Download - C4 Marking Schemes
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SERIES C4 Answers - Worksheet A 1 a = 1 + (−1)x + ( 1)( 2)
2− − x2 + ( 1)( 2)( 3)
3 2− − −
× x3 + … = 1 − x + x2 − x3 + …
b = 1 + ( 12 )x +
1 12 2( )( )
2− x2 +
31 12 2 2( )( )( )
3 2− −
× x3 + … = 1 + 1
2 x − 18 x2 + 1
16 x3 + …
c = 2[1 + (−3)x + ( 3)( 4)2
− − x2 + ( 3)( 4)( 5)3 2
− − −× x3 + …]
= 2 − 6x + 12x2 − 20x3 + …
d = 1 + ( 23 )x +
2 13 3( )( )
2− x2 +
2 1 43 3 3( )( )( )
3 2− −
× x3 + … = 1 + 2
3 x − 19 x2 + 4
81 x3 + …
e = (1 − 13)x = 1 + ( 1
3 )(−x) + 1 23 3( )( )
2− (−x)2 +
51 23 3 3( )( )( )
3 2− −
× (−x)3 + … = 1 − 1
3 x − 19 x2 − 5
81 x3 + …
f = (1 + x)−2 = 1 + (−2)x + ( 2)( 3)2
− − x2 + ( 2)( 3)( 4)3 2
− − −× x3 + …
= 1 − 2x + 3x2 − 4x3 + …
g = 14 (1 − x)−4 = 1
4 [1 + (−4)(−x) + ( 4)( 5)2
− − (−x)2 + ( 4)( 5)( 6)3 2
− − −× (−x)3 + …]
= 14 + x + 5
2 x2 + 5x3 + …
h = 3(1 − 12)x − = 3[1 + ( 1
2− )(−x) + 31
2 2( )( )2
− − (−x)2 + 3 51
2 2 2( )( )( )3 2
− − −× (−x)3 + …]
= 3 + 32 x + 9
8 x2 + 1516 x3 + …
2 a = 1 + ( 1
2 )(2x) + 1 12 2( )( )
2− (2x)2 +
31 12 2 2( )( )( )
3 2− −
× (2x)3 + … = 1 + x − 1
2 x2 + 12 x3 + …, | 2x | < 1 ∴ valid for | x | < 1
2
b = 1 + (−1)(−3x) + ( 1)( 2)2
− − (−3x)2 + ( 1)( 2)( 3)3 2
− − −× (−3x)3 + …
= 1 + 3x + 9x2 + 27x3 + …, | −3x | < 1 ∴ valid for | x | < 13
c = 1 + ( 12− )(−4x) +
312 2( )( )
2− − (−4x)2 +
3 512 2 2( )( )( )
3 2− − −
× (−4x)3 + … = 1 + 2x + 6x2 + 20x3 + …, | −4x | < 1 ∴ valid for | x | < 1
4
d = 1 + (−3)( 12 x) + ( 3)( 4)
2− − ( 1
2 x)2 + ( 3)( 4)( 5)3 2
− − −× ( 1
2 x)3 + … = 1 − 3
2 x + 32 x2 − 5
4 x3 + …, | 12 x | < 1 ∴ valid for | x | < 2
e = 1 + ( 13 )(−6x) +
1 23 3( )( )
2− (−6x)2 +
51 23 3 3( )( )( )
3 2− −
× (−6x)3 + … = 1 − 2x − 4x2 − 40
3 x3 + …, | −6x | < 1 ∴ valid for | x | < 16
f = 1 + (−4)( 14 x) + ( 4)( 5)
2− − ( 1
4 x)2 + ( 4)( 5)( 6)3 2
− − −× ( 1
4 x)3 + … = 1 − x + 5
8 x2 − 516 x3 + …, | 1
4 x | < 1 ∴ valid for | x | < 4
g = 1 + ( 32 )(2x) +
3 12 2( )( )
2 (2x)2 + 3 1 12 2 2( )( )( )
3 2−
× (2x)3 + … = 1 + 3x + 3
2 x2 − 12 x3 + …, | 2x | < 1 ∴ valid for | x | < 1
2
h = 1 + ( 43− )(−3x) +
743 3( )( )
2− − (−3x)2 +
7 1043 3 3( )( )( )
3 2− − −
× (−3x)3 + … = 1 + 4x + 14x2 + 140
3 x3 + …, | −3x | < 1 ∴ valid for | x | < 13
C4 SERIES Answers - Worksheet A page 2
Solomon Press
3 a = 1 + ( 1
2 )(−2x) + 1 12 2( )( )
2− (−2x)2 +
31 12 2 2( )( )( )
3 2− −
× (−2x)3 + … = 1 − x − 1
2 x2 − 12 x3 + …
b 0.98 = (1 − 122 )x when x = 0.01
∴ 0.98 ≈ 1 − (0.01) − 12 (0.01)2 − 1
2 (0.01)3 = 1 − 0.01 − 0.000 05 − 0.000 000 5 = 0.989 949 5
c 0.98 = 98100 = 49 2
100× = 7
10 2
∴ 2 ≈ 107 × 0.989 949 5 = 1.414 213 6 (8sf)
4 a = 2−1(1 + 1
2 x)−1 = 12 (1 + 1
2 x)−1
= 12 [1 + (−1)( 1
2 x) + ( 1)( 2)2
− − ( 12 x)2 + ( 1)( 2)( 3)
3 2− − −
× ( 12 x)3 + …]
= 12 − 1
4 x + 18 x2 − 1
16 x3 + …, | 12 x | < 1 ∴ valid for | x | < 2
b = 1 12 21
44 (1 )x+ = 2(1 + 121
4 )x
= 2[1 + ( 12 )( 1
4 x) + 1 12 2( )( )
2− ( 1
4 x)2 + 31 1
2 2 2( )( )( )3 2− −
× ( 14 x)3 + …]
= 2 + 14 x − 1
64 x2 + 1512 x3 + …, | 1
4 x | < 1 ∴ valid for | x | < 4
c = 3−3(1 − 13 x)−3 = 1
27 (1 − 13 x)−3
= 127 [1 + (−3)(− 1
3 x) + ( 3)( 4)2
− − (− 13 x)2 + ( 3)( 4)( 5)
3 2− − −
× (− 13 x)3 + …]
= 127 + 1
27 x + 281 x2 + 10
729 x3 + …, | − 13 x | < 1 ∴ valid for | x | < 3
d = 1 12 21
39 (1 )x+ = 3(1 + 121
3 )x
= 3[1 + ( 12 )( 1
3 x) + 1 12 2( )( )
2− ( 1
3 x)2 + 31 1
2 2 2( )( )( )3 2− −
× ( 13 x)3 + …]
= 3 + 12 x − 1
24 x2 + 1144 x3 + …, | 1
3 x | < 1 ∴ valid for | x | < 3
e = 1 13 38 (1 3 )x− = 2(1 −
133 )x
= 2[1 + ( 13 )(−3x) +
1 23 3( )( )
2− (−3x)2 +
51 23 3 3( )( )( )
3 2− −
× (−3x)3 + …] = 2 − 2x − 2x2 − 10
3 x3 + …, | −3x | < 1 ∴ valid for | x | < 13
f = 4−1(1 − 34 x)−1 = 1
4 (1 − 34 x)−1
= 14 [1 + (−1)(− 3
4 x) + ( 1)( 2)2
− − (− 34 x)2 + ( 1)( 2)( 3)
3 2− − −
× (− 34 x)3 + …]
= 14 + 3
16 x + 964 x2 + 27
256 x3 + …, | − 34 x | < 1 ∴ valid for | x | < 4
3
g = 1 12 23
24 (1 )x− −+ = 1231
2 2(1 )x −+
= 12 [1 + ( 1
2− )( 32 x) +
312 2( )( )
2− − ( 3
2 x)2 + 3 51
2 2 2( )( )( )3 2
− − −× ( 3
2 x)3 + …] = 1
2 − 38 x + 27
64 x2 − 135256 x3 + …, | 3
2 x | < 1 ∴ valid for | x | < 23
h = 3−2(1 + 23 x)−2 = 1
9 (1 + 23 x)−2
= 19 [1 + (−2)( 2
3 x) + ( 2)( 3)2
− − ( 23 x)2 + ( 2)( 3)( 4)
3 2− − −
× ( 23 x)3 + …]
= 19 − 4
27 x + 427 x2 − 32
243 x3 + …, | 23 x | < 1 ∴ valid for | x | < 3
2
C4 SERIES Answers - Worksheet A page 3
Solomon Press
5 a = 1 + (−1)(2x) + ( 1)( 2)
2− − (2x)2 + ( 1)( 2)( 3)
3 2− − −
× (2x)3 + … = 1 − 2x + 4x2 − 8x3 + … b = (1 − x)(1 + 2x)−1 = (1 − x)( 1 − 2x + 4x2 − 8x3 + …) = 1 − 2x + 4x2 − 8x3 − x + 2x2 − 4x3 + … = 1 − 3x + 6x2 − 12x3 + … 6 a = (1 + 3x)(1 − x)−1 = (1 + 3x)[1 + (−1)(−x) + ( 1)( 2)
2− − (−x)2 + ( 1)( 2)( 3)
3 2− − −
× (−x)3 + …] = (1 + 3x)(1 + x + x2 + x3 + …) = 1 + x + x2 + x3 + 3x + 3x2 + 3x3 + … = 1 + 4x + 4x2 + 4x3 + …, | −x | < 1 ∴ valid for | x | < 1
b = (2x − 1)(1 + 4x)−2 = (2x − 1)[1 + (−2)(4x) + ( 2)( 3)2
− − (4x)2 + ( 2)( 3)( 4)3 2
− − −× (4x)3 + …]
= (2x − 1)(1 − 8x + 48x2 − 256x3 + …) = 2x − 16x2 + 96x3 − 1 + 8x − 48x2 + 256x3 + … = −1 + 10x − 64x2 + 352x3 + …, | 4x | < 1 ∴ valid for | x | < 1
4
c = (3 + x)(2 − x)−1 = (3 + x) × 2−1(1 − 12 x)−1
= (3 + x) × 12 [1 + (−1)(− 1
2 x) + ( 1)( 2)2
− − (− 12 x)2 + ( 1)( 2)( 3)
3 2− − −
× (− 12 x)3 + …]
= (3 + x)( 12 + 1
4 x + 18 x2 + 1
16 x3 + …) = 3
2 + 34 x + 3
8 x2 + 316 x3 + 1
2 x + 14 x2 + 1
8 x3 + … = 3
2 + 54 x + 5
8 x2 + 516 x3 + …, | − 1
2 x | < 1 ∴ valid for | x | < 2
d = (1 − x)(1 + 122 )x − = (1 − x)[1 + ( 1
2− )(2x) + 31
2 2( )( )2
− − (2x)2 + 3 51
2 2 2( )( )( )3 2
− − −× (2x)3 + …]
= (1 − x)(1 − x + 32 x2 − 5
2 x3 + …) = 1 − x + 3
2 x2 − 52 x3 − x + x2 − 3
2 x3 + … = 1 − 2x + 5
2 x2 − 4x3 + …, | 2x | < 1 ∴ valid for | x | < 12
7 a 2
(1 )(1 2 )xx x
−− −
≡ 1
Ax−
+ 1 2
Bx−
x − 2 ≡ A(1 − 2x) + B(1 − x) x = 1 ⇒ −1 = −A ⇒ A = 1 x = 1
2 ⇒ 32− = 1
2 B ⇒ B = −3
∴ 2(1 )(1 2 )
xx x
−− −
≡ 11 x−
− 31 2x−
b 11 x−
= (1 − x)−1 = 1 + (−1)(−x) + ( 1)( 2)2
− − (−x)2 + ( 1)( 2)( 3)3 2
− − −× (−x)3 + …
= 1 + x + x2 + x3 + …, | −x | < 1 ∴ | x | < 1 3
1 2x− = 3(1 − 2x)−1 = 3[1 + (−1)(−2x) + ( 1)( 2)
2− − (−2x)2 + ( 1)( 2)( 3)
3 2− − −
× (−2x)3 + …]
= 3 + 6x + 12x2 + 24x3 + …, | −2x | < 1 ∴ | x | < 12
∴ 2(1 )(1 2 )
xx x
−− −
= (1 + x + x2 + x3 + …) − (3 + 6x + 12x2 + 24x3 + …)
= −2 − 5x − 11x2 − 23x3 + …, valid for | x | < 12
C4 SERIES Answers - Worksheet A page 4
Solomon Press
8 a 4
(1 )(1 3 )x x+ − ≡
1A
x+ +
1 3B
x−
4 ≡ A(1 − 3x) + B(1 + x) x = −1 ⇒ 4 = 4A ⇒ A = 1 x = 1
3 ⇒ 4 = 43 B ⇒ B = 3
∴ f(x) ≡ 11 x+
+ 31 3x−
11 x+
= (1 + x)−1 = 1 + (−1)x + ( 1)( 2)2
− − x2 + ( 1)( 2)( 3)3 2
− − −× x3 + …
= 1 − x + x2 − x3 + …, | x | < 1 3
1 3x− = 3(1 − 3x)−1 = 3[1 + (−1)(−3x) + ( 1)( 2)
2− − (−3x)2 + ( 1)( 2)( 3)
3 2− − −
× (−3x)3 + …]
= 3 + 9x + 27x2 + 81x3 + …, | −3x | < 1 ∴ | x | < 13
∴ f(x) ≡ (1 − x + x2 − x3 + …) + (3 + 9x + 27x2 + 81x3 + …) f(x) ≡ 4 + 8x + 28x2 + 80x3 + …, valid for | x | < 1
3 b 2
1 61 3 4
xx x−
+ − ≡ 1 6
(1 )(1 4 )x
x x−
− + ≡
1A
x− +
1 4B
x+
1 − 6x ≡ A(1 + 4x) + B(1 − x) x = 1 ⇒ −5 = 5A ⇒ A = −1 x = 1
4− ⇒ 52 = 5
4 B ⇒ B = 2
∴ f(x) ≡ 21 4x+
− 11 x−
21 4x+
= 2(1 + 4x)−1 = 2[1 + (−1)(4x) + ( 1)( 2)2
− − (4x)2 + ( 1)( 2)( 3)3 2
− − −× (4x)3 + …]
= 2 − 8x + 32x2 − 128x3 + …, | 4x | < 1 ∴ | x | < 14
11 x−
= (1 − x)−1 = 1 + (−1)(−x) + ( 1)( 2)2
− − (−x)2 + ( 1)( 2)( 3)3 2
− − −× (−x)3 + …
= 1 + x + x2 + x3 + …, | −x | < 1 ∴ | x | < 1 ∴ f(x) ≡ (2 − 8x + 32x2 − 128x3 + …) − (1 + x + x2 + x3 + …) f(x) ≡ 1 − 9x + 31x2 − 129x3 + …, valid for | x | < 1
4 c 2
52 3 2x x− −
≡ 5(1 2 )(2 )x x− +
= 1 2
Ax−
+ 2
Bx+
5 ≡ A(2 + x) + B(1 − 2x) x = 1
2 ⇒ 5 = 52 A ⇒ A = 2
x = −2 ⇒ 5 = 5B ⇒ B = 1 ∴ f(x) ≡ 2
1 2x− + 1
2 x+
21 2x−
= 2(1 − 2x)−1 = 2[1 + (−1)(−2x) + ( 1)( 2)2
− − (−2x)2 + ( 1)( 2)( 3)3 2
− − −× (−2x)3 + …]
= 2 + 4x + 8x2 + 16x3 + …, | −2x | < 1 ∴ | x | < 12
12 x+
= (2 + x)−1 = 2−1(1 + 12 x)−1 = 1
2 [1 + (−1)( 12 x) + ( 1)( 2)
2− − ( 1
2 x)2 + ( 1)( 2)( 3)3 2
− − −× ( 1
2 x)3 + …]
= 12 − 1
4 x + 18 x2 − 1
16 x3 + …, | 12 x | < 1 ∴ | x | < 2
∴ f(x) ≡ (2 + 4x + 8x2 + 16x3 + …) + ( 12 − 1
4 x + 18 x2 − 1
16 x3 + …) f(x) ≡ 5
2 + 154 x + 65
8 x2 + 25516 x3 + …, valid for | x | < 1
2
C4 SERIES Answers - Worksheet A page 5
Solomon Press
d 2
7 34 3
xx x
−− +
≡ 7 3( 1)( 3)
xx x
−− −
≡ 1
Ax −
+ 3
Bx −
7x − 3 ≡ A(x − 3) + B(x − 1) x = 1 ⇒ 4 = −2A ⇒ A = −2 x = 3 ⇒ 18 = 2B ⇒ B = 9 ∴ f(x) ≡ 9
3x − − 2
1x − ≡ 2
1 x− − 9
3 x−
21 x−
= 2(1 − x)−1 = 2[1 + (−1)(−x) + ( 1)( 2)2
− − (−x)2 + ( 1)( 2)( 3)3 2
− − −× (−x)3 + …]
= 2 + 2x + 2x2 + 2x3 + …, | −x | < 1 ∴ | x | < 1 9
3 x− = 9(3 − x)−1 = 9 × 3−1(1 − 1
3 x)−1
= 3[1 + (−1)(− 13 x) + ( 1)( 2)
2− − (− 1
3 x)2 + ( 1)( 2)( 3)3 2
− − −× (− 1
3 x)3 + …] = 3 + x + 1
3 x2 + 19 x3 + …, | − 1
3 x | < 1 ∴ | x | < 3 ∴ f(x) ≡ (2 + 2x + 2x2 + 2x3 + …) − (3 + x + 1
3 x2 + 19 x3 + …)
f(x) ≡ −1 + x + 53 x2 + 17
9 x3 + …, valid for | x | < 1 e 2
3 5(1 3 )(1 )
xx x+
+ + ≡
1 3A
x+ +
1B
x+ + 2(1 )
Cx+
3 + 5x ≡ A(1 + x)2 + B(1 + 3x)(1 + x) + C(1 + 3x) x = 1
3− ⇒ 43 = 4
9 A ⇒ A = 3 x = −1 ⇒ −2 = −2C ⇒ C = 1 coeffs of x2 ⇒ 0 = A + 3B ⇒ B = −1 ∴ f(x) ≡ 3
1 3x+ − 1
1 x+ + 2
1(1 )x+
31 3x+
= 3(1 + 3x)−1 = 3[1 + (−1)(3x) + ( 1)( 2)2
− − (3x)2 + ( 1)( 2)( 3)3 2
− − −× (3x)3 + …]
= 3 − 9x + 27x2 − 81x3 + …, | 3x | < 1 ∴ | x | < 13
11 x+
= (1 + x)−1 = 1 + (−1)x + ( 1)( 2)2
− − x2 + ( 1)( 2)( 3)3 2
− − −× x3 + …
= 1 − x + x2 − x3 + …, | x | < 1 2
1(1 )x+
= (1 + x)−2 = 1 + (−2)x + ( 2)( 3)2
− − x2 + ( 2)( 3)( 4)3 2
− − −× x3 + …
= 1 − 2x + 3x2 − 4x3 + …, | x | < 1 ∴ f(x) ≡ (3 − 9x + 27x2 − 81x3 + …) − (1 − x + x2 − x3 + …) + (1 − 2x + 3x2 − 4x3 + …) f(x) ≡ 3 − 10x + 29x2 − 84x3 + …, valid for | x | < 1
3
C4 SERIES Answers - Worksheet A page 6
Solomon Press
f
∴ 2
22 4
2 1x
x x+
+ − ≡ 1 + 2
52 1
xx x
−+ −
25
2 1x
x x−+ −
≡ 5(2 1)( 1)
xx x
−− +
≡ 2 1
Ax −
+ 1
Bx +
5 − x ≡ A(x + 1) + B(2x − 1) x = 1
2 ⇒ 92 = 3
2 A ⇒ A = 3 x = −1 ⇒ 6 = −3B ⇒ B = −2 ∴ f(x) ≡ 1 + 3
2 1x − − 2
1x + ≡ 1 − 3
1 2x− − 2
1 x+
31 2x−
= 3(1 − 2x)−1 = 3[1 + (−1)(−2x) + ( 1)( 2)2
− − (−2x)2 + ( 1)( 2)( 3)3 2
− − −× (−2x)3 + …]
= 3 + 6x + 12x2 + 24x3 + …, | −2x | < 1 ∴ | x | < 12
21 x+
= 2(1 + x)−1 = 2[1 + (−1)x + ( 1)( 2)2
− − x2 + ( 1)( 2)( 3)3 2
− − −× x3 + …]
= 2 − 2x + 2x2 − 2x3 + …, | x | < 1 ∴ f(x) ≡ 1 − (3 + 6x + 12x2 + 24x3 + …) − (2 − 2x + 2x2 − 2x3 + …) f(x) ≡ −4 − 4x − 14x2 − 22x3 + …, valid for | x | < 1
2
1 2x2 + x − 1 2x2 + 0x + 4
2x2 + x − 1 − x + 5
Solomon Press
SERIES C4 Answers - Worksheet B
1 a = 1 + ( 1
2 )(−x) + 1 12 2( )( )
2− (−x)2 +
31 12 2 2( )( )( )
3 2− −
× (−x)3 + … = 1 − 1
2 x − 18 x2 − 1
16 x3 + …
b when x = 0.01, (1 − 12)x ≈ 1 − 1
2 (0.01) − 18 (0.01)2 − 1
16 (0.01)3 = 1 − 0.005 − 0.000 012 5 − 0.000 000 062 5 = 0.994 987 437 5
(1 − 120.01) = 0.99 = 9 11
100× = 3
10 11
∴ 11 = 103 × 0.994 987 437 5 = 3.316 624 79 (9sf)
2 a a =
1 12 2( )( )
2− (8)2 = −8, b =
31 12 2 2( )( )( )
3 2− −
× (8)3 = 32
b when x = 0.01, (1 + 128 )x ≈ 1 + 4(0.01) − 8(0.01)2 + 32(0.01)3
= 1 + 0.04 − 0.000 8 + 0.000 032 = 1.039 232
(1 + 120.08) = 1.08 = 36 3
100× = 3
5 3
∴ 3 = 53 × 1.039 232 = 1.732 05 (5dp)
3 a =
1 12 22
39 (1 )x− = 3(1 − 122
3 )x
= 3[1 + ( 12 )(− 2
3 x) + 1 12 2( )( )
2− (− 2
3 x)2 + 31 1
2 2 2( )( )( )3 2− −
× (− 23 x)3 + …]
= 3 − x − 16 x2 − 1
18 x3 + … b let x = 0.05 8.7 ≈ 3 − 0.05 − 1
6 (0.05)2 − 118 (0.05)3
= 2.949 576 (7sf) 4 a = 1 + ( 1
3 )(6x) + 1 23 3( )( )
2− (6x)2 +
51 23 3 3( )( )( )
3 2− −
× (6x)3 + … = 1 + 2x − 4x2 + 40
3 x3 + …
b when x = 0.004, (1 + 136 )x ≈ 1 + 2(0.004) − 4(0.004)2 + 40
3 (0.004)3 = 1.007 936 853
(1 + 130.024) = 3 1.024 = 512 23
1000× = 4
53 2
∴ 3 2 = 54 × 1.007 936 853 = 1.259 921 (7sf)
5 a = 1 + (−3)(2x) + ( 3)( 4)
2− − (2x)2 + ( 3)( 4)( 5)
3 2− − −
× (2x)3 + … = 1 − 6x + 24x2 − 80x3 + …, | 2x | < 1 ∴ valid for | x | < 1
2 b = (1 + 3x)(1 + 2x)−3 = (1 + 3x)(1 − 6x + 24x2 − 80x3 + …) = 1 − 6x + 24x2 − 80x3 + 3x − 18x2 + 72x3 + … = 1 − 3x + 6x2 − 8x3 + … 6 2
4 2x
x+−
= (2 + x)(4 − 122 )x − = (2 + x) ×
124− (1 −
121
2 )x −
= (2 + x) × 12 [1 + ( 1
2− )( 12− x) +
312 2( )( )
2− − ( 1
2− x)2 + …] = (2 + x)( 1
2 + 18 x + 3
64 x2 + …) ∴ coeff of x2 = (2 × 3
64 ) + (1 × 18 ) = 7
32
C4 SERIES Answers - Worksheet B page 2
Solomon Press
7 a 2
2 111 5 4
xx x−
− + ≡
1A
x− +
1 4B
x−
2 − 11x ≡ A(1 − 4x) + B(1 − x) x = 1 ⇒ −9 = −3A ⇒ A = 3 x = 1
4 ⇒ 34− = 3
4 B ⇒ B = −1
b 22 11
1 5 4x
x x−
− + ≡ 3
1 x− − 1
1 4x−
31 x−
= 3(1 − x)−1 = 3[1 + (−1)(−x) + ( 1)( 2)2
− − (−x)2 + ( 1)( 2)( 3)3 2
− − −× (−x)3 + …]
= 3 + 3x + 3x2 + 3x3 + …, | −x | < 1 ∴ | x | < 1 1
1 4x− = (1 − 4x)−1 = 1 + (−1)(−4x) + ( 1)( 2)
2− − (−4x)2 + ( 1)( 2)( 3)
3 2− − −
× (−4x)3 + …
= 1 + 4x + 16x2 + 64x3 + …, | −4x | < 1 ∴ | x | < 14
∴ 22 11
1 5 4x
x x−
− + = (3 + 3x + 3x2 + 3x3 + …) − (1 + 4x + 16x2 + 64x3 + …)
= 2 − x − 13x2 − 61x3 + …, valid for | x | < 14
8 a 2
4 17(1 2 )(1 3 )
xx x−
+ − ≡
1 2A
x+ +
1 3B
x− + 2(1 3 )
Cx−
4 − 17x ≡ A(1 − 3x)2 + B(1 + 2x)(1 − 3x) + C(1 + 2x) x = 1
2− ⇒ 252 = 25
4 A ⇒ A = 2 x = 1
3 ⇒ 53− = 5
3 C ⇒ C = −1 coeffs of x2 ⇒ 0 = 9A − 6B ⇒ B = 3 ∴ f(x) ≡ 2
1 2x+ + 3
1 3x− − 2
1(1 3 )x−
b 21 2x+
= 2(1 + 2x)−1 = 2[1 + (−1)(2x) + ( 1)( 2)2
− − (2x)2 + ( 1)( 2)( 3)3 2
− − −× (2x)3 + …]
= 2 − 4x + 8x2 − 16x3 + … 3
1 3x− = 3(1 − 3x)−1 = 3[1 + (−1)(−3x) + ( 1)( 2)
2− − (−3x)2 + ( 1)( 2)( 3)
3 2− − −
× (−3x)3 + …]
= 3 + 9x + 27x2 + 81x3 + … 2
1(1 3 )x−
= (1 − 3x)−2 = 1 + (−2)(−3x) + ( 2)( 3)2
− − (−3x)2 + ( 2)( 3)( 4)3 2
− − −× (−3x)3 + …
= 1 + 6x + 27x2 + 108x3 + … f(x) = (2 − 4x + 8x2 − 16x3 + …) + (3 + 9x + 27x2 + 81x3 + …) − (1 + 6x + 27x2 + 108x3 + …) = 4 − x + 8x2 − 43x3 + … 9 a (1 + ax)b = 1 + b(ax) + ( 1)
2b b− (ax)2 + …
∴ ab = −6 (1) and 1
2 a2b(b − 1) = 24 (2)
(1) ⇒ a = − 6b
sub. (2) ⇒ 18b
(b − 1) = 24
18b − 18 = 24b b = −3 a = 2 b = ( 3)( 4)( 5)
3 2− − −
× (2)3 = −80
Solomon Press
SERIES C4 Answers - Worksheet C
1 a = 1 + ( 1
2 )(−4x) + 1 12 2( )( )
2− (−4x)2 +
31 12 2 2( )( )( )
3 2− −
× (−4x)3 + … = 1 − 2x − 2x2 − 4x3 + …, | −4x | < 1 ∴ valid for | x | < 1
4
b when x = 0.01, (1 − 124 )x ≈ 1 − 2(0.01) − 2(0.01)2 − 4(0.01)3
= 1 − 0.02 − 0.0002 − 0.000 004 = 0.979 796
(1 − 120.04) = 0.96 = 16 6
100× = 2
5 6
∴ 6 ≈ 52 × 0.979 796 = 2.44949 (6sf)
2 a 2
41 2 3x x+ −
≡ 4(1 3 )(1 )x x+ −
≡ 1 3
Ax+
+ 1
Bx−
4 ≡ A(1 − x) + B(1 + 3x) x = 1
3− ⇒ 4 = 43 A ⇒ A = 3
x = 1 ⇒ 4 = 4B ⇒ B = 1 ∴ f(x) = 3
1 3x+ + 1
1 x−
b 31 3x+
= 3(1 + 3x)−1 = 3[1 + (−1)(3x) + ( 1)( 2)2
− − (3x)2 + ( 1)( 2)( 3)3 2
− − −× (3x)3 + …]
= 3 − 9x + 27x2 − 81x3 + …, | 3x | < 1 ∴ valid for | x | < 13
11 x−
= (1 − x)−1 = 1 + (−1)(−x) + ( 1)( 2)2
− − (−x)2 + ( 1)( 2)( 3)3 2
− − −× (−x)3 + …
= 1 + x + x2 + x3 + …, | −x | < 1 ∴ valid for | x | < 1 ∴ f(x) = (3 − 9x + 27x2 − 81x3 + …) + (1 + x + x2 + x3 + …) = 4 − 8x + 28x2 − 80x3 + …, valid for | x | < 1
3 3 a = 2−2(1 − 1
2 x)−2 = 14 (1 − 1
2 x)−2
= 14 [1 + (−2)(− 1
2 x) + ( 2)( 3)2
− − (− 12 x)2 + ( 2)( 3)( 4)
3 2− − −
× (− 12 x)3 + …]
= 14 + 1
4 x + 316 x2 + 1
8 x3 + …
b 23
(2 )xx
−−
= (3 − x)(2 − x)−2 = (3 − x)( 14 + 1
4 x + 316 x2 + 1
8 x3 + …)
∴ coefficient of x3 = (3 × 18 ) + (−1 × 3
16 ) = 316
4 a f( 1
10 ) = 1
15
41+
= 1615
4 = 415
4 = 4 × 154 = 15
b = 4(1 + 122
3 )x − = 4[1 + ( 12− )( 2
3 x) + 31
2 2( )( )2
− − ( 23 x)2 + …]
= 4 − 43 x + 2
3 x2 + …
c 15 = f( 110 ) ≈ 4 − 4
3 × 110 + 2
3 ×( 110 )2 + …
= 4 − 215 + 1
150 = 1311503
d 15 = 3.872 98… 131
1503 = 3.873 33… 55
633 = 3.873 01…
∴ 15 < 55633 < 131
1503 , so 55633 is a more accurate approximation
C4 SERIES Answers - Worksheet C page 2
Solomon Press
5 a = 1 + ( 1
3 )(−x) + 1 23 3( )( )
2− (−x)2 + …
= 1 − 13 x − 1
9 x2 + …
b when x = 10−3, (1 − 13)x ≈ 1 − 1
3 (10−3) − 19 (10−3)2
= 0.999 666 555 6
(1 − 13310 )− = 3 0.999 = 27 373
1000× = 3
103 37
∴ 3 37 ≈ 103 × 0.999 666 555 6 = 3.332 221 85 (9sf)
6 a p =
3 25 5( )( )
2− (5)2 = −3
q = 3 725 5 5( )( )( )
3 2− −
× (5)3 = 7 b let x = 0.02
35(1.1) ≈ 1 + 3(0.02) − 3(0.02)2 + 7(0.02)3
= 1 + 0.06 − 0.0012 + 0.000 056 = 1.058 856 c
35(1.1) = 1.058 852 853…
% error = 1.058856 1.0588528531.058852853
− × 100% = 0.000 297 % (3sf) 7 a 8 − 6x2 ≡ A(2 + x)2 + B(1 + x)(2 + x) + C(1 + x) x = −1 ⇒ A = 2 x = −2 ⇒ −16 = −C ⇒ C = 16 coeffs of x2 ⇒ −6 = A + B ⇒ B = −8
b 2
28 6
(1 )(2 )x
x x−
+ + ≡ 2
1 x+ − 8
2 x+ + 2
16(2 )x+
21 x+
= 2(1 + x)−1 = 2[1 + (−1)x + ( 1)( 2)2
− − x2 + ( 1)( 2)( 3)3 2
− − −× x3 + …]
= 2 − 2x + 2x2 − 2x3 + … 8
2 x+ = 8(2 + x)−1 = 8 × 2−1(1 + 1
2 x)−1 = 4(1 + 12 x)−1
= 4[1 + (−1)( 12 x) + ( 1)( 2)
2− − ( 1
2 x)2 + ( 1)( 2)( 3)3 2
− − −× ( 1
2 x)3 + …] = 4 − 2x + x2 − 1
2 x3 + …
216
(2 )x+ = 16(2 + x)−2 = 16 × 2−2(1 + 1
2 x)−2 = 4(1 + 12 x)−2
= 4[1 + (−2)( 12 x) + ( 2)( 3)
2− − ( 1
2 x)2 + ( 2)( 3)( 4)3 2
− − −× ( 1
2 x)3 + …] = 4 − 4x + 3x2 − 2x3 + …
∴ 2
28 6
(1 )(2 )x
x x−
+ += (2 − 2x + 2x2 − 2x3 + …) − (4 − 2x + x2 − 1
2 x3 + …) + (4 − 4x + 3x2 − 2x3 + …)
= 2 − 4x + 4x2 − 72 x3 + …
C4 SERIES Answers - Worksheet C page 3
Solomon Press
8 a = 1 + ( 1
2 )(−2x) + 1 12 2( )( )
2− (−2x)2 + …
= 1 − x − 12 x2 + …
b when x = 0.0008, (1 − 122 )x ≈ 1 − 0.0008 − 1
2 (0.0008)2 = 1 − 0.0008 − 0.000 000 32 = 0.999 199 68
(1 − 120.0016) = 0.9984 = 256 39
10000× = 4
25 39
∴ 39 ≈ 254 × 0.999 199 68 = 6.244 998 (7sf)
9 a = 1 + ( 1
3 )(8x) + 1 23 3( )( )
2− (8x)2 + …
= 1 + 83 x − 64
9 x2 + …
b k = 531.08 = 5003
108 = 125327 = 5
3
