[Edu.joshuatly.com] N9 STPM Trial 2010 Maths T Paper 2 [w Ans] [19FCB291]

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Answer all the questions.

1. Given that θ is acute and t

1tan . Find, in terms of t, the value of .

2. Express in the form of with Hence find the set of values of , with , which satisfy the inequality

3. Four forces ,3

N

,

43

N

,

67

N

and ,

0N

act on a point. Find the numerical

values of λ and μ if a) the four forces are in equilibrium,

b) the resultant force of the four forces is 25N inclined at an angle sin-1(54 ) above the unit

vector i .

4. The position vectors of points are , and a b c respectively. The median

meet at the point with the condition that i) Express the position vectors of in terms of , and a b c . [2 marks]

ii) Show that the position vector of is given by 1 1 11 2 2

OP a b c

[2 marks]

iii) Find the position vector of in term of , , and a b c . Hence show that the position

vector of is 1 + 3

a b c [6 marks]

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5. In the diagram below , and are tangent to the circle at is produced to meet the tangent is parallel to and meets and produced at and

Prove that (i) angle and are equal.

(ii) (iii) is a cyclic quadrilateral

6. The length x, of a certain leaf, at time t (during a period of its growth) is proportional to the amount of water it contains. It may be assumed that during this period the leaf has a constant shape. The leaf absorbs water from the plant at a rate proportional to the length of the leaf and loses water by evaporation at a rate proportional to the area of the leaf. Show that the growth of

the leaf can be represented by the differential equation ,2bxaxdt

dx where a and b are positive

constants. Solve this equation, given that, when t = 0, x = b

a

2. Hence express x in terms of t

and deduce that the length of the leaf never exceeds the value b

a .

7. A and B are two independent events. Given P(A) and P(B) respectively

Diagram 1

A S

C B

Q P T


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.)(,61)(,

125)( qBAPBAPAP

Find the value of q.

8. Each customer at a supermarket pays by one of cash, cheque or credit card. The probability of a randomly selected customer paying by cash is 0.54 and by cheque is 0.18. a) Determine the probability of a randomly selected customer paying by credit card.

Three customers are selected at random. b) Find the probability of

(i) all three paying by cash.

(ii) exactly one paying by cheque.

(iii) one paying by cash, one by cheque and one by credit card. [3 marks]

9. The probability that a randomly chosen flight from Subang Airport is delayed by more than

x hours is , for No flight leaves early, and none is delayed for more than 10 hours. The delay, in hours, for a randomly chosen flight is denoted by

Find the cumulative distribution function, of and sketch the graph of

10. An athlete who is training for a marathon aims to complete many laps of a running track while running at a constant speed. In practice, the athlete’s judgement of pace is not perfect, and the time taken to complete a randomly chosen lap may be taken to be normally distributed with mean 65 s and standard deviation 0.8 s. The time taken for any lap may be assumed to be independent of the time taken for any other lap. (i) Find the probability that a randomly chosen lap will take more than 64 s.

[2 marks]

(ii) Deduce the probability that each of three randomly chosen laps will take more than


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64 s. [2 marks] (iii)The probability that the athlete will complete a randomly chosen lap in less than a s is

0.99. Find a [3 marks]

11. In a particular town, one person in 100, on average, has blood of type X. (a) If 200 blood donors are sampled at random, find, using a suitable approximation, (i) the probability that they include at least three people with blood type X

(ii) the most likely number of blood donors with blood type X and its probability.

(b) How many donors must be sampled in order that the probability of including at least

one donor of type X is 90% or more.

12. A student collected the heights, x cm, of 30 plants of a particular species. She chose to represent the data in a stem and leaf display, as shown below. Unit is 1 cm.

1 1 2 2 3 4 4 4 5 6 7 7 9 2 1 1 1 2 5 5 7 3 1 2 2 5 5 9 4 1 3 4 5 5

Key: 1│2 means 12

(a) (i) Explain why the data might be better represented by a two-part stem (two lines for each stem) and leaf display. (ii) Rewrite the above data in such a display. [1 mark]

(b) Calculate the mean height and standard deviation. [5 marks]

(c) Estimate the median and the quartiles of the data given in the display. [3 marks]


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(d) Draw a box plot to illustrate the distribution. [3 marks]

(e) State which of the mean and median would be a better measure of location for the

heights of these 30 plants. Give a reason for your answer. [2 marks]


1

Question Scheme Marks 1.

t cos2θ + sin2θ

= ]tan1tan2[]

tan1tan1[ 22

2

t

=

2

2

11

2)11(

t

ttt

= 2

22

11

t

t

t

t

= t

M1, M1 M1 A1

2. Let

………..(i)

………..(ii)

,

4 3< x < or < x <

6 3 3 2

4 3< x < or < x <6 3 3 2

M1(either one) M1(either one) A1 M1 A1

3.

