CATHOLIC JUNIOR COLLEGE - A Level Tuitiona-leveltuition.com/.../2011/12/2013-CJC-MA-H2-P2-Prelim-soln1.pdf · Page 3 of 20 (b)To apply the formula for the sum to infinity of a convergent

Page 1 of 20

CATHOLIC JUNIOR COLLEGE H2 MATHEMATICS JC2 PRELIMINARY EXAMINATION PAPER II 2013

Section A: Pure Mathematics [40 marks]

1 A.P. & G.P.

No.

Assessment Objectives Solution Feedback

To apply the formula for

the sum of a finite

arithmetic series.

To solve practical

problems involving

arithmetic series.

(i) Total volume: 850 m3

Volumes (in m3) filled: 4, 4.5, 5, 5.5, 6, … …

Assume that the container is completely filled at the nth

pumping action.

For the AP above,

2(4) ( 1)(0.5) 8502

n

nS n

2 15 3400 0n n ----- (*)

Considering 2 15 3400 0,n n

we have

215 15 4(1)( 3400)

2(1)n

,

i.e., 51.2899 or 66.2899n n (or use G.C.)

Hence from (*),

( 51.2899)( 66.2899) 0n n

66.2899 (NA, since 0) or 51.2899n n n at the

52th

pumping action, the container would be completely filled,

with some liquid overflowing.

52

522(4) 51(0.5)

2S

871

volume of liquid that overflows 3871 850 m 21 m3

[M1]

[M½]

[M½]

[A1]

Page 2 of 20

Alternative

2(4) ( 1)(0.5) 8502

n

nS n

Using G.C.,

n 2(4) ( 1)(0.5)2

nn

51

52

53

841.5

871

901

volume of liquid that overflows

3871 850 m

21 m3

[M1]

[M1]

[A1]


the sum of a finite

arithmetic series.


the sum of a finite

geometric series.

(ii)(a) For machine A,

50

502(4) 49(0.5)

2S

812.5

Volumes (in m3) filled by B:

25 5

5, 5 , 5 , ...6 6

[M1]

For the GP above, 10

10

55 1

6

51

6

S

315694125.1548 m

6239

total volume filled after 10th

pumping action by B 3(812.5 25.1548) m

3837.6548 m 3838 m

[M1]

[A1]

Page 3 of 20


the sum to infinity of a

convergent geometric

series.

To solve practical problems

involving geometric series.

[HOT]

(b) (For the GP above, since

5| | 1,

6r sum to infinity exists.)

Sum to infinity5

51

6

30

Volume of container to be filled by machine B

3 3850 812.5 m 37.5 m

Since the theoretical maximum volume that B alone can fill is only

30 m3, which is less than 37.5 m

3,

the container would never be completely filled.

Alternative:

Since the container can only be filled to a theoretical maximum total

volume of 812.5 + 30 = 842.5 m3 , which is less than 850 m

3, the

container would never be completely filled.

[M1]

[M1]

Page 4 of 20

2 Vectors

No.


To understand concepts of

scalar product and vector

product of vectors.

(i)

l: r ==

1

1

0 +

4

3

1 , (1)

p: r

3

0 = 26 (2)

If l is parallel to p, then normal of p l

i.e.

3

0 .

4

3

1 = 0

12 3 = 0

= 4

[M1]

[M1]

To understand the relationship

between direction vector of a

line and the normal of a plane

when they are parallel to each

other

(ii)

Let

ON be

1

1

0 + ’

3

4

0 for a particular .

Then [

1

1

0 + ’

3

4

0 ].

3

4

0 = 26

3 + 9’+ 4 + 16’ = 26

’ = 1

ON =

1

1

0 + 1

3

4

0 =

2

5

0

[M1]

[M1]

[M1]

[A1]

Page 5 of 20

To use the ratio theorem in

geometrical applications.

(iii) Using Ratio Theorem,

2

5

0 =

OA +

OB

2

OB = 2

2

5

0

1

1

0 =

5

9

0

Coordinates of B: (5, 9, 0)

[M1]

[A½]

[A½]

To calculate the area of a

triangle

(iv) Area of triangle OAB

= 1

2 |

OA

OB |

1 51

1 92

0 0

01

02

14

7

[M1]

[A1]

A

B

2,5,0

Page 6 of 20

3 Functions


To obtain the inverse of a

function.

