maths t pahang trial stpm 2008

CONFIDENTIAL*

PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP PEPERIKSAAN PERCUBAAN JPNP
























PEPERIKSAAN PERCUBAAN

SIJIL TINGGI PERSEKOLAHAN MALAYSIA

NEGERI PAHANG DARUL MAKMUR 2008

Instructions to candidates:

Answer all questions.

Answers may be written in either English or Bahasa Malaysia.

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.

950/1, 954/1 STPM 2008

Three hours

MATHEMATICS S

PAPER 1

MATHEMATICS T

PAPER 1

Pahang Trial STPM 2008 http://edu.joshuatly.com/ http://www.joshuatly.com/

CONFIDENTIAL*

2

Mathematical Formulae for Paper 1 Mathematics T / Mathematics S :

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

duvuvdx

dx

dvu

cxfdxxf

xf+=∫ )(ln

)(

)('

ca

x

adx

xa+

=

+

−

∫1

22tan

11

ca

xdx

xa+

=

−

−

∫1

22sin

1

Series:

N n where ∈++

++

+

+=+

−−− ,21

)( 221 nrrnnnnnbba

r

nba

nba

naba LL

1,!

)1()1(

!2

)1(1)1( 2

<++−−

++−

++=+ xxr

rnnnx

nnnxx

rn whereL

LL

Coordinate Geometry :

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

ratio m : n is

+

+

+

+

nm

myny

nm

mxnx 2121 ,

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

22

11

ba

cbyax

+

++

Numerical Methods :

Newton-Raphson iteration for 0)( =xf :

)('

)(1

n

n

nnxf

xfxx −=

+

Trapezium rule :

∫ +++++≈−

b

ann yyyyyhdxxf ])(2[

2

1)( 1210 L

n

abhrhafyr

−=+= and where )(

Trigonometry :

BAAABA sincoscossin)sin( ±=±

BABABA sinsincoscos)cos( m=±

BA

BABA

tantan1

tantan)tan(

m

±=±

AAAAA2222 sin211cos2sincos2cos −=−=−=

AAA3sin4sin33sin −=

AAA cos3cos43cos 3−=


CONFIDENTIAL*

3

1. Show that 07422=−−− xyy is a parabola.

Sketch the parabola and label its vertex and focus. [4 marks]

2. Given the complex numbers iz 321 += and iz 1252 +−= , verify that 1z is one

of the square roots of 2z .

Hence, simplify )*)(( 21 zz in terms of 1z . [5 marks]

3. Given the matrix

−=

16

2

k

kM , find the set of values of k for which the

inverse of M exists. [5 marks]

4. Given xy sin= , show that 014 4

2

23

=++ ydx

ydy . [5 marks]

5. Determine the solution set of the inequality 25

<+ x

x. [4 marks]

Hence, deduce the set of values of x that satisfy the inequality

25

<+ x

x. [2 marks]

6. The parametric equations of a curve are

2

,ln teytx == ( t > 0 )

(a) Express dx

dy in terms of t . [4 marks]

(b) Find the equation of the tangent to the curve at the point t = 1. [4 marks]


CONFIDENTIAL*

4

7. Expand y+1

1 as a series in ascending powers of y up to the term in 3y .

[2 marks]

Hence, by taking 2xxy += in your series, show that for small x ,

32

2 16

7

8

1

2

11

1

1xxx

xx+−−≈

++ [4 marks]

By taking 5

1=x , find the value of 31 correct to four decimal places.

[4 marks]

8. The points A(a , 0), B(-a , 0) and C(h , k) are the vertices of triangle ABC.

(a) State the perpendicular bisector of AB. [1 marks]

(b) Find the equation of the line which divides BC equally and also

perpendicular to BC. [4 marks]

Hence, (i) show that the centre of the circle that passes through the points A,

B and C is

−+

k

akh

2,0

222

, [2 marks]

(ii) determine the general equation of circle that passes through the

points (–2 , 0) , (2 , 0) and (1 , 1) [3 marks]

9. The polynomial qxpxxxf +++= 7)( 23 , where p, q are constants has a factor

of x – 1 . Given f '(1) = 20, find the values of p and q . [4 marks]

If RxQx

xf+=

+)(

1

)( ,

(a) state the value of R, [1 marks]

(b) show that Q(x) is positive for all real values of x. [5 marks]


CONFIDENTIAL*

5

10.(a) Express )21)(34(

17

xx

x

+−

+in partial fractions. [4 marks]

(b) The graph of )21)(34(

17

xx

xy

+−

+= is as shown below.