c let x = 0.01, 3 1.08 = 1 + 83 (0.01) − 64
9 (0.01)2 = 1.025 955 556 ∴ 3 5 = 5
3 × 1.025 955 556 = 1.710 (4sf) 10 a f(x) ≡ 6
( 1)( 3)x
x x− − ≡
1A
x − +
3B
x −
6x ≡ A(x − 3) + B(x − 1) x = 1 ⇒ 6 = −2A ⇒ A = −3 x = 3 ⇒ 18 = 2B ⇒ B = 9 f(x) ≡ 9
3x − − 3
1x −
b f(x) = 31 x−
− 93 x−
31 x−
= 3(1 − x)−1 = 3[1 + (−1)(−x) + ( 1)( 2)2
− − (−x)2 + ( 1)( 2)( 3)3 2
− − −× (−x)3 + …]
= 3 + 3x + 3x2 + 3x3 + … 9
3 x− = 9(3 − x)−1 = 9 × 3−1(1 − 1
3 x)−1 = 3(1 − 13 x)−1
= 3[1 + (−1)(− 13 x) + ( 1)( 2)
2− − (− 1
3 x)2 + ( 1)( 2)( 3)3 2
− − −× (− 1
3 x)3 + …] = 3 + x + 1
3 x2 + 19 x3 + …
∴ f(x) = (3 + 3x + 3x2 + 3x3 + …) − (3 + x + 13 x2 + 1
9 x3 + …) = 2x + 8
3 x2 + 269 x3 + …
∴ for small x, f(x) ≈ 2x + 83 x2 + 26
9 x3
C4 SERIES Answers - Worksheet C page 4
Solomon Press
11 a =
1 12 21
44 (1 )x+ = 2(1 + 121
4 )x = 2[1 + ( 12 )( 1
4 x) + 1 12 2( )( )
2− ( 1
4 x)2 + …] = 2 + 1
4 x − 164 x2 + …, | 1
4 x | < 1 ∴ valid for | x | < 4
b when x = 120 , (4 +
12)x ≈ 2 + 1
4 ( 120 ) − 1
64 ( 120 )2
= 2.012 460 938
(4 + 121
20) = 8120 = 81 5
100× = 9
10 5
∴ 5 ≈ 109 × 2.012 460 938 = 2.236 067 71 (9sf)
c 5 = 2.236 067 977… ∴ estimate is accurate to 7 significant figures 12 a = 1 + ( 1
2− )(2x) + 31
2 2( )( )2
− − (2x)2 + 3 51
2 2 2( )( )( )3 2
− − −× (2x)3 + …
= 1 − x + 32 x2 − 5
2 x3 + …
b 2 51 2
xx
−+
= (2 − 5x)(1 + 122 )x − = (2 − 5x)( 1 − x + 3
2 x2 − 52 x3 + …)
= 2 − 2x + 3x2 − 5x3 − 5x + 5x2 − 152 x3 + …
= 2 − 7x + 8x2 − 252 x3 + …
∴ for small x, 2 51 2
xx
−+
= 2 − 7x + 8x2 − 252 x3
c 2 − 5x = 3 × 1 2x+ = 3 6x+ (2 − 5x)2 = 3 + 6x 4 − 20x + 25x2 = 3 + 6x 25x2 − 26x + 1 = 0 (25x − 1)(x − 1) = 0 x = 1
25 , 1 d let x = 1
25
3 ≈ 2 − 7( 125 ) + 8( 1
25 )2 − 252 ( 1
25 )3 = 1.732 13 a = 1 + (−1)x + ( 1)( 2)
2− − x2 + ( 1)( 2)( 3)
3 2− − −
× x3 + … = 1 − x + x2 − x3 + … b = 1 − bx + b2x2 − b3x3 + … c 1
1axbx
++
= (1 + ax)(1 + bx)−1 = (1 + ax)(1 − bx + b2x2 − b3x3 + …)
= 1 − bx + b2x2 − b3x3 + ax − abx2 + ab2x3 + … = 1 + (a − b)x + (b2 − ab)x2 + (ab2 − b3)x3 + … ∴ a − b = −4 (1) and b2 − ab = 12 (2) (1) ⇒ a = b − 4 sub. (2) b2 − b(b − 4) = 12 4b = 12 b = 3, a = −1 d = ab2 − b3 = −9 − 27 = −36
Solomon Press
PARTIAL FRACTIONS C4 Answers - Worksheet A
1 a x = −4 ⇒ −12 = −6A ⇒ A = 2 b x = −2 ⇒ −5 = −5A ⇒ A = 1 x = 2 ⇒ −6 = 6B ⇒ B = −1 x = 1
2 ⇒ 10 = 52 B ⇒ B = 4
2 a 2 ≡ A(x + 3) + B(x + 1) b x − 3 ≡ A(x − 1) + Bx x = −1 ⇒ 2 = 2A ⇒ A = 1 x = 0 ⇒ −3 = −A ⇒ A = 3 x = −3 ⇒ 2 = −2B ⇒ B = −1 x = 1 ⇒ B = −2 c x + 1 ≡ A(x − 5) + B(x − 3) d x + 10 ≡ A(2 − x) + B(1 + x) x = 3 ⇒ 4 = −2A ⇒ A = −2 x = −1 ⇒ 9 = 3A ⇒ A = 3 x = 5 ⇒ 6 = 2B ⇒ B = 3 x = 2 ⇒ 12 = 3B ⇒ B = 4
e 4 1( 2)( 1)
xx x
−+ −
≡ 2
Ax +
+ 1
Bx −
f 9( 1)( 3)
xx x
−− −
≡ 1
Ax −
+ 3
Bx −
4x − 1 ≡ A(x − 1) + B(x + 2) x − 9 ≡ A(x − 3) + B(x − 1) x = −2 ⇒ −9 = −3A ⇒ A = 3 x = 1 ⇒ −8 = −2A ⇒ A = 4 x = 1 ⇒ 3 = 3B ⇒ B = 1 x = 3 ⇒ −6 = 2B ⇒ B = −3
3 a 8
( 1)( 3)x x− + ≡
1A
x − +
3B
x + b 1
( 2)( 3)x
x x−
+ + ≡
2A
x + +
3B
x +
8 ≡ A(x + 3) + B(x − 1) x − 1 ≡ A(x + 3) + B(x + 2) x = 1 ⇒ 8 = 4A ⇒ A = 2 x = −2 ⇒ A = −3 x = −3 ⇒ 8 = −4B ⇒ B = −2 x = −3 ⇒ −4 = −B ⇒ B = 4 ∴ 8
( 1)( 3)x x− + ≡ 2
1x − − 2
3x + ∴ 1
( 2)( 3)x
x x−
+ + ≡ 4
3x + − 3
2x +
c 10( 4)( 1)
xx x+ −
≡ 4
Ax +
+ 1
Bx −
d 5 7( 1)x
x x++
≡ Ax
+ 1
Bx +
10x ≡ A(x − 1) + B(x + 4) 5x + 7 ≡ A(x + 1) + Bx x = −4 ⇒ −40 = −5A ⇒ A = 8 x = 0 ⇒ A = 7 x = 1 ⇒ 10 = 5B ⇒ B = 2 x = −1 ⇒ 2 = −B ⇒ B = −2 ∴ 10
( 4)( 1)x
x x+ − ≡ 8
4x + + 2
1x − ∴ 2
5 7xx x
++
≡ 7x
− 21x +
e 2( 1)( 4)
xx x
+− −
≡ 1
Ax −
+ 4
Bx −
f 4 6( 3)( 3)
xx x
++ −
≡ 3
Ax +
+ 3
Bx −
x + 2 ≡ A(x − 4) + B(x − 1) 4x + 6 ≡ A(x − 3) + B(x + 3) x = 1 ⇒ 3 = −3A ⇒ A = −1 x = −3 ⇒ −6 = −6A ⇒ A = 1 x = 4 ⇒ 6 = 3B ⇒ B = 2 x = 3 ⇒ 18 = 6B ⇒ B = 3 ∴ 2
25 4
xx x
+− +
≡ 24x −
− 11x −
∴ 24 6
9x
x+−
≡ 13x +
+ 33x −
g 3 2( 6)( 4)
xx x
+− +
≡ 6
Ax −
+ 4
Bx +
h 38(4 )(3 )
xx x
−+ −
≡ 4
Ax+
+ 3
Bx−
3x + 2 ≡ A(x + 4) + B(x − 6) 38 − x ≡ A(3 − x) + B(4 + x) x = 6 ⇒ 20 = 10A ⇒ A = 2 x = −4 ⇒ 42 = 7A ⇒ A = 6 x = −4 ⇒ −10 = −10B ⇒ B = 1 x = 3 ⇒ 35 = 7B ⇒ B = 5 ∴ 2
3 22 24x
x x+
− − ≡ 2
6x − + 1
4x + ∴ 2
3812
xx x−
− − ≡ 6
4 x+ + 5
3 x−
C4 PARTIAL FRACTIONS Answers - Worksheet A page 2
Solomon Press
i 4 5
(2 1)( 3)x
x x−
+ − ≡
2 1A
x + +
3B
x − j 1 3
(3 4)(2 1)x
x x−
+ + ≡
3 4A
x + +
2 1B
x +
4x − 5 ≡ A(x − 3) + B(2x + 1) 1 − 3x ≡ A(2x + 1) + B(3x + 4) x = 1
2− ⇒ −7 = 72− A ⇒ A = 2 x = 4
3− ⇒ 5 = 53− A ⇒ A = −3
x = 3 ⇒ 7 = 7B ⇒ B = 1 x = 12− ⇒ 5
2 = 52 B ⇒ B = 1
∴ 4 5(2 1)( 3)
xx x
−+ −
≡ 22 1x +
+ 13x −
∴ 1 3(3 4)(2 1)
xx x
−+ +
≡ 12 1x +
− 33 4x +
k 1(1 3 )x
x x+−
≡ Ax
+ 1 3
Bx−
l 5(2 1)( 2)x x− +
≡ 2 1
Ax −
+ 2
Bx +
x + 1 ≡ A(1 − 3x) + Bx 5 ≡ A(x + 2) + B(2x − 1) x = 0 ⇒ A = 1 x = 1
2 ⇒ 5 = 52 A ⇒ A = 2
x = 13 ⇒ 4
3 = 13 B ⇒ B = 4 x = −2 ⇒ 5 = −5B ⇒ B = −1
∴ 21
3x
x x+
− ≡ 1
x + 4
1 3x− ∴ 2
52 3 2x x+ −
≡ 22 1x −
− 12x +
m 2 10(4 1)(2 3)
xx x
+− +
≡ 4 1
Ax −
+ 2 3
Bx +
n 3 7( 1)( 3)
xx x
−+ −
≡ 1
Ax +
+ 3
Bx −
2x + 10 ≡ A(2x + 3) + B(4x − 1) 3x − 7 ≡ A(x − 3) + B(x + 1) x = 1
4 ⇒ 212 = 7
2 A ⇒ A = 3 x = −1 ⇒ −10 = −4A ⇒ A = 52
x = 32− ⇒ 7 = −7B ⇒ B = −1 x = 3 ⇒ 2 = 4B ⇒ B = 1
2
∴ 22( 5)
8 10 3x
x x+
+ − ≡ 3
4 1x − − 1
2 3x + ∴ 2
3 72 3
xx x
−− −
≡ 52( 1)x +
+ 12( 3)x −
o 1 3(1 )(1 2 )
xx x−
+ − ≡
1A
x+ +
1 2B
x−
1 − 3x ≡ A(1 − 2x) + B(1 + x) x = −1 ⇒ 4 = 3A ⇒ A = 4
3 x = 1
2 ⇒ 12− = 3
2 B ⇒ B = 13−
∴ 21 3
1 2x
x x−
− − ≡ 4
3(1 )x+ − 1
3(1 2 )x−
4 a x = 4 ⇒ 84 = 21A ⇒ A = 4 x = −3 ⇒ −56 = 28B ⇒ B = −2 x = 1 ⇒ −12 = −12C ⇒ C = 1
b x = 13 ⇒ 20
3 = 209− A ⇒ A = −3
x = 2 ⇒ 30 = 15B ⇒ B = 2 x = −1 ⇒ −12 = 12C ⇒ C = −1
c x = −5 ⇒ 32 = 16A ⇒ A = 2 x = −1 ⇒ 12 = 4C ⇒ C = 3 coeffs of x2 ⇒ 1 = A + B ⇒ B = −1
d x = 3 ⇒ 196 = 49A ⇒ A = 4 x = 1
2− ⇒ 21 = 72− C ⇒ C = −6
coeffs of x2 ⇒ 20 = 4A + 2B ⇒ B = 2
C4 PARTIAL FRACTIONS Answers - Worksheet A page 3
Solomon Press
5 a 8x + 14 ≡ A(x + 1)(x + 3) + B(x − 2)(x + 3) + C(x − 2)(x + 1) x = 2 ⇒ 30 = 15A ⇒ A = 2 x = −1 ⇒ 6 = −6B ⇒ B = −1 x = −3 ⇒ −10 = 10C ⇒ C = −1
b 2x2 − 6x + 20 ≡ A(x + 2)(x − 6) + B(x + 1)(x − 6) + C(x + 1)(x + 2) x = −1 ⇒ 28 = −7A ⇒ A = −4 x = −2 ⇒ 40 = 8B ⇒ B = 5 x = 6 ⇒ 56 = 56C ⇒ C = 1
c 9x − 14 ≡ A(x − 1)2 + B(x + 4)(x − 1) + C(x + 4) x = −4 ⇒ −50 = 25A ⇒ A = −2 x = 1 ⇒ −5 = 5C ⇒ C = −1 coeffs of x2 ⇒ 0 = A + B ⇒ B = 2
d 3x2 − 7x − 4 ≡ A(x − 2)2 + B(x − 3)(x − 2) + C(x − 3) x = 3 ⇒ A = 2 x = 2 ⇒ −6 = −C ⇒ C = 6 coeffs of x2 ⇒ 3 = A + B ⇒ B = 1 6 a
22 4( 1)( 4)
xx x x
+− −
≡ Ax
+ 1
Bx −
+ 4
Cx −
2x2 + 4 ≡ A(x − 1)(x − 4) + Bx(x − 4) + Cx(x − 1) x = 0 ⇒ 4 = 4A ⇒ A = 1 x = 1 ⇒ 6 = −3B ⇒ B = −2
x = 4 ⇒ 36 = 12C ⇒ C = 3 ∴ 22 4
( 1)( 4)x
x x x+
− − ≡ 1
x − 2
1x − + 3
4x −
b 29
( 2)( 1)x x− + ≡
2A
x − +
1B
x + + 2( 1)
Cx +
9 ≡ A(x + 1)2 + B(x − 2)(x + 1) + C(x − 2) x = 2 ⇒ 9 = 9A ⇒ A = 1 x = −1 ⇒ 9 = −3C ⇒ C = −3 coeffs of x2 ⇒ 0 = A + B ⇒ B = −1 ∴ 2
9( 2)( 1)x x− +
≡ 12x −
− 11x +
− 23
( 1)x +
c 2 11 21
(2 1)( 2)( 3)x x
x x x+ −
+ − − ≡
2 1A
x + +
2B
x − +
3C
x −
x2 + 11x − 21 ≡ A(x − 2)(x − 3) + B(2x + 1)(x − 3) + C(2x + 1)(x − 2) x = 1
2− ⇒ 1054− = 35
4 A ⇒ A = −3 x = 2 ⇒ 5 = −5B ⇒ B = −1
x = 3 ⇒ 21 = 7C ⇒ C = 3 ∴ 2 11 21
(2 1)( 2)( 3)x x
x x x+ −
+ − − ≡ 3
3x − − 3
2 1x + − 1
2x −
d 210 9
( 4)( 3)x
x x+
− + ≡
4A
x − +
3B
x + + 2( 3)
Cx +
10x + 9 ≡ A(x + 3)2 + B(x − 4)(x + 3) + C(x − 4) x = 4 ⇒ 49 = 49A ⇒ A = 1 x = −3 ⇒ −21 = −7C ⇒ C = 3 coeffs of x2 ⇒ 0 = A + B ⇒ B = −1 ∴ 2
10 9( 4)( 3)
xx x
+− +
≡ 14x −
− 13x +
+ 23
( 3)x +
C4 PARTIAL FRACTIONS Answers - Worksheet A page 4
Solomon Press
e 2
24 5
( 1)( 2)x x
x x+ +
+ + ≡
1A
x + +
2B
x + + 2( 2)
Cx +
x2 + 4x + 5 ≡ A(x + 2)2 + B(x + 1)(x + 2) + C(x + 1) x = −1 ⇒ A = 2 x = −2 ⇒ 1 = −C ⇒ C = −1
coeffs of x2 ⇒ 1 = A + B ⇒ B = −1 ∴ 2
24 5
( 1)( 2)x x
x x+ +
+ + ≡ 2
1x + − 1
2x + − 2
1( 2)x +
f 16 2( 3)( 2)( 2)
xx x x
−− + −
≡ 3
Ax −
+ 2
Bx +
+ 2
Cx −
16 − 2x ≡ A(x + 2)(x − 2) + B(x − 3)(x − 2) + C(x − 3)(x + 2) x = 3 ⇒ 10 = 5A ⇒ A = 2 x = −2 ⇒ 20 = 20B ⇒ B = 1 x = 2 ⇒ 12 = −4C ⇒ C = −3 ∴ 2
16 2( 3)( 4)
xx x
−− −
≡ 23x −
+ 12x +
− 32x −
g 22 9
( 3)(2 1)x
x x−
− − ≡
3A
x − +
2 1B
x − + 2(2 1)
Cx −
2 − 9x ≡ A(2x − 1)2 + B(x − 3)(2x − 1) + C(x − 3) x = 3 ⇒ −25 = 25A ⇒ A = −1 x = 1
2 ⇒ 52− = 5
2− C ⇒ C = 1
coeffs of x2 ⇒ 0 = 4A + 2B ⇒ B = 2 ∴ 22 9
( 3)(2 1)x
x x−
− − ≡ 2
2 1x − + 2
1(2 1)x −
− 13x −
h 2
23 24 4( 1)( 4)
x xx x+ −+ −
≡ 1
Ax +
+ 4
Bx −
+ 2( 4)C
x −
3 + 24x − 4x2 ≡ A(x − 4)2 + B(x + 1)(x − 4) + C(x + 1) x = −1 ⇒ −25 = 25A ⇒ A = −1 x = 4 ⇒ 35 = 5C ⇒ C = 7
coeffs of x2 ⇒ −4 = A + B ⇒ B = −3 ∴ 2
23 24 4( 1)( 4)
x xx x+ −+ −
≡ 27
( 4)x − − 3
4x − − 1
1x +
i 29 2 12
( 3)( 2)x x
x x x− −
+ − ≡ A
x +
3B
x + +
2C
x −
9x2 − 2x − 12 ≡ A(x + 3)(x − 2) + Bx(x − 2) + Cx(x + 3) x = 0 ⇒ −12 = −6A ⇒ A = 2 x = −3 ⇒ 75 = 15B ⇒ B = 5
x = 2 ⇒ 20 = 10C ⇒ C = 2 ∴ 2
3 29 2 12
6x x
x x x− −
+ − ≡ 2
x + 5
3x + + 2
2x −
j 2
25 3 20
( 4)x xx x
+ −+
≡ Ax
+ 2Bx
+ 4
Cx +
5x2 + 3x − 20 ≡ Ax(x + 4) + B(x + 4) + Cx2
x = 0 ⇒ −20 = 4B ⇒ B = −5 x = −4 ⇒ 48 = 16C ⇒ C = 3
coeffs of x2 ⇒ 5 = A + C ⇒ A = 2 ∴ 2
3 25 3 20
4x x
x x+ −+
≡ 2x
− 25x
+ 34x +
k 2
213 3
(2 3)( 1)x
x x−
+ − ≡
2 3A
x + +
1B
x − + 2( 1)
Cx −
13 − 3x2 ≡ A(x − 1)2 + B(2x + 3)(x − 1) + C(2x + 3) x = 3
2− ⇒ 254 = 25
4 A ⇒ A = 1 x = 1 ⇒ 10 = 5C ⇒ C = 2
coeffs of x2 ⇒ −3 = A + 2B ⇒ B = −2 ∴ 2
213 3
(2 3)( 1)x
x x−
+ − ≡ 1
2 3x + − 2
1x − + 2
2( 1)x −
C4 PARTIAL FRACTIONS Answers - Worksheet A page 5
Solomon Press
l 226
( 1)( 3)( 5)x x
x x x− −
− + + ≡
1A
x − +
3B
x + +
5C
x +
26 − x − x2 ≡ A(x + 3)(x + 5) + B(x − 1)(x + 5) + C(x − 1)(x + 3) x = 1 ⇒ 24 = 24A ⇒ A = 1 x = −3 ⇒ 20 = −8B ⇒ B = 5
2−
x = −5 ⇒ 6 = 12C ⇒ C = 12 ∴
226( 1)( 3)( 5)
x xx x x
− −− + +
≡ 11x −
− 52( 3)x +
+ 12( 5)x +
7 a x2 ≡ A(x − 2)(x − 6) + B(x − 6) + C(x − 2) x = 2 ⇒ 4 = −4B ⇒ B = −1 x = 6 ⇒ 36 = 4C ⇒ C = 9 coeffs of x2 ⇒ A = 1
b 2 2 9
( 1)( 5)x xx x
+ +− +
≡ A + 1
Bx −
+ 5
Cx +
x2 + 2x + 9 ≡ A(x − 1)(x + 5) + B(x + 5) + C(x − 1) x = 1 ⇒ 12 = 6B ⇒ B = 2 x = −5 ⇒ 24 = −6C ⇒ C = −4 coeffs of x2 ⇒ A = 1
8 a
quotient: x + 3 remainder: −x + 4
b 3 2
24 2
2x xx x
+ −+ −
≡ x + 3 + 24
2x
x x−
+ −
4( 2)( 1)
xx x
−+ −
≡ 2
Ax +
+ 1
Bx −
4 − x ≡ A(x − 1) + B(x + 2) x = −2 ⇒ 6 = −3A ⇒ A = −2
x = 1 ⇒ 3 = 3B ⇒ B = 1 ∴ 3 2
24 2
2x xx x
+ −+ −
≡ x + 3 − 22x +
+ 11x −
9 a (x − 3)(x + 1) = x2 − 2x − 3
∴ 2 3
( 3)( 1)x
x x+
− + ≡ 1 + 2 6
( 3)( 1)x
x x+
− +
2 6( 3)( 1)
xx x
+− +
≡ 3
Ax −
+ 1
Bx +
2x + 6 ≡ A(x + 1) + B(x − 3) x = 3 ⇒ 12 = 4A ⇒ A = 3
x = −1 ⇒ 4 = −4B ⇒ B = −1 ∴ 2 3
( 3)( 1)x
x x+
− + ≡ 1 + 3
3x − − 1
1x +
x + 3 x2 + x − 2 x3 + 4x2 + 0x − 2
x3 + x2 − 2x 3x2 + 2x − 2 3x2 + 3x − 6 − x + 4
1 x2 − 2x − 3 x2 + 0x + 3
x2 − 2x − 3 2x + 6
C4 PARTIAL FRACTIONS Answers - Worksheet A page 6
Solomon Press
b
∴ 3 2
23 2
4x x x
x− − +
− ≡ x − 3 + 2
3 104
xx
−−
3 10( 2)( 2)
xx x
−+ −
≡ 2
Ax +
+ 2
Bx −
3x − 10 ≡ A(x − 2) + B(x + 2) x = −2 ⇒ −16 = −4A ⇒ A = 4
x = 2 ⇒ −4 = 4B ⇒ B = −1 ∴ 3 2
23 2
4x x x
x− − +
− ≡ x − 3 + 4
2x + − 1
2x −
c
∴ 2
22 7
6 8x x
x x+
+ + ≡ 2 + 2
5 166 8
xx x− −
+ +
5 16( 2)( 4)
xx x
− −+ +
≡ 2
Ax +
+ 4
Bx +
−5x − 16 ≡ A(x + 4) + B(x + 2) x = −2 ⇒ −6 = 2A ⇒ A = −3
x = −4 ⇒ 4 = −2B ⇒ B = −2 ∴ 2
22 7
6 8x x
x x+
+ + ≡ 2 − 3
2x + − 2
4x +
d 3(x + 1)(x − 1) = 3x2 − 3 (x − 4)(x + 5) = x2 + x − 20
∴ 3( 1)( 1)
( 4)( 5)x x
x x+ −
− + ≡ 3 + 57 3
( 4)( 5)x
x x−
− +
57 3( 4)( 5)
xx x
−− +
≡ 4
Ax −
+ 5
Bx +
57 − 3x ≡ A(x + 5) + B(x − 4) x = 4 ⇒ 45 = 9A ⇒ A = 5 x = −5 ⇒ 72 = −9B ⇒ B = −8 ∴ 3( 1)( 1)
( 4)( 5)x x
x x+ −
− + ≡ 3 + 5
4x − − 8
5x +
x − 3 x2 − 4 x3 − 3x2 − x + 2
x3 + 0x2 − 4x − 3x2 + 3x + 2 − 3x2 + 0x + 12 3x − 10
2 x2 + 6x + 8 2x2 + 7x + 0
2x2 + 12x + 16 − 5x − 16
3 x2 + x − 20 3x2 + 0x − 3
3x2 + 3x − 60 − 3x + 57
C4 PARTIAL FRACTIONS Answers - Worksheet A page 7
Solomon Press
e
∴ 3 2
23 7 4
4 3x xx x
+ ++ +
≡ 3x − 5 + 211 19
4 3x
x x+
+ +
11 19( 1)( 3)
xx x
++ +
≡ 1
Ax +
+ 3
Bx +
11x + 19 ≡ A(x + 3) + B(x + 1) x = −1 ⇒ 8 = 2A ⇒ A = 4
x = −3 ⇒ −14 = −2B ⇒ B = 7 ∴ 3 2
23 7 4
4 3x xx x
+ ++ +
≡ 3x − 5 + 41x +
+ 73x +
f
∴ 2
24 7 52 7 3
x xx x
− +− +
≡ 2 + 27 1
2 7 3x
x x−
− +
7 1(2 1)( 3)
xx x
−− −
≡ 2 1
Ax −
+ 3
Bx −
7x − 1 ≡ A(x − 3) + B(2x − 1) x = 1
2 ⇒ 52 = 5
2− A ⇒ A = −1
x = 3 ⇒ 20 = 5B ⇒ B = 4 ∴ 2
24 7 52 7 3
x xx x
− +− +
≡ 2 − 12 1x −
+ 43x −
g
∴ 2
222 3x
x x− − ≡ 2 + 2
4 62 3
xx x
+− −
4 6( 1)( 3)
xx x
++ −
≡ 1
Ax +
+ 3
Bx −
4x + 6 ≡ A(x − 3) + B(x + 1) x = −1 ⇒ 2 = −4A ⇒ A = 1
2−
x = 3 ⇒ 18 = 4B ⇒ B = 92 ∴
2
222 3x
x x− − ≡ 2 − 1
2( 1)x + + 9
2( 3)x −
3x − 5 x2 + 4x + 3 3x3 + 7x2 + 0x + 4
3x3 + 12x2 + 9x − 5x2 − 9x + 4 − 5x2 − 20x − 15 11x + 19
2 2x2 − 7x + 3 4x2 − 7x + 5
2x2 − 14x + 6 7x − 1
2 x2 − 2x − 3 2x2 + 0x + 0
2x2 − 4x − 6 4x + 6
C4 PARTIAL FRACTIONS Answers - Worksheet A page 8
Solomon Press
h
∴ 3 2
26 6 1
6 5x x x
x x− + +
− + ≡ x + 2
16 5
xx x
+− +
1( 1)( 5)
xx x
+− −
≡ 1
Ax −
+ 5
Bx −
x + 1 ≡ A(x − 5) + B(x − 1) x = 1 ⇒ 2 = −4A ⇒ A = 1
2−
x = 5 ⇒ 6 = 4B ⇒ B = 32 ∴
3 2
26 6 1
6 5x x x
x x− + +
− + ≡ x − 1
2( 1)x − + 3
2( 5)x −
i
∴ 3
29 27 23 4 4x xx x
− −− −
≡ 3x + 4 + 214
3 4 4x
x x+
− −
14(3 2)( 2)
xx x
++ −
≡ 3 2
Ax +
+ 2
Bx −
x + 14 ≡ A(x − 2) + B(3x + 2) x = 2
3− ⇒ 403 = 8
3− A ⇒ A = −5
x = 2 ⇒ 16 = 8B ⇒ B = 2 ∴ 3
29 27 23 4 4x xx x
− −− −
≡ 3x + 4 − 53 2x +
+ 22x −
10 a 5
( 1)(2 1)x
x x+
− + ≡
1A
x − +
2 1B
x + 11 a x(4x + 5) ≡ A(x + 2)2 + B(x − 1)(x + 2) + C(x − 1)
x + 5 ≡ A(2x + 1) + B(x − 1) x = 1 ⇒ 9 = 9A ⇒ A = 1 x = 1 ⇒ 6 = 3A ⇒ A = 2 x = −2 ⇒ 6 = −3C ⇒ C = −2 x = 1
2− ⇒ 92 = 3
2− B ⇒ B = −3 coeffs x2 ⇒ 4 = A + B ⇒ B = 3
∴ f(x) = 21x −
− 32 1x +
b x = −1 ∴ y = 12
b f ′(x) = −2(x − 1)−2 + 3(2x + 1)−2 × 2 f(x) = (x − 1)−1 + 3(x + 2)−1 − 2(x + 2)−2 = 2
6(2 1)x +
− 22
( 1)x − f ′(x) = −(x − 1)−2 − 3(x + 2)−2 + 4(x + 2)−3
SP: 26
(2 1)x + − 2
2( 1)x −
= 0 grad = 14− − 3 + 4 = 3
4
6(x − 1)2 − 2(2x + 1)2 = 0 ∴ y − 12 = 3
4 (x + 1) x2 + 10x − 2 = 0 4y − 2 = 3x + 3
x = 10 100 82
− ± + 3x − 4y + 5 = 0
x = −5 ± 3 3
x x2 − 6x + 5 x3 − 6x2 + 6x + 1
x3 − 6x2 + 5x x + 1
3x + 4 3x2 − 4x − 4 9x3 + 0x2 − 27x − 2
9x3 − 12x2 − 12x 12x2 − 15x − 2 12x2 − 16x − 16 x + 14
Solomon Press
PARTIAL FRACTIONS C4 Answers - Worksheet B
1 22 ≡ A(x + 4) + B(2x − 3) x = 3
2 ⇒ 22 = 112 A ⇒ A = 4
x = −4 ⇒ 22 = −11B ⇒ B = −2 2 x + 5 ≡ A(x − 3)2 + B(x + 1)(x − 3) + C(x + 1) x = −1 ⇒ 4 = 16A ⇒ A = 1
4 x = 3 ⇒ 8 = 4C ⇒ C = 2 coeffs of x2 ⇒ 0 = A + B ⇒ B = 1
4− 3 4x2 − 16x − 7 ≡ A(2x − 1)(x − 4) + B(x − 4) + C(2x − 1) x = 4 ⇒ −7 = 7C ⇒ C = −1 x = 1
2 ⇒ −14 = 72− B ⇒ B = 4
coeffs of x2 ⇒ 4 = 2A ⇒ A = 2 4 a f(1) = 3 + 11 + 8 − 4 = 18 f(−1) = −3 + 11 − 8 − 4 = −4 f(2) = 24 + 44 + 16 − 4 = 80 f(−2) = −24 + 44 − 16 − 4 = 0 ∴ (x + 2) is a factor ∴ f(x) = (x + 2)(3x2 + 5x − 2) = (3x − 1)(x + 2)2
b 16f ( )
xx
+ ≡ 3 1
Ax −
+ 2
Bx +
+ 2( 2)C
x +
x + 16 ≡ A(x + 2)2 + B(3x − 1)(x + 2) + C(3x − 1) x = 1
3 ⇒ 493 = 49
9 A ⇒ A = 3 x = −2 ⇒ 14 = −7C ⇒ C = −2 coeffs of x2 ⇒ 0 = A + 3B ⇒ B = −1 ∴ 16
f ( )x
x+ ≡ 3
3 1x − − 1
2x + − 2
2( 2)x +
5 2
1(2 1)x x −
≡ Ax
+ 2 1
Bx −
+ 2(2 1)C
x −
1 ≡ A(2x − 1)2 + Bx(2x − 1) + Cx x = 0 ⇒ A = 1 x = 1
2 ⇒ 1 = 12 C ⇒ C = 2
coeffs of x2 ⇒ 0 = 4A + 2B ⇒ B = −2 ∴ f(x) = 1
x − 2
2 1x − + 2
2(2 1)x −
3x2 + 5x − 2 x + 2 3x3 + 11x2 + 8x − 4
3x3 + 6x2 5x2 + 8x 5x2 + 10x − 2x − 4 − 2x − 4
C4 PARTIAL FRACTIONS Answers - Worksheet B page 2
Solomon Press
6 ∴ f(x) ≡ x − 2 + 2
2 17 10x
x x+
+ +
2 1( 2)( 5)
xx x
++ +
≡ 2
Bx +
+ 5
Cx +
2x + 1 ≡ B(x + 5) + C(x + 2) x = −2 ⇒ −3 = 3B ⇒ B = −1 x = −5 ⇒ −9 = −3C ⇒ C = 3 ∴ f(x) ≡ x − 2 − 1
2x + + 3
5x + [ A = −2, B = −1, C = 3 ]
7 a 4
( 1)( 1)x x+ − ≡
1A
x + +
1B
x −
4 ≡ A(x − 1) + B(x + 1) x = −1 ⇒ 4 = −2A ⇒ A = −2 x = 1 ⇒ 4 = 2B ⇒ B = 2 ∴ f(x) ≡ 2
1x − − 2
1x +
b 22 5
( 4)( 2)( 1)x x
x x x+ −
− − − ≡
4A
x − +
2B
x − +
1C
x −
2 + 5x − x2 ≡ A(x − 2)(x − 1) + B(x − 4)(x − 1) + C(x − 4)(x − 2) x = 4 ⇒ 6 = 6A ⇒ A = 1 x = 2 ⇒ 8 = −2B ⇒ B = −4 x = 1 ⇒ 6 = 3C ⇒ C = 2 ∴ g(x) ≡ 1
4x − − 4
2x − + 2
1x −
c 21x −
− 21x +
= 14x −
− 42x −
+ 21x −
42x −
− 14x −
− 21x +
= 0
4( 4)( 1) ( 2)( 1) 2( 2)( 4)( 2)( 4)( 1)
x x x x x xx x x
− + − − + − − −− − +
= 0
4(x2 − 3x − 4) − (x2 − x − 2) − 2(x2 − 6x + 8) = 0 x2 + x − 30 = 0 (x + 6)(x − 5) = 0 x = −6, 5
x − 2 x2 + 7x + 10 x3 + 5x2 − 2x − 19
x3 + 7x2 + 10x − 2x2 − 12x − 19 − 2x2 − 14x − 20 2x + 1
Solomon Press
VECTORS C4 Answers - Worksheet A
1 a) −p b) 2q c) 1
2 p d) p e) −q f) p + q
g) 12 p + 2q h) p − q i) 2q − p j) −p − 2q k) 1
2 p − q l) 12− p − 2q
2 a) u + v b) w − u c) u + v − w
3 a) q b) p + r c) r − q d) p + q + r e) −q − r f) q + r − p
4 a i = (a + 2b) + (a − 2b) = 2a ii = (a + 2b) − (a − 2b) = 4b b OA = 1
2 OS , OB = 14 TR
∴ R S A
B
O T
5 a i = 1
2 a ii = b − a iii = 1
2 (b − a) iv = a + 1
2 (b − a) = 12 (a + b)
v = 12 (a + b) − 1
2 a = 12 b
b they are parallel (and the magnitude of CD is half that of OB )
6 a parallel, 3p = 3
2 (2p) b not parallel c parallel, (p − 1
3 q) = 13 (3p − q)
d parallel, (4q − 2p) = −2(p − 2q) e parallel, (6p + 8q) = 8( 3
4 p + q) f not parallel
7 a = (2m + 3n) − (4m + 2n) = n − 2m b OM = 1
2 OC = m + 32 n
AM = (m + 32 n) − 4m = 3
2 n − 3m
∴ AM = 32 BC
∴ AM is parallel to BC
C4 VECTORS Answers - Worksheet A page 2
Solomon Press
8 a OM = 1
2 OA = 3u − 2v
AB = (3u − v) − (6u − 4v) = 3v − 3u ON = OA + 1
3 AB = (6u − 4v) + 13 (3v − 3u) = 5u − 3v
b CM = (3u − 2v) − (v − 3u) = 6u − 3v CN = (5u − 3v) − (v − 3u) = 8u − 4v ∴ CN = 4
3 CM
∴ CN and CM are parallel common point C ∴ C, M and N are collinear