(i) 2( 2) 1y x 2( 2) 1

2 1

2 1 (N.A.) or 2 1

x y

x y

x y y

1fD = fR 1,

1f : 2 1, , 1x x x x

[M½]

[A½]

[B1]

To illustrate in graphical terms

the relationship between f(x)

and f1

(x)

(ii)

O

y = x

y

x

(2, 1)

3

3

y = f 1

(x)

y = f (x)

(1, 2)

Shape [B2]

Intercepts [B½]

End points [B½]

Page 7 of 20

.

To determine the points of

intersection between y = f(x)

and y = f1

(x)

[HOT]

(iii) y = x 1f ( ) f ( ) f ( )x x x x

2

2

( 2) 1

5 3 0

5 25 4(1)(3)

2

5 13 5 13or

2 2

5 13 5 13 rejected <2

2 2

x x

x x

x

x

[B1]

[M1]

[A1]

To understand that a composite

function is a composition of

two functions

To determine whether a

composite function exists.

(iv)

For fg to exist, g fR D

gR = 0,

fD , 2

Since g fR D , fg does not exist.

For gf to exists, f gR D

fR = 1,

gD ,

Since f gR D , gf exists.

2gf( ) g ( 2) 1x x

2

2ln ( 2) 1 1x

4 3 2ln 8 22 24 10x x x x

gf fD D , 2

2

2gf : ln ( 2) 1 1 , , 2x x x x

[M½]

[A½]

[M½]

[A½]

[M½]

[A½]

[B1]

Page 8 of 20

4 Differential Equations

No

.


[HOT] (i) For increasing population,

d0

dt

P

0.02 100 0

100

P P

P

[B1]

[B1]

To solve differential equation

of the form d

f ( )d

yy

x from a

problem situation.

To use an initial condition to

find a particular solution.

(ii)

d0.02 100

d

PP P

t

1

d 0.02d100

P tP P

1 1 1d 0.02

100 100P t C

P P

ln ln 100 2 '

ln 100 ln 2 ''

P P t C

P P t C

100

ln 2 ''P

t CP

2 ''100e t CP

P

'' 2100e eC tP

P

2100e tP

AP

2

100

1 e tP

A

[M1]

[M1]

[M1]

[M1]

Page 9 of 20

Given P = 20 when t = 0, 100

201 A

1 5

4

A

A

2

100

1 4e tP

(shown)

[A1]

[HOT] (iii) P is an increasing function and for large values of t,

100P .

[B1]

20

100P

P

t Shape [B1]

Intercept [B½]

Asymptote [B½]

Page 10 of 20

Section B: Statistics [60 marks]

5 Sampling Methods

No

.


To describe the process of

conducting systematic

sampling in the given context

(i)

Obtain the list of all 1500 students (i.e. sampling

frame)

Arrange them in some order (e.g. alphabetical,

NRIC, class, level, etc)

Select a random starting point.

Subsequently select students at regular intervals

of

150030

50

students until the sample of 50 is obtained.

[M½]

[M½]

[M½]

[M½]

To state a disadvantage of

systematic sampling in the

given context

(ii) Systematic sampling of every 30th

student arranged

in by class may result in the smaller classes being

skipped over altogether.

OR

Any reasonable answer.

[M1]

To recommend/suggest an

alternative sampling method

which will yield a more

representative sample

To describe the procedure for

conducting the suggested

method

(iii) Stratified sampling.

Divide the students into non-overlapping strata

(e.g. JC1 and JC2).

Select the number of students to pick from each

stratum according to proportion.

JC1 JC2

80050 26 67 27

1500 .

70050 23 33 23

1500 .

Select required number of students randomly from

each stratum.

[B½]

[B½]

[B½]

[B½]

Page 11 of 20

6 Probability

No

.


To use Venn diagram to

identify and calculate required

probabilities.

(i)

P ( ' ) P ( ) P ( )A B A B A

P ( ) P ( ) P ( ' )A A B A B

7 11 41P ( )

8 36 72A

[M1]

[A1]

To recall and apply the

formula for conditional

probability.

To use Venn diagram to

identify and calculate required

probabilities.

