2009 srjc paper2 solutions

8/22/2019 2009 SRJC Paper2 Solutions

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SERANGOON JUNIOR COLLEGE

H2 MATHEMATICS PAPER 2

MARKING SCHEME

Section A: Pure Mathematics [40 marks]

1.

The diagram shows a vertical cross-section of a container in the form of an

inverted cone of height 24 cm and base radius 9 cm . The container is

initially empty. Water is poured into the container at a constant rate of3 130 cm s and the height of water level in the container at time t is h cm.

(i) Show that the volume of water in the container is given by33

64V h .

[2]

(ii) Find the value of h at the instant when the rate of increase in thedepth of water is 15.65 cms , leaving your answer in 2 decimal

places. [3]

Solution:

9 24

r h

9 3

24 8

h hr

21

3V r h

9cm

24cm

cmh


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SRJC 2009 JC2 Preliminary Examination

21 3

3 8

hV h

33

64V h

2d 9d 64

V hh

d d d

d d d

V V h

t h t

29

30 5.6564

h

3.47h Ans: 3.47 cm

2(a)

The graph of fy x is shown above. Sketch on separate diagrams, thegraphs of

(i) 1

fy

x

[2]

(ii) 'fy x [2]You are expected to indicate clearly all important features of the graphs in

each case.

4

2 0 1

3

y

x


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SRJC 2009 JC2 Preliminary Examination [Turn Over

(b) The graphs of 2 fy x and fy x are shown below.

2y

2y

Sketch the graph of fy x , indicating clearly all important features ofthe graph.

[2]

Solution:

0 2-2 -1 x

y

2

1

1

fy x

-1

1 1

x

y

2-2

2

1

0

2

f ( )y x


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a (i)

(ii)

(b)

2 1

1/3

2 1

1/4

y

1

f ( )y x

0 2-2 1 x1

2


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3.A curve Cis defined parametrically by the equations

1 1,x t y t

t t ,

0t .(i) Show that the equation of normal at a point with parametertis given

by

2 2

2

1 2( 1)

1

t t

y xt t

. [3](ii) Hence, determine the equation of normal at the point 2t . [1](iii) The normal at the point 2t cuts the curve Cagain at the point Q.

Determine the coordinates ofQ. [4]

Solution

(i) 1x t

t

2

d 11

d

x

t t

2

2

d 1

d

x t

t t

1y t

t

2

d 11

d

y

t t

2

2

d 1

d

y t

t t

2

2

2

2

1d

1d

ty t

tx

t

2

2

d 1

d 1

y t

x t

Gradient of normal =2

2

1

1

t

t

Equation of normal at point T2

2

1 1 1( )

1

ty t x t

t t t

2 2 2

2

1 1 1

1

t t ty x

t t t


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2 2

2

1 2( 1)

1

t ty x

t t

(ii) When 2t ,4 1 2(4 1)

4 1 2y x

5

53

y x

Or3 5 15y x

(iii) 3 5 15y x

1 13 5 15t t

t t

2 21 13 5 15

t t

t t

2 23 1 5 1 15t t t 2 2

3 3 5 5 15t t t 28 15 2 0t t

1, 2 ( )

8t t rej

When1

8t ,

1 1 1 1,1 18 8

8 8

x y

Q63 65

( , )8 8


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4. The polynomial P( )z has real coefficients. The equation P( ) 0z has a

root 3 ie where 0 .

(i) Show that a quadratic factor of P( )z is 2 cosz az b where a and bare real numbers to be determined. [3]

(ii) Solve the equation 4 81 0z , expressing the solution in the formire , giving the values of r and in exact form, where 0r , and

.

[4]

(iii) Hence or otherwise, show that 4 81z can be expressed in the form 2 22 2z p z q z p z q wherep and q are real numbers to

be determined.

[3]

(i) 3 3i iz e z e 2 2 03 3 3i iz ze ze e

2 3 9i iz z e e

2 3 cos sin cos sin 9z z i i

2 3 2cos 9z z 2 6 cos 9z z

6, 9a b

(ii) 4 81 0z 4

81z

4 81( 1)z 4 ( 2 )81( )i kz e , 0, 1, 2k

1

2 43i k

z e , 0, 1, 2k

33

4 44 43 ,3 ,3 ,3i ii i

z e e e e

(iii) From 4 81z 4 81 0z

33 44 4 43 3 3 3

ii i i

z e z e z e z e

=0

2 2 36 cos 9 6 cos 94 4

z z z z

=0

2 22 26 9 6 92 2

z z z z

=0


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2 23 2 9 3 2 9z z z z =0Thus, 4 2 281 3 2 9 3 2 9z z z z z

5.