9 a a = 5, b = 3 b 2 + b = 0 and a − 4 = 0 ∴ a = 4, b = −2
c −1 = b and 4a = −2 d 2a + 6 = 0 and b − a = 0 ∴ a = 1
2− , b = −1 ∴ a = −3, b = −3
10 a OC = 1
2 a
CB = b − 12 a
OD = 12 a + 1
2 ( b − 12 a) = 1
4 a + 12 b
b AB = b − a OE = OA + k AB ∴ OE = a + k(b − a) c OE = l OD ∴ a + k(b − a) = l( 1
4 a + 12 b)
∴ 1 − k = 14 l
and k = 12 l
adding 1 = 34 l
l = 43
∴ OE = 43 ( 1
4 a + 12 b) = 1
3 a + 23 b
d k = 12 l = 2
3
∴ AE = 23 AB
∴ AE : EB = 2 : 1
Solomon Press
VECTORS C4 Answers - Worksheet B
1 a 6i + j b −4i + 2j c −6i d 10i − 2j 2 a = 4(i − 3j) b = (4i + 2j) − (i − 3j) = 4i − 12j = 3i + 5j
c = 2(i − 3j) + 3(4i + 2j) d = 4(i − 3j) − 2(4i + 2j) = 14i = −4i − 16j 3 a = 9 16+ = 5 b = 2 1 4+ = 2 5
c p + 2q = 34
−
+ 2 12
= 50
d 3q − 2p = 3 12
− 2 34
−
= 314−
| p + 2q | = 5 | 3q − 2p | = 9 196+ = 205 = 14.3 (3sf) 4 a = tan−1 1
2 = 26.6° b = tan−1 3 = 71.6°
c 5p + q = 5(2i + j) + (i − 3j) = 11i + 2j d p − 3q = (2i + j) − 3(i − 3j) = −i + 10j angle = tan−1 2
11 = 10.3° angle = 180° − tan−1 10 = 95.7°
5 a 43
= 16 9+ = 5 b 724
−
= 49 576+ = 25
∴ 15
43
∴ 125
724
−
c 11−
= 1 1+ = 2 d 24
= 4 16+ = 20 = 2 5
∴ 12
11−
= 12 2 1
1−
∴ 12 5
24
= 15 5 1
2
6 a | 5i + 12j | = 25 144+ = 13 ∴ 26
13 (5i + 12j) = 10i + 24j
b | −6i − 8j | = 36 64+ = 10 ∴ 15
10 (−6i − 8j) = −9i − 12j
c | 2i − 4j | = 4 16+ = 20 = 2 5 ∴ 5
2 5(2i − 4j) = 5 (i − 2j)
7 a (2i + λj) + (µi − 5j) = 3i − j b 2(2i + λj) − (µi − 5j) = −3i + 8j 2 + µ = 3 and λ − 5 = −1 4 − µ = −3 and 2λ + 5 = 8 ∴ λ = 4, µ = 1 ∴ λ = 3
2 , µ = 7 8 a 6i + cj = 3(2i + j) b 6i + cj = 2
3− (−9i − 6j) ∴ c = 3 ∴ c = 4
c 36 + c2 = 102 = 100 d 36 + c2 = ( 53 )2 = 45 ∴ c2 = 64 ∴ c2 = 9 c > 0 ∴ c = 8 c > 0 ∴ c = 3
C4 VECTORS Answers - Worksheet B page 2
Solomon Press
9 a a(i + 3j) + b(4i − 2j) = −5i + 13j ∴ a + 4b = −5 (1) and 3a − 2b = 13 (2) (1) + 2×(2) ⇒ 7a = 21 ∴ a = 3, b = −2 b c(i + 3j) + (4i − 2j) = kj ∴ c + 4 = 0 ∴ c = −4 c (i + 3j) + d(4i − 2j) = k(3i − j) ∴ 1 + 4d = 3k and 3 − 2d = −k (1) + 2×(2) ⇒ 7 = k ∴ d = 5
10 a AB = 52
−
− 36
= 84
− −
b AB = 64 16+ = 80 = 4 5 c = OA + 1
2 AB
= 36
+ 12
84
− −
= 14−
d OC = AB
∴ pos. vector = 84
− −
11 a = 2 2 2(2 4) ( 3 0) (3 9)− + − − + − b = 2 2 2(7 11) ( 1 3) (3 5)− + − + + −
= 4 9 36+ + = 16 4 4+ + = 7 = 24 = 2 6 = 4.90 (3sf) 12 a = 16 4 16+ + b = 1 1 1+ + c = 64 1 16+ + d = 9 25 1+ + = 6 = 3 = 1.73 (3sf) = 9 = 35 = 5.92 (3sf) 13 a | 5i − 2j + 14k | = 25 4 196+ + = 15 ∴ 1
15 (5i − 2j + 14k)
b | 2i + 11j − 10k | = 4 121 100+ + = 15 ∴ 10
15 (2i + 11j − 10k) = 23 (2i + 11j − 10k)
c | −5i − 4j + 2k | = 25 16 4+ + = 45 = 3 5 ∴ 20
3 5(−5i − 4j + 2k) = 4
3 5 (−5i − 4j + 2k)
14 λ2 + 144 + 16 = 142 = 196 λ2 = 36 λ = ± 6
C4 VECTORS Answers - Worksheet B page 3
Solomon Press
15 a = 131
−
+ 242
1
−
= 91
1
−
b = 131
−
− 2
53
− −
= 32
2
−
c = 131
−
+ 42
1
−
+ 2
53
− −
= 363
−
d = 2131
−
− 342
1
−
+ 2
53
− −
= 12
178
− −
16 a −2i + λj + µk = 12− (4i + 2j − 8k) b −2i + λj + µk = 2
5 (−5i + 20j − 10k) ∴ λ = −1, µ = 4 ∴ λ = 8, µ = −4 17 a 2p − q = 2(i − 2j + 4k) − (−i + 2j + 2k) 18 a AB = (−4i + j + 8k) − (−2i + 7j + 4k) = 3i − 6j + 6k = −2i − 6j + 4k ∴ | 2p − q | = 9 36 36+ + = 9 pos. vec of mid-point = OA + 1
2 AB b (i − 2j + 4k) + k(−i + 2j + 2k) = (−2i + 7j + 4k) + 1
2 (−2i − 6j + 4k) = l(2i − 4j − 7k) = −3i + 4j + 6k ∴ 1 − k = 2l (1) b AC = (6i − 5j) − (−2i + 7j + 4k) −2 + 2k = −4l (2) = 8i − 12j − 4k 4 + 2k = −7l (3) AD = OA + 3
4 AC [(1) and (2) are the same equation] = (−2i + 7j + 4k) + 3
4 (8i − 12j − 4k) (2) − (3) ⇒ −6 = 3l = 4i − 2j + k ∴ l = −2 ∴ k = 5
19 a (λi − 2λj + µk) = k(2i − 4j − 3k) 20 a BC = 618
−
− 12
74
− −
= 6
84
− −
∴ λ = 2k (1) OM = OB + 12 BC
−2λ = −4k (2) = 12
74
− −
+ 12
684
− −
= 936
− −
µ = −3k (3) b OM = 32 OA
[(1) and (2) are the same equation] ∴ OM and OA are parallel 3×(1) + 2×(3) ⇒ 3λ + 2µ = 0 common point O b λ2 + (−2λ)2 + µ2 = ( 292 )2 ∴ O, A and M are collinear 5λ2 + µ2 = 116 µ = 3
2− λ ⇒ 5λ2 + 94 λ2 = 116
λ2 = 16 λ = ± 4 µ = 3
2− λ and µ > 0 ∴ λ = −4, µ = 6 21 a d2 = (9 − t)2 + (1 + 2t)2 + (5 − t)2 = 81 − 18t + t2 + 1 + 4t + 4t2 + 25 − 10t + t2 = 6t2 − 24t + 107 b d2 = 6(t2 − 4t) + 107 = 6[(t − 2)2 − 4] + 107 = 6(t − 2)2 + 83 ∴ closest when t = 2 min. d = 83 = 9.11 m (3sf)
Solomon Press
VECTORS C4 Answers - Worksheet C
Note: For this worksheet especially, there may be alternative correct answers
1 a y b y c y 2
O 2 x
O x −1 O x
d y e y f r = i − 2j + s(2i + 3j)
3 y
O 9 x O x O 7
3 x
72−
2 a r = −i + j + s(3i − 2j)
b r = 4j + si
c r = 3i − j + s(i + 5j) 3 a dirn = 3
1
− 10
= 21
b dirn = 11−
− 34
−
= 23
−
c dirn = 23
−
− 22
−
= 45
−
∴ r = 10
+ s 21
∴ r = 34
−
+ s 23
−
∴ r = 22
−
+ s 45
−
4 a −1 + 2λ = 5 ∴ λ = 3 3 + cλ = 3 + 3c = 0 ∴ c = −1
b ci + 2j = k(6i + 3j) ∴ k = 2
3 ∴ c = 4 5 a r = −i + sj b r = s(i + 2j) c r = j + s(i + 3j)
d r = −2j + s(4i + 3j) e r = 5j + s(2i − j) f y = 14 x + 2
∴ r = 2j + s(4i + j) 6 a x = 2 + 3λ, y = 1 + 2λ
b λ = 23
x − = 12
y −
2(x − 2) = 3(y − 1) 2x − 4 = 3y − 3 2x − 3y − 1 = 0
C4 VECTORS Answers - Worksheet C page 2
Solomon Press
7 a x = 3 + λ, y = 2λ b x = 1 + 3λ, y = 4 + λ c x = 4λ, y = 2 − λ λ = x − 3 =
2y λ = 1
3x − = y − 4 λ =
4x = 2 − y
2(x − 3) = y x − 1 = 3(y − 4) x = 4(2 − y) 2x − y − 6 = 0 x − 3y + 11 = 0 x + 4y − 8 = 0 d x = −2 + 5λ, y = 1 + 2λ e x = 2 − 3λ, y = −3 + 4λ f x = λ + 3, y = −2λ − 1 λ = 2
5x + = 1
2y − λ = 2
3x −−
= 34
y + λ = x − 3 = 12
y +−
2(x + 2) = 5(y − 1) 4(x − 2) = −3(y + 3) −2(x − 3) = y + 1 2x − 5y + 9 = 0 4x + 3y + 1 = 0 2x + y − 5 = 0 8 a 3
1 −
= 12− 6
2−
b 14
≠ k 41
c 24
= 23
36
∴ parallel ∴ not parallel ∴ parallel (1, 2) lies on first line (2, −5) lies on first line when x = 1 on second line when x = 2 on second line −2 − 6t = 1 ⇒ t = 1
2− −1 + 3t = 2 ⇒ t = 1 ⇒ y = 3 + 2( 1
2− ) = 2 ⇒ y = 1 + 6(1) = 7 parallel and common point ∴ (2, −5) not on second line ∴ identical ∴ parallel but not identical 9 a 1 + λ = 2 + 3µ (1) b 4 − λ = 5 + 2µ (1) c 2λ = 2 − µ (1) 2 = 1 + µ (2) 1 + λ = −2 − 3µ (2) 1 − λ = 10 + 3µ (2) (2) ⇒ µ = 1 (1) + (2) ⇒ 5 = 3 − µ (1) + 2×(2) ⇒ 2 = 22 + 5µ ∴ 5i + 2j µ = −2 µ = −4 ∴ i + 4j ∴ 6i − 2j
d −1 − 4λ = 2 − µ (1) e −2 − 3λ = −3 + 5µ (1) f 1 + 3λ = 3 + µ (1) 5 + 6λ = −2 + 2µ (2) 11 + 4λ = −7 + 3µ (2) 2 + 2λ = 5 + 4µ (2) 2×(1) + (2) ⇒ 3 − 2λ = 2 4×(1) + 3×(2) 2×(1) − 3×(2) λ = 1
2 ⇒ 25 = −33 + 29µ ⇒ −4 = −9 − 10µ ∴ −3i + 8j µ = 2 µ = 1
2− ∴ 7i − j ∴ 5
2 i + 3j
10 a r = 4i + k + s(i + 3j − 2k)
b r = 2i + j + sk
c r = −i + 4j + 2k + s(2i − 3j + 5k) 11 a AB = (6i − 3j + k) − (5i + j − 2k) = i − 4j + 3k
b r = (5i + j − 2k) + s(i − 4j + 3k)
c 5 + s = 3 ⇒ s = −2 when s = −2, r = (5i + j − 2k) − 2(i − 4j + 3k) = 3i + 9j − 8k ∴ l passes through (3, 9, −8)
C4 VECTORS Answers - Worksheet C page 3
Solomon Press
12 a direction = (5i + 4j + 6k) − (i + 3j + 4k) b direction = (i + 5j + 2k) − (3i − 2k) = 4i + j + 2k = −2i + 5j + 4k ∴ r = i + 3j + 4k + s(4i + j + 2k) ∴ r = 3i − 2k + s(−2i + 5j + 4k)
c r = s(6i − j + 2k) d direction = (4i − 7j + k) − (−i − 2j + 3k) = 5i − 5j − 2k ∴ r = −i − 2j + 3k + s(5i − 5j − 2k) 13 a 3 + 2λ = 9 ∴ λ = 3 −5 + aλ = −5 + 3a = −2 ∴ a = 1 1 + bλ = 1 + 3b = −8 ∴ b = −3
b 2i + aj + bk= k(8i − 4j + 2k) ∴ k = 1
4 ∴ a = −1, b = 1
2
14 a x = 2 + 3λ, b x = 4 + λ, c x = −1 + 4λ, y = 3 + 5λ, y = −1 + 6λ, y = 5 − 2λ, z = 2λ, z = 3 + 3λ, z = −2 − λ, (λ = ) 2
3x − = 3
5y − =
2z (λ = ) x − 4 = 1
6y + = 3
3z − (λ = ) 1
4x + = 5
2y −−
= 21
z +−
15 a s = 1
3x − = 4
2y + = z − 5 b s =
4x = 1
2y −−
= 73
z + c s = 54
x +−
= y + 3 = z
x = 1 + 3s, x = 4s, x = −5 − 4s, y = −4 + 2s, y = 1 − 2s, y = −3 + s, z = 5 + s, z = −7 + 3s, z = s, r = i − 4j + 5k + s(3i + 2j + k) r = j − 7k + s(4i − 2j + 3k) r = −5i − 3j + s(−4i + j + k) 16 4 + s = 7 − 3t (1) −2s = 2 + 2t (2) 3 + 2s = −5 + t (3) (2) + (3) ⇒ 3 = −3 + 3t t = 2, s = −3 check (1) 4 + (−3) = 7 − 3(2) true ∴ intersect point of intersection: (1, 6, −3) 17 2 + λ = 1 + µ (1) −1 + λ = 4 − 2µ (2) 4 + 3λ = 3 + µ (3) (1) − (2) ⇒ 3 = −3 + 3µ µ = 2, λ = 1 check (3) 4 + 3(1) = 3 + (2) false ∴ do not intersect (i + j + 3k) ≠ k(i − 2j + k) ∴ not parallel ∴ skew
C4 VECTORS Answers - Worksheet C page 4
Solomon Press
18 a 3 + 4λ = 3 + µ (1) b 213
− −
= 12−
426
−
1 + λ = 2 (2) ∴ parallel 5 − λ = −4 + 2µ (3) (2) ⇒ λ = 1 sub. (1) µ = 4 check (3) 5 − (1) = −4 + 2(4) true ∴ intersect
position vector of intersection: 724
c 8 + λ = −2 + 4µ (1) d 1 + λ = 7 + 2µ (1) 2 + 3λ = 2 − 3µ (2) 5 + 4λ = −6 + µ (2) −4 − 2λ = 8 − 4µ (3) 2 − 2λ = −5 − 3µ (3) (1) + (3) ⇒ 4 − λ = 6 2×(1) + (3) ⇒ 4 = 9 + µ λ = −2, µ = 2 µ = −5, λ = −4 check (2) 2 + 3(−2) = 2 − 3(2) check (2) 5 + 4(−4) = −6 + (−5) true ∴ intersect true ∴ intersect
position vector of intersection: 64
0
−
position vector of intersection: 311
10
− −
e 4 + 2λ = 3 + 5µ (1) f 6λ = −12 + 5µ (1) −1 + 5λ = −2 − 3µ (2) 7 − 4λ = −1 + 2µ (2) 3 − 3λ = 1 − 4µ (3) −2 + 8λ = 11 − 3µ (3) 3×(1) + 2×(3) ⇒ 18 = 11 + 7µ 2×(2) + (3) ⇒ 12 = 9 + µ µ = 1, λ = 2 µ = 3, λ = 1
2 check (2) −1 + 5(2) = −2 − 3(1) check (1) 6( 1
2 ) = −12 + 5(3) false ∴ do not intersect true ∴ intersect
253
−
≠ k534
− −
position vector of intersection: 352
∴ skew
Solomon Press
VECTORS C4 Answers - Worksheet D
1 a = 3 + 2 = 5 b = 12 − 5 = 7 c = −5 + 4 = −1 2 (i + 4j).(8i − 2j) = 8 − 8 = 0 ∴ perpendicular 3 a 3
1 −
.3c
= 0 b 21
. 3c
= 0 c 25
−
.4
c −
= 0
3c − 3 = 0 6 + c = 0 2c + 20 = 0 c = 1 c = −6 c = −10 4 a 4i − 3j = 16 9+ = 5, 8i + 6j = 64 36+ = 10 (4i − 3j).(8i + 6j) = 32 − 18 = 14 ∴ angle = cos−1 14
5 10× = cos−1 725 = 73.7°
b 7i + j = 49 1+ = 5 2 , 2i + 6j = 4 36+ = 2 10 (7i + j).(2i + 6j) = 14 + 6 = 20 ∴ angle = cos−1 20
5 2 2 10× = cos−1 1
5 = 63.4°
c 4i + 2j = 16 4+ = 2 5 , −5i + 2j = 25 4+ = 29 (4i + 2j).(−5i + 2j) = −20 + 4 = −16 ∴ angle = cos−1 16
2 5 29−×
= cos−1 (− 85 29
) = 131.6°
5 BA = (9i + j) − (3i − j) = 6i + 2j BC = (5i − 2j) − (3i − j) = 2i − j BA = 36 4+ = 2 10 , BC = 4 1+ = 5 BA . BC = (6i + 2j).(2i − j) = 12 − 2 = 10 ∴ ∠ABC = cos−1 10
2 10 5× = cos−1 1
2 = 45°
6 a = 3 + 2 + 8 = 13 b = 6 + 6 − 2 = 10
c = −5 + 0 − 6 = −11 d = −3 + 22 + 32 = 51
e = 27 − 28 − 1 = −2 f = 0 + 9 + 0 = 9 7 a = (2i + j − 3k).(i + 5j − k) = 2 + 5 + 3 = 10
b = (2i + j − 3k).(6i − 2j − 3k) = 12 − 2 + 9 = 19
c q + r = (i + 5j − k) + (6i − 2j − 3k) = 7i + 3j − 4k p.(q + r) = (2i + j − 3k).(7i + 3j − 4k) = 14 + 3 + 12 = 29 p.q + p.r = 10 + 19 = 29 ∴ p.(q + r) = p.q + p.r 8 a = p.q + p.r + p.q − p.r b = p.q + p.r + q.r − q.p = 2p.q = p.q + p.r + q.r − p.q = p.r + q.r = (p + q).r
C4 VECTORS Answers - Worksheet D page 2
Solomon Press
9 (5i − 3j + 2k).(3i + j − 6k) = 15 − 3 − 12 = 0 ∴ perpendicular 10 BA = (3i + 4j − 6k) − (i + 5j − 2k) = 2i − j − 4k BC = (8i + 3j + 2k) − (i + 5j − 2k) = 7i − 2j + 4k BA . BC = (2i − j − 4k).(7i − 2j + 4k) = 14 + 2 − 16 = 0 ∴ BA and BC are perpendicular ∴ ∠ABC = 90° 11 a (2i + 3j + k).(ci − 3j + k) = 0 b (−5i + 3j + 2k).(ci − j + 3ck) = 0 2c − 9 + 1 = 0 −5c − 3 + 6c = 0 c = 4 c = 3
c (ci − 2j + 8k).(ci + cj − 3k) = 0 d (3ci + 2j + ck).(5i − 4j + 2ck) = 0 c2 − 2c − 24 = 0 15c − 8 + 2c2 = 0 (c + 4)(c − 6) = 0 2c2 + 15c − 8 = 0 c = −4, 6 (2c − 1)(c + 8) = 0 c = −8, 1
2
12 a 122
−
= 1 4 4+ + = 3, 814
−
= 64 1 16+ + = 9, 122
−
.814
−
= 8 + 2 + 8 = 18
∴ cos θ = 183 9× = 2
3
b 412
−
= 16 1 4+ + = 21 , 2
36
− −
= 4 9 36+ + = 7, 412
−
.2
36
− −
= −8 + 3 + 12 = 7
∴ cos θ = 721 7×
= 121
or 121 21
c 121
−
= 1 4 1+ + = 6 , 17
2
−
= 1 49 4+ + = 3 6 , 121
−
.17
2
−
= 1 − 14 − 2 = −15
∴ cos θ = 156 3 6−×
= 56−
d 53
4
−
= 25 9 16+ + = 5 2 , 341
− −
= 9 16 1+ + = 26 , 53
4
−
.341
− −
= 15 + 12 − 4 = 23
∴ cos θ = 235 2 26×
= 2310 13
or 23130 13
13 a (3i − 4k) = 9 16+ = 5 b (2i − 6j + 3k) = 4 36 9+ + = 7 (7i − 4j + 4k) = 49 16 16+ + = 9 (i − 3j − k) = 1 9 1+ + = 11 (3i − 4k).(7i − 4j + 4k) = 21 + 0 − 16 = 5 (2i − 6j + 3k).(i − 3j − k) = 2 + 18 − 3 = 17 ∴ angle = cos−1 5
5 9× = cos−1 19 = 83.6° ∴ angle = cos−1 17
7 11× = 42.9°
c (6i − 2j − 9k) = 36 4 81+ + = 11 d (i + 5j − 3k) = 1 25 9+ + = 35 (3i + j + 4k) = 9 1 16+ + = 26 (−3i − 4j + 2k) = 9 16 4+ + = 29 (6i − 2j − 9k).(3i + j + 4k) = 18 − 2 − 36 = −20 (i + 5j − 3k).(−3i − 4j + 2k) = −3 − 20 − 6 = −29 ∴ angle = cos−1 20
11 26−×
= 110.9° ∴ angle = cos−1 2935 29
−×
= cos−1 ( 2935− ) = 155.5°
C4 VECTORS Answers - Worksheet D page 3
Solomon Press
14 a BA = (7 + 1)i + (2 − 6)j + (−2 + 3)k = 8i − 4j + k BC = (−3 + 1)i + (1 − 6)j + (2 + 3)k = −2i − 5j + 5k b BA = 64 16 1+ + = 9 BC = 4 25 25+ + = 3 6 BA . BC = (8i − 4j + k).(−2i − 5j + 5k) = −16 + 20 + 5 = 9 ∴ ∠ABC = cos−1 9
9 3 6× = cos−1 1
3 6 = 82.2°
c = 12 × 9 × 3 6 × sin 82.18° = 32.8
15 a AB = (4i + 3j − 2k) − (3i − 2j − k) = i + 5j − k AC = (2i − j) − (3i − 2j − k) = −i + j + k AB = 1 25 1+ + = 3 3 AC = 1 1 1+ + = 3 AB . AC = (i + 5j − k).(−i + j + k) = −1 + 5 − 1 = 3 cos (∠BAC) = 3
3 3 3× = 1
3
b sin2 (∠BAC) = 1 − ( 13 )2 = 8
9
sin (∠BAC) = 89 = 2
3 2
area = 12 × 3 3 × 3 × 2
3 2 = 3 2
16 a 44
2
−
= 16 16 4+ + = 6, 806
−
= 64 36+ = 10, 44
2
−
.806
−
= 32 + 0 − 12 = 20
∴ acute angle = cos−1 206 10× = cos−1 1
3 = 70.5°
b 61
18
− −
= 36 1 324+ + = 19, 4123
−
= 16 144 9+ + = 13, 61
18
− −
.4123
−
= 24 + 12 − 54 = −18
∴ acute angle = cos−1 1819 13
−× = cos−1 18
247 = 85.8°
c 11
3
−
= 1 1 9+ + = 11 , 25
3
−
= 4 25 9+ + = 38 , 11
3
−
.25
3
−
= 2 + 5 + 9 = 16
∴ acute angle = cos−1 1611 38×
= cos−1 1611 38
= 38.5°
d 46
7
− −
= 16 36 49+ + = 101 , 518
− −
= 25 1 64+ + = 90 , 46
7
− −
.518
− −
= −20 + 6 − 56 = −70
∴ acute angle = cos−1 70101 90
−×
= cos−1 70101 90
= 42.8°
C4 VECTORS Answers - Worksheet D page 4
Solomon Press
17 a AB = (6i + 5j + k) − (5i + 8j − k) = i − 3j + 2k ∴ r = 5i + 8k − k + λ(i − 3j + 2k) b 5 + λ = 4 − 5µ (1) 8 − 3λ = −3 + µ (2) −1 + 2λ = 5 − 2µ (3) 3×(1) + (2) ⇒ 23 = 9 − 14µ µ = −1, λ = 4 check (3) −1 + 2(4) = 5 − 2(−1) true ∴ intersect pos. vector of int. = 9i − 4j + 7k c (i − 3j + 2k) = 1 9 4+ + = 14 (−5i + j − 2k) = 25 1 4+ + = 30 (i − 3j + 2k).(−5i + j − 2k) = −5 − 3 − 4 = −12 acute angle = cos−1 12
14 30−×
= cos−1 6105
= 54.2° (1dp)
18 λ = 2
3x − =
2y = 5
6z +−
µ = 44
x −−
= 17
y + = 34
z −−
x = 2 + 3λ, y = 2λ, z = −5 − 6λ x = 4 − 4µ, y = −1 + 7µ, z = 3 − 4µ r = 2i − 5k + λ(3i + 2j − 6k) r = 4i − j + 3k + µ(−4i + 7j − 4k) (3i + 2j − 6k) = 9 4 36+ + = 7 (−4i + 7j − 4k) = 16 49 16+ + = 9 (3i + 2j − 6k).(−4i + 7j − 4k) = −12 + 14 + 24 = 26 ∴ acute angle = cos−1 26
7 9× = cos−1 2663 = 65.6°
19 a 7 + 2λ = −4 + 5µ (1) −λ = 7 − 4µ (2) −2 + 2λ = −6 − 2µ (3) (1) − (3) ⇒ 9 = 2 + 7µ µ = 1 ∴ A (1, 3, −8) b (2i − j + 2k) = 4 1 4+ + = 3 (5i − 4j − 2k) = 25 16 4+ + = 3 5 (2i − j + 2k).(5i − 4j − 2k) = 10 + 4 − 4 = 10 acute angle = cos−1 10
3 3 5× = cos−1 ( 2
9 5 ) = 60.2° (1dp)
c 7 + 2λ = 5 ⇒ λ = −1 sub. λ = −1 in eqn for l r = 7i − 2k − (2i − j + 2k) = 5i + j − 4k ∴ B lies on l d AB = (5i + j − 4k) − (i + 3j − 8k) = 4i − 2j + 4k AB = 16 4 16+ + = 6 ∴ dist. of B from m = 6 sin 60.20° = 5.21 (3sf)
C4 VECTORS Answers - Worksheet D page 5
Solomon Press
20 a AB = (11i + 5j + k) − (9i + 6j) = 2i − j + k OC = 9i + 6j + λ(2i − j + k) = OA + λ AB ∴ C lies on l b [(9 + 2λ)i + (6 − λ)j + λk].[2i − j + k] = 0 2(9 + 2λ) − (6 − λ) + λ = 0 λ = −2 c sub. λ = −2 in OC ∴ 5i + 8j − 2k 21 a [(−4 + λ)i + (2 + 3λ)j + (7 − 4λ)k].[i + 3j − 4k] = 0 (−4 + λ) + 3(2 + 3λ) − 4(7 − 4λ) = 0 λ = 1 ∴ (−3, 5, 3) b [(7 + 6λ)i + (11 − 9λ)j + (−9 + 3λ)k].[6i − 9j + 3k] = 0 6(7 + 6λ) − 9(11 − 9λ) + 3(−9 + 3λ) = 0 λ = 2
3 ∴ (11, 5, −7)
Solomon Press
VECTORS C4 Answers - Worksheet E
1 a 1 + 2λ = 7 ∴ λ = 3 2 a AB =
506
−
− 164
= 46
10
− −
p − 3λ = −1 ∴ p = 8 ∴ r = 164
+ s46
10
− −
−4 + qλ = −1 ∴ q = 1 b 1 + 4s = 5 + t (1) b 2i + j − 3k = 4 1 9+ + = 14 6 − 6s = −5 − 4t (2) −4i + 5j − 2k = 16 25 4+ + = 45 4×(1) + (2) ⇒ 10 + 10s = 15 (2i + j − 3k).(−4i + 5j − 2k) s = 1
2
= −8 + 5 + 6 = 3 ∴ pos. vector of C = 331
−
θ = cos−1 314 45
= 83.1° (1dp) c pos. vector of mid-point of AB
= OA + 12 AB
= 164
+ 12
46
10
− −
= 331
−
∴ C is mid-point of AB
3 a PQ = (3i + j) − (5i − 2j + 2k) 4 a 5 + 2λ = 7 − µ (1) = −2i + 3j − 2k −λ = −3 + µ (2) ∴ r = 5i − 2j + 2k + λ(−2i + 3j − 2k) 1 + 2λ = 7 − 2µ (3) b 5 − 2λ = 4 + 5µ (1) (1) + (2) ⇒ 5 + λ = 4 −2 + 3λ = 6 − µ (2) λ = −1, µ = 4 2 − 2λ = −1 + 3µ (3) check (3) 1 + 2(−1) = 7 − 2(4) (1) − (3) ⇒ 3 = 5 + 2µ true ∴ intersect µ = −1, λ = 3 pos. vector of int. = 3i + j − k check (2) −2 + 3(3) = 6 − (−1) b diagonals bisect each other true ∴ intersect let M be point of intersection pos. vector of int. = −i + 7j − 4k ∴ AM = (3i + j − k) − (9i − 2j + 5k) c −2i + 3j − 2k = 4 9 4+ + = 17 = −6i + 3j − 6k 5i − j + 3k = 25 1 9+ + = 35 OC = OA + 2 AM (−2i + 3j − 2k).(5i − j + 3k) = (9i − 2j + 5k) + 2(−6i + 3j − 6k) = −10 − 3 − 6 = −19 = −3i + 4j − 7k θ = cos−1 19
17 35− = 38.8° c area of triangle ABC = 1
2 × 54 = 27
AC = 2(−6i + 3j − 6k) = 6(−2i + j − 2k) AC = 6 4 1 4+ + = 18 let distance of B from l1 = d ∴ 1
2 × 18 × d = 27 d = 3
C4 VECTORS Answers - Worksheet E page 2
Solomon Press
5 a AB = (2i − j + 2k) − (4i + 2j − 4k) 6 a AB = 418
−
− 51
10
− −
= 1
22
−
= −2i − 3j + 6k ∴ r = 51
10
− −
+ λ1
22
−
∴ r = 4i + 2j − 4k + λ(−2i − 3j + 6k) b 5 − λ = 0 ⇒ λ = 5 b r = 4i − 7j − k + µ(6j − 2k) sub. λ = 5 in l
c −7 + 6µ = 2 ⇒ µ = 32 r =
090
∴ C (0, 9, 0)
sub. µ = 32 in l2 c OD =
51 2
10 2
λλλ
− − + − +
r = 4i − 7j − k + 32 (6j − 2k) OD .