(ii) P ( ' )P ( ' | )

P ( )

B AB A

A

P ( ' ) P ( ) P ( )B A A B B

7 3 11P ( ' )

8 5 40B A

11P ( ' ) 9940P ( ' | )

41P ( ) 20572

B AB A

A

[M1]

[A1]

A B

A B

Page 12 of 20

To recall and apply the

conditions for independent

events.

To apply the concept of

complement

(iii) If A and C are independent,

P ( ) P ( ) P ( )A C A C

P ( ' ) P ( ') P ( )A C A C

41 4P ( ' ) 1

72 7A C

31 4 31

72 7 126

[M1]

[A1]

Page 13 of 20

7 Permutations and Combinations

No

.


To arrange objects in a round

table without restriction

(i) No of ways for 10 students to sit in a round table

= (10 – 1)! =9! =362880

[B1]


table with restriction

(ii) Arrange the boys then slot in the girls.

No. of ways of arranging the 5 boys in a circle

= (5 – 1)! = 4! = 24

No. of ways of slotting the 5 girls into the 5 positions

= 5

5C = 1

No. of ways of arranging the 5 girls = 5! = 120

Total no. of ways = 24 1 120 2880

[M½]

[M½]

[A1]


table with numbered chairs

To group items together

(iii) Group the students from 2T39 and 2T40.

Obtain one group of 6 and one group of 4.

No. of ways of arranging the 2 groups in a circle

= (2 – 1)! = 1! = 1

No. of ways of arranging the group of 4 = 4! = 24


Total no. of ways = 1 24 720 10 172800

[M½]

[M½]

[A1]


table with numbered chairs

To group items together

(iv) Group the students from 2T39 and 2T40. Obtain one

group of 6 and one group of 3 with 1 empty chair.

No. of ways of arranging the 3 groups in a circle

= (3 – 1)! = 2! = 2



Total no. of ways = 2 6 720 10 86400

[M½]

[M½]

[A1]

Page 14 of 20

8 Binomial Distribution

No

.


To apply the concept that

P( ) 1 P( 1)X x X x , for

discrete random variable X.

Difficulty level: MK

(i) Let X be the r. v. denoting the number of faulty

Upads out of 25.

X ~ B(25, 0.03)

P( 2)X 1 P( 2)X

= 0.0379596464

= 0.0380

[M½]

[M½]

[A1]

To apply the idea of P and C

together with Binomial

distribution to find probability.

Difficulty level: SK

(ii) Required probability 2

P( 2) P( 1) 3X X

= 0.0148

[M1]

[A1]

To recall the conditions needed

for Poisson to approximate

Binomial.

To redefine the problem

statement in which failure

becomes success.


(iii) Let Y be the r. v. denoting the number of faulty

Upads out of 80.

Y ~ B(80, 0.03)

Since n = 80 is large, np = 2.4 < 5,

Y ~ P0(2.4) approximately.

P( 3) 0.779Y

[M½]

[M½]

[M½]

[M½]

[A1]

Page 15 of 20

9 Hypothesis Testing

No.


To calculate the unbiased estimate

of population mean and variance

(i) 330.4ˆ 330.4

15

332.153

x

2

22330.41

330.415 1 15

8.09838

xs x

[M½]

[A½]

[M½]

[A½]

To understand modeling

assumptions for t-test

(ii) To test

H0: μ = 330.4

H1: μ > 330.4 at 1% level of significance

332.153 330.4Value of test statistic,

8.09838 /15t

Degree of freedom = 14

p-value = 0.01586 > 0.01

We do not reject H0 as there is insufficient evidence at 1%

level of significance to conclude that the mean has

increased.

Assume that the lifetime of a light bulb is normally

distributed.

[B1]

[B½]

[B½]

[B1]

[B1]

To understand the conditions for

use of CLT

(iii) No, as n is large, Central Limit Theorem may be

applied.

[B1]

To understand that p-value is the

lowest level of significance at

which H0 can be rejected

(iv) p-value = 0.028906

least α to reject H0 = 2.90 (3s.f.)

[B1]

Page 16 of 20

10 Poisson Distribution

No.


To interpret the word ‘between’

to mean exclusion of endpoints.

They are able to express

(a ) ( ) ( 1)P X b P X b P X a

Difficulty Level: MK

(i) Let C be the random variable denoting the no. of

JC2 students on crutches in a month.

