
(e) x(t) = 2et + sin t − 2 cos t
y(t) = cos t − 2 sin t − 2et
(f) x(t) = −3 + et + 3e−t/3
y(t) = t − 1 − 12 et + 23 e−t/3
(g) x(t) = 2t − et + e−2t, y(t) = t − 27 + 3et + 12 e−2t
(h) x(t) = 3 cos t + cos(Ë3t)
y(t) = 3 cos t − cos(Ë3t)
(i) x(t ) = cos( ÷ 103 t ) + 43 cos( ÷6t )
y(t ) = 45 cos( ÷ 103 t ) − 14 cos( ÷6t )
(j) x(t ) = 13 e t + 23 cos 2t + 13 sin 2t
y(t ) = 23 e t − 23 cos 2t − 13 sin 2t
E (50 + s)s
7 I1 (s) = 2 1 4
(s + 10 )(s + 100)2
Es 2
I2 ( s ) = 2
4
(s + 10 )(s + 100)2
1
1
i2 (t ) = E(− 200
e −100t + 21 t e−100t + 200
cos 100t )
9 i1 (t ) = 20√ 17 e − t / 2 sin( 12 √7t )
−t
e − e (cos t − 3 sin t)
1
5
1109
James, Glyn, and Phil Dyke. Modern Engineering Mathematics 6th Edition PDF Ebook, Pearson Education, Limited, 2020. ProQuest Ebook Central,
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1110 A NSW ERS T O EXERCISES
10 x1 (t ) = − 10 cos( ÷3t ) − 10 cos( ÷13t )
21
x2 (t ) = − 101 cos( ÷3t ) + 10
cos(÷13t ), ÷3, ÷13
CHAPTER 12
11.5 Review exercises
1
2 ∞ cos(2n − 1)t
1 (a) f (t ) = − π − ∑
4
π n = 1 (2n − 1)2
3
7
Exercises
1 (a) x(t) = cos t + sin t − e−2t(cos t + 3 sin t)
(b) x(t ) = −3 + 137 e t + 157 e −2t / 5
∞
 3 sin(2n − 1)t sin 2nt 
+ ∑
−
2n − 1
2n 
n =1 
∞
2 ∞ cos(2n − 1)t
1
sin nt
−∑
(b) f (t ) = π + ∑
2
4
n
π n = 1 (2n − 1)
n =1
2 (a) e − t − 12 e −2t − 12 e − t (cos t + sin t)
(b) i(t) = 4e−t − 3e−2t
+ V [e − t − 12 e −2t − 12 e − t(cos t + sin t )]
3 x(t) = −t + 5 sin t − 2 sin 2t,
y(t) = 1 − 2 cos t + cos 2t
(c) f (t ) =
2 ∞ sin nt
∑
π n =1 n
4 15 (cos t + 2 sin t)
e−t[(x0 − 15 )cos t + (x1 + x0 − 35 )sin t]
1
Ë 5 , 63.4° lag
(d) f (t ) =
2
4 ∞ (−1) n + 1 cos 2nt
+ ∑
π π n =1
4n 2 − 1
(e) f (t ) =
2
4 ∞ (−1) n + 1 cos nt
+ ∑
π π n = 1 4n 2 − 1
(f) f (t ) =
1
4 ∞ cos(2n − 1)t
π − ∑
π n = 1 (2n − 1)2
2
s cos φ − ω sin φ
s2 + ω 2
s sin φ + ω (cos φ + sin φ )
(ii)
s 2 + 2ω s + ω 2
(b) 201 (cos 2t + 2 sin 2t) + 201 e−2t(39 cos 2t + 47 sin 2t)
6 (a) (i)
7 (a) e−2t(cos 3t − 2 sin 3t)
(b) y(t) = 2 + 2 sin t − 5e−2t
8 x(t) = e−8t + sin t, y(t) = e−8t − cos t
+
2 ∞  (−1) n − 1 (−1) n sinh π 
+
∑
 cos nt
π n = 1  n 2
n2 + 1 
1
1
9 q(t) = 500
(5e−100t − 2e−200t) − 505
(3 cos 100t −
sin 100t), current leads by approximately 18.5°
−
2 ∞ n(−1) n
sinh π sin nt
∑
π n =1 n 2 + 1
29 − t
445 − t / 5
+ 13 e −2t
e + 1212
e
10 x(t ) = 20
1
− 505 (76 cos 2t − 48 sin 2t)
1
4 ∞ cos(2n − 1)t 
3 q(t ) = Q  − 2 ∑
2 
 2 π n = 1 (2n − 1) 
E
12 i(t ) = [1 − e−nt(cos nt − sin nt)]
R
13 i1 (t ) =
E(4 − 3e − Rt / L − e −3 Rt / L )
, i2 (t ) → E / 3R
6R
4 f (t ) =
t
t
15 (a) (i) e (cos 3t + sin 3t)
(ii) e − e + 2te
(b) y(t) = 12 e−t(8 + 12t + t 3)
2t
e7t sin 2t
n 2i
i
(b)
, θ (t ) = (1 − e − Kt ) − ite − Kt
2
Ks (s + 2 Ks + n 2 )
K
16 (a)
5
2
17 (a) v1 = 250e−0.1t, v2 = (50 + 25t)e−0.1t
(b) t = 23.026
5 5
10 ∞ cos 2nt
+ sin t −
∑
π 2
π n = 1 4n 2 − 1
5 Taking t = 0 and t =
14 x1(t) = 13 [sin t − 2 sin 2t + Ë3 sin(Ë3t)]
x2(t) = 13 [sin t + sin 2t − Ë3 sin(Ë3t)]
−t
∞
1 2
cos nt
π + 4∑
3
n2
n =1
Taking t = gives the required result.
2 f (t ) =
1
11 (a) u(t) = 100
(4e−4t + 10te−4t − 4 cos 2t + 3 sin 2t)
1
(b) i1(t) = 7 (e4t + 6e−3t ), i2 = 17 (e−3t − e4t )
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∞
4 ∞ cos(2n − 1)t
sin 2nt
−∑
∑
2
π n = 1 (2n − 1)
n
n =1
1
1

