1

1. Mr Wang had a total of $37 000 in his savings accounts with Action Bank, Bonus
Bank and Champion Bank at the beginning of 2007. Saving accounts at Action Bank,
Bonus Bank and Champion Bank enjoy interest rates of 1%, 0.5% and 0.3% per
annum respectively. He had a total of $37 240 in the three banks at the end of 2007
and $37 481.89 at the end of 2008. Assuming that he did not deposit or withdraw any
money in 2007 and 2008, find the amount of money he had in his Action Bank
account at the end of 2008. [4]

2. Solve ( ) cos ln d x x x

. [5]

3. Solve the inequality ln
2
x
x > . Deduce the solution of x > 2 2 ln x. [5]

4. A curve is defined by the parametric equations
2
x t = and sin y t = , for t < .
(i) Sketch the curve, indicating clearly the axial intercepts. [2]
(ii) Find the equation of the normal to the curve that is parallel to the y-axis. [4]

5. The first three terms of a geometric progression are ( ln ) a b + , ( ln ) a b + and
3
( ln ) a b + , where b > 1. Show that a =
1
ln
4
b . Find the common ratio and deduce
that the geometric progression is convergent. Given that b = e
2
, find the sum to
infinity of the progression. [6]

6. Express
2x
2
+ 3x 1
(x 1)(x
2
+ 1)
in partial fractions.
Hence find the series expansion of
2x
2
+ 3x 1
(x 1)(x
2
+ 1)
in ascending powers of x, up to and
including the term in x
4
. Find the coefficient of x
2008
. [6]

7. Find the distance between a point (x, y) on the curve
2
x
y e = and the point (1, 1) in
terms of x. Hence find the coordinates of the point on the curve
2
x
y e = that is closest
to the point (1, 1), giving your answer correct to 3 decimal places. [6]

2
8. A communicable disease is spreading within a small community with a population of
1000 people. A scientist proposes that the infected population, x, at time t days after
the start of the spread of the disease, satisfies the differential equation
d
(1000 )
d
x
kx x
t
= , where k is a positive constant.
Initially one person in this community is infected and five days later, 12% of the
population is infected. Find the time taken for half the population to contract the
disease. State an assumption made by the scientist. [7]

9. A sequence is defined by u
1
= 0 and
n n
u n u n + = +
+1
) 1 ( for n
+
.
Prove by mathematical induction that
1
1
!
n
u
n
= for n
+
. [5]
Hence find the exact value of
1
1
2
r
r
r
u

. [2]

10. The sequence u
r
, r = 1, 2, 3, , is defined by
[ ] [ ]
2 2
( 1) ( 1)
r
u r r r r = + .
Find

=
n
r
r
u
1
. [2]
(i) By simplifying u
r
, deduce that
2
1
3
2
) 1 (
(

+
=

=
n n
r
n
r
. [2]
(ii) By considering

n
r
r
1
3
) 1 ( , show that
2
1
1
( 1)(2 1)
6
n
r
r n n n
=
= + +

. [5]

11. The diagram shows the graph of y = f(x):

On separate clearly labelled diagrams, sketch the graphs of
(i) y =
1
f(x)
,
(ii) y = f( | x | ),
(iii) y = f (x). [9]
(1, 0)
y = x
(1, 2)
y =
1
2

x
y

3
12. Given that
1
tan y x x

= , prove that
( )
2
2
2
d d
1 2 2 2 0
d d
y y
x x y
x x
+ + = .
By repeated differentiation, show that the first 2 non-zero terms of the Maclaurins
series for y is
2 4
1
3
x x . Hence evaluate
1
3
0
tan
lim
x
x x
x

. [9]

13. (i) Use the substitution 2 1 u x = + to find 2 1 d x x x +

. [3]
The region R is bounded by the curve 2 1 y x x = + , the y-axis and line
1
2
y = .
(ii) Find the exact area of region R using your result in part (i). [3]
(iii) Find the volume of the solid generated when R is rotated through four right
angles about the xaxis. [3]

14. The functions f and g are defined as follows:

2
f : 3 2 x x x , , x x k ,

4
g : , 0 4
x
x e x

.
State the largest value of k such that
1
f

exists, and find

1
f

in a similar form. [4]

(i) Show that the composite function gf does not exist. [1]
(ii) If h is a restriction of f, write down the maximal domain of h such that the
composite function gh exists. Define gh in a similar form and state its range.
[4]
(iii) Find the set of values of x such that g
1
g(x + 1) = g g
1
(x + 1). [3]