C ~ Po (4)

P(3 7)

P(4 6)

P(C 6) P(C 3)

0.45586 0.456

C

C

[M1]

[A1]

To recall the additive properties

of Poisson variables if they are

independent.

To be able to use the

probability density function of

Poisson Random Variable to

express an equation in terms of

.

To use G.C. to solve an

unfamiliar equation for .

Difficulty level: CK


(ii) Let J be the random variable denoting the no. of

JC1 students on crutches in a month.

J ~ Po() ~ Po(4 )C J

Given that P(C 2) 0.0027J

2(4 )

2 (4 )

(4 )0.0027

2!

(4 ) 0.0054

e

e

Using G.C. to solve the equation2 (4 )(4 ) 0.0054e , we have 6 .

We assume that the number of JC1 and JC2

students who are on crutches follow independent

Poisson Distributions.

[M1]

[A1]

[B1]

To find the conditional

probability in the context of a

Poisson Distribution model.

Difficulty Level: SK

(iii) P(J = 1C+ J = 4)

P( 3) P( 1)

P( 4)

0.1536

0.154

C J

C J

[M2]

[A1]

Page 17 of 20

To recall that the characteristics

of the Poisson distribution

model and able to explain in the

context of the question.

Difficulty Level: SK

(iv) The average number of students who are on

crutches may vary from month to month, for

example, on an average there will be more

students who suffer leg injuries during intensive

training just before sports competitions compared

to months when the competition is over.

[B1]

Page 18 of 20

11 Correlation & Linear Regression

No.


To use G.C to plot scatter diagram (i)

[B1]

To assess suitability of different

models (ii) A linear model is not likely to be appropriate as the

area covered would then increase continuously,

eventually to an infinite area.

[B1]

To interpret value of r (iii) D = 53: r = 0.99349

Since D = 53 gives a value of r closest to 1, it is

the most appropriate.

[B1]

[B1]

To use regression line to find

values (iv) a = 4.26272; b = 0.60899

Equation of regression line is

ln(D − A) = 4.26272 – 0.60899t

When t = 20, A = 52.99964 cm2

[B1]

[M1]

[A1]

To interpret a regression line in

context (v) D is the maximum area of the petri dish. [B1]

t

A

Page 19 of 20

12 Normal Distribution

No

.


To combine normal random

variables with new mean and

new variance.

To find unknown population

mean when given the

probability.

(i) Let X and Y be random variables denoting the waiting

times, in minutes, for one random order of Chili Crab

and one random order of Pepper Crab respectively.

Then X ~ N( , 32) and Y ~ N(10 , 2

2)

Y – X ~ N(10 – , 32 + 2

2)

Y – X ~ N(10 – , 13)

[M1]

P(Y – X > 0) = 0.290

P( Z > 0 – (10 – )

13 ) = 0.290

– 10

13 = 0.5533847152

= 11.99525697 12.0 (to 3 s.f.) (shown)

[M1]

[A1]

To combine normal random

variables involving constants

and multiples.

(ii) Carpark charges $0.05 per minute

Let T be the random variable denoting the carpark

charges of one random customer in cents.

T ~ N( 5 10 + 5 20 , 52 2

2 + 5

2 4

2 )

T ~ N( 150, 500)

P( T > $2) = P( T > 200 cents)

= 0.0126736174 0.0127

[M2]

[A1]

To find sample size n in a

sample mean distribution.

(iii) Since Y ~ N(10 , 22)

So, Y

~ N(10, 2

2

n )

Given P(Y

> 10.5) 0.0385

[M1]

Page 20 of 20

Method 1: Using G.C. table

n P(Y > 10.5)

50 0.03855

51 0.0371

52 0.03571

[M1]

Hence, least value of n is 51.

[A1]

Method 2: Using standardization

P(Y

> 10.5) 0.0385

p(Z > 10.5 – 10

2

n

) 0.0385

0.5 n

2 1.768364425

n 50.03380381

[M1]

Hence, least value of n is 51.

[A1]

To know when CLT is

applicable.

No assumption needed since Y is normally

distributed.

[B1]

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CATHOLIC JUNIOR COLLEGE - A Level Tuitiona-leveltuition.com/.../2011/12/2013-CJC-MA-H2-P2-Prelim-soln1.pdf · Page 3 of 20 (b)To apply the formula for the sum to infinity of a convergent