2011 HCI H2 Math Prelim Exam Paper 1 Solutions
Qn
1
Page 1
Solutions
Let x , y and z be the number of red, blue and green matchsticks respectively.
x  y  z  88    1
z
 x  y  6x  6 y  z  0     2
6
y z
x
  4    6 x  2 y  z  0     3
3 6
4
Solving the inequality
x  16 , y  24 and z  48
Number of blue matchsticks = 24
2(i)
x
h
x
( 2r )
r
2r
2
R
= x2 + x2
h2 = R 2 - r 2
Þ r 2 = R 2 - h2
Þ x 2 = 2r 2
1
2r 2
2
V = x2h =
h = h R 2 - h2
Shown
3
3
3
dV 2 2
2
2
= R - 2h2
V  R 2 h  h3 
dh 3
3
3
dV
R
2
= 0 , R 2  2h 2  h =
When
.
dh
3
3
(
(ii)
) (
)
d 2V
 4h  0 since h  0 .
dh 2
OR
h
æ Rö
çè ÷ø
3
dV
dh
>0
-
 V is maximum when h =
R
3
0
R
3
æ Rö
çè ÷ø
3
<0
.
+
2011 HCI H2 Math Prelim Exam Paper 1 Solutions
3
x 2  2 x  15 1

 x  5 x  1 2
5  x  2.4031 or 1  x  10.403
 5  x  2.40 or 1  x  10.4
ALTERNATIVELY (using analytical method)
x 2  2 x  15 1
x 2  8 x  25
  2
0
 x  5 x  1 2 x  4 x  5
Roots to x 2  8 x  25  0 are 2.4031 and 10.403.
Roots to x 2  4 x  5  0 are 1 and 5.

+
5
+

2.4031 1
+
10.4031
5  x  2.4031 or 1  x  10.4031
 5  x  2.40 or 1  x  10.4
Replace x by x  2
 x  2   2  x  2   15  1
 x  2  5 x  1  2  2
2

x 2  2 x  15 1

 x  7  x  1 2
5  x  2  2.4031 or 1  x  2  10.4031
 7  x  4.40 or  1  x  8.40
Page 2
2011 HCI H2 Math Prelim Exam Paper 1 Solutions
4
z 3  4  4 3 i
 4 1  3 i

Page 3

1  3 i  2
arg(1  3 i)  
 z  8e
3
2
3
 2 
i 

 3 
1  2

i 
 2 k 
3 3

z  2e
z  2e
 2 2 k 
i 


3 
 9
, where k  1, 0, 1
, where k  1, 0, 1
Im
2
2
3
–2
z2
8 
9
z1
4
9
O
2 
9
2
2
z3
–2
From the Argand diagram,
2 
1
Area of triangle  3  (2)(2)sin

3 
2
 3 3 (shown)
Re
2011 HCI H2 Math Prelim Exam Paper 1 Solutions
5(i)
yn 1 xn 1  2

yn
xn  2
1
xn  1  2
1 x 2 1
2
  n
  = constant
xn  2
2  xn  2  2
y1 , y2 , y3 ,
is a geometric sequence with common ratio
y1  x1  2  3
1
2
n 1
Therefore, yn  3  
1
xn  yn  2  3  
2
.
n 1
2
n 1
(ii)
Page 4
1
3   2
xn
1 x  1
2
 n  n   n   n 1
n
2 yn 2  yn  2
1
3 
2
1  2 n 1  1 1
 n 1   2    n 
2  3
 2 3
1
decreases to 0,
2n
x
1
 n n decreases to  .
3
2 yn
When n   ,
1
.
2
2011 HCI H2 Math Prelim Exam Paper 1 Solutions
6(a)
1 x
 1  3x
2
1
6x
1
dx  
dx
2

6 1  3x
1  3x 2
1
1
1
  ln 1  3 x 2   
dx
6
3 1/ 3  x 2
1
3
  ln 1  3 x 2  
tan 1 3 x  C
6
3
dx  
 
(b)
2
2
d  x2
xe  e  x  2 x 2e  x
dx
2
Using by parts with v '  1  2 x2 e x , u   x ,

 x  2x
2

 1 e x dx    x 1  2 x 2  e  x dx
2
2
  x 2 e x   xe x dx
2
2
2
1 2
  x 2e x  e x  C
2
Page 5
2011 HCI H2 Math Prelim Exam Paper 1 Solutions
7(i)
z  3i  iz  2  z  (3i)  i z 
Page 6
2
i
 z  (3i)  z  (2i)
And
4  3i  z  i  z  (i)  5
Im
4
O
Re
–1
5
–2.5
Locus of z
–6
7(ii)
arg( z  i)= arg( z  (i))
–1 A

