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ANNEX B SAJC H2 Math JC2 Preliminary Examination Paper 1

QN Topic Set Answers

1 Differentiation & Applications 2

1

3

cm /15

s2πλ

2 Maclaurin series -

3 Functions

4ab

π

4 Graphs and Transformation

i)

1. Scale by a factor of 1

a parallel to the x-axis,

2. Translate the resulting curve by b units in the

negative y-direction.

5 Vectors ii) 23

units4

iii) 33 units

12

6 Application of Integration

a)

i) 2

2

sin 1 e cose

1

xxn nx nx

cn n n n

+ + +

ii) ( )2

2

1e e

1 n

π π + +

b) 35units

288

π

7 Complex numbers ii)

i32ez

π − =

iii) 1 3 1 3

i or i2 2 2 2

w w= − = +

8 AP and GP i) n = 24

ii) ( )3 1 58n n− + , 5614 m away from the coach once he

finishes 8 km.

9 Application of Integration i)

32

2y x= +

ii) Q (0.421, 2.34)

iii) 22.77 units

10 Differential Equations iii)

0.1160.98140t

ex e−−=

iv) 40

11 Vectors ii) 3 6 2 90x y z+ + =

iii) 073.4θ =

iv)

4 3

9 6

24 2

λ− = +

r , λ ∈� .

v) The drill line will not exit from the side GCBF.

H2 Mathematics 2017 Prelim Exam Paper 1 Solution

1 3

2

4

3

d d4

d d

V r

V rr

t t

When V=20,

3

1

3

420

3

15

r

r

When

1

315r

, d

d

V

t .

2

3

2

3

15 d4

d

d

d 4 15

r

t

r

t

Surface Area, 24

d d8

d d

A r

A rr

t t

When

1

315r

,

2

3d

d 4 15

r

t

.

1 2

3

1

2

3

3

d d d

d d d

15 = 8

4 15

215

cm /s

A A r

t r t

2 (i)

BC = BD + DC

=

tan tan3 4

h h

x

(ii)

BC

= 3 tan tan

4

1 tan tan4

h h

x

x

= 3 (1 tan )

3 1 tan

h h x

x

3 (1 )

3 1

h h x

x

= 13(1 )(1 )

3

hh x x

= 23 ( 1)( 2)(1 )[1 ( 1) ) ...]

3 2!

hh x x x

= 23(1 )[1 ...]

3

hh x x x

= 23(1 2 2 ...)

3

hh x x

=23

1 2 + 23

h x x

3 (i)

(ii) 4

3

20

2

2 2

22

2

2

2

2

f ( ) d

1 d

cos= 1 ( sin ) d

= sin d

1 cos 2 d

2

sin 2θ= θ

2 2

2 2

4

a

a

a

x x

xb x

a

ab a

a

ab

ab

ab

ab

ab

4 (i)

2

2

e eaxaxy b b

If 2

f ( ) exx , then 2

f eax

ax and so

f f fy x y ax y ax b

Hence the sequence of transformations are:

1. Scale by a factor of 1

a parallel to the x-axis,

2. Translate the resulting curve by b units in the negative y-direction.

(ii)

y

x O

y=1/f(x)

y=0

y

y=f(x)

O

(0,1-b)

x

5 (i)

Since u v w is perpendicular to u v w ,

0

0

u v w

v

u v w u v w

u

u v v v w

u w v ww w

u u

Since 2 2 2, , u v wu u v v w w , and , , v v w w wu u u v wu v ,

2 2 22 0 u v w v w

Since , ,u v w are unit vectors, 1,1, 1 u v w ,

1 1 1 2 0

1

2

1cos

2

1cos

2

v w

v w

v w

Hence, 60

(ii)

2

Area of OVW

1sin 60

2

1 31 1

2 2

3 units

4

OV OW

(iii)

O V

W

Since u and v w are parallel, we have ,OV O OOU U W .

Volume of OUVW

3

1Area of OVW

3

1 31

3 4

3 units

12

OU

O

U

V

W

6 (a)

(i) Using integration by parts,

e cos dx nx x

sin ee sin d

xx nx

nx xn n

= sin 1 e cos e cos

e dx x

x nx nx nxx

n n n n

= 2

sin 1 e cos 1e e cos d

xx xnx nx

nx xn n n n

Rearranging,

2

11 e cos dx nx x

n

=

sin 1 e cose

xx nx nx

n n n

e cos dx nx x = 2

2

sin 1 e cose

1

xxn nx nx

cn n n n

where c is a constant

(ii) 2

e cos dx nx x

=

22

2

sin 1 e cose

1

xxn nx nx

n n n n

2

2

2

sin 2 1 cos 2 sin 1 cose e

1

n n n n n

n n n n n n n

For any positive integer n, 02sin n and 12cos n

If n is odd, 0sin n and 1cos n

2

e cos dx nx x

= 2

2

2 2 2

1 1e 0 e 0

1

n

n n n

2

2

1e e

1 n

(Ans)