(h) f (t ) =  π + sinh π 
2

π
(g) f (t ) = −
gives the required answers.
1
2 ∞ cos(4n − 2)t
π − ∑
π n = 1 (2n − 1)2
4
Taking t = 0 gives the required series.
6 f (t ) =
3
4 ∞ cos(2n − 1)t
+ 2∑
2 π n = 1 (2n − 1)2
Replacing t by t − 12 gives the following sine series
of odd harmonics:
7 f (t ) =
James, Glyn, and Phil Dyke. Modern Engineering Mathematics 6th Edition PDF Ebook, Pearson Education, Limited, 2020. ProQuest Ebook Central,
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AN S WE RS TO E XE RCI S E S
1  3
4 ∞ (−1)n sin(2n − 1)t

f t − π − = − 2 ∑

π n =1
2  2
(2n − 1)2
1
4A ∞
1
(2n − 1)π x
21 f ( x ) = − A − 2 ∑
cos
2
π n = 1 (2n − 1)2
l
8 f (t ) =
2l ∞ (−1)n + 1
nπ t
sin
∑
l
π n =1 n
22 T ( x ) =
8KL2 ∞
1
(2n − 1)π x
sin
∑
π 3 n = 1 (2n − 1)3
L
9 f (t ) =
2K ∞ 1
nπ t
∑ sin l
π n =1 n
23 f (t ) =
1 1
4 ∞
1
sin 2nπ t
+ cos π t + ∑ 2
2 2
π n = 1 4n − 1
10 f (t ) =
sin(2n – 1)π t
3 6 ∞
1
+ ∑
2 π n = 1 (2n – 1)
5
11 v(t ) =
∞
1
cos 2nω t 
A
π
ω
+
−
1
sin
t
2
∑