8

6

4

2

-2

-5 5 10

Show that the area of the region bounded by the curve )21)(34(

17

xx

xy

+−

+= ,

the line x = 3

1− and x =

2

1 is ( )3ln92ln19

6

1+ . [7 marks]

11.(a) The matrix A is defined by

−

−

−

=

1104

180

120

A .

Find (i) the matrix 2A , [1 mark]

(ii) the matrix B if IAAB 672+−= where I is the 3×3 identity

matrix. [2 marks]

Show that OIAB =− 24 where O is the 3×3 zero matrix.

Hence, deduce the inverse of matrix A. [4 marks]

(b) The points P(0 , 1), Q(0 , 4) and R(2 , 5) lie on the curve given by the equation

02222=+−−+ cbyaxyx .

Write a system of equations in the form of matrices that involve matrix A to

represent the above information.

Hence, find the values of a, b and c. [5 marks]


CONFIDENTIAL*

6

12. The function f is defined by

≥−

<−

−

=

1,22

1,1

1

)(

xx

xx

x

xf .

(a) Find )(1

limxf

x−

→ and )(

1

limxf

x+

→ . Hence, determine whether f is

continuous at x = 1. [4 marks]

(b) Sketch the graph of the function f . [3 marks]

Based on your graph,

(i) state the range of function f , [1 marks]

(ii) explain why the inverse of function f is not a function, [2 marks]

(iii) suggest the largest possible domain for function f so that

the inverse of f is a function. [1 marks]

(iv) sketch the graph of )(

1

xfy = . [3 marks]


CONFIDENTIAL*

























PEPERIKSAAN PERCUBAAN

SIJIL TINGGI PERSEKOLAHAN MALAYSIA

NEGERI PAHANG DARUL MAKMUR 2008

Instructions to candidates:

Answer all 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 7 printed pages.

954/2 STPM 2008

Three hours

MATHEMATICS T

PAPER 2


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

duvuvdx

dx

dvu

cxfdxxf

xf+=∫ )(ln

)(

)('

ca

x

adx

xa+

=

+

−

∫1

22tan

11

ca

xdx

xa+

=

−

−

∫1

22sin

1

Series:

N n where ∈++

++

+

+=+

−−− ,21

)( 221 nrrnnnnnbba

r

nba

nba

naba LL

Coordinate Geometry :

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

ratio m : 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

<++−−

++−

++=+ xxr

rnnnx

nnnxx

rn whereLL

L

...!

...!2

12

+++++=r

xxxe

rx

( )( )

11...,1

...32

1ln

132

≤<−+−

+−+−=+

+

xr

xxxxx

rr


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 L

n

abhrhafyr

−=+= and where )(

Correlation and regression :

Pearson correlation coefficient:

( )( )

( ) ( )∑∑

∑−−

−−=

22

yyxx

yyxxr

ii

ii

Regression line of y on x :

y = a + b x

where b = ( )( )

( )∑∑

−

−−

2

xx

yyxx

i

ii

xbya −=

Trigonometry

sin ( A ± B ) = sin A cos B ± cos A sin B

cos ( A ± B ) = cos A cos Bm sin A sin B

tan ( A ± B ) = BA

BA

tantan1

tantan

m

±

cos 2A = cos 2 A − sin 2 B = 2 cos 2 A − 1 = 1 − 2 sin 2 A

sin 3A = 3 sin A − 4 sin 3 A

cos 3A = 4 cos 3 A − 3 cos A

sin A + sin B = 2 sin

+

2

BAcos

−

2

BA

sin A − sin B = 2 cos

+

2

BAsin

−

2

BA

cos A + cos B = 2 cos

+

2

BAcos

−

2

BA

cos A − cos B = − 2 sin

+

2

BAsin

−

2

BA


CONFIDENTIAL*

4

1. By using the factor formulae, find all values of x, where °≤≤° 3600 x , which

satisfy the equation cos x − 2 cos2 x + cos3 x = 0. [4 marks]

2. (a) Express 3 cos x + sin x in the form of r cos (x − θ) where r > 0 and

0 <θ < 2

π, find r and θ. [3 marks]

(b) If y =1xsinxcos3 ++++++++

ππππ where

6

π < x <

2

π, show that

3

π< y<

2

π.