(b)

de cos

d

d sine

d

x

x

vu nx

x

u nxv

x n

de sin

d

d cose

d

x

x

vu nx

x

u nxv

x n

216

xy

x

2

2216

xy

x

Hence volume required 2

2 2

0dr h y x

2

2

20 2

2

2

20 2

212 2

0

3

22 d

12 16

2 22 d

12 2 16

1622

12 2 1

4 1 1

144 2 12 16

5units

288

xx

x

xx

x

x

7 (i) i

i

i

i i i2 2 2

i2

e

e

e

e e e

e

2isin2

cos i sin2 2

2isin2

1 1cot

2 2i 2

1 1cot i

2 2 2

r

r r

(ii)

i32ez

(iii) 2

2

2

4 44 0

( 1) 1

2 22 4 0

1 1

w w

w w

w w

w w

Let 2

1

wz

w

, then

2 2 4 0z z

From (ii) the solutions are i i

3 32e or 2e z z

Since

2

1

2

( 2)

wz

w

zw z w

w z z

2

zw

z

Part (i) result can be used as i

32e z

, where 2r with ,

3 3

.

1 1 1 1cot i or i cot

2 2 6 2 2 6w w

1 3 1 3i or i

2 2 2 2w w

8 (i) Distance travelled per lap is in AP:

a = 2(30) = 60, d = 2 3 = 6.

Given total distance travelled > 3000

2

n[ 2(60) + (n 1)6] > 3000

3n2 + 57n 3000 > 0

(n + 42.52)(n 23.52) > 0

n < 42.52 or n > 23.52

Since n , least n = 24

(ii)

1

0 1 2 2

2 2

1

1

3

Distance of the coach from just before the runner completes the th lap

=30 2(3 ) 2(3 ) 2(3 ) ... 2(3 )

30 2 1 3 3 .... 3

3 130 2

3 1

30 (3 1)

29r

r

r

r

r

rS

1

1

1

1 1

1

1

Distance covered by the athlete after laps

2 3

2 3 58

2 3 58

3 12 58

3 1

3 1 58

29n

r

r

n nr

r r

nr

r

n

n

n

n

n

n

When D = 8000m

3 1 58

From GC,

8.1254

8000 n n

n

Hence the athlete has run 8 complete laps.

The athlete has completed 7024 m

Hence he still have 8000-7024=976 m

On the 9th lap, the coach is 9 13 29 6590 m from S.

Hence the athlete would be 6590- 976 = 5614 m away from the coach once he finishes

8 km.

9 (i)

d2cos

d

x

,

d3 sin

d

y

d 3 sin 3

tand 2cos 2

y

x

When 6

, 2x ,

11

2y ,

d 1

d 2

y

x

Equation of normal : 11

2 22

y x

3

22

y x

(ii) 1 2sin (1)

4 3 cos (2)

x

y

Substitute equation (1) and (2) into 3

22

y x

34 3 cos 2(1 2sin )

2

13 cos 4sin

2

8sin 2 3 cos 1

At Point Q,

8sin 2 3 cos 1 (shown)

Using GC: 2.847916 or 0.52359 (Reject, same as ,point )6

P

Hence, using GC

coordinates of Q (0.42105, 2.3421)

Q (0.421, 2.34)

(iii)

Req

y

x

when 0.42105

0.42105 1 2sin

sin 0.289475

0.29368 or 2.8479 ( at point Q)

x

Required Area

0.42105 0.42105

2 2

0.4

1 2

26

0.29368

2

2105

d d

34 3 cos 2cos d 2 d

2

8.9613 6.1911

2.7702 2.77 units (3 s.f.)

y x y x

x x

10 (i)

d 40ln

d

xcx

t x

40ln

ln(40) ln( )

d 1

d

ux

x

u

x x

d d d

d d d

1 40ln

u u x

t x t

cxx x

cu

(ii)

d

d

1d d

ln

e

e e

e , = e

ct d

d ct

ct d

ucu

t

u c tu

u ct d

u

u

B B

Replace u with 40

lnx

40ln e ctB

x

e40e

ctB

x

e

e

40

e

40e

ctB

ctB

x

x

(iii)

When 0,t 15,x

15 =40Be

3e

8

8ln 0.98083 0.981

3

B

B

When 3, 20t x

3

3

e

3

3

20 40e

1e

2

1e ln

2

3 1ln e ln

8 2

c

c

B

Be

c

c

B

1ln

1 2ln 0.11572 0.116

33ln

8

c

0.1160.98140

tex e

(iv)

The population of foxes in the long run is 40.