122
−
= 0
= 4i + 2j − 4k ∴ A lies on l2 −(5 − λ) + 2(−1 + 2λ) + 2(−10 + 2λ) = 0 d −2i − 3j + 6k = 4 9 36+ + = 7 9λ − 27 = 0
6j − 2k = 36 4+ = 40 λ = 3, OD = 254
−
(−2i − 3j + 6k).(6j − 2k) ∴ D (2, 5, −4) = 0 − 18 − 12 = −30 d OD = 4 25 16+ + = 45 = 3 5 θ = cos−1 30
7 40− = 47.3° (1dp) CD = 4 16 16+ + = 6
area = 12 × 6 × 3 5 = 9 5
C4 VECTORS Answers - Worksheet E page 3
Solomon Press
7 a −6 + 4s = 6 ⇒ s = 3 8 a AB = (4i + 6j + 2k) − (4i + 5j + 6k) sub. s = 3 in l1 = j − 4k
r = 162
− −
+ 3041
−
= 165
−
∴ r = 4i + 5j + 6k + λ(j − 4k)
∴ P (1, 6, −5) lies on l1 b 4 = 1 + µ (1) b 1 = 4 + 3t ⇒ t = −1 5 + λ = 5 + µ (2) sub. t = −1 in l2 6 − 4λ = −3 − µ (3)
r = 441
− −
− 32
2
−
(1) ⇒ µ = 3
OQ = 123
− −
sub. (2) ⇒ λ = 3
c PQ = 0 64 4+ + = 68 = 2 17 check (3) 6 − 4(3) = −3 − (3)
32
2
−
= 9 4 4+ + = 17 true ∴ intersect
∴ OR = OQ ± 232
2
−
pos. vector of int. = 4i + 8j − 6k
= 5
27
− −
or 76
1
−
c (j − 4k) = 1 16+ = 17
(i + j − k) = 1 1 1+ + = 3 (j − 4k).(i + j − k) = 0 + 1 + 4 = 5 θ = cos−1 5
3 17 = 45.6° (1dp)
d let closest point be C OC = (1 + µ)i + (5 + µ)j + (−3 − µ)k AC = OC − OA = (−3 + µ)i + µj + (−9 − µ)k AC must be perpendicular to l2 ∴ AC .(i + j − k) = 0 (−3 + µ) + µ − (−9 − µ) = 0 µ = −2 ∴ OC = −i + 3j − k
Solomon Press
VECTORS C4 Answers - Worksheet F
1 a AB =
034
−
− 215
− −
= 2
41
−
2 a BA = (−4i + 2j − k) − (2i + 5j − 7k)
r = 215
− −
+ λ2
41
−
= −6i − 3j + 6k
b 2 − 2λ = 6 + aµ (1) BC = (6i + 4j + k) − (2i + 5j − 7k) −1 + 4λ = −5 − 3µ (2) = 4i − j + 8k −5 + λ = 1 + µ (3) BA = 36 9 36+ + = 9 (2) + 3×(3) ⇒ −16 + 7λ = −2 BC = 16 1 64+ + = 9 λ = 2, µ = −4 BA . BC = −24 + 3 + 48 = 27 sub. (1) 2 − 2(2) = 6 + a(−4) cos (∠ABC) = 27
9 9× = 13
−2 = 6 − 4a b AC = (6i + 4j + k) − (−4i + 2j − k) a = 2 = 10i + 2j + 2k point of intersection: (−2, 7, −3) OM = OA + 1
2 AC = (−4i + 2j − k) + 1
2 (10i + 2j + 2k) = i + 3j c BM = (i + 3j) − (2i + 5j − 7k) = −i − 2j + 7k BM . AC = (−i − 2j + 7k).(10i + 2j + 2k) = −10 − 4 + 14 = 0 ∴ BM perpendicular to AC d ∠ABC = cos−1 1
3 = 70.529 isosceles triangle ∴ ∠ACB = 1
2 (180 − 70.529) = 54.7° (1dp)
C4 VECTORS Answers - Worksheet F page 2
Solomon Press
3 a AB = 1173
−
− 953
−
= 220
4 a (7i − 5j − k) = 49 25 1+ + = 5 3
∴ r = 953
−
+ λ220
(4i − 5j + 3k) = 16 25 9+ + = 5 2
b OC = 9 25 2
3
λλ
+ + −
(7i − 5j − k).(4i − 5j + 3k) = 28 + 25 − 3 = 50
OC .220
= 0 cos (∠AOB) = 505 3 5 2×
= 26
= 13 6
2(9 + 2λ) + 2(5 + 2λ) + 0 = 0 b AB = (4i − 5j + 3k) − (7i − 5j − k) 8λ + 28 = 0 = −3i + 4k
λ = 72− , OC =
223
− −
AB . OB = (−3i + 4k).(4i − 5j + 3k)
c OC = 4 4 9+ + = 17 = −12 + 0 + 12 = 0
AC = 72
220
= 7 1 1+ = 7 2 ∴ AB perpendicular to OB
area = 12 × 17 × 7 2 = 20.4 c OC = 3
2 (4i − 5j + 3k) d AC = 7
2 AB = 6i − 152 j + 9
2 k
∴ area OAB : area OAC = 2 : 7 AC = (6i − 152 j + 9
2 k) − (7i − 5j − k) = −i − 5
2 j + 112 k
AC . OA = (−i − 52 j + 11
2 k).(7i − 5j − k) = −7 + 25
2 − 112 = 0
∴ AC perpendicular to OA d ∠CAO = 90° ∴ ∠ACO = 90° − ∠AOC = 90° − ∠AOB = 90° − cos−1 ( 1
3 6 ) = 54.7°
Solomon Press
DIFFERENTIATION C4 Answers - Worksheet A
1 a (5, 2)
b 4t
= −8 ∴ t = 12−
2 a (2, 0)
b 1 + sin t = 32 , sin t = 1
2 , t = π6 , 5π6
2 cos t = 3− , cos t = − 32
, t = 5π6 , 7π
6
∴ t = 5π6
3 a t =
3x ∴ y = (
3x )2 b t =
2x ∴ y =
2
1( )x c x2 = t 6, y3 = 8t 6
y = 19 x2 y = 2
x ∴ y3 = 8x2
d t = 4 − y e t = 12 (x + 1) f t = 1
x + 1
∴ x = 1 − (4 − y)2 ∴ y = 214
2( 1)x +
∴ y = 11
2 ( 1)x− + =
11
1 x−
y = 28
( 1)x + y =
1x
x −
4 a t = 1
2 (x − 1) ∴ y = 1
4 (x − 1)2
b y
(0, 14 )
O (1, 0) x 5 a cos2 θ + sin2 θ = 1 b cos 2θ = 1 − 2 sin2 θ c cos θ = 3
2x − , sin θ = 1
2y −
∴ x2 + y2 = 1 ∴ y = 1 − 2x2 cos2 θ + sin2 θ = 1 ∴ ( 3
2x − )2 + ( 1
2y − )2 = 1
(x − 3)2 + (y − 1)2 = 4
d sec θ = 2x , tan θ =
4y e sin 2θ = 2 sin θ cos θ f sec θ = 1
x
1 + tan2 θ = sec2 θ ∴ y = 4 sin2 θ cos2 θ 1 + tan2 θ = sec2 θ ∴ 1 + (
4y )2 = (
2x )2 y = 4 sin2 θ (1 − sin2 θ) ∴ 1 + y = ( 1
x)2
16 + y2 = 4x2 y = 4x2(1 − x2) y = 21x
− 1
y2 = 4x2 − 16
C4 DIFFERENTIATION Answers - Worksheet A page 2
Solomon Press
6 a cos θ = 1
3x − , sin θ = 4
3y −
cos2 θ + sin2 θ = 1 ( 1
3x − )2 + ( 4
3y − )2 = 1
(x − 1)2 + (y − 4)2 = 9
b centre (1, 4) radius 3
c y
O x 7 a x = 5 cos θ, y = 5 sin θ, 0 ≤ θ < 2π
b x = 6 + 2 cos θ, y = −1 + 2 sin θ, 0 ≤ θ < 2π
c x = a + r cos θ, y = b + r sin θ, 0 ≤ θ < 2π 8 a t =
2x b sin θ = 1 − x, cos θ = 2 − y
∴ y = 4(2x )(
2x − 1) cos2 θ + sin2 θ = 1 ∴ (2 − y)2 + (1 − x)2 = 1
y = x(x − 2) or (x − 1)2 + (y − 2)2 = 1
y y (0, 2) O
(0, 0) (2, 0) x
O x c t = x + 3 d t = x − 1 ∴ y = 4 − (x + 3)2 ∴ y = 2
1x −
y y (−5, 0) (−1, 0) O x O
(0, −5) x (0, −2)
θ = 0
θ = π4
θ = π2
θ = 3π
4
θ = π
θ = 5π4
θ = 3π
2 θ = 7π
4
Solomon Press
DIFFERENTIATION C4 Answers - Worksheet B
1 a d
dxt
= 1, ddyt
= 2t
b ddyx
= ddyt
÷ ddxt
= 2t ÷ 1 = 2t
2 a d
dxt
= 2t, ddyt
= 3 b ddxt
= 2t, ddyt
= 6t2 + 2t c ddxt
= 2 cos t, ddyt
= −6 sin t
ddyx
= ddyt
÷ ddxt
= 32t
ddyx
= ddyt
÷ ddxt
=26 22
t tt+ = 3t + 1 d
dyx
= ddyt
÷ ddxt
= 6sin2cos
tt
− = −3 tan t
d ddxt
= 3, ddyt
= t−2 e ddxt
= −2 sin 2t, ddyt
= cos t f ddxt
= et + 1, ddyt
= 2e2t − 1
ddyx
= ddyt
÷ ddxt
= 2
3t−
ddyx
= ddyt
÷ ddxt
= cos2sin 2
tt−
ddyx
= ddyt
÷ ddxt
= 2 1
12ee
t
t
−
+
= 21
3t = cos
4sin cost
t t− = 1
4− cosec t = 2et − 2
g ddxt
= 2 sin t × cos t, h ddxt
= 3 sec t tan t, i ddxt
= −(t + 1)−2,
ddyt
= 3 cos2 t × (−sin t) ddyt
= 5 sec2 t ddyt
= 21 ( 1) 1
( 1)t tt
× − − ×−
= −(t − 1)−2
ddyx
= ddyt
÷ ddxt
= 23cos sin
2sin cost t
t t− d
dyx
= ddyt
÷ ddxt
= 25sec
3sec tant
t t d
dyx
= ddyt
÷ ddxt
= 2
2( 1)( 1)tt
−
−− −− +
= 32− cos t = 5sec
3tantt
= 53 cosec t =
2
2( 1)( 1)tt
+−
= 21
1tt
+ −
3 a t = 1 ∴ x = 1, y = 3 b t = 2 ∴ x = −3, y = 0 d
dxt
= 3t2, ddyt
= 6t ddxt
= −2t, ddyt
= 2 − 2t
ddyx
= ddyt
÷ ddxt
= 263
tt
= 2t
ddyx
= ddyt
÷ ddxt
= 2 22
tt
−−
= 1 − 1t
grad = 2 grad = 12
∴ y − 3 = 2(x − 1) ∴ y − 0 = 12 (x + 3)
y = 2x + 1 y = 12 x + 3
2 c t = π3 ∴ x = 3 , y = −1 d t = 3 ∴ x = 0, y = 4
ddxt
= 2 cos t, ddyt
= 4 sin t ddxt
= − 14 t−
= 14t −
, ddyt
= 2t
ddyx
= ddyt
÷ ddxt
= 4sin2cos
tt = 2 tan t d
dyx
= ddyt
÷ ddxt
= 1
4
2
t
t
−
= 2t(t − 4)
grad = 2 3 grad = −6 ∴ y + 1 = 2 3 (x − 3 ) ∴ y − 4 = −6(x − 0) y = 2 3 x − 7 y = 4 − 6x
C4 DIFFERENTIATION Answers - Worksheet B page 2
Solomon Press
4 θ = π3 ∴ x = 2, y = 2 3 5 a d
dxt
= −t−2, ddyt
= −(t + 2)−2
dd
xθ
= sec θ tan θ, dd
yθ
= 2 sec2 θ ddyx
= ddyt
÷ ddxt
= 2
2( 2)t
t
−
−− +
−
ddyx
= dd
yθ
÷ dd
xθ
= 22sec
sec tanθ
θ θ =
2
2( 2)t
t + =
2
2t
t +
= 2sectan
θθ
= 2 cosec θ b t = 2 ∴ x = 12 , y = 1
4
grad = 43
∴ grad of normal = − 34 grad = 1
4 ∴ grad of normal = −4
∴ y − 2 3 = − 34 (x − 2) ∴ y − 1
4 = −4(x − 12 )
4y − 8 3 = − 3 x + 2 3 4y − 1 = −16x + 8 3 x + 4y = 10 3 16x + 4y − 9 = 0 6 a d
dxt
= 2 cos 2t 7 a dd
xθ
= −3 sin θ, dd
yθ
= 4 cos θ
ddyt
= 2 sin t × cos t = sin 2t ddyx
= dd
yθ
÷ dd
xθ
= − 4cos3sin
θθ
ddyx
= ddyt
÷ ddxt
= sin 22cos2
tt
= 12 tan 2t at (3 cos α, 4 sin α), θ = α
b t = π6 ∴ x = 32 , y = 1
4 ∴ grad = − 4cos3sin
αα
grad = 32 ∴ y − 4 sin α = − 4cos
3sinαα
(x − 3 cos α)
∴ y − 14 = 3
2 (x − 32 ) 3y sin α − 12 sin2 α = −4x cos α + 12 cos2 α
[ 3 x − 2y − 1 = 0 ] 3y sin α + 4x cos α = 12(cos2 α + sin2 α) 3y sin α + 4x cos α = 12 b at ( 3
2− , 2 3 ), 3 cos α = 3
2− ⇒ cos α = 12−
4 sin α = 2 3 ⇒ sin α = 32
∴ 3y × 32 + 4x × ( 1
2− ) = 12
4x − 3 3 y + 24 = 0 8 a x = 0 ⇒ t2 = 0 ⇒ t = 0 9 a d
dxθ
= 2 sin θ, dd
yθ
= 3 cos θ
y = 0 ⇒ t(t − 2) = 0 ⇒ t = 0, 2 ddyx
= dd
yθ
÷ dd
xθ
= 3cos2sin
θθ
or 32 cot θ
∴ (0, 0), (4, 0) b i 3cos2sin
θθ
= 0 ∴ cos θ = 0
b i ddxt
= 2t, ddyt
= 2t − 2 θ = π2 , 3π2 ∴ (1, 3), (1, −3)
ddyx
= ddyt
÷ ddxt
= 2 22t
t− = 1 − t−1 ii 3cos
2sinθθ
→ ∞ ∴ sin θ = 0
t = 12x ∴ d
dyx
= 1 − 12x− θ = 0, π ∴ (−1, 0), (3, 0)
ii y = 12x (
12x − 2) = x −
122x
ddyx
= 1 − 12x−
C4 DIFFERENTIATION Answers - Worksheet B page 3
Solomon Press
10 a x = 0 ⇒ sin θ = 0 ⇒ θ = 0 11 a t = π4 , x = 1
2 , y = 1
y = 0 ⇒ sin 2θ = 0 ⇒ θ = 0, π2 ddxt
= 2 sin t cos t , ddyt
= sec2 t
∴ (0, 0), (1, 0) ddyx
= ddyt
÷ ddxt
= 2sec
2sin cost
t t = 1
2 sec3 t cosec t
b dd
xθ
= cos θ, dd
yθ
= 2 cos 2θ grad = 12 × ( 2 )3 × 2 = 2
ddyx
= dd
yθ
÷ dd
xθ
= 2cos2cos
θθ
∴ y − 1 = 2(x − 12 )
2cos2cos
θθ
= 0 ∴ cos 2θ = 0 y = 2x
θ = π4 ∴ y = 1 when x = 0, y = 0 c y = sin 2θ = 2 sin θ cos θ ∴ passes through origin
cos θ = ± 21 sin θ− b y2 = tan2 t = 2
2sincos
tt
= 2
2sin
1 sint
t−
0 ≤ θ ≤ π2 ∴ cos θ = 21 sin θ− ∴ y2 = 1
xx−
∴ y = 22 1x x−
12 a t = 3, x = 10
3 , y = 83
ddxt
= 1 − t−2, ddyt
= 1 + t−2
ddyx
= ddyt
÷ ddxt
= 2
211
tt
−
−+−
= 2
211
tt
+−
grad = 54
∴ y − 83 = 5
4 (x − 103 )
[ 5x − 4y − 6 = 0 ] b sub. parametric eqns into tangent 5(t + 1
t) − 4(t − 1
t) − 6 = 0
5(t2 + 1) − 4(t2 − 1) − 6t = 0 t2 − 6t + 9 = 0 (t − 3)2 = 0 ∴ t = 3 (at P), no other solutions ∴ does not intersect again c x2 = (t + 1
t)2 = t2 + 2 + 2
1t
(1)
y2 = (t − 1t
)2 = t2 − 2 + 21t
(2)
(1) − (2) ⇒ x2 − y2 = 4 [k = 4]
Solomon Press
DIFFERENTIATION C4 Answers - Worksheet C
1 a = 4 d
dyx
b = 3y2 ddyx
c = 2 ddyx
cos 2y d = 2
3e y × 2y ddyx
= 2
6 e yy ddyx
2 a 2x + 2y d
dyx
= 0 b 2 − ddyx
+ 2y ddyx
= 0
2y ddyx
= −2x 2 = ddyx
(1 − 2y)
ddyx
= − xy
ddyx
= 21 2y−
c 4y3 ddyx
= 2x − 6 d 2x + 2y ddyx
+ 3 − 4 ddyx
= 0
ddyx
= 33
2x
y− 2x + 3 = d
dyx
(4 − 2y)
ddyx
= 2 34 2
xy
+−
e 2x − 4y ddyx
+ 1 + 3 ddyx
= 0 f cos x − ddyx
sin y = 0
2x + 1 = ddyx
(4y − 3) cos x = ddyx
sin y
ddyx
= 2 14 3
xy
+−
ddyx
= cossin
xy
g 6e3x − 2e−2y ddyx
= 0 h sec2 x − 2 ddyx
cosec 2y cot 2y = 0
6e3x = 2e−2y ddyx
sec2 x = 2 ddyx
cosec 2y cot 2y
ddyx
= 3
23ee
x
y− = 3e3x + 2y ddyx
= 2sec
2cosec2 cot 2x
y y
i 12x −
= 22 1y +
ddyx
ddyx
= 2 12( 2)
yx
+−
3 a = 1 × y + x × d
dyx
b = 2x × y3 + x2 × 3y2 ddyx
= y + x ddyx
= 2xy3 + 3x2y2 ddyx
c = cos x × tan y + sin x × ddyx
sec2 y d = 3(x − 2y)2 × (1 − 2 ddyx
)
= cos x tan y + ddyx
sin x sec2 y = 3(x − 2y)2(1 − 2 ddyx
)
C4 DIFFERENTIATION Answers - Worksheet C page 2
Solomon Press
4 a 2x × y + x2 ddyx
= 0 b 2x + 3 × y + 3x × ddyx
− 2y ddyx
= 0
x2 ddyx
= −2xy 2x + 3y = ddyx
(2y − 3x)
ddyx
= − 2yx
ddyx
= 2 32 3
x yy x
+−
c 8x − 2 × y − 2x × ddyx
+ 6y ddyx
= 0 d −2 sin 2x × sec 3y + cos 2x × 3 ddyx
sec 3y tan 3y = 0
8x − 2y = ddyx
(2x − 6y) 3 ddyx
cos 2x sec 3y tan 3y = 2 sin 2x sec 3y
ddyx
= 43
x yx y
−−
ddyx
= 2sin 23cos2 tan 3
xx y
= 23 tan 2x cot 3y
e ddyx
= 2(x + y) × (1 + ddyx
) f 1 × ey + x × ey ddyx
− ddyx
= 0
ddyx
[1 − 2(x + y)] = 2(x + y) ey = ddyx
(1 − xey)
ddyx
= 2( )1 2( )
x yx y+
− + d
dyx
= e1 e
y
yx−
g 2 × y2 + 2x × 2y ddyx
− 3x2 × y − x3 × ddyx
= 0 h 2y ddyx
+ 1 × ln y + x × 1y
ddyx
= 0
2y2 − 3x2y = ddyx
(x3 − 4xy) ddyx
(2y + xy
) = −ln y
ddyx
= 2 2
32 3
4y x yx xy
−−
ddyx
= − ln2 x
y
yy +
= − 2ln
2y yy x+
i 1 × sin y + x × ddyx
cos y + 2x × cos y + x2 × (−sin y) ddyx
= 0
sin y + 2x cos y = ddyx
(x2 sin y − x cos y)
ddyx
= 2sin 2 cos
sin cosy x y
x y x y+
−
5 a 2x + 2y d
dyx
− 3 ddyx
= 0 b 4x − 1 × y − x × ddyx
+ 2y ddyx
= 0
2x = ddyx
(3 − 2y) 4x − y = ddyx
(x − 2y)
ddyx
= 23 2
xy−
ddyx
= 42
x yx y
−−
grad = 4 grad = −1 ∴ y − 1 = 4(x − 2) ∴ y − 5 = −(x − 3) [ y = 4x − 7 ] [ y = 8 − x ]
c 4 ddyx
cos y − sec x tan x = 0 d 2 sec2 x × cos y + 2 tan x × (−sin y) ddyx
= 0
4 ddyx
cos y = sec x tan x 2 sec2 x cos y = 2 ddyx
tan x sin y
ddyx
= sec tan4cos
x xy
ddyx
= 2sec cos
tan sinx yx y
grad = 3
2
2 34
××
= 1 grad = 123
2
2
1
×
× = 2 3
3 3× = 2
3 3
∴ y − π6 = x − π3 ∴ y − π3 = 23 3 (x − π4 )
[ y = x − π6 ] [ 4 3 x − 6y + π(2 − 3 ) = 0 ]
C4 DIFFERENTIATION Answers - Worksheet C page 3
Solomon Press
6 a 2x + 4y d
dyx
− 1 + 4 ddyx
= 0 7 a 2x + 4 × y + 4x × ddyx
− 6y ddyx
= 0
ddyx
(4y + 4) = 1 − 2x 2x + 4y = ddyx
(6y − 4x)
ddyx
= 1 24( 1)
xy−
+ d
dyx
= 23 2x yy x+−
b grad = 18 grad = −4
∴ grad of normal = −8 ∴ y − 2 = −4(x − 4) ∴ y + 3 = −8(x − 1) [ y = 18 − 4x ] [ y = 5 − 8x ] b at Q, 2
3 2x yy x+−
= −4
x + 2y = −4(3y − 2x) x = 2y sub. into equation of curve ⇒ (2y)2 + 4y(2y) − 3y2 = 36 y2 = 4 y = 2 (at P) or −2 ∴ Q (−4, −2) 8 ln y = ln ax ln y = x ln a 1
yddyx
= ln a
ddyx
= y ln a = ax ln a
9 a = 3x ln 3 b = 62x ln 6 × 2 c = 51 − x ln 5 × (−1) d =
3
2x ln 2 × 3x2 = 2(62x) ln 6 = −(51 − x) ln 5 = 3x2(
3
2x ) ln 2 10 d
dNt
= 800 (1.04)t × ln 1.04
N = 4000 ∴ 4000 = 800(1.04)t (1.04)t = 5 d
dNt
= 800 × 5 × ln 1.04 = 157 (3sf)
∴ growing at rate of 157 per minute
Solomon Press
DIFFERENTIATION C4 Answers - Worksheet D
1 a i d
dyx
= 2x + 3 2 ddyt
= ddyx
× ddxt
ii ddxt
= 3(t − 4)2 ddyx
= 1 × 2 3x − + x × 121
2 (2 3)x −− × 2
b i ddyt
= ddyx
× ddxt
= (2 3)2 3
x xx
− +−
= 3( 1)2 3xx−−
t = 5, ddxt
= 3, x = 1, ddyx
= 5 x = 6, ddxt
= 0.3, ddyx
= 5
∴ ddyt
= 5 × 3 = 15 ∴ ddyt
= 5 × 0.3 = 1.5
ii x = 8, ddyx
= 19 ∴ y increasing at 1.5 units per second
8 = (t − 4)3 ∴ t = 6, ddxt
= 12
∴ ddyt
= 19 × 12 = 228
3 a d
dPt
= ddPr
× ddrt
4 a ddAt
= ddAr
× ddrt
ddrt
= 0.2, P = 2πr ∴ ddPr
= 2π A = πr2 ∴ ddAr
= 2πr
∴ ddPt
= 2π × 0.2 = 0.4π ddAt
= −0.5, r = 8 ∴ ddAr
= 16π
∴ perimeter increasing at 0.4π cm s−1 ∴ −0.5 = 16π × ddrt
b ddAt
= ddAr
× ddrt
ddrt
= − 132π = −0.00995 cm s−1
A = πr2 ∴ ddAr
= 2πr ∴ radius decreasing at 0.00995 cm s−1 (3sf)
r = 10, ddAr
= 20π b ddPt
= ddPr
× ddrt
∴ ddAt
= 20π × 0.2 = 4π cm2 s−1 P = 2πr ∴ ddPr
= 2π
c 20 = 2πr × 0.2 ∴ r = 50π = 15.9 cm (3sf) ∴ d
dPt
= 2π × − 132π = 1
16−
∴ perimeter decreasing at 0.0625 cm s−1
C4 DIFFERENTIATION Answers - Worksheet D page 2
Solomon Press
5 a d
dVt
= ddVl
× dd
lt
6 a r
ddVt
= 3.5, V = l3 ∴ ddVl
= 3l2 h tan 30° = 13
= rh
200 = l3 ∴ l = 3 200 = 5.848 30° ∴ r = 3
h
∴ ddVl
= 3 × (5.848)2 = 102.6 V = 13 πr2h = 1
3 π × 13 h2 × h = 1
9 πh3
∴ 3.5 = 102.6 × dd
lt
b t = 120, V = 565.06
dd
lt
= 3.5 ÷ 102.6 = 0.0341 cm s−1 ∴ h = 39 565.06
π× = 11.7 cm (3sf)
b 2 mm s−1 = 0.2 cm s−1 c ddVt
= ddVh
× ddht
3.5 = ddVl
× 0.2 ddVh
= 13 πh2
∴ ddVl
= 3.5 ÷ 0.2 = 17.5 h = 11.74, ddVh
= 144.4
∴ 17.5 = 3l2 ddVt
= 600 × (−0.0005)e−0.0005t
l = 17.53 = 2.415 = −0.3e−0.0005t
V = (2.415)3 = 14.1 cm3 (3sf) t = 120, ddVt
= −0.2825
∴ −0.2825 = 144.4 × ddht
ddht
= −0.00196
∴ depth decreasing at 0.00196 cm s−1 (3sf)
Solomon Press
DIFFERENTIATION C4 Answers - Worksheet E
1 6x + 1 × y + x × d
dyx
− 2y ddyx
= 0 2 a dd
xθ
= −a sin θ, dd
yθ
= a(cos θ − 1)
6x + y = ddyx
(2y − x) ddyx
= dd
yθ
÷ dd
xθ
= (cos 1)sin
aa
θθ−
− = 1 cos
sinθ
θ−
ddyx
= 62
x yy x
+−
= 2
2
2 2
1 (1 2sin )2sin cos
θ
θ θ
− − = 2
2
sincos
θ
θ = tan 2θ
b x = 0 ⇒ cos θ = 0 ⇒ θ = π2
∴ y = a(1 − π2 ), grad = 1
∴ y = x + a(1 − π2 )
3 a d
dxθ
= −sin θ, dd
yθ
= cos 2θ 4 a 2x − 4 × y − 4x × ddyx
+ 2y ddyx
= 0
ddyx
= dd
yθ
÷ dd
xθ
= cos2sin
θθ−
2x − 4y = ddyx
(4x − 2y)
= −cosec θ cos 2θ ddyx
= 2 44 2
x yx y
−−
= 22x y
x y−
−
b x = 0 ⇒ cos θ = 0 ⇒ θ = π2 , 3π2 b grad = 3
c θ = π2 , grad = −1 × (−1) = 1 ∴ y − 10 = 3(x − 2) [ y = 3x + 4 ]
θ = 3π2 , grad = 1 × (−1) = −1 c 2
2x y
x y−
− = 3
product of gradients = 1 × (−1) = −1 x − 2y = 3(2x − y) ∴ tangents are perpendicular y = 5x, sub. into eqn of curve d y = 1
2 sin 2θ = sin θ cos θ x2 − 4x(5x) + (5x)2 = 24 y2 = sin2 θ cos2 θ = cos2 θ (1 − cos2 θ) x2 = 4 ∴ y2 = x2(1 − x2) x = 2 (at P) or −2 ∴ (−2, −10) 5 a d
dxt
= 2t, ddyt
= 2t − 1 6 3x2 − 3 + 1 × y + x × ddyx
− 4y ddyx
= 0
ddyx
= ddyt
÷ ddxt
= 2 12t
t− 3x2 − 3 + y = d
dyx
(4y − x)
∴ 2 12t
t− = 0 d
dyx
= 23 34
x yy x− +
−
t = 12 grad = 1
3 ∴ ( 9
4 , 14− ) ∴ grad of normal = −3
b x = 3 ⇒ t 2 + 2 = 3 ⇒ t = ± 1 ∴ y − 1 = −3(x − 1) y = 2 ⇒ t 2 − t = 2 y = 4 − 3x t 2 − t − 2 = 0 (t − 2)(t + 1) = 0 t = −1 or 2 ∴ at (3, 2), t = −1 ∴ grad = 3
2 ∴ y − 2 = 3
2 (x − 3) 2y − 4 = 3x − 9 3x − 2y = 5
C4 DIFFERENTIATION Answers - Worksheet E page 2
Solomon Press
7 a d
dVt
= ddVh
× ddht
, ddVt
= 80 8 a ddxt
= 21 (1 ) 1
(1 )t tt
× + − ×+
= 21
(1 )t+
ddVh
= 40π × 0.1e0.1h = 4πe0.1h ddyt
= 21 (1 ) ( 1)
(1 )t t
t× − − × −
− = 2
1(1 )t−
h = 4, ddVh
= 4πe0.4 ddyx
= ddyt
÷ ddxt
= 21
(1 )t− ÷ 2
1(1 )t+
∴ 80 = 4πe0.4 × ddht
, ddht
= 4.27 = 2
2(1 )(1 )
tt
+−
= 21
1tt
+ −
∴ depth increasing at 4.27 cm s−1 (3sf) b t = 12 ∴ x = 1
