stpm trial 2009 matht2 q&a (pahang)

8/14/2019 STPM Trial 2009 MathT2 Q&A (Pahang)

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CONFIDENTIAL*

PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN

PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP


















PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN

JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP

PEPERIKSAAN PERCUBAAN

SIJIL TINGGI PERSEKOLAHAN MALAYSIA

NEGERI PAHANG DARUL MAKMUR 2009

Instructions to candidates:

Answerall questions. Answers may be written in either English or Malay.

All necessary working should be shown clearly.

Non-exact numerical answers may be given correct to three significant figures, or one

decimal place in the case of angles in degrees, unless a different level of accuracy is

specified in the question.

Mathematical tables, a list of mathematical formulae and graph paper are provided.

This question paper consists of 6 printed pages.

954/2 STPM 2009

Three hours

MATHEMATICS T

PAPER 2

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CONFIDENTIAL* 2

Mathematical Formulae for Paper 2 Mathematics T :

Logarithms :

a

xx

b

b

alog

loglog =

Series :

)1(2

1

1

+==

nnrn

r

)12)(1(6

1

1

2 ++==

nnnrn

r

22

1

3)1(

4

1+=

=

nnrn

r

Integration :

=

dxdx

du

vuvdxdx

dv

u

cxfdxxf

xf+= )(ln)(

)('

ca

x

adx

xa+

=

+

1

22tan

11

c

a

xdx

xa

+

=

1

22

sin1

Series:

Nnwhere ++

++

+

+=+ ,

21)( 221 nrrnnnnn bba

r

nba

nba

naba

Coordinate Geometry :

The coordinates of the point which divides the line joining (x1 ,y1) and (x2 ,y2) in

the ratiom :n is

++

++

nm

myny

nm

mxnx 2121 ,

The distance from ),( 11 yx to 0=++ cbyax is

22

11

ba

cbyax

+

++

Maclaurin expansions

1,!

)1()1(!2

)1(1)1(2


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CONFIDENTIAL* 3

Mathematical Formulae for Paper 2 Mathematics T :

Numerical Methods :

Newton-Raphson iteration for 0)( =xf :

)('

)(1

n

n

nnxf

xfxx =+

Trapezium rule :

+++++ b

ann yyyyyhdxxf ])(2[

2

1)(

1210

n

abhrhafyr

=+= andwhere )(

Correlation and regression :

Pearson correlation coefficient:

( )( )( ) ( )

=22

yyxx

yyxxr

ii

ii

Regression line ofy onx :

y =a +bx

where( )( )( ) = 2i ii xx yyxxb

xbya =

Trigonometry

BAAABA sincoscossin)sin( = BABABA sinsincoscos)cos( =

BA

BABA

tantan1

tantan)tan(

=

AAAAA 2222 sin211cos2sincos2cos === AAA 3sin4sin33sin = AAA cos3cos43cos 3 =

+=2

BAcos

2

BAsin2BsinAsin

+= 2BA

sin2

BAcos2BsinAsin

+=2

BAcos

2

BAcos2BcosAcos

+2

BAsin

2

BAsin2BcosAcos

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CONFIDENTIAL* 4

1. A passenger in an aeroplane flying at a certain height , H , sees two towns , Pekan

town and Kuantan town, directly to the left of the aeroplane. The angles of depression

to the Pekan towns and Kuantan town , are and respectively where < . If thetwo towns are d meter apart, show ( )eccossinsindH . [3]2.

The points A , B and C lie on the circumference of a circle as shown in the diagram

above with ABC= 80 and ACD=50. The tangent to the circle at point A meets thechord CD produced at point T.

(a) Show that AC = AT. [4]

(b) Show that the length of the chord CD is equal to the radius of the circle. [4]

3. The forces ( )Nj3i5F1 + , ( )Nj6i4F2 and ( )Nj7i2F3 + act at a point.(a) Calculate the magnitude of the resultant force. [2]

(b) Using the scalar product, calculate the angle between the resultant force and

force, ( )Nj3i5F4 + . [4]4. (a) A certain substance evaporates at a rate which is proportional to the amount of

substance left. Given that the initial amount of the substance is A and the amount which

has evaporated at time t is x , write a differential equation to show the rate of

evaporation. [1]

(b) Solve the differential equation and sketch the graph ofx against t. [6]

(c) Given that it took 2ln seconds for half the amount to be evaporated, find how

long it takes for4

3of the initial amount to be evaporated. [4]

5. The following table show the mean and standard deviation of the marks of the male

and female students who sat for a semester test.