(v)

x

15

t 0

x = 40

11 (i)

0 0 10

5 10 10

30 15 30

OP OQ OR

0 0 0

10 5 5

15 30 15

PQ OQ OP

10 0 10

10 5 5

30 30 0

PR OR OP

(ii)

A normal to p

0 2 3

1 1 6

3 0 2

Equation of plane

3 0 3

6 10 6

2 15 2

3

6 90

2

r

r

3 6 2 90x y z

Or any equivalent equation of plane

(iii)

A normal to the plane EFGH =

0

0

1

(or any equivalent vector)

cos

0 3

0 6

1 2 2

1 9 36 4 49

073.4

(iv)

0 10 51 1

10 10 = 102 2

15 30 122

2

OS OQ OR

5 0

4 10 5

1 3022

2

4 1

20 41

= 45 95

120 24

OT

1

H2 Mathematics 2017 Prelim Exam Paper 1 Question

Answer all questions [100 marks].

1 Without the use of a calculator, find the complex numbers z and w which satisfy the

simultaneous equations

i 3z w 2 6 3i 0z w

[6]

2 The function f is defined by 2

1f :

1x

x , ,x 1x .

(i) Show that 3

2 3 1

1 1

An B

n n n n n

, where A and B are constants to be found. [3]

(ii) Hence find 3

2

2 6n

r

r

r r

. [4]

(iii) Use your answer to part (ii) to find 2

2 10

1 2 3

n

r

r

r r r

. [1]

3 The function f is defined by 2

1f :

1x

x , ,x 1x .

(i) Find 1f ( )x

and write down the domain of 1f . [3]

(ii) On the same diagram, sketch the graphs of f ( )y x , 1f ( )y x and

1f f ( )y x

stating the equations of any asymptotes and showing the relationships between the

graphs clearly. [4]

(iii) State the set of values of x such that 1 1ff ( ) f f ( )x x . [1]

4 Referred to the origin O, the point A has position vector −5i + 2j + 2k and the point B

has position vector i + 3j − 2k. The plane has equation:

(1 2 ) (3 2 ) ( 2) r i j k where ,

(i) Find the vector equation of plane in scalar product form. [2]

(ii) Find the position vector of the foot of perpendicular, C, from A to . [3]

The line l1 passes through the points A and B.

The line l2 is the reflection of the line l1 about the plane . Find a vector equation of l2.

[3]

Blank Page

Blank Page

5

It is given that DEFG is a square with fixed side 2a cm and it is inscribed in the isosceles

triangle ABC with height AH , where AB = AC and angle BAH = .

(i) Taking tant , show that the area of the triangle ABC is given by

2 14 4S a t

t

[3]

(ii) Find the minimum area of S in terms of a when t varies. [4]

(iii) Hence sketch the graph showing the area of the triangle ABC as varies. [3]

6 (a) There are three yellow balls, three red balls and three blue balls. Balls of each colour

are numbered 1, 2, and 3. Find the number of ways of arranging the balls in a row such

that adjacent balls do not sum up to two. [2]

(b) In a restaurant, there were two round tables available, a table for five and a table for

six. Find the number of ways eleven friends can be seated if two particular friends are not

seated next to each other. [4]

7 For the events A and B, it is given that

P( ') 0.6, P( ') 0.83 and P( | ') 0.83A B A B A B

Find,

(i) P( )B [2]

(ii) P( )A B [2]

(iii) P( | ')B A [2]

Hence determine whether A and B are independent. [1]

8 A fairground game involves trying to hit a moving target with a gunshot. A round consists

of a maximum of 3 shots. Ten points are scored if a player hits the target. The round

ends immediately if the player misses a shot. The probability that Linda hits the target in

a single shot is 0.6. All shots taken are independent of one another.

(i) Find the probability that Linda scores 30 points in a round. [2]

The random variable X is the number of points Linda scores in a round.

(ii) Find the probability distribution of X. [3]

(iii) Find the mean and variance of X. [4]

D G

F E H C B

A

2a

3

(iv) A game consists of 2 rounds. Find the probability that Linda scores more points in

round 2 than in round 1. [2]

9 Six cities in a certain country are linked by rail to city O. The rail company provides the

information about the distance of each city to city O and the rail fare from that city to city

O on its website. Charles copied the table below from the website, but he had copied one

of the rail fares wrongly.

