RAFFLES INSTITUTION
H2 Mathematics 9740
2010 Year 6
____________________________
2010 H2 Maths Preliminary Exam Paper 1 Solutions
1
By writing z  x  iy, x, y  R , solve the simultaneous equations
w
z 2  zw  2  0 and z * 
, where z * is the conjugate of z .
1 i
Q1 [3]
[3]
Let z = x + iy. Substitute the second equation into the first.
z 2  z ( z * (1  i))  2  0
( x  iy ) 2  ( x 2  y 2 )(1  i)  2  0
2 x 2  2ixy  ( x 2  y 2 )i  2  0
On comparing real and imaginary parts,
2 x 2  2  0, 2 xy  x 2  y 2  ( x  y)2  0
x  1, y  1 .
When z  1  i , w = 2i
When z  1  i , w  2i
2
Sketch the graph of y  e x  x . By adding a suitable graph to the sketch, find the set of values of x that
2
satisfies 2 x  5  2e x  2 .
[4]
x
Q2 [4]
y
y  ex  x
y
α
βO
5 1

2 x2
x
x
2 x  5  2e x 
2
x2
2
 2e x  2 x
2
x
5 1
  ex  x
2 x2
 5

From the diagram,
x   or   x  0 or x  0
Using G.C.,
x  2.18 or  0.920  x  0 or x  0 (3s.f.)
Therefore solution set is (, 2.18]  [0.920, 0)  (0, ) .
__________________________________
2010 H2 Maths Preliminary Exam Paper 1
Page 1 of 14
3
Four friends returned from a trip to Europe and converted their foreign currencies back to Singapore
Dollars. The amounts of foreign currencies converted and the total amounts received in Singapore
Dollars are shown in the following table.
Donald
Leonard
Michael
Raphael
Sterling Pound
36
55
40
k
Euro Dollar
77
18
31
59
Swiss Franc
42
63
26
24
Total amount in
269.90
233.45
175.50
313.00
Singapore Dollars
Assuming that, for each foreign currency, the exchange rate quoted for each of the friends is the same,
calculate the value of k.
[4]
Q3 [4]
4
Let x, y, z be the exchange rate quoted for Sterling Pound, Euro Dollar and Swiss Franc,
respectively (i.e. 1 Sterling Pound = x Singapore Dollars, 1 Euro Dollar = y Singapore Dollars
and 1 Swiss Franc = z Singapore Dollars).
36 x  77 y  42 z  269.9
55 x  18 y  63z  233.45
40 x  31y  26 z  175.5
Using the GC, x = 2.15, y = 1.78, z = 1.32
kx  59 y  24 z  313
313  59(1.78)  24(1.32)
k 
 82
2.15
 1  2 x  x dx ,
x  x3
(b)
 (x  2)(1 x) dx .
1
Find (a)
[2]
2
2
[3]
2
Q4 [5]
(a)[2]

1
1 2x  x
2
dx 

1
2  ( x  1)
2
dx
 x 1 
 sin 1 
c
 2 
(b)[3]

x2  x  3
( x 2  2)(1  x)
dx 


1
1
dx
x2  2 1  x

 x 
tan 1 
  ln |1  x |  c
2
 2
1
__________________________________
2010 H2 Maths Preliminary Exam Paper 1
Page 2 of 14
A sequence u0 , u1, u2 ,
5
is defined by
u0  1 and un 1  run  d for n  0
where r and d are non-zero constants.
(i) If r = d = 1, write u n in terms of n.
(ii)
[1]
d  d