3 , y = 1 b after 5 seconds, V = 5 × 80 = 400 grad = 9 ∴ grad of normal = 1
9− ∴ 400 = 40π(e0.1h − 1) ∴ y − 1 = 1
9− (x − 13 )
h = 10 ln ( 10π + 1) = 14.31 27y − 27 = −3x + 1
∴ ddVh
= 4πe1.431 3x + 27y = 28
∴ 80 = 4πe1.431 × ddht
, ddht
= 1.52 c 31
tt+
+ 271
tt− = 28
∴ depth increasing at 1.52 cm s−1 (3sf) 3t(1 − t) + 27t(1 + t) = 28(1 − t 2) 26t 2 + 15t − 14 = 0 (13t + 14)(2t − 1) = 0 t = 1
2 (at P) or 1413−
∴ t = 1413− at Q
9 2 + 2x × y + x2 × d
dyx
− 2y ddyx
= 0 10 a dd
xθ
= a sec θ tan θ, dd
yθ
= 2a sec2 θ
2 + 2xy = ddyx
(2y − x2) ddyx
= dd
yθ
÷ dd
xθ
= 22 sec
sec tana
aθ
θ θ = 2 cosec θ
ddyx
= 22 22
xyy x+
− b θ = π4 , x = 2 a, y = 2a
∴ 22 22
xyy x+
− = 0, 2 + 2xy = 0 grad = 2 2
xy = −1, y = − 1x
∴ grad of normal = − 12 2
sub. 2x + x2(− 1x
) − (− 1x
)2 = 0 ∴ y − 2a = − 12 2
(x − 2 a)
2x − x − 21x
= 0 2 2 y − 4 2 a = −x + 2 a
x = 21x
, x3 = 1 x + 2 2 y = 5 2 a
x = 1 ∴ (1, −1) c y2 = 4a2 tan2 θ = 4a2(sec2 θ − 1) sec θ = x
a
∴ y2 = 4a2[( xa
)2 − 1]
y2 = 4(x2 − a2)
Solomon Press
DIFFERENTIATION C4 Answers - Worksheet F
1 a d
dxt
= 2t, ddyt
= −2t−2 2 x = 1 ∴ y = 4
ddyx
= ddyt
÷ ddxt
= 22
2tt
−− = − 31t
ddyx
= 4x ln 4
b t = 2 ∴ x = 4, y = 1 grad = 4 ln 4 = 4 ln 22 = 8 ln 2 grad = 1
8− ∴ y − 4 = (8 ln 2)(x − 1) ∴ grad of normal = 8 y = 4 + 8(x − 1) ln 2 ∴ y − 1 = 8(x − 4) y = 8x − 31 3 a d
dxθ
= sec θ tan θ, dd
yθ
= −2 sin 2θ 4 a 4x + 6 × y + 6x × ddyx
− 2y ddyx
= 0
ddyx
= dd
yθ
÷ dd
xθ
= 2sin 2sec tan
θθ θ
− 4x + 6y = ddyx
(2y − 6x)
= −4 sin θ cos θ × cos θ × cossin
θθ
ddyx
= 2 33
x yy x
+−
= −4 cos3 θ at P, grad = 1 b θ = π6 ∴ x = 2
3, y = 1
2 ∴ grad of normal = −1
grad = −4 × ( 32 )3 = 3
2 3− ∴ y + 5 = −(x − 2)
∴ y − 12 = 3
2 3− (x − 23
) x + y + 3 = 0
2y − 1 = 3 3− x + 6 b sub. y = −x − 3 into eqn of curve 3 3 x + 2y = 7 [k = 7] 2x2 + 6x(−x − 3) − (−x − 3)2 + 77 = 0 5x2 + 24x − 68 = 0 (5x + 34)(x − 2) = 0 x = 2 (at P) or 34
5− ∴ x = 456−
5 a y = 0 ⇒ cos θ = 0 ⇒ θ = π2 , 3π
2 6 a sin θ = 12
∴ ( π2 − 1, 0), ( 3π
2 + 1, 0) ∴ θ = π6
b dd
xθ
= 1 − cos θ, dd
yθ
= −sin θ b θ = π6 ∴ y = 43
ddyx
= dd
yθ
÷ dd
xθ
= sin1 cos
θθ
−−
dd
xθ
= cos θ, dd
yθ
= 2 sec θ × sec θ tan θ,
= − 2 22
2
2sin cos1 (1 2sin )
θ θ
θ− − d
dyx
= dd
yθ
÷ dd
xθ
= 22sec tan
cosθ θ
θ = 4
2sincos
θθ
= − 2
2
cossin
θ
θ = −cot 2θ ∴ grad = 16
9
c −cot 2θ = 0 ∴ y − 4
3 = 169 (x − 1
2 )
2θ = π2 , θ = π 9y − 12 = 16x − 8
∴ (π, −1) 16x − 9y + 4 = 0 c y = sec2 θ = 2
1cos θ
= 21
1 sin θ−
∴ y = 21
1 x−
C4 DIFFERENTIATION Answers - Worksheet F page 2
Solomon Press
7 a 2 cos x − 2 d
dyx
sec2 2y = 0 8 a ddxt
= −4 sin t, ddyt
= 3 cos t
2 cos x = 22
cos 2yddyx
ddyx
= ddyt
÷ ddxt
= 3cos4sin
tt−
= 34− cot t
ddyx
= cos x cos2 2y b y − 3 sin t = − 3cos4sin
tt
(x − 4 cos t)
b grad = 12 × ( 1
2 )2 = 18 4y sin t − 12 sin2 t = −3x cos t + 12 cos2 t
∴ y − π6 = 18 (x − π3 ) 3x cos t + 4y sin t = 12(sin2 t + cos2 t)
24y − 4π = 3x − π 3x cos t + 4y sin t = 12 3x − 24y + 3π = 0 c cos t =
4x , sin t =
3y
x − 8y + π = 0 cos2 t + sin2 t = 1 ∴ (
4x )2 + (
3y )2 = 1
9x2 + 16y2 = 144 9 a d
dxt
= 21 ( 1) 1
( 1)t tt
× + − ×+
= 21
( 1)t +
ddyt
= 22 ( 1) 2 1
( 1)t t
t× − − ×
− = 2
2( 1)t
−−
ddyx
= ddyt
÷ ddxt
= 22
( 1)t−−
÷ 21
( 1)t +
= 2
22( 1)( 1)
tt
− +−
= −221
1tt
+ −
b at O, t = 0 ∴ grad = −2 ∴ grad of normal = 1
2 ∴ y = 1
2 x
c 21
tt −
= 12 ×
1t
t +
4t(t + 1) = t(t − 1) 3t2 + 5t = 0 t(3t + 5) = 0 t = 0 (at O) or 5
3− ∴ ( 5
2 , 54 )
d x = 1
tt +
⇒ xt + x = t
x = t(1 − x) t =
1x
x−
∴ y = 2
1
1 1
xx
xx
−
− −
y = 2(1 )
xx x− −
y = 22 1
xx −
Solomon Press
INTEGRATION C4 Answers - Worksheet A
1 a ex + c b 4ex + c c ln | x | + c d 6 ln | x | + c 2 a = 2t + 3et + c b = 1
2 t 2 + ln | t | + c c = 13 t 3 − et + c d = 9t − 2 ln | t | + c
e = ∫ ( 7t
+ 12t ) dt f = 1
4 et − ln | t | + c g = ∫ ( 13t
+ t −2) dt h = 25 ln | t | − 3
7 et + c
= 7 ln | t | + 322
3 t + c = 13 ln | t | − t −1 + c
3 a = 5x − 3 ln | x | + c b = ln | u | − u−1 + c c = ∫ ( 2
5 et + 15 ) dt
= 25 et + 1
5 t + c
d = ∫ (3 + 1y
) dy e = ∫ ( 34 et +
123t ) dt f = ∫ (x2 − 2 + x−2) dx
= 3y + ln | y | + c = 34 et +
322t + c = 1
3 x3 − 2x − x−1 + c
4 f ′(x) = 24 4 1x x
x− + = 4x − 4 + 1
x
f(x) = ∫ (4x − 4 + 1x
) dx = 2x2 − 4x + ln | x | + c
(1, −3) ⇒ −3 = 2 − 4 + 0 + c ∴ c = −1 f(x) = 2x2 − 4x + ln | x | − 1 5 a = [ex + 10x] 1
0 b = [ 12 t 2 + ln | t |] 5
2 c = 4
1∫ ( 5x
− x) dx
= (e + 10) − (1 + 0) = ( 252 + ln 5) − (2 + ln 2) = [5 ln | x | − 1
2 x2] 41
= e + 9 = 212 + ln 5
2 = (5 ln 4 − 8) − (0 − 12 )
= 10 ln 2 − 152
d = 1
2
−
−∫ (2 + 13y
) dy e = [ex − 13 x3] 3
3− f = 3
2∫ (4 − 3r−1 + 6r−2) dr
= [2y + 13 ln | y |] 1
2−− = (e3 − 9) − (e−3 + 9) = [4r − 3 ln | r | − 6r−1] 3
2 = (−2 + 0) − (−4 + 1
3 ln 2) = e3 − e−3 − 18 = (12 − 3 ln 3 − 2) − (8 − 3 ln 2 − 3) = 2 − 1
3 ln 2 = 5 − 3 ln 32
g = [7u − eu] ln 4ln 2 h =
10
6∫ (2 + 9r−1) dr i = 9
4∫ (12x− + 3ex) dx
= (7 ln 4 − 4) − (7 ln 2 − 2) = [2r + 9 ln | r |] 106 = [
122x + 3ex] 9
4 = 7 ln 2 − 2 = (20 + 9 ln 10) − (12 + 9 ln 6) = (6 + 3e9) − (4 + 3e4) = 8 + 9 ln 5
3 = 3e9 − 3e4 + 2 6 =
2
0∫ (3 + ex) dx 7 = 4
1∫ (2x + 1x
) dx
= [3x + ex] 20 = [ x2 + ln | x |] 4
1 = (6 + e2) − (0 + 1) = (16 + ln 4) − (1 + 0) = e2 + 5 = 15 + 2 ln 2
C4 INTEGRATION Answers - Worksheet A page 2
Solomon Press
8 a = 1
0∫ (4x + 2ex) dx b = 4
2∫ (1 + 3x
) dx
= [2x2 + 2ex] 10 = [x + 3 ln | x |] 4
2 = (2 + 2e) − (0 + 2) = 2e = (4 + 3 ln 4) − (2 + 3 ln 2) = 2 + 3 ln 2
c = 1
3
−
−∫ (4 − 1x
) dx d = ln 2
0∫ (2 − 12 ex) dx
= [4x − ln | x |] 13
−− = [2x − 1
2 ex] ln 20
= (−4 − 0) − (−12 − ln 3) = 8 + ln 3 = (2 ln 2 − 1) − (0 − 12 ) = 2 ln 2 − 1
2
e = 12
2
∫ (ex + 5x
) dx f = 3
2∫ (x2 − 2x
) dx
= [ex + 5 ln | x |] 12
2 = [ 13 x3 − 2 ln | x |] 3
2
= (e2 + 5 ln 2) − (12e + 5 ln 1
2 ) = (9 − 2 ln 3) − ( 83 − 2 ln 2)
= e2 − 12e + 10 ln 2 = 19
3 − 2 ln 32
9 a 9 − 7x
− 2x = 0 10 a y
2x2 − 9x + 7 = 0 (2x − 7)(x − 1) = 0 O (ln a, 0) x x = 1, 7
2 ∴ (1, 0) and ( 7
2 , 0)
b = 72
1∫ (9 − 7x
− 2x) dx b = −ln
0
a
∫ (ex − a) dx = −[ex − ax] ln0
a
= [9x − 7 ln | x | − x2]721 = −[(a − a ln a) − (1 − 0)] = 1 − a + a ln a
= ( 632 − 7 ln 7
2 − 494 ) − (9 − 0 − 1) c 1 − a + a ln a = 1 + a
= 1411 − 7 ln 7
2 a ln a = 2a, ln a = 2, a = e2
11 a x = 3 ∴ y = e3 12 a ( 3x
− 4)2 = 0
ddyx
= ex ∴ grad = e3 x = 34
∴ y − e3 = e3(x − 3) [ y = e3(x − 2) ] x = 916 ∴ ( 9
16 , 0)
b at Q, y = 0 ∴ x = 2 b = 9
16
1
∫ ( 3x
− 4)2 dx
at R, x = 0 ∴ y = −2e3 = 9
16
1
∫ (9x−1 − 1224x− + 16) dx
∴ Q (2, 0), R (0, −2e3) = [9 ln | x | − 1248x + 16x] 9
16
1
c area under curve, 0 ≤ x ≤ 3 = (0 − 48 + 16) − (9 ln 916 − 36 + 9)
= 3
0∫ ex dx = [ex] 30 = e3 − 1 = −5 − 9 ln 9
16 ≈ 0.178
area of triangle under PQ = 1
2 × 1 × e3 = 12 e3
area of triangle above QR = 1
2 × 2 × 2e3 = 2e3
shaded area = (e3 − 1) − 1
2 e3 + 2e3 = 52 e3 − 1
(0, 1 − a)y = −a
Solomon Press
INTEGRATION C4 Answers - Worksheet B
1 a = 1
8 (x − 2)8 + c b = 1 12 4× (2x + 5)4
+ c c = 613 5× (1 + 3x)5
+ c d = 164× ( 1
4 x − 2)6 + c
= 18 (2x + 5)4 + c = 2
5 (1 + 3x)5 + c = 23 ( 1
4 x − 2)6 + c
e = 1 15 5− × (8 − 5x)5
+ c f = ∫ (x + 7)−2 dx g = ∫ 8(4x − 3)−5 dx h = ∫ 12 (5 − 3x)−3 dx
= 125− (8 − 5x)5 + c = −(x + 7)−1 + c = 81
4 4−× (4x − 3)−4 + c = 1 13 4−− × (5 − 3x)−2
+ c
= 41
2(4 3)x−
− + c = 2
112(5 3 )x−
+ c
2 a =
522
5 (3 )t+ + c b = ∫12(4 1)x − dx c = 1
2 ln2y + 1 + c
= 1 24 3×
32(4 1)x − + c
= 16
32(4 1)x − + c
d = 12 e2x − 3 + c e = 3 × 1
7− ln2 − 7r + c f = ∫13(5 2)t − dt
= 37− ln2 − 7r + c =
4331
5 4 (5 2)t× − + c
= 433
20 (5 2)t − + c
g = ∫12(6 )y −− dy h = 5
3− e7 − 3t + c i = 4 × 13 ln3u + 1 + c
= 122(6 )y− − + c = 4
3 ln3u + 1 + c
3 a f(x) = ∫ 8(2x − 3)3 dx b f(x) = ∫ 6e2x + 4 dx
= 12 × 2(2x − 3)4 + c = 3e2x + 4 + c
= (2x − 3)4 + c (−2, 1) ⇒ 1 = 3 + c (2, 6) ⇒ 6 = 1 + c ∴ c = −2 ∴ c = 5 f(x) = 3e2x + 4 − 2 f(x) = (2x − 3)4 + 5
c f(x) = ∫ 2 − 84 1x −
dx d f(x) = ∫ 8x − 3(3x − 2)−2 dx
= 2x − 8 × 14 ln4x − 1 + c = 4x2 + 1
3 × 3(3x − 2)−1 + c = 2x − 2 ln4x − 1 + c = 4x2 + (3x − 2)−1 + c ( 1
2 , 4) ⇒ 4 = 1 + c (−1, 3) ⇒ 3 = 4 − 15 + c
∴ c = 3 ∴ c = 45−
f(x) = 2x − 2 ln4x − 1 + 3 f(x) = 4x2 + 13 2x −
− 45
C4 INTEGRATION Answers - Worksheet B page 2
Solomon Press
4 a = [ 1 13 3× (3x + 1)3] 1
0 b = [ 1 12 4× (2x − 1)4] 2
1 c = 4
2∫ (5 − x)−2 dx
= 19 [(3x + 1)3] 1
0 = 18 [(2x − 1)4] 2
1 = [(5 − x)−1] 42
= 19 (64 − 1) = 1
8 (81 − 1) = 1 − 13
= 7 = 10 = 23
d = [ 12 e2x + 2] 1
1− e = 6
2∫12(3 2)x − dx f = [4 × 1
6 ln6x − 3] 21
= 12 (e4 − 1) = [
321 2
3 3 (3 2)x× − ] 62 = 2
3 [ ln6x − 3] 21
= 29 [
32(3 2)x − ] 6
2 = 23 (ln 9 − ln 3)
= 29 (64 − 8) = 2
3 ln 3 = 4
912
g = 1
0∫13(7 1)x −+ dx h = [ 1
5 ln5x + 3] 17
−− i = 1
8
7
4∫ (x − 4)3 dx
= [2331
7 2 (7 1)x× + ] 10 = 1
5 (ln 2 − ln 32) = 18 [ 1
4 (x − 4)4] 74
= 314 (4 − 1) = 1
5 (ln 2 − 5 ln 2) = 132 (81 − 0)
= 914 = 4
5− ln 2 = 17322
5 a =
4
3∫ e3 − x dx b = 3
2∫ (3x − 5)3 dx
= [−e3 − x] 43 = [ 1 1
3 4× (3x − 5)4] 32
= −e−1 − (−1) = 112 (256 − 1)
= 1 − 1e
= 1421
c = 4
1∫3
4 2x + dx d =
0
2−∫ (1 − 2x)−2 dx
= [3 × 14 ln4x + 2] 4
1 = [ 12− × −(1 − 2x)−1] 0
2− = 3
4 (ln 18 − ln 6) = 12 (1 − 1
5 ) = 3
4 ln 3 = 25
6 =
1
0∫ 12(2x + 1)−3 dx
= [ 12 × (−6)(2x + 1)−2] 1
0
= [ 23
(2 1)x−+
] 10
= 13− − (−3)
= 83
Solomon Press
INTEGRATION C4 Answers - Worksheet C
1 a 3 5
( 1)( 3)x
x x+
+ + ≡
1A
x + +
3B
x + 2 3
( 2)( 1)t t− + ≡
2A
t − +
1B
t +
3x + 5 ≡ A(x + 3) + B(x + 1) 3 ≡ A(t + 1) + B(t − 2) x = −1 ⇒ 2 = 2A ⇒ A = 1 t = 2 ⇒ 3 = 3A ⇒ A = 1 x = −3 ⇒ −4 = −2B ⇒ B = 2 t = −1 ⇒ 3 = −3B ⇒ B = −1 ∴ 3 5
( 1)( 3)x
x x+
+ + ≡ 1
1x + + 2
3x + ∴ ∫ 3
( 2)( 1)t t− + dt
b = ∫ ( 11x +
+ 23x +
) dx = ∫ ( 12t −
− 11t +
) dt
= lnx + 1 + 2 lnx + 3 + c = lnt − 2 − lnt + 1 + c
= ln 21
tt−+
+ c
3 a 6 11
(2 1)( 3)x
x x−
+ − ≡
2 1A
x + +
3B
x − b 2
142 8
xx x
−+ −
≡ 4
Ax +
+ 2
Bx −
6x − 11 ≡ A(x − 3) + B(2x + 1) 14 − x ≡ A(x − 2) + B(x + 4) x = 1
2− ⇒ −14 = 72− A ⇒ A = 4 x = −4 ⇒ 18 = −6A ⇒ A = −3
x = 3 ⇒ 7 = 7B ⇒ B = 1 x = 2 ⇒ 12 = 6B ⇒ B = 2 ∴ ∫ 6 11
(2 1)( 3)x
x x−
+ − dx ∴ ∫ 2
142 8
xx x
−+ −
dx
= ∫ ( 42 1x +
+ 13x −
) dx = ∫ ( 22x −
− 34x +
) dx
= 2 ln2x + 1 + lnx − 3 + c = 2 lnx − 2 − 3 lnx + 4 + c
c 6(2 )(1 )x x+ −
≡ 2
Ax+
+ 1
Bx−
d 21
5 14 8x
x x+
− + ≡
5 4A
x − +
2B
x −
6 ≡ A(1 − x) + B(2 + x) x + 1 ≡ A(x − 2) + B(5x − 4) x = −2 ⇒ 6 = 3A ⇒ A = 2 x = 4
5 ⇒ 95 = 6
5− A ⇒ A = 32−
x = 1 ⇒ 6 = 3B ⇒ B = 2 x = 2 ⇒ 3 = 6B ⇒ B = 12
∴ ∫ 6(2 )(1 )x x+ −
dx ∴ ∫ 21
5 14 8x
x x+
− + dx
= ∫ ( 22 x+
+ 21 x−
) dx = ∫ (12
( 2)x − −
32
5 4x −) dx
= 2 ln2 + x − 2 ln1 − x + c = 12 lnx − 2 − 3
10 ln5x − 4 + c
= 2 ln 21
xx
+−
+ c
4 a x2 − 6 ≡ A(x + 4)(x − 1) + B(x − 1) + C(x + 4) x = −4 ⇒ 10 = −5B ⇒ B = −2 x = 1 ⇒ −5 = 5C ⇒ C = −1 coeffs of x2 ⇒ A = 1 b = ∫ (1 − 2
4x + − 1
1x −) dx
= x − 2 lnx + 4 − lnx − 1 + c
C4 INTEGRATION Answers - Worksheet C page 2
Solomon Press
5 a 2
24
( 2)( 1)x x
x x− −
+ + ≡
2A
x + +
1B
x + + 2( 1)
Cx +
x2 − x − 4 ≡ A(x + 1)2 + B(x + 2)(x + 1) + C(x + 2) x = −2 ⇒ A = 2 x = −1 ⇒ C = −2 coeffs of x2 ⇒ 1 = A + B ⇒ B = −1
∴ 2
24
( 2)( 1)x x
x x− −
+ + ≡ 2
2x + − 1
1x + − 2
2( 1)x +
b = ∫ ( 22x +
− 11x +
− 22
( 1)x +) dx
= 2 lnx + 2 − lnx + 1 + 2(x + 1)−1 + c
6 a 2
23 5
1xx
−−
≡ A + 1
Bx +
+ 1
Cx −
3x2 − 5 ≡ A(x + 1)(x − 1) + B(x − 1) + C(x + 1) x = −1 ⇒ −2 = −2B ⇒ B = 1 x = 1 ⇒ −2 = 2C ⇒ C = −1 coeffs of x2 ⇒ A = 3
∴ ∫2
23 5
1xx
−−
dx = ∫ (3 + 11x +
− 11x −
) dx
= 3x + lnx + 1 − lnx − 1 + c = 3x + ln 11
xx
+−
+ c
b 2(4 13)
(2 ) (3 )x x
x x+
+ − ≡
2A
x+ + 2(2 )
Bx+
+ 3
Cx−
x(4x + 13) ≡ A(2 + x)(3 − x) + B(3 − x) + C(2 + x)2
x = −2 ⇒ −10 = 5B ⇒ B = −2 x = 3 ⇒ 75 = 25C ⇒ C = 3 coeffs of x2 ⇒ 4 = −A + C ⇒ A = −1 ∴ ∫ 2
(4 13)(2 ) (3 )
x xx x
++ −
dx = ∫ ( 33 x−
− 12 x+
− 22
(2 )x+) dx
= −3 ln3 − x − ln2 + x + 2(2 + x)−1 + c
c 2
21
3 10x x
x x− +
− − ≡ A +
5B
x − +
2C
x +
x2 − x + 1 ≡ A(x − 5)(x + 2) + B(x + 2) + C(x − 5) x = 5 ⇒ 21 = 7B ⇒ B = 3 x = −2 ⇒ 7 = −7C ⇒ C = −1 coeffs of x2 ⇒ A = 1
∴ ∫2
21
3 10x x
x x− +
− − dx = ∫ (1 + 3
5x − − 1
2x +) dx
= x + 3 lnx − 5 − lnx + 2 + c
d 2
22 6 5
(1 2 )x x
x x− +
− ≡ A
x + 2
Bx
+ 1 2
Cx−
2 − 6x + 5x2 ≡ Ax(1 − 2x) + B(1 − 2x) + Cx2
x = 12 ⇒ 1
4 = 14 C ⇒ C = 1
x = 0 ⇒ B = 2 coeffs of x2 ⇒ 5 = −2A + C ⇒ A = −2
∴ ∫2
22 6 5
(1 2 )x x
x x− +
− dx = ∫ ( 1
1 2x− − 2
x + 2
2x
) dx
= 12− ln1 − 2x − 2 lnx − 2x−1 + c
C4 INTEGRATION Answers - Worksheet C page 3
Solomon Press
7 3 5( 1)( 2)
xx x
−− −
≡ 1
Ax −
+ 2
Bx −
3x − 5 ≡ A(x − 2) + B(x − 1) x = 1 ⇒ −2 = −A ⇒ A = 2 x = 2 ⇒ B = 1
∴ 4
3∫3 5
( 1)( 2)x
x x−
− − dx =
4
3∫ ( 21x −
+ 12x −
) dx
= [2 lnx − 1 + lnx − 2] 43
= (2 ln 3 + ln 2) − (2 ln 2 + 0) = 2 ln 3 − ln 2
8 a 3( 1)x
x x++
≡ Ax
+ 1
Bx +
x + 3 ≡ A(x + 1) + Bx x = 0 ⇒ A = 3 x = −1 ⇒ 2 = −B ⇒ B = −2
∴ 3
1∫3
( 1)x
x x++
dx = 3
1∫ ( 3x
− 21x +
) dx
= [3 lnx − 2 lnx + 1] 31
= (3 ln 3 − 2 ln 4) − (0 − 2 ln 2) = 3 ln 3 − 2 ln 2
b 23 2
12x
x x−
+ − ≡
4A
x + +
3B
x −
3x − 2 ≡ A(x − 3) + B(x + 4) x = −4 ⇒ −14 = −7A ⇒ A = 2 x = 3 ⇒ 7 = 7B ⇒ B = 1
∴ 2
0∫ 23 2
12x
x x−
+ − dx =
2
0∫ ( 24x +
+ 13x −
) dx
= [2 lnx + 4 + lnx − 3] 20
= (2 ln 6 + 0) − (2 ln 4 + ln 3) = 2(ln 2 + ln 3) − 4 ln 2 − ln 3 = ln 3 − 2 ln 2
c 29
2 7 4x x− − ≡
2 1A
x + +
4B
x −
9 ≡ A(x − 4) + B(2x + 1) x = 1
2− ⇒ 9 = 92− A ⇒ A = −2
x = 4 ⇒ 9 = 9B ⇒ B = 1
∴ 2
1∫ 29
2 7 4x x− − dx =
2
1∫ ( 14x −
− 22 1x +
) dx
= [lnx − 4 − ln2x + 1] 21
= (ln 2 − ln 5) − (ln 3 − ln 3) = ln 2 − ln 5
d 2
22 7 7
2 3x xx x
− +− −
≡ A + 3
Bx −
+ 1
Cx +
2x2 − 7x + 7 ≡ A(x − 3)(x + 1) + B(x + 1) + C(x − 3) x = 3 ⇒ 4 = 4B ⇒ B = 1 x = −1 ⇒ 16 = −4C ⇒ C = −4 coeffs of x2 ⇒ A = 2
∴ 2
0∫2
22 7 7
2 3x xx x
− +− −
dx = 2
0∫ (2 + 13x −
− 41x +
) dx
= [2x + lnx − 3 − 4 lnx + 1] 20
= (4 + 0 − 4 ln 3) − (0 + ln 3 − 0) = 4 − 5 ln 3
C4 INTEGRATION Answers - Worksheet C page 4
Solomon Press
e 25 7
( 1) ( 3)x
x x+
+ + ≡
1A
x + + 2( 1)
Bx +
+ 3
Cx +
5x + 7 ≡ A(x + 1)(x + 3) + B(x + 3) + C(x + 1)2
x = −1 ⇒ 2 = 2B ⇒ B = 1 x = −3 ⇒ −8 = 4C ⇒ C = −2 coeffs of x2 ⇒ 0 = A + C ⇒ A = 2
∴ 1
0∫ 25 7
( 1) ( 3)x
x x+
+ + dx =
1
0∫ ( 21x +
+ 21
( 1)x + − 2
3x +) dx
= [2 lnx + 1 − (x + 1)−1 − 2 lnx + 3] 10
= (2 ln 2 − 12 − 2 ln 4) − (0 − 1 − 2 ln 3)
= 12 − 2 ln 2 + 2 ln 3
f 22
8 2x
x x+
− − ≡
4A
x+ +
2B
x−
2 + x ≡ A(2 − x) + B(4 + x) x = −4 ⇒ −2 = 6A ⇒ A = 1
3− x = 2 ⇒ 4 = 6B ⇒ B = 2
3
∴ 1
1−∫ 22
8 2x
x x+
− − dx =
1
1−∫ (23
2 x− −
13
4 x+) dx
= [ 23− ln2 − x − 1
3 ln4 + x] 11−
= (0 − 13 ln 5) − ( 2
3− ln 3 − 13 ln 3)
= ln 3 − 13 ln 5
9 a 2 2
1x a−
≡ Ax a+
+ Bx a−
1 ≡ A(x − a) + B(x + a) x = −a ⇒ 1 = −2aA ⇒ A = − 1
2a
x = a ⇒ 1 = 2aB ⇒ B = 12a
∴ 2 21
x a− ≡ 1
2 ( )a x a− − 1
2 ( )a x a+
b ∫ 2 21
x a− dx = 1
2a ∫ ( 1x a−
− 1x a+
) dx
= 12a
(lnx − a − lnx + a) + c
= 12a
ln x ax a
−+
+ c
c ∫ 2 21
a x− dx = − ∫ 2 2
1x a−
dx
= − 12a
ln x ax a
−+
+ c
= 12a
ln x ax a
+−
+ c
10 a = [ 16 ln 3
3xx
−+
] 11− b = 4[ 1
2 ln 11
xx
+−
]12
12− c = 3
2 [ 14 ln 2
2xx
−+
] 10
= 16 (ln 1
2 − ln 2) = 2(ln 3 − ln 13 ) = 3
8 (ln 13 − 0)
= 13− ln 2 = 4 ln 3 = − 3
8 ln 3
Solomon Press
INTEGRATION C4 Answers - Worksheet D
1 a = 2 sin x + c b = 1
4− cos 4x + c c = 2 sin 12 x + c d = −cos (x + π4 ) + c
e = 12 sin (2x − 1) + c f = 3 cos ( π
3 − x) + c g = sec x + c h = −cot x + c
i = 52 tan 2x + c j = −4 cosec 1
4 x + c k = ∫ 4 cosec2 x dx l = ∫ sec2 (4x + 1) dx
= −4 cot x + c = 14 tan (4x + 1) + c
2 a = [sin x]
π20 b = [ 1
2− cos 2x]π60 c = [4 sec 1
2 x]π20
= 1 − 0 = 1 = 14− − ( 1
2− ) = 14 = 4 2 − 4 = 4( 2 − 1)
d = [ 12 sin (2x − π3 )]
π30 e = [ 1
3 tan 3x]π3π4
f = [−cosec x]2π3π2
= 34 − (− 3
4 ) = 32 = 0 − ( 1
3− ) = 13 = − 2
3 − (−1) = 1 − 2
3 3
3 a tan2 θ = sec2 θ − 1
b ∫ tan2 x dx = ∫ (sec2 x − 1) dx = tan x − x + c 4 a cos (A + B) ≡ cos A cos B − sin A sin B let B = A ⇒ cos 2A ≡ cos2 A − sin2 A cos 2A ≡ cos2 A − (1 − cos2 A) cos 2A ≡ 2 cos2 A − 1 cos2 A ≡ 1
2 (1 + cos 2A)
b ∫ cos2 x dx = ∫ ( 12 + 1
2 cos 2x) dx = 12 x + 1
4 sin 2x + c 5 a = ∫ ( 1
2 − 12 cos 2x) dx b = ∫ (cosec2 2x − 1) dx
= 12 x − 1
4 sin 2x + c = 12− cot 2x − x + c
c = ∫ 12 sin 2x dx d = ∫ 1 sin
cos cosx
x x× dx
= 14− cos 2x + c = ∫ sec x tan x dx