2
π
2
n = 1 4n − 1 
1
4T 2 ∞ (−1)n
nπ t
12 f (t ) = T 2 + 2 ∑ 2 cos
π n =1 n
3
T
2 ∞ 1
2π nt 
E
13 e(t ) = 1 − ∑ sin
2
π n =1 n
T 
∞
15 f (t ) = −
8
1
cos(2n − 1)π t
2 ∑
π n = 1 (2n − 1)2
16 (a) f (t ) =
(b) f (t ) =
2
1 ∞ 1
1 ∞ 1
− 2 ∑ 2 cos 2nπ t + ∑ sin 2nπ t
3 π n =1 n
π n =1 n
1 ∞ 1
∑ sin 2nπ t
π n =1 n
+
−
2 ∞
1
sin(2n − 1)π t
∑
π n = 1 2n − 1
∞
(−1) n + 1
sin nt
n
n =1
26 (c) 1 + 4 ∑
12.6 Review exercises
1 f (t ) =
∞
1 2
2
π + ∑ 2 (−1) n cos nt
6
n
n =1
∞

 π
4
sin(2n − 1)t
+ ∑
−
3
2
n
1
2
n
1
(
)
−
−
π
n =1 

∞
π
−∑
sin 2nt
2
n =1 n
Taking T = gives the required sum.
2 f (t ) =
2 ∞  1
4

+
∑
π n = 1  2n − 1 π 2(2n − 1)3 
× sin(2n − 1) t
2
4 ∞ (−1)n + 1
cos nπ t
(c) f (t ) = + 2 ∑
3 π n =1 n2
1
π
9
2 ∞ 1 
1
1

+ ∑ 2 cos nπ − [2 + (−1)n ] cos nt ; 29 π
π n =1 n 
3
3

3 (a) f (t ) =
2T ∞ (−1)n + 1
2 (2n − 1)π t
sin
∑
π 2 n = 1 (2n − 1)2
T
(b) − 14 T
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1
∞
1
1
17 f (t ) = π 2 − ∑ 2 cos 2nt
6
n =1 n
8 ∞
1
f (t ) = ∑
sin(2n − 1)t
π n = 1 (2n – 1)3
18 f ( x ) =
8a ∞ (−1)n + 1
(2n − 1)π x
sin
∑
π 2 n = 1 (2n – 1)2
l
∞
19 f ( x ) =
2l
(−1)n + 1
2(2n − 1)π x
sin
2 ∑
2
π n = 1 (2n – 1)
l
(c) Taking t = 14 T gives S = 8 π 2
4 y=
4P ∞
1
sin(2n − 1) α sin(2n − 1) x
∑
πα n = 1 (2n − 1)2
6 f (t ) =
4 ∞ (−1)n sin(2n − 1)t
∑ (2n − 1)2
π n =1
8 f ( x) =
4 ∞ cos(2n − 1) x
∑
π n = 1 (2n − 1)2
Taking x = 0 gives
∞
1
4 ∞ n (−1)n + 1
20 f (t ) = sin t + ∑
sin 2nt
2
π n = 1 4n 2 − 1
1111
1
(
n
− 1) 2
2
n =1
π2 = 8∑
James, Glyn, and Phil Dyke. Modern Engineering Mathematics 6th Edition PDF Ebook, Pearson Education, Limited, 2020. ProQuest Ebook Central,
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Created from monash on 2022-06-03 01:16:25.
1112 A NSW ERS T O EXERCISES
∞
1
2(−1)n+1 