–2.5
C
From diagram, sin  

5
B
1.5 3

5 10


 3
 3
   sin 1   and       sin 1  
2
2
 10 
 10 

 3
Hence     arg( z  i)   sin 1  
2
 10 
 3
 3
   sin 1    arg( z  i)   sin 1  
 10 
 10 
2.84  arg( z  i)  0.305
OR
7
(iii)
D
–1 A
1
–2.5
E
Greatest value of z 1 is the line segment DE that passes through A, the centre of the
circle.
 greatest z  1  DE  DA  AE  2  5
2011 HCI H2 Math Prelim Exam Paper 1 Solutions
8(i)
Page 7
1
1
1
 
r (r  1) r r  1
n 1

1

n 1

1
1 
  4r  2  r (r  1)    4r  2   r  r  1 

 r 1 
1
1


 4 r   2    

r 1 
r 1
r 1
r 1  r
n 1
1 1
4
1  (n  1)  2(n  1)   
2
 1 2
1 1
 
2 3
1 1
 
3 4
 ...
1
1

 
n 1 n 
1
 2( n  1)(1  (n  1)  1)  1 
n
1
1
2
 2( n  1)( n  1)  1   2n  2  1 
n
n
1
2
 2n  3 
n
1
un 1  un  4n  2 
n(n  1)
1
un 1  un  4n  2 
n(n  1)
n 1
n 1

1 
 ur 1  ur     4r  2 


r (r  1) 
r 1
r 1 
n 1
1
 ur 1  ur   2n 2  3 

n
r 1
1
u2  u1  2n 2  3 
n
1
1
un  u1  2n 2  3   un  3  2n 2  3 
 u3  u2
n
n
 u4  u3
1
2
 ...
 un  2n 
(Shown)
n
 un  un 1
r 1

n 1
8(ii)
8
(iii)
n 1
n 1
Let Pn be the statement un  2n 2 
1
for all n 
n
When n  1 , LHS  u1  3
1
RHS  2(1) 2   3  LHS
1

.
2011 HCI H2 Math Prelim Exam Paper 1 Solutions
Page 8
 P1 is true.
Assume Pk is true for some k   ,
1
i.e. uk  2k 2 
------- (*)
k
To prove Pk+1 is also true, i.e. uk 1  2(k  1) 2 
1
.
k 1
1
(k )(k  1)
1
1
from (*)
 2k 2   4k  2 
k
(k )(k  1)
1
1
 2k 2  4k  2  
k (k )(k  1)
k  1 1
 2(k 2  2k  1) 
(k )(k  1)
1
 2(k  1) 2 
= RHS
k 1
Thus Pk is true  Pk 1 is true .
Since P1 is true, and Pk is true  Pk 1 is true , by Mathematical Induction, Pn is true for all
LHS  uk 1  uk  4(k )  2 
n
9
(a)

.
y  ux 2
dy
du
 2 xu  x 2
dx
dx
du 

2
2 2
x 2  2 xu  x 2
  2 x  ux    ux 
dx 

du
x4
 u 2 x4
dx
du
 u2
dx
1
 u 2 du   1dx
1
  xC
u
x2
  xC
y
y
x2
xC
2011 HCI H2 Math Prelim Exam Paper 1 Solutions
(b)
Page 9
dv
 k  mv , where m is positive.
dt
0  k  m 1
k m
dv
 k  kv  k 1  v 
dt
1
 1  v dv   k dt
 ln 1  v  kt  C
ln 1  v  kt  C
1  v  Be  kt  v  1  Be  kt where k  0.
The rate of discharge is always less than the rate of water poured into the device  v is
always increasing. This implies that B > 0.
v
v  1  e kt  where B  1
1
t
O
2011 HCI H2 Math Prelim Exam Paper 1 Solutions
10.
a(i)
y
Page 10
y  3( x  1)
2
1,1
x
O
1, 1
2
y   3( x  1)
2
(ii)
2
 y
The curve     x  1  1 is an ellipse with centre at (1, 0). The horizontal distance
h
from its centre is 1 and its vertical distance from its centre is |h|.
The curve y  3( x  1)2  1 is the part of the curve in (i) that lies above the x-axis.
y
y  3( x  1)
3
1,1
|h|
x
O
1 1,0 
 3
y   3( x  1)
From observation, if h  1(i.e. h  1) , there will be exactly one point of intersection. In
order for the curves to intersect at two points, h  1 or h  1 .
2011 HCI H2 Math Prelim Exam Paper 1 Solutions
Page 11
b(i)
y x 1
1
2
1

y
x
1
2
1
O
f ( x)
b(ii)
y
y 2  f ( x)
2
1
 2
O
x
2011 HCI H2 Math Prelim Exam Paper 1 Solutions
11
(i)
Page 12
1
 x 2
4  2 x  2 1  
 2
 

