1

Section A: Pure Mathematics [35 Marks]

1 The points P(4,12) and Q(6,48) lie on curve C, y = 1

4x

3 − x.

(a) Find the equation of the normal to curve C at P, giving your answer in the form ax + by = c, where a, b and c are integers. [3]

(b) Find the set of values of x for which the curve C is strictly decreasing. [2] (c) Find the exact area of the region bounded by the curve C and the line PQ. [3]

Solutions*

y = 1

4x

3 − x

y' = 3

4x

2 − 1

When x = 4, y' = 11 Equation of the normal is

y − 12 = −(x − 4)/11 x + 11y = 136

y' = 3

4x

2 − 1 < 0

x = ± 2

3(roots)

The required values are −2

3 < x <

2

3

Or −1.15 < x < 1.15

63

4

1d

4x x x

−

∫

=

6

4 2

4

1 1

16 2x x

−

= 63 − 8 = 55 Area of trapezium

=(12+48)(6−4)/2 =60 Area of required region

= 60 − 55 = 5

Equation of PQ is

Y = 18x − 60

Q

P

4 6 x

y

O

2

2 Oil is poured into a container at a rate of 60 m3s−1. The volume, V m3, of the oil in the container is given by V = ln (0.01x

3 + 3.2x2 + 4.5x) where x m is the depth of the oil in the

container. Find the rate of increase in the depth of oil when x = 10. [3] Solutions*

V = ln(0.01x3 + 3.2x

2 + 4.5x)

d

d

V

x =

2

3 2

0.03 6.4 4.5

0.01 3.2 4.5

x x

x x x

+ ++ +

When x = 10,

d

d

V

x =

71.5

375 =

143

750 = 0.1906666666

Given d

d

V

t = 60.

By chain rule, d

d

V

t=

d

d

V

x×

d

d

x

t

60 = (0.1906666666) d

d

x

t

d

d

x

t = 60/0.1906666666 = 315 (3sf)

31498

143 or

45000

143

3 (a) A curve C is given by the equation g(x) = 6

3

ax

b x

−−

. Find the values of a and b for which

y = 1

3 and x = −2 are equations of asymptotes of C. Sketch the curve C, showing clearly

the asymptotes and axial intercepts. [4]

Solutions*

(a)

b − 3x = 0 ⇒ b = 3x = − 6

6

3

ax

b x

−−

= 6 /

( / ) 3

a x

b x

−−

→ 3

a

− =

1

3 as x →∞ i.e. a = − 1

b = − 6, a = − 1

Shape & Axial intercepts Asymptotes

0 x

y

x = −2

y = 1

3 1

−6

3

(b) Given that f(x) = 1

1 3x−, find

(i) f ( )dx x∫ . [2]

(ii) f ' (x). Hence obtain the exact value of m for which f ' (m) = 3

16. [4]

Solutions*

(b)f(x) = 1

1 3x−= ( )

1

21 3x−

−

f ( )dx x∫ = ( )1

21 3 dx x−

−∫

= ( )

( )

1

21 3

13

2

x

− −

+ c = −2

1 33

x− + c

f(x) = 1

1 3x−= ( )

1

21 3x−

−

f ' (x) = −1

2( )

3

21 3x−

− (−3) = 3

2( )

3

21 3x−

−

f ' (m) = 3

2( )( )

3

21 3 m−

− = 3

16

( )( )3

21 3 m−

− = 1

8

( )( )3

21 3 m− = 8

1 − 3m = ( )2

38 = 4

m = −1

4

4 (a) Find in exact form, the range of values of m for which (3x − 2m)2 > x2 − 5 for all real values of x. [3]

(b) Without using a calculator, solve the simultaneous equations 3x + 2y = 27 ln x + ln (x + y) = ln 5 [4]

Solutions*

(a)(3x − 2m)2 > x2 − 5

9x2 − 12xm + 4m

2 > x2 − 5

8x2 − 12mx + 4m

2 + 5 > 0

Apply b2 − 4ac < 0,

(12m)2 − 4(8)(4m2 + 5) < 0

144 m2 − 128 m

2 − 160 < 0

16 m2 − 160 < 0

m2 − 10 < 0

(m − √10)( m + √10) < 0

− √10 < m < √10

(b) 3x + 2y = 27 ⇒ x + 2y = 3

ln x + ln (x + y) = ln 5 ⇒ x (x + y) = 5

(3 − 2y) (3 − 2y + y) = 5

9 − 9y + 2y2 = 5

2y2 − 9y + 4 = 0

(y − 4) (2y − 1) = 0

y = 4 or 0.5

When y = 4, x = 3 − 2(4) = −5 (reject as ln(x + y) is undefined)

When y = 0.5, x = 3 − 2(0.5) = 2

5

5 In an experiment, the population of bacteria, in millions follows an exponential growth

modelled by f(t) = 36e−0.1t + 5e0.05t, t ≥ 0. The time t, in hours, is taken to be zero at the start of

the experiment.

