2010 IJC MA H2 P2 Prelim Soln

7/31/2019 2010 IJC MA H2 P2 Prelim Soln

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2010 JC2 Prelim 2 H2 Maths Paper 2

IJC/2010/JC2 9740/02/S/10

Section A : Pure Mathematics [40 marks]

1 Three space probes, Probe A, Probe B and Probe Cwith mass 600 kg, 630 kg and 900 kg

respectively were sent into space to find the gravitational pull on the planets Venus, Mars and

Saturn. The sum of the weights of ProbeB on the three planets is 13860 N. The weight of Probe

C on Saturn is 2880 N more than the weight of Probe A on Venus. The average weight of

ProbeA on Saturn, ProbeB on Venus and Probe Con Mars is 4870 N.

Find the gravitational pull on each of the planets. [3]

Hence find the weight of ProbeD,which has mass 500 kg, on planet Saturn. [1]

[Weight (in N) = Mass (in kg) gravitational pull 2ms ]Total: 4 marks

Solution

1 Let V2

ms

,M2

ms

, S2

ms

be the gravitational pull on each planet Venus, Mars andSaturn respectively

630S+ 630V+ 630M= 13860 ---------- (1)

900S 600V = 2880 ---------- (2)

600S+ 630V+ 900M = 3(4870) = 14610 --------- (3)

630 630 630 13860

900 600 0 2880

600 630 900 14610

S

V

M

From GC,

M= 3.8, V= 9, S= 9.2

Hence the weight of ProbeD on Saturn is

500 9.2 N = 4600 N


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3/18

4

3 Given that 1 2cosy x , show that

d

sin 2 0d

yy x

x .

Show that

3

3

d0

dy

x when 0x . Hence write down the first two non-zero terms in the Maclaurin

series for 1 2cos x . [8]Total: 8 marks

Solution

3 1 2cosy x 2

cos xy

d

sin 2

d

yy x

x

d

sin 2 0d

yy x

x (shown)

22

2

d dsin cos 2

dd

y yy y

xx

3 2

3 2

22

2

d d dsin cos

dd d

d d d dcos 2 sin 0

d d dd

y y yy y

xx x

y y y yy y

x x xx

33 2

3 2

d d d dsin 3 cos sin 0

d dd d

y y y yy y y

x xx x

When 0x ,

20cos 1

y

d0

d

y

x

2

2

d 2d

y

x

3

3

d0

d

y

x

22

2...02

!20

2x

xy


4/18

5

4 The curve Chas equation 4 5

2 42

ay ax a

x

, where a is a constant.

(i) Given that curve Chas turning points, show that5

4a or 0a . [3]

(ii) Sketch curve Cfor the case when 1a , indicating clearly the coordinates of the turningpoints and the equations of any asymptotes. [4]

(iii) Hence by sketching an appropriate graph on the same diagram, solve

96

2x x

x

. [3]

Total: 10 marks

Solution

4(i)

4 52 4

2

ay ax a

x

2

d (4 5)( 1)

d 2

y aa

x x

For turning points,d

0d

y

x

2

4 50

2

aa

x

2

2 4 5a x a

24 4 4 5 0a x x a

2 4 5 0ax ax

Since there are turning points, there are two distinct real roots.

Thus, 0D

2

4 4( )( 5) 0a a

24 5 0a a

4 5 0a a

Thus,5

4a or 0a . (Shown)

5

4

0


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6

Alternatively

For turning points,d

0d

y

x

2

4 50

2

aa

x

2 4 5

2a

xa

Since there are turning points, there are two distinct real roots.

Thus,4 5

0a

a

Thus,5

4a or 0a . (Shown)

4(ii)When 1a ,

96

2y x

x

6

1

2

6

14

5 x

y

6y x

2x

96

2y x

x

5

4 0

++


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4(iii) 96

2x x

x

9

62

xx

x

From GC, the curves intersect atx = 0.823 (to 3 s.f)

From graph, solution is 2x or 0.823x .

6

1

2

6

14

5 x

y

6y x

2x

96

2y x

x

y x


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8

5 (a) The complex numberz is given by

3

2

3 i

1 iz

p

, where 0p .