City A B C D E F

Distance, x km 100 270 120 56 289 347

Rail fare, $y 11.1 17.1 6.44 7.62 17.9 18.8

(i) Give a sketch of the scatter diagram for the data as shown on your calculator. On

your diagram, circle the point that Charles has copied wrongly. [2]

For parts (ii), (iii) and (iv) of this question you should exclude the point for which Charles

has copied the rail fare value wrongly.

(ii) Find, correct to 4 decimal places, the product moment correlation coefficient

between

(a) ln x and y,

(b) 2x and y. [2]

(iii) Using parts (i) and (ii), explain which of the cases in part (ii) is more appropriate

for modelling the data. [2]

(iv) By using the equation of a suitable regression line, estimate the rail fare when the

distance is 210 km. Explain if your estimate is reliable. [3]

10 A factory manufactures round tables in two sizes: small and large. The radius of a small

table, measured in cm, has distribution N2(30,2 ) and the radius of a large table, measured

in cm, has distribution N2(50,5 ) .

(i) Find the probability that the sum of the radius of 5 randomly chosen small tables is less

than 160 cm. [2]

(ii) Find the probability that the sum of the radius of 3 randomly chosen small tables is

less than twice the radius of a randomly chosen large table. [2]

(iii) State an assumption needed in your calculation in part (ii). [1]

A shipment of 12 large tables is to be exported. Before shipping, a check is done and the

shipment will be rejected if there are at least two tables whose radius is less than 40 cm.

(iv) Find the probability that the shipment is rejected. [3]

The factory decides now to manufacture medium sized tables. The radius of a medium

sized table, measured in cm, has distribution N2( , ) . It is known that 20% of the

medium sized tables have radius greater than 44 cm and 30% have radius of less than 40

cm.

(v) Find the values of and . [4]

11 The Kola Company receives a number of complaints that the volume of cola in their cans

are less than the stated amount of 500 ml. A statistician decides to sample 50 cola cans to

investigate the complaints. He measures the volume of cola, x ml, in each can and

summarised the results as follows:

24730x , 2 12242631x .

Blank Page

Blank Page

− End Of Paper −

(i) Find unbiased estimates of the population mean and variance correct to 2 decimal places and carry out the test at the 1% level of significance. [6]

(ii) One director in the company points out that the company should test whether the

volume of cola in a can is 500 ml at the 1% significance level instead. Using the

result of the test conducted in (i), explain how the p-value of this test can be obtained

from p-value in part (i) and state the corresponding conclusion. [2]

The head statistician agrees the company should test that the volume of cola in a can is

500 ml at the 1% level of significance. He intends to make a simple rule of reference for

the production managers so that they will not need to keep coming back to him to conduct

hypothesis tests. On his instruction sheet, he lists the following: 1. Collect a random sample of 40 cola cans and measure their volume.

2. Calculate the mean of your sample, x and the variance of your sample, 2

xs .

3. Conclude that the volume of cola differs from 500 ml if the value of x lies…..

(iii) Using the above information, complete the decision rule in step 3 in terms of .xs

[4]

A party organiser has n cans of cola and 2n packets of grape juice. Assume now that the

volume of a can of cola has mean 500 ml and variance 144 ml2, and the volume of a packet

of grape juice has mean 250 ml and variance 25 ml2. She mixes all the cola and grape

juice into a mocktail, which she pours into a 120-litre barrel. Assume that n is sufficiently

large and that the volumes of the cans of cola and packets of grape juice are independent.

(iv) Show that if the party organiser wants to be at least 95% sure that the barrel will not

overflow, n must satisfy the inequality 1000 22.9 120,000 0n n . [4]

ANNEX B SAJC H2 Math JC2 Preliminary Examination Paper 2

QN Topic Set Answers

1 Complex numbers 2iz = or 3iz = −

2 3w i= + or 3 3iw = − +

2 Sigma Notation and Method of Difference

i) 3

3n

n n

+−

ii) 4 2

31n n

− ++

iii)5 4 2

6 2 3n n− +

+ +

3 Functions i) ( )1 1

f 1xx

− = + ; ( )1fD x− = ( )0,∞

iii) 1x >

4 Vectors

i)

2

1 3

4

= −

r �

ii)

371

137

10

−

; 2

1 46

: 3 9 ,

2 20

l t t

− = + − ∈ −

r �

5 Differentiation & Applications

ii) 28a

6 P&C, Probability a) 151200

b) 1064448

7 P&C, Probability i) 0.277

ii) 0.107

iii) 0.580; Events A & B are not independent.