[3 marks]

3. B

In the figure above, OAB is an equilateral triangle with OA = 1 unit. M is the mid-

point of AB and P divides the line segment OA in the ratio 2:1 . Q is a point on OB

such that PQ intersects OM at G and PG: GQ = 4:3 . Given that →

OA = a and →

OB = b,

(a) Find →

OM in terms of a and b. [1 mark]

(b) Given that OQ : QB = k : ( l − k ) .

(i) Find →

OG in terms of k, a and b.

(ii) Use ABOG o to find k and show that →

PQ = a3

2b

2

1−−−− . [5 marks]

(iii) Using results from (ii) above, find →

PQ . [4 marks]

4. Two forces of magnitude P Newton each act on a particle. The resultant force

on the particle is 6 Newton. When the magnitude of one of the forces acting on the

particle is increased by 4 times, the angle between the two forces remains unchanged

but the resultant force is increased to 18 Newton. Calculate the value of P and the

angle between the forces. [7 marks] Pahang Trial STPM 2008 http://edu.joshuatly.com/ http://www.joshuatly.com/

CONFIDENTIAL*

5

5 C

An octagon ABCDEFGH is inscribed in a circle, centre O, as shown in the figure

above. K and J are the feet of the perpendiculars from O to the sides AB and DE

respectively. Chords AD and BE intersect at I.

Given that AB = BC = GH = HA = 3 units, and CD = DE = EF = FG = 2 units.

(a) Show that E, O, and A are collinear. [2 marks]

(b) Show that OK = 21

EB and OJ = 21

AD. [4 marks]

(c) Prove that ∆DCB is congruent to ∆DIB. [4 marks]

6. A cylindrical container has a height of 200 cm. The container was initially

full of a chemical but there is a leak from a hole in the base. When the leak is noticed,

the container is half-full and the level of the chemical is dropping at a rate of 1 cm per

minute. It is required to find for how many minutes the container has been leaking.

To model the situation it is assumed that, when the depth of the chemical remaining is

x cm, the rate at which the level is dropping is proportional to x .

Show that differential equation is . x 0.1 dt

dx−−−−==== Solve the differential equation,

and hence show that the container has been leaking for about 80 minutes (rounded up

to one significant figure). [13 marks]


CONFIDENTIAL*

6

7. The average weekly incomes in RM, of households in 11 states of Malaysia are

given below.

255.80 252.00 270.60 398.40

362.30 297.20 266.80 261.70

247.50 259.10 220.60

(a) Find the median and upper and lower quartiles. [3 marks]

(b) Draw a box and whiskers plot to represent the data. [2 marks]

(c) Describe the skewness of this distribution and identify a possible outlier.

[4 marks]

8. The table below shows the age distribution of accidents victims under the age of 16

in a country during one calendar year.

Age ( x , years ) Frequency

5 ≤ x < 8 99

8 ≤ x < 11 143

11 ≤ x < 13 130

13 ≤ x < 14 78

14 ≤ x < 15 84

15 ≤ x < 16 91

(a) Draw a histogram to represent the data. [2 marks]

(b) Calculate estimates of the mean and standard deviation of this distribution.

[5 marks]

(c) Calculate estimates of the median and mode. [4 marks]

(d) State which measure of location, mean or median would be chosen to represent the

data. Give a reason for your choice. [2 marks]

9. A Personal Identification Number (PIN) consists of 4 digits in order, each of which

is one of the digits 0 , 1 , 2 , 3 , …, 9. Irene has difficulty remembering her PIN. She

tries to remember her PIN and writes down what she thinks it is. The probability that

the first digit is correct is 0.8 and the probability that the second digit is correct is

0.86. The probability that the first two digits are both correct is 0.72. Find

(a) the probability that the first digit is correct and the second digit is incorrect.

[2 marks ]

(b) the probability that the second digit is incorrect given that the first digit is

incorrect. [2 marks ]


CONFIDENTIAL*

7

10. A bag consists four green apples and three red apples. A boy takes out apples one

by one randomly , without replacement until he gets a red apple. When he gets a red

apple, he stops. Let X represents the number of apples he takes out.

(a) Tabulate the probability distribution of X. [3 marks]

(b) Calculate E(X) and Var(X) . [3 marks]

11. The lifespan T ( in working hours ) of a certain type of drill used in oil exploration

drilling machinery is a continuous random variable with probability density function

defined by :

f ( t ) = >

−

otherwise

te t

0

0 µλ

where λ and µ are constants.

