H2 MATHS (9740) JC2 PRELIMINARY EXAM 2010
PAPER 1 SUGGESTED SOLUTIONS
Qn
1
Solution
Method of Differences
N
N 
r 1
r
2 
f
(
r
)








 r  1!  r  2 ! 
r 2
r 2  r !
1 2 2
  
 2! 1! 0!
2 3 2
  
3! 2! 1!
3 4 2
  
4! 3! 2!
4 5 2
  
5! 4! 3!


N 3
N 2
2


 N  2  !  N  3 !  N  4 !

N 2
N 1
2


 N  1!  N  2 !  N  3!

N 1
N
2 



N !  N  1 !  N  2 !
 2  2  2 
N 2
N
N 1


 N  1!  N  1! N !
N  N  2   N 2   N  1
 2
N!
N 1
 2
N!
N 1
 0 , hence the series is convergent.
N!
As N  ,

 f (r )  2
r 2
Qn
2
(i)
(ii)
Solution
Integration (by part)


2
d x2 1
e
 2 xe x 1
dx
 x e dx
1
=   2 xe   x  dx
2
3 x2 1
x 2 1
2


2
1 2 x2 1
x e   2 xe x 1dx
2
2
1 2 x2 1 x2 1
e x 1 2
x e e
c 
=
 x  1  c
2
2
=

(iii)

1
0
x e

3 x 2 1

 e 2 dx
1
2
2
1

=   x 2e x 1  e x 1   e2 x 


2
0
1
 1

=   e 2  e 2   e 2     e  
2
 2

2
= e  0.5e
Qn
3
(i)
Solution
Recurrence Relation
As n  , xn   & xn1  

3
92
 2 
2
2
  2  2     9 2  0
  2    7   0    0 or   7.
Thus   7 since   0.
(ii)
Method 1 [Graphical]
3xn
xn 1  xn 
 xn
2  xn
Sketch y  xn 1  xn 
3xn
 xn
2  xn

From the graph:
For 0  xn   : xn1  xn  0  xn1  xn
Therefore the sequence is strictly increasing.
For xn   : xn1  xn  0  xn 1  xn
Therefore the sequence is strictly decreasing. [Shown]
Method 2 [Graphical]
3x
Sketch y  x & y 
2 x
xn
xn 1

xn 1
xn
From the graph:
For 0  xn   : xn1  xn
Therefore the sequence is strictly increasing.
For xn   : xn1  xn
Therefore the sequence is strictly decreasing.
[Shown]
Method 3 [Algebraic]
3x  x 2  xn
3xn
xn 1  xn 
 xn  n n
2  xn
2  xn


xn 3  2  xn

2  xn
For 0  xn   :
2  xn  0, xn  0 & 3  2  xn  0 since 2  xn  9  3
Thus: xn1  xn  0  xn1  xn
Therefore the sequence is strictly increasing.
For xn  
2  xn  0, xn  0 & 3  2  xn  0 since 2  xn  9  3
Thus: xn1  xn  0  xn 1  xn
Therefore the sequence is strictly decreasing.
[Shown]
Qn
4
(i)
Solution
Binomial Expansion
x4
A
B


Let f  x  
 x  1 3x  2 x  1 3x  2
 x  4  A  3x  2  B  x  1
Using cover up rule,
2
2
1
For x   ,   4  B    B  14
3
3
3
For x  1 , 1  4  A  1  A  5
f  x 
5
14

x  1 3x  2
5
14

x  1 3x  2
1
1
 5 1  x   14  3 x  2 
f  x 
 5 1  x 
1
 5 1  x 
1
  3x  
 14  2 1   
2 
 
14  3 x 
 1  
2
2 
 5 1  x  x 2  x 3 
1
1

2
3

 3 x   1 2   3 x   1 2  3  3 x 
7 1   1   
  
  

2!
3!
 2 
 2 
 2 

 3x 9 x 2 27 x3


 
 5 1  x  x 2  x 3   7 1  
2
4
8


11
43 2 149 3
 2  x  x 
x 
2
4
8
(ii)
Expansion of 1  x 
1
 3x 
Expansion of 1  
2 




is valid for 1  x  1
1
is valid for 
2
2
x
3
3
Therefore, the range of values of x for the expansion of f(x) to be valid is
2
2
 x .
3
3
(iii)
Coefficient of x   1 5   1
n
n
Qn
5
n 1
n
n
n 
3
3 
7     1 5  7   
2
 2  

