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 dx x x . [5]

3. Solve the inequality ln2

xx . Deduce the solution of x > 2 2 ln x. [5]

4. A curve is defined by the parametric equations 2x t and siny 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

ln4

b . Find the common ratio and deduce

that the geometric progression is convergent. Given that b = e2, find the sum to

infinity of the progression. [6]

6. Express 2x2 + 3x – 1

(x – 1)(x2 + 1) in partial fractions.

Hence find the series expansion of 2x2 + 3x – 1

(x – 1)(x2 + 1) in ascending powers of x, up to and

including the term in x4. Find the coefficient of x2008. [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

xkx 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 u1 = 0 and nn unun 1)1( for n .

Prove by mathematical induction that 1

1!nu

n for n . [5]

Hence find the exact value of 1

1

2r

rr

u

. [2]

10. The sequence ur , r = 1, 2, 3, …, is defined by 2 2( 1) ( 1)ru r r r r .

Find

n

rru

1

. [2]

(i) By simplifying ur , deduce that 2

1

3

2

)1(

nnr

n

r

. [2]

(ii) By considering

n

r

r1

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 1tany x x , prove that 2

22

d d1 2 2 2 0

d d

y yx x y

x x .

By repeated differentiation, show that the first 2 non-zero terms of the Maclaurin’s

series for y is 2 41

3x x . Hence evaluate

1

30

tanlimx

x x

x

. [9]

13. (i) Use the substitution 2 1u x to find 2 1 dx x x . [3]

The region R is bounded by the curve 2 1y x x , the y-axis and line 1

2y .

(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 x–axis. [3]

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

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

4g : , 0 4xx e x .

State the largest value of k such that 1f exists, and find 1f 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–1g(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) 1.012a + 1.0052b + 1.0032c = 37 481.89 –––(3) From GC, a = 15 000, b = 12 000, c = 10 000. So at the end of 2008, he had 1.012 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 xx

2

cos ln d

1 1cos 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 xx

2 2

cos ln d

1 1 1 1cos 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 1cos ln d cos ln sin ln

4 2 42 1

cos ln d cos ln sin ln5 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 x2: x2

2 > ln x2

x > 2 2 ln x From above, 0 < x2 < 4.42806 or x2 > 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 cosd d

x yt 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

0d

y

x

cos0 cos 0

2

or 2 2

tt

t

t

When 2

, 2 4

t x

Equation of normal is 2

4x

5 Since the 1st 3 terms are in G.P., 21 1

( ln ) ( ln )( ln )2 3

a b a b a b

2 2 2 21 4 1ln (ln ) ln (ln )

4 3 3a a b b a a b b

1 1ln

3 12a b

1ln

4a b

Common ratio, r =

1 1ln ln 12 4

3ln 3ln4

a b b

a b b

Since 13

1r , the G.P. is convergent.

Given that b = e2, 21 1 1

ln 2ln4 4 2

a e e

2ln 3 1 9( 2)

1 2 2 413

a eS

6 Let

2x2 + 3x – 1(x – 1)(x2 + 1) =

Ax – 1 +

Bx + Cx2 + 1

2x2 + 3x – 1 = A(x2 + 1) + (Bx + C)(x – 1) Let x = 1: 4 = 2A A = 2 Let x = 0: –1 = 2 – C C = 3 Compare coefficients of x2 : 2 = 2 + B B = 0

2 O

6

2x2 + 3x – 1(x – 1)(x2 + 1) =

2x – 1 +

3x2 + 1

= –2(1 – x)–1 + 3(1 + x2)–1

= –2(1 + x + x2 + x3 + x4 + ...) + 3(1 – x2 + x4 – ...) = 1 – 2x – 5x2 – 2x3 + x4 +... Coefficient of x2008 = –2 + 3 = 1

7 Distance between (1, 1) and (x, y), 2 2

1 1W x y

Since (x, y) lies on the curve, then 2

x

y e

Thus, 22 21 1

xW x e

222 21 1

xW x e

Differentiating w.r.t. x, 2 2d

2 2 1 1d

x xWx e e

x

2 2d 1

1 1d 2

x xWx e e

x

When 2 2d 1

0, 1 1 0d 2

x xWx 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 1.4202y e The point is (0.702, 1.420) [to 3 d.p.]

8 d(1000 )

d

xkx x

t

1d d

(1000 )x k t

x x

Method 1(Partial Fraction):1 1

+ d1000 1000(1000 )

x kt Cx x

1

[ln ln(1000 )]1000

x x kt C

1

ln1000 1000

xkt C

x

7

Method 2(Complete the sq.): 2 2

1d

(500) ( 500)x kt C

x

1

ln1000 1000

xkt C

x

1000e1000

ktxA

x

1000(1000 )e ktx A x

When t = 0, x = 1, 1

999A

When t = 5, x = 120, 1 2997

ln5000 22

k

When x = 500, 1000999 e kt

ln 999

7.03 days = 168 hrs 39 mins1000

tk

Assumption: No one leaves and enters the community; birth rate = death rate. 9 Let Pn be the statement:

11

!nun

for n .