Student Number of students Mean Standard deviation

Male 80 52. 9 5. 3

Female 100 61. 4 4. 1

Calculate the mean and standard deviation of the marks of all the students.[6]

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CONFIDENTIAL* 5

6. Two bowlers , Adam Lambert and Kris Allen , take turns to throw, with bowler

Adam Lambert taking the first throw. The probabilities of bowlers Adam Lambert and

Kris Allen scoring a strike are3

2and

5

4respectively. Find the probabilities of bowler

Adam Lambert scoring a strike first. [3]

7. There are one or two flowers on the faces of 50 cents stamps. 90% of all these 50 cents

stamps have two flowers while the rest of the stamps have single flower.

From the stamps which have single flower, 95% of these stamps have a flower at the

centre of the stamps while the rest have a flower on the left side of the stamps.

(a) By using a suitable approximation, determine the probability that between 5 and 15

stamps inclusive have one flower , out of a random sample of 100 pieces of 50 cents

stamps. [5]

(b) By using a suitable approximation, determine the probability that less than 3 stamps

have only one flower on the left side of the stamp out of a random sample of 100 pieces

of 50 cents stamps. [4]

8. X is a continuous random variable with a probability density function,fdefined as

( )


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CONFIDENTIAL* 6

10. (a) Prove that 2

tan2sin6sin4sin2

2sin6sin4sin2 =+

.

Hence, find the value of 15tan2 , leaving your answer in surd form. [5](b) Solve the equation

2cos

2sin6sin4sin2

2sin6sin4sin2 =+

for 3600 . [5]

11. (a) At noon, a man A is travelling at a speed of 20 km/h in the direction of N 25 W.Another man , B is 2 km to the north of A and travelling at the speed of 15 km/h in the

direction of N 60 W. Find the magnitude and the direction of the velocity of A relativeto B. Hence, find the shortest distance between man A and man B . [9]

(b) Find the course man A must travel in order to intercept B if man A maintains hisspeed but changes its course. [4]

12. (a) A machine is used to fill up bottles with a mean volume of 550 ml. Suppose that

the volume of water delivered by the machine follows a normal distribution with mean , ml and standard deviation 4 ml. Find the range of values of mean , , if it is requiredthat not more than 1% of the bottles contain less than 550 ml. [4]

(b) The mass of a box of chocolate cereal is distributed normally with mean 100 g and

standard deviation 2 g.

(i) Calculate the probability that three boxes of chocolate cereal chosen at

random, each has a mass less than 98 g. [2]

(ii) Calculate the probability that three boxes of chocolate cereal chosen at

random, have a total mass exceeding 305 g. [2]

(iii) Calculate the probability that out of three boxes of chocolate cereal chosen at

random, exactly two have a mass greater than 98 g while the mass of the remaining box

is greater than 105 g. [3]

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1

Marking Scheme

PEPERIKSAAN PERCUBAAN STPM NEGERI PAHANG 2009

Mathematics T Paper 2 ( 954 / 2 )

Question Scheme Marks

1

ABK ,AK

Hsin = AK = sin

H M1

AKP , Sine Rule :

( )sin dsinAK , ( )

sin

d

sin

sin

H

( )eccossinsindH M1

A1


2. (a) ADC = 180 80 ( Opposite angles of a cyclic= 100 quadrilateral are supplementary)TDA = 180 100 ( Angles on a straight line )= 80TAD = ACD = 50 ( Angles in the alternate segment)

ATC , DTA or CTA = 180 ( 80 + 50)= 50

{ The sum of angles in a triangle is 180 )

Since CTA = ACT = 50, ATC is an isoscelestriangle because the base angles are the same.AC = AT

M1

M1

M1

A1

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2

2. (b) Join A and C to the centre ,

then OA=OC=radius

AOC = 2 ABC= 160

( Angle at the center = 2 angle at the circumference.)-------------------------------------------------------------------

OCA = ( )1601802

1= 10

OAC is an isosceles triangle ( OA = OC = radius )-------------------------------------------------------------------

ODC, OCD =OCA + ACD= 50+10= 60 OCD =ODC = 60OCD is an isosceles triangle ( OA = OD = radius )

COD= 180 ( 60 + 60 ) = 60Since OCD =ODC = COD = 60 , thereforeOCD is an equilateral triangle andCD = OC = OD = radius of circle.

M1

M1

M1

A1


3. (a) Resultant force, RF = ( )j3i5 + + ( )j6i4 + ( )j7i2 + = ( )j4i7 +

Magnitude of the resultant force,

RF =22 47 + = 65 = 8.062 N

B1

B1

3. (b)

4R

4R

FF

FF

cos

= ,( ) ( )

j3i5j4i7

j3i5j4i7

cos ++

= ( ) ( )2222

3547

3457cos +

=

3465

47cos =

'131

M1

M1

M1

A1

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3


4. (a) Amount which has evaporated at time t = x

Amount which has not evaporated at time t = A x( )xA

dt

dx , ( )xAkdt

dx B14. (b) Interpret initial condition t=0 ,x = 0

( ) = t0x0 dtkdxxA 1 separate variables M1( )] [ ] t0x0 tkxAln = correct integration