A plane 1 has equation

2

. 1 3

5

r . The pointsA andB have

coordinates 2 , 1 , 3 and 4 ,1 , 3 respectively.

(i) Find the exact length of projection of the vector ABonto the

normal of the plane 1 . [2]

(ii) Given that plane 1 meets another plane 2 with equation1

. 1 2

1

r

along the line l, find a vector equation of the line lin the form

r a b, . [2](iii) A third plane

3 contains the line land passes through the point

with position vector i + 2j + 3k. Find the vector equation of the

plane 3 and express it in the form ax by cz d . [4](iv) Deduce, or prove otherwise, that the system of equations

2 5 3

2

4 3 3 1

x y z

x y z

x y z

has an infinite number of solutions, stating your reason(s) clearly. [2]

(v) The plane 4 is parallel to 1 and has equation 2 5 5x y z .Comment on the geometrical representation of the 3 planes 2 , 3

,

and 4 .[1]


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Solution:

(i)

4 2 2

1 1 2

3 3 6

AB

Length of projection =

2 2

2 1

6 5

2

1

5

36 6 30530

(ii) Method 1:

Using GC, the vector equation of the line lis

5 6

, .: 7 7

0 1

l

r

Method 2:

Direction vector of l =

2 1 6

1 1 7

5 1 1

.

2 5 3 (1)

2 (2)

x y z

x y z

Let 0x


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(1)+ (2):6 5

5

6

z

z

,

06

7: 7

61

5

6

.l

r

Method 3:

Direction vector of l =

2 1 6

1 1 7

5 1 1

2 5 3 (1)

2 (2)

x y z

x y z

Let 0y

(1)+ (2):7 7

1

z

z

1x

4

5y

1 6

: 0 7

1 1

, .l

r

Method 4: Using rref


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6 5 (1)

7 7 (2)

x z

y z

Let z

5 6

7 7

x

y

z

Thus,

5 6

, .: 7 7

0 1

l

r

(iii)

A vector parallel to plane 3 =

1 5

2 7

3 0

=

6

3

5

Normal vector of 3 =

6 6 16

7 12 4

3 12

4

5 3

1 3

.

Thus 3

4 1 4

3 3

3 3 3

: 2 1

r .

and Cartesian equation of 3 is 4 3 3 1x y z .


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Or

3

4 5 4

3 7 3

3 0

:

3

1

r

For Method 2:

,

06

7: 7

61

5

6

.l

r

A vector parallel to plane 3 =

07

26

35

1

6

=

15

6

13

6

Normal vector of 3 =5

7 12

16 16 4

3

1 313

46

12

6

.

Thus 3

4 1 4

3 3

3 3 3

: 2 1

r .

Or 3

04 4

73 36

3

: 1

35

6

r



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For Method 3:

1 6

: 0 7

1 1

, .l

r

A vector parallel to plane3

=

1 1

2 0

3 1

=

0

2

2

Normal vector of3 =

0 6 16 4

3

2 1

2 7 12 4

1 32

.

Thus 3

4 1 4

3 3

3 3 3

: 2 1

r .

Or 3

4 1 4

3 0 3

3 1 3

: 1

r


(iv) Method 1:

The system of linear equations represents the intersection between

1 2 3, a .nd

(Keywords: Intersection of 3 planes)

Since l is a common line to 1 2 3, a ,nd thus the system has an infinite

number of solutions.

(Keywords: 3 Planes intersect along l)


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Method 2:

Using GC, the rref is

From above, the last row is zero, suggesting infinite solution.

Since l is a common line to 1 2 3, a ,nd thus the system has an infinite

number of solutions.

(Keywords: 3 Planes intersect along l)

(v) Since 4 is parallel to 1 , the planes 2 3 4, and will form

an infinite triangular prism


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Section B: Statistics [60 marks]

6. A presidential candidate in country USS wanted to compare the preference ofregistered voters in the north-eastern states. The candidate hired a professional

pollster to randomly choose 1000 registered voters to be interviewed on their

preference.(i) Describe in detail how the method of quota sampling can be used toselect the 1000 registered voters for the interview.

[2]

(ii) The pollster decided to use stratified sampling. List any two pieces ofinformation he should consider acquiring. [2]

Solution:

(i) Mutually Exclusive subgroups defined.

(e.g. Gender, Age group)Number for specific subgroups quoted.

(e.g. 400 males and 600 females)

(ii) Students should include the following as answers:(1) List of voters particulars(2) Give example of strata Proportion of gender for the population of registered voters Proportion of the category of age groups for the registered voters

7. A box contains 20 balls that are labeled from 1 to 20. Raju and Kelvin take turns

to pick a ball from the box with Kelvin drawing a ball first. The game will stop

when a prime number (2, 3, 5, 7, 11, 13, 17, 19) is drawn.