(i) What will be the population of bacteria after 5 hours? [1]

(ii) When will the population double from the initial population size? [2]

(iii) Find the exact rate of growth when t = 20. [2]

(iv) Suppose the amount of nutrients consumed by the bacteria, in g, after t hours is given by

0

1d

2 1

t

tt +∫ . Find the exact time taken if 0.001 kg of nutrients was consumed. [2]

Solutions*

(i) f(t) = 36e−0.1t + 5e0.05t, t ≥ 0

When t = 5, f(5) = 36e−0.1(5) + 5e0.05(5) f(5) = 28.2552 = 28.3 millions(3sf)

(ii) f(0) = 36e−0.1(0) + 5e0.05(0) = 41

f(t) = 36e−0.1t + 5e0.05t = 2(41) = 82 Using GC, t = 55.9 h

(iii) f(t) = 36e−0.1t + 5e0.05t, t ≥ 0

f ' (t) = −3.6e−0.1t + 0.25e0.05t, t ≥ 0

f ' (20) = (−3.6e−2 + 0.25e1) millions/h

(iv) 0

1d

2 1

t

tt +∫ = 1

[ln(2t+1)/2] 0

t = 1

ln(2t+1) = 2

t = (e2 −1)/2 h

6

Section B: Statistics [60 Marks]

6 C and D are events such that P(C ∪ D) = 113

140, P(C ∩ D) =

1

4 and P(C D) =

7

10.

Find P(D) and P(C). Are events C and D independent? [5]

Solutions*

P(C D) = 7

10

P(C ∩ D) = 7

10 P(D)

1

4 =

7

10 P(D)

P(D) = 1

4 ×

10

7 =

5

14

P(C ∪ D) = P(C) + P(D) − P(C ∩ D)

113

140 = P(C) +

5

14 −

1

4

P(C) = 113

140 −

5

14 +

1

4 =

7

10

P(C) × P(D) = 7

10 ×

5

14 =

1

4 = P(C ∩ D)

∴ events C and D are independent.

Verifying

7

7 In order to find out the willingness of local Secondary Four students doing voluntary work

after the O-level Examinations, four undergraduates, majoring in Social Studies, decide to

survey 100 local Secondary Four students. The undergraduates propose that each of them will

survey 25 students from schools in Clementi only.

(i) Name the sampling model chosen by these four undergraduates. [1]

(ii) Identify one disadvantage of their proposed method and suggest how it can be improved.

[2]

Solutions* (i)Quota sampling

(ii) Disadvantages Suggestion[

The sample is not representative of schools in Singapore.

Survey students from number of schools from various parts of Singapore. This allows a wider spread of students selected.

The sample is interviewer biased and not random

8

8 Three white balls, five red balls and four black balls are placed in a box. All balls are of

identical size. Three balls are randomly picked and placed in bag A.

Find the probability that

(i) there are no white balls in bag A, [2]

(ii) all balls in bag A are of the same colour, [2]

(iii) there are exactly three white balls in bag A, given that at least one white ball has been

drawn and put in bag A. [3]

Suppose now bag A has exactly two white balls and one black ball and that all the remaining balls are put in a bag B. A bag is randomly selected by tossing a fair coin and one ball is drawn from it. Find the probability that the ball drawn is white. [2]

Solutions* (i)P(no white balls in bag A)

= 9 8 7

12 11 10× × =

21

55

(ii) P(all balls in bag A are the same colour)

= 3 2 1

12 11 10× × +

5 4 3

12 11 10× × +

4 3 2

12 11 10× ×

= 3

44

(iii)Let X denotes the number of white balls drawn.

