CAPE PURE MATHEMATICS UNIT 2 MODULE 1 PRACTICE QUESTIONS

Challenge questions to prepare you for CAPE exam

1. 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 3
cos7 64cos 112cos 56cos 7cos       

(b) Use de Moivre’s theorem to evaluate  
8
1i
(c) Express  
2
cos3 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
dx in terms of x and y where 3 3 2
3 6 4x x y y x    
4. (a) Find the derivative of the function
12ln( )
( ) cot( ) sin( )cos( ) cos ( ) 9
ln(2 )
x
h x x x x x x
x

    
(b) The curve C has equation 2
cos( )
x
y 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 0
z
f x y z xyz xy x e xz y    
i. Determine x
f ,y
f ,z
f
ii. Determine xy
f , yx
f ,yz
f
(b) Given that 24
2 4 18
x
p xv v x
v
   

2. i. Determine p
v

 and p
x


ii. Determine 2
p
xv

 and 2
p
vx


6. (a) Integrate with respect to x
i. 2
10
1
x
x
ii. 2
15
1
x
x
iii. 2
28
1
x
x


(b) (i) Express the function 4 3 2
32
4 9 17 12
()
44
x x x x
hx
x x x
   

 as partial fractions
(ii) Hence, evaluate 4
4 3 2
32
3
4 9 17 12
44
x x x x
dx
x x x
   

(c) Determine 1
2
1
tan
1
x dx
x



7. Using the substitution secx ,find 2
2
11
1
x
dx
xxx



8. (a) Show that 43
1 (1 )x x x  

(b) Given that 1
3
0
(1 )
n
n
I x x dx
 , show that 1
3
32
nn
n
II
n



(c) Use your reduction formula to evaluate 4
I .
9. Given that sin(2 1)
sin( )
m
mx
J dx
x


 ,
(a) Show that 1
sin2
mm
mx
JJ
m


(b) Hence find 5
J .

3. 10. Use the trapezium rule using 4 strips to estimate 3
0
1 tan( ) x dx


 giving your answer to
3 significant figures.

4. By Carlon R. Baird

5. 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:
ii
ii
   
   
  
    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
ii
ii
      
      
      
  
   
   
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
57
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 3
cos7 64cos 112cos 56cos 7cos        



6. 2
(cos3 sin3 )
cos(6 ) sin(6 )
cos sin
cos7 sin7
q i q
q q i q q
q i q
q i q

     

 (b)
8
22
1
Let ( 1 )
Let 1
( 1) (1) 2
tan 1
tan (1)
4
zi
pi
rp




  
  
    


arg
4
3
4
p



  


8
8
8
Rewriting in polar form: (cos sin )
33
2 cos( ) sin( )
44
33
2 cos( ) sin( )
44
Now applying de Moivre's theorem:
33
( 2) (cos(8 ) sin(8 ))
44
24 24
16(cos( ) sin( )
44
p p r i
pi
zp
zi
zi
zi







  




  


    
   )
16(cos(6 ) sin(6 ))
16(1 (0))
16.
zi
zi
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 11
1 2 1 2
22
(cos( ) sin( ))
zr
i
zr
      
 α
Im z
Re z
arg p
1
1
Recall that cos( ) cos
and sin( ) sin( )  

7. 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
22
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
   
     
     
     
     

        

        
    

22
22
22
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
xy
xy
     
       
    


(b)

8. 3. 3 3 2
22
22
2
2
3 6 4
:3 1 3 3 8
(3 3) 8 1 3
8 1 3

33
x x y y x
d dy dy
x y x
dx dx dx
dy
y x x
dx
dy x x
dx y
    
   
   




4. (a)
  
 
12
12
2
22
ln( )
( ) cot( ) sin( )cos( ) cos ( ) 9
ln(2 )
ln( ) 1
( ) sin( )cos( ) cos ( ) 9
ln(2 ) tan( )
12
ln(2 ) ln( )
tan( ) 0 1 sec
2
'( )
(ln(2 )) tan
(cos )(cos ) (sin )(
x
h x x x x x x
x
x
h x x x x x
xx
xx
xx
xx
hx
xx
x x x


    
     
   

    
     

   
2
2
22
22
2
2
22
22
2
2
22
22
2
1
sin ) 18
1
1
(ln(2 ) ln( ))
sec 1
'( ) cos sin 18
ln (2 ) tan 1
2
ln( )
sec 1
= cos sin 18
ln (2 ) tan 1
ln(2) sec 1
'( ) = cos sin 18
ln (2 ) tan 1
xx
x
xx
x
x
h x x x x
xx x
x
xx
x x x
x x x x
x
h x x x x
x x x x