Sum of the forces = 0 when in equilibrium (λ + 4)i + (μ – 5 )j = 0i + 0j

M1


2

Therefore, λ = - 4 and μ = 5.

R = 25 cosθ i + 25 sinθ j where θ is the angle between resultant and positive i axis.

= 25 ( 53 )i + 25 (

54 ) j

= 15i + 20j λ + 4 = 15, and μ – 5 = 20 λ = 11 μ = 25

A1

M1 M1 A1

4.

Y Using ratio theorem,

1(1 ) ( )2

1 1 11 2 2

OP a b c

OP a b c

Similarly, Comparing From

B1 B1 M1 A1 B1 B1(any one) B1(all 3 correct) M1

1(1 ) ( )2

1 1 11 2 2

OP b a c

OP b a c

12: ...(1)

1 112: ...(2)

1 11 12 2: ... 3

1 1

a

b

c

Let OA , , and OC1 1( ) ( )2 2

a OB b c

OY a c OX b c


3

From

Hence,

M1 A1

5.

from (alternate segment) (alternate angles)

(angles in st line)

B1 B1 B1 B1 B1 B1 B1 B1 B1 B1 B1

6. 2,dx dx

x xdt dt

B1

12

OP a b c


4

2bxaxdt

dx

dtbxax

dx

)(

dtdxbxaa

b

ax]

)(1[

ctbxaa

xa

ln1ln1

ctbxa

x

a

ln1

(for substituting t=0 and x=2a

b)

ba

c1ln1

ba

tbxa

x

a

1ln1ln1

b

e

bxa

x at

(for getting rid of ln)

)1( at

at

eb

aex

)11(

ateb

ax

When t→ , b

ax

eat ,01

B1 B1 M1A1 A1 M1 A1 M1 M1 A1 B1 M1 A1


5

7.

61)( BAP

)()()()( BAPBPAPBAP

61

52

125

q

2013

M1 A1 M1 A1

8

P(credit card) = 0.28 P(all three paying by cash) = 0.543 = 0.157

P(exactly one paying by cheque) = 21

3 )46.0()54.0( C

= 0.343

P(one paying by cash, one by cheque and one by credit card) = !3)28.0()18.0()54.0(

= 0.163

B1 M1 A1 B1 M1 A1 B1 M1 A1

9.

M1 A1 B1(all correct) D1

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10.

Let X be the time (in seconds) taken to complete a randomly chosen lap.

P(X > 64) =

A1

P(time for each of three randomly chosen laps > 64) = P(X1 > 64, X2 > 64, X3 > 64) = [P(X > 64)]3 = (0.8944)3 = 0.71548

P(X < a) = 0.99 P(X > a) = 0.01

M1

M1 A1 M1 A1 M1 M1 A1

11.

X ~ B(200, 0.01) Since n is big and p is small, Poisson approximation is used.

P(X ≥ 3) = 1 – P (X<3)

= 1 – e-2 [ !2

2!1

2121

]

= 0.3233

1

2)(

)1(

xxXP

xXP

If P(X=x+1) > P(X=x), x < 1 then P(X=1) > P(X=0) If P(X=x+1) < P(X=x), x > 1 then P(X=2) > P(X=3) > P(X=4) > … Therefore the most likely number of blood donors with blood type X is 1 or 2. P(X=1) = e-2 (2) = 0.2707 or P(X=2) = e-2 (22/ 2!) = 0.2707

B1 M1 M1 A1 M1 M1(show both) A1 A1

64 65 ( 1.25) 1 (1.25)0.8

1 0.1056 0.8944

P Z P Z Q

65 0.010.8

65 650.01 2.3270.8 0.8

66.862

aP Z

a aQ

a


7

P ( X ≥ 1) > 0.9 1 – P(X<1) > 0.9 P ( X = 0) < 0.1 0.99n < 0.1

n > 99.0ln1.0ln

n > 229.1

230 donors must be sampled.

M1 M1 M1 A1

12

Easier to see the spread / The data should be represented by at least 5 classes.

1 1 2 2 3 4 4 4 1 5 6 7 7 9 2 1 1 1 2 2 5 5 7 3 1 2 2 3 5 5 9 4 1 3 4 4 5 5 Key: 1│2 means 12

Mean = 9.2430746

n

x

Standard deviation = 2

22

30746

3022700

x

n

x

= 11.8 Median = 21.5 First quartile 15 Third quartile = 35 Lower boundary = -15, Upper boundary = 65

Median The distribution is skewed.

B1(either one) B1 M1 A1

B1 M1 A1 B1 B1 B1 D1D1D1 B1B1 [12 marks]

| | | | | 21.5 11 15 45 35


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[Edu.joshuatly.com] N9 STPM Trial 2010 Maths T Paper 2 [w Ans] [19FCB291]