Pure Mathematics 3 Question & Workbook answers

1 Cambridge International AS & A Level Mathematics – Pure Mathematics 3 Question & Workbook © Greg Port 2018

Answers1 Further algebra

1.1 The general binomial expansion (Page 2)

1 (i) (1 ) 2+ −x

1 2 2( 2 1)

2!2( 2 1)( 2 2)

3! ...

1 2 2( 3)2!

2( 3)( 4)3! ...

1 2 3 4 ...

2 3

2 3

2 3

= + − + − − − + − − − − − +

= − + − − + − − − +

= − + − +

x x x

x x x

x x x Valid for x 1<

(ii) (1 4 )12+ x

1 12(4 )

12

12 1

2! (4 )

12

12 1 1

2 2

3! (4 ) ...

1 212

12

2! (16 )

12

12

32

3! (64 ) ...

1 2 2 4 ...

2

3

2

3

2 3

( )( )( )

( )( )( )

= + +−

+− −

+

= + +−

+− −

+

= + − + −

x x

x

x x

x

x x x

Valid for x x4 1 14< ⇒ <

2 (i) x( )+ −2 4

2 1 12

2 1 12

116 1 1

2

116 1 4 1

24( 4 1)

2!12 ...

116 1 2 4( 5)

2!14 ...

116 1 2 5

2 ...

116

18

532 ...

Valid for 12 1 2

4

44

4

2

2

2

2

( )( )

( )( )

( )( ) ( )

( )( )

= +

= +

= +

= + − + − − − +

= − + − − +

= − + +

= − + −

< ⇒ <

−

−−

−

x

x

x

x x

x x

x x

x x

x x

(ii) x−

14

( )( )

( )( )

( )

( )( )

)( ) ( )(

( )

= −

= −

= −

= −

= + − − +− − −

− +

= + +− −

+

= + + +

= + + +

−

−

− −

−

x

x

x

x

x x

x x

x x

x x

(4 )

4 1 14

4 1 14

12 1 1

4

12 1 1

214

12

12 1

2!14 ...

12 1 1

8

12

32

2!1

16 ...

12 1 1

83

128 ...

12

116

3256 ...

12

12

12

12

12

2

2

2

2

Valid for 14 1 4< ⇒ <x x

3 (i) (1 )+ ax b

b ax b b ax

abx b b a x

ab a ba b b a b b

a

b b b

bb b

b b b

b b b

b bb b

b b

a b

( )( )

= + + − +

= + + − +

= − ⇒ = −

− = ⇒ − =

− − =

− =

− =

− =

= += +

≠ = −

= − − − =

1 ( ) ( 1)2! ( ) ...

1 ( 1)2! ( ) ...

10 10

( 1)2 75 ( 1) 150

Substituting for ,

10 ( 1) 150

100 ( 1) 150

100 ( 1) 150

100 100 150

0 50 1000 50 ( 2)

0 so 210 = 10

2 5

2

2 2

22

2

2

2

2 2

2

(ii) Valid for x x5 1 15< ⇒ <

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9781510458444_Answer.indb 1 11/8/18 10:33 AM

2

1Furtheralgebra

Cambridge International AS & A Level Mathematics – Pure Mathematics 3 Question & Workbook © Greg Port 2018

4 (i) x−

11 2 2

x

x x

x x

( )= −

= + − − +− − −

− +

= + + +

−(1 2 )

1 12( 2 )

12

12 1

2! ( 2 ) ...

1 32 ...

212

2 2 2

2 4

Valid for x x2 1 12

2 < ⇒ <

(ii) xx

+−

11 2 2

x x x

x x x x x

x x x x x

(1 ) 1 32 ...

1 32

32 ...

1 32

32 ...

2 4

2 4 3 5

2 3 4 5

( )= + + + +

= + + + + + +

= + + + + + +

5 (i) (1 3 )− x n

n x n n x

nx n n x

nx n n x

n n n

n n nn n n

n nn n

n n

1 ( 3 ) ( 1)2! ( 3 ) ...

1 3 2 (9 ) ...

1 3 9( )2 ...

3 9( )2

6 9( )6 9 90 9 30 3 (3 1)

0 so 13

2

2 2

22

2

2

2

2

= + − + − − +

= + − + − +

= + − + − +

− = −

− = −− = −

= −= −

≠ =

(ii) x( )−1 313

1 13( 3 )

13

13 1

2! ( 3 )

13

13 1 1

3 2

3! ( 3 ) ...

113

23

2! (9 )13

23

53

3! ( 27 ) ...

1 53 ...

2

3

2 3

2 3

( )

( )( )

( ) ( )( )

= + − +−

−

+− −

− +

= − +−

+− −

− +

= − − − −

x x

x

x x x

x x x

6 (i) x ax( )( )− + −1 3 1 2

x ax ax

x ax a x

xa x x ax

a x axa a x

xa a

a aa a

(1 3 ) 1 2( ) 2( 2 1)2! ( ) ...

(1 3 ) 1 2 3 ...

Just considering the terms,3 3 2

3 6(3 6 )

Coefficient of 03 6 0

3 ( 2) 00 so 2

2

2 2

2

2 2

2 2 2

2 2

2

2

= − + − + − − − +

= − − + +

− × −= += +

=⇒ + =

+ =≠ = −

x ax ax

x ax a x

xa x x ax

a x axa a x

xa a

a aa a

(1 3 ) 1 2( ) 2( 2 1)2! ( ) ...

(1 3 ) 1 2 3 ...

Just considering the terms,3 3 2

3 6(3 6 )

Coefficient of 03 6 0

3 ( 2) 00 so 2

2

2 2

2

2 2

2 2 2

2 2

2

2

= − + − + − − − +

= − − + +

− × −= += +

=⇒ + =

+ =≠ = −

(ii) (1 3 )(1 2 ) 2− − −x x

xx x

x

x x x x

x

x x x

x

(1 3 )1 2( 2 ) 2( 2 1)

2! ( 2 )

2( 2 1)( 2 2)3! ( 2 ) ...

(1 3 ) 1 4 12 32 ...

The term is given by

1 32 3 12

4

2

3

2 3

3

3 2

3

= −+ − − + − − − −

+ − − − − − −

= − + + + +

× + − ×

= −

7 (i) (a + x)−2

1 1

1 1

1 1 1

1 1 2 1 2( 2 1)2!

1 ...

1 1 2 3 ...

1 2 3 ...

2

22

2

2

2

2

2 22

2 3 42

( )( )

( )( )

( )( ) ( )

= +

= +

= +

= + − + − − − +

= − + +

= − + +

−

−−

−

a a x

a a x

a a x

a a x a x

a ax

ax

a ax

ax

(ii) (1 − x)(a + x)−2

(1 ) 1 2 3 ...

Considering the terms,

1 3 2

3 2

3 2

3 2 0

3 2 032

2 3 42

2

42

3

42

32

4 32

4 3

( )

( )

= − − + +

× + − × −

= +

= +

+ =

+ =

= −

xa a

xa

x

x

ax x

ax

ax

ax

a ax

a aa

a

8 (i) yx x

=− − −

11 2 1

x xx xx x

x xx xx x

x

x x x

11 2 1

1 2 11 2 1

1 2 11 2 (1 )1 2 1

1 ( 1 2 1 )

=− − −

× − + −− + −

= − + −− − −

= − + −−

= − − + −

9781510458444_Answer.indb 2 11/8/18 10:33 AM

3

1Furtheralgebra


(ii) 1 ( 1 2 1 )= − − + −y x x x

x

x

x x

x x

1 2

(1 2 )

1 12( 2 )

12

12 1

2! ( 2 ) ...

1 12 ...

12

2

2

( )

−

= −

= + − +−

− +

= − − −

x

x

x x

x x

1

(1 )

1 12( )

12

12 1

2! ( ) ...

1 12

18 ...

12

2

2

( )

−

= −

= + − +−

− +

= − − −

y x x x x x

x x x

x x

x

1 1 12 1 1

218 ...

1 2 32

58 ...

2 32

58 ...

Coefficient of the term is 58

2 2

2

( )( )

= − − − + − − −

= − − − −

= − + + +

1.2 Review of algebraic fractions (Page 6)

1 (i) 34

89

23

2

2× =ab

ba

ba

(ii) 166 8

( 4)( 4)( 4)( 2)

42

2

2−

− += − +

− − = +−

ff f

f ff f

ff

(iii) h h h h− ÷ − +44

4 48

3 2

44

84 4

( 4)4

8( 2)( 2)

( 2)( 2)4

8( 2)( 2)

2 ( 2)2

3

2

2

= − ×− +

= − × − −

= − + × − −

= +−

h hh h

h hh h

h h hh h

h hh

2 (i) m m m m m+ = + =2 34

84

34

114

(ii) p

pp−

++2 23

pp

p pp

pp

p pp

p pp

=−

++

=−

++

=+ −

3( 2)3

( 2)3

3 63

23

5 63

2

2

(iii) r

r r r+ + −5

43

52

2 3

= + + −

= + + −

= + −

5 ( 5) 3(4 ) 2(20)20

5 25 12 4020

17 25 4020

2

3

2 2

3

2

3

r r rr

r r rr

r rr

1.3 Partial fractions (Page 7)

1 (i) 5 7( 1)( 2) 1 2

xx x

Ax

Bx

++ + ≡ + + +

5 7 ( 2) ( 1)1 22 3 3

So 5 7( 1)( 2)

21

32

x A x B xx Ax B B

xx x x x

+ ≡ + + += − ⇒ == − ⇒ − = − ⇒ =

++ + ≡ + + +

(ii) 3 24

3 2( 2)( 2)2

xx

xx x

−−

= −− +

3 2( 2)( 2) 2 23 2 ( 2) ( 2)

2 4 4 12 8 4 2

So 3 2( 2)( 2)

12

22

xx x

Ax

Bx

x A x B xx A Ax B B

xx x x x

−− + ≡ − + +− ≡ + + −

= ⇒ = ⇒ == − ⇒ − = − ⇒ =

−− + ≡ − + +

(iii) 4 4( 1)2x x x x−

=−

4( 1) 1

4 ( 1)1 40 4 4

So 4( 1)

4 41

x xAx

Bx

A x Bxx Bx A A

x x x x

− ≡ + −≡ − += ⇒ == ⇒ = − ⇒ = −

− ≡ − + −

(iv) 42 18( 1)( 2)( 4) 1 2 4

xx x x

Ax

Bx

Cx

−+ − − ≡

++

−+

−

42 18 ( 2)( 4) ( 1)( 4)( 1)( 2)

2 6 6 11 60 15 4

4 30 10 3

So 42 18( 1)( 2)( 4)

41

12

34

x A x x B x xC x x

x B Bx A Ax C C

xx x x x x x

− ≡ − − + + −+ + −

= ⇒ = − ⇒ = −= − ⇒ = ⇒ == ⇒ − = ⇒ = −

−+ − − ≡ + − − − −

9781510458444_Answer.indb 3 11/8/18 10:33 AM

4

1Furtheralgebra


(v) 2 4 32 3

2 5 32 3

2 5 3( 3)( 1)

3 2

2 2x x x

x xx x

x x

x xx x

− − −− −

= + −− −

= + −− +

5 3( 3)( 1) 3 15 3 ( 1) ( 3)

3 12 4 31 8 4 2

So 5 3( 3)( 1)

33

21

and 2 4 32 3

2 33

21

3 2

2

xx x

Ax

Bx

x A x B xx A Ax B B

xx x x x

x x xx x

x x x

−− + ≡ − + +− ≡ + + −

= ⇒ = ⇒ == − ⇒ − = − ⇒ =

−− + ≡ − + +

− − −− −

= + − + +

2 (i) x xx x

Ax

Bx

Cx

+ −−

≡ + +−

3 1( 1) 1

2

2 2

x x Ax x B x Cxx Cx B B

x A C A

x xx x x x x

+ − ≡ − + − += ⇒ == ⇒ − = − ⇒ =

⇒ = + ⇒ = −

+ −−

≡ − + +−

3 1 ( 1) ( 1)1 30 1 1

terms 1 2

So 3 1( 1)

2 1 31

2 2

2

2

2 2

(ii) x x

Ax

Bx

Cx−

≡ +−

+−

1( 1) 1 ( 1)2 2

A x Bx x Cxx Cx A

x A B B

x x x x x

x x

≡ − + − += ⇒ == ⇒ =

⇒ = + ⇒ =

−≡ +

−+

−

= +−

1 ( 1) ( 1)1 10 1

terms 1 0

So 1( 1)

1 01

1( 1)

1 1( 1)

2

2

2 2

2

3 (i) x xA

xBx Cx+ +

≡+

+ ++

1( 1)( 1) 1 ( 1)2 2

A x Bx C x

x A A

x A B B

x A C C

x x x

x

x

xx

x

≡ + + + +

= − ⇒ = ⇒ =

⇒ = + ⇒ = −

= ⇒ = + ⇒ =

+ +≡

++

− +

+=

++ −

+

1 ( 1) ( )( 1)

1 1 2 12

terms 0 12

0 1 12

So 1( 1)( 1)

12

1

12

12

( 1)1

2( 1)1

2( 1)

2

2

2 2

2

(ii) x x

Ax

Bx Cx

A x Bx C x

x A A

x A B B

x A C C

− +≡

−+ +

+≡ + + + −

= ⇒ = ⇒ =

⇒ = + ⇒ = −

= ⇒ = − ⇒ = −

10( 4)( 4) 4 ( 4)

10 ( 4) ( )( 4)

4 10 20 12

terms 0 12

0 10 4 4 2

2 2

2

2

x xA

xBx Cx

A x Bx C x

x A A

x A B B

x A C C

− +≡

−+ +

+≡ + + + −

= ⇒ = ⇒ =

⇒ = + ⇒ = −

= ⇒ = − ⇒ = −

10( 4)( 4) 4 ( 4)

10 ( 4) ( )( 4)

4 10 20 12

terms 0 12

0 10 4 4 2

2 2

2

2

So 10

( 4)( 4)

12

4

12 2

( 4)1

2( 4)4

2( 4)

2 2

2

x x x

x

x

xxx

− +≡

−+

− −

+

=−

− ++

1.4 Using partial fractions with the binomial expansion (Page 8)

1 (i) 2 11(1 2 )(2 )

xx x−

+ − = Ax

Bx+ + −1 2 2

2 11 (2 ) (1 2 )2 20 5 4

12

152

52 3

So 2 11(1 2 )(2 )

31 2

42

x A x B xx B B

x A A

xx x x x

− ≡ − + += ⇒ − = ⇒ = −

= − ⇒ = ⇒ =

−+ − = + − −

(ii) 11 2 (1 2 )

1 1(2 ) 1( 1 1)2! (2 ) ...

1 2 4 ...

1

2

2

x x

x x

x x

+ = +

= + − + − − − +

= − + −

−

Valid for x x2 1 12.< ⇒ <

(iii) 12 (2 ) 1

x x− = − −

2 1 12

2 1 12

12 1 1

2

12 1 1 1

21( 1 1)

2!12 ...

12 1 1

214 ...

12

14

18 ...

1

11

1

2

2

2

x

x

x

x x

x x

x x

( )

( )( )

( )( ) ( )

( )

= −

= −

= −

= + − − + − − − − +

= + + +

= + + +

−

−−

−

Valid for x 1<

(iv) 2 11(1 2 )(2 )

31 2

42

xx x x x−

+ − = + − −

3 1 2 4 ... 4 12

14

18 ...

3 6 12 2 12 ...

1 7 232 ...

2 2

2 2

2

x x x x

x x x x

x x

( )( )= − + − − + + +

= − + − − − +

= − +

Valid for x 12<

9781510458444_Answer.indb 4 11/8/18 10:33 AM

5

1Furtheralgebra


2 (i) 8( 2)( 1) 2 1

xx x

Ax

Bx

−− + = − + +

8 ( 1) ( 2)2 6 3 2

1 9 3 3

So 8( 2)( 1)

22

31

x A x B xx A Ax B B

xx x x x

− ≡ + + −= ⇒ = ⇒ == − ⇒ = − ⇒ = −

−− + = − − +

22

2( 2)

2 2 1 12

2( 2) 1 12

1 1 12

1( 1 1)2!

12 ...

1 12

14 ...

1 12

14 ...

31

3(1 )

3 1 1 1( 1 1)2! ...

3 1 ...

3 3 3 ...2

23

1 1 12

14 ...

3 3 3 ...

4 52

134 ...

1

1

11

2

2

2

1

2

2

2

2

2

2

xx

x

x

x x

x x

x x

xx

x x

x x

x x

x x x x

x x

x x

( )

( )

( )( )( ) ( )

( )

( )

( )

( )

−= −

= − −

= − −

= − + − − + − − − − +

= − + + +

= − − − −

+= +

= + − + − − − +

= − + −

= − + −

− − + = − − − −

− − + −

= − + − +

−

−

−−

−

Valid for x 1<

(ii) 9 2(1 2 )(1 ) 1 2 1 (1 )

2

2 2x x

x xA

xB

xC

x+ +

− +≡

−+

++

+

9 2 (1 ) (1 2 )(1 ) (1 2 )1 6 3 2

12

274

94 3

terms 1 2 1

So 9 2(1 2 )(1 )

31 2

11

2(1 )

31 2 3(1 2 )

3(1 2 4 ...)

3 6 12 ...1

1 (1 )

1 ...

2 2

2

2

2 2

1

2

2

1

2

x x A x B x x C xx C C

x A A

x A B B

x xx x x x x

x x

x x

x x

x x

x x

+ + ≡ + + − + + −= − ⇒ − = ⇒ = −

= ⇒ = ⇒ =

⇒ = − ⇒ =

+ +− +

≡−

++

−+

− = −

= + + +

= + + +

+ = +

= − + +

−

−

2(1 )

2 1

2(1 2 3 ...)

2 4 6 ...3

1 21

12

(1 )

(3 6 12 ...) (1 ...)

(2 4 6 ...)

2 9 7 ...

22

2

2

2

2 2

2

2

xx

x x

x x

x x x

x x x x

x x

x x

( )+

= +

= − + −

= − + −

−+

+−

+= + + + + − + +

− − + −

= + + +

−


3 (i) 4 14(1 )(2 )(1 ) 1 2 1

xx x x

Ax

Bx

Cx

+− + + = − + + + +

4 14 (2 )(1 ) (1 )(1 )(1 )(2 )

1 18 6 32 6 3 21 10 2 5

So 4 14(1 )(2 )(1 )

31

22

51

x A x x B x xC x x

x A Ax B Bx C C

xx x x x x x

+ ≡ + + + − ++ − +

= ⇒ = ⇒ == − ⇒ = − ⇒ = −= − ⇒ = ⇒ =

+− + + = − − + + +

31 3(1 )

3(1 ...)

3 3 3 ...2

2 2(2 )

2 2 1 12

2 2 1 12

2 12 1 1

2

1 12

1 12

14 ...

51 5(1 )

5(1 ...)

5 5 5 ...

1

2

2

1

1

11

1

1

2

1

2

2

( )( )

( )( )

( )( )

− = −

= + + +

= + + +

+ = +

= +

= +

= +

= +

= − + −

+ = +

= − + −

= − + −

−

−

−

−−

−

−

−

x x

x x

x x

x x

x

x

x

x

x x

x x

x x

x x

x x x

x x x x

x x

x x

So 31

22

51

(3 3 3 ...) (1 12

14 ...)

(5 5 5 ...)

7 32

314 ...

2 2

2

2

− − + + +

= + + + − − + −

+ − + −

= − + −

9781510458444_Answer.indb 5 11/8/18 10:33 AM

6

1Furtheralgebra


4 (i) 3 2(1 ) (1 4 ) 1 (1 ) 1 4

2

2 2x

x xA

xBx

Cx

++ −

≡+

++

+−

3 2 (1 )(1 4 ) (1 4 ) (1 )1 5 5 1

14

258

2516 2

terms 2 4 0

So 3 2(1 ) (1 4 )

1(1 )

21 4

2 2

2

2

2 2

x A x x B x C xx B B

x C C

x A C A

xx x x x

+ ≡ + − + − + += − ⇒ = ⇒ =

= ⇒ = ⇒ =

⇒ = − + ⇒ =

++ −

≡+

+−

(ii) 3 2(1 ) (1 4 )

1(1 )

21 4

2

2 2x

x x x x+

+ −=

++

−

1(1 )

(1 )

1 2 3 ...2

1 4 2(1 4 )

2(1 4 16 ...)

2 8 32 ...

So 1(1 )

21 4

(1 2 3 ...) (2 8 32 ...)

3 6 35 ...

22

2

1

2

2

2

2 2

2

xx

x x

x x

x x

x x

x x

x x x x

x x

+= +

= − + −

− = −

= + + +

= + + +

++ −

= − + − + + + +

= + + +

−

−

Further practice (Page 11)

1 (i) x( )− −1 2 1

x x

x

x x x

x x x

1 1( 2 ) 1( 1 1)2! ( 2 )

1( 1 1)( 1 2)3! ( 2 ) ...

1 2 1( 2)2! (4 ) 1( 2)( 3)

3! ( 8 ) ...

1 2 4 8 ...

2

3

2 3

2 3

= + − − + − − − −

+ − − − − − − +

= + + − − + − − − − +

= + + + +


(ii) x( )− −1 913

1 13( 9 )

13

13 1

2! ( 9 )

13

13 1 1

3 2

3! ( 9 ) ...

1 313

43

2! (81 )

13

43

73

3! ( 729 ) ...

1 3 18 126 ...

2

3

2

3

2 3

x x

x

x x

x

x x x

( )( )( )

( )( )( )

= + − − +− − −

−

+− − − − −

− +

= + +− −

+− − −

− +

= + + + +


2 (i) x( )−9 312

( )( )

( )( )

( )( ) ( )

( )( )

( )( )

= −

= −

= −

= + − +−

− +

= − +−

+

= − − +

= − − −

x

x

x

x x

x x

x x

x x

9 1 13

9 1 13

3 1 13

3 1 12

13

12

12 12!

13 ...

3 1 16

12

12

2!19 ...

3 1 16

172 ...

3 12

124 ...

12

12

12

12

2

2

2

2


(ii) xx

32 +

3 (2 )

3 2 1 12

3 2 1 12

3 12 1 1

2

32 1 1

2

32 1 1 1

21( 1 1)

2!12 ...

32 1 1

21( 2)2!

14 ...

32 1 1

214 ...

32

34 ...

1

1

11

1

1

2

2

2

2

( )( )

( )( )

( )( )

( ) ( )( )

( )

= +

= +

= +

= +

= +

= + − + − − − +

= − + − − +

= − + +

= − +

−

−

−−

−

−

x x

x x

x x

x x

x x

x x x

x x x

x x x

x x


3 (i) x( )+113

x x

x x

x x

( )( )

= + +−

+

= + +−

+

= + − +

1 13

13

13 1

2! ...

1 13

13

23

2! ...

1 13

19 ...

2

2

2

9781510458444_Answer.indb 6 11/8/18 10:33 AM

7

1Furtheralgebra


(ii)(a) ( )+ x8 1613

x

x

x

x x

x x

x x

8 1 2

8 1 2

2 1 2

2 1 13(2 ) 1

9(2 ) ...

2 1 23

49 ...

2 43

89 ...

13

13

13

13

2

2

2

( )( )

[ ]( )

( )

( )

= +

= +

= +

= + − +

= + − +

= + − +

(b) Valid for x x2 1 12< ⇒ <

4 (i) ax( )+ −1 3

ax ax

ax a xa a

a aa a

a a

1 ( 3)( ) 3( 3 1)2! ( ) ...

1 3 6 ...3 60 6 30 3 2 1

Since 0, 12

2

2 2

2

2

( )

= + − + − − − +

= − + +− =

= += +

≠ = −

(ii) x( )−−

1 12

3

x x

x x

1 ( 3) 12

3( 3 1)2!

12 ...

1 32

32 ...

