-
CAPE Pure Mathematics Unit 2
Practice Questions
By Carlon R.Baird
MODULE 1: COMPLEX NUMBERS AND CALCULUS II
1. (a) Use de Moivres theorem to prove the trigonometric
identity:
7 5 3cos7 64cos 112cos 56cos 7cos
(b) Use de Moivres 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 ( ) 9
ln(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 12
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 )nnI x x dx , show that 13
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 1sin 2
m 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 lets consider 7(cos sin )i
Now, by de Moivres 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 0 337 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 1 1 2 1 22 2
(cos( ) sin( ))z r
iz r
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
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(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 88 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 lets apply a little bit of trigonometry:
x
1A
B
C
By Pythagorass 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 ) nn
I x x dx Employing integration by parts:
Let 3(1 ) and n
dvu x x
dx
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
5sin(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}