Drilling is planned to take place continuously for one six-hour shift each day.

(a) If the mean life of the drill is 20 hours, find the values of the constant λ and µ.

[5 marks]

(b) If a new drill is used for a shift, what is the probability that it will fail during the

shift ? [2 marks]

12. (a), A machine packs sugar into bags marked 2 kg. The weight produced by the

machine is normally distributed. The standard deviation of the measures produced by

the machine is 0.03 kg . At least 95% of the bags must be over 2 kg in weight , find

the limit of the value of µ, mean of the measures produced by the machine.

[3 marks]

(b) When a telephone call is made in the country “ABC” , the probability of getting

the intended number is 0.95.

(i) Ten independent calls are made. Find the probability of getting eight or

more of the intended numbers . [2 marks]

(ii) Three hundred independent calls are made. Find the probability of failing

to get the intended number on at least ten but not more than twenty of the calls.

[3 marks]

(iii) Four hundred independent calls are made. For each call the probability of

getting “number unobtainable” is 0.004. Find the probability of getting “number

unobtainable” less than three times. [3 marks]


Marking Scheme

PEPERIKSAAN PERCUBAAN STPM NEGERI PAHANG 2008

Mathematics T Paper 1 (954/1)/ Mathematics S Paper 1 (950/1)

1.

)24( )1(

84 )1(

74 2

0742

2

2

2

2

+=−

+=−

+=−

=−−−

xy

xy

xyy

xyy

This is a standard equation of parabola, therefore the curve is a parabola. A1

x

y

-3 -2 -1 0 1 2 3 4

-4

-3

-2

-1

0

1

2

3

4

[4]

2. (2 + 3i)2 = -5 + 12i M1

It is verified that 2 + 3i is one of the square roots of -5 + 12i. A1

(2 + 3i)* = 2 – 3i B1

(2 – 3i) ( 2 + 3i)2 = 13 1z M1A1 [5]

3. M =

−16

2

k

k

Determinant of M is M = k( k – 1) – 12 M1

= k2 – k – 12 A1

The inverse for M exists if M ≠ 0,

k2 – k – 12 ≠ 0 M1

(k – 4) (k + 3) ≠ 0

k ≠ -3 , k ≠ 4 A1

∴The set of values of m for which the inverse of M exists is

{ 4,3,: ≠−≠∈ kkRkk } A1 [5]

(y-1)2 = 4(x + 2)

M1

A1

D1 : correct

curve

x

y

(-2,1) (-1,1)

D1 : both vertex

and focus

curve


4. y2 = sin x

xdx

dyy cos2 = M1

xdx

dy

dx

ydy sin22

2

2

2

−=

+ M1A1

xy

x

dx

ydy sin

2

cos22

2

2

2

−=

+

xy

x

dx

ydy sin

4

sin122

2

2

2

2

−=

−+

2

2

4

2

2

2

12 y

y

y

dx

ydy −=

−+

44

2

23 214 yy

dx

ydy −=−+ M1

014 4

2

23

=++ ydx

ydy A1 [5]

5. 025

<−+ x

x

05

210<

+

−−

x

xx M1

05

10<

+

−−

x

x A1

∴{x : x > -5 , x < -10} M1A1

10,5 −<−> xx M1

Hence, {x : 0≥x } A1 [6]


6. tdt

dx 1= and

2

2 tte

dt

dy= M1A1

2

2

221

2 tt

et

t

te

dx

dy== M1A1

When t = 1,

x = 0 , y = e M1

edx

dy2=∴ A1

Equation of tangent, y - e = 2e (x – 0) M1

y = 2ex + e A1 [8]

7. 2

1

)1(1

1 −+=

+y

y

= L+

−

−

−

+

−

−

+−32

!3

2

5

2

3

2

1

!2

2

3

2

1

2

11 yyy M1

= L+−+−32

16

5

8

3

2

11 yyy A1 [2]

When y = x + x2,

...)(16

5)(

8

3)(

2

11

1

1 32222

2++−+++−=

++xxxxxx

xx M1A1

= ......)(16

5)2(

8

3

2

1

2

11 34322

++−+++−− xxxxxx M1

32

16

7

8

1

2

11 xxx +−−≈ A1 [4]