Given that 2z , find the value ofp. [3]

Show that 5

arg6

z

. [3]

(b) Find the exact roots of the equation

52 0z

in the form ier , where 0r and . [5]

Total: 11 marks

Solution


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9

5(a)

3 3

2 2

3 i 3 i

1 i 1 iz

p p

3

22

3 1

1 p

2

8

1 p

2z

2

82

1 p

24 1

3 or 3 (rejected)

p

p

3

2

3 iarg arg

1 iz

p

3 2

arg 3 i arg 1 3i

3arg 3 i 2arg 1 3i 7

3 26 3 6

For arg( )z ,

7 5arg( ) 2

6 6z

(Shown)

5b 5 2z

i

2 i2e2e , 0, 1, 2

kk

By De Moivres Theorem21

510i

2 e , 0, 1, 2

k

z k

3 31 1 1 1 1

5 5 5 510 10 10 10 10i i i i

i2 e , 2 e , 2 e , 2 e , 2 ez

1 1 3 1 1 1 3

10 5 10 5 10 10 5 10 5i i i i

2 e , 2 e , 2 , 2 e , 2 ez

5

6

7

6


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10

Section B: Statistics [60 marks]

6 A companys director wants to obtain his employees views on flexible working

hours. The company has 600 employees. Describe clearly how you would choose a systematic

random sample of 30 employees. Describe briefly one disadvantage of this sampling method inthis context. [3]

Total: 3 marks

Solution

6

- Use a list with all the employees names arranged in alphabetical orderand

number them from 1 to 600

- Determine the sampling interval :600

2030

k

- Randomly select the first person from the first 20

people on the list, then select every 20

th

person subsequently until a sample of 30employees is obtained.

The sample obtained might be over-represented by a particular department (or

gender or ethnic group) of the company which has the greatest proportion of

employees, hence the systematic random sample obtained is not a good

representative of the population.


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11

7 BoxA contains 6 balls numbered 1, 2, 2, 2, 5, 7.

BoxB contains 4 balls numbered 1, 4, 4, 7.

Box Ccontains 3 balls numbered 3, 4, 6.

One ball is removed at random from each box.

(i) Show that the probability that each of the three numbers obtained is greater than 3 is1

6.

[1]

(ii) Find theprobability thateach ofthe three numbers obtained is greater than 3 or the sum

of the three numbers obtained is 13 (or both). [4]

All the balls are now placed in a single container.

A game is played by a single player. The player takes balls, one by one and with replacement,

from the container, continuing until either a number 1 results or a prime number results. Theplayer wins if the number on the last ball chosen is 1 and loses otherwise.

Find the probability that a player wins a particular game. [3]

Total : 8 marks

Solution

7(i) P(all are greater than 3)

=2 3 2

. .6 4 3

=1

6

7(ii) LetXbe the event : each of the 3 numbers is greater than 3Ybe the event : sum of the 3 numbers is equal to 13

P(Y) = P((2,7,4), (5,4,4))

=3 1 1 1 2 1

. . . .6 4 3 6 4 3

=5

72

Required probability = P( )X Y

= P( ) P( ) P( )X Y X Y

=1 5 1 2 1

. .6 72 6 4 3

=5

24

P(a player wins a particular game)

=

22 4 2 4 2

. . ......13 13 13 13 13

=

22 4 4

1 ......13 13 13

=4

13

2 1.

13 1

= 29


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12

8 Magnolia cows are milked by hand and Daisy cows are milked by machine. The time taken to

milk a randomly chosen Magnolia cow may be taken to have a normal distribution with mean

30 minutes and standard deviation 2 minutes. The time taken to milk a randomly chosen Daisy

cow may be taken to have an independent normal distribution with mean 5.5 minutes and

standard deviation 0.5 minutes.

(i) The probability that it will take less than a minutes to milk a randomly chosen Magnolia

cow is 0.85. Find a. [1]

(ii) Using an appropriate approximation, find the probability that out of 50 randomly chosen

Magnolia cows, there are more than 10 but at most 40 which take less than a minutes to

milk. [4]

(iii) Find the probability that the total time taken to milk two randomly chosen Magnolia cows

exceeds eleven times the time taken to milk a randomly chosen Daisy cow by at least 3

minutes. [3]

Total: 8 marks

Solution

8 LetMminutes be the time taken to milk a Magnolia cow,

andD minutes be the time taken to milk a Daisy cow.