(a) If the balls are drawn without replacement, find the probability that Kelvinis the first to pick a prime number given that the game ends on the second

draw of theplayer. [3](b)If the balls are now drawn with replacement, find the probability that Rajuis the first to pick a prime number. [3]

Solution

(a) P(Kelvin is the first to pick a prime | game ends on the second draw)

P Kelvin picks a prime on his second draw

P Game ends on the second draw

P Kelvin picks a prime on his second draw

P Kelvin picks a prime on his second draw +P Raju picks a prime on his second draw

12 11 8

20 19 18

12 11 8 12 11 10 8

20 19 18 20 19 18 17

17

27 (or 0.630 )

(b) P(Raju picks a prime first) =

3 2 3 3 3 2 3 3 3 3 3 2...

5 5 5 5 5 5 5 5 5 5 5 5


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=

26 6 9 6 9

...25 25 25 25 25

=

6

259

1 25

=3

8

8. In a factory, a machine automatically cuts out metal caps. It is given that 0.5%of the caps are dented. Find the probability that out of 200 caps, at least 4 are

dented.

A box containing 200 of such metal caps are analyzed. The box is considered

substandard if it contains at least 4 dented caps. Using a suitableapproximation, calculate the probability that at most 6 out of 300 boxes will be

`substandard. [6]

(i) Let X be the random variable number of metal caps out of 200 that are dented.

T ask for it ~ 200,0.005X B ( 4) 1 ( 3)P X P X

( 4) 0.018681P X

0.0187

(ii) Y number of boxes out of 300 fail (ie each contains 4 dented caps)~ (300, 0.018681)Y B

300n , n is large5.6043np >5

294.3957nq >5

~ 5.6043, 5.4997Y N approximately

P(at most 6 of 300 boxes) implies P( 6)Y

Using GC, .P( 6) P( 6.5) 0.649c cY Y

9. The random variableXhas a normal distribution with mean and variance 2 .

The random variable Sis the sum of 10 independent observations ofX, and the

random variable Tis the sample mean of 4 independent observations ofX. It isgiven P( 510) 0.132S and P( 53) 0.983T . Find and .

Solution:Method 1:

2~ ,X N


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2

1 2 10... ~ 10 , 10S X X X N

Given P( 510) 0.132S

1-P( 510) 0.132S

P( 510) 0.868S

510 10P( ) 0.868

10Z

510 101.11699

10

510 10 1.11699 10

51 0.35322 -----------------(1)

2

1 2 3 4 ~ ,4 4

X X X XT N

P( 53) 0.983T

53P( ) 0.983

2

Z

532.12007

2

53 1.0600 --------------------(2)

(2)(1): 2 0.70677

2.829 2.83

50

Method 2:

2

1 2 3 4' ~ 4 , 4T X X X X N

P( 53) 0.983T

P( ' 212) 0.983T

212 4P( ) 0.983

2Z

53 1.0600 ------------(as in equation 2)

10. A fruit grower produces a large number of apples every day. A small proportionp of these apples is infected. A check is carried out each day by taking a

random sample of 50 apples and examining them for infection. The numberX

of infected apples in the sample of 50 apples may be assumed to have anapproximate Poisson distribution.


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(i) State an inequality satisfied byp. [1](ii) The probability that none of the 50 apples is infected is 0.3. Show that

the probability that not more than 2 apples are infected is 0.879. [3](iii) If exactly 3 apples are infected, a further random sample of 10 apples is

taken. The days production is accepted as satisfactory in either of the

following two cases :a) the number of infected apples in the sample of 50 is at most 2;b) the number of infected apples in the sample of 50 is 3 and the

number of infected apples in the sample of 10 is 0 or 1.

Find the probability that the days production is accepted as satisfactory. [4]

Solution

i. ~ B( , )X n p

For Poisson distribution to approximate Binomial Distribution,

5

1

10

np

p

ii ~ Po(50 )X p

P( 0) 0.3X

50

e 0.350 1.20397

1.20397

50

0.0240794

0.02408

p

p

p

p

p

~ Po(1.20397)X

P( 2) 0.87863

0.879 (3.s.f)

X

iii P( 3) 0.087261X

Let Ybe the random variable of the number of infected apples in the sample of

10.

~ B(10, 0.0240794)Y

Probability that the days production is satisfactory


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= 0.87863 + 0.087261 x P( 1)Y

= 0.964

11. (a) In how many ways can 7 packs of chocolates be distributed among 20children, if no child can get more than one pack?

[1]

(b)In how many ways can 4 different gifts be distributed among 20 children ifeach child can get any number of gifts?