Let x = 5

1,

31

5

25

31

1

25

1

5

11

1==

++

B1

32

5

1

16

7

5

1

8

1

5

1

2

11

31

5

+

−

−≈∴ M1A1

5648.531 = ( 4 d.p. ) A1 [4]


8. (a) x = 0 B1 [1]

(b) Midpoint of BC =

−

2,

2

kah B1

mBC = ah

k

+

Equation of line

−−

+−=−

22

ahx

k

ahky B1M1

k

akhx

k

ahy

2

222−+

+

+−= A1 [4]

(i) x = 0 -----------------------------(1)

k

akhx

k

ahy

2

222−+

+

+−= ----------(2)

Centre of circle is

−+

k

akh

2,0

222

M1A1 [2]

(ii) radius = 5

centre = (0, -1) B1

equation of circle

222 )5()1()0( =++− yx M1

x2 + y

2 + 2y – 4 = 0 A1 [3]

9. f(x) = x3 + px

2 + 7x + q

f(1) = 1 + p + 7 + q = 0

p + q = -8 M1

f '(x) = 3x2 + 2px + 7

f '(1) = 3 + 2p + 7 = 20 M1

p = 5 A1

q = -13 A1 [4]

f(x) = x3

+ 5x2

+ 7x -13

R = f(-1) = -16 B1

Q(x) = x2 + 6x + 13 B1

= (x + 3)2 + 4 M1A1

Since (x + 3)2 > 0, xxQ ∀>∴ 0)( M1A1 [6]


10. Let x

B

x

A

xx

x

2134)21)(34(

17

++

−≡

+−

+

17 + x ≡ A(1 + 2x) + B(4 – 3x) M1

Substituting x = 2

1− :

=

2

11

2

33B M1

B = 3

Substituting x = 3

4 :

=

3

11

3

55A

A = 5 A1(both A&B)

Thus xxxx

x

21

3

34

5

)21)(34(

17

++

−=

+−

+ A1(stating)

Area = ∫ − +−

+2

1

3

1)21)(34(

17dx

xx

x B1

= dxxx∫ −

++

−

2

1

3

1

21

3

34

5 M1

=2

1

3

1

)21ln(2

3)34ln(

3

5

−

++−− xx M1A1

=

+−−+

−

3

1ln

2

35ln

3

52ln

2

3

2

5ln

3

5 M1A1

=

−+

−

3

1ln

2

32ln

2

3

2

5ln5ln

3

5

= 3ln2

32ln

2

32ln

3

5++

= 3ln2

32ln

6

19+

= )3ln92ln19(6

1+ A1 [11]


11(a) (i)

−−

−−

−−

=

−

−

−

−

−

−

13784

7544

164

1104

180

120

1104

180

120

B1

(ii) B = A2 – 7A + 6I

=

+

−

−

−

−

−−

−−

−−

100

010

001

6

1104

180

120

7

13784

7544

164

M1

=

−

−

−

0832

044

682

A1

(iii) AB – 24I =

−

100

010

001

24

2400

0240

0024

= O M1A1

AB = 24I

A-1

AB = 24 A-1

I

24

1B = A

-1 M1

−

−

−

=−

03

1

3

4

06

1

6

14

1

3

1

12

1

1A A1 [7]


(b) 1- 2b + c = 0

6 – 8b + c = 0

-4a – 10b + c = 0 B1B1 (all correct)

B1B0 (any two

correct)

2b – c = 1 B0B0 ( others )

8b – c = 16

4a + 10b – c = 29

=

−

−

−

29

16

1

1104

180

120

c

b

a

M1

−

−

−

=

29

16

1

0832

044

682

24

1

c

b

a

M1

=

4

2

25

A = 2, b = 2

5, c = 4 A1 [5]

12. (a)1

1lim)(lim

11 −

−=

−−→→ x

xxf

xx

= )1(

1lim

1 −−

−−

→ x

x

x

= -1 M1

)22(lim)(lim11

−=++

→→

xxfxx

= 0 A1

Since )(lim)(lim11

xfxfxx

+−→→

≠ , )(lim1

xfx→

does not exist. M1

So f is not continuous at x = 1. A1


(b)