2

2

( ) ( )= + − − + − − − − +

= + + +

5 (i) 65

1825

65

2518

53

2 2

2

2 2

2

2c de

de

c de

ed

c ed

÷ = × =

(ii) 3 14

13 4 1

3 14

1(3 1)( 1)

14

2g g

g gg g

g g+

×+

+ +=

+×

++ +

=

(iii) 4 4

2 2j kj k

j kj k

−+

×+−

( )( )

( )( )( )

( )

2 2 2 2

2 2

2 2

2 2

2

j k j kj k

j kj k

j k j k j kj k

j kj k

j k

=− +

+×

+−

=− + +

+×

+−

= +

6 (i) n n n n n− + = − + = −53

14

2012

3( 1)12

17 312

(ii) q

q q+ − −2

14

1

q q qq q

q q qq q

q qq q

2 ( 1) 4( 1)( 1)( 1)

2 2 4 4( 1)( 1)

2 6 4( 1)( 1)

2

2

= − − ++ −

= − − −+ −

= − −+ −

(iii) 49 3

2

2ss

ss

+−

−−

4( 3)( 3) 3

4 ( 3)( 3)( 3)

4 3( 3)( 3)

4 3( 3)( 3)

2

2

2 2

ss s

ss

s s ss s

s s ss s

ss s

= +− + − −

= + − +− +

= + − −− +

= −− +

7 (i) 2 14( 1)( 3) 1 3

xx x

Ax

Bx

+− + ≡ − + +

2 14 ( 3) ( 1)1 16 4 4

3 8 4 2

So 2 14( 1)( 3)

41

23

x A x B xx A Ax B B

xx x x x

+ ≡ + + −= ⇒ = ⇒ == − ⇒ = − ⇒ = −

+− + ≡ − − +

(ii) 16 36

16 3( 3)( 2)2

xx x

xx x

−+ −

= −+ −

16 3( 3)( 2) 3 216 3 ( 2) ( 3)

2 10 5 23 25 5 5

So 16 3( 3)( 2)

23

52

xx x

Ax

Bx

x A x B xx A Ax B B

xx x x x

−+ − ≡ + + −− ≡ − + += ⇒ = ⇒ == − ⇒ = − ⇒ = −

−+ − ≡ + − −

(iii) 102 5 3

10(2 1)( 3)2

xx x

xx x

−− −

= −+ −

10(2 1)( 3) 2 1 3

10 ( 3) (2 1)3 7 7 1

12

212

72 3

So 10(2 1)( 3)

32 1

13

xx x

Ax

Bx

x A x B xx B B

x A A

xx x x x

−+ − ≡ + + −

− ≡ − + += ⇒ − = ⇒ = −

= − ⇒ − = − ⇒ =

−+ − ≡ + − −

9781510458444_Answer.indb 7 11/8/18 10:33 AM

8

1Furtheralgebra


(iv) 5 20 3216

5 20 32( 16)

5 20 32( 4)( 4)

2

3

2

2

2x xx x

x xx x

x xx x x

+ −−

= + −−

= + −− +

5 20 32( 4)( 4) 4 4

5 20 32 ( 4)( 4) ( 4)( 4)

0 32 16 24 128 32 4

4 32 32 1

So 5 20 32( 4)( 4)

2 44

14

2

2

2

+ −− + ≡ + − + +

+ − ≡ − + + ++ −

= ⇒ − = − ⇒ == ⇒ = ⇒ == − ⇒ − = ⇒ = −

+ −− + ≡ + − − +

x xx x x

Ax

Bx

Cx

x x A x x Bx xCx x

x A Ax B Bx C C

x xx x x x x x

(v) x xx

xx

+ −−

= + −−

3 2 204

3 2 84

2

2 2

2 84

2 8( 2)( 2) 2 2

2 8 ( 2) ( 2)2 4 4 1

2 12 4 3

So 2 8( 2)( 2)

12

32

and 3 2 204

3 32

12

2

2

2

xx

xx x

Ax

Bx

x A x B xx A Ax B B

xx x x x

x xx x x

−−

= −− +

≡−

++

− ≡ + + −= ⇒ − = ⇒ = −= − ⇒ − = − ⇒ =

−− + ≡ − − + +

+ −−

= ++

−−

8 (i) 3 2 3( 1)( 1) 1 1 ( 1)

2

2 2x x

x xA

xB

xC

x+ −

− +≡

−+

++

+

3 2 3 ( 1) ( 1)( 1) ( 1)

1 2 4 12

1 2 2 1

terms 3 52

So 3 2 3( 1)( 1)

12

1

52

11

( 1)1

2( 1)5

2( 1)1

( 1)

2 2

2

2

2 2

2

+ − ≡ + + − + + −

= ⇒ = ⇒ =

= − ⇒ − = − ⇒ =

⇒ = + ⇒ =

+ −− +

≡−

++

++

=−

++

++

x x A x B x x C x

x A A

x C C

x A B B

x xx x x x x

x x x

(ii) 1( 1) 1 ( 1) ( 1)3 2 3

xx

Ax

Bx

Cx

−+

≡+

++

++

x A x B x Cx C

x Ax A B C B

xx x x x

x x

− ≡ + + + += − ⇒ − =

⇒ == ⇒ − = + + ⇒ =

−+

≡+

++

+ −+

=+

−+

1 ( 1) ( 1)1 2

terms 00 1 1

So 1( 1)

01

1( 1)

2( 1)

1( 1)

2( 1)

2

2

3 2 3

2 3

(iii) 2 4( 1)( 1) 1 1 ( 1)

2 4 ( )( 1) ( 1)( 1)

( 1)1 2 2 1

( )( 1) ( 2 )( 2 )

( 1)( 1) ( 1)

terms 00 4 3

terms 0 2

Since 3, 3 2 0 32

0 32

3 32

So 2 4( 1)( 1)

32

321

32

11

( 1)3 3

2( 1)3

2( 1)1

( 1)

2 2 2 2

2 2

2

2 3 2

2 3 2

3

2

2 2 2 2

2 2

− ++ −

≡ ++

+−

+−

− + ≡ + − + + −

+ += ⇒ = ⇒ =

+ − = + −+ − +

+ − = − + −

⇒ = += ⇒ = − + ⇒ − =

⇒ = − −

− = − = ⇒ =

+ = ⇒ = −

− = ⇒ =

− ++ −

≡+

++

−

−+

−

= ++

−−

+−

xx x

Ax Bx

Cx

Dx

x Ax B x C x x

D xx D D

Ax B x Ax B A xA B x B

C x x C x x x

x A Cx B C D B C

x B A C

B C A A

A C C

B C B

xx x

x

x x xxx x x

(iv) + +− +

≡−

+ ++

2 3 1( 1)( 2) 1 ( 2)

2

2 2x x

x xA

xBx Cx

x x A x Bx C xx A A

x A B Bx A C C

x xx x x x

2 3 1 ( 2) ( )( 1)1 6 3 2

terms 2 00 1 2 3

So 2 3 1( 1)( 2)

21

32

2 2

2

2

2 2

+ + ≡ + + + −= ⇒ = ⇒ =

⇒ = + ⇒ == ⇒ = − ⇒ =

+ +− +

≡−

++

9 2 3 3(1 )(1 ) 1 (1 )

2

2 2x x

x xA

xBx C

x− +

+ +≡

++ +

+

2 3 3 (1 ) ( )(1 )1 8 2 4

terms 2 20 3 1

So 2 3 3(1 )(1 )

41

2 1(1 )

41

2 1(1 )

41 4(1 )

4(1 ...)4 4 4 ...

2 1(1 )

(2 1)(1 )

(2 1)(1 ...)2 2 2 1 ...

2 2

2

2

2 2

2

1

2

2

22 1

2 4

3 5 2 4

x x A x Bx C xx A A

x A B Bx A C C

x xx x x

xx

xx

x

x x

x xx x

xx

x x

x x xx x x x x

− + ≡ + + + += − ⇒ = ⇒ =

⇒ = + ⇒ = −= ⇒ = + ⇒ = −

− ++ +

≡+

+ − −+

=+

− ++

+ = +

= − + −= − + −

++

= + +

= + − + −= − + + − + +

−

−

9781510458444_Answer.indb 8 11/8/18 10:33 AM

9

1Furtheralgebra


x

xx

x x

x x x x x

x x

So 41

2 1(1 )

(4 4 4 ...)

(2 2 2 1 ...)

3 6 5 ...

2

2

3 5 2 4

2

+ − ++

= − + −

− − + + − + +

= − + −

Valid for x x1 12 < ⇒ <

10 (i) 2 4 3(2 )(2 ) 2 2

2

2 2− −− +

≡−

+ ++

x xx x

Ax

Bx Cx

2 4 3 (2 ) ( )(2 )2 18 6 3

terms 3 00 2 2 2 4

So 2 4 3(2 )(2 )

32

42

42

32

2 2

2

2

2 2

2

− − ≡ + + + −= ⇒ − = ⇒ = −

⇒ − = − ⇒ == ⇒ = + ⇒ =

− −− +

≡ −−

++

=+

−−

x x A x Bx C xx A A

x A B Bx A C C

x xx x x x

x x

(ii) 32 3 2

3 2 1 12

3 2 1 12

3 12 1 1

2

32 1 1

232 1 1

214

18 ...

32

34

38

316 ...

42

4(2 )

1

1

11

1

1

2 3

2 3

22 1

x x

x

x

x

x

x x x

x x x

xx

( )( )

( )( )

( )( )( )

( )− = −

= −

= −

= −

= −

= + + + +

= + + + +

+= +

−

−

−−

−

−

−

4(2 )

4 2 1 12

4 12 1 ( 1) 1

21( 1 1)

2!12

2 1 12

14 ...

2 12 ...

So 2 4 32 (2 )

42

32

2 12 ...

32

34

38

316 ...

12

34

118

316 ...

2 1

1 21

2 22

2 4

2 4

2

2 2

2 4

2 3

2 3

x

x

x x

x x

x x

x xx x x x

x x

x x x

x x x

( )( )( )

( ) ( )( )

( )( )

( )

= +

= +

= + − + − − −

= − + +

= − + +

− −− +

=+

−−

= − + +

− + + + +

= − − − +

−

−−

Past exam questions (Page 12)

1 (i) 5(1 )(2 ) (1 ) 2

2

2 2x xx x

Ax

Bx Cx

−+ +

≡+

+ ++

5x − x2 ≡ A(2 + x2) + (Bx + C)(1 + x)

x = −1 ⇒ −6 ≡ 3A ⇒ A= -2

x = 0 ⇒ 0 ≡ 2A + C ⇒ C = 4

x2 terms ⇒ −1 ≡ A + B ⇒ B = 1

So 5(1 )(2 )

21

42

2

2 2x xx x x

xx

−+ +

≡ −+

+ ++

(ii)

…

�

�

�

�

�

2(1 ) ( 4)(2 )

2(1 ) ( 4) 2 1 12

2(1 ) ( 4) 12 1 1

2

2(1 )

2 1 ( 1) ( 1)( 2)2!

( 1)( 2)( 3)3!

2(1 )

2 2 2 2

12( 4) 1 1

2

12( 4) 1 ( 1) 1

2( 1)( 2)

2!12

12( 4) 1 1

212( 4) 1

4 ( 4)

12 2 1

4

1 2 1

1 21

1 21

1

2 3

2 3

2 3

21

2 22

2

2

3 2

( )

( )

( )( )

( )( ) ( )

( )

− + + + +

= − + + + +

= − + + + +

− +

= − + − + − − + − − − +

= − − + − +

≈ − + − +

+ +

= + + − + − −

= + − +

= + − + +

= + − − +

− −

−−

−−

−

−

x x x

x x x

x x x

x

x x x

x x x

x x x

x x

x x x

x x

x x x

x x x

The expansion is −2 + 2x – 2x2 + 2x3 + 12x + 2 – 14x3– x2

52 3 7

4 ...2 3x x x= − + +

2 (i) p(x) = ax3 − x2 − 4x − a

(2x − 1) a factor ⇒ p ( ) =12 0

a a

a a

a

aa

12

12 4 1

2 0

18

14 2 0

74

78 0

14 7 02

3 2( ) ( ) ( )⇒ × − + − =

− + − =

− =

− ==

)2 1 2 4 22

(2 )4 2

(4 2)0

3 2

2

3 2

− − + −+

− −−

− −

x x x xx

x xxx

So p(x) = (2x − 1)(x2 + 2)

9781510458444_Answer.indb 9 11/8/18 10:33 AM

10

1Furtheralgebra


(ii) −− +

≡−

+ ++

− ≡ + + + −

= ⇒ − ≡ ⇒ = −

= ⇒ − = − ⇒ =

⇒ = + ⇒ =− ≡ −

−+ +

+

8 13(2 1)( 2) 2 1 2

8 13 ( 2) ( )(2 1)12 9 9

4 4

0 13 2 5

term 0 2 2

So 8 13p( )

42 4

2 52

2 2

2

2

2

xx x

Ax

Bx Cx

x A x Bx C x

x A A

x A C C

x A B Bx

x xx

x

3 11 2x−

x

x x

x

x x x

1 2

1 12( 2 )

12

12 1

2! ( 2 )

12

12 1 1

2 2

3! ( 2 ) ...

1 32

52 ...

12

2

3

2 3

( )( )( )

( )= −

= + − − +− − −

−

+− − − − −

− +

= + + + +

−


4 f ( ) 5 7 4(3 2)( 5)

2

2x x xx x

= − ++ +

(i) 5 7 4(3 2)( 5) 3 2 5

2

2 2x x

x xA

xBx Cx

− ++ +

=+

+ ++

5 7 4 ( 5) ( )(3 2)23

989

499 2

terms 5 3 10 4 5 2 3

So 5 7 4(3 2)( 5)

23 2

35

2 2

2

2

2 2

x x A x Bx C x

x A A

x A B Bx A C C

x xx x x

xx

− + ≡ + + + +

= − ⇒ = ⇒ =

⇒ = + ⇒ == ⇒ = + ⇒ = −

− ++ +

≡+

+ −+

(ii) 2 3 2 1x( )+ −

2 2 1 32

2 2 1 32

2 12 1 3

2

1 32

1 32

94 ...

1

11

1

1

2

( )( )

( )( )

( )( )( )

= +

= +

= +

= +

= − + −

−

−−

−

−

x

x

x

x

x x

35

( 3)( 5)

3 5 1 15

3 15 1 ( 1) 1

51( 1 1)

2!15

3 15 1 1

5125 ...

3 15

125

1125 ...

35

15

325 ...

22 1

1 21

2 22

2 4

2 4

2

( )( )( )

( )( ) ( )

( )( )

( )

( )

( )

( )

−+

= − +

= − +

= − + − + − − −

= − − + +

= − − + +

= − + + +

−

−−

xx

x x

x x

x x x

x x x

x x x

x x

So 5 7 4(3 2)( 5)

23 2

35

1 32

94 ...

+ 35

15

325 ...

25

1310

237100 ...

2

2 2

2

2

2

x xx x x

xx

x x

x x

x x

( )( )

− ++ +

≡+

+ −+

= − + −

− + + +

= − + +

Stretch and challenge (Page 13)

1 (i) 4 3 4 312x x( )− = −

4 3

4 1 34

4 1 34

2 1 12

34

12

12 12!

34 ...

2 1 38

9128 ...

2 34

964 ...

12

12

12

12

2

2

2

( )

( )( )

( ) ( )( )

( )

( )−

= −

= −

= + − +−

− +

= − − −

= − − −

x

x

x

x x

x x

x x

(ii) (1 )(4 3 )12+ −ax x

(1 ) 2 34

964 ...

terms are 964

34

964

34

964

34

9 4864

So 9 4864

11164

9 48 11148 120

52

2

2 2

2 2

2

2

ax x x

x x ax x

x a x

a x

a x

a

aa

a

( )

( )( )

= + − − −

− + × −

= − + −

= − −

= − −

− − =

− − =− =

= −

9781510458444_Answer.indb 10 11/8/18 10:34 AM

11

2Furthercalculus


2 1 31

2

2x

kx( )( )

++

1 3 1

(1 6 9 )(1 )

(1 6 9 ) 1 ( 2)( ) 2( 2 1)2! ( )

(1 6 9 )(1 2 3 ...)

terms are 3 6 2 9

(3 12 9)

So 3 12 9 105

4 32 0( 4)( 8) 0

4 or 8

2 2

2 2

2 2

2 2 2

2 2 2 2

2 2

2

2

( )

( ) ( )= + +

= + + +

= + + + − + − − −

= + + − + −

+ × − +

= − +

− + =

− − =+ − =

= −

−

−

x kx

x x kx

x x kx kx

x x kx k x

x k x x kx x

k k x

k k

k kk k

k

3 You want to find n such that

( 1)( 2)3!

( 1)( 2)( 3)4!

24 ( 1)( 2) 6 ( 1)( 2)( 3)24 6( 3)24 6 1842 6

7

n n n n n n n

n n n n n n nn

nn

n

− − < − − −

− − < − − −< −< −<>

4 (i) ax bax b b ax

abx a b b x

b1 1 ( 1)2! ( ) ...

1 ( 1)2 ...

2

22

( )+ = + + − +

= + + − +

ab a b

a b b b b b

bb b

b b

b bb

b

a

( )= − ⇒ = −

− = ⇒− −

=

−=

− =

− =− =

= −

= −−

=

1 1

( 1)2 2

1 ( 1)

2 2

1 ( 1)

2 2

1( 1) 4

1 41 3

13

113

3

22

2

(ii) Valid for x x3 1 13< ⇒ <

2 Further calculus2.1 Differentiating tan−1x (Page 14)

1 tan 1y x= −

tansincos

dd

cos cos ( sin )(sin )cos

cos sincos

coscos

sincos

1 tan

1

2

2 2

2

2

2

2

2

2

2

y xyy x

xy

y y y yy

y yy

yy

yy

y

x

=

=

=× − −

=+

= +

= +

= +

So

dd

11 2

yx x

=+

2 (i) 2 tan 81y x= −

dd

21 (8 )

8 161 642 2

yx x x

=+

× =+

(ii) y x12 tan ( )1 3= −

dd

12

11 ( )

3 32 23 2

22

6yx x

x xx

= ×+

× =+

(iii) tan sin21y x( )= −

dd

2cos21 sin 22

yx

xx

=+

(iv) lncos

e

2lncos

e

tan

2

tan

1

1

( )( )

=

=

−

−

yx

yx

x

x

2 ln(cos ) ln(e )

2 ln(cos ) tan

dd

2 sincos

11

2tan 21

tan

1

2

2

1y x

y x x

yx

xx x

xx

x= −

= −

= − −+

= − −+

−

−

9781510458444_Answer.indb 11 11/8/18 10:34 AM

12

2Furthercalculus


3 y xtan (e )1= −

dd

e1 e

When 0,dd

e1 e

12

When 0, tan (e ) tan 1 4

412 0 4

So 12 4

4 22 4 0

2

0

2 0

1 0 1

yx

xyx

x y

y mx c c c

y x

y xx y

x

x=+

= =+

=

= = = = π

= + ⇒ π = × + ⇒ = π

= + π

= + π− + π =

×

− −

2.2 Integration by substitution (Page 15)

1 (i) 2 1 d 2du x u x= + ⇒ =

x x u u

u u

u c

u c

x c

2 1 d d2

12 d

12 21414 2 1

2

2

2

∫ ∫∫

( )

( )

+ =

=

= × +

= +

= + +

(ii) u x u x x x ux1 d 2 d d d

22= − ⇒ = − ⇒ = −

x x x x u ux

u u

u c

u c

u c

x c

x c

1 d d2

12 d

12 3

2

12

23

1313(1 )

13 (1 )

2

32

32

32

232

2 3

∫ ∫∫

( )− = −

= −

= −

+

= −

+

= − +

= − − +

= − − +

(iii) u x u x x u

x x x

u u u

u u u

u u c

4 d d and 4

2 4 d

2( 4) d

2 ( 4 )d

2 52

432

32

12

52

32

∫∫∫

= + ⇒ = = −

+

= −

= −

= −

+

u u c

u u c

x x c

x x c

2 25

83

45

163

4( 4)5

16( 4)3

4 ( 4)5

16 ( 4)3

52

32

52

32

52

32

5 3

= −

+

= − +

= + − + +

= + − + +

(iv) 1 d du x u x= − ⇒ = Also 1x u= +

x x x

u u u

u u u u

u u u u

u u u c

u u u c

x x x c

x x x c

1 d

( 1) d

( 2 1) d

( 2 )d

72

252

32

27

45

23

2( 1)7

4( 1)5

2( 1)3

2 ( 1)7

4 ( 1)5

2 ( 1)3

2

2

2

52

32

12

72

52

32

72

52

32

72

52

32

7 5 3

∫∫∫∫

−

= +

= + +

= + +

= + + +

= + + +

= − + − + − +

= − + − + − +

2 (i) 1 d 4 d d d4

4 33u x u x x x u

x= + ⇒ = ⇒ =

x u x u1 2, 0 1= ⇒ = = ⇒ =

1 d

d4

14 d

14 3

2

14

23

14

2 23

2 13

0.305 (3 s.f.)

3 4

0

1

331

2

1

2

32

1

2

32

1

2

32

32

∫∫∫

( ) ( )

+

=

=

=

=

= × − ×

=

=

=

x x x

x u ux

u u

u

u

u

u

9781510458444_Answer.indb 12 11/8/18 10:34 AM

13

2Furthercalculus


(ii) u x x u x x2 1 d 4 1 d2 ( )= − + ⇒ = − x u x u3 16, 0 1= ⇒ = = ⇒ =

x x x x

x u ux

u u

u

u

u

u

∫∫∫

( ) ( )

( )

( )

− − +

= − −

=

=

=

= −

=

=

=

4 1 2 1 d

4 1 d4 1

d

32

23

2 163

2 13

42

2

0

3

1

16

1

16

32

1

16

3

1

16

3 3

(iii) 2 d 2 2 d, 1 3

1( 2 )

d

1( )

d2 2

12

1 d

12 d

12 4

12

14

12

14

14 3

12

1324

1648 0.00154

2

2 51

53

53

5

3

4

3

43

4 4

u x x u x xx u x u

xx x

x

xu

ux

uu

u u

u

u

u

u

∫

∫

∫∫

( )( )

( )= + ⇒ = += ∞ ⇒ = ∞ = ⇒ =

++

= ++

=

=

= −

= −

= −× ∞

− −×

=

= ≈

∞

=

=∞

∞

−∞

− ∞

∞

(iv) u x u x x1 d 2 d2= − ⇒ = x u x u2 1, 1 0= ⇒ = = ⇒ =

( 1) d

d2

12 d

12 1 d

12 ( ) d

3

1

2 2 6

3

0

1 6

2

0

1 6

0

1 6

7 6

0

1

∫∫∫∫∫

( )

( )

( )

−

=

=

= +

= +

=

=

=

=

x x x

x u ux

x u u

u u u

u u u

u

u

u

u

12 8 7

12

18

17

0

15112 0.134

8 7

0

1

8 7( )= +

= + −

= ≈

u u

3 1 1 d 1 d d d22u

xu

xx x x u= + ⇒ = − ⇒ = −

∫

∫

∫

( )

( )

( )

( )

= ⇒ = = ⇒ =

+

= −

= −

= −

= − −

= − −

= ≈

=

=

x u x u

xx

x

ux

x u

u u

u

u

u

2 32 , 1 2

1 1d

( d )

d

4

324

24

8164

164

17564 2.73

3

21

2

3

22

2

32

3

2

32

4

2

32

4

4

4 = + ⇒ = + ⇒ = −u x u x x u2 2 22 2

d 2 d=x u u

7 3, 1 1= ⇒ = = − ⇒ =x u x u

2d

( 2) 2 d

2 ( 4 4)d

2 54

3 4

2 35

4 33 4 3 1

54 1

3 4 1

2 1235

4315

65215

2

1

7

2 2

1

3

4 2

1

3

5 3

1

3

5 3 5 3

∫∫∫

( ) ( )( ) ( )

+

= −

= − +

= − +

= − × + × − − × + ×

= −

=

−

=

=

xx

x

uu u u

u u u

u u u

u

u

9781510458444_Answer.indb 13 11/8/18 10:34 AM

14

2Furthercalculus


2.3 Integrals involving exponentials and natural logarithms (Page 16)

1 (i) 3 d 2 d2= + ⇒ =u x u x x

8 e d

8 e d2

4 e d

4e

4e

3

3

2

2

∫∫∫

=

=

= +

= +

+

+

x x

x ux

u

c

c

x

u

u

u

x

(ii) = − ⇒ =u u xx xe 1 d e d

ee 1

d

e de

1 d

ln

ln e 1

∫∫∫

−

=

=

= +

= − +

x

uu

u u

u c

c

x

x

x

x

x

(iii) 1 d 1 d d d22u

xu

xx x x u= ⇒ = − ⇒ = −

e d

e ( d )

e d

e

e

1

2

22

1

∫∫∫

= −

= −

= − +

= − +

xx

xx u

u

c

c

x

u

u

u

x

(iv) 1 cos d sin d= − ⇒ =u x u x x

e sin d

e sin dsin

e d

e

e

1 cos

1 cos

∫∫∫

=

=

= +

= +

−

−

x x

x ux

u

c

c

x

u

u

u

x

2 = + ⇒ =u x u x2 1 d 2d

1 3, 0 1= ⇒ = = ⇒ =x u x u

2 1 d

12 d

214

1 d

14 1 1 d

0

1

1

3

1

3

1

3

∫

∫∫∫ ( )

+

=−

= −

= −

=

=

xx x

u

uu

uu u

u u

u

u

14 ln

14 (3 ln3) (1 ln1)

14 (3 ln3) 1

14(2 ln3)

1

3

[ ]

[ ]

= −

= − − −

= − −

= −

u u

3 = + ⇒ =u u xx x1 e d 2e d2 2

ln2 1 e 1 e 5

0 1 e 1 1 2

2ln2 ln4

2 0

= ⇒ = + = + =

= ⇒ = + = + =×

x u

x u

32e (1 e ) d2

0

ln2 2 4∫ + xx x

32e d2e

16 d

16 5

16 55

25

9897.6

2

2

5 42

4

2

5

5

2

5

5 5

∫∫

( )=

=

=

= −

=

=

=u u

u u

u

x

u

u

x

4 2 ln d 1d d d= + ⇒ = ⇒ =u t u t t t t u e 3, 1 2= ⇒ = = ⇒ =t u t u

1(2 ln )

d

1( )

d

1 d

d

1

1

13

12

16

e

3

3

3

21

22

22

2

2

1

2

3

2

3

∫

∫

∫∫

( ) ( )

+

=

=

=

= −

= −

= − − −

=

=

=

−

−

t tt

t ut u

uu

u u

u

u

u

u

5 e1 e

4

4=+

yx

x

(i) P is where = ⇒ =+

=x y0 e1 e

12

0

0

(ii) yx

x x x x

x

x

x

= + − ×+

=+

dd

4e (1 e ) 4e e(1 e )

4e(1 e )

4 4 4 4

4 2

4

4 2

When 0,dd

4e(1 e )

44

10

0 2= =+

= =xyx

9781510458444_Answer.indb 14 11/8/18 10:34 AM

15

2Furthercalculus


(iii) Area = e1 e

d4

40

1

∫ +x

x

x

14

4e1 e

d

14 ln 1 e

14 ln 1 e ln 1 e

14 ln 1 e ln2

14 ln

1 e2

4

40

1

4

0

1

4 0

4

4

∫

( )( )

( )

=+

= +

= + − +

= + −

=+

xx

x

x

2.4 Integrals involving trigonometrical functions (Page 19)

1 (i) d 2 d

sin( )d

sin( )d2

12 sin( )d

12cos( )

12cos( )

2

2

2

∫∫∫

= ⇒ =

=

=

= − +

= − +

u x u x x

x x x

x u ux

u u

u c

x c

(ii) cos d sin d= ⇒ = −u x u x x

sin e d

sin e dsin

e d

e

e

cos

cos

∫∫∫

= −

= −

= − +

= − +

x x

x ux

u

c

c

x

u

u

u

x

(iii) = ⇒ =sin d cos du x u x x

∫∫∫∫

=

=

=

= += +

x x

xx x

xu

ux

u u

u cx c

cot d

cossin d

cos dcos

1 d

lnln sin

(iv) = ⇒ = −cos d sin du x u x x

cos sin d

sin dsin

d

6cos

6

5

5

5

6

6

∫∫∫

= −= −

= − +

= − +

x x x

u x ux

u u

u c

x c

2 = − ⇒ =1 cos d sin du x u x x

sin cos (1 cos ) d

sin cos ( ) dsin

(1 ) d

( )d

4 5(1 cos )

4(1 cos )