If r  1, prove by induction that un  r n 1 
for all non-negative integral values of

 r 1  r 1
n.
[4]
(iii) If r  1, state the limit of the sequence as n   .
5 [6]
(i) [1]
un  n  1
(ii) [4]
d  d

for n  Z0 .
Let Pn be the statement un  r n 1 

 r 1  r 1
When n = 0,
LHS: u0  1 (given)
[1]
d  d

1
RHS: r 0 1 

 r 1  r 1
 P0 is true.
Assume Pk is true for some k  Z  ,
d  d

i.e. uk  r k 1 

 r 1  r 1
Prove that Pk 1 is true,
d  d

i.e. to prove uk 1  r k 1 1 

 r 1  r 1
uk 1  ruk  d
 
d  d 
=r  r k  1 

d
  r 1  r 1 
d  rd

=r k 1 1 
d

 r 1  r 1
d   rd  rd  d

=r k 1 1 

r 1
 r 1 
d  d

=r k 1 1 

 r 1  r 1
Since Pk is true implies Pk 1 is true, and P0 is true, by mathematical induction, Pn is true for all
n Z0 .
(iii) [1]
As n  , un 
d
1 r
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2010 H2 Maths Preliminary Exam Paper 1
Page 3 of 14
6
The diagram shows the graph of y  f ( x) . The curve has a maximum point at (0, 2) and it cuts the xaxis at the points (  a, 0) and (a, 0) where a is a positive constant. The lines x  2 , x  2 and y  2
are asymptotes to the curve.
y
y  f(x)
2
y2
a O
a
x  2
Sketch, on separate diagrams, the graphs of
(i)
y 2  f ( x),
(ii)
y
x
x2
[3]
1
.
f (2 x)
[3]
Your sketches should show clearly the equations of asymptotes, stationary points and intersection with
the axes, where applicable.
y
Q6 [6]
(i) [2]
y 2  f(x)
2
a O
y 2
x
a
 2
x  2
(ii)[2]
y 2
x2
y
y
1
2
|
1
x   a2
O
1
f  2x
y  12
|
1
x
x  a2
__________________________________
2010 H2 Maths Preliminary Exam Paper 1
Page 4 of 14
The curve C has equation y  f (x)  2 x  1 
7
a
where a and b are positive constants.
bx  1
(i)
Show that C has exactly two stationary points.
[3]
(ii)
Given that C passes through the point (0,3) , find the value of a.
[1]
Hence sketch the graph of y  f '(x) , indicating clearly any intersection with the axes and the
equation(s) of any asymptote(s) of the curve in terms of b.
Q7[7]
(i)[3]
y  2x  1 
[3]
a
bx  1
ab
dy
2
2
dx
 bx  1
When
dy
 0,
dx
ab
 bx  1
2
 bx  1
2
ab
2
1
ab 
x   1 

b
2 
2

Since a and b are positive constants, ab > 0 and thus there are 2 distinct real solutions for x.
Hence, C has exactly two stationary points. (shown)
(ii)[1]
Given that C passes through the point (0,3) ,
3  1 a
a2
[3]
y
y  f ' (x)
y2
2


b
1
|
b 1
 O
__
2  2b
x
8
|

b
1
b 1

x
1
b
A cylindrical water tank has a horizontal base with a fixed area of A m 2 and is initially empty. Water is
poured into the tank at a constant rate of 5 m 3s 1 , and leaks out through a small hole in the base at a rate
which is proportional to the depth of water in the tank. The depth of water in the tank is x metres at
time t seconds. Show that
A
dx
 5  kx , where k is a positive constant.
dt
[2]
Solve the differential equation, expressing x in terms of A, k and t .
[5]
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2010 H2 Maths Preliminary Exam Paper 1
Page 5 of 14
8 [7]
V  Ax
[2]

dV
dx
A
dt
dt
dV
 5  kx , where k is a positive constant.
dt
dx
 A  5  kx (shown)
dt
dx
 5  kx
dt
1
 A
dx   dt
5  kx
A
  ln 5  kx  t  c1 , where c1 is an arbitrary constant
k
k
 ln 5  kx   t  c, where c is an arbitrary constant
A
A
[5]
 5  kx  e
k
 t c
A
 5  kx  Be
k
 t
A
, where B is an arbitrary constant
k
 t
A
1
(5  Be )
k
When t  0, x  0  B  5
 x
 x
9
k
 t
5
(1  e A )
k
(a)
Solve the equation z 6  4 2(1  i) , expressing the solutions in the form re i , where r  0 and
[3]
     .
(b)
(i)
The two complex numbers 2  3 + 2i and 2  3 + 2i satisfy the cubic equation
a3 z 3  a2 z 2  a1 z  a0  0 . Explain clearly whether it is possible for all the coefficients
a3 , a2 , a1 , a0 to be real numbers.
(ii)
Q9 [7]
In an Argand diagram, the point A represents the complex number 2  3  i . If A, B, and C
are the vertices of an equilateral triangle taken in clockwise order, and these three points lie
on a circle with centre at the origin, find the complex number represented by B in the form
p  iq where p, q  R , giving the exact values of p and q.
[3]
z 6  4 2(1  i)  8e
(a)[3]
z  2e
(b)(i)[1]
[1]
i (8 k 1)
24
,
i