= sec x + c
e = ∫ (2 + 2 cos 6x) dx f = ∫ (1 + 2 sin x + sin2 x) dx
= 2x + 13 sin 6x + c = ∫ (1 + 2 sin x + 1
2 − 12 cos 2x) dx
= ∫ ( 32 + 2 sin x − 1
2 cos 2x) dx
= 32 x − 2 cos x − 1
4 sin 2x + c
g = ∫ (sec2 x − 2 sec x tan x + tan2 x) dx h = ∫ 1 cossin 2 sin
xx x
× dx
= ∫ (sec2 x − 2 sec x tan x + sec2 x − 1) dx = ∫ 1 cos2sin cos sin
xx x x
× dx
= ∫ (2 sec2 x − 2 sec x tan x − 1) dx = ∫ 12 cosec2 x dx
= 2 tan x − 2 sec x − x + c = 12− cot x + c
C4 INTEGRATION Answers - Worksheet D page 2
Solomon Press
i = ∫ (cos2 x)2 dx
= ∫ [ 12 (1 + cos 2x)]2 dx
= 14 ∫ (1 + 2 cos 2x + cos2 2x) dx
= 14 ∫ [1 + 2 cos 2x + 1
2 (1 + cos 4x)] dx
= 18 ∫ (3 + 4 cos 2x + cos 4x) dx
= 18 (3x + 2 sin 2x + 1
4 sin 4x) + c = 3
8 x + 14 sin 2x + 1
32 sin 4x + c
6 a = π2
0∫ (1 + cos 2x) dx b = π4
0∫ 12 sin 4x dx
= [x + 12 sin 2x]
π20 = [ 1
8− cos 4x]π40
= ( π2 + 0) − (0 + 0) = π2 = 1
8 − ( 18− ) = 1
4
c = π2π3∫ (sec2 1
2 x − 1) dx d = π4π6∫ 1 cos2
sin 2 sin 2x
x x× dx
= [2 tan 12 x − x]
π2π3 =
π4π6∫ cosec 2x cot 2x dx
= (2 − π2 ) − ( 23
− π3 ) = [ 12− cosec 2x]
π4π6
= 2 − 23 3 − π6 = 1
2− − (− 13
) = 13 3 − 1
2
e = π4
0∫ (1 − 4 sin x + 4 sin2 x) dx f = π3π6∫ 2 2
1 1cos sinx x
× dx
= π4
0∫ [1 − 4 sin x + 2(1 − cos 2x)] dx = π3π6∫ 21
2
1( sin 2 )x
dx
= π4
0∫ (3 − 4 sin x − 2 cos 2x) dx = π3π6∫ 4 cosec2 2x dx
= [3x + 4 cos x − sin 2x]π40 = [−2 cot 2x]
π3π6
= ( 3π4 + 2 2 − 1) − (0 + 4 − 0) = 2
3 − (− 2
3)
= 3π4 + 2 2 − 5 = 4
3 3 7 a sin (A + B) ≡ sin A cos B + cos A sin B (1) sin (A − B) ≡ sin A cos B − cos A sin B (2) (1) + (2) ⇒ sin (A + B) + sin (A − B) ≡ 2 sin A cos B sin A cos B ≡ 1
2 [sin (A + B) + sin (A − B)]
b = ∫ ( 12 sin 4x + 1
2 sin 2x) dx = 18− cos 4x − 1
4 cos 2x + c 8 a = ∫ (cos 4x − cos 6x) dx b = ∫ ( 1
2 cos 3x + 12 cos x) dx
= 14 sin 4x − 1
6 sin 6x + c = 16 sin 3x + 1
2 sin x + c
c = ∫ [2 sin 5x + 2 sin (−3x)] dx d = ∫ [ 12 sin (2x + π6 ) − 1
2 sin π6 ] dx
= ∫ (2 sin 5x − 2 sin 3x) dx = ∫ [ 12 sin (2x + π6 ) − 1
4 ] dx
= 25− cos 5x + 2
3 cos 3x + c = 14− cos (2x + π6 ) − 1
4 x + c
Solomon Press
INTEGRATION C4 Answers - Worksheet E
1 a u = x2 + 1 ∴ d
dux
= 2x b u = sin x ∴ ddux
= cos x
∫ 2x(x2 − 1)3 dx = ∫ u3 du ∫ sin4 x cos x dx = ∫ u4 du
= 14 u4 + c = 1
5 u5 + c = 1
4 (x2 + 1)4 + c = 15 sin5 x + c
c u = 2 + x3 ∴ ddux
= 3x2 d u = x2 ∴ ddux
= 2x
∫ 3x2(2 + x3)2 dx = ∫ u2 du ∫2
2 exx dx = ∫ eu du
= 13 u3 + c = eu + c
= 13 (2 + x3)3 + c =
2ex + c
e u = x2 + 3 ∴ ddux
= 2x f u = cos 2x ∴ ddux
= −2 sin 2x
∫ 2 4( 3)x
x + dx = ∫ 1
2 u−4 du ∫ sin 2x cos3 2x dx = ∫ 12− u3 du
= 16− u−3 + c = 1
8− u4 + c
= − 2 31
6( 3)x + + c = 1
8− cos4 2x + c
g u = x2 − 2 ∴ ddux
= 2x h u = 1 − x2 ∴ ddux
= −2x
∫ 23
2x
x − dx = ∫ 3
2u du ∫ 21x x− dx = ∫ 1
2−12u du
= 32 lnu + c = 1
3−32u + c
= 32 lnx2 − 2 + c =
3221
3 (1 )x− − + c
i u = sec x ∴ ddux
= sec x tan x j u = x2 + 2x ∴ ddux
= 2x + 2
∫ sec3 x tan x dx = ∫ u2 du ∫ (x + 1)(x2 + 2x)3 dx = ∫ 12 u3 du
= 13 u3 + c = 1
8 u4 + c = 1
3 sec3 x + c = 18 (x2 + 2x)4 + c
2 a i u = 3 ii u = 4 b u = x2 + 3 ∴ d
dux
= 2x
1
0∫ 2x(x2 + 3)2 dx = 4
3∫ u2 × ddux
dx
= 4
3∫ u2 du
c 1
0∫ 2x(x2 + 3)2 dx = 4
3∫ u2 du
= [ 13 u3] 4
3 = 64
3 − 9 = 1312
C4 INTEGRATION Answers - Worksheet E page 2
Solomon Press
3 a u = x2 − 3 ∴ d
dux
= 2x b u = sin x ∴ ddux
= cos x
x = 1 ⇒ u = −2 x = 0 ⇒ u = 0 x = 2 ⇒ u = 1 x = π6 ⇒ u = 1
2
2
1∫ x(x2 − 3)3 dx = 1
2−∫ 12 u3 du
π6
0∫ sin3 x cos x dx = 12
0∫ u3 du
= [ 18 u4] 1
2− = [ 14 u4]
120
= 18 (1 − 16) = 1
4 ( 116 − 0)
= 158− = 1
64
c u = x2 + 1 ∴ ddux
= 2x d u = tan x ∴ ddux
= sec2 x
x = 0 ⇒ u = 1 x = − π4 ⇒ u = −1
x = 3 ⇒ u = 10 x = π4 ⇒ u = 1
3
0∫ 24
1x
x + dx =
10
1∫2u
du π4π4−∫ tan2 x sec2 x dx =
1
1−∫ u2 du
= [2 lnu] 101 = [ 1
3 u3] 11−
= 2 ln 10 − 0 = 13 [1 − (−1)]
= 2 ln 10 = 23
e u = x2 − 3 ∴ ddux
= 2x f u = x3 + 2 ∴ ddux
= 3x2
x = 2 ⇒ u = 1 x = −2 ⇒ u = −6 x = 3 ⇒ u = 6 x = −1 ⇒ u = 1
3
2∫ 2 3
x
x − dx =
6
1∫121
2 u− du 1
2
−
−∫ x2(x3 + 2)2 dx = 1
6−∫ 13 u2 du
= [12u ] 6
1 = [ 19 u3] 1
6−
= 6 − 1 = 19 [1 − (−216)]
= 1924
g u = 1 + e2x ∴ ddux
= 2e2x h u = x2 − 4x ∴ ddux
= 2x − 4
x = 0 ⇒ u = 2 x = 3 ⇒ u = −3 x = 1 ⇒ u = 1 + e2 x = 5 ⇒ u = 5
1
0∫ e2x(1 + e2x)3 dx = 21 e
2
+
∫ 12 u3 du
5
3∫ (x − 2)(x2 − 4x)2 dx = 5
3−∫ 12 u2 du
= [ 18 u4]
21 e2+ = [ 1
6 u3] 53−
= 18 [(1 + e2)4 − 16] = 1
6 [125 − (−27)] = 1
8 (1 + e2)4 − 2 = 1325
C4 INTEGRATION Answers - Worksheet E page 3
Solomon Press
4 a u = 4 − x2 ∴ ddux
= −2x
x = 0 ⇒ u = 4 x = 2 ⇒ u = 0
2
0∫ x(4 − x2)3 dx = 0
4∫ u3 × ( 12− d
dux
) du
= 4
0∫ 12 u3 du
b = [ 18 u4] 4
0 = 1
8 (256 − 0) = 32 5 a u = 2 − x2 ∴ d
dux
= −2x b u = 1 + cos x ∴ ddux
= −sin x
x = 0 ⇒ u = 2 x = 0 ⇒ u = 2 x = 1 ⇒ u = 1 x = π2 ⇒ u = 1
1
0∫22e xx − dx =
1
2∫ 12− eu du
π2
0∫sin
1 cosx
x+ dx =
1
2∫ − 1u
du
= 2
1∫ 12 eu du =
2
1∫1u
du
= [ 12 eu] 2
1 = [lnu] 21
= 12 (e2 − e) = ln 2 − 0
= 12 e(e − 1) = ln 2
6 a u = sin x ∴ d
dux
= cos x
∫ cot x dx = ∫ cossin
xx
dx
= ∫ 1u
× ddux
dx
= ∫ 1u
du
= lnu + c = lnsin x + c
b ∫ tan x dx = ∫ sincos
xx
dx
u = cos x ∴ ddux
= −sin x
∫ sincos
xx
dx = ∫ 1u
× (− ddux
) dx
= ∫ − 1u
du
= −lnu + c = −lncos x + c = ln (cos x)−1 + c = lnsec x + c
c = [ 12 lnsec 2x]
π60
= 12 (ln 2 − 0)
= 12 ln 2
C4 INTEGRATION Answers - Worksheet E page 4
Solomon Press
7 a = 1
4 (x3 − 2)4 + c b = esin x + c c = 12 ∫ 2
21
xx +
dx
= 12 lnx2 + 1 + c
[ = 12 ln (x2 + 1) + c ]
d = 13 (x2 + 3x)3 + c e = 1
2 ∫1222 ( 4)x x + dx f = − ∫ cot3 x (−cosec2 x) dx
= 12 ×
3222
3 ( 4)x + + c = − 14 cot4 x + c
= 3221
3 ( 4)x + + c
g = ln1 + ex + c h = 12 ∫ 2cos 2
3 sin 2xx+
dx i = 14 ∫
3
4 24
( 2)x
x − + c
[ = ln (1 + ex) + c ] = 12 ln3 + sin 2x + c = 1
4 × [−(x4 − 2)−1] + c
= − 41
4( 2)x − + c
j = 14 (ln x)4 + c k = 2
3 ∫31
2 2 232 (1 )x x+ dx l = 1
2− ∫1222 (5 )x x −− − dx
= 23 ×
32 31
3 (1 )x+ + c = 12− ×
1222(5 )x− + c
= 32 32
9 (1 )x+ + c = − 25 x− + c
8 a = −
π2
0∫ (−sin x)(1 + cos x)2 dx b = 12−
0
1−∫2
22e
2 e
x
x−
− dx
= −[ 13 (1 + cos x)3]
π20 = 1
2− [ln2 − e2x] 01−
= 13− (1 − 8) = 1
2− [0 − ln (2 − e−2)] = 7
3 = 12 ln (2 − e−2)
c = −π4π6∫ (−cot x cosec x) cosec3 x dx d = 1
2
4
2∫ 22 2
2 8x
x x+
+ + dx
= −[ 14 cosec4 x]
π4π6
= 12 [lnx2 + 2x + 8] 4
2
= 14− (4 − 16) = 1
2 (ln 32 − ln 16) = 3 = 1
2 ln 2
9 u = x + 1 ∴ x = u − 1, d
dux
= 1
∫ x(x + 1)3 dx = ∫ (u − 1)u3 du
= ∫ (u4 − u3) du
= 15 u5 − 1
4 u4 + c = 1
5 (x + 1)5 − 14 (x + 1)4 + c
= 120 (x + 1)4[4(x + 1) − 5] + c
= 120 (4x − 1)(x + 1)4 + c
C4 INTEGRATION Answers - Worksheet E page 5
Solomon Press
10 a u = 2x − 1 ∴ x = 1
2 (u + 1), ddux
= 2 b u2 = 1 − x ∴ x = 1 − u2, dd
xu
= −2u
∫ x(2x − 1)4 dx = ∫ 12 (u + 1)u4 × 1
2 du ∫ 1x x− dx = ∫ (1 − u2)u × (−2u) du
= 14 ∫ (u5 + u4) du = 2 ∫ (u4 − u2) du
= 14 ( 1
6 u6 + 15 u5) + c = 2( 1
5 u5 − 13 u3) + c
= 14 [ 1
6 (2x − 1)6 + 15 (2x − 1)5] + c = 2[ 1
5 (1 − 52)x − 1
3 (1 − 32)x + c
= 1120 (2x − 1)5[5(2x − 1) + 6] + c = 2
15 (1 − 32)x [3(1 − x) − 5] + c
= 1120 (10x + 1)(2x − 1)5 + c = − 2
15 (2 + 3x)(1 − 32)x + c
c x = sin u ∴ dd
xu
= cos u d x = u2 ∴ dd
xu
= 2u
∫ 322
1
(1 )x− dx = ∫ 3
1cos u
× cos u du ∫ 11x −
dx = ∫ 11u −
× 2u du
= ∫ sec2 u du = ∫ 2( 1) 21
uu− +−
du
= tan u + c = ∫ (2 + 21u −
) du
= sincos
uu
+ c = 2u + 2 lnu − 1 + c
= 21
x
x− + c = 2 x + 2 ln x − 1 + c
e u = 2x + 3 ∴ x = 12 u − 3
2 , ddux
= 2 f u2 = x − 2 ∴ x = u2 + 2, dd
xu
= 2u
∫ (x + 1)(2x + 3)3 dx ∫2
2xx −
dx = ∫2 2( 2)u
u+ × 2u du
= ∫ ( 12 u − 1
2 )u3 × 12 du = 2 ∫ (u4 + 4u2 + 4) du
= 14 ∫ (u4 − u3) du = 2( 1
5 u5 + 43 u3 + 4u) + c
= 14 ( 1
5 u5 − 14 u4) + c = 2[ 1
5 (x − 522) + 4
3 (x − 322) + 4(x −
122) ] + c
= 14 [ 1
5 (2x + 3)5 − 14 (2x + 3)4] + c = 2
15 (x − 122) [3(x − 2)2 + 20(x − 2) + 60] + c
= 180 (2x + 3)4[4(2x + 3) − 5] + c = 2
15 (3x2 + 8x + 32)(x − 122) + c
= 180 (8x + 7)(2x + 3)4 + c
C4 INTEGRATION Answers - Worksheet E page 6
Solomon Press
11 a x = sin u ∴ d
dxu
= cos u b u = 2 − x ∴ x = 2 − u, ddux
= −1
x = 0 ⇒ u = 0 x = 0 ⇒ u = 2 x = 1
2 ⇒ u = π6 x = 2 ⇒ u = 0
12
0∫ 2
1
1 x− dx =
π6
0∫1
cosu× cos u du
2
0∫ x(2 − x)3 dx = 0
2∫ (2 − u)u3 × (−1) du
= π6
0∫ du = 2
0∫ (2u3 − u4) du
= [u]π60 = [ 1
2 u4 − 15 u5] 2
0
= π6 − 0 = (8 − 325 ) − (0)
= π6 = 85
c x = 2 sin u ∴ dd
xu
= 2 cos u d x = 3 tan u ∴ dd
xu
= 3 sec2 u
x = 0 ⇒ u = 0 x = 0 ⇒ u = 0 x = 1 ⇒ u = π6 x = 3 ⇒ u = π4
1
0∫24 x− dx
3
0∫2
2 9x
x + dx =
π4
0∫2
29 tan9sec
uu
× 3 sec2 u du
= π6
0∫ 2 cos u × 2 cos u du = 3π4
0∫ tan2 u du
= π6
0∫ 4 cos2 u du = 3π4
0∫ (sec2 u − 1) du
= π6
0∫ (2 + 2 cos 2u) du = 3[tan u − u]π40
= [2u + sin 2u]π60 = 3[(1 − π4 ) − (0)]
= ( π3 + 3
2 ) − (0) = 34 (4 − π)
= 16 (2π + 3 3 )
Solomon Press
INTEGRATION C4 Answers - Worksheet F
1 u = x, d
dux
= 1; dd
vx
= cos x, v = sin x
∫ x cos x dx = x sin x − ∫ sin x dx = x sin x + cos x + c 2 a u = x, d
dux
= 1; dd
vx
= ex, v = ex b u = 4x, ddux
= 4; dd
vx
= sin x, v = −cos x
∫ xex dx = xex − ∫ ex dx ∫ 4x sin x dx = −4x cos x − ∫ −4 cos x dx
= xex − ex + c = −4x cos x + ∫ 4 cos x dx
= ex(x − 1) + c = −4x cos x + 4 sin x + c
c u = x, ddux
= 1; dd
vx
= cos 2x, v = 12 sin 2x d u = x, d
dux
= 1; dd
vx
= 12( 1)x + , v =
322
3 ( 1)x +
∫ x cos 2x dx = 12 x sin 2x − ∫ 1
2 sin 2x dx ∫ 1x x + dx = 322
3 ( 1)x x + − ∫322
3 ( 1)x + dx
= 12 x sin 2x + 1
4 cos 2x + c = 322
3 ( 1)x x + − 524
15 ( 1)x + + c
= 322
15 ( 1)x + [5x − 2(x + 1)] + c
= 215 (3x − 2)(x +
321) + c
e u = x, ddux
= 1; dd
vx
= e−3x, v = 13− e−3x f u = x, d
dux
= 1; dd
vx
= sec2 x, v = tan x
∫ 3e xx dx = 1
3− xe−3x − ∫ 13− e−3x dx ∫ x sec2 x dx = x tan x − ∫ tan x dx
= 13− xe−3x + ∫ 1
3 e−3x dx = x tan x + ∫ sincos
xx
− dx
= 13− xe−3x − 1
9 e−3x + c = x tan x + lncos x + c = 1
9− e−3x(3x + 1) + c
3 i u = x, d
dux
= 1; dd
vx
= (2x + 1)3, v = 18 (2x + 1)4
∫ x(2x + 1)3 dx = 18 x(2x + 1)4 − ∫ 1
8 (2x + 1)4 dx
= 18 x(2x + 1)4 − 1
80 (2x + 1)5 + c = 1
80 (2x + 1)4[10x − (2x + 1)] + c = 1
80 (8x − 1)(2x + 1)4 + c
ii u = 2x + 1 ∴ x = 12 (u − 1), d
dux
= 2
∫ x(2x + 1)3 dx = ∫ 12 (u − 1)u3 × 1
2 du
= 14 ∫ (u4 − u3) du
= 14 ( 1
5 u5 − 14 u4) + c
= 14 [ 1
5 (2x + 1)5 − 14 (2x + 1)4] + c
= 180 (2x + 1)4[4(2x + 1) − 5] + c
= 180 (8x − 1)(2x + 1)4 + c, as for part i
C4 INTEGRATION Answers - Worksheet F page 2
Solomon Press
4 u = x, ddux
= 1; dd
vx
= e−x, v = −e−x
2
0∫ xe−x dx = [−xe−x] 20 −
2
0∫ −e−x dx = [−xe−x] 20 +
2
0∫ e−x dx
= [−xe−x − e−x] 20 = (−2e−2 − e−2) − (0 − 1)
= 1 − 3e−2 5 a u = x, d
dux
= 1; dd
vx
= cos x, v = sin x b u = x, ddux
= 1; dd
vx
= e2x, v = 12 e2x
π6
0∫ x cos x dx = [x sin x]π60 −
π6
0∫ sin x dx 1
0∫ xe2x dx = [ 12 xe2x] 1
0 − 1
0∫ 12 e2x dx
= [x sin x + cos x]π60 = [ 1
2 xe2x − 14 e2x] 1
0
= ( π12 + 3
2 ) − (0 + 1) = ( 12 e2 − 1
4 e2) − (0 − 14 )
= 112 (π + 6 3 − 12) = 1
4 (e2 + 1)
c u = x, ddux
= 1; dd
vx
= sin 3x, v = 13− cos 3x
π4
0∫ x sin 3x dx = [ 13− x cos 3x]
π40 −
π4
0∫ 13− cos 3x dx
= [ 13− x cos 3x]
π40 +
π4
0∫ 13 cos 3x dx
= [ 13− x cos 3x + 1
9 sin 3x]π40
= [− π12 (− 1
2) + 1
9 ( 12
)] − (0)
= 172 2 (3π + 4)
6 a u = x2, ddux
= 2x; dd
vx
= ex, v = ex
∫ x2ex dx = x2ex − ∫ 2xex dx
for ∫ 2xex dx, u = 2x, ddux
= 2; dd
vx
= ex, v = ex
∫ 2xex dx = 2xex − ∫ 2ex dx
= 2xex − 2ex + c ∴ ∫ x2ex dx = x2ex − (2xex − 2ex) + c
= ex(x2 − 2x + 2) + c
b u = ex, ddux
= ex; dd
vx
= sin x, v = −cos x
∫ ex sin x dx = −ex cos x − ∫ −ex cos x dx
= −ex cos x + ∫ ex cos x dx
for ∫ ex cos x dx, u = ex, ddux
= ex; dd
vx
= cos x, v = sin x
∫ ex cos x dx = ex sin x − ∫ ex sin x dx
∴ ∫ ex sin x dx = −ex cos x + ex sin x − ∫ ex sin x dx
2 ∫ ex sin x dx = −ex cos x + ex sin x + c
∫ ex sin x dx = 12 ex(sin x − cos x) + c
C4 INTEGRATION Answers - Worksheet F page 3
Solomon Press
7 a u = x2, d
dux
= 2x; dd
vx
= sin x, v = −cos x
∫ x2 sin x dx = −x2 cos x − ∫ −2x cos x dx
= −x2 cos x + ∫ 2x cos x dx
for ∫ 2x cos x dx, u = 2x, ddux
= 2; dd
vx
= cos x, v = sin x
∫ 2x cos x dx = 2x sin x − ∫ 2 sin x dx = 2x sin x + 2 cos x + c ∴ ∫ x2 sin x dx = −x2 cos x + (2x sin x + 2 cos x) + c
= (2 − x2)cos x + 2x sin x + c
b u = x2, ddux
= 2x; dd
vx
= e3x, v = 13 e3x
∫ x2e3x dx = 13 x2e3x − ∫ 2
3 xe3x dx
for ∫ 23 xe3x dx, u = 2
3 x, ddux
= 23 ; d
dvx
= e3x, v = 13 e3x
∫ 23 xe3x dx = 2
9 xe3x − ∫ 29 e3x dx
= 29 xe3x − 2
27 e3x + c
∴ ∫ x2e3x dx = 13 x2e3x − ( 2
9 xe3x − 227 e3x) + c
= 127 e3x(9x2 − 6x + 2) + c
c u = e−x, ddux
= −e−x; dd
vx
= cos 2x, v = 12 sin 2x
∫ e−x cos 2x dx = 12 e−x sin 2x − ∫ − 1
2 e−x sin 2x dx
= 12 e−x sin 2x + ∫ 1
2 e−x sin 2x dx
for ∫ 12 e−x sin 2x dx, u = 1
2 e−x, ddux
= 12− e−x; d
dvx
= sin 2x, v = 12− cos 2x
∫ 12 e−x sin 2x dx = 1
4− e−x cos 2x − ∫ 14 e−x cos 2x dx
∴ ∫ e−x cos 2x dx = 12 e−x sin 2x − 1
4 e−x cos 2x − 14 ∫ e−x cos 2x dx
54 ∫ e−x cos 2x dx = 1
2 e−x sin 2x − 14 e−x cos 2x + c
∫ e−x cos 2x dx = 15 e−x(2 sin 2x − cos 2x) + c
8 a 1
x
b u = ln x, ddux
= 1x
; dd
vx
= 1, v = x
∫ ln x dx = x ln x − ∫ 1x
× x dx
= x ln x − ∫ dx
= x ln x − x + c = x(ln x − 1) + c
C4 INTEGRATION Answers - Worksheet F page 4
Solomon Press
9 a u = ln 2x, d
dux
= 1x
; dd
vx
= 1, v = x b u = ln x, ddux
= 1x
; dd
vx
= 3x, v = 32 x2
∫ ln 2x dx = x ln 2x − ∫ 1x
× x dx ∫ 3x ln x dx = 32 x2 ln x − ∫ 1
x× 3
2 x2 dx
= x ln 2x − ∫ dx = 32 x2 ln x − ∫ 3
2 x dx
= x ln 2x − x + c = 32 x2 ln x − 3
4 x2 + c = x(ln 2x − 1) + c = 3
4 x2(2 ln x − 1) + c
c u = (ln x)2, ddux
= 2(ln x) × 1x
; dd
vx
= 1, v = x
∫ (ln x)2 dx = x(ln x)2 − ∫ 2 ln x dx
for ∫ 2 ln x dx, u = ln x, ddux
= 1x
; dd
vx
= 2, v = 2x
∫ 2 ln x dx = 2x ln x − ∫ 2 dx
= 2x ln x − 2x + c ∴ ∫ (ln x)2 dx = x(ln x)2 − (2x ln x − 2x) + c
= x[(ln x)2 − 2 ln x + 2] + c 10 a u = x + 2, d
dux
= 1; dd
vx
= ex, v = ex b u = ln x, ddux
= 1x
; dd
vx
= x2, v = 13 x3
0
1−∫ (x + 2)ex dx 2
1∫ x2 ln x dx
= [(x + 2)ex] 01− −
0
1−∫ ex dx = [ 13 x3 ln x] 2
1 − 2
1∫ 13 x2 dx
= [(x + 2)ex − ex] 01− = [ 1
3 x3 ln x − 19 x3] 2
1 = (2 − 1) − (e−1 − e−1) = ( 8
3 ln 2 − 89 ) − (0 − 1
9 ) = 1 = 8
3 ln 2 − 79
c u = 2x, ddux
= 2; dd
vx
= e3x − 1, v = 13 e3x − 1 d u = ln (2x + 3), d
dux
= 22 3x +
; dd
vx
= 1, v = x
13
1
∫ 2xe3x − 1 dx 3
0∫ ln (2x + 3) dx
= [ 23 xe3x − 1] 1
3
1 − 13
1
∫ 23 e3x − 1 dx = [x ln (2x + 3)] 3
0 − 3
0∫2
2 3x
x + dx
= [ 23 xe3x − 1 − 2
9 e3x − 1] 13
1 = [x ln (2x + 3)] 30 −
3
0∫(2 3) 3
2 3x
x+ −
+ dx
= ( 23 e2 − 2
9 e2) − ( 29 − 2
9 ) = [x ln (2x + 3)] 30 −
3
0∫ (1 − 32 3x +
) dx
= 49 e2 = [x ln (2x + 3) − x + 3
2 ln2x + 3] 30
= (3 ln 9 − 3 + 32 ln 9) − (0 − 0 + 3
2 ln 3) = 15
2 ln 3 − 3
C4 INTEGRATION Answers - Worksheet F page 5
Solomon Press
e u = x2, d
dux
= 2x; dd
vx
= cos x, v = sin x
∫ x2 cos x dx = x2 sin x − ∫ 2x sin x dx
for ∫ 2x sin x dx, u = 2x, ddux
= 2; dd
vx
= sin x, v = −cos x
∫ 2x sin x dx = −2x cos x − ∫ −2 cos x dx
= −2x cos x + ∫ 2 cos x dx
= −2x cos x + 2 sin x + c
∴ π2
0∫ x2 cos x dx = [x2 sin x + 2x cos x − 2 sin x]π20
= ( 14 π2 + 0 − 2) − (0 + 0 − 0)
= 14 π2 − 2
f u = e3x, ddux
= 3e3x; dd
vx
= sin 2x, v = − 12 cos 2x
∫ e3x sin 2x dx = − 12 e3x cos 2x − ∫ − 3
2 e3x cos 2x dx
= − 12 e3x cos 2x + ∫ 3
2 e3x cos 2x dx
for ∫ 32 e3x cos 2x dx, u = 3
2 e3x, ddux
= 92 e3x; d
dvx
= cos 2x, v = 12 sin 2x
∫ 32 e3x cos 2x dx = 3
4 e3x sin 2x − ∫ 94 e3x sin 2x dx
∴ ∫ e3x sin 2x dx = − 12 e3x cos 2x + 3
4 e3x sin 2x − ∫ 94 e3x sin 2x dx
134 ∫ e3x sin 2x dx = − 1
2 e3x cos 2x + 34 e3x sin 2x + c
∴ π4
0∫ e3x sin 2x dx = 413 [− 1
2 e3x cos 2x + 34 e3x sin 2x]
π40
= 413 [(0 +
3π43
4 e ) − ( 12− + 0)]
= 113 (
3π43e + 2)
Solomon Press
INTEGRATION C4 Answers - Worksheet G
1 a x = 0 ⇒ t = 2 2 a x = 0 ⇒ cos θ = 0 ⇒ θ = π2 , 3π
2
x = 2 ⇒ t = 3 for y > 0, θ = π2 at A
b area = 2
0∫ y dx y = 0 ⇒ sin θ = 0 ⇒ θ = 0, π
x = 2t − 4 ∴ ddxt
= 2 for x > 0, θ = 0 at B
∴ area = 3
2∫1t
× 2 dt b x = 4 cos θ ∴ dd
xθ
= −4 sin θ
= 3
2∫2t
dt ∴ area = π2
0
∫ 2 sin θ × −4 sin θ dθ
c = [2 lnt] 32 =
π2
0∫ 8 sin2 θ dθ
= 2 ln 3 − 2 ln 2 c shaded area = π2
0∫ (4 − 4 cos 2θ ) dθ
= 2 ln 32 = [4θ − 2 sin 2θ ]
π20
d t = 42
x + = (2π − 0) − (0 − 0)
∴ y = 24x +
= 2π
∴ area = 2
0∫2
4x + dx area of ellipse = 4 × 2π
= [2 lnx + 4] 20 = 8π