sin (2n − 1) x
1+

π (2n − 1) 
n = 1 (2n − 1) 
9 f ( x) = ∑
∞
−∑
n =1
1
sin 2nx
2n
CHAPTER 13
Exercises
5 (a) 8
7
∞
6
Number
50(1 − e −1⋅ 2 )
10 V = 253 (1 − e −1⋅ 2 ) + ∑
2 2
n = 1 9 + 25n π
× (3 cos 5n t + 5n sin 5n t)
Amplitude of the nth harmonic is
4 ∞ (−1)n + 1
sin(2n − 1) x
∑
π n = 1 (2n − 1)2
f ( x) =
1
2 ∞ cos 2 (2n − 1) x
π − ∑
π n = 1 (2n − 1) 2
4
4
8
7
3
50(1 − e −1⋅ 2 )
50(1 − e −1⋅ 2 )
2 ⋅ 22
2 2
÷(9 + 25n π )
n
5nπ
13 f ( x ) =
5
5
2
1
1
0
0
4
1
1
2
1
2
2
8 10 12 14 16 18 20 22 24 26
Ambient temperature in ºC
6
16
14
∞
2
16 f (t ) = 3 +
17 (a)
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2 π2
π2

+ 
− 2 cos t − 12 cos 2t

24 π  4
2 π2
2
− 
−  cos 3t + 18 cos 4t
π  12 27 
18 (a) f (t ) =
(b) g(t ) =
∞
1
4 ∞
1
π − ∑
cos(2n – 1)t
π n = 1 (2n − 1)2
2
4 ∞
1
sin(2n – 1)t
∑
π n = 1 2n − 1
4
1
sin(2n – 1)t
∑
π n = 1 2n − 1
f(t) = 1 + g(t)
Number
10
8
16
6
4
4 ∞ (−1)n + 1
cos nπ t
∑
π 2 n =1 n2
(i) a constant term and cosine terms with
even harmonics
(ii) constant, cosine and sine terms
present
(iii) a constant term and sine terms with odd
harmonics
(b) f (t ) =
12
6
2
5
(b)
5
1
2
0
10
15
20
Ambient temperature in ºC
100
Cumulative percentage
2
sin nt
n =1 n
1
4 ∞
1
cos(2n – 1)t
(b) f (t ) = π + ∑
π n = 1 (2n − 1)2
2
15 (a) f (t ) = ∑
30
25
Engine A
Engine B
80
60
40
20
0
20
22
24
26
28
Running time in minutes
19 g(t ) =
James, Glyn, and Phil Dyke. Modern Engineering Mathematics 6th Edition PDF Ebook, Pearson Education, Limited, 2020. ProQuest Ebook Central,
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Created from monash on 2022-06-03 01:16:25.
30
AN S WE RS TO E XE RCI S E S
6 (a) A > B
(c) S − A
30 (a) 19
(b) A < B
(d) S − (A > B)
(c) 0.102
31 1 − exp(−x /2a), 0.0804
2
7 (a) {car, bicycle, motorcycle, boat}
(b) {train}
(c) {car, motorcycle, boat}
32 mean = 4.5, P(less than 5 days) = 0.6
8 (a) 0.7
33 mean = 1.8, median = 2,
standard deviation = 1.34
(b) 0.8
(c) 0.5
16
9 2652
34 Average length = 5.88
1
6
10 P(same values) = , P(differ by at most 1) =
12 (a)
1
26
13 (a)
1
9
(b)
4
13
(b)
5
18
1
2
(c)
(c)
4
9
1
13
(d)
35 mean = 5, median = 3
standard deviation = 4.47
36 mean = 30 minutes, standard deviation = 17.3 min
5
6
38 (a) 0.47
(b) mX = 30, sX = 30
(c) median = 20.8, q3 − q1 = 33.0
14 P(total = 7 7 or 10) = 23
15 (a) 43
(b) 7 to 1
39 0.969
16 (a) 12
(b) 13
41 24 hours, 3.32 hours
17 65
18 (a) 0.15
(b) 0.55
(c) 0.357
42 (a) X = 2.28, SX = 0.60, SX,n−1 = 0.63
(b) sample median = 2.1, range = 2.2
19 0.6
43 X = 5.44, SX = 0.81, median = 5.45, range = 3.2
20 0.381
44 Ā = 24.2, SA = 1.76, T = 16.7, ST = 4.00
B = 25.4, SB = 1.66, U= 18.1, SU = 4.93
where A, T are time, temperature for A, and B, U are
time, temperature for B
21 (a) P(A) + P(b) − P(A > B)
P(C | A) P( A) − P(C | A B) P ( A B)
(b)
P( A) − P( A B)
P (C | A) P ( A) + P (C | B) P ( B) − P (C | A B) P ( A B)
(c)
P ( A) + P ( B) − P ( A B)
22 0.149
2r 