To prove that P1 is true: When n = 1, LHS = u1 = 0

RHS = 1–1

1!= 0 = LHS

P1 is true.

Assume that Pk is true for some k , i.e. 1

1!ku

k .

To prove that Pk+1 is true, i.e. to prove 1

11

( 1)!kuk

When n = k + 1, LHS = k

u

k

ku k

k

111

= 1 1

11 !

kk k

= 1 ( 1) ! 1

1 !

k k

k k

= ( 1)! 1 1

1( 1)! ( 1)!

k

k k

= RHS

Thus, Pk is true Pk+1 is true Since P1 is true and Pk is true Pk+1 is true, by the Principle of Mathematical Induction, the statement is true for all n .

1 1

1

1/2

11 (1 )1 !

2 2

1

2 !

1

rr r

r r

rr

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

rr

u

n n n n

= 2)1( nn

10i Now, 2222 )1()1( rrrrur

= 2 2 3( 1 1)[ 1 ( 1)] (2 )(2) 4r r r r r r r r

Thus, 21

3

1

)1(4

nnrun

r

n

rr

2

1

3

2

)1(

nnr

n

r

10ii Method 1:

n

r

n

r

rrrr1

23

1

3 )133()1(

1

3

1

n

r

r

= nrrr

n

r

n

r

n

r

11

2

1

3 33

2 22

1

( 1) ( 1) 33 ( 1)

2 2 2

n

r

n n n n nr n n

3 3( 1)

2

nn n n

= )12)(1(2

)132(2

2 nnn

nnn

)12)(1(6

1

1

2

nnnrn

r

Method 2:

n

r

n

r

rrrr1

23

1

3 )133()1(

= 3 2

1 1 1 1

3 3 1n n n n

r r r r

r r r

2 3 3

1 1 1

3 [ ( 1) ] 3n n n

r r r

r r r r n

= [13 – 03 + 23 – 13 :

+ n3 – (n – 1)3 ] + 3

( 1)2

nn n

3 3( 1)

2

nn n n

9

= )12)(1(2

)132(2

2 nnn

nnn

)12)(1(6

1

1

2

nnnrn

r

Method 3:

n

r

n

r

rrrr1

23

1

3 )133()1(

3 3

1

( )n

r

r n

= nrrrn

r

n

r

n

r

11

2

1

3 33

2 3

1

33 ( 1)

2

n

r

nr n n n

= )12)(1(2

)132(2

2 nnn

nnn

)12)(1(6

1

1

2

nnnrn

r

11i

11ii

11iii y = f (x)

12 1tany x x

–1

y = 1

1

(1, ½)

x = –1

(0, 0)

y = 2

y = 1

f(x)

y = 1

2

(–1, 0) (1, 0)

y = f(– | x | )

10

12

2 2 1

22 1

2

22

2

dtan

d 1d

1 1 tand

d d2 1 2 tan 1 1

d d

d d1 2 2 2 0

d d

y xx

x xy

x x x xx

y yx x x x

x x

y yx x y

x x

3 2 2

23 2 2

d d d d d1 2 2 2 2 0

d d d d d

y y y y yx x x

x x x x x

3 22

3 2

4 3 22

4 3 2

d d1 4 0

d d

d d d1 6 4 0

d d d

y yx x

x x

y y yx x

x x x

When x = 0, y = 0

d

0d

y

x

2

2

d2

d

y

x

3

3

d0

d

y

x

4

4

d8

d

y

x

Maclaurin’s Series for y is 2 4 2 482 1

... ...2! 4! 3

y x x x x

Method 1: 1 2 1

3 40 0

tan tanlim limx x

x x x x x

x x

2 2 4

40

0

1...

3lim

1lim ...

31

3

x

x

x x x

x

Method 2:

31

3 30 0

0

1...

tan 3lim lim

1 1lim ...

3 3

x x

x

x x xx x

x x

13i 2 1 2

duu x

dx

11

3 12 2

5 32 2

5 32 2

5 32 2

1 12 1 d d

2 21

d4

1 1

4 4

1 1 (2 1) (2 1)

10 6

ux x x u u

u u u

u uC

x x C

13ii Let

12 1

2x x . From GC, x =

1

2

Area of R

1/2

0

1 12 1 d

2 2x x x

1/25 3

2 2

0

2 1 2 12

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

21

2 2

0

1 12 1 d

22x x x

1/24 3

0

3

4 2 3

17 = 0.556 units

96

x x

14

f(x) = 3 – 2x – x2 = 4 – (x + 1)2 For 1f to exist, f must be one–one. k = 1 To find 1f : Let y = 4 – (x + 1)2

yx 41

(1, 4)

3 1

12

x

y 2 1y x x

12

Since x 1, yx 41

Thus, 1f : x x 41 , 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) = 24 4 ( 1)xe =

2( 1)xe = | 1|xe = 1xe since x [ –3, –1] gh : x 1xe , x [3, 1] Range of gh = [1, e2] = [1, 7.39]

14iii 1 ( 1)g g xD

= [ -1, 3] and 1 ( 1)gg xD

= [0, e2 –1]

Hence set of values of x = [0, 3]