[1]

(c)9 people go to a restaurant with the layout given by the diagram below. Inhow many ways can the 9 people be seated on the chairs markedA,B, C, ,H, I? [1]

Find the number of ways in which the 9 people can be seated if

(i) 3 particular people do not want to be seated on any of the 4 chairsA,D,EandI, in front of windows.

[2]

(ii) 2 particular people do not want to be seated next to each other on thesame side of the table.

[3]

Solution:

(a) 2077520

7

(b) 420 160000 (c) Number of ways 9 people can be seated = 9! = 362880

(i) Method 1

Number of ways the 3 particular people can be seated = 5 3 60P


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Number of ways the 6 other people can be seated = 6!

Total ways = 60 (6!) = 43200

Method 2

No. of ways the 6 other people can take up the window seats 6 4 4! 360C

Total ways = 360 (5!) = 43200(ii) Method 1

No. of ways the two person are seated together on the side with seats labeled A

to D = 3 2! 7!

No. of ways the two person are seated together on the side with seats labeled E

to I = 4 2! 7!

Total ways that the 2 particular people are not seated next to each other on the

same side = 9! 3 2! 4 2! 7! 292320.

Method 2

No. of ways the two person are seated separately on the side with seats labeled

A to D 4 2 2! 3 2! 7! 30240.C

No. of ways the two person are seated separately on the side with seats labeled E

to I 5 2 2! 4 2! 7! 60480.C

No. of ways that the 2 particular people are seated on different side

5 41 1 2! 7! 201600C C

Total ways that the 2 particular people are not seated next to each other on the

same side 201600 30240 60480 292320


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12. A film distributor claimed that the tickets sales for a particular movie hit an

average of 1185 tickets a day. A random sample of 56 days was taken and the

sales,x tickets, for each day were recorded. It was found that

1000 9976x

and

2

43286.1x x .(i) Find, correct to 2 decimal places, the unbiased estimate of the population

mean and variance of the ticket sales. [2](ii) Test at 2% significance level whether the distributor was overstating his

claim. [5]

(iii) Explain what is meant by the phrase at 2% significance level in thecontext of this question. [1]

(iv) The same sample of 56 days is used in another test to determine whetherthe average sale of tickets is 1185. If the null hypothesis is rejected,

determine an inequality satisfied by the significance level % , of this

test. [2]

Solution

i ( 1000)1000

99761000

56

1178.14

xx

n

2

2( )

1

43286.1

55

787.02

x x

s n

ii H : 1185o

1H : 1185

By CLT, 787.02~ (1185, )56

X N approximately

p-value = 0.0336 > 0.02

Do not reject HO


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Insufficient evidence that mean ticket sale is less than 1185 (or equivalent but

must be contextual) at 5% level of significance.

iii The probability of 0.02 that the test indicates the distributor was overstating

his claim when in fact he is not.

iv H : 1185o

1H : 1185

p-value = 0.0673 (from GC)

If Ho is rejected, then100

p

6.73

Alternative solution

If Ho is rejected, then 2100

p

wherep is the value obtained in part ii).

6.72


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13. The temperature of the body, oC , is observed to vary with time t. At a

particular environmental temperature o0 C , the table below shows a set of data

relating 0( )Y and time t.

Y( o C) 60 46 33 21 15 11 8 7

t(min) 1 4 7 10 13 16 19 22

(i) Sketch a scatter diagram ofYagainst tfor the dataand, by comparing the product moment correlation

coefficients of the models below, deduce which of

the following models is more appropriate.

A:

2

Y a bt

B: ln( )Y a bt [4]

(ii) For the selected model, find the values ofa and b. [2](iii) If the environmental temperature is o0 28 C ,

find the temperature of the body, correct to 1

decimal place, half an hour after the start of theexperiment. Comment on why it is unwise to use

the selected equation in (i) to estimate this

temperature of the body. [2]

(iv) Explain why it is reasonable to use the selectedregression line in (i) to estimate twhen 50Y . [2]

Solution

Axes labeling and scale1 markCorrect plotting of the points1 mark


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For model A, 0.837r while for model B, 0.995r .Thus model B is the more appropriate model.

(ii) From G.C, 0ln( ) 4.201 0.1088t So a = 4.20 and b = 0.109

(iii) When 0 = 28 and t= 30,

So ln( 28) 4.201 0.1088(30)

30.6 (1 d.p) Since t= 30 is out of the range oftfor the given data

(which is [1, 22]), the behavior of ln( 0 ) against tmay

no longer be linear and so the estimate may be inaccurate.

(iv) Time is a controlled variable and so the temperature ofthe body is dependent on it.

2009 srjc paper2 solutions

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