( )P 3 1X X= ≥

= ( )

( )P 3 1

P 1

X X

X

= ∩ ≥

≥

( )( )

P 3

P 1

X

X

==

≥

= 3 2 1

12 11 10

× ×

÷ 21

155

−

= 1

220 ÷÷÷÷

34

55=

1

136

(iv) Required probability

= ( ) ( )P bag A and white ball P bag B and white ball+

= 1 2 1 1

2 3 2 9

× + ×

= 7

18

9

9 A survey was conducted by the Ministry of Public Transport (MPT) in July 2009 to determine

the waiting time, denoted as X minutes, of a bus passenger. The bus company claims the mean

waiting time for each bus passenger was 30 minutes. Data from 90 randomly selected bus

passengers are collected and their waiting times are summarised as shown below:

( )30x −∑ = 90, ( )230x −∑ = 2037.

(i) Calculate the unbiased estimate of the mean and variance of X. What do you understand

by the term ‘unbiased estimate’? [3]

(ii) An audit company suspects that the bus company is underestimating the mean waiting

time. Conduct an appropriate test at 5% level of significance. [4]

(iii) State, giving a reason, whether it is necessary for the waiting time to follow a normal

distribution for the test to be valid. [1]

(iv) A year later, MPT wants to find out if there is a change in the mean waiting time. Data

from 100 randomly selected bus passengers are collected and their waiting times has a

sample mean of 29.3 minutes and a sample standard deviation is 4.5.

Find the set of values of α if MPT were to conclude that there is a change in the mean

waiting time at α % level of significance. [3]

10

Solutions*

(i) x = 90

90 + 30 = 31

s2 =

( )2901

203789 90

−

= 1947

89 or 21.9 (to 3 sig figs)

unbiased estimate means the mean of the estimate is expected to equal to the parameter when collected infinitely(many times).

E(θ) = p where θ is estimate and p is parameter

(ii) Test 0 1: 30 vs : 30 H Hµ µ= > and perform a upper-tailed

z-test at 5% significance level.

Under 0H , X ~ (30, 2

90

σ).

As σ2 is unknown, estimate with s2 = 1947

89

The test Statistic, Z = s

n

X µ− ~ N(0,1).

Using GC, p-value = 0.0212644905,

Since p-value = 0.0213 < 0.05, we reject 0H and conclude that

there is sufficient evidence at 5% significance level to say that that the bus company is underestimating the mean waiting time.

H1: µ > 30

(iii)It is not necessary for the waiting times to follow a normal

distribution, as 90n = is large, Central Limit Theorem states that

the sampling distribution of X is approximately normal.

(iv) Test 0 : 30 H µ = against 1 : 30 H µ ≠ .

Perform a 2-tailed Z-test at α % significance level.

Under 0H , X ~ (30, 2

100

σ).

As σ2 is unknown, estimate with s2 = 100

99(4.52) = 20.4545454545

The test statistic Z = s

n

X µ− ~N(0,1) (approximately)

x = 29.3, s2 = 20.4545454545

Using GC, z = −1. 548 and p-value = 0.12168805432.

To reject 0H , p-value <100

α. ∴ 12.1 < α (3 sf)

11

10 Brook News Corporation has 150 printing presses. The probability of a printing press needing

minor repair weekly is p. The number of printing presses needing minor repair weekly is

denoted by X. Given that 75Var(X) = 2{E(X)}2, show that p = 1

5, [3]

(i) Find the probability that less than 28 printing presses need minor repair in a week. [2]

(ii) Find the least possible value of n such that the probability of at most n printing presses

needs minor repair weekly exceeds 0.6. [2]

(iii) Using a suitable approximation, find the probability that at most 37 printing presses

needs minor repair in a week. [3]

(iv) Find the probability that the mean number of printing presses needing minor repair

weekly is between 25.5 and 32(inclusive) over a period of 50 weeks? [3]

Solutions* X ~ B(150, p)

E(X) = 150 p, Var(X) = 150 p(1 − p) 75Var(X) = 2{E(X)}2

75{150 p(1 − p)} = 2{150p}2

75(1 − p) = 300p

1 − p = 4p ⇒ p = 1

5

(i)Let X denotes the random variable of the number of printing

presses that needs minor repair weekly. 1

B 150,5

X

∼

Required probability

= ( )P 28X < = ( )P 27X ≤

= 0.310 (to 3 sig figs)

(ii) ( )P 0.6X n≤ >

Using GC, 31.n =

(iii)Since 60n = is large, 30 5np = > , 120 5nq = >

( )N 30,24 approxX ∼

( ) ( )P 37 P 37.5 (cc)X X≤ = ≤ = 0.937.(to 3 sig figs) (ans)

(iv)Let X denotes the random variable of the mean number of printing presses needs minor repair weekly

1 2 50...

