CAPE Pure Mathematics Unit 2

Practice Questions

By Carlon R.Baird

MODULE 1: COMPLEX NUMBERS AND CALCULUS II

1. (a) Use de Moivre’s theorem to prove the trigonometric identity:

7 5 3cos7 64cos 112cos 56cos 7cos

(b) Use de Moivre’s theorem to evaluate 8

1 i

(c) Express

2cos3 sin3

cos sin

q i q

q i q

in the form cos sinkq i kq where k is an integer to be

determined.

2. If | 6 | 2 | 6 9 |z z i ,

(a) Use an algebraic method to show that the locus of z is a circle, stating its centre and its

radius.

(b) Sketch the locus of z on an Argand diagram.

3. Find dy

dxin terms of x and y where

3 3 23 6 4x x y y x

4. (a) Find the derivative of the function

1 2ln( )

( ) cot( ) sin( )cos( ) cos ( ) 9ln(2 )

xh x x x x x x

x

(b) The curve C has equation 2 cos( )xy e x

i. Show that the stationary points on C occur when tan( ) 2x

ii. Find an equation of the tangent to C at the point where x=0

5. (a) Given that8( , , ) 4 cos( ) sin(4 ) tan 0zf x y z xyz xy x e xz y

i. Determine xf , yf , zf

ii. Determine xyf , yxf , yzf

(b) Given that 2 42 4 18x

p xv v xv

i. Determine p

v

and

p

x

ii. Determine 2 p

x v

and

2 p

v x

6. (a) Integrate with respect to x

i. 2

10

1

x

x

ii. 2

15

1

x

x

iii. 2

2 8

1

x

x

(b) (i) Express the function 4 3 2

3 2

4 9 17 12( )

4 4

x x x xh x

x x x

as partial fractions

(ii) Hence, evaluate

4 4 3 2

3 2

3

4 9 17 12

4 4

x x x xdx

x x x

(c) Determine 1

2

1tan

1x dx

x

7. Using the substitution secx ,find 2

2

1 1

1

xdx

xx x

8. (a) Show that 4 31 (1 )x x x

(b) Given that

1

3

0

(1 )n

nI x x dx , show that 1

3

3 2n n

nI I

n

(c) Use your reduction formula to evaluate 4I .

9. Given that sin(2 1)

sin( )m

m xJ dx

x

,

(a) Show that 1

sin 2m m

mxJ J

m

(b) Hence find 5J .

10. Use the trapezium rule using 4 strips to estimate 3

0

1 tan( ) x dx

giving your answer to

3 significant figures.

By Carlon R. Baird

1. (a) First let’s consider 7(cos sin )i

Now, by de Moivre’s theorem

7

7

(cos sin ) cos7 sin7

cos7 sin7 (cos sin )

Using binomial expansion:

i i

i i

7 7 6 7 5 2 7 4 3

1 2 3

7 3 4 7 2 5 7 6 7

4 5 6

7 6 5 2 2 4 3 3

cos7 sin7 cos (cos )( sin ) (cos )( sin ) (cos )( sin )

(cos )( sin ) (cos )( sin ) (cos )( sin ) ( sin )

cos 7(cos )( sin ) 21(cos )( sin ) 35(cos )( sin )

i C i C i C i

C i C i C i i

i i i

3 4 4 2 5 5 6 6 7 7

7 6 5 2 4 3

3 4 2 5 6 7

35(cos )( sin ) 21(cos )( sin ) 7(cos )( sin ) sin

cos 7cos sin 21cos sin 35cos sin

35cos sin 21cos sin 7cos sin sin

i i i i

i i

i i

Now equating real parts:

7 5 2 3 4 6

7 5 2 3 2 2 2 3

7 5 7 3 2 4

3 3 0 3 2 3 1

0 1 2

cos7 cos 21cos sin 35cos sin 7cos sin

cos 21cos (1 cos ) 35cos (1 cos ) 7cos (1 cos )

cos 21cos 21cos 35cos 1 2cos cos

7cos (1) ( cos ) (1) ( cos ) (1) ( cC C C

2 3 0 3

3

7 5 7 3 5 7

2 4 6

7 5 7 3 5 7 3

5 7

7 7 7

os ) (1) ( cos )

cos 21cos 21cos 35cos 70cos 35cos

7cos 1 3cos 3cos cos

cos 21cos 21cos 35cos 70cos 35cos 7cos 21cos

21cos 7cos

cos 21cos 35cos

C

7 5 5 5 3

3

7cos 21cos 70cos 21cos 35cos

21cos 7cos

7 5 3cos7 64cos 112cos 56cos 7cos

2(cos3 sin3 )cos(6 ) sin(6 )

cos sin

cos7 sin7

q i qq q i q q

q i q

q i q

(b)