8 DRV i) 0.216

iii) 11.76, 137.7024

iv) 0.358

9 Correlation & Linear Regression

iia) 0.9996r =

iib) 0.9514r =

iv) 15.73

10 Normal Distribution i) 0.987

ii) 0.828

iv) 0.0294

v) 2.93; 41.5σ µ≈ ≈

11 Hypothesis Testing i) 2 494.60; 228.02x s= =

ii) p-value 0.00572 0.01= ≤ , reject 0H .

1

H2 Mathematics 2017 Prelim Exam Paper 2 Solution

1 Method 1

i 3z w− =

⇒ 3

3i ii

zw z

−= = − --- (1)

Substitute (1) into 2 6 3i 0z w− + + = 2 (3i i) 6 3i 0z z− − + + =

⇒ 2 i 6 0z z+ + =

⇒ ( )2

i i 4(1)(6) i 1 24

2 2z

− ± − − ± − −= =

i 5i

2

− ±=

∴ 2iz = or 3iz = −

⇒ 3i (2i)iw = − 3i ( 3i)iw = − −

2 3i= + 3 3i= − +

Method 2

i 3z w− =

⇒ 3 iz w= + …. (1)

Substitute (1) into 2 6 3i 0z w− + + =

( )23 i 6 3i 0w w+ − + + =

⇒ 29 6 i 6 3i 0w w w+ − − + + =

⇒ 2 (1 6i) 15 3i 0w w− − − + + =

⇒ 2 (1 6i) 15 3i 0w w− − − + + =

⇒ 2 (1 6i) 15 3i 0w w+ − − − =

⇒

2(1 6i) (1 6i) 4(1)( 15 3i)

2w

− − ± − − − −=

1 6i 1 12i 36 60 12i

2

− + ± − − + +=

1 6i 25

2

− + ±=

∴ 2 3iw = + or 3 3iw = − +

⇒ 3 (2 3i)i 2iz = + + = 3 ( 3 3i)i 3iz = + − + = −

Method 3

i 3

i 3i

w z

w z

= −⇒ = − +

∴ 2 ( i 3i) 6 3i 0z z− − + + + =

⇒ 2 i 6 0z z+ + =

Let iz a b= + where ,a b∈�

2( i) i( i) 6 0a b a b+ + + + =

⇒ 2 2 2 i i 6 0a b ab a b− + + − + =

⇒ 2 2 6 (2 )i 0a b b ab a− − + + + =

By comparing the real and imaginary parts,

2

2 2 6 0a b b− − + = … (1)

2 0ab a+ = … (2)

From (2), 0a = or 12

b = −

When 0a = , 2 6 0b b+ − =

( 2)( 3) 0b b− + =

2b = or 3b = −

Hence 2i, i(2i) 3i 2 3iz w= = − + = +

or 3i, i( 3i) 3i 3 3iz w= − = − − + = − +

When 12

b = − , 2 251 14 2 4

6a = − − = −

There is no real solution for a.

2 (i)

( ) ( ) ( )( ) ( ) ( )( )( )( )

( ) ( ) ( )2 2 2

3

3

2 3 1

1 1

2 1 3 1 1 1

1 1

2 2 3 3

3

n n n

n n n n n n

n n n

n n n n n

n n

n

n n

− +− +

+ − − + + −=

− +

+ − − + −=

−+=−

(ii)

32

32

2

2 6

32

2 3 12

1 1

2 3 1

1 2 3

2 3 1

2 3 4

2 3 1

2 3 4 5

...

2 3 1

2 1

2 3 1 +

1 1

n

r

n

r

n

r

r

r r

r

r r

r r r

n n n

n n n

=

=

=

+−

+=−

= − + − +

− + + − + + − +

= + + − + − − + − − +

∑

∑

∑

3

2 3 2 1 3 12

1 2 2 1

3 2 12

2 1

4 23

1

n n n

n n

n n

= − + + − + +

= − + +

= − ++

(iii)

( )( )( )2

2 10

1 2 3

n

r

r

r r r=

++ + +∑

Let 2 2r p r p+ = ⇒ = −

( ) ( )( )2

2 2

2

34

2 3

3 32 2

2 6

1 1

2 6

2 6 2 6

4 2 4 23 3

2 3 3 4

5 4 2

6 2 3

p n

p

n

p

n

p p

p

p p p

p

p p

p p

p p p p

n n

n n

− =

− =

+

=

+

= =

+=− +

+=−+ += −− −

= − + − − + + +

= − ++ +

∑

∑

∑ ∑

3 (i)

2

2

2

1f :

1

1Let

1

11

11

xx

yx

xy

xy

−

=−

= +

= ± +

a

Since 1x > , 1

1xy

= +

( )1 1f 1x

x

− = + =1 x

x

+ .