5

3

3

3

3 4

4 5

4 5

∫∫∫∫

−

=

= −

= −

= − +

= − − − +

x x x x

x x u ux

u u u

u u u

u u c

x x c

3 x xθ θ θ= ⇒ =2sin d 2cos d

1 sin 12

π6

0 sin 0 0

θ θ

θ θ

= ⇒ = ⇒ =

= ⇒ = ⇒ =

x

x

x x∫∫∫∫∫∫∫∫∫

θ θ θ

θ θ θ

θ θ θ

θ θ θ

θ θ θ

θ θ

θ θ

θ θ

θ θ

( )

( )

( ) ( )( )

( )

( )

( )

−

= −

= −

= −

=

= ×

=

= +

= +

= +

= × π + π − × +

= π + π −

= + π −

= + π

θ

θ

=

= π

π

π

π

π

π

π

π

π

4 d

4 2sin 2cos d

4 4sin 2cos d

4(1 sin ) 2cos d

4cos 2cos d

2cos 2cos d

4cos d

4 12 cos2 1 d

2 cos2 1 d

2 12 sin2

2 12 sin2 6 6

12 sin2 0 0

2 12 sin 3 6 0

2 34 6 0

32 3

2

0

1

2

0

6

2

0

6

2

0

6

2

0

6

0

6

2

0

6

0

6

0

6

0

6

9781510458444_Answer.indb 15 11/8/18 10:34 AM

16

2Furthercalculus


4 (i) sin d 2sin cos d2θ θ θ θ= ⇒ =x x

1 sin 1 20 sin 0 0

2

2

x

x

θ θ

θ θ

= ⇒ = ⇒ = π

= ⇒ = ⇒ =

1 d

1 sinsin

2sin cos d

cossin

2sin cos d

cossin 2sin cos d

2cos d

0

1

2

20

2

2

20

2

0

2

2

0

2

xx x∫

∫

∫

∫

∫

θθ

θ θ θ

θθ

θ θ θ

θθ θ θ θ

θ θ

−

= −

=

=

=

θ

θ

=

= π

π

π

π

(ii) 1 d0

1

∫ − xx x

2cos d

2 12(cos2 1) d

cos2 1 d

12 sin2

12 sin2 2 2

12 sin2 0 0

12 sin 2 0

2

2

0

2

0

2

0

2

0

2

∫

∫

∫

θ θ

θ θ

θ θ

θ θ

( )

( ) ( )( )

( )

=

= +

= +

= +

= × π + π − × +

= π + π −

= π

π

π

π

π

5 θ θ θ )(= = = −2sec 2cos 2 cos 1x

dd

2 cos sin 2sincos

22

xθ

θ θ θθ

( )= − × − =−

∫∫

∫∫

∫∫∫

θ θθθ

θ

θ θθθ

θ

θ θθθ

θ

θ θθθ

θ

θ θθθ

θ

θθθ

θθ

θ

−

=−

=−

=−

=

=×

=×

14

d

1(2sec ) (2sec ) 4

2sincos

d

14sec 4sec 4

2sincos

d

14sec 4(sec 1)

2sincos

d

14sec 4 tan

2sincos

d

14sec 2tan

2sincos

d

14

cos2sincos

2sincos

d

2 2

2 2 2

2 2 2

2 2 2

2 2 2

2 2

22

x xx

∫∫

∫∫

∫∫∫

θ θθθ

θ

θ θθθ

θ

θ θθθ

θ

θ θθθ

θ

θ θθθ

θ

θθθ

θθ

θ

−

=−

=−

=−

=

=×

=×

14

d

1(2sec ) (2sec ) 4

2sincos

d

14sec 4sec 4

2sincos

d

14sec 4(sec 1)

2sincos

d

14sec 4 tan

2sincos

d

14sec 2tan

2sincos

d

14

cos2sincos

2sincos

d

2 2

2 2 2

2 2 2

2 2 2

2 2 2

2 2

22

x xx

cos8sin

2sincos

d

14 cos d

14 sin

2sec 2cos cos 2 cos 2

So 14

d 14

sin cos 2

3

2

1

2 21

∫∫

∫

θθ

θθ

θ

θ θ

θ

θ θ θ θ

( )

=

=

= +

= = ⇒ = ⇒ =

−= +

−

−

c

x x x

x xx

xc

6 (i) x y xy

y= ⇒ = −1 d 1 d2

x xx

y yy

y

yy

y

yy

y

y

yy

y

y

yy

∫∫

∫

∫

∫

∫

−

=−

−

= −−

= −−

= −−

= −−

11

d

11 1 1

1 d

11 1

d

11

d

11

d

11

d

2

22

2

2

2

2

2

(ii) x x

xy

y11

d 11

d2 2∫ ∫

−= −

−

y y

yy

c

y c

x c

sin d cos d

So 11

d

11 sin

cos d

1cos

cos d

d

sin

sin 1

2

2

2

1

1

∫

∫∫∫

θ θ θ

θθ θ

θθ θ

θ

θ

( )

= ⇒ =

−−

= −−

= −

= −

= − +

= − +

= − +

−

−

9781510458444_Answer.indb 16 11/8/18 10:34 AM

17

2Furthercalculus


7 (i) 21

d 2tan21

xx x c∫ +

= +−

(ii) x

xx

x

xx

x c

x c

19

d 19 1 1

9

d

19

1

1 3

d

19 tan 3 3

13 tan 3

2 2

2

1

1

∫ ∫

∫ ( )( )( )

( )

( )

+=

+

=+

= × +

= +

−

−

(iii) ∫ ∫ ( )+=

+

=

× +

=

+

=

+

−

−

−

xx

xx

x c

x c

x c

12 3

d 12 1 3

2

d

12 tan 3

223

22 3

tan 32

66 tan 3

2

2 2

1

1

1

2.5 The use of partial fractions in integration (Page 22)

1 (i) ∫ ∫( )++ − = + + −

xx x x x x x7 8

( 2)( 1)d 22

51 d

2ln 2 5ln 1= + + − +x x c

(ii) 4 8( 1) ( 1)

d2

2x x

x xx∫ −

− +

x x xx

x x x c

∫( )=−

−−

++

= − + − + + +

11

2( 1)

31

d

ln 1 21 3ln 1

2

2 (i) −−

=+

−−

51

31

212

xx x x

(ii) 51

d 31

21 d22

3

2

3

∫ ∫ ( )−−

= + − −xx

x x x x

3ln 1 2ln 1

ln 1 ln 1

ln1

1

ln3 13 1

ln2 12 1

ln 42

ln31

ln16 ln27

ln1627

2

3

3 2

2

3

3

2

2

3

3

2

3

2

3

2

3

2

= + − −

= + − −

=+

−

= +−

− +

−

= −

= −

=

x x

x x

x

x

3 (i) 2 1( 3)2

xx

+−

= 23

7( 3)2x x−

+−

(ii) ∫ +−

xx

x2 1( 3)

d24

10

23

7( 3)

d

2ln 3 73

2ln 10 3 710 3 2ln 4 3 7

4 3(2ln7 1) (2ln1 7)

2ln7 6

24

10

4

10

∫ ( )

( ) ( )

=−

+−

= − − −

= − − − − − − −= − − −= +

x xx

x x

4 (i) = − ⇒ = −−

u x ux

x2 d 12 2

d

= − −x x ud 2 2 d

2 22 0, 1 1

32

d

32

( 2 2 d )

62

d

6(2 )(1 ) d

2 2

1

2

21

0

21

0

0

1

∫∫∫

∫

( )

= − ⇒ = −= ⇒ = = ⇒ =

=+ −

=− +

− −

= −− +

= − +

=

=

u x x ux u x u

Ix x

x

u ux u

uu u

u

uu u u

u

u

(ii) ∫ − +u

u u u6(2 )(1 ) d

0

1

u u u

u u

∫ ( )= − − +

= − − − +

=− − − +

− − − − +

= − − − − −= − − −= −==

42

21 d

4ln(2 ) 2ln(1 )

( 4ln(2 1) 2ln(1 1))( 4ln(2 0) 2ln(1 0))

( 4ln1 2ln2) ( 4ln2 2ln1)(0 ln4) ( ln16)ln16 ln4ln42ln2

0

1

0

1

2.6 Integration by parts (Page 24)

1 (i) = ′ =′ = = −

u x v xu v x

sin1 cos

∫ ∫= − − −

= − + +

x x x x x x x

x x x c

sin d cos cos d

cos sin

(ii) = ′ =′ = =

u x vu v

x

x4 e4 e

4 e d 4 e 4e d

4 e 4e∫ ∫= −

= − +

x x x x

x c

x x x

x x

9781510458444_Answer.indb 17 11/8/18 10:34 AM

18

2Furthercalculus


(iii) = ′ =

′ = =

u x v x

u x v x

ln

13

2

3

∫ ∫

∫

= − ×

= −

= − +

x x x x x xx x

x x x x

x x x c

ln d 3 ln 13 d

3 ln 3 d

3 ln 9

23 3

3 2

3 3

2 (i) = − ′ =

′ = =

u x v

u v

x

x

2 1 e

2 12e

2

2

∫∫∫

−

= −

− ×

= −

−

= − −

= −

= −

= − − −

=

x x

x x

x x

x

x

x

x

x x

x x

x x

x x

x

(2 1)e d

(2 1)12e 2 1

2e d

(2 1)12e e d

(2 1)12e 1

2e

e e

e ( 1)

e (2 1) e (1 1)

e

2

1

2

2

1

2

1

2 2

2

1

22

1

2

2 2

1

2

2 21

2

21

2

4 2

4

(ii) = ′ =

′ = =

u x v x

u v x

cos2

1 12 sin2

cos2 d

12 sin2 1

2 sin2 d

12 sin2 1

4 cos2

12 6 sin 3

14 cos 3

12 0sin0 1

4 cos0

12 6

32

14

12 0 1

4

324

18

14

324

18

0

6

0

6

0

6

0

6

∫

∫

( )( )

( )( )

( )( )

=

−

= +

=× π π + π

− × +

= × π × + × − +

= π + −

= π −

π

π π

π

x x x

x x x x

x x x

3 (i) cos3y x x= P and Q are when y = 0

==

= π π π

= π π π

cos3 0cos3 0

3 2 , 32 , 5

2 ,...

6 , 2 , 56 ,...

x xx

x

x

( ) ( )π πP is 6 , 0 and Q is 2 , 0

(ii) = × + − ×

= −

yx x x x

x x x

dd 1 cos3 3sin3

cos3 3 sin3

= π = × π − × π × πAt 6 ,dd cos3 6 3 6 sin3 6x

yx

= π − π π

= − π

= − π

cos 2 2 sin 2

0 2

2(iii) = ′ =

′ = =

u x v x

u v x

cos3

1 13sin3

Area = ∫π

cos3 d0

6 x x x

∫

( )( )

( ) ( )( ) ( )

=

−

= +

=× π × π + × π

− × × + ×

= π π + π − +

= π − +

= π −

= π −

π π

π

13 sin3 1

3sin3 d

13 sin3 1

9cos3

13 6 sin3 6

19cos3 6

13 0sin3 0 1

9cos3 0

18 sin 219cos 2 0 1

9

18 0 19

1819

218

0

6

0

6

0

6

x x x x

x x x

4 (i) x

xx

x x x x

xx x x

xx x

x

dd

ln1 2 ln

( )2 ln

(1 2ln )

2

2

2 2

4

4

( ) =× − ×

= −

= −

9781510458444_Answer.indb 18 11/8/18 10:34 AM

19

2Furthercalculus


(ii) = ′ =

′ = = −

u x vx

u x v x

ln 1

1 1

2

ln d

1 ln d

1 ln 1 1 d

1 ln 1 d

1 ln 1

1 (ln 1)

2

2

2

∫∫

∫∫ ( )

( )=

= − − × −

= − − −

= − − +

= − + +

xx

x

xx x

x x x x x

xx

xx

x x x c

x x c

5 = ′ =

′ = =

u x v x

u x v x

ln

15

4

5

x x x

x x xx x

x x x x

x x x

ln d

5 ln 15 d

5 ln 5 d

5 ln 25

e5 lne e

2515 ln1 1

25

e5

e25

125

4e 125

4

1

e

5

1

e 5

1

e

5

1

e 4

1

e

5 5

1

e

5 5 5 5

5 5

5

∫∫

∫

( )( )

( ) ( )( ) ( )

=

− ×

=

−

= −

= − − −

= − − −

= +

6 (i) ddx (ecos x ) = x x−( sin ) ecos

(ii) u x v x

u x v

x

x

= ′ =

′ = − = −

cos sin e

sin e

cos

cos

cos sin e d

cos sin e d

cos e sin e d

cos e e

(1 cos ) e

1 cos12 e (1 cos0)e

e (1 1) e1

cos

0

cos

0

cos0

12 cos

0

cos cos0

12

cos0

12

cos12 cos0

0 cos0

12

12

12

∫

∫

∫

( )

= ×

= − −

= − +

= −

= − π − −

= − −=

π

π

π

π

π

π

π

x x x

x x x

x x x

x

x

x

x

x x

x x

x

7 e

e

= ′ =

′ = = −

u v x

u v x

x

x

sin

cos

e

e e

e e

∫∫∫

=

= − − −

= − +

I x x

x x x

x x x

x

x x

x x

sin d

cos cos d

cos cos d

e

e

= ′ =

′ = =

u v x

u v x

x

x

cos

sin

cos sin sin d

cos sin

2 cos sin

(sin cos )

(sin cos )2

∫( )= − + −

= − + − +

= − + +

= − +

= − +

I x x x x

x x I c

I x x c

x x c

I x x c

x x x

x x

x x

x

x

e e e

e e

e e

e

e


1 (i) = + ⇒ = ⇒ =u x u x x x ux1 d 2 d d d

22

∫∫∫

+

=

=

= +

= + +

x x x

x u ux

u u

u c

x c

2 ( 1) d

2 ( ) d2

d

5( 1)

5

2 4

4

4

5

2 5

(ii) 3 2 d 6 d d d6

3 22= + ⇒ = ⇒ =u x u x x x u

x

∫∫∫

+

=

=

= +

= + +

3 2d

( )d

616

1 d

16 ln

16 ln 3 2

2

3

2

2

3

xx

x

xu

ux

u u

u c

x c

(iii) 4 d d d dAlso 4

= − ⇒ = − ⇒ = −= −

u x u x x ux u

∫∫∫∫( )

−

= − −

= −

= −

xx

x

uu

u

uu

u

uu u

u

2(4 )

d

2(4 )( )

( d )

2 4 d

2 4 d

5

5

5

5 5

9781510458444_Answer.indb 19 11/8/18 10:34 AM

20

2Furthercalculus


∫= −

= − − −

+

= − +

+

= − + +

= −−

+−

+

− −

− −

u u u

u u c

u uc

u uc

x xc

2 ( 4 )d

2 34

4

2 13

1

23

2

23(4 )

2(4 )

4 5

3 4

3 4

3 4

3 4

2 (i) 2 d d , also 2u x u x x u= + ⇒ = = − = ⇒ = = ⇒ =x u x u2 4, 1 3

(2 )d

2 d

( 2 )d

12

2

1 1

14

14

13

13

316

29

5144

31

2

33

4

2 3

3

4

1 2

3

4

23

4

2 2

∫

∫∫

( )( )( ) ( )

+

= −

= −

= − − −

= − +

= − + − − +

= − − −

=

=

=

− −

− −

xx

x

uu

u

u u u

u u

u u

u

u

(ii) u x u x x= + ⇒ =1 d 2 d2

= ⇒ = = ⇒ =x u x u3 4, 0 1

1 1 ( 1)2 2 4 2= + ⇒ = − ⇒ = −u x x u x u

x x x

x u ux

x u u

u u u

u u u u

u u u u

u u u

u u u

u

u

u

u

1 d

d2

12 d

12 ( 1) d

12 ( 2 1) d

12 2 d

12 7

2

252

32

12

27

45

23

5

0

3 2

5

1

4

4

1

4

2

1

4

2

1

4

52

32

12

1

4

72

52

32

1

4

72

52

32

1

4

∫∫∫∫∫∫ ( )

+

=

=

= −

= − +

= − +

= − +

= − +

=

=

=

=

12

2 47

4 45

2 43

2 17

4 15

2 13

12

1712105

16105

848105 8.08

72

52

32

72

52

32

( )( )

=× − × + ×

− × − × + ×

= −

= ≈

3 (i) u x x u x x= − ⇒ = −d (1 2 ) d2

∫∫∫

−−

= −−

=

= +

= − +

xx x

x

xu

ux

u u

u c

x x c

1 2 d

1 2 d1 2

1 d

ln

ln

2

2

(ii) 1 e d 3e d3 3= + ⇒ =u u xx x

x

uu

uu

u c

uc

uc

c

x

x

x

x

x

e(1 e )

d

e( )

d3e

13

1 d

13 213

12

16

16(1 e )

3

3 3

3

3 3

3

2

2

2

3 2

∫

∫∫

( )( )

+

=

=

= − +

= − +

= − +

= −+

+

−

(iii) 1 3 d 9 d3 2u x u x x= − ⇒ = −

∫∫

∫

−

=−

= −

= − +

= − − +

61 3

d

6 d9

23

1 d

23 ln

23 ln 1 3

2

3

2

2

3

xx

x

xu

ux

u u

u c

x c

9781510458444_Answer.indb 20 11/8/18 10:35 AM

21

2Furthercalculus


4 (i) 3 1 d 3du x u x= − ⇒ = = ⇒ = = ⇒ =x u x u1 2, 2

3 1

Area = xx x∫ −3 1 d

2

23

1

u

uu

u u

uu

u uu u

u u u

u u u

u

u1

3 d3

2 19 d

3

127

2 1 d

127 2 1 d

127 2 2 ln

127

22 2 2 ln2 1

2 2 1 ln1

127 (6 ln2) 5

21

2772 ln2 0.155

2

1

2

2

1

2

2

1

2

1

2

2

1

2

2 2

∫

∫∫∫

( ) ( )

( )

( )

( )( )

( )=

+

=+ +

= + +

= + +

= + +

= + × + − + × +

= + −

= + ≈

=

=

(ii) = − −−

dd

2 (3 1) 3(3 1)

2

2yx

x x xx

3 2(3 1)

(3 2)(3 1)

2

2

2

= −−

= −−

x xx

x xx

Stationary point when =yx

dd 0

(3 2)(3 1)

0 (3 2) 0

0 or 23

When 0, (0)3(0) 1 0 (0, 0)

When 23 ,

23

3 23 1

49

23 , 4

9

2

2

2

( )( )( )

−−

= ⇒ − =

=

= = − = ⇒

= =−

= ⇒

x xx

x x

x

x y

x y

5 (i) u x u x x= + ⇒ =1 sin3 d 3cos3 d

cos3 (1 sin3 ) d

cos3 ( ) d3cos3

13 d

13 4

12(1 sin3 )

12

3

3

3

4

4

4

∫∫∫

+

=

=

=

+

= +

=+

+

x x x

x u ux

u u

u c

u c

xc

(ii) u x u x x= ⇒ = −cos4 d 4sin4 d

tan4 d

sin4cos4 d

sin4 d4sin4

14

1 d

14 ln

14 ln cos4

∫∫∫

∫

=

= −

= −

= − +

= − +

x x

xx x

xu

ux

u u

u c

x c

(iii) u x u x x= ⇒ =sin3 d 3cos3 d

x x x

u x ux

u u

u c

x c

∫∫∫( )

=

=

= +

= +

2sin 3 cos3 d

2 cos3 d3cos3

23 d

23 2sin 3

3

3

3

3

4

4

(iv) 2 d 2du uθ θ= ⇒ =

1cos 2

d

1cos

d2

12

1cos

d

12 sec d

12 tan

12 tan2

2

2

2

2

∫∫∫∫

θθ

θ

=

=

=

= +

= +

uu

uu

u u

u c

c

6 θ θ θ= ⇒ = −cos d sin dx x

θ θ

θ θ

= ⇒ = ⇒ =

= ⇒ = ⇒ = π1 cos 1 0

0 cos 0 2

x

x

∫

∫

∫

∫

∫

θθ

θ θ

θθ

θ θ

θθ θ θ

θ θ

−

=−

−

= −

= −

= −

θ

θ

= π

=

π

π

π

1d

cos1 cos

( sin d )

cossin

( sin d )

cossin ( sin d )

cos d

2

20

1

2

22

0

2

22

0

2

2

0

2

2

0

xx

x

9781510458444_Answer.indb 21 11/8/18 10:35 AM

22

2Furthercalculus


∫

∫

θ θ

θ θ

θ θ

( )

( ) ( )( )

( )

= − +

= +

= +

= × π + π − × +

= π + π −

= + π

= π

π

π

π

12(cos2 1) d

12 (cos2 1) d

12

12 sin2

12

12 sin2 2 2

12 sin2 0 0

12

12 sin 2 0

12 0 2

4

2

0

0

2

0

2

7 θ θ θ= ⇒ =tan d sec d2x x

x

x

1 tan 1 40 tan 0 0

θ θ

θ θ

= ⇒ = ⇒ = π

= ⇒ = ⇒ =

∫

∫

∫

∫

∫∫

θθ θ

θθ θ

θθ

θ θ

θ θ

θ θ

( )

( ) ( )( )( )

+

=+

=

=

=

= +

= +

= × π + π − × +

= π + π −

= + π −

= + π

θ

θ

=

= π

π

π

π

π

π

1( 1)

d

1(tan 1)

sec d

1(sec )

sec d

1sec

d

cos d

12(cos2 1) d

12

12 sin2

12

12 sin2 4 4

12 sin2 0 0

12

12 sin 2 4 0

12

12 4 0

14 8

2 20

1

2 22

0

4

2 22

0

4

20

4

2

0

4

0

4

0

4

xx

8 x = tan θ x θ θ⇒ =d sec d2

θ θ

θ θ

= ⇒ = ⇒ = π

= ⇒ = ⇒ = π3 tan 3 3

1 tan 1 4

x

x

∫

∫

∫

θθ

θ θ

θθ

θ θ

−+

= −+

= − −

θ

θ

= π

= π

π

π

11

d

1 tan1 tan

sec d

1 (sec 1)sec

sec d

2

21

3

2

22

4

3

2

22

4

3

xx

x

(2 sec )d

2 tan

2 3 tan 3 2 4 tan 423 3 2 1

6 1 3

2

4

3

4

3

∫ θ θ

θ θ

( ) ( )( ) ( )

= −

= −

= × π − π − × π − π

= π − − π −

= π + −

π

π

π

π

9 x xx x

xx

xx

x

x x c

5 12 1( 3)( 1)

d 13

41

d

ln 3 2ln( 1)

2

2 2

2

∫ ∫( )+ ++ +

=+

++

= + + + +

10 (i) + + ++

= + ++

x x xx

x xx

2 3 9 124

2 34

3 2

2 2

A = 2, B = 3, C = 1, D = 0

(ii) ∫ + + ++

x x xx

x2 3 9 124

d3 2

21

3

x xx

x

x x x

2 34

d

3 12 ln( 4)

3 3 3 12 ln(3 4)

1 3 1 12 ln(1 4)

18 12 ln (13) 4 1

2 ln (5)

14 12 ln 13 ln5

14 12 ln 13

5

14 ln 135

21

3

2 2

1

3

2 2

2 2

∫ ( )

( )( )

( ) ( )( )

= + ++

= + + +

=+ × + +

− + × + +

= + − +

= + −

= +

= +

11 θ θ θ= ⇒ =u usin d cos d

θ θ= π ⇒ = = ⇒ =2 1, 0 0u u

cossin 5sin 6

d

cos5 6

dcos

15 6

d

1( 3)( 2) d

13

12 d

ln 3 ln 2

20

2

20

1

20

1

0

1

0

1

0

1

∫∫∫∫∫

θθ θ

θ

θθ

( )

− +

=− +

=− +

= − −

= − − −

= − − −

π

=

=

u uu

u uu

u u u

u u u

u u

u

u

9781510458444_Answer.indb 22 11/8/18 10:35 AM

23

2Furthercalculus


= −

−

= −− − −

−

= −

=

=

uuln 3

2

ln 1 31 2 ln 0 3

0 2

ln2 ln 32

ln2.32

ln 43

0

1

12 (i) u = x ux

x x x u⇒ = ⇒ =d 12

d d 2 d

∫∫∫∫

+

=+

=+

= +

x xx

u ux u

u uu u

u u u

1(1 )

d

1(1 )

2 d

1(1 )

2 d

2(1 ) d

2

2

(ii) 1(1 )

d1

9

x xx∫ +

9 3, 1 1

1(1 )

d

2(1 ) d

2 21 d

2ln 2ln 1

ln 1

ln 31 3 ln 1

1 1

ln 916 ln 1

4

ln 94

1

9

1

3

1

3

13

2

1

3

2 2

∫∫∫ ( )

( )( ) ( )

[ ]

= ⇒ = = ⇒ =

+

= +

= − +

= − +

= +

= + − +

= −

=

=

=

x u x u

x xx

u u u

u u u

u u

uu

u

u

13

( )

+ −+ −

= − ++ −

= − +− +

= − − + +

= − − − +

x xx x

xx x

xx x

x x

x x

4 4 172 5 3

2 6 112 5 3

2 6 11(2 1)( 3)

2 42 1

13

2 42 1

13

2

2 2

x xx x

x

x x x

x x x

x x x

x x x

∫∫ ( )

( )( )

[ ]

( )

( )

( )

+ −+ −

= − − − +

= − − − +

= − − + +

= − − + = − × − − ×= − − +

= +

= −

4 4 172 5 3

d

2 42 1

13 d

2 2ln(2 1) ln 3

2 ln(2 1) ln( 3)

2 ln(2 1) ( 3)

4 ln(9 5) 2 ln(1 4)4 ln45 2 ln4

2 ln 445

2 ln 454

2

21

2

1

2

12

21

2

21

2

14 (i) dd (sec ) d

d1

cosx x x x( )=

0 cos ( sin ) 1cos

sincos

1cos

sincos

sec tan

2

2

x xx

xx

xxx

x x

= × − − ×

=

= ×

=

(ii) d 3 d3 2u x u x x= ⇒ =

sec tan d2 3 3x x x x∫

∫∫

=

=

= +

= +

x u u ux

u u u

u c

x c

sec tan d3

13 sec tan d

13sec

13sec

22

3

(iii) ∫ +π

(sec tan ) d2

0

3 x x x

x x x x x

x x x x x

x x x x

x x x

(sec 2sec tan tan ) d

(sec 2sec tan sec 1) d

(2sec 2sec tan 1) d

2tan 2sec

2tan 3 2sec 3 3 (2tan0 2sec0 0)

2 3 2 2 3 (2)

2 3 2 3

2( 3 1) 3

2 2

0

3

2 2

0

3

2

0

3

03

∫

∫

∫

( )( )

= + +

= + + −

= + −

= + −

= π + π − π − + −

= + × − π −

= + − π

= + − π

π

π

π

π

9781510458444_Answer.indb 23 11/8/18 10:35 AM

24

2Furthercalculus


15 (i) dd 8sin cos cos 3cos sin 4sin3 2 2y

x x x x x x x= × + ×− ×

= −

= −

=

= ⇒ =

= ⇒ = π

8sin cos 12cos sin

4sin cos (2cos 3sin )dd 0 when

sin 0 0 (not M)

cos 0 2 (not M)

4 2 3

2 2 2

2

x x x x

x x x xyx

x x

x x

− =

− =

− =

=

= ±

=

x x

xx

xx xx

x

x

x

2cos 3sin 0

2coscos

3sincos

0cos

2 3tan 0

tan 23

tan 23

0.685 (3 s.f.)