(i) ),0[}1{ ∞∪−∈y B1

(i) From the graph , f is many to one, so it is not one to one, M1

therefore the inverse is not a function. A1

(iii) }1:{ ≥xx A1

(iv)

x

y

1

– 1

D1 : line y = 2x – 2

D1 : line y = –1

D1 : correct and

x

y

x = 1

D1 : correct

D1 : correct


1

Marking Scheme

PEPERIKSAAN PERCUBAAN STPM NEGERI PAHANG 2008

Mathematics T Paper 2 (954/2)

1. °°°°≤≤≤≤≤≤≤≤====++++ 360x0,0x3cosx2cos2-xcos

0=x2cos2-xcosx2cos2 factor formula [M1]

0=)2-xcos2(x2cos [A1]

2

2=xcosor,0=x2cos [A1]

oooooo315,45=xor,630,450,270,90=x2

oooo

315,225,135,45=x.ie [A1 all correct]

2(a) )-xcos(xsinxcos3 θθθθr ≡≡≡≡++++

xsinsinrxcoscosr θθθθθθθθ ++++≡≡≡≡

3=θcosr,1=θsinr (can be implied ) [B1 both]

3

1=θtan

2221+)3(=r [M1 either one]

6

π=θ 2=r [A1 both]

(b) 1+)

6

π-xcos(2

π=

1+xsin+xcos3

π=y [M1]

36-x0

2x

6

ππππππππππππππππ<<<<<<<<⇒⇒⇒⇒<<<<<<<<Q

3cos)

6-xcos(0cos

ππππππππ>>>>>>>> [M1]

2

1)

6-xcos(1 >>>>>>>>

ππππ 1)

6-xcos(22 >>>>>>>>

ππππ 21)6

-xcos(23 >>>>++++>>>>ππππ

2

1

1)6

-xcos(2

1

3

1<<<<

++++

<<<<ππππ

21)

6-xcos(2

3

ππππ

ππππ

ππππππππ<<<<

++++

<<<< [A1]

3(a) 2

b+a=OM [B1]

kb=OQanda

3

2=OP)i((b)

4+3

)kb(4+)a3

2(3

=OG [M1]

7

kb4+a2=OG∴ [A1]

0=a)-(b•7

4kb+2a

0=AB.•OGsocollinear,areMandGO,Since(ii)

[M1]

0=)1(2-)2

1)(k4-2(+)1(k4

2

1=60cos)1)(1(=b•a

0=a2-b•a)k4-2(+bk4

22

22

o

2

1=k [A1]

b2

1=OQ

a3

2b

2

1PQ −−−−==== [A1]

2

1=60cos)1)(1(=b•a)c( o (relax on dot) [B1]

)a3

2b

2

1()a

3

2b

2

1(PQ

2

−−−−

−−−−==== o

[B1]

22

22

2

)1(9

4)

2

1(

3

2)1(

4

1

a9

4ba

3

2 b

4

1

)a3

2b

2

1()a

3

2b

2

1(PQ

++++−−−−====

++++••••−−−−====

−−−−

−−−−==== o

[M1]

36

13=

6

13=PQ [A1]

3

300

B

4300

a

b

2 1

k

1-k


2

4. Correct arrows & labels (angles &magnitude) in parallelogram or triangle of forces

[correct form/shape & ≥2 labels in either diagram D1] [all correct D1]

)180cos()P(P2PP6 222 αααα−−−−°°°°−−−−++++==== ⇒⇒⇒⇒ ααααcosP2P236 22++++==== (eqn1) substitute into either eqn [M1]

)180cos()P4(P2P4P18 222 αααα−−−−°°°°−−−−++++==== ⇒⇒⇒⇒ ααααcosP8P17324 22++++==== (eqn2) both eqn correct [A1]

Eqn2 −−−− 4 ×××× eqn 1 , 2P9180 ==== , N472.420P ======== , solving eqn [M1] , ans for P [A1]

From eqn 1 , ααααcos)20(2)20(236 ++++====

°°°°°°°°°°°°====

−−−−====

−−−− [email protected]@'449540

4cos 1αααα answer for αααα [A1]

5(a)Arcs ED, DC, CB and BA make up half of a whole circle, therefore

ECA is a semicircle. [B1]

collinearareAand,O,Ehence,180EOAo

====∠∠∠∠ [B1]

(b) EA is a diameter, therefore o90EBA ====∠∠∠∠ [B1]

Given OK is perpendicular to AB & K midpoint of AB

OK is parallel to EB [B1]

EB2

1=OK ( midpoint theorem ) [B1]

Similarly, AD2

1=OJ [B1]