4


i   2 k 

 8e  4
k  3, 2, 1, 0,1, 2
Since the two complex roots are non-real, and not conjugate, for all the coefficients a3 , a2 , a1 , a0
to be real, the degree of the polynomial has to be at least four. Since the degree is only three, it is
not possible.
__________________________________
2010 H2 Maths Preliminary Exam Paper 1
Page 6 of 14
(ii)[3]
Since the points A, B and C are on a circle centered at the origin, and they form an equilateral
2
2
triangle, angle AOB is
. OA rotated by
clockwise about O will give OB. Let b be the
3
3
complex number represented by B.
be

2
i
3
(2  3  i)
 1
3
b     i
 (2  3  i)
2 
 2
b  3  1  i(1  3)
10
The curve C has equation y  xe x for x  0 and P(a, ae a ) is a point on C .
(i)
Sketch the curve C .
[1]
(ii)
Find, in terms of a , the equation of the tangent to the curve at P .
[2]
This tangent cuts the y  axis at the point Q(0, h) .
Using differentiation, find, as a varies, the exact maximum value of h .
[6]
(i)[1]
(ii)[8]
[2]
y  xe x
dy
 x(e  x )  e  x  e  x (1  x)
dx
dy
 e  a (1  a)
At P , x  a, y  ae  a ,
dx
Equation of tangent to the curve at P is
y  ae a  e a (1  a)( x  a)

 y  e a (1  a)( x  a)  ae a
At Q , x  0, y  h
 h  e a (1  a)(0  a)  ae a
[6]
 e a (a  1)(a)  ae a  a 2e a
dh

 2ae  a  a 2 (e  a )  ae  a (2  a)
da
dh
0
For stationary values of h ,
da
 a  0 (N.A. since a  0 ) or a  2
dh
da
Tangent
2
2
2
0
0
0
Maximum value of h  4e 2
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2010 H2 Maths Preliminary Exam Paper 1
Page 7 of 14
11
The equations of two planes p1 , p2 are
2x  5y + 3z = 3,
3x + y + 6z =  ,
respectively, where  and  are constants.
(i) Given that the two planes intersect in a line l, with a vector equation given by
 4   2 
r =  2   s  1  , s  R,
 5   3 
   
show that the value of  is 12 and find the value of  .
[3]
A third plane p3 has equation given by
5x + 8y + tz = 12,
where t is a constant.
(ii) With the values of  and  found in (i), find the exact value of t if the three planes have no point
in common.
[2]
(iii) The plane p4 contains the line l and the point (1, 1, 2). Find the cartesian equation of p4 and the
acute angle between p1 and p4.
[5]
Q11[10]
(i)[3]
Since the two planes intersect in a line l, with a vector equation given by
 4   2 
r =  2   s  1  , s  R,
 5   3 
   
The line l lies in p2.
 2   3 
So  1       0  6    18  0    12
 3  6
   
 4
 
 2  satisfies equation of p2.
 5 
 
 4   3
So  2           12  2  30  12  24  30  6
 5   6 
   
(ii)[2]
Since p1 and p2 intersect in line l, and the 3 planes have no common point of intersection, l does
not intersect p3.
Hence l is parallel to p3, but l does not lie in p3.
 2   5 
2
   