2 2~ N 30,2 , ~ N 5.5,0.5M D

8(i) P 0.85M a

From GC, a = 32.073 = 32.1 (3 s.f.)

8(ii) Let Xbe the number of cows which take less than a minutes to milk, out of 50 Magnolia

cows.

~ B 50,0.85X

Since n = 50 is large, 42.5 5, 7.5 5np nq ,

~ N 42.5,6.375X approx.

Reqd prob = P 10 40X

= P 10.5 40.5X (using c.c.)= 0.21415

= 0.214 (3 s.f.)

8(iii) 1 2E 11 2 30 11 5.5 0.5M M D

2 2 21 2Var 11 2 2 11 0.5 38.25M M D

1 2 11 ~ N 0.5, 38.25M M D

Reqd prob = 1 2P 11 3M M D = 0.28573

= 0.286 (3 s.f.)


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13

9 An electronic game called Wishful Thinking is played with 5 boxes arranged in a row. When a

button is pressed, each of the five boxes displays a picture of a fruiteither an apple, orange or

pear. A possible result of the game is shown below.

Events Success and Windfall are defined as follows.

Success : Exactly 3 boxes display the same fruit.

Windfall : All 5 boxes display the same fruit.

Find

(i) the total number of possible results, [1]

(ii) the number of ways of obtaining Windfall, [1]

(iii) the total number of ways ofnotobtaining Success. [3]

Ten electronic game machines with distinct serial numbers are sent to 3 different game centres.

In how many ways can the game machines be distributed if each centre must have at least 3

machines? [3]Total: 8 marks

Solution

9(i)Total number of possible results = 53 243

9(ii) Number of ways of obtaining Windfall = 3

9(iii) No. of ways to obtain a success

=5! 5!

2 3 1203!2 ! 3!

Total number of ways 243 120 123

9 Reqd no. of ways

= 10 7 43 3 43!

126002!

C C C

Box 1 Box 2 Box 3 Box 4 Box 5


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14

10 Observations of a cactus graft were made under controlled environmental conditions. The table

gives the observed heightsx cm of the graft at tweeks after grafting.

t 1 2 3 4 5 6 8 10

x 2.0 2.4 2.5 5.1 6.7 9.4 18.3 35.1

(i) Calculate the product moment correlation coefficient between tandx. [1]

(ii) Draw a scatter diagram for the data. [2]

(iii) Using your answer in part (i) and the scatter diagram in part (ii), explain why it is

advisable to draw a scatter diagram first before interpreting the value of the product

moment correlation coefficient. [1]

(iv) Explain why the scatter diagram may be consistent with a model of the form eat b

x . [1]

(v) For the model eat bx , show that the relation between lnx and t is linear. Hencecalculate the equation of the appropriate regression line. [3]

(vi) Use the regression line in part (v) to predict the height of the cactus graft 20 weeks after

grafting. Hence explain in the context of the question why it is unwise to extrapolate. [2]

Total: 10 marks

Solution

10(i) From GC, r= 0.92378 = 0.924 (3 s.f.)

10(ii)

10(iii) From (i) and (ii), we see that though ris close to +1, the scatter diagram indicates acurvilinear relationship between tandx, instead of a positive linear relationship. Thus it

is advisable to draw a scatter diagram first before interpreting the value of the product

moment correlation coefficient.

10(iv) As tincreases,x increases but by increasing amounts. Thus the scatter diagram may be

consistent with a model of the form eat bx .

x

t

1 10

2.0

35.1


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10(v) eat bx lnx at b , where a and b are constants

Thus the relation between lnx and tis linear.

Find regression line of lnx on t.

t 1 2 3 4 5 6 8 10

lnx 0.69315 0.87547 0.91629 1.6292 1.9021 2.2407 2.9069 3.5582

From GC, eqn of regression line of lnx on tis:

ln 0.20723 0.33498x t ln 0.207 0.335x t (3 s.f.)