Solution
Complex Numbers (Loci)
Im
z  5  3  7i
p
6  3i
Re
5
4  3i
q
z  6  3i  z  4  3i
Cartesian equation of (i):
( x  5)2  y 2  42
---- (1)
Gradient of line segment joining 6  3i and 4  3i
=

3  3
64

3
Cartesian equation of (ii):
y
1
 x  5
3
---- (2)
Using GC to solve (1) & (2), we get
p  1.5359  2i
q  8.4641  2i
 4  5
 arg( p  q)    sin 1   
 2.62 (3 s.f.)
8 6
Alternative:
p  q  6.9282  4i
 arg  p  q     tan 1
4
 2.62 (3s.f.)
6.9282
Or using GC,
arg  p  q   2.62 (3s.f.)
Qn
6
Solution
Differential Equations
d
 k   30  , k  0
dt
1
   30 d    k dt
ln   30   kt  C
  30  e  kt C
  30  e kt eC  Ae kt (where A  eC )
  30  Ae kt
when t  0,   90
90  30  A
A  60
  30  60e  kt (shown)
when t  8,   55
55  30  60e 8 k
5
12
1 5
k   ln
8 12
e 8 k 
when   35,
1 5 
 ln t
12 
35  30  60e 8
1 5 
 ln t
 8 12 
1
12
t  22.707
e

 additional time needed  22.707  8  14.7 min (1d.p.)
Qn
7
(a)
Solution
AP and GP
1
 d  a
2
Sum of first n even-numbered terms
ar  a  3d

n 1 

2  a   (n  1)   a  

2 2 

n
 2a  an
2
an  2  n 

2
Amount owed at the beginning of the third month

(b)(i)
  50000  x  (1.035)  x  (1.035)
(b)(ii)
 50000(1.0352 )  (1.035  1.0352 ) x
Amount owed at the beginning of the nth month
 50000(1.035n 1 )  (1.035  1.0352 
 1.035n 1 ) x
For the repayment to be completed during the nth payment,
50000(1.035n1 )  (1.035  1.0352 
 1.035n1 ) x  x
50000(1.035n 1 )  (1  1.035  1.0352 
1.035n  1
x
1.035  1
1.035n  1

x
0.035

Thus x 
1750(1.035n 1 )
.
1.035n  1
 1.035n 1 ) x
Qn
8
(i)
Solution
Transformation of Graphs + System of Linear Equations
3

A '  , 75 
2

y
C '  0,12 
1 
B ' , 3
2 
0
x
y  f 1  2 x 

A " 2, 72
y

C "1, 3
(ii)
B " 0, 0 


0
C "' 1,  3

A "' 2,  72

Let the curve be y  ax3  bx 2  cx  d .
Since the points  2, 75 ,  0,3 and 1,12  lie on the curve.
Using  0,3 , d  3
Using  2, 75 ,
a  2   b  2   c  2   d  75
3
2
8a  4b  2c  72
Using 1,12  ,
a  b  c  d  12
abc  9
(1)
(2)
Since 1,12  is a maximum point,
3ax 2  2bx  c  0
3a  2b  c  0
(3)
dy
 0.
dx
x
Using GC to solve (1),(2) and (3),
a  8, b  7, c  10
Thus the equation of the curve is y  8x3  7 x 2  10 x  3 .
Qn
9
Solution
Vectors (line and planes)
(i)
Vector equation of the line l
1
1
 
 
r = 0    1 ,

1
 1
 
 
Angle between OA and the line l
01
  
 1   1 
 2   1 
3
3
  
cos  


5 3
15
15
o
o
  39.232  39.2 1 d.p.
(ii)
Let the shortest distance from the origin to the line l be x.
x
5
x  1.41 3 s.f.
sin  
(iii)
Since
1
 