END OF PAPER

4
2008 C1 H2 Mathematics Promotional Examination Solution:
1 Let a, b, c be the amount in his Action Bank, Bonus Bank and Champion Bank accounts at
the beginning of 2007.
a + b + c = 37 000 (1)
1.01a + 1.005b + 1.003c = 37 240 (2)
1.01
2
a + 1.005
2
b + 1.003
2
c = 37 481.89 (3)
From GC, a = 15 000, b = 12 000, c = 10 000.
So at the end of 2008, he had 1.01
2
15 000 = $15 301.50 in his Action Bank account.
2
( )
( )
2
cos ln '
1 1
' sin ln
2
u x v x
u x v x
x
= =
= =

( )
( ) ( )
2
cos ln d
1 1
cos ln sin ln d
2 2
x x x
x x x x x = +

( )
( )
2
sin ln '
1 1
' cos ln
2
u x v x
u x v x
x
= =
= =

( )
( ) ( ) ( )
2 2
cos ln d
1 1 1 1
cos ln sin ln cos ln d
2 2 2 2
x x x
x x x x x x x
(
= +
(

( ) ( ) ( )
( ) ( ) ( )
2 2
2 2
5 1 1
cos ln d cos ln sin ln
4 2 4
2 1
cos ln d cos ln sin ln
5 5
x x x x x x x
x x x x x x x C
= +
= + +

3

From GC, 0 < x < 4.42806 or x > 13.706
i.e. 0 < x < 4.42 or x > 13.8
Replacing x by x
2
:
x
2
2
> ln x
2
x > 2 2 ln x
From above, 0 < x
2
< 4.42806 or x
2
> 13.706
0 < x < 2.1043 or x > 3.702
0 < x < 2.10 or x > 3.71

5
4(i)






4(ii) d d
2 cos
d d
x y
t t
t t
= =
d d d cos
d d d 2
y y t t
x t x t
= =
Normal // y-axis Tangent // x-axis
d
0
d
y
x
=
cos
0 cos 0
2
or
2 2
t
t
t
t

= =
=

When
2
,
2 4
t x

= =
Equation of normal is
2
4
x

=
5
Since the 1st 3 terms are in G.P.,
2
1 1
( ln ) ( ln )( ln )
2 3
a b a b a b + = + +
2 2 2 2
1 4 1
ln (ln ) ln (ln )
4 3 3
a a b b a a b b + + = + +
1 1
ln
3 12
a b =
1
ln
4
a b =
Common ratio, r =
1 1
ln ln
1
2 4
3
ln 3
ln
4
a b b
a b
b
+
= =
+

Since 1
3
1
< = r , the G.P. is convergent.

Given that b = e
2
,
2
1 1 1
ln 2ln
4 4 2
a e e = = =
2
ln 3 1 9
( 2)
1
2 2 4
1
3
a e
S

+
= = + =

6
Let
2x
2
+ 3x 1
(x 1)(x
2
+ 1)
=
A
x 1
+
Bx + C
x
2
+ 1

2x
2
+ 3x 1 = A(x
2
+ 1) + (Bx + C)(x 1)
Let x = 1: 4 = 2A A = 2
Let x = 0: 1 = 2 C C = 3
Compare coefficients of x
2
: 2 = 2 + B B = 0

2
O


6
2x
2
+ 3x 1
(x 1)(x
2
+ 1)
=
2
x 1
+
3
x
2
+ 1

= 2(1 x)
1
+ 3(1 + x
2
)
1
= 2(1 + x + x
2
+ x
3
+ x
4
+ ...) + 3(1 x
2
+ x
4
...)
= 1 2x 5x
2
2x
3
+ x
4
+...
Coefficient of x
2008
= 2 + 3 = 1
7
Distance between (1, 1) and (x, y), ( ) ( )
2 2
1 1 W x y = +
Since (x, y) lies on the curve, then
2
x
y e =
Thus, ( )
( )
2
2
2
1 1
x
W x e = +
( )
( )
2
2
2
2
1 1
x
W x e = +
Differentiating w.r.t. x, ( )
( )
2 2
d
2 2 1 1
d
x x
W
x e e
x
= +
( )
( )
2 2
d 1
1 1
d 2
x x
W
x e e
x
= +
When ( )
( )
2 2
d 1
0, 1 1 0
d 2
x x
W
x e e
x
= + =