10(vi) When t= 20, ln 0.20723 0.33498 20 6.90683x x = 999

Height of the cactus graft 20 weeks after grafting is 999 cm.

It is impossible that the cactus can grow to that height after 20 weeks. Thus it is unwise

to extrapolate.


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11 A company claims that the guitar strings that the company manufactures have a tensile strength

of 430 kpsi (kilo-pounds per square inch) on average. An engineer obtained a sample of 8 guitar

strings and the tensile strength of each guitar string, x kpsi, is measured. The data is summarised

by

2

430 23, 430 211x x .

Test, at the 2% significance level, whether the company has overstated the average tensile

strength of a guitar string. State any assumptions that you have made. [7]

The engineer will be able to conclude that the company has overstated the average tensile

strength of a guitar string if he conducts the same test at % significance level. State the

smallest possible integer value of . [1]

The engineer takes a sample of 20 guitar strings manufactured by a rival company, whose guitar

strings have tensile strength that is normally distributed with mean kpsi and standard

deviation 4.7 kpsi. The null hypothesis 430 is being tested against the alternative hypothesis

430 at 5% level of significance. Find the range of values of the sample mean for which the

null hypothesis is rejected, giving 2 decimal places in your answer. [3]

Total: 11 marks

Solution

11 The t-test will be used.

Assume a normal distribution for tensile strength of guitar strings.

2

2 231 2117 8

20.696

s

23 4308

427.125

x

0

1

: 430

: 430

H

H

Level of significance = 2%

Under 0H : 7430

~X

T tS n

From GC, 1.79t p-value 0.0585 0.02

Do not reject 0H and conclude that there is insignificant evidence at the 2% significance

level that the manufacturer has overstated its claim.

11 6


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17

11 4.7

0

1

: 430

: 430

H

H

Level of significance = 5%

Under 0H :430

~ (0,1)4.7 20

XZ N

To reject 0H ,

4301.96

4.7 20

x or

4301.96

4.7 20

x

427.94x or 432.06x


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18

12 A roller-coaster ride has a safety system to detect faults on the track.

State a condition under which a Poisson distribution would be a suitable probability model for

the number of faults detected on the track on a randomly chosen day. [1]

Faults on the track are detected at an average rate of 0.16 per day. Find the probability that on arandomly chosen day, the number of faults detected on the track is between 2 and 6 inclusive. [2]

Find the probability that in a randomly chosen period of 20 days, there are not more than 4 faults

detected on the track. [2]

There is a probability of at least 0.15 that the mean number of faults detected on the track per day

over a randomly chosen long period ofn days is at least 0.2. Find the greatest value ofn. [3]

There is also a separate safety system to detect faults on the roller-coaster train itself. Faults are

detected by this system at an average rate of 0.05 per day, independently of the faults detected on

the track.

Find the probability that in a randomly chosen period of 20 days, the number of faults detected

on the track is at most 1 given that the total number of faults detected is 5. [4]

Total: 12 marks

Solution

12 Suitable condition

The mean number of faults detected is a constant from day to day. The faults are detected independently of one another.

LetXbe the number of faults detected on the track in a day. ~ Po 0.16X

P 2 6X = P 6 P 1X X = 0.011513 = 0.0115 (3 s.f.)

Let Ybe the number of faults detected on the track in 20 days.

~ Po 3.2Y

P 4Y 0.78061 = 0.781 (3 s.f.)

1 2 nX X X

Xn

Since n is large, by Central Limit Theorem,0.16

~ N 0.16,Xn

approx.

P 0.2 0.15X 0.2 0.16

P 0.150.16

Z

n

P 0.1 0.85Z n From GC, 0.1 1.0364n

n 107.41Greatest value ofn is 107.


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12 Let W be the number of faults detected on the train in 20 days.

~ Po 1W

~ Po 4.2Y W

P 0 P 5 P 1 P 4P 1| 5P 5

Y W Y W Y Y WY W

=0.00212437

0.163316

= 0.0130 (3 s.f.)

2010 IJC MA H2 P2 Prelim Soln

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