0
1
 
1
 
 2   4  A point on l lies on  1
 3
 
 1  1
   
and  1   2   0  l is parallel to  1
 1   3 
   
The line l lies on the plane (shown)
Alternative Solution (1):
Since
0
 
 -1
2
 
1
 
0
1
 
1
 
 2   4  Point A on l lies on  1
 3
 
1
 
 2   4  Point B on l lies on  1
 3
 
 line l lies on  1
Alternative Solution (2):
Since




1
 1 
 
 
 0     1 
1
 1  
 
 
1     1 

 
     2
1     3 

 
1
 
 2
 3
 
 1    2  3  3
4
 line l lies on  1
(iv)
(v)
 1  1  5 
     
normal of  2   1    2    4 
 1  3   1 
     
 5  1  5 
 2 : r  4    0   4 
 1  1  1 
     
 5
 2 : r  4   6
1
 
 1   5   14 
1
    

 
normal of  3   2    4    14   14  1 
 3   1   14 
 1
    

 
OR
1
 
normal of  3  director vector of l   1 
 1
 
1
 3 : r  1   d
 1
 
perpendicular distance from origin to  3 
d
d

3
1
 
1
 1
 
d
 3
 d  3
3
1
 3 : r  1   3
or
 1
 
3 : r
1
 
 1   3
 1
 
Qn Solution
10 Functions + Maclaurin’s Series
(i) R g  [ ln 2, ) and Df 
Since R g  Df , the composite function fg exists.
(ii)
1 

fg( x)  cos  ln

 1 x 
 cos   ln 1  x  

x 2 x3 
 cos   x   
2 3

2
4

x 2 x3 
x 2 x3  
 

x



x

 

2 3
2 3 



 1
4!
2!
2
2

x 2   x3   x3  
x2 
1 
 1    x    2   x         
2  3   3  
2
2 



1
 1   x 2  x3  
2
1
1
 1  x 2  x3 
2
2
(iii)
h(1)  1 so h 1 (1)  1 .
Let y   ln(1  x) . Then x  e  y  1 .
For 1  x  1 , h( x)   ln(1  x) so h( x)   ln 2 .
Thus h 1 ( x)  e x  1 for x   ln 2 .
Thus
e x  1 for x   ln 2,
1
h :x
1
for
x  1.
Qn
11
(a)
(b)
Solution
Complex Numbers



z  3  i  2  cos  i sin 
6
6

n
n 

z n  2n  cos
 i sin

6
6 

n
n 

z n   z *   2i Im( z n )  2i  2n sin
0
6 

n
sin
0
6
n
 k , k  
6
The set of values of n is n : n  6k , k 
zw

 
3 1  i 1  3



Im


W
i 3
i 1 3
P
i
Z
Re
1
3 1
O
3
Note that OWPZ is a parallelogram,
3 2
arg( w)    tan 1

1
3
2  
 
3 6 2
OZ= z  2
WOZ 
OW= w  1  i 3  2
Since WOZ 

2
and OZ  OW , OWPZ is a square. (shown)
arg( z  w)  arg( z )  POZ 
Also, arg( z  w)  tan 1

5
3 1
 tan 1
12
3 1
5
3 1
 tan


12
3 1


6


4

5
12
3 1
3 1

3 1
3 1
2

42 3
 2  3 (deduced)
2
Qn
12
(i)
Solution
Graphing + tangent/normal
2 x 2  3x  1
Let y 
x2  4
yx 2  4 y  2 x 2  3x  1
 2  y  x 2  3x   4 y  1  0
For real roots of x,
Discriminant  0
 3
2
 4  2  y  4 y  1  0


9  4 8 y  2  4 y2  y  0
16 y 2  28 y  1  0
For 16 y 2  28 y  1  0
y

28 
 28  4 16 
2 16 
2
28  720
32
73 5
8
2
16 y  28 y  1  0

y
73 5
8
y
or
73 5
8
Hence, C cannot lie between
73 5
73 5
and
8
8
(ii)
y
y
2 x 2  3x  1
x2  4
y2
 0.764,0.0365 5.23,1.71
1
1 2

4
O
x  2
(iii)
At x  5,
Using GC,
y
dy
dx
22
7
 0.40136
x 5
x
1
x2
Equation of tangent at x  5 is
22
 0.40136  x  5
7
y  0.40136 x  5.149657
When x  0, y  5.149657  5.15 (to 3 s.f)
The coordinates of P are  0,5.15 (shown)
y
Equation of normal at x  5 is
22
1

 x  5
7 0.40136
y  2.4915 x  9.3148
When x  0, y  9.3148  9.31 (to 3 s.f)
The coordinates of Q are  0, 9.31 (shown)
y
1
 5.15  9.31 5 
2
= 36.2 units2
Area of PQR =