 1    8   0  10  8  3t  0  t  3
 3  t
   
 
 4  5
   2
Also  2  .  8    12 which means (4, 2, 5) is not in p3 and so l does not lie in p3.
 5   2  3
  
3
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2010 H2 Maths Preliminary Exam Paper 1
Page 8 of 14
(iii)[5]
 2 
Since p4 contains l,  1  is parallel to p4.
 3
 
Since p4 contains the points (1, 1, 2) and (4, 2, 5).
 4 1  3
Hence  2    1   1  is parallel to p4.
 5   2   7 
     
 3   2   4 
A vector normal to p4 is  1    1    5  .
 7   3   1 
     
 4  1   4
Equation of p4 is given by : r  5    1   5   1  4 x  5 y  z  1
1  2  1
     
Let  be the angle between p1 and p4.
cos  =
 2   4
   
 5    5 
 3  1
14
   

2
2
2
2
2
2
38 42
2  5  3 4  5 1
  = 110.514
Acute angle between p1 and p4 = 180  
= 69.5 (1 d.p.)
12
Relative to the origin O, three points A, B and P have position vectors given by
a = 14i + 8j + 6k, b = 11i + tj + 12k and p = 12i  4j + 10k respectively, where t is a constant.
(a) If P divides the line segment AB in the ratio  : 2, find the values of  and t.
[3]
(b) It is given that t = 2.
(i) Find the exact length of projection of PB onto AB .
Deduce the exact shortest distance of P from line AB.
[4]
(ii) BAPQ forms a parallelogram. Find the position vector of the point Q and the area of the
parallelogram correct to 2 decimal places.
[4]
Q12[11]
(a) [3]
By Ratio Theorem,
2OA   OB
OP 
2
 12 
 14 
 11 
2  
  
 
  4  
 8 
 t 
 10    2  6    2  12 
 
 
 
So 12 =
28
11

 12  24  28  11    4
 +2   2
Also,  4 =
16
t
16 4t

 4    t  10
 +2   2
6 6
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2010 H2 Maths Preliminary Exam Paper 1
Page 9 of 14
(b) [8]
(i)[4]
 11   12   1
     
PB  OB  OP   2    4    6 
12   10   2 
     
 11  14   3 
     
AB  OB  OA   2    8    6 
12   6   6 
     
Length of projection of PB onto AB
 1  3 
   
 6    6 
2  6
3  36  12 21 7
   




9 3
81
32  62  62
Shortest distance of P from AB
2
49
320 8 5
7
 BP 2     41 


3
9
3
3
 
(ii)[4]
BAPQ forms a parallelogram.
Hence AB  PQ
 3   12   9 
    

 OQ  AB  OP   6    4    10 
 6   10   16 
    

Area of parallelogram BAPQ
 AB  AP
 3   2 
  

  6    12 
 6  4 
  

 48 
 
0
 24 
 
 482  02  242
= 53.67 (to 2 dec places)
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2010 H2 Maths Preliminary Exam Paper 1
Page 10 of 14
4
1
,x  .
3
(1  3x) x  4
Express f ( x) as a series expansion in ascending powers of x, up to and including the term in x 3 .
[4]
Given that f ( x) 
13
(i)
2
(ii)
Find the range of values of x for which the expansion in (i) is valid.
(iii)
A curve is given by the equation
13[11]
(i)[4]
[2]
2 tan y  f (x).
By differentiation, find the series expansion for y in ascending powers of x, up to an including the
term in x 2 .
[5]
4
f ( x) 
(1  3 x) x 2  4
  x2 
 4(1  3 x) 1  4 1   
4 
 
 x2 
4

(1  3 x) 1 1  
4 
4



1
2
1
2
 1  x2 

 2 1  3 x  (3 x) 2  (3 x)3  ... 1     ...
 2 4 



 1

 3

 2 1  3 x     9  x 2     27  x 3  ...
 8

 8



71
213 3
 2  6x  x2 
x  ...
4
4
(ii)[2]
For expansion in (i) to be valid,
x2
3x  1 and
1
4
1
and x 2  4
3
1
 x
and x  2
3
 1 1
Range of values of x    , 
 3 3