     

    

     


9. (b) i) 2
cos
x
y e x    
 
 
22
22
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( )

xx
xx
x
x
x
dy
x e e x
dx
e x e x
e x x
dy
dx
e x x
e x x
xx
xx
     
   



  
   


sin( )
2=
cos( )
tan( ) 2
x
x
x



ii) When0x ,2(0)
cos(0) 1ye
We have co-ordinates (0,1)
 
 
2(0)
0
11
2cos(0) sin(0) 2
Gradient of tangent at x=0 is 2
So equation of tangent : ( )
1 2( 0)
2 1
x
dy
e
dx
y y m x x
yx
yx

  

  
  


5. (a) 8
( , , ) 4 cos( ) sin(4 ) tan( )
z
f 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
y
f xz xy x x y
xz x x y
    
  

10.  
88
88
8
4 0 [ 4 cos(4 ) (sin(4 )(8 )] 0
4 4 cos(4 ) 8 sin(4 )
4 4 ( cos(4 ) sin(4 ))
zz
z
zz
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( )
xy
f z x x x
z x x x
    
  
 4 cos( ) ( )(0) (sin( ))( ) 0
4 cos( ) sin( )
yx
f z x xy x x
z x x x
    
  
4 0 0
4
yx
fx
x
  


(b) 24
2 4 18
x
p xv v x
v
   
i)
2
2
2 2 4 0
4
22
p
xv xv
v
x
xv
v

   

  

23
23
4
0 72
4
72
p
vx
xv
vx
v

   

  

ii)
2
2
2
4
20
4
2
p
v
x v v
v
v

  


2
2
2
2 4 0
4
2
p
vv
vx
v
v

  


6 (a) i)
22
2
10 2
5
11
=5ln 1
xx
dx dx
xx
xc





11. ii) 1
22
2
15
15 (1 )
1
x
dx x x dx
x



Recall that if some function 1
22
( ) (1 )f x x
1
22
1
22
1
'( ) (2 )(1 )
2
'( ) (1 )
f x x x
f x x x


  
  
So
11
2222
1
22
15 (1 ) 15 (1 )
15(1 )
x x dx x x dx
xC

  
  


iii) 2 2 2
2 8 2 8
1 1 1
xx
dx dx dx
x x x


    
22
12
12
24
11
2tan ( ) 4ln 1
x
dx dx
xx
x x C



   


(b) i) 4 3 2
32
4 9 17 12
()
44
x x x x
hx
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
xx
     
  

2
32
5 17 12
()
44
xx
h x x
x x x

  

2
2
2
2
5 17 12
( 4 4)
5 17 12
( 2)
xx
x
x x x
xx
x
xx






Let 2
2
5 17 12
()
( 2)
xx
qx
xx


  2
2 ( 2)
A B C
x x x


Multiplying out both sides by 2
( 2)xx gives
2
5 17 12xx 2
( 2) ( 2)A x Bx x Cx     

12. 22
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
AB
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)
qx
x x x
   

So ( ) ( )h x x q x
2
3 2 1
()
2 ( 2)
h x x
x x x
   



ii) Hence, 44
4 3 2
3 2 2
33
4 9 17 12 3 2 1
4 4 2 ( 2)
x x x x
dx x dx
x x x x x x
   
   
   

13. 4 4 4 4
2
3 3 3 3
11
3 2 ( 2)
( 2)
x dx dx dx x dx
xx

   
      
44
21
44
33
33
2 2 1 1
32
( 2)
3 ln 2 ln 2
21
4 3 (4 2) (3 2)
3 ln(4) ln(3) 2 ln(4 2) ln(3 2)
2 2 1 1
9 4 1
8 3 ln( ) 2ln(2)
2 3 2
7 4 1
ln( ) ln(2)
2 3 2
64
3 ln( 4)
27
25
3 ln(
xx
xx


    
        
      
   
    
          
   
   
   
    
   
   
   
  

6
)
27

(c) 11
22
11
tan ( ) tan ( )
11
x dx x dx dx
xx

  
  
11
tan ( ) tan ( )x dx x



Let 1
tan ( )I x dx



1
(1)(tan ( ))x dx



Let 1
tan and 1
dv
ux
dx


2
1
1
du
dx x

 vx
Using integration by parts:
 
1
2
1
2
1
2
12
1
tan ( ) ( )( )
1
tan ( )
1
12
tan ( )
21
1
tan ( ) ln |1 |
2
I x x x dx
x
x
x x dx
x
x
x x dx
x
x x x