23 (a) 1 − 

d
24 P(2) = 361
2
 2r 
( b) 1 −  
d
2
26 (a) 0.488
6
28 (a) 14
(c) 12
29 P(X < 30) = 0.28
48 47.1 and 46.3
50 0.998
3
8
27 (b) P(−3) = , P(−1) =
P(1) = 83 , P(3) = 18
(c) P(−3) = 271 , P(−1) = 276
8
P(1) = 12
27 , P(3) = 27
0

(b) FX ( x ) =  12 ÷x
1

46 median = Ë[2a ln 2], mode = Ëa
a = 6: mean = 3.07, median = 2.88, mode = 2.45,
q3 − q1 = 2.22
49 P(4 boys) = 0.273
(b) 0.3123
1
8
45 2.19
47 q/( p + q − pq)
P(3) = 362 , p, P(7) = 36
5
P(8) = 36
, p, P(12) = 361
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(b) 43
1113
( x 0)
( 0 x 4)
( x 4)
51 0.677
52 (a) 0.1271
(c) 0.1413
(b) 0.3594
(d) 0.5876
53 4 engines
54 (a) 0.957
(b) 0.0071
55 0.027
56 P(8 or more) = 0.249
57 (a) 0.050
(b) 0.224
(c) 0.084
58 0.986
59 6.09
James, Glyn, and Phil Dyke. Modern Engineering Mathematics 6th Edition PDF Ebook, Pearson Education, Limited, 2020. ProQuest Ebook Central,
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1114 A NSW ERS T O EXERCISES
61 0.144
6 ;5.66 × 10 −5
62 0.011
9 0.407
63 46
64 0.3%, 0.0258
65 (a) 0.102
(b) 0.128
(c) 0.011
66 Warning 9.5, action 13.5, sample 12
UCL = 11.4, sample 9
67 UK sample 28, US sample 25
68 P(at least one such area) = 0.133
69 P(at least one such area) = 0.688
13.8 Review exercises
1 (a) 3
(c) 18
25
12λ
25
1
=
hours when λ =
4
3
10 E(minimum) =
0
(b) FX ( x ) = 
1 − x −3
(d) 23
for x 1
for x 1
(e) Ë3/2
2 0.0159
3 60, 6342 hours
4 P(10, 5, 3, 2) = 0.009
13 (a) single
k = 4: n = 7, P(error) = 0.0020
k = 8: n = 12, P(error) = 0.0062
k = 16: n = 21, P(error) = 0.0185
k = 32: n = 38, P(error) = 0.0555
k = 64: n = 71, P(error) = 0.1588
double
k = 4: n = 11, P(error) = 0.0002
k = 8: n = 17, P(error) = 0.0006
k = 16: n = 26, P(error) = 0.0022
k = 32: n = 43, P(error) = 0.0092
k = 64: n = 77, P(error) = 0.0424
(b) single: k = 8, so total 96 bits
double: k = 64, so total 77 bits
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5 e−l(1 + l) > 0.9, proportion = 0.0053
N
k
= 0.196 N when k = 11
12 E(number of analyses) = N [1 − (1 − p)k ] +
James, Glyn, and Phil Dyke. Modern Engineering Mathematics 6th Edition PDF Ebook, Pearson Education, Limited, 2020. ProQuest Ebook Central,
http://ebookcentral.proquest.com/lib/monash/detail.action?docID=6036640.
Created from monash on 2022-06-03 01:16:25.