Substitute upper limits & lower limits

M1

M1

tkA

xAln

tkeA

xA =

M1

( )tke1Ax A1Graph

D1

4. (c) ( )tke1Ax ,Substitute , 2lnt = ,2

Ax = , M1

k = 2 A1( )t2e1Ax , Substitute A4

3x = M1

2lnt = A1


5 Mean =FM

FM

nn

xx

++

, Mean =( ) ( )

10080

4.611009.5280

+

Mean =180

61404232 +, Mean =

180

10372= 57.622

M1

A1

Male : ( ) [ ]222M nx + ( ) [ ]222

M 9.523.580x + = 226 120Female : ( ) [ ]222F 4.611.4100x + = 378 677

B1

B1

Standard deviation =

22

F

2

M

180

10372

180

xx

+

=

2

180

10372

180

604797

= 6.2978

M1

A1

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4


6

P(A) + P(A K A ) + P(A K A K A ) + =

3

2+

3

2

5

1

3

1+

3

2

5

1

3

1

5

1

3

1+

=

15

11

3

2

Infinite Geometric Series

=7

5

M1

M1

A1


7. (a) X represents the number of stamps with one flower

X B ( 100 , 0.1 )Normal Approximation : mean=10 , variance = 9

( )15X5P =

3

105.15Z

3

105.4P

= ( )833.1Z833.1P = ( )0334.021 = 0. 9332

B1

B1

M1 Standardize

M1 Continuity

correction

A1

7. (b) Y represents the number of stamps with only one flower

on the left side. 005.005.01.0p = Y B ( 100 , 0. 005 )Poisson Approximation : mean, = 0.5

)3X(P < = )0X(P = + )1X(P = + )2X(P = =

+

2

5.05.01e

25.0

= 1.625 5.0e = 0.9856

B1

B1

M1

A1


8. (a)

1dxe

4

12

xk

0

= , 1e2

1k

0

2

x =

Integration M1

k = 2 ln 3 A1

8. (b)

3ln2x0 < , dxe4

1)x(F 2

xx

0Integration

x

0

2

x

e2

1)x(F

= ,

1e

2

1)x(F 2

x

M1

The cumulative distribution function is

>


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6


10. (a)LHS,

2sin6sin4sin2

2sin6sin4sin2

+ ( )( )2sin6sin4sin2 2sin6sin4sin2 ++= ( )( )2cos4sin24sin2 2cos4sin24sin2 +=

( )( )2cos14sin2 2cos14sin2 += [ ][ ]1cos21 sin211 22

=

2

2

cos2

sin2= 2tan Q.E.D

Factor

formulae M1

Double

Angle M1

A1

Substitute 15 into

2sin6sin4sin2

2sin6sin4sin2tan2 +

=

= 30sin90sin60sin230sin90sin60sin2

15tan2

2

11

2

32

2

11

2

32

+

=

332

332

+=

( )( )332 332 += ( )( )332 332 347

Values of

sin andcosM1

A1

10. (b)

2cos2sin6sin4sin2

2sin6sin4sin2 =+

, 2costan2 = 1cos2

cos

sin 22

2

242 coscos2sin 242 coscos2cos1

4

1cos4 = , 8409.0cos

2.3278.212,2.147,8.32

M1

M1

cos

sintan =

M1 Basic

Identity

M1

A1

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11. (a)

2

BAV = ( )( ) 35cos152021520 22 BAV = ( )( ) 35cos152021520 22 BAV =11.5546 km/h

The magnitude of the velocity of

A relative to B is 11.6 km/h .

D1 arrows

D1

directions/angles

M1

A1

35sin5546.11

sin

15

5546.1135sin15

sin=

'848 , '82325'848 or 23.13The direction of the velocity of A relative to B

is N 23 8 E.

M1

A1

d = '823sin2 d = 0.7857 km

The shortest distancebetween A and B is 0.786 km

D1 diagram

M1

A1

11. (b)

120sin20

sin

15

20

120sin15sin

=

'3040 '3040120180

'3019 In order to intercept B , man A must travel in the

direction of N 40 30 W . or (bearing 319 30)

D1 arrows

D1

directions/angles

M1

A1

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12 (a) ( ) 01.0550XP < , 01.04

550ZP 12

300305ZP

= ( )443.1ZP > = 0.0745

M1 Standardize

A1

(iii) 3 ( )98XP > ( )98XP > ( )105XP > = 3 2

2

10098ZP

>

>

2

100105ZP

= 3 ( )]21ZP ( )5.2ZP > = 3 [ ]28413.0 0.00621= 0.0132

M1 Standardize

M1

A1

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stpm trial 2009 matht2 q&a (pahang)

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