  




14. 1 1 2 1
2
11
tan ( ) tan ( ) ln |1 | tan ( )
12
x dx x x x x C
x
  
      


7. 2
2
11
1
x
dx
xxx



Using the substitution 1
sec
cos
x


2
2
[cos ][0] [1][ sin ]
cos
sin
cos
sin 1
cos cos
dx
d










tan sec
tan sec
dx
d
dx d


  


 
22
22
1 1 1 sec 1
tan sec
sec1 sec sec 1
1
sec
x
dx d
xxx

  




    





2
2
1
tan tan sec
tan
  




 
2

1
tan tan
tan
1 tan
tan
d
d

  










tan

2
2

1 tan
sec
tan
d
d
d
C









Remember in the question that 1
sec
cos
x


1
cos
adj
x hyp
  

15. Now let’s apply a little bit of trigonometry:
θ
x
1
A
B
C
By Pythagoras’s theorem:
2 2 2
22
2
1
AB BC AC
BC AB AC
BC x


  
So 2
Opp 1
tan
Adj 1
BC x
AC


  
Now we can replace 2
tan with 1x 
2
2
2
11
1
1
x
dx x C
xxx

    


8 (a) R.T.S.  
43
1 1 x x x  

R.H.S.:



(b) 1
3
0
(1 )
n
n
I x x dx

Employing integration by parts:
Let 3
(1 ) and
n dv
u x x
dx
    
33
4
4
1 1 (1 )

x x x x x
x x x
x
    

  


16. 3 1 2
2 3 1
(1 ) ( 3 )
3 (1 )
n
n
du
n x x
dx
nx x


  
   and 2
2
x
v
  
1
122
1
3 2 3
0
0
1
4 3 1
0
1
3 3 1
0
11
3 1 3 3 1
00
1
(1 ) 3 1
22
3
0 (1 )
2
Using the identity in
3
1 (1 ) (1 )
2
33
(1 ) (1 ) (1 )
22
33
2
n
n
n
n
n
n
n
nn
n
nn
xx
I x nx x dx
n
I x x dx
n
I x x x dx
nn
I x x dx x x x dx
nn
II





    
         
   
  
    

      
  




1
3
0
1
1
1
1
1
(1 )
2
33
22
33
22
23 3
22
(2 3 ) 3
3
32

n
n n n
n n n
nn
n
nn
nn
x x dx
nn
I I I
nn
I I I
I nI n
I
n I nI
n
II
n






  
  


  




17. (c)











1
30
0
0
1
0
1
2
0
(1 )
(1)
2
1
0
2
1
2
I x x dx
x dx
x












 4
12 9 6 3 1 243

14 11 8 5 2 1540
I      

43
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
II
I
I
I
I
I
I












  


  

18. 9. sin((2 1) )

sin( )
m
mx
J dx
x



(a)






Recall: sin( ) sin( ) 2cos sin
22
   

   

   
   

1
(2 1 2 1) ((2 1) (2 1))
2cos sin
22

sin( )
42
2cos sin
22
=
sin( )
2cos 2 sin( )
=
mm
m m x m m x
J J dx
x
mx x
dx
x
mx x

        
   
   
  
   
   
   



sin( )x

=2 cos(2 )
= 2
dx
mx dx


1
2
sin(2 )
sin(2 )
=

mx
m
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( )
mm
mxmx
J J dx
xx
m x m x
dx
xx
m x m x
dx
xx
m x m x
dx
x


  
  



  






19. (b)









0
sin(2(0) 1)

sin( )
1
x
J dx
x
dx
x






5
sin(10 ) sin(8 ) sin(6 ) sin(4 ) sin(2 )

5 4 3 2 1
x x x x x
J x C       
10. 3
0
1 tan( ) x dx



0
3
width of strips= where n is the number o f strips
4 12
ba
h
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
st
x dx

  


1
54
3
2
1
0
sin(2 )
sin(2(5))
5
sin 2(4)sin(10 )
54
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
mm
mx
JJ
m
x
JJ
xx
J
x x x
J
x x x x
J
x x x x x
J
x



  
   
    
     

0
(8 ) sin(6 ) sin(4 ) sin(2 )
4 3 2 1
x x x x
J   

20.  
 
 
1
1 2(1.126032 1.25593 1.41421) 1.65289
2 12
1 7.592344 1.65289
24
10.245234
24
1.3410979...
1.34 {3 sig. fig}




    


  