4

From graph of f, ( )fR 0,= ∞

∴ ( )1fD x− = ( )0,∞ .

(ii)

(iii)

Since 1 1ff ( ) f f( )x x x− −= = have the same rule, we investigate the domain

( )1f f1,D − = ∞ ( )1ff

0,D − = ∞

Taking the intersection of these domains,

Range of values is 1x > .

x =1

y =0

x =1

y =1

y = f(x)

=1

y =0

x =0

5

4 (i)

Equation of plane is

1 1 2

3 2 0 , ,

2 0 1

λ µ λ µ−

= + − + ∈ −

r �

A normal vector to plane is

1 2 2

2 0 1

0 1 4

− − − × = − −

Hence vector equation of the plane is

2 1 2

1 3 1

4 2 4

2

1 3

4

= −

= −

r

r

� �

�

(ii)

5 2

: 2 1 ,

2 4ACl s s

− = + ∈

r �

Thus

5 2

2 1 for some

2 4

OC s s

− = + ∈

uuur� .

Since C lies on the plane:

5 2 2

2 1 1 3

2 4 4

2( 5+2 )+(2+ ) 4(2 4 ) 3

3

21

s

s s s

s

− + = −

− + + = −

= −

�

Thus

32 5

2137

3 12 13

21 710

34 2

21

OC

− − − = − + = − +

uuur

(iii)

Using mid-point theorem

6

' 2

37 5 392 1

13 2 127 7

10 2 6

OA OC OA= −− − − = − =

uuur uuur uuur

B is the point of intersection of l1 and π .

' '

39 11

12 37

6 2

461

97

20

BA OA OB= −− = − −

− = −

uuur uuur uuur

2

39 461

: 12 9 ,7

6 20

l t t

− − = + − ∈

r � or

2

1 46

: 3 9 ,

2 20

l t t

− = + − ∈ −

r �

5 (i)

The height of triangle ADG is tan

a a

tθ= .

Hence 1

2 2a

AH a at t

= + = +

.

2 tan (2 1)BH BE EH a a a tθ= + = + = +

Area 1

( )( )2

S AH BC=

( )12 2 (2 1)

2

aS a t

t

= + +

2 12 (2 1)S a t

t

= + +

2 14 4S a t

t

= + +

(ii)

2

2

d 14

d

Sa

t t

= −

When d

0d

S

t= ,

2 1

4t =

⇒ 1

2t = ±

7

Reject 1

tan2

t θ= = − as θ is acute

22

2 3

d 2

d

Sa

t t

=

When 1

2t = ,

23

32

d 216 0

d 1

2

Sa a

t

= = >

.

Hence the minimum value of S occurs when 1

2t = .

Minimum ( )2 24 2 2 8S a a= + + = .

(iii)

To sketch the graph of

2 14 4 tan

tanS a θ

θ = + +

6 (a)

Since adjacent balls do not sum up to two, balls numbered ‘1’ needs be separated.

Number of ways of arranging the other balls with no restriction = 6!

Slotting in the balls numbered ‘1’, permutation is done as balls are of different colour =7

33!C ×

No of ways 7

33!6!

151200

C= ×=

×

(b)

Method 1

Table of 5 Table of 6

Case 1 – 2 friends are seated together at table of 5

2

πθ = 0θ =

1 21tan ,8

2a−

S

θ

2 friends

X

X

X (6-1)!

8

No. of ways to select 3 other friends and arrange them at the table of 5 = 9

3(4 1)!C × −

No. of ways to arrange the 2 friends = 2!

No. of ways to sit the remaining friends at the table of 6

= (6-1)! = 5! = 120

Total no. of ways = 9

3(4 1)! 2! 5!C × − × × =120960

Case 2 – 2 friends are seated together at table of 6

Table of 5 Table of 6

No. of ways to select 4 other friends and arrange them at the table of 6 = 9

4(5 1)!C × − =

3024

No. of ways to sit the 2 friends at the table of 6 = 2!

No. of ways to sit the remaining friends at the table of 5

= (5-1)! = 4! = 24

Total no. of ways = 9

4(5 1)! 2! 4!C × − × × = 145152

No of ways to arrange 11 friends without restrictions

= 11

5(5 1)! (6 1)!C × − × − = 1330560

Total no. of ways of arranging 11 people such that 2 particular friends are not seated

together

= 1330560 – 120960 – 145152 = 1064448

Method 2

Alternative Method

Case 1: Two particular friends seated at table of 5

No of ways 9

3C 2! 3 2

1209 0

5!