2 2

2

2

2

2 2

2

2

(ii) Area = ∫π

4sin cos d2 3

0

2 x x x

= ⇒ =u x u x xsin d cos d

= π ⇒ = = ⇒ =2 1, 0 0x u x u

∫∫∫∫∫∫

( )

=

=

= −

= −

= −

= −

= −

−

=

=

π

4sin cos d

4 cos dcos

4 cos d

4 (1 sin )d

4 (1 )d

4 ( )d

4 3 5

4 13

15 0

4 215

815

2 3

0

2

2 3

0

1

2 2

0

1

2 2

0

1

2 2

0

1

2 4

0

1

3 5

0

1

3 5

x x x

u x ux

u x u

u x u

u u u

u u u

u u

16 (i) yx x x x x= × + × − ×

−dd 1 ln 1

2(ln ) 112

ln 12 ln

xx

= −

When x = 2,

dd

ln2 12 ln2

2ln2 12 ln2

0.232 (3 s.f.)= − = − =yx

2 ln2 2ln2 12 ln2

2

2 ln2 4ln2 22 ln2

4ln2 (4ln2 2)2 ln2

22 ln2

1ln2

2ln2 12 ln2

1ln2

= +

⇒ = − × +

= − −

= − −

=

=

= − +

y mx c

c

c

y x

(ii) ln 0x x =

⇒ = = ⇒ =x x x0 or ln 0 1

Volume ∫∫∫

= π

= π

= π

d

( ln ) d

ln d

2

1

e

2

1

e

2

1

e

y x

x x x

x x x

= ′ =

′ = ′ =

u x v x

u x v x

ln ,

1 , 3

2

3

x x xx x

x x x x

x x x

∫

∫

( ) ( )

( )( )

( ( ))

= π

− π ×

= π

− π

= π −

= π − − −

= π − − −

= π +

= π + ≈

Volume3 ln 1

3 d

3 ln 3 d

3 ln 9

e3 lne e

913 ln1 1

9

e3

e9

19

2e9

19

2e 19 14.4

3

1

e 3

1

e

3

1

e 2

1

e

3 3

1

e

3 3 3 3

3 3

3

3

17 (i) ∫ −

= − − × − +

= − +

x x

x c

x c

sin(1 4 )d

cos(1 4 ) 14

14 cos(1 4 )

(ii) ee

ee

e

∫∫

+

=+

= + +

x

x

c

x

x

x

x

x

1d

12

21

d

12 ln 1

2

2

2

2

2

9781510458444_Answer.indb 24 11/8/18 10:35 AM

25

2Furthercalculus


(iii) d

1dd 3

dd3

13 d

1313e

2 1

3 2

2 1 22

1

3

3

3

∫

∫ ∫∫

= − ⇒ = −

=−

= −

= − +

= − +

−

−

−

e

e e

e

e

x x

u xux x

x x xux

u

c

c

x

x u

u

u

x

(iv) ∫= ′ =

′ = =

x x x

u x v x

u v x

2 cos2 d

2 cos2

2 12 sin2

∫ ∫∫

( ) ( )= − ×

= −

= + +

x x x x x x x

x x x x

x x x c

2 cos2 d 2 12 sin2 2 1

2 sin2 d

sin2 sin2 d

sin2 12cos2

(v) ∫( )

+

= +−

xx

x c

125

d

15 tan 5

2

1

(vi) 24

d

4 d d2

4d

2(4 ) d

2 4 d

2 44 3

2 1 13

2 23

24

23 4

5

5

5

5 4

4 3

4 3

4 3

4 3

∫

∫

∫∫

( )( )

( )( )

( )

( )

( ) ( )

−= − ⇒ = −

−

= − −

= − −

= − − − − +

= − − + +

= − +

=−

−−

+

− −

− −

xx

x

u x u xxx

x

uu

u

u u u

u u c

u uc

u uc

x xc

(vii) ∫ −−−

− + ≡ − + +− ≡ + + −= ⇒ − = ⇒ = −= − ⇒ − = − ⇒ =

xx

x

xx x

Ax

Bx

x A x B xx A Ax B B

51

d

5( 1)( 1) 1 1

5 ( 1) ( 1)1 4 2 2

1 6 2 3

2

5( 1)( 1)

21

31

51

d 31

21

d

3ln 1 2ln 1

2∫ ∫( )−

− + ≡ −− + +

−−

=+

−−

= + − − +

xx x x x

xx

xx x

x

x x c

(viii) ∫

∫ ∫∫( )

= ′ =

′ = = −

= − − × −

= − +

= − − +

−

−

−

−

ln d

ln1

ln d 1 ln 1 d

1 ln d

1 ln 1

2

2

1

21

2

xx

x

u x v x

u x v x

xx

xx

xx

x x

x x x x

x x x c


1 (i) x xA

xB

x+ + ≡ + + +2

( 1)( 3) 1 3 2 ≡ A(x + 3) + B(x + 1)

x = −1 ⇒ 2 ≡ 2A ⇒ A = 1 x = −3 ⇒ 2 ≡ −2B ⇒ B = −1

+ + = + − +x x x xSo 2

( 1)( 3)1

11

3

(ii)

( )

( )( )

+ +

= + − +

=+

−+ +

++

=+

−+

−+

++

=+

−+

++

++

2( 1)( 3)

11

13

1( 1)

2( 1)( 3)

1( 3)

1( 1)

11

13

1( 3)

1( 1)

11

13

1( 3)

2

2

2 2

2 2

2 2

x x

x x

x x x x

x x x x

x x x x

(iii) 4( 1) ( 3)

d

( 1) 11

13 ( 3) d

11 ln( 1) ln( 3) 1

3

12 ln2 + ln4 1

4 1 ln1 + ln3 13

34 ln2 4

3+ ln3

712 ln 2

37

12 ln 32

2 20

1

2 2

0

1

0

1

∫

∫( )

( ) ( )( ) ( )

+ +

= + − + + + + +

= − + − + + + + − +

= − − − − − − −

= − + − −

= +

= −

− −

x xx

x x x x x

x x x x

9781510458444_Answer.indb 25 11/8/18 10:35 AM

26

2Furthercalculus


2 (i)

ee

( )

=

= + × = +

= + = ⇒ = −

=−

y x xyx x x x x x x

yx x x

x

lndd 3 ln 1 3ln 1

dd 0 when 3ln 1 0 ln 1

3

or 1

3

2 3 2

13

3

(ii) y = x3 lnx

y = 0 when ln x = 0 so x = 1

ln

41

3

4

′ = =

= ′ =

u x v x

u x v x

∫=A xSo 3

1

2 lnx dx

x x x

x x x

4 ln 41 d

4 ln 4 d

4 ln 16

24 ln 2 2

1614 ln1 1

16

(4ln2 1) 116

4 ln 2 1516

4

1

2 4

1

2

4

1

2 3

1

2

4 4

1

2

4 4 4 4

∫

∫

( ) ( )( )

=

− ×

=

−

= −

= − − −

= − − −

= −

xx

x

x x x

3 (i)

=

x

x

xx

xx

2sin 1 2sin 60 2sin 0

dd 2cos

d 2cos d

So4

d

4sin4 4sin

2cos d

4sin4(1 sin )

2cos d

4sin4(cos )

2cos d

4sin2cos 2cos d

4sin d

2

20

1

2

20

6

2

20

6

2

20

6

2

0

6

2

0

6

∫

∫

∫

∫

∫∫

θ θ θθ θ

θ θθ θ

θθ

θ θ

θθ

θ θ

θθ

θ θ

θθ θ θ

θ θ

= = ⇒ = π

= ⇒ =

=

=

−

−

=−

=

=

=

π

π

π

π

π

(ii) θ θ

θ θ

= −

= −

cos2 1 2sin

sin 12(1 cos2 )

2

2

ISo 4 12(1 cos2 ) d

2 (1 cos2 ) d

2 12 sin2

2 63

4 0

33

2

0

6

0

6

60

∫∫

θ θ

θ θ

θ θ

= −

= −

= −

= π −

−

= π −

π

π

π

4 (i)

vu

vu

x

x

x x

x

∫

∫∫

∫

[ ]

[ ]

[ ]

=

= ′ =′ = = −

= − − −

= − +

= ′ =′ = = −

= − + − − −

= − − −

= − + + = − + × + − − + × += − +

= −

−

−

−

− −

− −

−

−

− − −

− − −

−

−

−

A x x

u xx v

x x x

x x x

uv

x

x x

x x

e d

e2 e

e 2 e d

e 2 e d

2 e2 e

e 2 e 2e d

e 2 e 2e

e ( 2 2)

( e (3 2 3 2)) ( e (0 2 0 2))e (17) 2

2 17e

x

x

x

x

x

x

x

x x x

x x x

x

2

0

3

2

203

0

3

203

0

3

203

0

3

20

3

20

3

3 2 0 2

3

3

(ii) Maximum point when =dd 0y

x

⇒ + − =

− ==

− − −

−

x x

x xx

x x

x

2 e ( e ) 0

e (2 ) 02

2

(iii) Equation of tangent is y = mx = c and c = 0

= − ×

= −

= −

= −=

− −

− − −

− −

−

x x x x

x x x

x x

x xx

x x

x x x

x x

x

e e (2 )

e 2 e e

0 e e

0 e (1 )So 1

2

2 2 3

2 3

2

9781510458444_Answer.indb 26 11/8/18 10:35 AM

27

2Furthercalculus


5 (i) tan

tan 4 1 tan 0 0

d sec d

d dsec

d cos d

(tan tan )d

tan (tan 1)d

(sec )cos d

d

1

11

01

11

2

22

2

0

4

2

0

4

2

0

1 2

0

1

1

0

1

1 1

u x

u u

u x x

x ux

x x u

x x x

x x x

u x x x

u u

un

n n

n

n n

n

n

n

n

n n

∫∫∫∫

=

= π = = =

=

= ⇒ =

+

= +

=

=

= +

= + − += +

+π

π

+

+ +

(ii)(a) ∫ −π

x(sec sec ) d4 2

0

4 x x

∫∫∫

= −

= +

= +

= =

π

π

π

x

x

sec (sec 1) d

(1 tan ) tan d

(tan tan ) d

13 ( 2)

2 2

0

4

2

0

4 2

4 2

0

4

x x x

x x

x

n

(b)

I x x x

x x x

x x x

∫∫

∫

= +

+ +

+ +

= + × + =

π

π

π

(tan tan ) d

4 (tan tan )d

(tan tan )d

18 4 1

614

2524

9 7

0

4

7 5

0

4

5 3

0

4

tan 5tan 5tan tantan tan 4(tan tan )

tan tan

9 7 5 3

9 7 7 5

5 3

+ + += + + +

+ +

x x x xx x x x

x x

6 (i) y xx

=+1

2

3

dd

2 (1 ) 3 ( )(1 )

2(1 )

2(1 )

0 2 0

(2 ) 00 or 2

Since clearly the -coordinate is > 0,2

3 2 2

3 2

4

3 2

4

3 24

3

3

3

= + −+

= −+

−+

= ⇒ − =

− =⇒ =

=

yx

x x x xx

x xx

x xx

x x

x xx

xx

(ii) xx

xp∫ +

=1

d 12

31

13

31

d 1

13 ln 1 1

13 ln 1 1

3 ln 1 1 1

13 ln 1 1

3 ln 2 1

13 ln 1

21

ln 12

3

12 e

1 2e

2e 1 3.40 (3 s.f.)

2

31

3

1

3 3

3

3

3

33

3 3

33

∫ +=

+

=

+ − + =

+ − =

+ =

+ =

+ =

+ =

= − =

xx

x

x

p

p

p

p

p

p

p

p

p

7 ∫π

sin2 d2

0

12 x x x

sin 2

2 12 cos2

sin d

( 12 cos2 ) 2 ( 1

2 cos2 ) d

12 cos2 cos2 d

cos2

1 12 sin 2

So sin d

12 cos2 1

2 sin 2 12 sin 2 d

12 cos2 1

2 sin 2 14 cos2

So sin d

12 cos2 1

2 sin 2 14 cos2

12

12 ( 1) 1

4 ( 1) 0 14 (1)

814

14

812

2

2

2

2

2

2

2

2

0

12

2

0

12

2

2

2

∫∫

∫

∫∫

∫

( )( )( ) ( )

( )

= ′ =

′ = = −

= − − −

= − +

= ′ =

′ = =

= − + −

= − + +

= − + +

= − π − + − − +

= π − −

= π −

π

π

u x v x

u x v x

x x x

x x x x x

x x x x x

u x v x

u v x

x x x

x x x x x x

x x x x x

x x x

x x x x x

9781510458444_Answer.indb 27 11/8/18 10:35 AM

28

2Furthercalculus


8

I xx x

x∫= −+

12( )

d1

4

(i)

∫

∫∫

= ⇒ = ⇒ = =

= ⇒ = = ⇒ =

= −+

= −+

= −+

dd

12

d 2 d 2 d

4 2, 1 1

12( )

2 d

12 ( 1) 2 d

11 d

21

2

1

2

1

2

u x ux x

x x u u u

x u x u

I uu u

u u

uu u u u

uu u

(ii) I uu u∫= −

+11d

1

2

Using division,

1 21 d

2ln( 1)

(2 2ln3) (1 2ln2)1 2ln 3 2ln 21 ln 9 ln 41 ln 4 ln 9

1 ln 49

1

2

1

2

∫ ( )= − +

= − + = − − −= − += − += + −

= +

I u u

u u

9 y x= (ln )2

(i) dd 2ln 1

When e, dd

2e

When e, 1

Gradient of normal at P = e2

1 e2 e

1 e2

So Q is when 0 e2 1 e

20 e 2 e

2 ee 3.45 (3 s.f.)

2

2

2

2

= ×

= =

= =

−

= + ⇒ = − × +

= +

= − + +

= − + +

= + ≈

yx x x

x yx

x y

y mx c c

c

x

x

x

(ii)

ln d 1 ln d

ln 11

ln d ln 1 d

ln 1d

ln

∫ ∫

∫ ∫∫

= ×

= ′ =

′ = =

= − ×

= −

= − +

x x x x

u x v

u x v x

x x x x x x x

x x x

x x x c

(iii) Area = area of triangle + area under curve

12 e 2

e e 1 (ln ) d

1e (ln ) d

Let e dd e

(ln ) d e d

Integrate by parts (twice)

e

2 e

e d e 2 e d

2 e

2 e

e d e 2 e 2e d

e 2 e 2e

e 2 e 2e

e ( 2 2)

e 2

2

1

e

2

1

e

2

1

e 2

0

1

2

2

0

1 20

1

0

1

2

0

1 20

1

0

1

20

1

0

1

20

1

20

1

x x

x x

x xu

x x u u

u u v

u u v

u u u u u

u u v

u v

u u u u u

u u

u u

u u

u u

u

u

u

u u u

u

u

u u u u

u u u

u u u

u

∫∫

∫ ∫

∫ ∫

∫ ∫

( )= + − × +

= +

= ⇒ =

=

= ′ =

′ = =

= −

= ′ =

′ = =

= − −

= − −

= − +

= − + = −

Area = + −1e e 2


1 (i) u = 1 + x u x⇒ =d d = ⇒ = = ⇒ =1 2, 0 1x u x u

1 d

( 1) d

3 3 1 d

3 3 1 d

33

2 3 ln

23

3 22 3 2 ln 2

13

3 12 3 1 ln 1

83 6 6 ln2 1

332 3 0

83 ln2 11

656 ln2

3

0

1

3

1

2

3 2

1

2

2

1

2

3 2

1

2

3 2

3 2

xx x

uu u

u u uu u

u u u u

u u u u

u

u

∫∫∫∫

( )( )

( )

( ) ( )( ) ( )

+

= −

= − + −

= − + −

= − + −

=− × + × −

− − × + × −

= − + − − − + −

= − −

= −

=

=

9781510458444_Answer.indb 28 11/8/18 10:35 AM

29

2Furthercalculus


(ii)(a) yx x x x

x= + + +dd 2 ln(1 ) 1

2

(b) Area = ln(1 )d2

0

1

∫ +x x x

u x v x

u x v x

= + ′ =

′ = + =

ln(1 )

11 3

2

3

ln(1 )d

3 ln(1 ) 11 3 d

3 ln(1 ) 13 1 d

3 ln(1 ) 13

56 ln 2

13 ln 2 0 1

356 ln 2

23 ln 2 5

18 0.184

2

0

1

3

0

1 3

0

1

3

0

1 3

0

1

3

0

1

∫

∫

∫

( )( ) ( )

+

= +

− + ×

= +

− +

= +

− −

= − − −

= − ≈

x x x

x x xx x

x x xx x

x x

2 (i) = ′ = cosu x v xn

sin1u nx v xn′ = =−

cos d

sin sin d

sin

( 1) cos

sin( cos )

( 1) ( cos )d

sin cos

( 1) cos d

2 sin 2 2 cos 2

0 sin0 0 cos0( 1)

2 ( 1)

0

12

0

12 1

0

12

1

2

0

12

1

2

0

12

102

2

0

12

1

12

2

∫

∫

∫

∫( ) ( )

( )

=

= −

= ′ =

′ = − = −

= −−

− − −

= +

− −

=π π + π π

− +

− −

= π − −

π −

−

−

π−

−

−π

−

−

−−

−

π

π

π

π

I x x x

x x nx x x

u nx v x

u n n x v x

x xnx x

n n x x x

x x nx x

n n x x x

n

nn n I

n n I

nn

n n

n

n

n

n

n

n n

n

n n

n nn

n

n

OR

cossin

cos d sin sin d

sin( cos )

( 1) cos d

sin cos ( 1) cos d

12 0 ( 1) cos d

12 ( 1)

1

0

12

0

12 1

0

12

0

12

1

2

0

12

10

12 2

0

12

2

0

12

2

u x v xu nx v x

x x x x x nx x x

x x nx x

n x x x

x x nx x n n nx x x

n n x x x

n n I

n

n

n n n

nn

n

n n n

nn

n

n

∫ ∫

∫∫

∫( )( )

= ′ =′ = =

= −

= −× −

− − × −

= + − −

= π + − −

= π − −

−

π −

π−

−

π

− π −

−π

−

π π

π

cossin

cos d sin sin d

sin( cos )

( 1) cos d

sin cos ( 1) cos d

12 0 ( 1) cos d

12 ( 1)

1

0

12

0

12 1

0

12

0

12

1

2

0

12

10

12 2

0

12

2

0

12

2

u x v xu nx v x

x x x x x nx x x

x x nx x

n x x x

x x nx x n n nx x x

n n x x x

n n I

n

n

n n n

nn

n

n n n

nn

n

n

∫ ∫

∫∫

∫( )( )

= ′ =′ = =

= −

= −× −

− − × −

= + − −

= π + − −

= π − −

−

π −

π−

−

π

− π −

−π

−

π π

π

(ii) ( )= π − × ×12 4 34

4

2I I

( )= π − × ×12 2 12

2

0I I

∫= = = π − =π π

cos d sin sin12 sin0 10 0

12

0

12I x x x

( )= π − × × = π −12 2 1 1 1

4 22

22I

( ) ( )= π − × × π −

= π − π +

12 4 3 1

4 2

116 3 24

4

42

4 2

I

OR

( )= π − − −12 4(4 1)4

4

4 2I I

( )

( )( )

( )( )

( )( )( )

= π −

= π − π − −

= π − π −

= π − π − π −

= π − π −

= π − π +

116 12

116 12 1

2 2(2 1)

116 12 1

4 2

116 12 1

4 2 12 0

116 12 1

4 2 1

116 3 24

42

42

0

4 20

4 20

4 2

4 2

I

I

I

3 (i)

x u a x a u

=

u = u

I =

= =

a

a a a

0

0

aaa

a a

a a

0

∫

∫∫∫

∫∫

−

− = −

= = = =

−

= = −

− = − − = − +

I a x x

a x, x

u u

u u x x

x

x x x

x x x

Consider f ( ) d .

Let then dd 1,

and when 0, ; when , 0, so

f ( )d

f ( )d f ( )d

ORLetF( ) be an antiderivative of f

f ( )d F( ) F( ) F( )

f ( )d F( ) F( ) F( )

and hence result.

a

0

00

0 0

0 0

9781510458444_Answer.indb 29 11/8/18 10:35 AM

30

2Furthercalculus


(ii) Using the result in 3(i) with = π2 ,a

xx x

x

x

x xx

xx x

x

xx x

x

x x

xx x

x

xx x

x xx x

x

x xx x

x

sinsin cos

d

sin

sin cosd

coscos sin

d

cossin cos

d

since sin cos and so

sinsin cos

d

12

sinsin cos

d cossin cos

d

12

sin cossin cos

d 12 2 4

0

2

2 20

0

0

2

0

0 0

0

2

2

2

2

2

2 2

2

∫

∫

∫

∫

∫

∫ ∫

∫

( )( ) ( )

( )

( )( ) ( )

+

=−

− + −

=+

=+− =

+

=+

++

= ++

= π = π

π

π π

π

π

π

π

π

π

π π

π

n

n n

n

n n

n

n n

n

n n

n

n n

n

n n

n

n n

n n

n n

4 (i) yx

x x xx

xx

yx

xx

x x xx

x x x x

x x x

x x x x

x x

x x x x

yx

xx

dd

2 (1 ) 2(1 )

2(1 )

Whendd

12

2(1 )

12

2 4 1 0But 1 is a solution, so by division or otherwise

( 1)( 3 1) 0

and any other solutions are from 3 1 0

Let g( ) 3 1, then

g 14

1 4 48 6464

1164 0

g 12

1 2 12 88

78 0

Hence there is a root in the interval 14 , 1

2 .

g( ) 0 for some 14

12 .

OR

f ( ) 2 4 1

f 14

33256 0.1289 0

f 12

716 0.4375 0

Hence there is a root in the interval 14 , 1

2 .

ORdd

2(1 )

2 3

2 2 2 2

2 2

4 2

3 2

3 2

3 2

4 2

2 2

=

=

( )( )

( )

( )( )

( )

= + −+

=+

=

+=

+ − + ==

− + + − =

+ + − =

+ + −

= + + − = − <

= + + − = >

= < <

+ − +

= = >

= − = − <

=+

x

yx

xyx

yx

yx x

When 14 ,

dd

12

1 116

128289

12

and when 12 ,

dd

1

1 14

1625

12

Hence there is a value ofdd

12 in the interval 1

4 , 12 .

dd

12 for some 1

412

2

2

( )

( )

( )

=+

= <

= =+

= >

=

= <

=

<

(ii)

x y xx

V x y

xx

x

V

V

Vy

y y u yu yy u

V uu u

u u

u u

First translate the function1unit in the negative

direction to get1

.

