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 . [5]

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

4. A curve is defined by the parametric equations and , for .

(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 , and

, where b > 1. Show that a = . 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 in partial fractions.

Hence find the series expansion of 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 and the point (1, 1) in

terms of x. Hence find the coordinates of the point on the curve that is closest

to the point (1, 1), giving your answer correct to 3 decimal places. [6]

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

1

the start of the spread of the disease, satisfies the differential equation

, 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 for .

Prove by mathematical induction that for . [5]

Hence find the exact value of . [2]

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

Find . [2]

(i) By simplifying ur , deduce that . [2]

(ii) By considering , show that . [5]

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

On separate clearly labelled diagrams, sketch the graphs of

(i) y = ,(ii) y = f(– | x | ),

(iii) y = f (x). [9]

12. Given that , prove that .

2

(–1, 0)

y = x

(1, 2)y =

x

y

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

series for y is . Hence evaluate . [9]

13. (i) Use the substitution to find . [3]

The region R is bounded by the curve , the y-axis and line .

(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:

, ,

.

State the largest value of k such that exists, and find 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

3

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

3

From GC, 0 < x < 4.42806 or x > 13.706i.e. 0 < x < 4.42 or x > 13.8

Replacing x by x2: > ln x2

x > 2ln xFrom above, 0 < x2 < 4.42806 or x2 > 13.706

0 < x < 2.1043 or x > 3.7020 < x < 2.10 or x > 3.71

4(i)

4

2O

4(ii)

Normal // y-axis Tangent // x-axis

When

Equation of normal is

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

Common ratio, r =

Since , the G.P. is convergent.

Given that b = e2,

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

= –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 = 17 Distance between (1, 1) and (x, y),

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

5

Thus,

Differentiating w.r.t. x,

When

From GC, x = 0.70160785

0.70160785- 0.70160785 0.70160785+

-ve 0 +ve

\ – /Thus, W is minimum when x = 0.70160785When x = 0.70160785, The point is (0.702, 1.420) [to 3 d.p.]

8

Method 1(Partial Fraction):

Method 2(Complete the sq.):

6

When t = 0, x = 1,

When t = 5, x = 120,

When x = 500,

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

Let Pn be the statement: for n .

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

RHS = 1– = 0 = LHS

P1 is true.

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

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

When n = k + 1, LHS =

=

=

= = RHS

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

10

=

7

10i Now, =

Thus,

10ii Method 1:

=

=

Method 2:

=

= [13 – 03

+ 23 – 13

:

+ n3 – (n – 1)3 ] +

=

Method 3:

=

8

=

11i

11ii

11iii y = f (x)

12

9

–1

y = 1

1

(1, ½)

x = –1

(0, 0)

y = 2

y =

y =

(–1, 0) (1, 0)

y = f(– | x | )

When x = 0, y = 0

Maclaurin’s Series for y is

Method 1:

Method 2:

13i

10

13iiLet . From GC, x =

Area of R

13iiiVolume of solid

14 f(x) = 3 – 2x – x2

= 4 – (x + 1)2

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

Since x 1, Thus, : x , 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) = = = = since x [ –3, –1] , x [3, 1]Range of gh = [1, e2] = [1, 7.39]

14iii = [ -1, 3] and = [0, e2 –1]

11

(1, 4)

3 1

12

x

y2 1y x x

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

12