= 2 ln 6 − 2 ln 4 = 2 ln 3
2
3 a y = 0 ⇒ sin 2t = 0 ⇒ t = 0, π2
x = 2 sin t ∴ ddxt
= 2 cos t
area above x-axis
= π2
0∫ 5 sin 2t × 2 cos t dt
= π2
0∫ 10 sin 2t cos t dt
area enclosed by curve
= 2π2
0∫ 10 sin 2t cos t dt
= π2
0∫ 20 sin 2t cos t dt
b = 40π2
0∫ sin t cos2 t dt
= −40π2
0∫ (−sin t)cos2 t dt
= −40[ 13 cos3 t]
π20
= 403− (0 − 1)
= 1313
Solomon Press
INTEGRATION C4 Answers - Worksheet H
1 a = 1
2 × 15 (2x − 3)5 + c b = −2 cot 1
2 x + c = 1
10 (2x − 3)5 + c
c = 12 e4x − 1 + c d 2( 1)
( 1)x
x x−+
≡ Ax
+ 1
Bx +
, 2(x − 1) ≡ A(x + 1) + Bx
x = 0 ⇒ A = −2, x = −1 ⇒ B = 4 ∫ 2( 1)
( 1)x
x x−+
dx = ∫ ( 41x +
− 2x
) dx
= 4 lnx + 1 − 2 lnx + c e = ∫ 3 sec2 2x dx f = 1
2 ∫ 2x(x2 + 3)3 dx
= 32 tan 2x + c = 1
2 × 14 (x2 + 3)4 + c
= 18 (x2 + 3)4 + c
g = ∫ (sec x tan x)sec3 x dx h = 12 ×
322
3 (7 2 )x+ + c
= 14 sec4 x + c =
321
3 (7 2 )x+ + c
i u = x, ddux
= 1; dd
vx
= e3x, v = 13 e3x j 2
( 3)( 1)x
x x+
− +≡
3A
x −+
1B
x +, x + 2 ≡ A(x + 1) + B(x − 3)
∫ xe3x dx = 13 xe3x − ∫ 1
3 e3x dx x = 3 ⇒ A = 54 , x = −1 ⇒ B = 1
4−
= 13 xe3x − 1
9 e3x + c ∫ 22
2 3x
x x+
− − dx = ∫ (
54
3x − −
14
1x +) dx
= 19 e3x(3x − 1) + c = 5
4 lnx − 3 − 14 lnx + 1 + c
k = 1
4 × ( 12− )(x + 1)−2 + c l = ∫ (sec2 3x − 1) dx
= − 21
8( 1)x + + c = 1
3 tan 3x − x + c
m = ∫ [2 + 2 cos (4x + 2)] dx n = 32− ∫ 2
21
xx
−−
dx
= 2x + 12 sin (4x + 2) + c = 3
2− ln1 − x2 + c
o u = x, ddux
= 1; dd
vx
= sin 2x, v = 12− cos 2x p = ∫ ( 2) 2
2x
x+ +
+ dx
∫ x sin 2x dx = ∫ (1 + 22x +
) dx
= 12− x cos 2x + ∫ 1
2 cos 2x dx = x + 2 lnx + 2 + c
= 12− x cos 2x + 1
4 sin 2x + c
C4 INTEGRATION Answers - Worksheet H page 2
Solomon Press
2 a 2
1∫ 6e2x − 3 dx b π3
0∫ tan x dx = −π3
0∫sin
cosx
x− dx
= [3e2x − 3] 21 = −[lncos x] 3
π
0 = 3(e − e−1) = −(ln 1
2 − 0) = ln 2
c 2
2−∫2
3x − dx d 6
(4 )(1 )x
x x+
− +≡
4A
x−+
1B
x+, 6 + x ≡ A(1 + x) + B(4 − x)
= [2 lnx − 3] 22− x = 4 ⇒ A = 2, x = −1 ⇒ B = 1
= 0 − 2 ln 5 3
2∫ 26
4 3x
x x+
+ − dx =
3
2∫ ( 24 x−
+ 11 x+
) dx
= −2 ln 5 = [−2 ln4 − x + ln1 + x] 32
= (0 + ln 4) − (−2 ln 2 + ln 3) = 4 ln 2 − ln 3
e 2
1∫ (1 − 2x)3 dx f π3
0∫ sin2 x sin 2x dx = π3
0∫ 2 sin3 x cos x dx
= [ 12− × 1
4 (1 − 2x)4] 21 = [ 1
2 sin4 x] 3π
0
= 18− (81 − 1) = 1
2 ( 32 )4 − 0
= −10 = 932
3 a x = 3 sin u ∴ d
dxu
= 3 cos u b u = 1 − 3x ∴ x = 13 (1 − u), d
dux
= −3
x = 0 ⇒ u = 0 x = 0 ⇒ u = 1 x = 3
2 ⇒ u = π6 x = 1 ⇒ u = −2
32
0∫ 2
1
9 x− dx =
π6
0∫1
3cosu× 3 cos u du
1
0∫ x(1 − 3x)3 dx = 2
1
−
∫ 13 (1 − u)u3 × ( 1
3− ) du
= π6
0∫ du = 19
1
2−∫ (u3 − u4) du
= [u]π60 = 1
9 [ 14 u4 − 1
5 u5] 12−
= π6 − 0 = 19 [( 1
4 − 15 ) − (4 + 32
5 )]
= π6 = 2320−
c x = 2 tan u ∴ dd
xu
= 2 sec2 u d u2 = x + 1 ∴ x = u2 − 1, dd
xu
= 2u
x = 2 ⇒ u = π4 x = −1 ⇒ u = 0
x = 2 3 ⇒ u = π3 x = 0 ⇒ u = 1
2 3
2∫ 21
4 x+ dx =
π3π4∫ 2
14sec u
× 2 sec2 u du 0
1−∫2 1x x + dx =
1
0∫ (u2 − 1)2u × 2u du
= 12
π3π4∫ du =
1
0∫ 2u2(u4 − 2u2 + 1) du
= 12 [u]
π3π4
= 1
0∫ (2u6 − 4u4 + 2u2) du
= 12 ( π
3 − π4 ) = [ 27 u7 − 4
5 u5 + 23 u3] 1
0
= 124 π = ( 2
7 − 45 + 2
3 ) − (0) = 16
105
C4 INTEGRATION Answers - Worksheet H page 3
Solomon Press
4 a = 2
3− ln5 − 3x + c b = 12 ∫ (2x + 2)
2 2ex x+ dx
= 12
2 2ex x+ + c
c = ∫31
2 2(2 1)2 1xx
− + ++
dx d = 12 ∫ (sin 5x + sin x) dx
= ∫ (32
2 1x + − 1
2 ) dx = 12 (− 1
5 cos 5x − cos x) + c
= 34 ln2x + 1 − 1
2 x + c = 110− (cos 5x + 5 cos x) + c
e u = 3x, ddux
= 3; dd
vx
= (x − 1)4, v = 15 (x − 1)5 f
23 6 2( 1)( 2)
x xx x
+ ++ +
≡ 3 + 1
Ax +
+ 2
Bx +
∫ 3x(x − 1)4 dx 3x2 + 6x + 2 ≡ 3(x + 1)(x + 2) + A(x + 2) + B(x + 1)
= 35 x(x − 1)5 − ∫ 3
5 (x − 1)5 dx x = −1 ⇒ A = −1, x = −2 ⇒ B = −2
= 35 x(x − 1)5 − 1
10 (x − 1)6 + c ∫2
23 6 2
3 2x xx x
+ ++ +
dx = ∫ (3 − 11x +
− 22x +
) dx
= 110 (x − 1)5[6x − (x − 1)] + c = 3x − lnx + 1 − 2 lnx + 2 + c
= 110 (5x + 1)(x − 1)5 + c
g = ∫135(2 1)x −− dx h = 1
3 ∫ 3cos2 3sin
xx+
dx
= 12 ×
2315
2 (2 1)x − + c = 13 ln2 + 3 sin x + c
= 2315
4 (2 1)x − + c
i = 13 ∫
122 33 ( 1)x x −− dx j = ∫ (4 − 4 cot x + cot2 x) dx
= 13 ×
1232( 1)x − + c = ∫ (4 − 4 cos
sinxx
+ cosec2 x − 1) dx
= 323 1x − + c = 3x − 4 lnsin x − cot x + c
k 26 5
( 1)(2 1)x
x x−
− − ≡
1A
x − +
2 1B
x − + 2(2 1)
Cx −
l u = x2, ddux
= 2x; dd
vx
= e−x, v = −e−x
6x − 5 ≡ A(2x − 1)2 + B(x − 1)(2x − 1) + C(x − 1) ∫ x2e−x dx = −x2e−x + ∫ 2xe−x dx
x = 1 ⇒ A = 1, x = 12 ⇒ C = 4 u = 2x, d
dux
= 2; dd
vx
= e−x, v = −e−x
coeffs of x2 ⇒ B = −2 ∫ x2e−x dx = −x2e−x − 2xe−x + ∫ 2e−x dx
∫ 26 5
( 1)(2 1)x
x x−
− − dx = −x2e−x − 2xe−x − 2e−x + c
= ∫ ( 11x −
− 22 1x −
+ 24
(2 1)x −) dx = −e−x(x2 + 2x + 2) + c
= lnx − 1 − ln2x − 1 − 2(2x − 1)−1 + c
= ln 12 1xx−−
− 22 1x −
+ c
C4 INTEGRATION Answers - Worksheet H page 4
Solomon Press
5 a 4
2∫1
3 4x − dx b
π4π6∫ cosec2 x cot2 x dx = −
π4π6∫ (−cosec2 x)cot2 x dx
= [ 13 ln3x − 4] 4
2 = −[ 13 cot3 x]
π4π6
= 13 (ln 8 − ln 2) = − 1
3 [1 − ( 3 )3]
= 23 ln 2 = 3 − 1
3
c 2
27
(2 ) (3 )x
x x−
− − ≡
2A
x− + 2(2 )
Bx−
+ 3
Cx−
d u = x, ddux
= 1; dd
vx
= cos 12 x, v = 2 sin 1
2 x
7 − x2 ≡ A(2 − x)(3 − x) + B(3 − x) + C(2 − x)2
π2
0∫ x cos 12 x dx
x = 2 ⇒ B = 3, x = 3 ⇒ C = −2 = [2x sin 12 x]
π20 −
π2
0∫ 2 sin 12 x dx
coeffs of x2 ⇒ A = 1 = [2x sin 12 x + 4 cos 1
2 x]π20
1
0∫2
27
(2 ) (3 )x
x x−
− − dx = [π( 1
2) − 4( 1
2)] − [0 + 4]
= 1
0∫ ( 12 x−
+ 23
(2 )x− − 2
3 x−) dx = 1
2 2 (π − 4) − 4
= [−ln2 − x+ 3(2 − x)−1 + 2 ln3 − x] 10
= (0 + 3 + 2 ln 2) − (−ln 2 + 32 + 2 ln 3)
= 32 + 3 ln 2 − 2 ln 3
e 5
1∫1
4 5x + dx f
π6π6−∫ 2 cos x cos 3x dx
= [ 14 ×
122(4 5)x + ] 5
1 = π6π6−∫ [cos 4x + cos (−2x)] dx
= 12 (5 − 3) =
π6π6−∫ (cos 4x + cos 2x) dx
= 1 = [ 14 sin 4x + 1
2 sin 2x]π6π6−
= [ 14 ( 3
2 ) + 12 ( 3
2 )] − [ 14 (− 3
2 ) + 12 (− 3
2 )]
= 34 3
g 2
0∫22 1x x + dx = 1
4
2
0∫24 2 1x x + dx h
= 14 [
3222
3 (2 1)x + ] 20
= 16 (27 − 1)
= 134
1
0∫2 1
2xx
+−
dx = 1
0∫ (x + 2 + 52x −
) dx
= [ 12 x2 + 2x + 5 lnx − 2] 1
0 = ( 1
2 + 2 + 0) − (0 + 0 + 5 ln 2) = 5
2 − 5 ln 2
x + 2 x − 2 x2 + 0x + 1
x2 − 2x 2x + 1 2x − 4
5
C4 INTEGRATION Answers - Worksheet H page 5
Solomon Press
i u = x − 2, ddux
= 1; dd
vx
= (x + 1)3, v = 14 (x + 1)4
1
0∫ (x − 2)(x + 1)3 dx = [ 14 (x − 2)(x + 1)4] 1
0 − 1
0∫ 14 (x + 1)4 dx
= [ 14 (x − 2)(x + 1)4 − 1
20 (x + 1)5] 10
= (−4 − 85 ) − ( 1
2− − 120 )
= 1205−
6 a = 2
1∫ 2 3( 2)x
x + dx b =
4
2∫ ln x dx
= 12
2
1∫ 2 32
( 2)x
x + dx u = ln x, d
dux
= 1x
; dd
vx
= 1, v = x
= 12 [ 1
2− (x2 + 2)−2] 21 = [x ln x] 4
2 − 4
2∫ dx
= 14− ( 1
36 − 19 ) = [x ln x − x] 4
2 = 1
48 = (4 ln 4 − 4) − (2 ln 2 − 2) = 6 ln 2 − 2
7 6
3∫2ax bx+ dx =
6
3∫ (ax + bx
) dx 8 a 6 − 2ex = 0
= [ 12 ax2 + b ln | x |] 6
3 x = ln 3 ∴ (ln 3, 0)
= (18a + b ln 6) − ( 92 a + b ln 3) b =
ln3
0∫ (6 − 2ex) dx
∴ 272 a + b ln 2 = 18 + 5 ln 2 = [6x − 2ex] ln3
0 a, b rational = (6 ln 3 − 6) − (0 − 2) ∴ b = 5, 27
2 a = 18 = 6 ln 3 − 4 a = 4
3 , b = 5 9 u = cot x ∴ d
dux
= −cosec2 x 10 a y = 0 ⇒ 4 − t2 = 0
x = π6 ⇒ u = 3 t = ± 2
x = π4 ⇒ u = 1 x = t + 1 ∴ ddxt
= 1
cosec2 x = 1 + cot2 x = 1 + u2 ∴ area = 2
2−∫ y × 1 dt
∴ π4π6∫ cot2 x cosec4 x dx =
2
2−∫ (4 − t2) dt
= 1
3∫ u2(1 + u2) × (−1) du b = [4t − 13 t3] 2
2−
= 3
1∫ (u2 + u4) du = (8 − 83 ) − (−8 + 8
3 )
= [ 13 u3 + 1
5 u5] 31 = 2
310
= ( 3 + 95 3 ) − ( 1
3 + 15 )
= 145 3 − 8
15
= 215 ( 21 3 − 4)
C4 INTEGRATION Answers - Worksheet H page 6
Solomon Press
11 a d
dx(x2 sin 2x + 2kx cos 2x − k sin 2x) 12 curve meets x-axis when 2
ln xx
= 0 ∴ x = 1
= 2x sin 2x + 2x2 cos 2x + 2k cos 2x area = 2
1∫ 2ln xx
dx
−4kx sin 2x − 2k cos 2x u = ln x, ddux
= 1x
; dd
vx
= x−2, v = −x−1
= 2x2 cos 2x + (2 − 4k)x sin 2x area = [− ln xx
] 21 +
2
1∫ x−2 dx
b let k = 12 = [− ln x
x − x−1] 2
1
ddx
(x2 sin 2x + x cos 2x − 12 sin 2x) = ( 1
2− ln 2 − 12 ) − (0 − 1)
= 2x2 cos 2x = 12 − 1
2 ln 2
∴ ∫ x2 cos 2x dx = 12 (1 − ln 2)
= 12 (x2 sin 2x + x cos 2x − 1
2 sin 2x) + c = 1
4 (2x2 sin 2x + 2x cos 2x − sin 2x) + c 13 a f(1) = 18, f(2) = 80, f(−1) = −4, f(−2) = 0 ∴ (x + 2) is a factor ∴ 3x3 + 11x2 + 8x − 4 = (x + 2)(3x2 + 5x − 2)
= (3x − 1)(x + 2)2 b 3 2
163 11 8 4
xx x x
++ + −
≡ 3 1
Ax −
+ 2
Bx +
+ 2( 2)C
x +
x + 16 ≡ A(x + 2)2 + B(3x − 1)(x + 2) + C(3x − 1) x = 1
3 ⇒ 493 = 49
9 A ⇒ A = 3 x = −2 ⇒ 14 = −7C ⇒ C = −2 coeffs of x2 ⇒ 0 = A + 3B ⇒ B = −1
∴ f(x) ≡ 33 1x −
− 12x +
− 22
( 2)x +
c = 0
1−∫ ( 33 1x −
− 12x +
− 22
( 2)x +) dx
= [ln3x − 1 − lnx + 2 + 2(x + 2)−1] 01−
= (0 − ln 2 + 1) − (ln 4 − 0 + 2) = −1 − ln 2 − ln 22 = −1 − 3 ln 2 = −(1 + 3 ln 2)
3x2 + 5x − 2 x + 2 3x3 + 11x2 + 8x − 4
3x3 + 6x2 5x2 + 8x 5x2 + 10x − 2x − 4 − 2x − 4
Solomon Press
INTEGRATION C4 Answers - Worksheet I
1 = π
12
2
∫ ( 2x
)2 dx 2 = π2
0∫ (x2 + 3)2 dx
= π12
2
∫ 4x−2 dx = π2
0∫ (x4 + 6x2 + 9) dx
= π[−4x−1] 12
2 = π[ 15 x5 + 2x3 + 9x] 2
0
= π[−2 − (−8)] = π[( 325 + 16 + 18) − (0)]
= 6π = 2025 π ≈ 127
3 a = π
1
0∫ ( 22ex)2 dx b = π
1
2
−
−∫ ( 23x
)2 dx
= π1
0∫ 4ex dx = π1
2
−
−∫ 9x−4 dx
= π[4ex] 10 = π[−3x−3] 1
2−−
= π(4e − 4) = π(3 − 38 )
= 4π(e − 1) = 218 π
c = π9
3∫ (1 + 1x
)2 dx d = π2
1∫ (3x + 1x
)2 dx
= π9
3∫ (1 + 2x−1 + x−2) dx = π2
1∫ (9x2 + 6 + x−2) dx
= π[x + 2 ln | x | − x−1] 93 = π[3x3 + 6x − x−1] 2
1 = π[(9 + 2 ln 9 − 1
9 ) − (3 + 2 ln 3 − 13 )] = π[(24 + 12 − 1
2 ) − (3 + 6 − 1)] = π( 2
96 + 2 ln 3) = 552 π
e = π6
2∫ ( 12x +
)2 dx f = π1
1−∫ (e1 − x)2 dx
= π6
2∫1
2x + dx = π
1
1−∫ e2 − 2x dx
= π[lnx + 2] 62 = π[ 1
2− e2 − 2x] 11−
= π(ln 8 − ln4) = 12− π(1 − e4)
= π ln 2 = 12 π(e4 − 1)
4 a =
2
0∫4
2x + dx 5 = π
3
1∫ (122x +
12x− )2 dx
= [4 lnx + 2] 20 = π
3
1∫ (4x + 4 + x−1) dx
= 4(ln 4 − ln 2) = 4 ln 2 = π[2x2 + 4x + ln | x |] 31
b = π2
0∫ ( 42x +
)2 dx = π[(18 + 12 + ln 3) − (2 + 4 + 0)]
= π2
0∫ 16(x + 2)−2 dx = π(24 + ln 3)
= π[−16(x + 2)−1] 20
= π[−4 − (−8)] = 4π
C4 INTEGRATION Answers - Worksheet I page 2
Solomon Press
6 a y 7 a y = 0 ∴ x = 1
3 ∴ ( 1
3 , 0)
b = 13
3
∫ (3 − 1x
) dx
= [3x − ln | x |] 13
3
= (9 − ln 3) − (1 − ln 13 )
b = π3
0∫ (3x − x2)2 dx = 8 − 2 ln 3
= π3
0∫ (9x2 − 6x3 + x4) dx c = π13
3
∫ (3 − 1x
)2 dx
= π[3x3 − 32 x4 + 1
5 x5] 30 = π
13
3
∫ (9 − 6x−1 + x−2) dx
= π[(81 − 2432 + 243
5 ) − (0)] = π[9x − 6 ln | x | − x−1] 13
3
= 8110 π = π[(27 − 6 ln 3 − 1
3 ) − (3 − 6 ln 13 − 3)]
= π( 2326 − 12 ln 3)
8 a x − 1
x = 0
x2 = 1 x = ± 1 ∴ (−1, 0) and (1, 0)
b = 3
1∫ (x − 1x
) dx
= [ 12 x2 − ln | x |] 3
1 = ( 9
2 − ln 3) − ( 12 − 0)
= 4 − ln 3
c = π3
1∫ (x − 1x
)2 dx
= π3
1∫ (x2 − 2 + x−2) dx
= π[ 13 x3 − 2x − x−1] 3
1 = π[(9 − 6 − 1
3 ) − ( 13 − 2 − 1)]
= 163 π
(0, 0) (3, 0)
O x
Solomon Press
INTEGRATION C4 Answers - Worksheet J
1 a d
dyx
= 2x, grad = 2 2 a = e
1∫ (4x + 9x
) dx
∴ grad of normal = 12− = [2x2 + 9 ln | x |] e
1 ∴ y − 2 = 1
2− (x − 1) = (2e2 + 9) − (2 + 0) [ y = 5
2 − 12 x ] = 2e2 + 7
b y = 0 ∴ x = 5 b = πe
1∫ (4x + 9x
)2 dx
∴ (5, 0) = πe
1∫ (16x2 + 72 + 81x−2) dx
c volume 0 ≤ x ≤ 1 = π[ 163 x3 + 72x − 81x−1] e
1
= π1
0∫ (x2 + 1)2 dx = π[( 163 e3 + 72e − 81e−1) − ( 16
3 + 72 − 81)]
= π1
0∫ (x4 + 2x2 + 1) dx = 869 (3sf)
= π[ 15 x5 + 2
3 x3 + x] 10
= π[( 15 + 2
3 + 1) − (0)] = 2815 π
volume 1 < x ≤ 5 = volume of cone = 1
3 × π × 22 × 4 = 163 π
total volume = 28
15 π + 163 π
= 365 π
3 a = π
π3π6∫ cosec2 x dx b = π
4
1∫ ( 32
xx
++
)2 dx
= π[−cot x]π3π6
= π4
1∫32
xx
++
dx
= −π( 13
− 3 ) = π4
1∫( 2) 1
2x
x+ +
+ dx
= π( 3 − 13 3 ) = π
4
1∫ (1 + 12x +
) dx
= 23 π 3 = π[x + lnx + 2] 4
1 = π[(4 + ln 6) − (1 + ln 3)] = π(3 + ln 2)
c = ππ4
0∫ (1 + cos 2x)2 dx d = π2
1∫ (12x e2 − x)2 dx
= ππ4
0∫ (1 + 2 cos 2x + cos2 2x) dx = π2
1∫ xe4 − 2x dx
= ππ4
0∫ ( 32 + 2 cos 2x + 1
2 cos 4x) dx u = x, ddux
= 1; dd
vx
= e4 − 2x, v = − 12 e4 − 2x
= π[ 32 x + sin 2x + 1
8 sin 4x]π40 = π{[− 1
2 xe4 − 2x] 21 +
2
1∫ 12 e4 − 2x dx}
= π[( 38 π + 1 + 0) − (0)] = π[− 1
2 xe4 − 2x − 14 e4 − 2x] 2
1
= 18 π(3π + 8) = π[(−1 − 1
4 ) − (− 12 e2 − 1
4 e2)] = 1
4 π(3e2 − 5)
C4 INTEGRATION Answers - Worksheet J page 2
Solomon Press
4 volume = π1
0∫ (12e xx − )2 dx 5 a =
π2
0∫ (2 sin x + cos x) dx
= π1
0∫ x2e−x dx = [−2 cos x + sin x]π20
u = x2, ddux
= 2x; dd
vx
= e−x, v = −e−x = (0 + 1) − (−2 + 0)
∫ x2e−x dx = −x2e−x + ∫ 2xe−x dx = 3
u = 2x, ddux
= 2; dd
vx
= e−x, v = −e−x b = ππ2
0∫ (2 sin x + cos x)2 dx
∫ x2e−x dx = −x2e−x − 2xe−x + ∫ 2e−x dx = ππ2
0∫ (4 sin2 x + 4 sin x cos x + cos2 x) dx
= −x2e−x − 2xe−x − 2e−x + c = ππ2
0∫ (2−2 cos 2x + 2 sin 2x + 12 + 1
2 cos 2x) dx
volume = π[−e−x(x2 + 2x + 2)] 10 = π
π2
0∫ ( 52 − 3
2 cos 2x + 2 sin 2x) dx
= π[−e−1(1 + 2 + 2)] − [−1(2)] = π[ 52 x − 3
4 sin 2x − cos 2x]π20
= π(2 − 5e−1) = π[( 54 π − 0 + 1) − (0 − 0 − 1)]
= 14 π(5π + 8)
6 a x = 0 ⇒ θ = 0 7 a y = 0 ⇒ t = 0, −1 x = 1 ⇒ θ = π4 x = 0 ⇒ t = ± 1
b x = tan θ ∴ dd
xθ
= sec2 θ t ≥ 0 ∴ t = 0, 1
∴ volume = ππ4
0∫ (sin 2θ )2 × sec2 θ dθ b x = t2 − 1 ∴ ddxt
= 2t
= ππ4
0∫ (4 sin2 θ cos2 θ × 21
cos θ) dθ ∴ volume = π
1
0∫ [t(t + 1)]2 × 2t dt
= 4ππ4
0∫ sin2 θ dθ = 2π1
0∫ t3(t2 + 2t + 1) dt
c = 4ππ4
0∫ ( 12 − 1
2 cos 2θ ) dθ = 2π1
0∫ (t5 + 2t4 + t3) dt
= 4π[ 12 θ − 1
4 sin 2θ ]π40 = 2π[ 1
6 t6 + 25 t5 + 1
4 t4] 10
= 4π[( 18 π − 1
4 ) − (0)] = 2π[( 16 + 2
5 + 14 ) − (0)]
= 12 π(π − 2) = 49
30 π
Solomon Press
INTEGRATION C4 Answers - Worksheet K 1 a y = ∫ (x + 2)3 dx b y = ∫ 4 cos 2x dx
y = 14 (x + 2)4 + c y = 2 sin 2x + c
c x = ∫ 3e2t + 2 dt d ddyx
= 12 x−
x = 32 e2t + 2t + c y = ∫ 1
2 x− dx
y = −ln2 − x + c
e N = ∫ 2 1t t + dt f y = ∫ xex dx
N = 12 ∫
1222 ( 1)t t + dt u = x, d
dux
= 1; dd
vx
= ex, v = ex
N = 12 ×
3222
3 ( 1)t + + c y = xex − ∫ ex dx
N = 3221
3 ( 1)t + + c y = xex − ex + c [ y = ex(x − 1) + c ] 2 a y = ∫ e−x dx b y = ∫ tan3 t sec2 t dt
y = −e−x + c y = 14 tan4 t + c
y = 3 when x = 0 y = 1 when t = π3
∴ 3 = −1 + c ∴ 1 = 14 ( 3 )4 + c
c = 4 c = 1 − 94 = 5
4− ∴ y = 4 − e−x ∴ y = 1
4 tan4 t − 54 [ y = 1
4 (tan4 t − 5) ]
c ddux
= 24
3x
x − d y = ∫ 3 cos2 x dx
u = ∫ 24
3x
x − dx = 2 ∫ 2
23
xx −
dx y = 32 ∫ (1 + cos 2x) dx
u = 2 lnx2 − 3 + c y = 32 (x + 1
2 sin 2x) + c = 34 (2x + sin 2x) + c
u = 5 when x = 2 y = π when x = π2
∴ 5 = 0 + c ∴ π = 34 (π + 0) + c
c = 5 c = π4
∴ u = 2 lnx2 − 3 + 5 ∴ y = 34 (2x + sin 2x) + π4
[ y = 14 (6x + 3 sin 2x + π) ]
3 a 2
86
xx x
−− −
≡ 3
Ax −
+ 2
Bx +
b ddyx
= 28
6x
x x−
− −
x − 8 ≡ A(x + 2) + B(x − 3) y = ∫ 28
6x
x x−
− − dx = ∫ ( 2
2x + − 1
3x −) dx
x = 3 ⇒ −5 = 5A ⇒ A = −1 y = 2 lnx + 2 − lnx − 3 + c x = −2 ⇒ −10 = −5B ⇒ B = 2 y = ln 9 when x = 1 2
86
xx x
−− −
≡ 22x +
− 13x −
∴ ln 9 = 2 ln 3 − ln 2 + c
c = ln 2 (ln 9 = ln 32 = 2 ln 3) ∴ y = 2 lnx + 2 − lnx − 3 + ln 2 when x = 2, y = 2 ln 4 − 0 + ln 2 = ln (42 × 2) = ln 32
C4 INTEGRATION Answers - Worksheet K page 2
Solomon Press
4 a ∫ 1
2 3y + dy = ∫ dx b ∫ cosec2 2y dy = ∫ dx
12 ln2y + 3 = x + c 1
2− cot 2y = x + c [ y = 1
2 (ke2x − 3) ] [ cot 2y = k − 2x ]
c ∫ 1y
dy = ∫ x dx d ∫ 1y
dy = ∫ 11x +
dx
lny = 12 x2 + c lny = lnx + 1 + c
[ y = 21
2e xk ] [ y = k(x + 1) ]
e ∫ y dy = ∫ (x2 − 2) dx f ∫ sec2 y dy = ∫ 2 cos x dx
12 y2 = 1
3 x3 − 2x + c tan y = 2 sin x + c [ y2 = 2
3 x3 − 4x + k ]
g ∫ e3 − y dy = ∫12x− dx h y d
dyx
= x(y2 + 3)
−e3 − y = 122x + c ∫ 2 3
yy +
dy = ∫ x dx
[ y = 3 − ln (k − 2 x ) ] 12 ∫ 2
23
yy +
dy = ∫ x dx
12 lny2 + 3 = 1
2 x2 + c
[ y2 = 2
exk − 3 ]
i ∫ 1y
dy = ∫ x sin x dx j ddyx
= 2e
e
x
y
u = x, ddux
= 1; dd
vx
= sin x, v = −cos x ∫ ey dy = ∫ e2x dx
lny = −x cos x + ∫ cos x dx ey = 12 e2x + c
lny = sin x − x cos x + c [ y = ln ( 12 e2x + c) ]
[ y = kesin x − x cos x ]
k ∫ 3( 1)y
y y−−
dy = ∫ x dx l ∫ y−2 dy = ∫ ln x dx
3( 1)y
y y−−
≡ Ay
+ 1
By −
u = ln x, ddux
= 1x
; dd
vx
= 1, v = x
y − 3 ≡ A(y − 1) + By −y−1 = x ln x − ∫ dx
y = 0 ⇒ A = 3, y = 1 ⇒ B = −2 −y−1 = x ln x − x + c ∫ ( 3
y − 2
1y −) dy = ∫ x dx [ y = 1
lnx x x k− + ]
3 lny − 2 lny − 1 = 12 x2 + c
C4 INTEGRATION Answers - Worksheet K page 3
Solomon Press
5 a ∫ 2y dy = ∫ x dx b ∫ (y + 1)−3 dy = ∫ dx
y2 = 12 x2 + c 1
2− (y + 1)−2 = x + c y = 3 when x = 4 (y + 1)−2 = k − 2x ∴ 9 = 8 + c y = 0 when x = 2 c = 1 ∴ 1 = k − 4 ∴ y2 = 1
2 x2 + 1 k = 5 ∴ (y + 1)−2 = 5 − 2x [ (y + 1)2 = 1
5 2x− ]
c ∫ 1y
dy = ∫ cot2 x dx d ∫ 12y +
dy = ∫ 11x −
dx
∫ 1y
dy = ∫ (cosec2 x − 1) dx lny + 2 = lnx − 1 + c