50

X X XX

+ + +=

Since 50n = is large, by Central Limit Theorem,

24N 30,

50X

∼ approx

( )P 25.5 32 0.998.X≤ ≤ = (to 3 sig figs)

1125

5625

12

11 Consider the data obtained from a chemical process where the yield of the process, y, is determined by the reaction temperature, x.

Temperature (oC), x 5 55 60 6 7 7 8 8

�

(i) Calculate the correlation coefficient, and the line of regression of y on x. [3] (ii) It was discovered that one of the results may be wrong. Give a sketch of the scatter

diagram. Giving a reason, identify a data pair which is most likely to be incorrect. [2]

(iii) The corrected data gave the correlation coefficient as 0.9964053552. State, giving a

reason, which of the least squares regression lines, y on x or x on y, should be used to

express a possible linear relation between y and x. [1]

(iv) Using the appropriate regression line, estimate the reaction temperature when the yield is 178 g/dm3, giving your answer to the nearest oC. Comment on its reliability. [2]

Solutions* (i) the correlation coefficient, r = 0.939 the line of regression of y on x is y = 50.4 + 1.69x

SetStatDiagnostics M

(ii)

From the scatter diagram, x = 55, y = 160 (55,160) may be wrong (an outlier) because the data pair is farthest from the regression line of y on x.

Label axis & Plot

(iii)As x is the independent variable, y on x should be used.

13

(iv) Regression line is y on x, y = 29.2 + 1.96x

178 = 29.2 + 1.96x ⇒ x = 75.918 ≈ 76 (to nearest oC) r is close to 1 and y = 178 is within data range(interpolated) the estimate for x is reliable.

14

12 A chef bought two kinds of fishes in a wet market. The weight of groupers and weight of

pomfrets bought are modelled as having independent normal distributions with the following

means and standard deviations.

Means (kg) Standard deviations (kg)

Weight of grouper µ σ

Weight of pomfret 1.8 0.5

The weight W, in kg, of a randomly chosen grouper has mean, µ kg and standard deviation,

σ kg.

(i) Given that P(W < 3.3) = P(W > 5.3) = 0.15, state the value of µ.

Show that σ = 0.965, correct to 3 significant figures. [3]

(ii) Find the probability that out of three randomly chosen groupers, two weigh between 3.3

kg and 5.3 kg each and one weighs more than 5.3 kg. [3]

The price of groupers is $8.50 per kg and the price of pomfrets is $6.30 per kg (iii) Find the mean of the total price of four groupers and six pomfrets and show that variance

of the total price is 328.7, correct to 4 significant figures. [4]

(iv) Find the probability that the total price of four groupers and six pomfrets lies between

$220 and $300. [1]

15

Solutions*

By symmetry, µ = 3.3 5.3

2

+ = 4.3

(i) P(W < 3.3) = 0.15

3.3 4.3P( )Z

σ−

< = 0.15

−1

σ = −1.03643338

σ = 0.964847349 ≈ 0.965

(ii) W ~ N(4.3, 0.9652) Required probability

= P(3.3 < W < 5.3) × P(3.3 < W < 5.3) × P(W > 5.3) × 3

= (0.7) × (0.7) × (0.15) × 3 = 0.2205

(iii) T = 8.5(W1 + W2 + W3 + W4) + 6.3(F1 + F2 + … + F6) Let T denotes the total price of four groupers and six pomfrets. E(T) = 8.5(E(W1) + E(W2) + E(W3) + E(W4)) + 6.30(E(F1) + E(F2) + E(F3) + E(F4) + E(F5) + E(F6))

= 8.5(4 × 4.3) + 6.30(6 × 1.8) = 214.24

VAR(T) = 8.52(VAR(W1) + VAR(W2) + VAR(W3) + VAR(W4)) + 6.302 (VAR(F1) + VAR(F2) + VAR(F3) + VAR(F4) + VAR(F5) + VAR(F6))

= 8.52 (4 × 0.9652) + 6.302 (6 × 0.52) = 269.90388876 + 59.535 = 328.659025 = 328.7

(iv)P(220 < T < 300) = 0.375341846 = 0. 375

End of Paper