x
__________________________________
2010 H2 Maths Preliminary Exam Paper 1
Page 11 of 14
(iii)[5]
2 tan y  f ( x) ----- (1)
Differentiate (1) with respect to x,
dy
2sec 2 y
 f '( x) ----- (2)
dx
Differentiate (2) with respect to x,
2

d2 y 
 dy 
2
2  2sec y tan y    sec 2 y 2   f "( x) ----- (3)
dx 
 dx 

When x  0,
from (i), f (0)  2, f '(0)  6 and f ''(0) 
(1) gives
2 tan y  2
(2) gives
2(2)
(3) gives
Hence y 
dy
6
dx
 y
71
2
π
4
dy 3

dx 2
2

d 2 y  71
3
2  2(2)(1)    2 2  
dx  2
2


d 2 y 35


dx 2
8
2
π 3
π 3
35
 35  x
 x     ...   x  x 2  ...
4 2
4 2
16
 8  2!
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2010 H2 Maths Preliminary Exam Paper 1
Page 12 of 14
14 (a)
The sum of the first n terms of a series, Sn , is given by
pn
 5 , where p is a non-zero constant
5 n 1
and p  5 .
Obtain an expression for Tn , the nth term of the series and prove that this is a geometric series.
Find the range of values of p for the sum to infinity to exist.
(b)
[5]
An arithmetic progression is grouped into sets of numbers as follows:
{2} , {6 , 10} , {14 , 18 , 22} , {26 , 30 , 34 , 38} , …
where the number of terms is 1 for the first set, 2 for the second set, 3 for the third set, 4 for the
fourth set and so forth.
(i) Show that the first term in the nth set is 2n 2  2n  2 .
(ii) Find, in terms of n, a simplified expression for the last term in the nth set.
(iii) Hence, find the sum of all the terms in the nth set.
Q14[10]
(a)[5]
[5]
pn
Given S n  n 1  5.
5
p n 1

Sn 1  n  2  5.
5
Tn  Sn  Sn 1

 p n 1

pn
 n 1  5   n 2  5 
5
5

n- 1
æp ö
= çç ÷
÷
÷
èç 5 ø
é
æ 1 öù
êp - çç - 1 ÷
ú
÷
÷ú
çè5 ø
êë
û
n- 1
æp ö
= ( p - 5)çç ÷
÷
çè 5 ÷
ø
n- 2
æp ö
Now, Tn- 1 = ( p - 5)çç ÷
÷
÷
çè 5 ø
n- 1

Tn
Tn- 1
æp ö
( p - 5)ççç ÷÷÷
è5 ø
=
n- 2
æp ö
÷
( p - 5)ççç ÷÷
è5 ø
=
p
5
(constant)
 series is a geometric series.
For the sum to infinity to exist,
(shown)
p
1
5
.
 5  p  5, p  0
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2010 H2 Maths Preliminary Exam Paper 1
Page 13 of 14
(b)(i)[5]
The first terms are: 2 , 6 , 14 , 26 , …
The difference of the numbers in the pattern follows an AP with a = 4 and d = 4.
The summation of the common difference gives the subsequent terms in the number pattern.
Thus the first term in the nth set
 n 1 
 2
  2(4)   n  2  (4) 
 2 
 2   n  1 4  2n  4   2n2  2n  2
(ii)
The last terms are: 2 , 10 , 22 , 38 , …
Method 1:
The difference of the numbers in the pattern follows an AP with a = 8 and d = 4.
The summation of the common difference gives the subsequent terms in the number pattern.
Thus the last term in the nth set
 n 1 
 2
  2(8)   n  2  (4) 
 2 
 2   n  1 2n  4  2n2  2n  2
Method 2:
The last term in the nth set
 First term in the (n+1)th set  4
 2(n  1)2  2(n  1)  2  4
 2n 2  2n  2
(iii)
Sum of all terms in nth set
n
    2n 2  2n  2    2n 2  2n  2  
2
n
    4n 2   2n3
2
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2010 H2 Maths Preliminary Exam Paper 1
Page 14 of 14