8

2 2

1

Let ( 1 )

Let 1

( 1) (1) 2

tan 1

tan (1)4

z i

p i

r p

arg

4

3

4

p

8

8

8

Rewriting in polar form: (cos sin )

3 32 cos( ) sin( )

4 4

3 32 cos( ) sin( )

4 4

Now applying de Moivre's theorem:

3 3( 2) (cos(8 ) sin(8 ))

4 4

24 2416(cos( ) sin( )

4 4

p p r i

p i

z p

z i

z i

z i

)

16(cos(6 ) sin(6 ))

16(1 (0))

16.

z i

z i

z

(c)

2(cos3 sin3 ) cos(2(3) ) sin(2(3) )

cos sin cos( ) sin( )

cos6 sin6

cos( ) sin( )

q i q q i q

q i q q i q

q i q

q i q

Recall that 1 11 2 1 2

2 2

(cos( ) sin( ))z r

iz r

α

Im z

Re z

arg p

1

1

Recall that cos( ) cos

and sin( ) sin( )

C(-10,12)

O

12

-10

y

x

7k

2. (a)

2 2 2 2

2 2 2 2

2 2 2 2

2 2 2 2

2 2

6 2 6 9

6 2 6 9

( 6) 2 ( 6) ( 9)

( 6) 2 ( 6) ( 9)

( 6) 4 ( 6) ( 9)

12 36 4 12 36 18 81

12 36 4 48 144 4 72 324

3 60 3 72 432 0

ou

z z i

x iy x iy i

x iy x y i

x y x y

x y x y

x x y x x y y

x x y x x y y

x x y y

2 2

2 2

2 2

t by 3

20 24 144 0

By completing the square

( 10) 100 ( 12) 144 144 0

( 10) ( 12) 100

The locus of z is a circle with radius 10 and centre (-10,12)

x x y y

x y

x y

(b)

3.

3 3 2

2 2

2 2

2

2

3 6 4

: 3 1 3 3 8

(3 3) 8 1 3

8 1 3

3 3

x x y y x

d dy dyx y x

dx dx dx

dyy x x

dx

dy x x

dx y

4. (a)

1 2

1 2

2

2 2

ln( )( ) cot( ) sin( )cos( ) cos ( ) 9

ln(2 )

ln( ) 1( ) sin( )cos( ) cos ( ) 9

ln(2 ) tan( )

1 2ln(2 ) ln( )

tan( ) 0 1 sec2'( )

(ln(2 )) tan

(cos )(cos ) (sin )(

xh x x x x x x

x

xh x x x x x

x x

x xx xx x

h xx x

x x x

2

22 2

2 2 2

22 2

2 2 2

22 2

2 2 2

1sin ) 18

1

1(ln(2 ) ln( ))

sec 1'( ) cos sin 18

ln (2 ) tan 1

2ln( )

sec 1 = cos sin 18

ln (2 ) tan 1

ln(2) sec 1'( ) = cos sin 18

ln (2 ) tan 1

x xx

x xxxh x x x x

x x x

xxx x x x

x x x x

xh x x x x

x x x x

(b) i) 2 cosxy e x

2 2

2 2

2

2

2

cos( ) 2 sin( )

=2 cos( ) sin( )

= 2cos( ) sin( )

At stationary pts. 0

2cos( ) sin( ) 0

0 and 2cos( ) sin( ) 0

2cos( ) sin( ) 0

2cos( ) sin( )

x x

x x

x

x

x

dyx e e x

dx

e x e x

e x x

dy

dx

e x x

e x x

x x

x x

sin( ) 2=

cos( )

tan( ) 2

x

x

x

ii) When 0x ,2(0) cos(0) 1y e

We have co-ordinates (0,1)

2(0)

0

1 1

2cos(0) sin(0) 2

Gradient of tangent at x=0 is 2

So equation of tangent : ( )

1 2( 0)

2 1

x

dye

dx

y y m x x

y x

y x

5. (a) 8( , , ) 4 cos( ) sin(4 ) tan( )zf x y z xyz xy x e xz y

i)

8

8

4 (cos )( ) ( )( sin( )) (sin(4 )(0) ( )(4 cos(4 )) 0

4 cos( ) sin( ) 4 cos(4 ).