6

==

× × × ×

9

3C : Selection of friends to be seated at table of 5. This automatically selects friends to

be seated at table of 6.

(3-1)!: Arranging the 3 other friends in table of 5.

3

2P : Slotting in the 2 particular friends

5!: Arranging the 6 other friends in table of 6.

Case 2: Two particular friends seated at table of 6

No of ways 9

4C 4! 3! 4

21772

3

8

× × ×=

×=

9

4C : Selection of friends to be seated at table of 5. This automatically selects friends to

be seated at table of 6.

(5-1)!: Arranging the 5 friends in table of 5.

(5-1)! X

2 friends

X X

X

9

4!: Arranging the 5 friends in table of 6.

4

2P : Slotting in the 2 particular friends

Case 3: Two particular friends seated at separate tables

No of ways 9

4C 4! 5!

7 5 60

2

2 7

× ×=

×=

9

4C : Selection of friends to be seated at table of 5. This automatically selects friends to

be seated at table of 6.

(5-1)!: Arranging the 5 friends in table of 5.

(6-1)!: Arranging the 6 friends in table of 6.

x2: The 2 particular friends can switch tables

Total no. of ways

120960 217728 725760

1064448

= + +=

7 (i)

Given P( | ') 0.83A B =

P( ')0.83

P( ')

0.60.83

1 P( )

( ) 1 0.72289 0.27711 0.277

A B

B

B

P B

∩⇒ =

⇒ =−

⇒ = − = =

(ii)

Let P( A B∩ ) = x

P( ) ( ') ( )

0.6 0.27711

0.87711

A B P A B P B

x x

∪ = ∩ += + + −=

x

A B

0.6 0.27711-x

0.12289

10

P( ) ' 1 0.87711 0.12289A B∪ = − =

Since P( ') 0.83A B∪ =

0.6 0.12289 0.83

0.10711

x

x

∴ + + =⇒ =

∴P( A B∩ ) = 0.107 .

(iii)

P( ')P( | ')

P( ')

0.27711 0.10711

1 (0.6 0.10711)

0.17

0.29289

0.58042

0.580

B AB A

A

∩=

−=− +

=

==

Since P( | ')B A P( )B≠ ⇒ B is not independent of A’

∴ A and B are not independent.

8 (i)

P(Linda scores 30 points) = ({hit, hit, hit})P

= 0.63

= ��

��� (0.216)

(ii)

Let X be the number of points scored by Linda in a round.

X 0 10 20 30

P(X=x) 0.4 0.6×0.4

=0.24

0.62×0.4

=0.144

0.216

(iii)

E(X) = 0×0.4 + 10×0.24 + 20×0.144 + 30×0.216

=11.76

E(X2) = 02×0.4 + 102×0.24 + 202×0.144 + 302×0.216

= 276

Var(X) = E(X2) – [E(X)]2

= 276 – 11.762 = 137.7024

(iv)

Let X1 be the number of points scored by Linda in Round 1 and let X2 be the number of

points scored by Linda in Round 2.

P(Linda scores more in round 2 than in round 1)

11

1 2

1 2

1 2

1 2

1 2

1 2

( 0 & 10)

( 10 & 20)

( 20 & 30)

( 0) ( 10)

( 10) ( 20)

( 20) ( 30)

0.4 (1 0.4)

0.24 (0.144 0.216) 0.144 0.216

0.357504 0.358 (3 s.f.)

P X X

P X X

P X X

P X P X

P X P X

P X P X

= = ≥+ = ≥+ = == = ≥+ = ≥+ = == × −+ × + + ×= =

9 (i)

(ii) (a)

Product moment correlation coefficient , 0.99959r =

(b)

Product moment correlation coefficient, 0.95137r =

(iii)

From the scatter diagram, as x increases, the value of y increases at a decreasing rate,

that seems to fit model (a) better. Also, the value of r for model (a) is closer to 1 as

compared to model (b).

(iv)

We use the regression line y on ln x

( )6.1619 ln 17.223 6.16ln 17.2y x x= − ≈ −

When 210x = ,

( )6.1619 ln 210 17.223 15.725 15.7y = − = ≈

As the value of r is close to 1 and 210x = is within the given data range, the

estimation may be reliable.

10 (i)

Let S be the random variable “radius of a small table in cm’.

Let L be the random variable “radius of a large table in cm’.