The volume of revolution is then

d

and1

, so rearrange to get 1

and 11 1 d

ln|1 |

ln 12

12

ln(2) 12 units

OR to integrate using substitution,

1 d Letd d

1 ( d )

1 1 d

ln|

(0 1) ln 12

12

ln2 12 units

2

2

02

22

0

0

3

0

11

1

3

12

12

12

12

12

12

12

=

=

y= yy

y y

y y

|

1

1

2∫

∫

∫

∫∫

( )

( )

( )

( )( )

( )

( )( )

+

π

+= −

= π − −

= π − − −

= π − −

= π −

= π − = −= −= −

= π − −

= π −

= π −

= π − − −

= π −

9781510458444_Answer.indb 30 11/8/18 10:35 AM

31

3Differentialequations


3 Differential equations3.1 Forming differential equations from rates of change (Page 32)

1 (i) dd

vt k=

(ii) ddBt kB=

(iii) dd

3ht k t=

(iv) Vt k V=d

d

(v) dd

rt

kr

=

(vi) = −dd ( )P

t k N P

(vii) ddVt kS=

(viii) dd

2At kr=

3.2 Solving differential equations (Page 33)

1 (i) dd 1 2y

x x= −

∫ ∫= −

= − +

d (1 2 ) d

2

y x x

y x x c

(ii) =dd

4yx

xy

∫ ∫=

= +

= +

= ± +

d 4 d

2 2

4

4

22

1

2 22

22

y y x x

yx c

y x c

y x c

(iii) dd 0.01A

t A=

1 d 0.01d

ln 0.01

e

e e

e

0.01

0.01

0.01

∫ ∫=

= +

=

= ×

=

+

A A t

A t c

A

A

A K

t c

t c

t

(iv) dd sinyx y y x= −

dd (1 sin )

1 d (1 sin )d

ln cosee e

e

cos

cos

cos

∫ ∫= −

= −

= + +== ×=

+ +

+

+

yx y x

y y x x

y x x cyyy A

x x c

x x c

x x

dd (1 sin )

1 d (1 sin )d

ln cosee e

e

cos

cos

cos

∫ ∫= −

= −

= + +== ×=

+ +

+

+

yx y x

y y x x

y x x cyyy A

x x c

x x c

x x

2 (i) dd e

dd e e

=

= ×

+yxyx

x y

x y

e d e d

e e0, 2

e e e 1

e e e 1

e e e 1

ln( e e 1)

ln( e e 1)

2 0 2

2

2

2

2

∫ ∫=

− = += = ⇒

− = + ⇒ = − −

− = − −

= − + +

− = − + +

= − − + +

−

−

− −

− −

− −

−

−

y x

cx y

c c

y

y

y x

y x

y x

y x

x

x

(ii) + =(1 )dd 22x

yx xy

1 d 2

(1 )d

ln ln(1 )0, 2

ln2 ln(1 0 ) ln2

ln ln(1 ) ln2

ln ln(2(1 ))

e e

2(1 )

2

2

2

2

2

ln ln(2(1 ))

2

2

∫ ∫=+

= + += = ⇒

= + + ⇒ =

= + +

= +

=

= +

+

y y xx

x

y x cx y

c c

y x

y x

y x

y x

3 (i) =sec

cos 2dd

22

2yx

yx

sec d 2cos 2 d

sec d (cos4 1)d

tan 14 sin4

2 2

2

∫∫∫∫

=

= +

= + +

y y x x

y y x x

y x x c

(ii) π = + + ⇒ =tan 14

14 sin0 0 1c c

tan 1

4 sin 4 1

tan 14 sin 4 1

6

So tan 14 sin 4

6 6 1

1.05 (3.s.f )

1

1

( )

( )

= + +

= + +

= π

= π + π +

=

−

−

y x x

y x x

x

y

9781510458444_Answer.indb 31 11/8/18 10:36 AM

32



4 (i) dd

620

ht

h= −

h h t

h t c

h t c

t hc c

h t

h t

h t

t h

206 d d

20ln 6

ln 6 120

0, 1ln 6 1 ln5

ln 6 120 ln5

ln 6 ln5 120

ln 65

120

20ln 65

1

2

2 2

∫ ∫− =

− − = +

− = − +

= = ⇒− = ⇒ =

− = − +

− − = −

− = −

= − −

h h t

h t c

h t c

t hc c

h t

h t

h t

t h

206 d d

20ln 6

ln 6 120

0, 1ln 6 1 ln5

ln 6 120 ln5

ln 6 ln5 120

ln 65

120

20ln 65

1

2

2 2

∫ ∫− =

− − = +

− = − +

= = ⇒− = ⇒ =

− = − +

− − = −

− = −

= − −

(ii) t 20ln 6 25 4.46 years= − − =

(iii) h10 20ln 65= − −

h

h

h

h

12 ln 6

5

e 65

5e 6

6 5e 2.97 m

12

12

12

− = −

= −

= −

= − =

−

−

−

(iv) t h20ln 65= − −

= −

= −→ ∞ →

−

−

e 65

6 5eAs , 6

20

20

h

ht h

t

t

Maximum possible height is 6 m.

5 t kθ θ= −d

d (160 )

k t

kt ct c

ktt k

k

t

t

t

t

t

t

∫ ∫θ θ

θθ

θθ

θ

θθθθ

θ

− =

− = += = ⇒ =

− = += = ⇒ = +

= −

= −− = − +

− =

− = ×

− =

= −

= ⇒ = − = °

− +

−

−

−

− ×

1160 d d

ln 1600, 20 ln140

ln 160 ln1405, 65 ln95 5 ln140

ln95 ln1405

0.0776ln 160 0.0776 ln140

160 e

160 e e

160 140e

160 140e

10 160 140e 95.5 C

0.0776 ln140

0.0776 ln140

0.0776

0.0776

0.0776 10

6 (i) ( )=dd

2yx

yx

1 d 1 d

1 1

1 1

1

1

2 2∫ ∫=

− = − +

− = − +

− = − +

= −

yy

xx

y x c

ycx

x

y xcx

y xcx

1 d 1 d

1 1

1 1

1

1

2 2∫ ∫=

− = − +

− = − +

− = − +

= −

yy

xx

y x c

ycx

x

y xcx

y xcx

(ii) 1 21 2 1 2 2 1

2= − ⇒ − = ⇒ = −c c c

1 12

22

2 82 8

1610

85

=+

=+

= ×+ = − = −

y xx

xx

y

7 (i) ddxt k x=

(ii) =dd

10 000yt y

d 10 000d

32

10 000

23 10 000

0, 900 2 9003 18 000

23 10 000 18 000

15 000 27 000

(15 000 27 000)

After 10 minutes,

(15 000 10 27 000) 3152.45 3150

32

3

3

3

3

23

23

∫ ∫=

= +

= +

= = ⇒ = =

= +

= +

= +

= × + = ≈

y y t

yt c

yt c

t y c

yt

y t

y t

y

8 dd 10e

12v

tt

=−

(i) ∫− +

⇒ − ⇒

−

−

−

−

v t

ct v c c

v

10e d

eWhen

So e

12

12

12

t

t

t

=

==

=

20= 0, 0 0 = 20 + = 20

20 20

(ii) → ∞ → ⇒ →−

As , e 0 2012t v

t

So long term speed is 20 m s−1

(iii) w w w w− + = − − +1

( 4)( 5)1

9( 4)1

9( 5)

9781510458444_Answer.indb 32 11/8/18 10:36 AM

33



(iv) +

=

=

+

+

+

4)( 5)

4)( 5)

4) + 5)

4) + 5) =

4+5 =

= 0, =10 =

∫ ∫∫ ∫)(

= − −

⇒ − + −

⇒ − − −

⇒ − − −

⇒ − −

⇒ =

⇒ −+ = −

⇒ −+ = = =− + − −

wt w w

ww w t

w w w t

w w t c

ww t c

t w c

ww t

ww

dd

12(

d(

12d

19(

19( d 1

2d

19 ln( 1

9 ln( 12

19 ln 1

2

When 19 ln 6

1519 ln 2

5

ln 45

92 ln 2

545 e 2

5 e 0.4et t tln 4.592

25

92

(v) → ∞ → ⇒ − →−

−

t wAs , e 0

So long term speed is m s .

t4.5

1

4 0

4

9 θ : temperature,: time in hours from 4 p.m.t

t k

k t

kt ct c c

t k k

t

t

t

t

t

t

t

t

∫∫

θ θ

θ θ

θθ

θ

θ

θ

θ

( )

( )( )( )

( )( )

( )

( )

= −

− =

− = += = ⇒ − = ⇒ =

= = ⇒ − = + ⇒ =

− = +

− = ×

= +

= +

=

=

=

= = −

dd ( 6)

1( 6)d d

ln 60, 16 ln 16 6 ln10

1, 8 ln 8 6 ln10 ln15

ln 6 ln15 ln10

6 e e

10 15 6

37 10 15 6

31 10 15

3.1 15

ln3.1 ln 15

ln3.1ln 1

5

0.703

ln15 ln10

i.e. 42 minutes ago = 3.18 p.m.

10 dd

1

12

∫ ∫∫∫

=

=

=−

Vt k V

VdV kdt

V dV kdt

12

20, 900 2 900 601, 841 2 841 60 2

2 2 602 0 2 60

2 6030

12

= +

= += = ⇒ = + ⇒ == = ⇒ = + ⇒ = −

= − += − +==

V kt c

V kt ct V kt c ct V k k

V tt

tt

The tank would empty in 30 days.

11 y: amount of substance A in grams

dd

2=yt ky

1 d d

d d

11

1

0, 60 60 1 160

1, 10 10 11

60

112

11

121

60

605 1

When 2, 605 2 1 5.45 g

2

2

1

∫ ∫∫ ∫

=

=

− = +

− = +

= − +

= = ⇒ = − ⇒ = −

= = ⇒ = −−

⇒ = −

= −− −

= +

= = × + =

−

−

yy k t

y y k t

y kt c

y kt c

y kt c

t y c c

t yk

k

yt t

t y


1 (i) =dd e2x

t x t

∫ ∫=

= +

=

=

+

1 d e d

ln 12e

e e

e

2

2

ln12e

12e

2

2

x x t

x c

x A

t

t

x ct

t

(ii) dd

cose

2yt

yt=

∫ ∫∫ ∫

=

=

= − +

= − +

−

−

− −

1cos

d 1e

d

sec d e d

tan e

tan ( e )

2

2

1

yy t

y y t

y c

y c

t

t

t

t

9781510458444_Answer.indb 33 11/8/18 10:36 AM

34



2 (i) (x2 − 5x + 6) ÷ (x − 1) = ( 4) 21x x− + −

(ii)(a) ( )−− +

= −15 6

dd

52x

x xyx

y

15 d 5 6

1 d

15 d 4 2

1 d

ln 5 2 4 2ln 1

5 e

5 e

5 e e e

( 1) e 5

2

2

2 4 2ln 1

2 4 ln( 1)

2 4 ln( 1)

2 2 4

2

2 2

22

2

y y x xx x

y y x x x

y x x x c

y

y

y

y A x

x x x c

x x x c

x x x c

x x

∫ ∫∫ ∫( )

− = − +−

− = − + −

− = − + − +

− =

− =

− = × ×

= − +

− + − +

− + − +

− −

−

(b) = − +− ×

7 (8 1) e 5282 4 82

A

= +

=

= − +

= − +

=

−

− ×

7 49 e 52

49

249( 1) e 5

249(6 1) e 5

5.00

0

2 2 4

262 4 6

2

2

A

A

y x

y

x x

3 (i) dd 4 8

13x

t x( )= − −

xx t

x x t

x t c

x t c

t x c c

x t

t x

t x

x

t

1

( 8)d 4 d

( 8) d 4 d

( 8)23

4

3( 8)2 4

0, 72 3(72 8)2 24

3( 8)2 4 24

4 24 3( 8)2

6 3( 8)8

When 35,

6 3(35 8)8

218 2.625 mins

13

13

23

23

23

23

23

23

23

∫ ∫

∫ ∫−

= −

− = −

− = − +

− = − +

= = ⇒ − = ⇒ =

− = − +

= − −

= − −

=

= − − = =

−

(ii) 6 3( 8)8

23

t x= − −

When x = 0,

6 3(0 8)8 4.5 mins

23

t = − − =

4 (i) 1(2 1)( 1)

22 1

11x x x x+ + = + − +

(ii) dd (2 1)( 1)

yx

yx x= + +

∫ ∫∫ ∫ ( )

( )

= + +

= + − += + − + +

=++

+

= = ⇒ = + ⇒ =

= ++ +

=

= ×

= ++ = +

+

++ +

++

y y x x x

y y x x x

y x x c

yx

xc

x y c c

y xx

y

y xx

xx

yx

x

xx

1 d 1(2 1)( 1)d

1 d 22 1

11 d

ln ln 2 1 ln 1

ln ln2 1

1

0, 2 ln2 ln1 ln2

ln ln 2 11 ln2

e e

e e

2 2 11

4 21

ln ln 2 11 ln2

ln 2 11 ln2

5 (i) − + = − + +

= + + −− +

= + + −= ⇒ = ⇒

− ⇒ − ⇒ −

− + = − − +

y yAy

By

A y B yy y

A y B yy A Ay B B

= 1= 1 3 = 3 = 1

3( 2)( 1) 2 1

( 1) ( 2)( 2)( 1)

3 ( 1) ( 2)2 3 3

3( 2)( 1)

12

11y y y y

(ii) dd ( 2)( 1)2y

x x y y= − +

x c

∫∫∫∫( )

( )

⇒ − + =

⇒ − − + =

⇒ − − + = +

⇒−− = +

⇒−+ =

= × =

+

y y y x x

y y y x x

y y x cyyyy

A

3( 2)( 1) d 3 d

12

11 d 3 d

ln( 2) ln( 1)

ln2121 e

e e e

x c

x c x

2

2

3

3

3

3 3

6 (i) y = cosec x = 1sinx

yx

xx x

xx

x x

= − = − ×

= −

dd

cossin

1sin

cossin

cosec cot

2

9781510458444_Answer.indb 34 11/8/18 10:36 AM

35



(ii) = −dd sin tan cotx

t x x t

x x x t t

x x xtt t

x t c

x t c

xt c

x tc

c c

xt

xt

∫ ∫∫ ∫

( )

− =

− =

= +

= +

=+

= π = π ⇒ π =π +

= + ⇒ =

=+

=+

−

1sin tan d cot d

cosec cot dcossin d

cosec ln sin1

sin ln sin

sin 1ln sin

16 , 1

2 sin16

1ln sin 1

212

1ln1 2

sin 1ln sin 2

sin 1ln sin 2

1

7 (i) 1(3 )(6 )

13(3 )

13(6 )x x x x− − = − − −

(ii)(a) dd (3 )(6 )x

t k x x= − −

x x x k t

x x x k t

x x kt c

x x kt c

xx kt c

∫ ∫∫ ∫

( )

− − =

− − −

=

− − + − = +

− − − = +

−− = +

1(3 )(6 ) d d

13(3 )

13(6 ) d d

13ln(3 ) 1

3ln(6 )

13 ln(6 ) ln(3 )

13ln(6 )

(3 )

(b) ( )−− = +1

3ln(6 )(3 )

13ln 5

413ln2x

x t

13ln(6 )

(3 )13ln 5

4 2 13ln2

ln(6 )(3 ) 2ln 5

4 ln2

63 e

63 e e

2ln54 ln2

2ln54 ln2

( )−− = × +

−− = +

−− =

−− = ×

( )( )

+

xxxx

xxxx

63 2e

63 2 25

1663

258

8(6 ) 25(3 )48 8 75 25

17 272717 1.59

ln 54

2

−− =

−− = ×

−− =

− = −− = −

=

= ≈

( )xxxxxxx xx xx

x

8 yx x

yx x

y xx x

x c

y

y

y

y

∫ ∫

+ =

= −

= −

= +

edd tan 0

edd tan

e d sincos d

12e ln cos

2

2

2

2

x y c c

x

x

y x

y x

y

y

( )

( )

= = ⇒ = + ⇒ =

= +

= +

= +

= +

0, 0 12e ln cos0 1

212e ln cos 1

2e 2ln cos 1

2 ln ln(cos ) 112 ln ln(cos ) 1

0

2

2

2

2

9 (i) dd 3 2A

t k A= −

(ii) ∫ ∫−=1

3 2d d

AA k t

A A k t

A kt c

A kt c

t A k c c

t A k k

A t

A

A

AAA

A

∫ ∫− =

− × = +

− = +

= = ⇒ × − = × + ⇒ =

= = ⇒ × − = + ⇒ =

− = +

− = × +

− =

− =− =− =

=

−(3 2) d d

(3 2)12

13

2 3 23

0, 6 2 3 6 23 0 8

3

4, 17 2 3 17 23 4 8

312

2 3 23

12

83

2 3 23

12 8 8

32 3 2

3203

2 3 2 203 2 103 2 100

34 m

12

12

2

t x k c c

t x k

k

k

k

( )

= = ⇒ = × + ⇒ =

= = ⇒ = +

⇒ = −

⇒ = −

⇒ =

0, 0 13ln 6

3 0 13ln2

1, 1 13ln 5

213ln2

13ln 5

213ln2

13 ln 5

2 ln2

13ln 5

4

9781510458444_Answer.indb 35 11/8/18 10:36 AM

36



10 dd 12 3xy y

x y= −

yy

yx

x

y x c

x y c c

y x

y x

y x

y x

y x

∫ ∫

( )

( )

−=

− = +

= = ⇒ − = + ⇒ = −

− = −

− = −

− =

− =

= +

1d 1 d

13 ln 1 ln

2, 0 13 ln 0 1 ln 2 ln2

13 ln 1 ln ln2

ln 1 3 ln ln2

ln 1 3ln 2

ln 1 ln 2

8 1

2

3

3

3

3

3

3

33

33


1 (i) dd

yx kxy=

x = 1, y = 2 ⇒ 4 = k × 1 × 2 ⇒ k = 2

∫ ∫⇒ =

=

dd 2

1 d 2 d

yx xy

y y x x

ln y = x2 + C ln 2 =12 + C lC = ln 2 - 1 ln y = x2 + ln2 - 1

=

= × ×

+ −

−

y

y

x

x

e

e e e

( ln2 1)

ln2 1

2

2

(ii) = × − × = −dd 2 ( 1) 2 4y

x

x

y

O

2e

2 (i) = −dd 1.2e 0.02 0.5N

t Nt

∫ ∫

∫ ∫

=

=

= × − +

= − +

= = ⇒ = − +

−

− −

−

−

− ×

1 d 1.2e d

d 1.2 e d

12

1.2 10.02e

2 60e

0, 100 2 100 60e

0.50.02

12 0.02

12 0.02

0.02

0.20 0

NN t

N N t

N C

N C

t N C

t

t

t

t

∫ ∫

∫ ∫

=

=

= × − +

= − +

= = ⇒ = − +

−

− −

−

−

− ×

1 d 1.2e d

d 1.2 e d

12

1.2 10.02e

2 60e

0, 100 2 100 60e

0.50.02

12 0.02

12 0.02

0.02

0.20 0

NN t

N N t

N C

N C

t N C

t

t

t

t

20 = -60 + C ⇒ C = 80

⇒ = − +

= −

= −

−

−

−

2 60e 80

40 30e

(40 30e )

0.02

0.02

0.02 2

N

N

N

t

t

t

(ii) As → ∞ →−, e 00.02t t

= =So 40 16002N

3 (i) dd

ee

xt

xk

t

t=+

−

−

∫ ∫=+

= − + +

= − + += − + +

= + + = +

⇒ = − + + +

− = − + + +

−

−

−

−

−

1 d ee

d

ln ln( e )

ln10 ln( e )ln10 ln( 1)

ln10 ln( 1) ln10( 1)

ln ln( e ) ln10( 1)OR

ln ln10 ln( e ) ln( 1)

0

x xk

t

x k c

k ck c

c k k

x k k

x k k

t

t

t

t

t

(ii)

( )− = − + + +

= ++

++

=

+ = +

+ = +

= − = −

−

−

−

−

−

−

ln20 ln10 ln( e ) ln( 1)

ln2 ln 1e

1e

2

1 2( e )

1 2 2e

1 2e 1 2e

1

1

1

1

1

1

k k

kk

kk

k k

k k

k

(iii) As → ∞ →−, e 0t t

So

( ) ( )

− = − + + += − + + +

= − − + − + +

==

ln ln10 ln( 0) ln( 1)ln ln ln( 1) ln10

ln ln 1 2e ln 1 2

e 1 ln10

ln 3.8679...47.844...

x k kx k k

x

xx

So x will never reach 48.

4 (i) (3 cos2 )dd sin2x xθ θ θ+ =

x x

x c

c

c

c

x

x

∫ ∫ θθ θ

θ

θ

θ

θ

θ

( )( )

= +

= − + +

= − + × π +

= − +

=

= − + +

=

= ×

= ×

= + ×

=+

= +

θ

θ

θ

− + +

− +

+

−

−

1 d sin23 cos2 d

ln 12 ln(3 cos2 )

ln3 12 ln 3 cos2 4

ln3 12 ln 3

32 ln3

so ln 12 ln(3 cos2 ) 3

2 ln3

e

e e

e e

(3 cos2 ) 3

33 cos2

273 cos2

12ln(3 cos2 ) 3

2 ln3

12 ln(3 cos2 ) 3

2 ln3

ln(3 cos2 ) ln3

12

32

3

12

329781510458444_Answer.indb 36 11/8/18 10:36 AM

37



x x

x c

c

c

c

x

x

∫ ∫ θθ θ

θ

θ

θ

θ

θ

( )( )

= +

= − + +

= − + × π +

= − +

=

= − + +

=

= ×

= ×

= + ×

=+

= +

θ

θ

θ

− + +

− +

+

−

−

1 d sin23 cos2 d

ln 12 ln(3 cos2 )

ln3 12 ln 3 cos2 4

ln3 12 ln 3

32 ln3

so ln 12 ln(3 cos2 ) 3

2 ln3

e

e e

e e

(3 cos2 ) 3

33 cos2

273 cos2

12ln(3 cos2 ) 3

2 ln3

12 ln(3 cos2 ) 3

2 ln3

ln(3 cos2 ) ln3

12

32

3

12

32

(ii) Least value is when cos2 1θ =

= + = ≈273 1

3 32 2.60 (3 s.f.)x

5 (i) Since = 2 ,V h V

tht

Vt

h

h ht

h ht

th

h

th

h

t h

Th

h

∫ ∫∫

∫

=

= −

= −

− =

− =

=−

= −=

=−

dd 2d

ddd volume in volume out

1 0.2

1 0.2 2dd

5 10dd

d 105

d

105

d

To find when 4,

105

d0

4

Vt

ht

Vt

h

h ht

h ht

th

h

th

h

t h

Th

h

∫ ∫∫

∫

=

= −

= −

− =

− =

=−

=−

=

=−

dd 2d

ddd volume in volume out

1 0.2

1 0.2 2dd

5 10dd

d 105

d

105

d

To find when 4,

105

d0

4

(ii)

∫∫

∫

( ) ( )

= − ⇒ = −

= − −= − +

= ⇒ = = ⇒ =

= − +

= − −

= −

= − = − − − =

=

=

=

=

5 dd

12

d 2(5 ) dd ( 10 2 ) d

4 3, 0 5

10 ( 10 2 ) d

20 5 d

20 5 d

20 5ln

20 5ln5 5 5ln3 3

11.1 (3 s.f.)

5

3

5

3

3

5

u h uh hh u uh u u

h u h u

T u u u

uu u

uu u

u u

u

u

u

u

u

u

6 (i) x: amount of A y: amount of B

∫ ∫∫ ∫

( )( )

= − ×+

×

= −+

= − +

= + +

= =

= + +

= −

= − =

= + +

=

= ×

=

=

=

=

=

( )( )

−

+ +

+

+

− +

+ −

+ − ++

− +

dd 0.2 10

(1 )1 d 2

(1 )d

1 d 2 (1 ) d

ln 21

When 0, 100

ln100 21 0ln100 2

ln100 lne ln 100e

ln 21 ln 100

e

e

e e

100e

e

100e e

100e

100e

100e

2

2

2

22

2

21 ln 100

e

21

ln 100e

2

21

22

1

21 2

21

2(1 )1

21

2

2

yt t

y

y yt

t

y y t t

y t c

t y

c

c

c

y t

y t

t

t

t

t

tt

t

tt

(ii) As → ∞ −+ → −, 2

1 2t tt

= =

→ ∞+

→

−100e 100e

The mass of B approaches 100e

.

As , 10(1 )

0

The mass of A approaches 0.

22

2

2

y

tt


1 (i) 1 1x a kt( )= + −

( )( )= − +

= −

= −

−dd 1 2

2

2

xt ka kt

ka xa

kxa

9781510458444_Answer.indb 37 11/8/18 10:36 AM

38



(ii) = = ⇒ =When 0, 2.5t x a a

= =

⇒ = ++ =

=

When 1, 1.6

1.6 2.51

1 1.56250.5625

t x

kkk

(iii) In the long term, 0x →

(iv)y y y y

Ay

By−

=−

= +−

12

1(2 ) 22

( )

( )

⇒ = − +

= ⇒ = ⇒ =

= ⇒ = ⇒ =

−= +

−

1 2

0 2 1 12

2 1 2 12

So 12

12

12 22

A y By

y A A

y B B

y y y y

(v) ∫∫ −=1

2d d2y y

y t

y y y t

y y t c

t y c cy y t

yy t

yyy y

y yy

y

t

t t

t t

t t

t

t t

∫ ∫( )⇒ + − =

⇒ − − = += = ⇒ − = + ⇒ =

⇒ − − =

⇒ − =

− =

⇒ = −⇒ + =⇒ + =

⇒ =+

=+ −

12

12(2 ) d d

12 ln 1

2 ln(2 )

When 0, 1 0 0 0 0ln ln(2 ) 2

ln 2 2

2 e

2e ee 2e

(1 e ) 2e2e

1 e2

1 e

2

2 2

2 2

2 2

2

2 2

(vi) As , e 0 22t yt→∞ → ⇒ →−

So the long-term population is 2000.

2 (i) 700

5 2e1000

2P =

+=

−

(ii) → ∞ →−

As , e 02tt

So 7005 140P → =

(iii) P P

P

t

t

t

=+

⇒ + =

= −

−

−

−

700

5 2e5 2e 700

e 350 52

2

2

2

t

PPt

P PPt

t

( )− = −

− =

−

Differentiating with respect to ,12e 350 d

d12

350 52

350 dd

22

2

( )

( )

− =

= −

= −

700350 5

2dd

dd 2 280dd 2 1 140

2

2

PP

Pt

Pt

P P

Pt

P P

(iv) When t = 0, P = 100 ⇒

dd

1002 1 100

140100

7 14.3Pt ( )= − = =

3dd

1yx ym= +

d d

as 0

1( )

1( )

1y y x

ym x c m

y m k x

ym k x

m

m

m

m

∫∫( )

=

− = + ≠

= −=

−

− −

−

Using the chain rule,

x y y yyx ny y nyn n n m n m= = =− + +d

d ( ) dd ( )

dd ,1 1

as required.

4 (i) xt

=dd

02

2

Integrating with respect to t, ddxt C=

But when = = α0, cos ,t v Vx so = αdd cosx

t V

dd

,2

2y

tg= −

Integrating with respect to = − +, ddt y

t gt K

But when = = α0, sint v Vy , so = αsinK V

and = − + αdd sin

yt gt V

Integrating again,

= αdd cosx

t V so = +αcosx Vt M

But 0x = when 0t = , so 0M = and = αcosx Vt

= − + αdd sin

yt gt V so =

−+ +α2 sin

2y

gtVt N

But 0y = when 0, so 0 andt N= =

From = αcosx Vt we get = αcost xV

and substituting into =−

+ α2 sin2

ygt

Vt gives

( )( )=

−+α

α αcos2 cos sin

2

yg x

V V xV so

= −α αtan2

sec2

22y x

gxV

or

= − +α αtan2

(1 tan )2

22y x

gxV

=−

+ α2 sin2

ygt

Vt

9781510458444_Answer.indb 38 11/8/18 10:37 AM

39



(ii)(a) If the centre passes through the point ( , ),kh h then, from the above

( )

= − +α αtan2

(1 tan )2

22h kh

g khV

which is a quadratic in αtan so, rearranging,

and, since 0≠h

− + + =α αtan 2 tan 2 02 2 2 2 2ghk V k V ghk

This has two distinct real solutions if and only if 4 02 − >b ac .