From GC, x = 0.70160785
0.70160785
-
0.70160785 0.70160785
+
d
d
W
x
-ve 0 +ve
\ /
Thus, W is minimum when x = 0.70160785
When x = 0.70160785,
0.70160785
2
1.4202 y e = =
The point is (0.702, 1.420) [to 3 d.p.]
8 d
(1000 )
d
x
kx x
t
=
1
d d
(1000 )
x k t
x x
=

Method 1(Partial Fraction):
1 1
+ d
1000 1000(1000 )
x kt C
x x
= +

1
[ln ln(1000 )]
1000
x x kt C = +

1
ln
1000 1000
x
kt C
x
| |
= +
|

\


7
Method 2(Complete the sq.):
2 2
1
d
(500) ( 500)
x kt C
x
= +

1
ln
1000 1000
x
kt C
x
| |
= +
|

\

1000
e
1000
kt
x
A
x
=

1000
(1000 )e
kt
x A x =
When t = 0, x = 1,
1
999
A =
When t = 5, x = 120,
1 2997
ln
5000 22
k
| |
=
|
\

When x = 500,
1000
999 e
kt
=

ln999
7.03 days = 168 hrs 39 mins
1000
t
k
= =
Assumption: No one leaves and enters the community; birth rate = death rate.
9

Let P
n
be the statement:
1
1
!
n
u
n
= for n
+
.
To prove that P
1
is true:
When n = 1, LHS = u
1
= 0
RHS = 1
1
1!
= 0 = LHS
P
1
is true.
Assume that P
k
is true for some k
+
, i.e.
1
1
!
k
u
k
= .
To prove that P
k+1
is true, i.e. to prove
1
1
1
( 1)!
k
u
k
+
=
+

When n = k + 1, LHS =
k
u
k
k
u
k
k
+
+
+
=
+
1 1
1

=
1 1
1
1 !
k
k k
(
+
(
+


=
1 ( 1) ! 1
1 !
k k
k k
+ (
(
+


=
( 1)! 1 1
1
( 1)! ( 1)!
k
k k
+
=
+ +
= RHS
Thus, P
k
is true P
k+1
is true
Since P
1
is true and P
k
is true P
k+1
is true, by the Principle of Mathematical Induction, the
statement is true for all n
+
.

1 1
1
1/ 2
1
1 (1 )
1
!
2 2
1

2 !
1
r
r r
r r
r
r
u
r
r
e

= =

=
=
=

8
10
2 2
1
2 2 2 2
2 2 2 2
2 2 2 2
1 (2 ) 0
2 (3 ) 1 (2 )
3 (4 ) 2 (3 )
...
( 1) ( 1)
n
r
r
u
n n n n
=
=
+
+
+
+ +

= [ ]
2
) 1 ( + n n
10i
Now,
2 2 2 2
) 1 ( ) 1 ( r r r r u
r
+ =
=
2 2 3
( 1 1)[ 1 ( 1)] (2 )(2) 4 r r r r r r r r + + + = =
Thus, [ ]
2
1
3
1
) 1 ( 4 + = =

= =
n n r u
n
r
n
r
r

2
1
3
2
) 1 (
(

+
=

=
n n
r
n
r

10ii Method 1:

= =
+ =
n
r
n
r
r r r r
1
2 3
1
3
) 1 3 3 ( ) 1 (

1
3
1
n
r
r

= n r r r
n
r
n
r
n
r
+

= = = 1 1
2
1
3
3 3
2 2
2
1
( 1) ( 1) 3
3 ( 1)
2 2 2
n
r
n n n n n
r n n
=
+ ( (
= + +
( (

3
3
( 1)
2
n
n n n = + +
= ) 1 2 )( 1 (
2
) 1 3 2 (
2
2
+ + = + + n n
n
n n
n

) 1 2 )( 1 (
6
1
1
2
+ + =

=
n n n r
n
r

Method 2:

= =
+ =
n
r
n
r
r r r r
1
2 3
1
3
) 1 3 3 ( ) 1 (
=
3 2
1 1 1 1
3 3 1
n n n n
r r r r
r r r
= = = =
+


2 3 3
1 1 1
3 [ ( 1) ] 3
n n n
r r r
r r r r n
= = =
= +


= [1
3
0
3

+ 2
3
1
3

:
+ n
3
(n 1)
3
] +
3
( 1)
2
n
n n +

3
3
( 1)
2
n
n n n = + +

9
= ) 1 2 )( 1 (
2
) 1 3 2 (
2
2
+ + = + + n n
n
n n
n

) 1 2 )( 1 (
6
1
1
2
+ + =

=
n n n r
n
r

Method 3:

= =
+ =
n
r
n
r
r r r r
1
2 3
1
3
) 1 3 3 ( ) 1 (

3 3
1
( )
n
r
r n
=

= n r r r
n
r
n
r
n
r
+

= = = 1 1
2
1
3
3 3
2 3
1
3
3 ( 1)
2
n
r
n
r n n n
=
= + +

= ) 1 2 )( 1 (
2
) 1 3 2 (
2
2
+ + = + + n n
n
n n
n

) 1 2 )( 1 (
6
1
1
2
+ + =

=
n n n r
n
r

11i

11ii

11iii y = f (x)

12
1
tan y x x

=
1
y = 1
1
(1, )
x = 1
(0, 0)
y = 2
y =
1
f(x)

y =
1
2

(1, 0) (1, 0)
y = f( | x | )

10
( ) ( )
( )
( )
1
2
2 2 1
2
2 1
2
2
2
2
d
tan
d 1
d
1 1 tan
d
d d
2 1 2 tan 1 1
d d
d d
1 2 2 2 0
d d
y x
x
x x
y
x x x x
x
y y
x x x x
x x
y y
x x y
x x

= +
+
+ = + +
+ + = + +
+ + =


( )
3 2 2
2
3 2 2
d d d d d
1 2 2 2 2 0
d d d d d
y y y y y
x x x
x x x x x
+ + + + =
( )
( )
3 2
2
3 2
4 3 2
2
4 3 2
d d
1 4 0
d d
d d d
1 6 4 0
d d d
y y
x x
x x
y y y
x x
x x x
+ + =
+ + + =

When x = 0, y = 0

d
0
d
y
x
=

2
2
d
2
d
y
x
=

3
3
d
0
d
y
x
=

4
4
d
8
d
y
x
=
Maclaurins Series for y is
( )
2 4 2 4
8 2 1
... ...
2! 4! 3
y x x x x

= + + = +
Method 1:
1 2 1
3 4
0 0
tan tan
lim lim
x x
x x x x x
x x



=

2 2 4
4
0
0
1
...
3
lim
1
lim ...
3
1
3
x
x
x x x
x

| |
+
|
\
=
= +
=

Method 2:

3
1
3 3
0 0
0
1
...
tan 3
lim lim
1 1
lim ...
3 3
x x
x
x x x
x x
x x

| |
+
|

\
=
= + =

13i

2 1 2
du
u x
dx
= + =

11


3 1
2 2
5 3
2 2
5 3
2 2
5 3
2 2
1 1
2 1 d d
2 2
1
d
4
1 1

4 4
1 1
(2 1) (2 1)
10 6
u
x x x u u
u u u
u u
C
x x C

+ =
=
= +
= + + +

13ii
Let
1
2 1
2
x x + = . From GC, x =
1
2


Area of R

1/ 2
0
1 1
2 1 d
2 2
x x x = +

( ) ( )
1/ 2
5 3
2 2
0
2 1 2 1 2
4 10 6
x x
(
+ +
(
=
(


2
2 2 2 2 1 1
( )
4 5 3 10 6
11 2 1 1
(11 2 4) units
60 15 60
(
=
(

= =

13iii
Volume of solid ( )
2
1
2 2
0
1 1
2 1 d
2 2
x x x
| | | |
= +
| |
\ \


1/ 2
4 3
0
3
4 2 3
17
= 0.556 units
96
x x (
= +
(

=

14



f(x) = 3 2x x
2

= 4 (x + 1)
2

For
1
f

to exist, f must be oneone.

k = 1
To find
1
f

:
Let y = 4 (x + 1)
2

y x = 4 1
(1, 4)
3 1
1
2

x
y
2 1 y x x = +

12
Since x 1, y x = 4 1
Thus,
1
f

: x x 4 1 , x (, 4]
14i

Range of f = (, 4], domain of g = [0, 4]
Since range of f / domain of g, gf does not exist.
14ii Maximal range of h for gh to exist is [0, 4],
therefore maximal domain of h is [3, 1]
gh(x) =
2
4 4 ( 1) x
e
+ +
=
2
( 1) x
e
+
=
| 1| x
e
+
=
1 x
e

since x [ 3, 1]
gh : x
1 x
e

, x [3, 1]
Range of gh = [1, e
2
] = [1, 7.39]
14iii
1
( 1) g g x
D

+
= [ -1, 3] and
1
( 1) gg x
D

+
= [0, e
2
1]
Hence set of values of x = [0, 3]