(c) DB is common to both triangles. [B1]

DC=DE & CB=AB the corresponding arcs subtend equal angles at the

circumference [B1]

Hence αααα====∠∠∠∠====∠∠∠∠====∠∠∠∠ IBDEBDCBD & θθθθ====∠∠∠∠====∠∠∠∠====∠∠∠∠ IDBADBCDB [B1]

DIBDCB ∆∆∆∆∆∆∆∆ ≡≡≡≡ (by using ASA) [B1]

xk dt

dx.6 −−−−==== Rate of change [B1]

x =100 and 1dt

dx−−−−==== ⇒ k = 0.1.Solving for k [B1] [M1] [A1]

∴ x1.0dt

dxisDE −−−−==== For correct DE [A1]

dxx 2

1

∫∫∫∫−−−−

= ∫∫∫∫−−−− dt 1.0 ⇒⇒⇒⇒ 2

1

x2 = ct 1.0 ++++−−−−

Separation of variables and their integration [B1] [ M1]

for correct RHS ∫∫∫∫ ⇒⇒⇒⇒ 2

1

x2 [A1]

for correct LHS ( kt ±±±± ) [A1]

arbitrary constant C included

or equivalent statement of both pairs of definite integral

limits [B1]

0t,200x ======== ⇒⇒⇒⇒ 2002c ==== for evaluating C [M1]

2002t1.01002,100x ++++−−−−======== for evaluating t [M1]

)figure tsignificanone to ( min 80 about

for leaking been has container the Hence.utesmin84.82=t

For correct value [email protected] mins seen before rounding [A1]

2

2 2

2

2

3

3

3

C 2 3

3

α

αθ

θ

P N

P N

P N

P N

P N

18 N

4P N

6 N

180-α α

α

180-α


3

7. (a) median = RM261.70 Q 1 = RM252.00 Q 3 = RM297.20 B1 B1 B1

7. (b) Box = D1 Whiskers = D1 * Must be drawn on graph paper.

Q 1 Q 2 Q 3 outlier

••••

Income (RM)

220 260 300 340 380 420

7.(c) Q 2 −−−− Q1 = 9.70 and Q 3 −−−− Q 2 = 35.50

Q 3 −−−− Q 2 > Q 2 −−−− Q1 , therefore the distribution is positively skewed. M1 A1

Upper boundary = Q 3 + 1.5 ( Q 3 −−−− Q1 ) = RM 365.00 , Q 3 −−−− Q1 = 45.20

Lower boundary = Q1 −−−− 1.5 ( Q 3 −−−− Q1 ) = RM 184.20 M1 ( find boundaries )

∴∴∴∴ The outlier is RM 398.40 A1

8. (a)

Age ( x , years ) Mid-point Frequency Class width Frequency density

5 ≤≤≤≤ x <<<< 8 6.5 99 3 33

8 ≤≤≤≤ x <<<< 11 9.5 143 3 47.4

11 ≤≤≤≤ x <<<< 13 * 12 130 2 65

13 ≤≤≤≤ x <<<< 14 13.5 78 1 78

14 ≤≤≤≤ x <<<< 15 14.5 84 1 84

15 ≤≤≤≤ x <<<< 16 15.5 91 1 91

Suitable Scale = D1

Correct Rectangles = D1

8. (b) mean = 625

5.7243

M1

mean = 11. 5896

= 11.59 @ 11.6 A1

variance = 625

75.89547 −−−−

2

625

5.7243

B1 for 89547.75

standard deviation =

2

625

5.7243

625

75.89547

− M1

= 2.992 or 2.99 A1


4

8.(c)

median = 11 +

( )

143

22426252

1×

−

M1

= 11.986

= 11.99 or 12.0 A1

mode = 8 + 31344

44×

+ M1

mode = 10.3158

= 10.32 or 10.3 A1

8. (d) Median because the distribution is negatively skewed. B1 B1

9. (a) A = the event that the first digit is correct P ( A ) = 0.8

B = the event that the second digit is correct P ( B ) = 0.86, P( A ∩∩∩∩ B ) =0.72