V k ghk V ghk

V k ghk V ghk

V ghV g h k k

− + >

− + >

− − > >

(2 ) 4 (2 ) 0

4 4 (2 ) 0

2 0 ( 0)

2 2 2 2 2

4 2 2 2 2

4 2 2 2 2 2

The LHS of this is a quadratic in V 2, and has the form (V 2- a) (V 2- b) where a, b arise from

V

gh gh g h k

Vgh gh k

=± +

= ± +

2 (2 ) 42

11

22 2 2 2

22

V gh k

k

V ghV g h k

V gh k

= ± +

− + <

− − >

> + +

(1 1 )

But 1 1 0

So 4 8 4 0 when

(1 1 )

2 2

2

2 2 2 2 2

2 2

OR

=−

+ α2 sin2

ygt

Vt and = αcosx Vt

Since it passes through the point ( ),kh h ,

=−

+ α2 sinhgt

Vt and = αcoskh Vt

( )( )( )( )

= +

= +

− = +

− = + +

α

α

sin 2

sin 2

1 24( ) 4 4

2

2 2 22 2

2 2 2 2 2

2 2 2 2 2 2 2 4

Vt hgt

V t hgt

V t khVt h

gt

V t k h h ght g t

g t gh V t h k+ − + + =4( ) 4 (1 ) 02 4 2 2 2 2

To get two different values of t we need

4 02 − >b ac gh V g h k

gh V g h k

gh V gh k gh V gh k

V gh k

V gh k

k

− − + >

− − + >

− < − + − > +

> + +

< − +

− + <

16( ) 16 (1 ) 0

( ) (1 ) 0

1 or 1

So (1 1 )

(Since (1 1 ) is not possible as

1 1 0).

2 2 2 2 2

2 2 2 2 2

2 2 2 2

2 2

2 2

2

(b) Use

( ) ( )− + + =α αtan 2 tan 2 02 2 2 2 2g kh V kh V h g kh

Let α αtan , tan1 2 be the two roots, then by the sum and the product of the roots:

+ = =α αtan tan 2 21 2

2

2 2

2V khgk h

Vgkh

and

× =+

=+

α αtan tan2 2

1 2

2 2 2

2 2

2 2

2V h gk h

gk hV gk h

gk h

and, since + =+

− ×α αα α

α αtan( )tan tan

1 tan tan1 21 2

1 2,

+ =− +

=− −

= −

α αtan( )

2

12

22

1 2

2

2 2

2

2

2 2 2

Vgkh

V gk hgk h

kVgk h V gk h

k

and + = −α α −tan ( ).1 21 k

5 Let y be the number of tigers at time t.

yt ky yyt ky y

y y y k t

u y u y y y y u

y u y u k t

u u k t

u kt cy kt c

y

y At y A

y

t y

k

y

y

y

yt

y

kt c

kt

kt

k

k

t

t

t

t

∫ ∫

∫ ∫∫ ∫

=

= −

− =

= ⇒ = ⇒ =

− =

− =

− − = +− = − +

− =

− == = ⇒ =

− =

= = ⇒ − =

==

− =

= −

= ×

==

= ≈

− +

−

−

−

−

−

−

−

−

−

−

−

− ×

dd ln180

dd (ln180 ln )

1(ln180 ln ) d d

ln d 1 d d d

1(ln180 ) d d

1(ln180 ) d d

ln(ln180 )ln(ln180 ln )

ln180 ln e

ln180 ln e0, 10 ln18

ln180 ln ln18 e

2, 30 ln180 ln30 ln18 e

ln6 ln18 e0.239

ln180 ln ln18 e

ln ln180 ln18 e

e e

180 eWhen 8,

180 e 117

1

2

2

2

0.239

0.239

ln180 ln18e

ln18e

ln18e

2

0.239

0.239

0.239 8

( ) ( )− + + =α αtan 2 tan 2 02 2 2 2 2g kh V kh V h g kh

9781510458444_Answer.indb 39 11/8/18 10:37 AM

40

4Vectors


6 A t: amount of alcohol (litres) in the tank at time .

At

A

A

A

A A t

A t c

A t c

A

A K

A Kt A K K

A

A

t c

t

t

t

∫ ∫

= −

= − ×

= −

= −

− =

− − = +

− = − +

− =

− =

= −= = ⇒ = − ⇒ =

= −

= − =

− +

−

−

−

− ×

dd amount in amount out

2 40 4

2 1020

1010

20 d d

10ln 20

ln 20 110

20 e

20 e

20 e0, 8 8 20 e 12

20 12e

20 12e 17.3 litres

1

2

110

110

110

0

110

110 15

2

4 Vectors4.1 Vectors in two dimensions (Page 44)

1 (i)

(ii)(a) =−

a262

(b) − =−

b50

(c) − =−

c12

12

2a

−b

12—c

(iii)(a) + =

a b21

(b) − =

b c218

(c) + =−

c a12 2

50

2 (i) 4i − 5j

(ii) −i + 2j(iii) g = −6j h = 4i − 5j g + h = − 6j + 4i − 5j = 4i − 11j

3 (i) ( 3) 2 13 3.61 (3 s.f.)2 2= − + = =l

(ii) = + − = =l 2 ( 1) 5 2.24 (3 s.f.)2 2

(iii)AB b a= − =−

−−

=−

40

21

61

= − + = =AB ( 6) 1 37 6.08 (3 s.f.)2 2

a

b

c

9781510458444_Answer.indb 40 11/8/18 10:37 AM

41

4Vectors


4

A

B

C

y

xO

(i) OA =−

32

(ii) BC = c − b =

−−

=−

04

43

47

(iii) CA = a − c =−

−

=−−

32

04

32

(iv) AD =−

=−

347

1221

OD = OA + AD =−

+−

=−

32

1221

1523

So D is (−15, 23).

5 (i) OB m n= +2 12

(ii)OE m n= +52

52

(iii) BD m n= +2 32

(iv) EC m n= −12

32

6 (i) ( 4) 3 52 2= − + =l

Unit vector is −

=−

=−

15

43

4535

0.80.6

(ii) EF f e= − =−

−−

=−

21

62

41

= + − =EF 4 ( 1) 172 2

Unit vector is −

=−

117

41

417117

or i j( )−117

4

4.2 Vectors in three dimensions (Page 47)

1 (i) 28

18−

(ii) 38

13−

2 AB = b − a

= (i + 2j − 3k) − (2i − 3j + k)

= −i + 5j − 4k

3 (i) OD = 3i + 4k

(ii) OF = 5i + 6j + 4k

(iii) CF = 5i + 4k

(iv) BF = −3i + 4k

(v) GA = 5i − 6j − 4k

(vi) DF = 2i + 6j

4 CD = d c− =−

−−

=−

325

203

528

= − + + =CD ( 5) 2 8 932 2 2

9.64 (3 s.f.)=

4.3 Vector calculations (Page 48)

1 (i) PQ = q − p =−

− −

=−

123

113

036

(ii)QR = r − q = −

−−

= −

149

123

06

12

(iii)SinceQR=−2PQ

● they are parallel

● QR is twice the length

● QR goes in the opposite direction.

(iv)RP = p − r = −

− −

=−

113

149

036

RS = 2RP = 2 −

=−

036

06

12

OS = OR + RS = −

+−

=−

149

06

12

123

So S is (1, 2, −3).

9781510458444_Answer.indb 41 11/8/18 10:38 AM

42

4Vectors


2 EF f e= − = −

− −

= −−

334

216

122

= + − + − = =

=

= −−

= −−

EF 1 ( 2) ( 2) 9 3

EG 6 EF

EG 6122

61212

2 2 2

= +

= −

+ −−

= −−

OG OE EG

216

61212

813

6

So G is (8, −13, −6).

3 l = − + − + =( 1) ( 2) 2 32 2 2

Unit vector is

13(−i − 2j + 2k) = −1

3 i 23− j + 23 k

4 AB l k= − = −

− −

=

51

10

216

304

= + + =

= =

=

= + = −

+

= −

= + − + =

AB 3 0 4 5

AM 3AB 3304

90

12

OM OA AM216

90

12

111

18

OM 11 ( 1) (18) 446

2 2 2

2 2 2

The unit vector is −

= −

1446

111

18

0.5210.0470.852

4.4 The angle between two vectors (Page 49)

1 (i)−

−

= × − + − × = −•34

12

3 1 4 2 11

l

lθ

θ

θ

= + − =

= − + =− = × ×

= −×

= °

3 ( 4) 5

( 1) 2 511 5 5 cos

cos 115 5169.7 (1 d.p.)

12 2

22 2

(ii)

−

= × + × − = −•37

16

3 1 7 6 39

l

lθ

θ

θ

= + =

= + − =− = × ×

= −×

= °

3 7 58

1 ( 6) 3739 58 37 cos

cos 3958 37

147.2 (1 d.p.)

12 2

22 2

2 a b= −−

=−

231

and052

−−

−

= × + − × + − × − = −•

231

052

2 0 3 5 1 2 13

l

lθ

θ

θ

= + − + − =

= + + − =− = × ×

= −×

= °

2 ( 3) ( 1) 14

0 5 ( 2) 2913 14 29 cos

cos 1314 29

130.2 (1 d.p.)

12 2 2

22 2 2

3

−

−−

= × + × − + − × − =•

235

321

2 3 3 2 5 1 5

l

lθ

θ

θ

= + + − =

= + − + − == × ×

=×

= °

2 3 ( 5) 38

3 ( 2) ( 1) 145 38 14 cos

cos 538 14

77.5 (1 d.p.)

12 2 2

22 2 2

4 g.h = 0

cc

−

−−

=•

8

231

0

c cc

c

+ − + =+ =

= −

8 3 2 05 2 0

25

5 (i) OM = 14i + 20k

C′M = 30i − 20k

MB′ = −14i + 16j + 20k

CB′ = 12i +16j + 40k

9781510458444_Answer.indb 42 11/8/18 10:38 AM

43

4Vectors


(ii) Angle between OM and C′M:

−

= × + × + × −=

•

140

20

300

2014 30 0 0 20 2020

l

lθ

θ

θ

= + + =

= + + − == × ×

=×

= °

14 0 20 596

30 0 ( 20) 130020 596 1300 cos

cos 20596 1300

88.7 (1 d.p.)

12 2 2

22 2 2

Angle between MB′ and CB′:

−

= − × + × + ×=

•

141620

121640

14 12 16 16 20 40888

l

lθ

θ

θ

= − + + =

= + + == × ×

=×

= °

( 14) 16 20 852

12 16 40 2000888 852 2000 cos

cos 888852 2000

47.1 (1 d.p.)

12 2 2

22 2 2

4.5 The vector equation of a line (Page 51)

1 (i) r = 3i − j + t(−2i + 5j)

= (3 − 2t)i + (−1 + 5t)j

t31

25

=−

+

−

(ii)AB =

−−

=

08

21

27

r = −2i + j + t(2i + 7j)

= (−2 + 2t)i + (1 + 7t)j

t=−

+

21

27

(iii) r = 4i − 3j + k + λ(i − 4k)

λ= −

+−

431

104

(iv) AB =−

−−

= −

201

195

194

r = 2i − k + λ(i − 9j + 4k)

= (2 + λ)i − 9λj + (−1 + 4λ)k

λ=−

+ −

201

194

2 (i) We need to find λ such that 2 − 3λ = −7 −1 − 2λ = −7 1 + λ = 4

When λ = 3, all equations are satisfied, hence the point lies on the line.

(ii) We need to find µ such that 1 + µ = 0 −2µ = 2 −6 + µ = −8 If µ = −1, the first two equations are satisfied, but

the last equation is not. Hence the point does not lie on the line.

3 r = 3i − 4j + 2k + t(i + 3j − k)

t= −

+−

342

131

3 + t = a ① −4 + 3t = b ② 2 − t = 0 ③ From ③, t = 2 Hence a = 5, b = 2

4.6 The intersection of two lines (Page 52)

1 (i) Equating i, j and k components,

4 + 2λ = 3 − 5µ ① −1 − 2λ = −3 + 2µ ② 1 + 3λ = 7 ③ From ③, λ = 2

From ②, µ µ− − × = − + ⇒ = −1 2 2 3 2 1

Substituting into ①,

+ × = − × −=

4 2 2 3 5 18 8

So the lines intersect. Position vector of point of intersection is

411

2223

857

−

+ −

= −

Point is (8, −5, 7).(ii) Equating i, j and k components,

1 − 2λ = −1 + 4µ ① 3λ = 1 − 5µ ② 2 + λ = −3 + µ ③ Solving ① and ② simultaneously gives

µλ = − =3, 2

Substituting into ③, + − = − +

− = −2 3 3 2

1 1

So the lines intersect.

9781510458444_Answer.indb 43 11/8/18 10:38 AM

44

4Vectors


Position vector of point of intersection is

− × −

× −−

= −−

1 2 33 32+ 3

791

Point is (7, −9, −1).

(iii) Equating i, j and k components.

−4 − s = 3 + 11t ① 4 = 1 − 3t ② −5 + 3s = 8 + t ③ From ②, t = −1

From ①, 4 3 11 1 4s s− − = + × − ⇒ =

Substituting into ③,

− + × = + −=

5 3 4 8 ( 1)7 7



r = (−4 − 4)i + 4j + (−5 + 3 × 4)k

r = −8i + 4j + 7k

or ( )−8, 4, 7

2 (i) (3i − 4j + 2k) • (2i − j − 5k)

3 2 4 1 2 50

= × + − × − + × −=

(ii) Equating i, j and k components,

5 + 3s = 2 + 2t ① −2 − 4s = −2 − t ② −2 + 2s = 7 − 5t ③ Solving ① and ② simultaneously gives

s t35 , 12

5= =


− + × = − ×

− ≠ −

2 2 35 7 5 12

545 5

Hence there is no point of intersection.

The lines are not parallel so the lines are skew.

3 Equating i, j and k components,

1 − λ = µ ① 2 + 2λ = 6 ② −1 + 3λ = 3 − 2µ ③ From ②, λ = 2

From ①, µ µ− = ⇒ = −1 2 1


− + × = − × −=

1 3 2 3 2 15 5



r =

−

+−

=−

121

2123

165

Point is ( )−1, 6, 5 .

4 (i) PQ =

−

=−

=−

1392

347

1055

5211

Equation is

r = λ

+−

347

211

RS k k=

−

= −

10

6

123

92

3

Equation is

r = µ k

+ −

123

92

3

Equating i, j and k components,

3 + 2λ = 1 + 9µ ① 4 + λ = 2 + µ(k − 2) ② 7 − λ = 3 + 3µ ③ Solving ① and ③ simultaneously gives

λ = =2, 23µ

Substituting into ②,

( )

( )

+ = + −

= −

=

4 2 2 23 2

4 23 2

8

k

k

k

(ii) r =

+−

=

347

2211

765

A is (7, 6, 5).

5 (i) a

aa

−

=

+ − ==

•

2

1

226

0

4 2 6 01


2t = 3 + 2s ① 1 + at = 2s ② 1 + t = −1 − 6s ③

9781510458444_Answer.indb 44 11/8/18 10:38 AM

45

4Vectors


Solving ① and ③ simultaneously gives

t s1, 12= = −

From ②,

1 2 1 2 12 2at s a a+ = ⇒ + = × − ⇒ = −


−3 + λ = 7 + aµ ① 2 + 2λ = 3 + bµ ② 3 + λ = 3 − 2µ ③ ① − ③ gives

a µ

a µ

µ a

− = + +− = +

= − +

6 4 ( 2)10 ( 2)

102

② − 2 × ③ gives b µ

b µ

µ b

− = − + +− = +

= − +

4 3 ( 4)1 ( 4)

14

So a b− + = − +10

21

4

b ab a

a b

+ = ++ = +

− =

10( 4) 210 40 2

10 38

(ii) ab

−

=•

121 2

0

2 2 0 2 2a b a b+ − = ⇒ + =

Solving simultaneously,

a b8, 3= = −

(iii) 102

108 2 1µ a= − + = − + = −

Position vector of the point is

r = 7i + 3j + 3k + −(8i − 3j − 2k)

= −i + 6j + 5k

4.7 The angle between two lines (Page 56)

1 −

−

= −•

123

102

7

θθ

θ

− + + =+ + − =

− =

= −

= °

( 1) 2 3 141 0 ( 2) 5

7 14 5cos

cos 714 5

146.8

2 2 2

2 2 2

° − ° = °Acute angle is 180 146.8 33.2

2 (i) bc

c c

= ⇒ + = ⇒ = −•

3 401

0 12 0 12

bc

b c

= ⇒ + + =•

3 432

0 12 3 2 0

⇒ bbb

+ + − =− + =

=

12 3 2( 12) 012 3 0

4

(ii)

=•

401

432

18

θ

θ

θ

+ + =+ + ==

=

= ° = °

4 0 1 174 3 2 2918 17 29cos

cos 1817 29

35.8 36 (nearest degree)

2 2 2

2 2 2

3 a( ) ( ) ( )− − −A 1, 3, 2 , B 12, 2, 4 and C 5, 1,

AC a a

= −

−−

= −+

51

132

44

2

BC a a

= −

− −

=−

−

51

824

31

4

AC • BC = 0 ⇒

4 OB = i + j, OD = 12

i + 12

j + k

=•

110

0.50.5

11

= = + + =OB 2, OD 0.5 0.5 1 62

2 2 2

1 2 62 cos

54.7

θ

θ

=

= °

a aa a

a a

a aa a

a

−+

−

−

=

− − + + − =

− − + − − =

− − =+ − =

= −

•

44

2

31

40

12 4 ( 2)( 4) 0

12 4 2 8 0

2 24 0( 4)( 6) 0

4 or 6

2

2

9781510458444_Answer.indb 45 11/8/18 10:38 AM

46

4Vectors


4.8 The perpendicular distance from a point to a line (Page 57)

1 (i) Let M be the point on the line that gives the minimum distance.

AM ttt

ttt

=−

− ++

− −−

=−++

1 31 41 2

073

1 36 44 2

tttt t t

tt

−++

−

=

− + + + + + =+ =

= −

−

+ −−

= −−

= − + − − − + − − − =

•

1 36 44 2

342

0

3 9 24 16 8 4 029 29 0

1

M is111

( 1)342

451

AM (4 0) ( 5 7) ( 1 3) 242 2 2

(ii) Let M be the point on the line that gives the minimum distance.

λλλ

λλ

λ

λλ

λλ λ λ

λλ

=+

− −−

−−−

=+−

− −

+−

− −

−−

=

+ − + + + =+ =

= −

−

+ − −−

=

= − − + − − + − =

•

BM4 2

21 2

167

5 246 2

5 246 2

212

0

10 4 4 12 4 018 9 0

2

M is421

( 2)212

005

BM ( 1 0) ( 6 0) (7 5) 412 2 2

2 (i) PQ = (4i + 4j − 6k) − (5i + 2j − 9k)

= −i + 2j + 3k

r t=−

+−

529

123

(ii) OT is r = s

+−

000

121

−

−

= ⇒•

123

121

0 lines are perpendicular.

(iii) Equating i, j and k components,

5 − t = s ① 2 + 2t = 2s ② −9 + 3t = −s ③

Solving ① and ③ simultaneously gives

= =2, 3t s


+−

=−

000

3121

363

(iv) + + =3 6 3 542 2 2

3 (i) Equation of plane's path is

r = t

+−−

1.20.80

211

Let M be the point closest to O.

tt

tt t t

tt

−−

=

−−

−−

=

− + − + + =− + =

=

− ×−

=

= + + =

•

•

OM211

0

1.2 20.8

211

0

2.4 4 0.8 03.2 6 0

0.533 mins (32 s)

M is1.2 2 0.533

0.8 0.5330.533

0.1330.2670.533

OM 0.133 0.267 0.533 0.611 km2 2 2

(ii) Height is 0.533 km or 533 m.

Time is 0.533 min or 32 seconds.

4 (i) Rabbit: r = t

+

520

0

210

t

t t

+

=

+ = ⇒ =

520

0

210

2530

05 2 25 10

It would take the rabbit 10 seconds.

9781510458444_Answer.indb 46 11/8/18 10:38 AM

47

4Vectors


(ii) Eagle: r = t−

+−

102050

51

10 Eagle reaches the ground when

− = ⇒ =50 10 0 5t t

Eagle is then at −

+−

=

102050

551

10

1525

0

At t = 5 the rabbit is at

+

=

520

05

210

1525

0

So the eagle catches the rabbit.


1 (i) OC = 2OA = 2 −

= −

223

446

OD = 3OB = 3 −

=−

405

120

15

CD = − =−

− −

=−

d c12

015

446

84

21

(ii) AB = b a− =−

− −

=−

405

223

228

AE = 12 AB =−

=−

12

228

114

OE = OA + AE = −

+−

= −−

223

114

311

EC = − = −

− −−

= −

c e446

311

137

2 GH = − = −

− −

=−

h g910

41

12

50

12

= + + − = =GH 5 0 ( 12) 169 132 2 2

Unit vector is −

=

−

113

50

12

513

01213

or i k( )−113 5 12

3 (i) −

= × + × + − × = −•

132

605

1 6 3 0 2 5 4

l

lθ

θ

θ

= + + − =

= + + =− = × ×

= −×

= °

1 3 ( 2) 14

6 0 5 614 14 61 cos

cos 414 61

97.9 (1 d.p.)

12 2 2

22 2 2

(ii) CD = − =

−−

= −

d c605

132

537

−

−

= × + × − + − × = −•

132

537

1 5 3 3 2 7 18

l

lθ

θ

θ

= + + − =

= + − + =− = × ×

= −×

= °

1 3 ( 2) 14

5 ( 3) 7 8318 14 83 cos

cos 1814 83

121.9 (1 d.p.)

12 2 2

22 2 2

(iii) If perpendicular, then CD • OF = 0

CD • OF = a

aa a−

= + − +•

537

32

5 9 14

+ − + =5 9 14 0a a

− =19 9 0a

= 919a

4 First find the vector PQ:

PQ =

− −

=

101

12

214

828

so

PR = 12 PQ =

=

12

828

414

Now OR = OP + PR = −

+

=

214

414

608

= + + = =OR 6 0 8 100 102 2 2

So the unit vector in the direction of OR is

110

608

0.60

0.8

=

9781510458444_Answer.indb 47 11/8/18 10:39 AM

48

4Vectors


5 When two vectors are perpendicular then a • b = 0 so − − =

− − == −

2 6 3 06 0

6

p ppp

6 (i) CD =−−

−

=−−

345

230

575

r = 2i + 3j + λ(−5i − 7j + 5k)

= (2 − 5λ)i + (3 − 7λ)j + 5λk

λ=

+−−

230

575

(ii) r = 2i + 3j − 2k + λ(8i − 2j + k)

= (2 + 8λ)i + (3 − 2λ)j + (−2 + λ)k

λ=−

+ −

232

821


1 + 2s = 5 − t ① −1 + 4s = 2 + 3t ②

3 + s = 1 + 2t ③ Solving ① and ② simultaneously gives

s t32 , 1= =


+ = + ×

≠

3 32 1 2 1

92 3

So the lines do not intersect.

Since the direction vectors are not multiples, the vectors are skew.


1 + 4λ = 1 − 8µ ① 8 − λ = 2µ ②

2λ = 12 − 4µ ③ Solving ① and ② simultaneously gives

33 = 1 so no point of intersection.

The direction vectors are multiples of each other so the lines are parallel.

(iii) Equating i, j and k components, −5 + 3t = −1 + 2s ①

3 = 4 − s ② 4 − t = 2 ③

From ②, s = 1

From ③, t = 2


− + × = − + ×

=5 3 2 1 2 1

1 1 So the lines intersect.

r = −

+−

=

534

2301

132

Point of intersection is (1, 3, 2).

8 (i) PQ =

−−−

=

124

513

631

PQ is s=

+

r124

631

l is t=

+ −

r135

122

To intersect,

+ = ++ = −+ = +

+ =

=

= −

1 6 12 3 3 24 5 2

6 4 812

14

s ts ts ts

s

t


⇒ ( )+ × = + −

≠

1 6 12 1 1

4

4 34

So the lines do not intersect.

(ii) ttt

ttt

ttt

=

−+−+

=−

− +− −

=

−− +− −

=

•

•

AQ124

13 25 2

1 21 2

AQ QP 0

1 21 2

631

0

− − + − − =− − =

= −

6 3 6 1 2 04 2 0

2

t t ttt

⇒ + − + − × − + − ×(1 ( 2),3 2 2,5 2 2)

A is (−1, 7, 1).

① ② ③ ②+ ③

9781510458444_Answer.indb 48 11/8/18 10:39 AM

49

4Vectors


9 To intersect,

k

kk

λλλλλ

+ = −+ = − ++ = − ++ =

==

+ × = − += −

µµµ

µ

2 3 112 4

1 2 1 25 8 21

23

2 2 4 35

Point is (11 + 3 × -1, -4 + 3 × 1, -1 + 3 × 2) = (8, -1, 5)

10 (i) −6i + 8j −2k = −2(3i + cj + k) 8 = −2c ⇒ c = −4

(ii) Equating i, j and k components, 3 + t = 2 + 3u ①

−8 + 3t = 1 + cu ② 2t = 3 + u ③

Solving ① and ③ simultaneously gives t = 2, u = 1 Substituting into ②,

cc

c

− + × = + ×− = +

= −

8 3 2 1 12 1

3

11 (i) =

−

= −

AB704

513

211

Equation is

r = λ

+ −

513

211

(ii) The position vector for any point on AB is

r = λλ

λλ

+ −

=+−+

513

211

5 213

λλλ

λ λ λλλ

=

+−+

−

=

+ − + + + =+ =

= −

=+ × −− −+ −

=

•

•

OP AB 0

5 213

211

0

10 4 1 3 012 6 0

2

OP5 2 21 ( 2)3 ( 2)

131

So P is (1, 3, 1).