lny = −cot x − x + c y = 6 when x = 3 y = 1 when x = π2 ∴ ln 8 = ln 2 + c
∴ 0 = 0 − π2 + c c = ln 4
c = π2 ∴ lny + 2 = lnx − 1 + ln 4
∴ lny = π2 − cot x − x [ y + 2 = 4(x − 1) ⇒ y = 4x − 6 ]
e ∫ cot y dy = ∫ x2 dx f ddyx
= 3
yx +
∫ cossin
yy
dy = ∫ x2 dx ∫12y− dy = ∫
12( 3)x −+ dx
lnsin y = 13 x3 + c
122y =
122( 3)x + + c
y = π6 when x = 0 y = 3x + + k
∴ ln 12 = 0 + c y = 16 when x = 1
c = −ln 2 ∴ 4 = 2 + k ∴ lnsin y = 1
3 x3 − ln 2 k = 2
[ 2 sin y = 31
3e x ] ∴ y = 3x + + 2
[ y = ( 3x + + 2)2 ]
g ∫ sin y dy = ∫ xe−x dx h ∫ sin1 cos
yy+
dy = ∫ 12 x−2 dx
u = x, ddux
= 1; dd
vx
= e−x, v = −e−x − ∫ sin1 cos
yy
−+
dy = ∫ 12 x−2 dx
−cos y = −xe−x + ∫ e−x dx −ln1 + cos y = 12− x−1 + c
−cos y = −xe−x − e−x + c ln1 + cos y = 12 x−1 + k
cos y = (x + 1)e−x + k y = π3 when x = 1
y = π when x = −1 ∴ ln 32 = 1
2 + k ∴ −1 = 0 + k k = ln 3
2 − 12
k = −1 ∴ ln1 + cos y = 12 x−1 + ln 3
2 − 12
∴ cos y = (x + 1)e−x − 1 [ (1 + cos y)2 = 1
94 e
xx−
]
Solomon Press
INTEGRATION C4 Answers - Worksheet L
1 a ( 4)
(1 )(2 )xx x
++ −
≡ 1
Ax+
+ 2
Bx−
2 ddyx
= ey × ex cos x
x + 4 ≡ A(2 − x) + B(1 + x) ∫ e−y dy = ∫ ex cos x dx
x = −1 ⇒ A = 1, x = 2 ⇒ B = 2 u = ex, ddux
= ex; dd
vx
= cos x, v = sin x
∴ ( 4)(1 )(2 )
xx x
++ −
≡ 11 x+
+ 22 x−
I = ∫ ex cos x dx = ex sin x − ∫ ex sin x dx
b ∫ 1y
dy = ∫ ( 11 x+
+ 22 x−
) dx u = ex, ddux
= ex; dd
vx
= sin x, v = −cos x
lny = ln1 + x − 2 ln2 − x + c I = ex sin x − (−ex cos x + ∫ ex cos x dx) y = 2 when x = 3 2I = ex sin x + ex cos x + C ∴ ln 2 = ln 4 − 0 + c −e−y = 1
2 ex(sin x + cos x) + c c = −ln 2 e−y = k − 1
2 ex(sin x + cos x) ∴ lny = ln1 + x − 2 ln2 − x − ln 2 y = 0 when x = 0 [ y = 2
12(2 )
xx
+−
] ∴ 1 = k − 12
k = 32
∴ e−y = 32 − 1
2 ex(sin x + cos x) [ 2e−y = 3 − ex(sin x + cos x) ] 3 d
dyx
= kx
4 a ddNt
= kN
ddyx
= 53 when x = 3 b ∫ 1
N dN = ∫ k dt
∴ 53 =
3k , k = 5 ln N = kt + c
y = ∫ 5x
dx N = ekt + c = ec × ekt
y = 5 ln | x | + c N = Aekt y = 4 when x = 3 c t = 0, N = 40 ∴ A = 40 ∴ 4 = 5 ln 3 + c t = 5, N = 60 ∴ 60 = 40e5k c = 4 − 5 ln 3 ∴ k = 1
5 ln 32 = 0.0811 (3sf)
∴ y = 5 ln | x | + 4 − 5 ln 3 d t = 12 ∴ N = 40e0.08109 × 12 y = 5 ln
3x + 4 = 106 (3sf)
C4 INTEGRATION Answers - Worksheet L page 2
Solomon Press
5 a let side length be l 6 a − d
dmt
= km
A = 6l2 ∴ l = 6A ∫ 1
m dm = ∫ −k dt
V = l3 = (6A )3 =
3 32 26 A− ln m = −kt + c
∴ ddVA
= 3 12 23
2 6 A−× × = k A m = e−kt + c = ec × e−kt
b ddVt
= ddVA
× ddAt
m = Ae−kt
ddVt
= cA t = 0, m = 24 ∴ A = 24
∴ cA = k A × ddAt
m = 24e−kt
ddAt
= d A b t = 20, m = 22.6 ∴ 22.6 = 24e−20k
A = 100, ddAt
= 5 ∴ d = 12 ∴ k = 1
20− ln 22.624 = 0.00301 (3sf)
ddAt
= 12 A c d
dmt
= −km = −0.003005 × 22.6
∫122A− dA = ∫ dt = −0.0679 (3sf)
124A = t + C ∴ decreasing at 0.0679 grams per day
t = 10, A = 100 ∴ C = 30 d m = 12 ∴ 12 = 24e−0.003005t
12A = 1
4 (t + 30) t = − 10.003005 ln 1
2 A = 1
16 (t + 30)2 = 231 days (nearest day)
7 a − d
dPt
= k P 8 a − ddVt
= aV where a is a positive constant
∫12P− dP = ∫ −k dt d
dVt
= ddVh
× ddht
122P = −kt + c if angle at vertex = 2θ, tan θ = r
h
P = 12 c − 1
2 kt = a − bt ∴ r = bh where b is a positive constant
∴ P = (a − bt)2 V = 13 πr2h = 1
3 πb2h3 ∴ ddVh
= πb2h2
b t = 0, P = 400 ∴ 400 = a − 0 ∴ −πb2h2 × ddht
= a × 13 πb2h3
a = 20 ddht
= −kh where k is a positive constant
c t = 30, P = 100 ∴ 100 = 20 − 30b b ∫ 1h
dh = ∫ −k dt
b = 13 lnh = −kt + c
∴ P = (20 − 13 t)2 h = e−kt + c = ec × e−kt = Ae−kt
t = 50 ∴ P = (20 − 503 )2 t = 0, h = 12 ∴ A = 12
= 1911 t = 20, h = 10 ∴ 10 = 12e−20k
∴ k = − 120 ln 5
6 ∴ h = 12e−kt, k = 0.00912 (3sf) c 6 = 12e−0.009116t t = − 1
0.009116 ln 12 = 76.0 (3sf)
C4 INTEGRATION Answers - Worksheet L page 3
Solomon Press
9 a 1
(1 )(1 )x x+ − ≡
1A
x+ +
1B
x−
1 ≡ A(1 − x) + B(1 + x) x = −1 ⇒ A = 1
2 , x = 1 ⇒ B = 12
1(1 )(1 )x x+ −
≡ 12(1 )x+
+ 12(1 )x−
b t = 0, m = 0, ddmt
= 0.5
∴ 0.5 = k × 1 k = 0.5 c ∫ 1
(1 )(1 )m m+ − dm = ∫ 1
2 e−t dt
∫ (12
1 m+ +
12
1 m−) dm = ∫ 1
2 e−t dt
12 ln1 + m − 1
2 ln1 − m = 12− e−t + c
ln1 + m − ln1 − m = C − e−t t = 0, m = 0 ∴ 0 − 0 = C − 1 C = 1 ln1 + m − ln1 − m = 1 − e−t for 0 ≤ m < 1, 1 + m > 0 and 1 − m > 0 ∴ ln (1 + m) − ln (1 − m) = 1 − e−t
ln 11
mm
+ −
= 1 − e−t
d m = 0.1 ∴ ln 1.10.9
= 1 − e−t
t = −ln (1 − ln 119 ) = 0.2240 hrs
= 13.4 minutes
e t → ∞, ln 11
mm
+ −
→ 1
∴ limiting value of m is given by 1
1mm
+−
= e
1 + m = e(1 − m) m(1 + e) = e − 1 m = e 1
1 e−+
= 0.4621
∴ max. produced ≈ 462 g
Solomon Press
INTEGRATION C4 Answers - Worksheet M
1 a x 1 3 5 x ln (x + 1) ln 2 3 ln 4 5 ln 6 ∴ integral ≈ 1
2 × 2 × [ln 2 + 5 ln 6 + 2(3 ln 4)] = 18.0 (3sf)
b x π6 π
3 π2
cot x 3 13
0
∴ integral ≈ 12 × π6 × [ 3 + 0 + 2( 1
3)] = 0.756 (3sf)
c x −2 −1 0 1 2
2
10ex
1.492 1.105 1 1.105 1.492 ∴ integral ≈ 1
2 × 1 × [1.492 + 1.492 + 2(1.105 + 1 + 1.105)] = 4.70 (3sf)
d x 0 0.25 0.5 0.75 1 arccos (x2 − 1) 3.142 2.786 2.419 2.024 1.571 ∴ integral ≈ 1
2 × 0.25 × [3.142 + 1.571 + 2(2.786 + 2.419 + 2.024)] = 2.40 (3sf)
e x 0 0.1 0.2 0.3 0.4 0.5 sec2 (2x − 1) 3.4255 2.0602 1.4680 1.1788 1.0411 1 ∴ integral ≈ 1
2 × 0.1 × [3.4255 + 1 + 2(2.0602 + 1.4680 + 1.1788 + 1.0411)] = 0.796 (3sf)
f x 0 1 2 3 4 5 6 x3e−x 0 0.368 1.083 1.344 1.172 0.842 0.535 ∴ integral ≈ 1
2 × 1 × [0 + 0.535 + 2(0.368 + 1.083 + 1.344 + 1.172 + 0.842)] = 5.08 (3sf)
2 a 2 − 1
sin x = 0
sin x = 12
x = π6 , π − π6
x = π6 , 5π6
b x π6 π
3 π2 2π
3 5π6
2 − cosec x 0 0.8453 1 0.8453 0 ∴ area ≈ 1
2 × π6 × [0 + 0 + 2(0.8453 + 1 + 0.8453)] = 1.41 (3sf)
3 a x −1 0 1 2 f(x) 0 0.5236 1.0472 2.0944 ∴ I ≈ 1
2 × 1 × [0 + 2.0944 + 2(0.5236 + 1.0472)] = 2.62 (3sf)
b x −1 −0.5 0 0.5 1 1.5 2 f(x) 0 0.2709 0.5236 0.7763 1.0472 1.3717 2.0944 ∴ I ≈ 1
2 × 0.5 × [0 + 2.0944 + 2(0.2709 + 0.5236 + 0.7763 + 1.0472 + 1.3717)] = 2.52 (3sf)
C4 INTEGRATION Answers - Worksheet M page 2
Solomon Press
4 a x 1 1.5 2 2.5 3 3.5 4 4.5 5 ln x 0 ln 1.5 ln 2 ln 2.5 ln 3 ln 3.5 ln 4 ln 4.5 ln 5 i ≈ 1
2 × 2 × [0 + ln 5 + 2(ln 3)] = 3.807 (3dp) ii ≈ 1
2 × 1 × [0 + ln 5 + 2(ln 2 + ln 3 + ln 4)] = 3.983 (3dp) iii ≈ 1
2 × 0.5 × [0 + ln 5 + 2(ln 1.5 + ln 2 + ln 2.5 + ln 3 + ln 3.5 + ln 4 + ln 4.5)] = 4.031 (3dp) b 2 → 4 strips, increase = 0.176 4 → 8 strips, increase = 0.048 e.g. suggest 8 → 16 strips, increase ≈ 0.013 16 → 32 strips, increase ≈ 0.004 32 → 64 strips, increase ≈ 0.001 ∴ area ≈ 4.031 + 0.013 + 0.004 + 0.001 = 4.049 c u = ln x, d
dux
= 1x
; dd
vx
= 1, v = x
5
1∫ ln x dx = [x ln x] 51 −
5
1∫1x
× x dx
= [x ln x − x] 51
= (5 ln 5 − 5) − (0 − 1) = 5 ln 5 − 4 = 4.047 (3dp) 5 volume = π
0
4−∫ (ex − x)2 dx
let I = 0
4−∫ (ex − x)2 dx
x −4 −3 −2 −1 0 (ex − x)2 16.147 9.301 4.560 1.871 1 ∴ I ≈ 1
2 × 1 × [16.147 + 1 + 2(9.301 + 4.560 + 1.871)] = 24.306 ∴ volume ≈ 24.306 × π = 76.4 (3sf)
Solomon Press
INTEGRATION C4 Answers - Worksheet N
1 = [ 1
4 × 8 ln4x − 3] 72 2 d
dyx
= 3cos sinx
y y
= 2[ln4x − 3] 72 ∫ cos y sin3 y dy = ∫ x dx
= 2(ln 25 − ln 5) 14 sin4 y = 1
2 x2 + c = 2 ln 25
5 sin4 y = 2x2 + k
= 2 ln 5 y = π4 when x = 1
= ln 52 ∴ ( 12
)4 = 2 + k
= ln 25 14 = 2 + k
k = − 74
∴ sin4 y = 2x2 − 74
3 a x 0 0.5 1 1.5
2 1ex − 0.3679 0.4724 1 3.4903 ∴ integral ≈ 1
2 × 0.5 × [0.3679 + 3.4903 + 2(0.4724 + 1)] = 1.70 (3sf) b x 0 0.25 0.5 0.75 1 1.25 1.5
2 1ex − 0.3679 0.3916 0.4724 0.6456 1 1.7551 3.4903 ∴ integral ≈ 1
2 × 0.25 × [0.3679 + 3.4903 + 2(0.3916 + 0.4724 + 0.6456 + 1 + 1.7551)] = 1.55 (3sf) 4 a 2
3(2 )(1 2 ) (1 )
xx x
−− +
≡1 2
Ax−
+ 2(1 2 )B
x−+
1C
x+ 5 a d
dNt
= kN
3(2−x) ≡ A(1−2x)(1+x)+B(1+x)+C(1−2x)2 ∫ 1N
dN = ∫ k dt
x = 12 ⇒ 9
2 = 32 B ⇒ B = 3 ln N = kt + c
x = −1 ⇒ 9 = 9C ⇒ C = 1 N = ekt + c = ec × ekt coeffs x2 ⇒ 0 = −2A + 4C ⇒ A = 2 ∴ N = Aekt f(x) ≡ 2
1 2x− + 2
3(1 2 )x−
+ 11 x+
b t = 0, N = 200 ∴ A = 200
b = 2
1∫ ( 21 2x−
+ 23
(1 2 )x− + 1
1 x+) dx t = 2, N = 3000 ∴ 3000 = 200e2k
= [−ln1 − 2x + 32 (1 − 2x)−1 + ln1 + x] 2
1 ∴ k = 12 ln 15 = 1.354
= (−ln 3 − 12 + ln 3) − (0 − 3
2 + ln 2) ∴ N = 200e1.354t = 1 − ln 2 ∴ 10 000 = 200e1.354t t = 1
1.354 ln 50 = 2.889 hours = 2 hours 53 minutes c 5 per second = 18 000 per hour d
dNt
= 200 × 0.1354e1.354t
∴ 18 000 = 270.8e1.354t t = 1
1.354 ln 18000270.8 = 3.099 hours
= 3 hours 6 minutes
C4 INTEGRATION Answers - Worksheet N page 2
Solomon Press
6 a = 4
0∫12(2 1)x −+ dx 7 u2 = x + 3 ∴ x = u2 − 3, d
dxu
= 2u
= [ 12 ×
122(2 1)x + ] 4
0 x = 0 ⇒ u = 3
= [12(2 1)x + ] 4
0 x = 1 ⇒ u = 2
= 3 − 1 = 2 1
0∫ 3x x + dx = 2
3∫ (u2 − 3)u × 2u du
b = π4
0∫ ( 12 1x +
)2 dx = 2
3∫ (2u4 − 6u2) du
= π4
0∫1
2 1x + dx = [ 2
5 u5 − 2u3] 23
= π[ 12 ln2x + 1] 4
0 = ( 645 − 16) − ( 2
5 × 9 3 − 2 × 3 3 )
= 12 π(ln 9 − 0) = π ln
129 = 16
5− − ( 125 3− )
= π ln 3 = 45 ( 3 3 − 4) [ k = 4
5 ]
8 a sin (A + B) ≡ sin A cos B + cos A sin B 9 a 2 22
( 2)( 4)x
x x−
+ −≡ A +
2B
x + +
4C
x −
sin (A − B) ≡ sin A cos B − cos A sin B x2 − 22 ≡ A(x+2)(x−4)+B(x−4)+C(x+2) adding x = −2 ⇒ −18 = −6B ⇒ B = 3 2 sin A cos B ≡ sin (A + B) + sin (A − B) x = 4 ⇒ −6 = 6C ⇒ C = −1 b y = 0 ⇒ sin 4t = 0 ⇒ t = 0, π4 , π2 , 3π
4 coeffs x2 ⇒ A = 1
curve in 1st quadrant: 0 ≤ t ≤ π4 b = 2
0∫ (1 + 32x +
− 14x −
) dx
x = 2 sin 2t ∴ ddxt
= 4 cos 2t = [x + 3 lnx + 2 − lnx − 4] 20
area in 1st quadrant = (2 + 3 ln 4 − ln 2) − (0 + 3 ln 2 − ln 4)
= π4
0∫ sin 4t × 4 cos 2t dt = 2 + 6 ln 2 − ln 2 − 3 ln 2 + 2 ln 2
total area = 4 × area in 1st quadrant = 2 + 4 ln 2
= π4
0∫ 16 sin 4t cos 2t dt = 2 + ln 16
c = 8π4
0∫ (sin 6t + sin 2t) dt
= 8[− 16 cos 6t − 1
2 cos 2t]π40
= 8[(0 − 0) − (− 16 − 1
2 )] = 135
10 a = 12 ∫ (1 − cos 2x) dx = 1
2 (x − 12 sin 2x) + c = 1
4 (2x − sin 2x) + c
b u = x, ddux
= 1; dd
vx
= sin2 x, v = 12 x − 1
4 sin 2x
∫ x sin2 x dx = x( 12 x − 1
4 sin 2x) − ∫ ( 12 x − 1
4 sin 2x) dx
= 12 x2 − 1
4 x sin 2x − ( 14 x2 + 1
8 cos 2x) + c = 1
4 x2 − 14 x sin 2x − 1
8 cos 2x + c = 1
8 (2x2 − 2x sin 2x − cos 2x) + c
c = ππ
0∫ (12 sinx x )2 dx = π
π
0∫ x sin2 x dx
= 18 π[2x2 − 2x sin 2x − cos 2x] π
0 = 1
8 π[(2π2 − 0 − 1) − (0 − 0 − 1)] = 14 π3
Solomon Press
INTEGRATION C4 Answers - Worksheet O 1 a 2
13 2x x− +
≡ 1
Ax −
+ 2
Bx −
2 = π6
0∫ 12 [cos 4x + cos (−2x)] dx
1 ≡ A(x − 2) + B(x − 1) = π6
0∫ ( 12 cos 4x + 1
2 cos 2x) dx
x = 1 ⇒ 1 = −A ⇒ A = −1 = [ 18 sin 4x + 1
4 sin 2x]π60
x = 2 ⇒ B = 1 = ( 18 × 3
2 + 14 × 3
2 ) − (0)
∴ 213 2x x− +
≡ 12x −
− 11x −
= 316 3
b 4
3∫ 213 2x x− +
dx =4
3∫ ( 12x −
− 11x −
) dx
= [lnx − 2 − lnx − 1] 43
= (ln 2 − ln 3) − (0 − ln 2) = 2 ln 2 − ln 3 = ln 22
3 = ln 4
3 [ a = 4, b = 3 ]
3 a 4 a = π4
1∫ (2 − 1x
)2 dx
= π4
1∫ (4 − 124x− + x−1) dx
= π[4x − 128x + ln | x |] 4
1 = π[(16 − 16 + ln 4) − (4 − 8 + 0)] = π(4 + ln 4) = π(4 + ln 22) quotient: x + 2, remainder: 1 = π(4 + 2 ln 2) = 2π(2 + ln 2)
b ∫2 1
1x x
x+ −
− dx = ∫ (x + 2 + 1
1x −) dx b = 103 × 2π(2 + ln 2)
= 12 x2 + 2x + lnx − 1 + c = 2000π(2 + ln 2) = 16 900 cm3
5 a u = ln x, ddux
= 1x
; dd
vx
= x, v = 12 x2 6 a =
π3
0∫ sec x tan x dx
∫ x ln x dx = 12 x2 ln x − ∫ 1
2 x dx = [sec x] 3π
0
= 12 x2 ln x − 1
4 x2 + c = 2 − 1 = 1
4 x2(2 ln x − 1) + c = 1
b ∫ 1y
dy = ∫ x ln x dx b u = cos θ ∴ dd
uθ
= −sin θ
lny = 14 x2(2 ln x − 1) + c θ = 0 ⇒ u = 1
y > 0 ∴ ln y = 14 x2(2 ln x − 1) + c θ = π4 ⇒ u = 1
2
y = 4 when x = 2 π4
0∫ 4sin
cosθθ
dθ = 21
1∫ 41
u× (−1) du
∴ ln 4 = 2 ln 2 − 1 + c, c = 1 = 12
1
∫ u−4 du
∴ ln y = 14 x2(2 ln x − 1) + 1 = [ 1
3− u−3] 12
1
when x = 1, ln y = 14 (0 − 1) + 1 = 3
4 = 13− (1 − 2 2 )
∴ y = 34e = 1
3− + 23 2 [ a = 1
3− , b = 23 ]
x + 2 x − 1 x2 + x − 1
x2 − x 2x − 1 2x − 2 1
C4 INTEGRATION Answers - Worksheet O page 2
Solomon Press
7 a x = 0 ⇒ t = 1
2− 8 a u = 6x, ddux
= 6; dd
vx
= cos 3x, v = 13 sin 3x
x = 3 ⇒ t = 1 ∫ 6x cos 3x dx = 2x sin 3x − ∫ 2 sin 3x dx
b x = 2t + 1 ∴ ddxt
= 2 = 2x sin 3x + 23 cos 3x + c
∴ area = 12
1
−∫1
2 t− × 2 dt b x = 2 sin u ∴ d
dxu
= 2 cos u
= 12
1
−∫2
2 t− dt x = 0 ⇒ u = 0
= [−2 ln | 2 − t |] 12
1− x = 3 ⇒ u = π3
= −2(0 − ln 52 )
3
0∫ 2
1
4 x− dx =
π3
0∫1
2cosu× 2 cos u du
= 2 ln 52 =
π3
0∫ du
c = π12
1
−∫ ( 12 t−
)2 × 2 dt = [u]π30
= 2π12
1
−∫ (2 − t)−2 dt = π3 − 0
= 2π[(2 − t)−1] 12
1− = π3
= 2π(1 − 25 ) = 6
5 π
9 a A = πr2 ∴ r = πA 10 a 5 1
(1 )(1 2 )x
x x+
− + ≡
1A
x− +
1 2B
x+
P = 2πr = 2ππA = 2 πA 5x + 1 ≡ A(1 + 2x) + B(1 − x)
ddAt
= cP = c × 2 πA = 2 πc × A x = 1 ⇒ 6 = 3A ⇒ A = 2
∴ ddAt
= k A x = 12− ⇒ 3
2− = 32 B ⇒ B = −1
b ∫12A− dA = ∫ k dt f(x) ≡ 2
1 x− − 1
1 2x+
122A = kt + C b =
12
0∫ ( 21 x−
− 11 2x+
) dx
A = 12 kt + 1
2 C = pt + q = [−2 ln1 − x − 12 ln1 + 2x]
120
∴ A = (pt + q)2 = (−2 ln 12 − 1
2 ln 2) − (0 − 0)
c t = 0, A = 25 ∴ 25 = 0 + q = 2 ln 2 − 12 ln 2 = 3
2 ln 2 q = 5 c (1 − x)−1 = t = 20, A = 40 ∴ 40 = 20p + 5 1 + (−1)(−x) +
( 1)( 2)2
− − (−x)2 +
( 1)( 2)( 3)3 2
− − −× (−x)3
+…
p = 2 10 520
− = 1 + x + x2 + x3 + …
A = 50 ⇒ 50 = ( 2 10 520
− t + 5)2 (1 + 2x)−1 =
t = ( 5 2 − 5) ÷ 2 10 520
− 1 + (−1)(2x) + ( 1)( 2)
2− − (2x)2
+ ( 1)( 2)( 3)
3 2− − −
× (2x)3 +…
= 31.3 minutes (3sf) = 1 − 2x + 4x2 − 8x3 + … f(x) = 2(1 + x + x2 + x3) − (1 − 2x + 4x2 − 8x3) + … f(x) = 1 + 4x − 2x2 + 10x3 + …
Solomon Press
INTEGRATION C4 Answers - Worksheet P
1 a when y = 3, x = 1
3 area 0 ≤ x ≤ 1
3 = 3 × 13 = 1
area 13 < x ≤ 3 =
13
3
∫ 1x
dx
= [ln | x |] 13
3
= ln 3 − ln 13 = ln 1
3
3 = ln 9
total area = 1 + ln 9 b volume 0 ≤ x ≤ 1
3 = volume of cylinder = π × 32 × 1
3 = 3π
volume 13 < x ≤ 3 = π
13
3
∫ ( 1x
)2 dx
= π13
3
∫ x−2 dx = π[−x−1] 13
3
= π[ 13− − (−3)] = 8
3 π total volume = 3π + 8
3 π = 173 π
2 a x 0 0.5 1 1.5 2 2.5 3 3.5 4 x sec ( 1
3 x) 0 0.507 1.058 1.709 2.545 3.718 5.552 8.901 17.004 i ≈ 1
2 × 2 × [0 + 17.004 + 2(2.545)] = 22.1 (3sf) ii ≈ 1
2 × 1 × [0 + 17.004 + 2(1.058 + 2.545 + 5.552)] = 17.7 (3sf) iii ≈ 1
2 ×0.5×[0 + 17.004 + 2(0.507 + 1.058 + 1.709 + 2.545 + 3.718 + 5.552 + 8.901)] = 16.2 (3sf) b e.g. halving the interval width leads to the estimate reducing by 4.4, then 1.5 further halving might lead to reductions of approximately 0.5 and 0.2 so I ≈ 15.5 3 a − d
dtθ = kθ 4 a sin4 x ≡ (sin2 x)2
∫ 1θ
dθ = ∫ −k dt ≡ [ 12 (1 − cos 2x)]2
ln θ = −kt + c ≡ 14 (1 − 2 cos 2x + cos2 2x)
θ = e−kt + c = ec × e−kt ≡ 14 − 1
2 cos 2x + 14 [ 1
2 (1 + cos 4x)] ∴ θ = Ae−kt ≡ 3
8 − 12 cos 2x + 1
8 cos 4x t = 0, θ = 25 − 10 = 15 ∴ p = 3
8 , q = − 12 , r = 1
8
∴ A = 15 b = π2
0∫ ( 38 − 1
2 cos 2x + 18 cos 4x) dx
∴ θ = 15e−kt = [ 38 x − 1
4 sin 2x + 132 sin 4x]
π20
b t = 30, θ = 20 − 10 = 10 = ( 316 π − 0 + 0) − (0 − 0 + 0)
∴ 10 = 15e−30k = 316 π
k = − 130 ln 2
3 = 0.0135 (3sf) c θ = 15 − 10 = 5 ∴ 5 = 15e−0.01352t t = − 1
0.01352 ln 13
= 81.3 minutes (3sf)
C4 INTEGRATION Answers - Worksheet P page 2
Solomon Press
5 a ∫ y−3 dy = ∫ x dx 6 a x = eu ∴ u = ln x, d
dxu
= eu
12− y−2 = 1
2 x2 + c ∫ 22 ln x
x+ dx = ∫ 2
2(e )u
u+ × eu du
y−2 = k − x2 = ∫ (2 + u)e−u du
y2 = 21
k x− b x = eu
b y = 12 when x = 1 x = 1 ⇒ u = 0
∴ 14 = 1
1k − x = e ⇒ u = 1
k = 5 ∴ 1
e∫ 2
2 ln xx
+ dx = 1
0∫ (2 + u)e−u du
∴ y2 = 21
5 x− v = 2 + u, d
dvu
= 1; ddwu
= e−u, w = −e−u
= [−(2 + u)e−u] 10 +
1
0∫ e−u du
= [−(2 + u)e−u − e−u] 10
= (−3e−1 − e−1) − (−2 − 1) = 3 − 4e−1
7 a x = 0 ⇒ cos 2t = 0 ⇒ t = π4 8 a 2
26 2
( 1) ( 3)x
x x−
+ + ≡ 2( 1)
Ax +
+ 1
Bx +
+ 3
Cx +
y = 0 ⇒ tan t = 0 ⇒ t = 0 6 − 2x2 ≡ A(x + 3) + B(x + 1)(x + 3) + C(x + 1)2 x = cos 2t ∴ d
dxt
= −2 sin 2t x = −1 ⇒ 4 = 2A ⇒ A = 2
shaded area = π4
0
∫ tan t × (−2 sin 2t) dt x = −3 ⇒ −12 = 4C ⇒ C = −3
= π4
0∫sincos
tt
× 4 sin t cos t dt coeffs x2 ⇒ −2 = B + C ⇒ B = 1
= π4
0∫ 4 sin2 t dt b x = 0 ⇒ y = 2
b = π4
0∫ (2 − 2 cos 2t) dt f(x) = 2(x + 1)−2 + (x + 1)−1 − 3(x + 3)−1
= [2t − sin 2t]π40 f ′(x) = −4(x + 1)−3 − (x + 1)−2 + 3(x + 3)−2
= ( π2 − 1) − (0 − 0) ∴ f ′(0) = −4 − 1 + 1
3 = 143−
= π2 − 1 ∴ y − 2 = 143− (x − 0)
c i sin2 A = 12 (1 − cos 2A) 3y − 6 = −14x
ii cos2 A = 12 (1 + cos 2A) 14x + 3y = 6
y2 = tan2 t = 2
2sincos
tt
= 1212
(1 cos2 )(1 cos2 )
tt
−+
c = 1
0∫ ( 22
( 1)x + + 1
1x + − 3
3x +) dx
∴ y2 = 11
xx
−+
= [−2(x + 1)−1 + lnx + 1 − 3 lnx + 3] 10
= (−1 + ln 2 − 3 ln 4) − (−2 + 0 − 3 ln 3) = 1 − 5 ln 2 + 3 ln 3