P( A ∩∩∩∩ B′′′′ ) = P ( A ) −−−− P( A ∩∩∩∩ B ) = 0.8 −−−− 0.72 = 0.08 M1 A1

9. (b) P ( B′′′′ A′′′′ ) = )'(

)''(

AP

BAP ∩=

)(1

)(1

AP

BAP

−

∪−=

[ ])(1

)()()(1

AP

BAPBPAP

−

∩−+−

= 8.01

]72.086.08.0[1

−

−+− =

2.0

06.0 = 0.3 M1 A1

10. (a) X = x 1 2 3 4 5 Total B3 if 5 values correct

P(X =

x) 7

3

7

2

35

6

35

3

35

1

1 B2 if 4 values correct

B1 if 3 values correct

P(X=1) =7

3 P(X=2) =

7

4××××

6

3 =

7

2 P(X=3) =

7

4××××

6

3 ××××

5

3 =

35

6

P(X=4) = 7

4××××

6

3 ××××

5

2××××

4

3 =

35

3 P(X=5) =

7

4××××

6

3 ××××

5

2××××

4

1 ×××× 1 =

35

1

10(b) E(X) = 1 ××××

7

3 + 2 ××××

7

2 + 3 ××××

35

6 + 4 ××××

35

3 + 5 ××××

35

1 = 2 B1

E(X 2 ) = 1 2××××

7

3 + 2 2

××××

7

2 + 3 2

××××

35

6 + 4 2

××××

35

3 + 5 2

××××

35

1 =

5

26

Var(X) = 5

26 −−−− ( 2 ) 2 =

5

6 their E(X 2 ) M1√ A1

11.(a)

dtet

∫∞

−

0

µλ = 1

∞

−

−

0

te

µ

µ

λ = 1 M1

0 + µ

λ = 1

λλλλ = µµµµ A1

Given E ( T ) = 20

dtet

∫∞

−

0

t µλ = 20 B1

dtet

∫∞

−

0

t λλ = 20 because µµµµ = λλλλ

[ ] ∞−−

0

tet

λ + dte

t

∫∞

−

0

λ

= 20 M1 “by part”

0 +

∞

−

−

0

1 te λ

λ = 20

0 +

−−

λ

10 = 20 ;

λ

1 = 20 ,

∴∴∴∴λλλλ = 0.05 and µµµµ = 0.05 A1 Pahang Trial STPM 2008 http://edu.joshuatly.com/ http://www.joshuatly.com/

5

11.(b)

f ( t ) = >

−

otherwise

te t

0

0 05.0 05.0

P ( T < 6 ) = dtet

∫−

6

0

05.0 05.0

= [ ] 6

0

05.0 te

−− M1

= 1 −−−− 3.0 −e

= 0.2592 or 0.259 A1

12.(a) X ∼∼∼∼ N ( µµµµ , 0.03 2 ) and P ( X > 2 ) ≥0.95

−>

03.0

2 µZP ≥ 0.95 . M1 standardized

03.0

2 µ− ≤ −−−−1.645 B1 for ( −−−−1.645 )

µµµµ ≥ 2.049 kg or µµµµ ≥ 2.05 kg A1

12.(b) (i)

X represents the number of the intended calls obtained

X ∼∼∼∼ B ( 10 , 0.95 ) x = 0,1,2,…10

P ( X ≥≥≥≥ 8 ) = P ( X = 8 ) + P ( X = 9 ) + P ( X = 10 )

= 8

10C (0.95) 8 (0.05) 2 + 9

10C (0.95) 9 (0.05) 1 + 10

10C (0.95) 10 (0.05) 0 M1

= 0.9885 or 0.989 A1

(ii) Y represents the number of unsuccessful calls.

Y ∼∼∼∼ B ( 300 , 0.05 ) x = 0,1,2,…300

Normal Approximation : µµµµ = np = 15

σσσσ2 = n p q = 14.25 B1 ( both correct) may be implied

P ( 10 ≤≤≤≤ Y ≤≤≤≤ 20 ) =

−≤≤

−

25.14

155.20

25.14

155.9ZP M1√(continuity correction & standardise)

= P ( −−−−1.457 ≤≤≤≤ Z ≤≤≤≤ 1.457 )

= 1 −−−− 0.0725 −−−− 0.0725

= 0.855 A1

(iii) T represents the number of unobtainable calls .

T ∼∼∼∼ B ( 400 , 0.004 ) x = 0,1,2,…400

Poisson Approximation : λ = np = 1.6 B1 may be implied

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

= 6.1−e

++

!2

6.1

!1

6.1

!0

6.1 210

M1 √ their λ

= 0.783 A1


maths t pahang trial stpm 2008

Documents