 




1 1 3
2
3
1 1
1 x
2
2  x   ...
2  x  2
 2 1     2
 
 
 2 2

2!
3!
2
2


1
1
1
x  x 2  x3  ...
2
16
64
x
Expansion is valid if
 1  x  2  2  x  2.
2
d
dy
: ey
 3cos 3 x .
dx
dx
 2
(ii)


2
d
d2 y
y  dy 
: ey
 e    9sin 3x  9 e y  1
2
dx
 dx 
dx
2
d 2 y  dy 
 2     9e  y  9  0
dx  dx 
Shown  .
2
d d3 y
 dy   d y 
 y dy
:

2

9e
 0.




dx dx3
dx
 dx   dx 2 
dy
d3 y
d2 y
 3,


9
 27 .
,
dx
dx 3
dx 2
3
9 2 27 3
9
9
y  x
x 
x  ...  3 x  x 2  x3  ... .
1!
2!
3!
2
2
2
1

x
 e

ln 
 2 1  x  ln 1  sin 3 x 
 1  sin 3 x 


When x  0 , y  0 ,
(iii)
2 1 x  4  4x  4  2 2x
 2
1
1
1
1
1
2
3
 2 x    2 x    2 x   ...  2  x  x 2  x3
2
16
64
4
8
 4  4 x  ln 1  sin 3 x 
1
1
9
9

 

  2  x  x 2  x 3  ...    3 x  x 2  x 3  ... 
4
8
2
2

 

17
35
 2  2 x  x 2  x 3  ...
4
8
2
1

x
 e

0.1
0.1
17 2 35 3
 0 ln  1  sin 3x  dx   0 2  2 x  4 x  8 x dx


 0.19131
 to 5 dp 
2011 HCI H2 Math Prelim Exam Paper 1 Solutions
12
(i)
 5     1

  
 9      1  0
 6    1

  
 5    9    6    0    8
ALTERNATIVELY
 x
 
Let r   y  in r  5i  9 j  6k    i  j  k   and substitute x  5   , y  9   ,
z
 
z  6   into the equation x  y  z  0 .
 5    9    6    0    8
 5  8   3 

  
 OP   9  8    1 
 6  8   2 

  
(ii)
The point P which is on l2 also lies on  --- (1).
  1   0    1
 1   1  0   1
      
       
  0   2  1   1    0   1  2  1  1
 1  1  1   1
  1  1   1
       
  1  1  2(1  1)  0
 l2 is parallel to  --- (2)
(1) and (2)  l2 lies on  .
ALTERNATIVELY
 1
Sub r in r  3i  j  2k     i  k   2   j  k   into r. 1
, we obtain
 1
 
 3    1

  
 1  2   . 1
 2    2  1

  
 3    1  2  2    2
0
This means that all points on l2 are on .
Page 13
2011 HCI H2 Math Prelim Exam Paper 1 Solutions
(iii)
Page 14
  1   0   1 
      
  0   2  1    1 
  1  1    1
1 1 0 1
1 0
       
   
   0    1   2  1   1     0   2  1 
 1  1  1   1
1 1
       
   
 3    1   0  
      
 1  .    0   2  1    3  2   2  4   6  
 2   1 1
      
  1   0 
    
 r    0   2  1    6   (Shown)
  1   1  
(iv)
 1   1   0  
2
2
      
2
1    0   2  1    3   4    2  3
 1   1   1  
      
2
2 2
  2    2    3  2  4    2 
9


3  2  4   2  2  4   2  4  4 
2
 8 2  40  32  0
   1  4   0
   1 or   4
 1  0  
 1
    
 
If   1: r   0    2    7  r   2   7
1
 1  2  
 
 4   0  
 4 
    
 
If   4 : r   0    2    10  r   2   10
 2 
 4   2  
 