c4 marking schemes

Solomon Press

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


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


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


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


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 + …


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


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


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 +


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 −


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


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


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


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)


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


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


0

−


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

− −


∴ 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°


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°


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)


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


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


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


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


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


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


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


∫ 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


Solomon Press

7 a u = x2, d

dux

= 2x; dd

vx


∫ 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


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


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


∫ 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


= π2


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


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


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


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)


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


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

c4 marking schemes

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