12 Second line has direction

−

=−

−

782

975

213

θθ

−

−

−

= − + − + − = −

− = × ×= °

•

352

213

6 5 6 17

17 38 14 cos42.5

13 Let M be the point on the line that gives the minimum distance.

s

ss

ss

s= −

−

−−

=− +

−−

CM5

13 2

112

1 5

5 2

ss

ss s s

s

s

− +−−

−−

=

− + + − + =− + =

=

+ −−

=

•

1 5

5 2

512

0

5 25 10 4 015 30 0

12

M is013

12

512

2.50.5

2

CM (2.5 1) (0.5 1) (2 2)4.30

2 2 2= − + − + − −=

14 (i) P is

θ

θ

θ

+ −

=

−

=

= × ×

=×

= °

•

311

1112

403

403

112

10

10 5 6 cos

cos 105 635.3

θ

θ

θ

+ −

=

−

=

= × ×

=×

= °

•

311

1112

403

403

112

10

10 5 6 cos

cos 105 635.3

(ii) Q is ttt

+−+

31

1 2

ttt

t t tt

t

+−+

−

=

+ − + + + =+ =

= −

•

31

1 2

112

0

3 1 2 4 04 6 0

23

① ② ③ 2 × ①+ ③

9781510458444_Answer.indb 49 11/8/18 10:39 AM

50

4Vectors


( )

−

− −

+ −

=

−

=−

Q is

3 23

1 23

1 2 23

735313

13

751

(iii) ( ) ( ) ( )= + + − =OQ 73

53

13

5 33

2 2 2


1 (i) = + =

+ −

=

OB OA AB133

31

1

424

= + + =OB 4 2 4 62 2 2

Unit vector is

=

+ +16

424

231323

or 23 i 1

3 j 23 k

(ii) = + =−−−

+ −

= −−

AC AO OC133

311

242

= + − + − =AC 2 ( 4) ( 2) 242 2 2

= −−

= − − = −• •AC OB242

424

8 8 8 8

Angle between diagonals:

θ

θ

θ

− = × ×

= −×

= °

8 6 24 cos

cos 86 24105.8

Acute angle is ° − ° = °180 105.8 74.2

(iii) = + + =OA 1 3 3 192 2 2

= + − + =OC 3 ( 1) 1 112 2 2

Perimeter = + = + =2 OA 2 OC 2 19 2 11 15.4

= + = + =2 OA 2 OC 2 19 2 11 15.4

2 (i)

θ

θ

θ

• = × − + × + − × − =

= + + − =

= − + + − == × ×

=×

= °

OA OB 2 3 1 2 3 4 8

OA 2 1 ( 3) 14

OB ( 3) 2 ( 4) 298 14 29 cos

cos 814 29

66.6

2 2 2

2 2 2

(ii) p p= + = + = −

+−

BC BO OC BO OA324

213

ppp

p p p=+

− +−

+ + − + + −i j k3 2

24 3

or (3 2 ) ( 2 ) (4 3 )

(iii) ppp

p p p+

− +−

−

= + + − + + − −•

3 22

4 3

213

2(3 2 ) ( 2 ) 3(4 3 )

p p pp p p

p

p

+ + − + + − − =+ − + − + =

− + =

= =

2(3 2 ) ( 2 ) 3(4 3 ) 06 4 2 12 9 0

8 14 08

1447

3 (i) PQ = 3i + 6j − 3k

RQ = −3i + 8j + 3k

(ii)

θ

θ

θ

• = × − + × + − × =

= + + − =

= − + + == × ×

=×

= °

PQ RQ 3 3 6 8 3 3 30

PQ 3 6 ( 3) 54

RQ ( 3) 8 3 8230 54 82 cos

cos 3054 82

63.2

2 2 2

2 2 2

4 Lines are

r = λ−

+−

321

121

and r = ab

+−

µ442 1

(i) At intersection,

ab

λλλ

− = +− + = +

+ = −

µµµ

3 42 2 4

1 2

Adding ① + ③, a a= + − ⇒ = − −µ µ4 6 ( 1) 2

1 ② - 2 × ③

b b= + ⇒ = − +µ µ4 ( 2) 42

− − = − ++ = −

= −− =

21

42

2 4 4 48 4 2

2 4

a bb a

a ba b

(ii) Perpendicular lines ab⇒

−

−

=•

121 1

0

a b a ba b

a b

− + − = ⇒ − + =− =

= =

2 1 0 2 1Solving simultaneously, with 2 4,

3, 2

① ② ③

9781510458444_Answer.indb 50 11/8/18 10:39 AM

51

4Vectors


(iii) Point of intersection is

( )

= − − = −

=

+ −−

=

+ +

µ

r

i j k

23 1 1

442

( 1)321

123

Point is 1, 2, 3 or 2 3

5 (i) Perpendicular means

pp

pp

p p p p

p p p p pp

p

( )( ) ( )

−−

−

=

− − + − + =

− − + + − + =− =

=

•

62 6

1

4 2

20

6 4 2 2 6 2 0

4 2 24 12 2 6 2 010 22 0

2.2

2 2

(ii) If OAB is a straight line, one vector is a scalar multiple of the other. Looking at the x-coordinates,

p p

=⇒ − = −

OB 2OA4 2 2( 6)

p ppp

− = −==

=−

=−

= − + + =

4 2 2 124 16

4

OA221

, OB442

OA ( 2) 2 1 32 2 2

6 (i) p p=

=−−

OP0

2, XP

45

2

p p

p p

p pp p

p

•−−

=

− + =

− + =− − =

=

0

2

45

20

( 5) 4 0

5 4 0( 1)( 4) 0

1 or 4

2

(ii) XP442

=−

Unit vector is

− + +

−

=−

=

−

1( 4) 4 2

442

16

442

232313

2 2 2

(iii) AG4

152

=−

i j k

XQ AG4

152

10

23

So XQ 23

415

2

83

1043

or 83 10 4

3

= =−

=

⇒ =

=−

=

−

− + +

k ka

bk


1 abc

a b c

−

= ⇒ − + + =•

113

0 3 0 ①

abc

a c

= ⇒ + =•

205

0 2 5 0 ②

− + + =2 2 6 0a b c ①× 2 − ③ + =2 11 0b c ② + ③ There are infinitely many solutions to this system of

2 equations with 3 unknowns. Choosing the whole-number solutions,

b c a11, 2, 5= = − =

The vector is −

5112

(or any scalar multiple).

2 = ⇒ + + = + +3 4 32 2 2 2 2 2p q a b a

+ + = + +

== ±

9 16 9

164

2 2 2

2

a b a

bb

ab a a ab

= ⇒ + + =•

3

4

30 4 9 0

If b = 4, + + = ⇒ = −4 4 9 0 98a a a

If b = −4, there are no solutions for a.

So = − =98 and 4a b

3 v

v

v vk kkkk

B

A

A B

= + − + =

= + =

= = −

= −

−

−120 ( 60) 40 140 m s

140 35 175 m s

1206040

1206040

2 2 2

1

1

9781510458444_Answer.indb 51 11/8/18 10:39 AM

52

4Vectors


v k k kA = + − + =(120 ) ( 60 ) (40 ) 1752 2 2

19 600 175251654

2 2

2

=

=

=

k

k

k

vA i j k= −

= −

= − +54

1206040

1507550

150 75 50

4 (i) tt

sss

r r=−

− −− −

=− −− −− −

121

24 31 3

1 2

st s

t s=

− −− + −

−

−−

=•

PQ1

2 33

PQ011

0

2 3 3 02 6 2

3 1

− + − + =− + = −

− =

t s t st st s ①

−−−

=•PQ133

0

1 6 3 9 3 9 07 19 6 0

6 19 7

+ + − + − + =+ − =− + = −

s t s t ss t

t s ② Solving gives s = −1, t = −2

− − − − − − − = −− − − − − − − − − = − −

P is ( 1, 2 ( 2), 1 ( 2) ( 1, 0,1)Q is ( 2 ( 1), 4 3( 1), 1 3( 1)) ( 1, 1, 2)

(ii) PQ 1 1 1 0 2 1

2 1.41 km

2 2 2( )( ) ( ) ( )= − − − + − − + −

= ≈ r1

r2P

Q

5 Let Q be the point on the line that gives the minimum distance.

Q1 2

3

PQ2 2

1

=− +

−−

=− +

−− −

ta t

t

tt

t

tt

tt t t

t

t

−−

=

− +−

− −

−

−

=

− + + + + =− + =

=

•

•

PQ211

0

2 2

1

211

0

4 4 1 03 6 0

12

Q 1 2 12

12 3 1

2

0 12

52

PQ 1 0 12 4 5

2142

22 2

( )( )

( ) ( )( )

= − + × , − , −

= , − ,

= − + − − + −

=

a

a

a a

a is any real number.

9781510458444_Answer.indb 52 11/8/18 10:39 AM

53

5 Complex numbers


5 Complex numbers5.1 Working with complex numbers (Page 65)

1 (i) = × = −i i i i3 2

(ii) 2i 2 i

2 12

6 2 3

3

( )( )

− = −

= − −=

(iii) 3i 3i

9i9

3 2 2

2

( ) ( )= −

== −

(iv) ( ) ( )− + − = − − + − −=

i i i i 1 i 1 i0

4 3 2

2 (i) +9i 2

(ii) = −12i 122

(iii) + = − +7i i 1 7i2

(iv) 15 5i 6i 2i 15 i 217 i

2− + − = + += +

3 (i) z

z

zz

9 0

9

9i3i

2

2

2 2

+ =

= −

== ±

(ii) 4 8 0

4 4 4 1 82 1

4 162

4 16i2

4 4i2

2 2i

2

2

2

( ) ( )( )( )

− + =

=± − −

= ± −

= ±

= ±

= ±

z z

z

(iii)

( )− + =

=± − − × ×

×

= ± −

= ±

= ±

= ±

2 6 5 0

6 6 4 2 52 2

6 44

6 4i4

6 2i4

32

12 i

2

2

2

z z

z

(iv)

( ) ( )( )( )

+ − =

=− ± − −

= − ± +

= − ±

= − +

= − ±

2i 5 0

2i 2i 4 1 52 1

2i 4i 202

2i 162

2i 42

i 2

2

2

2

z z

z

zSo i or 2 i= 2 − − −

4 z w5 2i 3 7i= − , = +

(i) z w ( ) ( )− = − − += − − −= −

4 3 4 5 2i 3 3 7i20 8i 9 21i11 29i

(ii) z w* ( )( )= + +

= + + += + −= +

5 2i 3 7i

15 35i 6i 14i15 41i 141 41i

2

(iii) z z* ( )( )= + −

= − + −= +=

5 2i 5 2i

25 10i 10i 4i25 429

2

(iv) z z*( ) ( )( )( )

− = − − +

= −=

Re Re 5 2i 5 2i

Re 4i0

(v) w*

[ ]( )

[ ]

( ) = −

= − +

= − −= −

Im Im 3 7i

Im 9 42i 49i

Im 40 42i42

2 2

2

(vi) z w* * *

*

[ ] [ ][ ][ ]

( )

( )

( ) = + + +

= +

= −

= − += − −

+ 5 2i 3 7i

8 9i

8 9i

64 144i 81i17 144i

2 2

2

2

2

5 (i)

( )( )+ ∈

= + −

= − + −

= +

*

a b a bzz a b a b

a ab ab b

a b

zLet = i, ,i i

i i i2 2 2

2 2

(ii) z z a b a ba

*+ = + + −=

i i2

9781510458444_Answer_40-69.indd 53 11/8/18 10:57 AM

54

5Complexnumbers


5.2 Dividing and finding square roots of complex numbers (Page 67)

1 (i) × = = −2i

ii

2ii

2i2

(ii) ( ) = = −15i

225i

2252

2

(iii) − × ++ = +

−

= +

= +

31 i

1 i1 i

3 3i1 i3 3i

232

32 i

2

(iv) 2i3 i

3 i3 i

6i 2i9 i2 6i10

15

35i

2

2− × ++ = +

−

= − +

= − +

(v) 5 3i4 3i

4 3i4 3i

20 15i 12i 9i16 9i

29 3i25

2925

325i

2

2++ × −

− = − + −−

= −

= −

(vi) ( )−+ = − +

+

= − −+ × −

−

= − + − +−

= − −

= − −

2 3i3 i

4 12i 9i3 i

5 12i3 i

3 i3 i

15 15i 36i 12i9 i

27 21i10

2710

2110 i

2 2

2

2

2 (i) a b

a ab b

a b ab

( i) 5 12i

2 i i 5 12i

( ) 2 i 5 12i

2

2 2 2

2 2

+ = − −

+ + = − −

− + = − −

a⇒ − = − = − ⇒ = −a b ab b5 and 2 12 62 2

a a

aa

a a

a a

a a

6 5

36 5

36 5

5 36 0

9 4 0

22

22

4 2

4 2

2 2

( )

( )( )

− − = −

− = −

− = −

+ − =

+ − =

⇒ = ⇒ = −⇒ = −

= − −

a aba b

4 2 or 23 or 3

So ( , ) (2, 3) or ( 2, 3)

2

(ii) a b

a b ab

a b ab a b

b b

bb

b b

b b

b b

b ba

+ i 24 10i

2 i 24 10i

24 and 2 =10 5

5 24

25 24

25 24

24 25 0

25 1 0

25 5 or 51 or 1

2

2 2

2 2

22

22

4 2

4 2

2 2

2

( )

( )( )

( ) = − +

− + = − +

⇒ − = − ⇒ =

− = −

− = −

− = −

− − =

− + =

⇒ = ⇒ = −= −

− −a bSo ( , ) = (1, 5) or ( 1, 5)

3 (i) z zx y x yx y x y

x yx y yz

*− = ++ − − = ++ − + = +

+ = +⇒ = = ⇒ =

= +

2 3 6i2( i ) ( i ) 3 6i

2 2 i i 3 6i3 i 3 6i

3, 3 6 2So 3 2i

(ii) z zx y x y

x y x y x yx y x y x y

x x y

*+ + = −+ + + − = −

+ + − + − = −− + − + + = −

+ − = −

i (2 i) 10 2ii( i ) (2 i)( i ) 10 2i

i i 2 2 i i i 10 2ii 2 2 i i 10 2i

2 (2 2 )i 10 2i

2 2

2 10 52 2 2 2 5 2 2

2 126

= ⇒ =− = − ⇒ × − = −

− = −=

x xx y y

yy

So 5 6i= +z

4 z z

a b a b

a ab b a b

a b ab a b

a b ab a b

ab a b a b

ab a a b bab a

b

a

a

aa

*

( )

=

+ = −

+ + = −

− + = −

− + = −

− + − = −

− = − = −= −= −

− − = − −

− =

== ±

i 4

i( i) 4( i)

i 2 i i 4( i)

i( 2 i) 4 4 i

i i 2 i 4 4 i

2 ( )i 4 4 i

2 4 and 422

( 2) 4( 2)

4 8

1212

2

2

2 2 2

2 2

2 2 2

2 2

2 2

2 2

2

2

9781510458444_Answer.indb 54 11/8/18 10:40 AM

55

5Complexnumbers


zab a a

b b

b bb b

b

= − − −− = ⇒ =

− = −

−= −=

So 12 2i or 12 2i2 4 0

So 4

0 = 40 ( 4)

0 or 4

2

2

So z = 0 + 0i or 4i

5 = − × ++

= +−

= − ++

= −+

++

4i3i

3i3i

4 i 12i9i

12 4 i9

129

49

i

2

2 2

2

2 2

u aaa

aa

aa

aa

a

5.3 Representing complex numbers geometrically and finding the modulus (Page 69)

1

2

3

1

4

5

–2

–1

–3

–4

–5

2 31 4 5–2–3–4–5 –1

Im(z)

Re(z)

−4 − 3i

−4

1 − i

−5 + 2i

4i 3 + 4i

0

2 (i) = +1 2iz(ii) + = +2 3 2iz

(iii) − = −3i 1 iz

(iv) z* = −1 2i

(v) = − +i 2 iz

(vi) = + += − += − −

i i(1 4i 4i )i( 3 4i)

4 3i

2 2z

2

3

1

4

5

–2

–1

–3

–4

–5

2 31 4 5–2–3–4–5 –1

Im(z)

Re(z)

−4 − 3i

1 − i

3 + 2i1 + 2i

1 − 2i

−2 + i

0

3 (i) 3 i 3 ( 1)= 2

2 2− = + −

(ii) 3 4i 3 ( 4)5

2 2− = + −=

(iii) 5 12i 5 12= 13

2 2+ = +

(iv) 4i 4− =

(v) − + = − +

=

2 5i ( 2) ( 5)

7

2 2

(vi) 12 12− =

4 (i) 1 1 = 22 2= +u

(ii) 3 2i

3 2

13

2 2( ) ( )+ = − −

= − + −

=

u v

(iii) w* = − =2i 2

(iv) uw*

( ) ( )

= +−

= +− ×

= +−

= − +

= − +

= − +

= = =

1 i2i

1 i2i

2i2i

2i 2i4i

2 2i4

12

12i

12

12

12

12

22

2

2

2 2

uw*

( ) ( )

= +−

= +− ×

= +−

= − +

= − +

= − +

= = =

1 i2i

1 i2i

2i2i

2i 2i4i

2 2i4

12

12i

12

12

12

12

22

2

2

2 2

9781510458444_Answer.indb 55 11/8/18 10:40 AM

56

5Complexnumbers


5 (i) 3 4i 3 ( 4)5

2 2− = + −=

(ii) x x

x x

x x

− + = − +

= − + +

= − +

1 i ( 1) 1

2 1 1

2 2

2 2

2

2

(iii) x y

x y

x x y y

x y x y

+ + +

= + + +

= + + + + +

= + + + +

( 1) i( 1)

( 1) ( 1)

2 1 2 1

2 2 2

2 2

2 2

2 2

(iv) θ θ

θ θ

+

= +==

cos isin

cos isin1

1

2 2

5.4 Sets of points in an Argand diagram (Page 71)

1 (i)

2

420

–2

–4 –2

–4

4

Im

Re

(ii)

2

420

–2

–4 –2

–4

4

Im

Re

(iii)

2

420

–2

–4 –2

–4

4

Im

Re

(iv)

2

420

–2

–4 –2

–4

4

Im

Re

2 (i)

2

420

–2

–4 –2

–4

4

Im

Re

(ii)

2

420

–2

–4 –2

–4

4

Im

Re

(iii)

2

420

–2

–4 –2

–4

4

Im

Re

(iv)

2

420

–2

–4 –2

–4

4

Im

Re

3 If the < or > inequality signs had been used, the boundary curves or lines would need to be dashed, representing the fact that the actual boundary curve or line is not included in the region.

4 w = 1 − 2i

(i) (1 2i)(1 2i)

1 4i 4i1 4i 4

3 4i

3 4 5

2

2

2 2 2( ) ( )

= − −

= − += − −= − −

= − + − =

w

w

9781510458444_Answer.indb 56 11/8/18 10:40 AM

57

5Complexnumbers


(ii)

( )

−

− − −3 4i 5

2 2z w w

z

�

�

2

4 62 80

–2

–8

–4–6 –2–8

–4

–6

4

6

8

Im

Re

5 (i)

2

420

–2

–4 –2

–4

4

Im

Re

Minimum value of is 13 2Maximum value of is 13 2

−+

zz

(ii)

2

420–4 –2

–4

Im

Re

4

–2

Minimum value of is 3 2

Maximum value of is 3 2

−

+

z

z

6

2

420

–2

–4 –2

–4

4

Im

Re

2 2i

Minimum value of is 1 1 22 2

( )= − +

+ =

z z

z

5.5 The modulus−argument form of complex numbers

(Page 74)

1 (i) 2i = 2 cos 2 isin 2( )− − π + − π

(ii) 4 3i 5(cos 0.644 isin 0.644)− = − + −

(iii) 1 3 i 2 cos 3 isin 3( )+ = π + π

(iv) 8 8 cos0 isin0( )= +

2 (i) 2 cos45 isin45 = 2 2i( )° + ° +

(ii) cos 3 isin 312

32 i( ) ( )− π + − π = −

(iii) 0.5 cos 90 isin 90 0.5 i( )− ° + − ° = −

(iv) 2(cos135 isin135 ) 1 i° + ° = − +

3

2

3

1

4

–2

–1

–3

–4

2 31 4–2–3–4 –1

Im

ReA

BC

O

OABC is a parallelogram.

(ii) uu*

= +−

= +− × +

+

= + +−

= +

= +

2 i2 i2 i2 i

2 i2 i

4 4i i4 i

3 4i5

35

45 i

2

2

(iii) uu u u*

*

( ) ( )( ) ( )

= −

= − −

=

− − −

− −

arg arg arg

tan4535

tan 12 tan 1

2

tan 43 2tan 1

2

1 1 1

1 1

9781510458444_Answer.indb 57 11/8/18 10:40 AM

58

5Complexnumbers


5.6 Sets of points using the polar form (Page 76)

1 (i)

2

420

–2

–4 –2

–4

4

Im

Re

(ii)

2

420

–2

–4 –2

–4

4

Im

Re

(iii)

2

420

–2

–4 –2

–4

4

Im

Re

(iv)

2

420

–2

–4 –2

–4

4

Im

Re

2 (i) 4 2i 2− − =z

(ii) arg 4 2i 0z( )− − =

(iii) 4 2cos 4 4 2= − π = −a

2 2sin 4 2 2

is 4 2, 2 2( )= + π = +

− +

b

P

(iv) 0 arg 4 2i 34 and 4 2i 2( )< − − < π − − <z z

3

23

1

421 3 50

–2–1

–3

–5

–4 –2–3–5 –1

–4

45

Im

Re

67

6 7–6–7

–6–7

4 (i)

2

3

4

1

21 30

–2

–1

–3

–2–3 –1

Im

Re4–4

–4

(ii) cos 4 2 2π = ⇒ =x x

5 (i)

2

3

1

421 3 50

–2

–1

–3

–5

–4 –2–3–5 –1

–4

4

5Im

Re

(ii)(a) Maximum value of z is 10 2+

Minimum value of z is 10 2−

(b) Maximum value of − 3z is 3

Minimum value of − 3z is 1

(c) sin 210

39.23θ θ= ⇒ = °

tan 13 18.43α α= ⇒ = °

Maximum value of arg(z) is

39.23° − 18.43° = 20.8°

Minimum value of arg(z) is

−(39.23° + 18.43°) = −57.7°

9781510458444_Answer.indb 58 11/8/18 10:40 AM

59

5Complexnumbers


5.7 Working with complex numbers in polar form

(Page 80)

1 4 cos 30 , 2cos60 , cos( 90 )= ° = ° = − °u v w

(i) u* [ ]= − ° + − °4 cos( 30 ) isin( 30 )

(ii) v w [ ]× = − ° + − °2 cos( 30 ) isin( 30 )

(iii) u w*

[ ] [ ][ ]

×= ° + ° × ° + °= ° + °

4 cos30 isin30 cos90 isin904 cos120 isin120

(iv) 12 cos30 isin30[ ]= ° + °v

u

(v) uv*

[ ]

[ ]

=° + °

− ° + − °= ° + °

4 cos30 isin302cos( 60 ) isin( 60 )2 cos 90 isin90

(vi) 1 1 cos0 isin02 cos60 isin60

[ ][ ]=

° + °° + °v

(vii) cos ) i sin( 90cos ) i sin( 3601

4w = (−90° + − °)= (−360° + − °)=

(viii) 2i(cos i sin60 )4(cos i sin30 )12(cos i sin30 )

12(icos sin30 )

ivu = 60° + °

30° + °

= 30° + °

= 30° − °

2 (i) T

(iii) T

(v) T

(ii) F

(iv) T

(vi) T

3 i i

i i

( )( ) ( )( )

= − − − π + − π

= π + π =

α

β

3 = 2 cos 56 sin 5

6

4 cos 12 sin 1

2 4

(i)

= − πα

α

= 2

arg 56

(ii) =

= π

β

β

4

arg 12

(iii)

i i

i

( )( )

( ) ( ) ( )( ) ( )− π + − π × π + π

= − π + − π

=

= − π

αβ

αβ

αβ

= 2 cos 56 sin 5

6 4 cos 12 sin 1

2

8 cos 13 sin 1

3

8

arg 13

(iv)

( )( )

( )

( ) ( )( ) ( )

( ) ( )( )=

− π + − π

π + π

= − π + − π

= π + π

=

αβ

αβ

2 cos 56 isin 5

6

4 cos 12 isin 1

212 cos 4

3 isin 43

12 cos 2

3 isin 23

12

( ) = πα

βarg 23

–1–1

–2

–3

–4

–5

–6

–7

–2–3 1

1

2

2

4

3

3(i)

(ii)

(iii)

(iv)

4

5

Im

Re

5

6

6

7

7

–4–5–6–7 0

4 Let iz x y= +

z x yx yx y

x yx yx y

x yzz*

= + ×−−

=−

−

=−+

=

1 1i

ii

iii

2 2 2

2 2

2

5 5 12i2z = −

(i) z [ ]= − + − − °13 cos( 1.176) isin( 1.176) or 13 cis( 67.4 )2

(ii) z

z

[ ]= − = − + −

= − + −

5 12i 13 cos( 1.176) isin( 1.176)

(13 cos( 1.176) isin( 1.176))

2

12

(iii) z ( )= − + −13 [cos 0.588 isin( 0.588)]

(iv) z = +13cos2.554 isin2.554

(v) 3 2i or 3 2iz = − + −

9781510458444_Answer.indb 59 11/8/18 10:40 AM

60

5Complexnumbers


6 = 32 i+u a

(i) 32 i

2 i2 i

6 3 i4 i6 3 i4

64

34

i

2 2

2

2 2

u aaa

aaa

a

aaa

= + × −−

= −−

= −+

=+

−+

(ii)(a) ua

aa

ua a

a

a

a

a

a

a

a

*

*

( )( )

( )

=+

++

+

+

=

=

= π

= π

=

=

−

−

−

−

64

34

i

arg = tan3464

tan 36

tan 2

tan 2 4

2 tan 4

2 1

2

2 2

12

2

1

1

1

(b) ua

aa

aa

aa

a a

a a a

a a a

a a

a a

( ) ( )=+

++

= ++

++

=

+ = +

+ = + +

= +

= + −

= − +

64

34

36 9(4 )

36 9(4 )

2

36 9 2(4 )

36 9 2(16 8 )

36 + 9 32 + 16 2

0 2 7 4

0 (2 1)( 4)

2

2

2

2

2

2 2

2

2 2

2 2 2

2 2 4

2 2 4

4 2

2 2

2 1 0 or 4 012 (no solution)

So 12

2 2

2

a a

a

a

⇒ − = + =

=

= ±

5.8 Complex exponents (Page 83)

1 (i) π

2e 2 i

(ii) − + =π

1 i 2 e34 i

(iii) 2 3i 13 e 0.983i− = −

(iv) − =− π

5i 5e 2 i

2 (i) = +π

e 12

32 i3 i

(ii) 2e 1.08 1.68ii = −−

(iii) = −π4e 4i

(iv) −−π

3e = 3 32

32i6 i

–1–2–3 1 2 43 5–4–5 0–1

–2

–3

–4

–5

1

2

3

4

5Im

Re

(i)

(ii)

(iii)

(iv)

3 (i) ( )( ) ( )× π + π

= ×

=

π

π π

π

4e 3 cos 6 isin 6

4e 3e

12e

3 i

3 i 6 i

2 i

(ii)(a) 2i + 3 i = 3 i− + −

(b) ( ) ( )− = −π + −π

3 i 2 cos 6 isin 6

5.9 Complex numbers and equations (Page 85)

1 z z z kz

− + + == +

−

2 4 01 2i is a root

So 1 2i is a root.