z

x

z

f yz x y xy x xz e z xz

yz y x xy x ze xz

2

2

4 [( )(0) (cos )( )] 0 sec

4 cos( ) sec

yf xz xy x x y

xz x x y

8 8

8 8

8

4 0 [ 4 cos(4 ) (sin(4 )(8 )] 0

4 4 cos(4 ) 8 sin(4 )

4 4 ( cos(4 ) sin(4 ))

z z

z

z z

z

f xy e x xz xz e

xy xe xz e xz

xy e x xz xz

ii)

4 [ ][ sin( )] [cos( )][1] 0

4 sin( ) cos( )

xyf z x x x

z x x x

4 cos( ) ( )(0) (sin( ))( ) 0

4 cos( ) sin( )

yxf z x xy x x

z x x x

4 0 0

4

yxf x

x

(b) 2 42 4 18x

p xv v xv

i)

2

2

2 2 4 0

42 2

pxv xv

v

xxv

v

2 3

2 3

40 72

472

pv x

x v

v xv

ii)

2

2

2

42 0

42

pv

x v v

vv

22

2

2 4 0

42

pv v

v x

vv

6 (a) i)

2 2

2

10 25

1 1

=5ln 1

x xdx dx

x x

x c

ii) 1

2 2

2

1515 (1 )

1

xdx x x dx

x

Recall that if some function

1

2 2( ) (1 )f x x

1

2 2

1

2 2

1'( ) (2 )(1 )

2

'( ) (1 )

f x x x

f x x x

So

1 1

2 22 2

1

2 2

15 (1 ) 15 (1 )

15(1 )

x x dx x x dx

x C

iii) 2 2 2

2 8 2 8

1 1 1

x xdx dx dx

x x x

2 2

1 2

1 22 4

1 1

2tan ( ) 4ln 1

xdx dx

x x

x x C

(b) i) 4 3 2

3 2

4 9 17 12( )

4 4

x x x xh x

x x x

This algebraic fraction is improper so we shall use algebraic long division:

3 2 4 3 2

4 3 2

2

4 4 4 9 17 12

4 4 0

5 17 12

x

x x x x x x x

x x x

x x

2

3 2

5 17 12( )

4 4

x xh x x

x x x

2

2

2

2

5 17 12

( 4 4)

5 17 12

( 2)

x xx

x x x

x xx

x x

Let 2

2

5 17 12( )

( 2)

x xq x

x x

22 ( 2)

A B C

x x x

Multiplying out both sides by 2( 2)x x gives

25 17 12x x 2( 2) ( 2)A x Bx x Cx

2 2

Let 0;

5(0) 17(0) 12 (0 2)

4 12

3

x

A

A

A

2

2 2 2

Comparing terms:

5

5

3 5

2

x

Ax Bx x

A B

B

B

Comparing terms:

17 4 2

17 4 2

17 4(3) 2(2)

17 12 4

C= 17 16 1

x

x Ax Bx Cx

A B C

C

C

2

3 2 1( )

2 ( 2)q x

x x x

So ( ) ( )h x x q x

2

3 2 1( )

2 ( 2)h x x

x x x

ii) Hence,

4 44 3 2

3 2 2

3 3

4 9 17 12 3 2 1

4 4 2 ( 2)

x x x xdx x dx

x x x x x x

4 4 4 4

2

3 3 3 3

1 1 3 2 ( 2)

( 2)x dx dx dx x dx

x x

4 42 1

4 4

3 33 3

2 2 1 1

3 2

( 2)3 ln 2 ln 2

2 1

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

2 2 1 1

9 4 18 3 ln( ) 2ln(2)

2 3 2

7 4 1ln( ) ln(2)

2 3 2

643 ln( 4)

27

253 ln(

x xx x

6

)27

(c) 1 1

2 2

1 1tan ( ) tan ( )

1 1x dx x dx dx

x x

1 1tan ( ) tan ( )x dx x

Let 1tan ( )I x dx

1(1)(tan ( ))x dx

Let 1tan and 1

dvu x

dx

2

1

1

du

dx x

v x

Using integration by parts:

1

2

1

2

1

2

1 2

1tan ( ) ( )( )

1

tan ( )1

1 2tan ( )

2 1

1tan ( ) ln |1 |

2

I x x x dxx

xx x dx

x

xx x dx

x

x x x

1 1 2 1

2

1 1tan ( ) tan ( ) ln |1 | tan ( )

1 2x dx x x x x C

x

7. 2

2

1 1

1

xdx

xx x

Using the substitution 1

seccos

x

2

2

[cos ][0] [1][ sin ]

cos

sin

cos

sin 1

cos cos

dx

d

tan sec

tan sec

dx

d

dx d

2 2

2 2

1 1 1 sec 1 tan sec

sec1 sec sec 1

1

sec

xdx d

xx x

2

2

1tan tan sec

tan

2

1tan tan

tan

1 tan

tan

d

d

tan

2

2

1 tan

sec

tan

d

d

d

C

Remember in the question that 1

seccos

x

1

cosadj

x hyp

Now let’s apply a little bit of trigonometry:

θ

x

1A

B

C

By Pythagoras’s theorem:

2 2 2

2 2

2 1

AB BC AC

BC AB AC

BC x

So 2Opp 1

tanAdj 1

BC x

AC

Now we can replace 2tan with 1x

2

2

2

1 11

1

xdx x C

xx x

8 (a) R.T.S. 4 31 1 x x x

R.H.S.:

(b)

1

3

0

(1 ) n

nI x x dx

Employing integration by parts:

Let 3(1 ) and n dv

u x xdx

3 3

4

4

1 1 (1 )

x x x x x

x x x

x

3 1 2

2 3 1

(1 ) ( 3 )

3 (1 )

n

n

dun x x

dx

nx x

and

2

2

xv

1 12 2

13 2 3

00

1

4 3 1

0

1

3 3 1

0

1 1

3 1 3 3 1

0 0

1

(1 ) 3 12 2

3 0 (1 )

2

Using the identity in

31 (1 ) (1 )

2

3 3(1 ) (1 ) (1 )

2 2

3 3

2

nn

n

n

n

n

n

n n

n

n n

x xI x nx x dx

nI x x dx

nI x x x dx

n nI x x dx x x x dx

n nI I

1

3

0

1

1

1

1

1

(1 )2

3 3

2 2

3 3

2 2

2 3 3

2 2

(2 3 ) 3

3

3 2

n

n n n

n n n

n n

n

n n

n n

x x dx

n nI I I

n nI I I

I nI nI

n I nI

nI I

n

(c)

1

3 0

0

0

1

0

12

0

(1 )

(1)

2

10

2

1

2

I x x dx

x dx

x

4

12 9 6 3 1 243

14 11 8 5 2 1540I

4 3

2

2

1

1

0

0

3(4)

3(4) 2

12 3(3) =

14 3(3) 2

12 9 =

14 11

12 9 3(2) =

14 11 3(2) 2

12 9 6 =

14 11 8

12 9 6 3(1) =

14 11 8 3(1) 2

12 9 6 3 =

14 11 8 5

I I

I

I

I

I

I

I

9. sin((2 1) )

sin( )

m

m xJ dx

x

(a)

Recall: sin( ) sin( ) 2cos sin2 2

1

(2 1 2 1) ((2 1) (2 1))2cos sin

2 2

sin( )

4 22cos sin

2 2 =

sin( )

2cos 2 sin( ) =

m m

m m x m m x

J J dxx

mx x

dxx

mx x

sin( )x

=2 cos(2 )

= 2

dx

mx dx

1

2sin(2 )

sin(2 ) =

mxm

mx

m

1

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

sin( ) sin( )

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

sin( ) sin( )

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

sin( ) sin( )

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

sin( )

m m

m xm xJ J dx

x x

m x m xdx

x x

m x m xdx

x x

m x m xdx

x

(b)

0

sin(2(0) 1)

sin( )

1

xJ dx

x

dx

x

5

sin(10 ) sin(8 ) sin(6 ) sin(4 ) sin(2 )

5 4 3 2 1

x x x x xJ x C

10. 3

0

1 tan( ) x dx

03width of strips= where n is the number of strips

4 12

b ah

n

x 0 12

6

4

3

y 1 1.126032 1.25593 1.41421 1.65289

Using the trapezium rule:

3

0

1 1 tan( ) (width of strips)(1 height+2(sum of all middle heights)+last height)

2

stx dx

1

5 4

3

2

1

0

sin(2 )

sin(2(5))

5

sin 2(4)sin(10 )

5 4

sin(10 ) sin(8 ) sin(2(3) )

5 4 3

sin(10 ) sin(8 ) sin(6 ) sin(2(2) )

5 4 3 2

sin(10 ) sin(8 ) sin(6 ) sin(4 ) sin(2(1) )

5 4 3 2 1

sin(10 ) sin

5

m m

m xJ J

m

xJ J

xxJ

x x xJ

x x x xJ

x x x x xJ

x

0

(8 ) sin(6 ) sin(4 ) sin(2 )

4 3 2 1

x x x xJ

11 2(1.126032 1.25593 1.41421) 1.65289

2 12

1 7.592344 1.6528924

10.24523424

1.3410979...

1.34 {3 sig. fig}