S ~ N 2(30, 2 )

L ~ N 2(50, 5 )

18.8

56 345

y

7.62

x

12

( )

2

1 2 3 4 5

1 2 3 4 5

+ + + + ~ N(5 30, 5 2 )

+ + + + ~ N 150, 20

S S S S S

S S S S S

× ×

1 2 3 4 5P( + + + + 160) 0.98733 0.987S S S S S < = ≈

(ii)

( )2 2 2

1 2 3

1 2 3

+ + 2 ~ N(3 30 2 50, 3 2 2 5 )

+ + 2 ~ N 10, 112

S S S L

S S S L

− × − × × + ×− −

1 2 3 1 2 3( + + 2 ) ( + + 2 0) 0.82765 0.828P S S S L P S S S L< = − < = ≈

(iii)

The radii of the large and small round tables are independent of one another.

(iv)

Let X be the random variable “number of large tables, out of 12, with radius less than

40 cm”. ~ B(12, P( 40 ))

~ B(12, 0.022750)

X L

X

<

( )P( 2 ) 1 P 1

1 0.97064

0.029357

0.0294

X X≥ = − ≤= −=≈

(v)

Let Y be the random variable “radius of a medium sized table in cm”

( )

P( 44 ) 0.20

P( 44 ) 0.80

440.80

440.84162

44 0.84162 1

Y

Y

P Zµ

σµ

σµ σ

≥ =< =

− < =

− =

= − − − − −

( )

P( 40 ) 0.30

400.30

400.52440

40 0.52440 2

Y

P Zµ

σµ

σµ σ

< =− < =

− = −

= + − − − −

Solving (1) and (2),

44 0.84162 40 0.5244

4 1.3660

2.9283 2.93

41.535 41.5

σ σσ

σµ

− = +== ≈= ≈

11 (i)

Unbiased estimate of population mean,

13

24730 494.60

50x = =

Unbiased estimate for population variance, 2

2 1 2473012242631 228.02

49 50s

= − =

Let X be the volume of beer in one beer can in ml and µ be the population mean

volume of beer of the beer cans.

0

1

: 500

: 500

H

H

µµ

=<

Under 0H , since n = 50 is large, by the Central Limit Theorem,

2

~ 500,50

sX N

approximately.

Use a left-tailed z-test at the 1% level of significance.

Test statistic: 500

~ (0,1)

50

XZ N

s

−= .

Reject 0H if p-value 0.01≤ .

From the sample,

value 0.0057248 0.00572p − = =

Since p-value 0.00572 0.01= ≤ , we reject 0H . There is sufficient evidence at the 1%

level of significance to conclude that the volume of cola in a can is less than 500 ml.

(iii)

Let X be the volume of cola in one can in ml and µ be the population mean volume of

cola of the cans.

0

1

: 500

: 500

H

H

µµ

=≠

Unbiased estimate of population variance,

( )22 40

39xs s=

Under 0H , since n = 40 is large, by the Central Limit Theorem,

2

~ 500,39

xsX N

approximately.

Use a two-tailed z-test at the 1% level of significance.

Test statistic: 500

~ (0,1)

39

x

XZ N

s

−=

Critical values: (1) (2)2.5758 2.5758

crit critz z= − = .

Reject 0H if

2.5758 or 2.5758cal calz z≤ − ≥ .

Since 0H is rejected,

14

2 2

2 2

2.5758 or 2.5758

500 5002.5758 or 2.5758

39 39

500 2.5758 or 500 2.575839 39

500 0.41246 or

cal cal

x x

x x

x

z z

x x

s s

s sx x

s x

− ≤ ≥− −− ≤ ≥

− ≤ ≥ +

− ≤ 500 0.41246

500 0.412 or 500 0.412

x

x x

x s

s x x s

≥ +− ≤ ≥ +

Hence the decision rule should read:

Conclude that the volume of cola differs from 500 ml if the value of x lies within this

range : 500 0.412 or 500 0.412x xs x x s− ≤ ≥ + .

(iv)

Let X be the volume of cola in one can in ml.

since n is large, by the Central Limit Theorem,

( )1 2 .... ~ 500 ,144nX X X N n n+ + + approximately.

Let Y be the volume of grape juice in one packet in ml.

since 2n is large, by the Central Limit Theorem,

( )1 2 2.... ~ 500 ,50nY Y Y N n n+ + + approximately.

1 2 1 2 2.... .... ~ (1000 ,194 )n nX X X Y Y Y N n n+ + + + + + +

1 2 1 2 2( .... .... 120,000) 0.95

120,000 10000.95

194

120,000 10001.6449

194

120,000 1000 1.6449 194

1000 22.9 120,000 0

n nP X X X Y Y Y

nP Z

n

n

n

n n

n n

+ + + + + + + ≤ ≥

− ≤ ≥

− ≥

− ≥

+ − ≤