3 2

1 2i 1 2i 2 0

1 2i 1 2i 2 0

(( 1) 4i ) 2 0

( 2 5)(2 ) 0

2 4 2 10 5 0

2 4 2 10 5 0

2 2

2

3 2 2

3 2

z z z c

z z z c

z z c

z z z c

z cz z cz z c

z c z c z c

( )( )( ) ( ) ( )( )( )( )

( )

( ) ( )

− + − − − =− − − + − =

− − − =

− + − =

− − + + − =

+ − − + + − =

4 1 32 10 4 3

5 3 15

c cc c

k

− − = − ⇒ = −+ = ⇒ = −

= − × − =

9781510458444_Answer.indb 60 11/8/18 10:41 AM

61

5Complexnumbers


2 10 37 0

6 10 6 37 6 078

3 2

3 2

z z z p

pp

− + + =

− × + × + == −

10 37 786 4 13

4 13 0 = 4 (4) 4 1 132

= 4 6i2

= 2 3iThe other roots are 2 + 3i and 2 3i

3 22

22

z z zz z z

z z z

⇒ − + −− = − +

− + = ⇒ ± − × ×

±

±−

3 (i) i 2i 0

1 ( 1) 4 i 2i2i

1 1 8i2i

1 32i

42i or 2

2i2i or 1

iii

2i or i

2

2

2

z z

z

− + =

= ± − − × ×

= ± −

= ±

= −

= − ×= −

(ii) The coefficients of the polynomial are not real numbers.

4 (i) 16 0( 4)( 4) 0

4 42 = 4

4i2i

2, 2, 2i, 2i

4

2 2

2 2

2

zz z

z zz z

z

− =− + =

= = −= ± ± −

= ±= ±

⇒ = − −

(ii)

23

1

21 30

–2–1

–3

–2–3 –1

Im

Re

The solutions are all 90° apart.

(iii)

(1 i) (1 i) (1 i)

=(1 2i i )(1 2i i )= 2i 2i

= 4i4

4

4 2 2

2 2

2

z k=

+ = + +

+ + + +×

= −

4Other solutions are 1 i, 1 i, 1 i

k⇒ = −− + − − −

5 04 3 2+ + + + =x Ax Bx Cx D

(i) Roots 2 i , 2 i, 2i, 2i+ − −

(ii) ( (2 i))( (2 i))( ( 2i))( 2i) 0( 2 i)( 2 i)( 2i)( 2i) 0

(( 2) i )( 4i ) 0

( 4 5)( 4) = 0

4 4 16 5 20 0

4 9 16 20 0

2 2 2

2 2

4 2 3 2

4 3 2

x x x xx x x x

x x

x x x

x x x x x

x x x x

− + − − − − − =− − − + + − =

− − − =

− + +

+ − − + + =

− + − + = 4, 9, 16, 20A B C D= − = = − =

6 (i) 3 3 7 3 1527 9 21 150

So 3 is a root.

3 2

z

+ − × −= + − −=

=

By long division or synthetic division,

3 1 1 −7 153 12 15

1 4 5 0

the quotient is 4 52 + +x x .

4 4 4 1 52 1

4 42

4 4i2

4 2i2

2 i

2

2

= − ± − × ××

= − ± −

= − ±

= − ±

= − ±

x

(ii) Im

0–2 –1 1 2 Re3

–1.0

–0.5

0.5

1.0

7

n

a b

1324 24

1224 2

So the roots are 2 apart.

2

24

cos 24 isin 24 cos 4 24 isin 4 24cos 6 isin 6

32

12 i

32 , 1

2

4( ) ( ) ( )

π − π = π = π

π

⇒ = ππ =

π + π = × π + × π

= π + π

= +

= =

9781510458444_Answer.indb 61 11/8/18 10:41 AM

62

5Complexnumbers



1 (i) (a) 3 (1 2i) 3(3 i)1 2i 9 3i

8 i

z w− = − − −= − − += − +

(b) (1 2i)(3 i)

3 i 6i 2i3 7i 21 7i

2

zw = − −

= − − += − −= −

(c) 1 13 i

3 i3 i

3 i9 i3 i103

101

10i

2

w = − × ++

= +−

= +

= +

(d) 3 i1 2i

1 2i1 2i

3 6i i 2i1 4i

1 7i5

15

75 i

2

2

wz* = −

+ × −−

= − − +−

= −

= −

(ii)(a) 1 2i3 i

1 4i 4i3 i

3 4i3 i

3 i3 i

9 3i 12i 4i9 i

5 15i10

12

32i

2 2

2

2

2

u zw

( )= = −−

= − +−

= − −− × +

+

= − − − −−

= − −

= − −

(b)

2

3

1

4

–2

–1

–3

–4

2 31 4–2–3–4 –1

Im

Re0

2 6iu w− = 13( 6i) 13

6i 13 02

uwu u

u u

=− =

− − =

6i ( 6i) 4 1 132

6i 36i 522

6i 162

6i 42

3i 2

2

2

u = ± − − × × −

= ± +

= ±

= ±

= ±

2 3i, 2 3iand 2 3i, 2 3i

u wu w

⇒ = + = −= − + = − −

3 (i),(ii)and(iii)(a)

2

420

–2–4 –2

–4

4

Im

Re

(iii)(b) sin 23 41.8θ θ= ⇒ = °

90 41.8 48.2α⇒ = ° − ° = °

3i2

3

θα

4 1 2iu = +

(i) (1 2i) 1 4i 4i 3 4i

(1 2i) (1 2i)( 3 4i)

( 3 2i 8i ) 11 2i

(1 2i) ( 3 4i)( 3 4i)

9 24i 16i 7 24i 7 24i3( 11 2i) 5( 3 4i) (1 2i) 107 24i 33 6i 15 20i 1 2i 10

0

2 2

3

2

4

2

+ = + + = − +

+ = + − +

= − − + = − −

+ = − + − +

= − + = − − − −− − − + − + − + −

= − − + + − + − − −=

So 1 2i is a root.Also 1 2i is a root.

u = +−

9781510458444_Answer.indb 62 11/8/18 10:41 AM

63

5Complexnumbers


(ii) − + − −x x( (1 2i))( (1 2i))

x xx

x xx x

x x x x

x x x x xx x

x xx x x x x

x x x x

= − − − += − −= − + −= − +⇒ = − +

= − + − −− +

= − −⇒ = − + − −

= − + + −

( 1 2i)( 1 2i)( 1) (2i)

2 1 4i2 5

p( ) ( 2 5)Q( )

Q( ) 3 5 102 5

2p( ) ( 2 5)( 2)

( 2 5)( 1)( 2)

2 2

2 2

2

2

4 3 2

2

2

2 2

2

The other two roots are x = −1 and 2

5 (i) If one of the roots is z = 3, then p(3) = 0

= + + −= + + × −= + + −=

p( ) 4 48p(3) 3 3 4 3 48

27 9 12 480

3 2

3 2z z z z

So the real root is 3. Using synthetic or long division,

β γ

− = + +

+ + =

= − ± − × ××

= − ± −

= − ±

= − ±

= − ±= − + = − −

zz z z

z z

z

p( )3 4 16

Solving 4 16 0,

4 4 4 1 162 1

4 482

4 48i2

4 4 3i2

2 2 3iSo 2 2 3i and 2 2 3i

2

2

2

2

(ii) and (iii)

0

2

3

1

4

5

–2

–1

–3

–4

–5

2 31 4 5–2–3–4–5 –1

Im

Re

6 (i) wwwww

w

i ( 3 3i)i 9 18i 9ii 18i

1818i

18i

2 2

2 2

2

2

2 2

= − += − += −= −== ±

(ii)(a) − −(3 3i) 2z �

23

1

421 3 50

–2–1

–3

–5

–4 –2–3–5 –1

–4

45

Im

Re

(b)

θ γ

2

2

18

3 3 182 2+ = 18 2, 18 2p q= − = +

sin 218

28.1θ θ= ⇒ = °

z

z� �

� �

γ γ

γ θ( ) ( )( )

= ⇒ = °

⇒ = − − = − ° − ° = − °= − ° + ° = − °

− +− ° − °

βα

tan 33 45

45 28.1 16.945 28.1 73.1

So 18 2 18 2and 73.1 arg 16.9

7 p( ) 273 2z z z kz= − + −

(i) zkk

kk

− ⇒ =− + − =− + − =

3 is a factor p(3) 03 3 3 27 0

27 9 3 27 03 = 9

= 3

3 2

(ii)(a) By long division or synthetic division,

3 1 −1 3 −273 6 27

1 2 9 0

⇒ z z z z

z

z

p( ) ( 3)( 2 9) 0

2 2 4 1 92

2 32i2

1 2 2i

So 3, 1 2 2i, 1 2 2i

2

2

2

= − + + =

= − ± − × ×

= − ±

= − ±= − + − −

9781510458444_Answer.indb 63 11/8/18 10:41 AM

64

5Complexnumbers


(b) p( ) 0

3 or 1 2 2i or 1 2 2 i

3 ( i) 1 2 2i ( i) 1 2 2i

1 12 2 2 2 2 2

2 2

2 1 2 1

2 1 2 1

2 So 1 and 2

2 0

( 2)( 1) 0

2

2 2 2

2 2

2 2 2 2

22

22

22

22

4 2

4 2

2 2

z

z z z

z a b a b

a b a bab ab

b a b a

a a a a

aa

aa

a a a b

a a

a a

( ) ( )

=

⇒ = = − + = − −

= ± + = − + + = − −

− = − − = −= = −

= = −

⇒ − = − ⇒ − − = −

− = − − = −

− = − = ± =

+ − =

+ − =

aa b= − ⇒

= ± = ±− + − − − − +

2 no solutionSo 1 and 2So the roots are 3, 3, 1 2 i, 1 2 i, 1 2 i, 1 2 i

2


1 (i) 2 2i= +u

2 2 8 2 2

arg π4

2 2= + = =

=

u

u

(ii)

23

1

21 30

–2–1

–3

–2–3 –1

Im

Re

4

4

u

–4

–4

(iii) 1 8

7

2 2 2a

a( )+ =

=

2 (i) 1 i 2 cos 34 i sin 3

4w ( )= − + = π + π

2 cos34 isin3

4 2 cos 32 isin3

2

2 cos 32 isin3

2

2 cos34 isin3

4 2 cos 94 isin9

4

2 2 cos 4 isin 4

22 2

33 3

( )

( )

( ) ( )( )

( ) ( )( )

= π + π = π + π

= π + π

= π + π = π + π

= π + π

w

w

⇒ 2 arg 22 2 arg 4

2 2

3 3

w w

w w

= = − π

= = π

(ii)

1

2

3

–1

–2

–3

1 2 3–1–2–3

Im

Re

w

w2

0

w w

z

( ) ( )

( )

= − + − + = − −

= + =

⇒ =

− − − =

midpoint 1 02 , 2 1

212 , 1

2

to 1 3 10

radius 102

Equation of the circle is 12

12i 10

2

2 2 2

9781510458444_Answer.indb 64 11/8/18 10:41 AM

65

5Complexnumbers


3 (i) i 2 3i 02 + − =z z

2 2 4i( 3i)2i

2 4 12i2i

2 82i

2 8i2i

2 8i2i

1i

82

1i

ii

2 22

i 2

2

2

2

z = − ± − −

= − ± +

= − ± −

= − ±

= − ±

= − ±

= − × ±

= ±

So 2 i or 2 iz = + − +

(ii)(a) 4 3iz z= − −

(4 3i)z z= − +

which is the perpendicular bisector between (0, 0) and (4, 3)

2

1

4

5

3

6

–2

–1

–3

–5

–4

–6

2 31 4 65–2 –1–4 –3–6 –5

Im

Re

4 + 3i

0

(b) The point where z is least is the point W on the diagram.

2

1

4

5

3

6

–2

–1

–3

–5

–4

–6

2 31 4 65–2 –1–4 –3–6 –5

Im

Re

4 + 3i

W

0

W is the midpoint between (0, 0) and (4, 3)

Hence W is (2, 1.5)

2 1.5 2.5

arg( ) tan 1.52 36.9 or 0.64

2 2

1

w

w ( )= + =

= = °−

4 ( 2) ( 6)i= −z

(i) = + = ≈2 6 8 2.832 2z

z = −

= − ° − π−arg( ) tan 6

260 or 3

1

(ii)(a) z z* ( )+ = − + +

= +

2 ( 2) ( 6)i 2 ( 2) ( 6)i

(3 2) ( 6)i

(b) zz*

( )= +−

= +−

= ++

× −−

= + −−

= +

= +

= +

i( 2) ( 6)i

i ( 2) ( 6)i

( 2) ( 6)i( 2)i ( 6)i( 2) ( 6)i( 6) ( 2)i

( 6) ( 2)i( 6) ( 2)i

( 12) 4i 12i6 2i

(2 12) 4i8

(4 3) 4i8

32

12i

2

2

2

(iii) z z* = + = +( 2) ( 6)i, i ( 6) ( 2)i

1

A

B2

3

3

–1

–1

Im(z)

Re(z)0 1 2

A Bz z*

∠ = −= −

= −

= −

= π − π

= π

− −

− −

AOB arg argarg argi

tan 62

tan 26

tan 3 tan 13

3 6

6

1 1

1 1

9781510458444_Answer.indb 65 11/8/18 10:41 AM

66

5Complexnumbers


5 (i) u u*= − = +2 i, 2 i is also a root

3 0

(2 i) (2 i) 0

2 i) 2 i) 0

( 2 i 2 4 2i i 2i i )( ) 0

( 4 5)( ) 0

3 2

2 2

2

x ax x b

x x x k

x x x k

x x x x x x k

x x x k

( )( )( )( )

( )( )

+ − + =− − − + − =

− + − − − =

− − − + + + − − − =

− + − =

Looking at the term,4 5 3 2

So ( 4 5)( 2) 02, 10

2

xk k

x x xa b

+ = − ⇒ = −

− + + == − =

(ii) − < ⇒ − − <1 (2 i) 1z u z

⇒ inside of a circle, centre 2 − i, radius 1

z z z z< + ⇒ < − −i ( i)

⇒ all points closer to (0, 0) than −i

–1

1

1 3 4

–1

–2

–3

–2

Im

Re0 2


1 e 3i =+x y

y y

y y

y y y k k

yy y

y y

x

x y

x

x x

x x

x

x

x

× =

+ =

+ =

= = ⇒ = π ∈

=• = − π − π π π π = −

⇒ − =• = − π − π π π π =

⇒ = ⇒ =

e e 3

e (cos isin ) 3

e cos ie sin 3

e cos 3 and e sin 0 ,

Solving e cos 3,If ... 3 , , , 3 , 5 ... then cos 1

e 3 which has no solutionsIf 4 , 2 , 0, 2 , 4 , 6 ... then cos 1

e 3 ln3

i

�

2 (i) zz

z z

i in n n nn n n n

n

nn

n n

n nθ θ θ θθ θ θ θθ θ θ θ

θ

+ = +

= + + += + + − + −= + + −=

−

−

1

(cos sin ) (cos sin )cos isin cos ( ) i sin( )cos isin cos isin2 cos

1

cos i sin (cos i sin )2 i sin ( )

zz

z z

n n n nn

nn

n n

θ θ θ θθ

− = −

= + − −=

−

(ii) z z z z z z z z z zz z

z zz

zz

zθ θ θ

θ θ θ

θ θ θ

θ θ

( )

( ) ( ) ( ) ( )+ = + + + +

= + + + +

= + + + +

= += +

= +

= +

1 4 1 6 1 4 1 1

4 6 4 1

1 4 1 6

(2cos ) 2 cos 4 + 4(2cos2 ) 616 cos 2 cos 4 + 8cos 2 6

cos 18 cos 4 +1

2cos 2 38

18(cos 4 +4cos 2 3)

44 3 2

2 3

4

4 22 4

44

22

4

4

4

(iii)(a) 1

i 12i sin

i 2cos

2sin2costan

z zz z

θθ

θθ

θ

( ) ( )−

+=

=

=

(b)

z zz zz zz z

zz

zz

zz

zz

θθ

θθθθ

θ θθ

θ θθ

θ

θ

( )( )

( )( )

−+

=− −

+

+ −+

=

−+ −

+ +

++ −

+ +

=+ −

+− −

+

=+ + −

++ − −

+

=

=

−

−

−

−

1 tan1 tan

1i( )

1i( )

1

1 2

i ( 1 2)

1

1 2

i ( 1 2)

1 2cos2 22cos2 2

1 2cos2 22cos2 2

2cos2 2 2cos2 22cos2 2

2cos2 2 (2cos2 2)2cos2 2

4cos24

cos 2

2

2

1

1

2

1

1

2

22

2 22

22

2 22

3 (i) 3 i 2 cos π6 isin π

6

2 cos π6 isin π

6

2 cos 12 π6 isin 12 π

64096 cos ( 2π) i sin( 2π)4096

1212

12

w

w

( ) ( )( ) ( )( ) ( )

[ ]

= − = − + −

= − + −

= × − + × −

= − + −=

9781510458444_Answer.indb 66 11/8/18 10:41 AM

67

5Complexnumbers


(ii) 1

32 i 132 cos 2 i sin 21

32 i 132 cos 2 i sin 2

132 cos 2 i sin 2

132 cos 1

5 2 isin 15 2

12 cos10 isin10

5

15

15

z

z ( )

( )( )

( )( ) ( ) ( )

( )

= π + π

= = π + π

= π + π

= × π + × π

= π + π

The other four solutions are 25 apart

12 cos10 isin10 , 1

2 cos 2 i sin 2 ,

12 cos 9

10 isin 910 ,

12 cos 3

10 isin 310 , 1

2 cos 710 isin 7

10( ) ( )

( ) ( )( )

( ) ( ) ( ) ( )

π

⇒ π + π π + π

π + π

− π + − π − π + − π

(iii)(a) (cos +isin )

cos 5cos (isin ) 10cos (isin )

+ 10 cos (isin ) 5cos (isin ) + (isin )

cos 5icos sin 10cos sin

10icos sin 5cos sin isin

5

5 4 3 2

2 3 4 5

5 4 3 2

2 3 4 5

θ θ

θ θ θ θ θ

θ θ θ θ θ

θ θ θ θ θ

θ θ θ θ θ

= + +

+

= + −

− + +

(b)(cos isin ) = cos5 isin 5sin5 5cos sin 10cos sin + sin

5 (1 sin ) sin 10(1 sin ) sin + sin5 (1 2sin + sin )sin 10sin 10sin sin5sin 10sin 5sin 10sin 10sin sin16sin 20sin 5sin

5

4 2 3 5

2 2 2 3 5

2 4 3 5 5

3 5 3 5 5

5 3

θ θ θ θθ θ θ θ θ θ

θ θ θ θ θθ θ θ θ θ θ

θ θ θ θ θ θθ θ θ

+ +⇒ = −= − − −= − − + += − + − + += − +

(c)16 20 5 1 0Let sin

5 3x x xx θ

− + − ==

16sin 20sin 5sin 1sin5 1

5 sin (1)

5 3

1

θ θ θθ

θ

− + ==

= −

θ

θ

= π π π − π − π

= π π π − π − π

5 2 , 52 , 9

2 , 72 , 3

2

10 , 510 , 9

10 , 710 , 3

10

x ( ) ( )⇒ = π π π − π − πsin10 , sin 2 , sin 910 , sin 7

10 , sin 310

4 (i) ( ) ( )( )

( )( ) ( ) ( )

Γ Γ − = ππ

Γ

= π

Γ = π

Γ = Γ = × Γ = π

12 1 1

2 sin 12

12

1252

32

32

32

12

12

34

2

(ii) The equation has real coefficients, so 2 3i= −c

is also a root. Since all the terms are even powers of x, so are

2 3i− = − −c and 2 3i− = − +c

These roots give factors of the polynomial

( 2 3i)( 2 3i)( 2 3i)( 2 3i)

( 2 2 5)( 2 2 5)

2 25

2 2

4 2

x x xx

x x x x

x x

− − − + + ++ −

= − + + +

= + + By observation, we find that

( 2 25)( ) 04 2 2x x x k+ + − =

is a factorisation of the original equation.

The roots are , , , andc c c k k− − − .

It is possible to find the last two roots by inspection, by substituting 2 =x k .

(iii) Some quick geometry gives that the distance between two roots is 2 and the centre is at

a b ( ) ( )= + = + + +C i 1 12

1 12

i

Shifting the points to be centred on the origin, the new points are the roots of 08 8+ =x r .

r ( ) ( )= + + = + + +

= +

12

1 12

12 1 2 1

2

2 2

2 2

From the diagram we find

2 2 08 4x ( )+ + =

Now shift the centre from the origin to

( ) ( )= + + +C 1 12

1 12

i

z z a b r

z( )( ) ( ) ( )

= − + +

= − + − + + +

p( ) ( ( i))

1 12

1 12

i 2 2

8 8

8 4

In the original question, as well as the obvious n = 8, we have

a b ( ) ( ) ( )( )+ = + + + = + +i 1 12

1 12

i 1 i 1 12

and

2 2 68 48 24( )= + = +q

(cos isin ) = cos5 isin 5sin5 5cos sin 10cos sin + sin

5 (1 sin ) sin 10(1 sin ) sin + sin5 (1 2sin + sin )sin 10sin 10sin sin5sin 10sin 5sin 10sin 10sin sin16sin 20sin 5sin

5

4 2 3 5

2 2 2 3 5

2 4 3 5 5

3 5 3 5 5

5 3

θ θ θ θθ θ θ θ θ θ

θ θ θ θ θθ θ θ θ θ θ

θ θ θ θ θ θθ θ θ

+ +⇒ = −= − − −= − − + += − + − + += − +

9781510458444_Answer.indb 67 11/8/18 10:41 AM

68

5Complexnumbers


5 1 05 − =z ,

( 1)( 1) 04 3 2z z z z z− + + + + =

Roots of 1 05 − =z are

cos 25 i sin 2

5 , 0, 1, 2, 3, 4z k k k( ) ( )= π + π

=

and 0 gives 1, so= =k z

14 3 2+ + + +z z z z

cos 25 i sin 2

5 cos 45 isin 4

5 cos 65 isin 6

5 cos 85 isin 8

5

cos 25 isin 2

5 cos 45 isin 4

5 cos 45 isin 4

5 cos 25 isin 2

5

z z z z

z z z z

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

= − π + π

− π + π

− π + π

− π + π

= − π π

− π + π

− − π + − π

− − π + − π

cos 25 i sin 2

5 cos 25 isin 2

5 cos 25 isin 2

5 cos 25 isin 2

5

cos 45 isin 4

5 cos 45 isin 4

5 cos 45 isin 4

5 c s 45 i sin 4

5

2

2

z z

z z o

( )( )( )( )

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

− π + π

+ − π + − π

+ π + π

− π − π

= − π π

+ − π + − π

+ π + π

− π + − π

But

cos 25 i sin 2

5 cos 25 isin 2

5 2cos 25 and

cos 45 isin 4

5 cos 45 isin 4

5 2cos 45

( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

π + π

+ − π + − π

= π

π + π

+ − π + − π

= π

Also

cos 25 i sin 2

5 cos 25 isin 2

5 cos(0) 1

and cos 45 isin 4

5 cos 45 isin 4

5 1

So

14 3 2z z z z

( ) ( ) ( ) ( )( ) ( ) ( ) ( )

π + π

− π + − π

= =

π + π

− π + − π

=

+ + + + =

z z z z( )( )( ) ( )− π + − π +2cos 25 1 2cos 4

5 12 2

Comparing coefficients of z2,

1 1 1 4cos 2

5 cos 45 and

cos 25 cos 4

514

( ) ( )( ) ( )

= + + π π

π π = −

6 (i) z a a b a a b

a b a

z a a ba b

( ) ( )( )

= + + − − + +

+ + −

= + + = +

12

12

2i 14

2 2 2i 2 i

2 2 2 2 2

2 2 2

2

(ii)(a) For 1 i2 ,

f (0) 1 i2 1 i

274

3i2

2 22

c

c c ( ) ( )= +

= + = + + +

= +

= + = >

= +

f (0) 14( 49 36) 85

4 2,

so 1 i2 is not part of the set.

2

z

9781510458444_Answer.indb 68 11/8/18 10:41 AM

69

5Complexnumbers


( )=

= + = + = − +

= <

c

c c

For i,

f (0) i i 1 i

and f (0) 2 2, so OK so far.

2 2 2

2

= − + + = − + = −

= <

f (0) ( 1 i) i 2i i 1

and f (0) 1 2, so OK so far.

3 2

3

( )= − + = − +f (0) i i 1 i4 2

and so the process will loop continually, with

<f (0) 2n for all n, and i=z is part of the set.

(b) c c c a b c

c a a a a

a a a

a b a

a a a a( )

( ) ( )( )

( )

( ) ( )

= = + = + +

= + + − +

= + = +

= + + +

= + + +

f (0) , f (0) i12 4 i 1

2 4

12 3 i 1

2 2 3 i

f (0) i 2 3 i

32 i 3 2

2 2

2 2

2

a ( )( )

= = + + +

= + + +

and when 18 , f (0) 1

83

16 i 38

14

1 2 38 i 3 2

8

2

= + + +

= + + +

= +

= +

So f (0) 18 (1 2 3) ( 3 2)

18 (13 4 3) (7 4 3)

18 20 8 3

14 5 2 3

2 2 2

9781510458444_Answer.indb 69 11/8/18 10:42 AM

Pure Mathematics 3 Question & Workbook answers

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