2009-percubaan matematik tambahan+skema [sarawak].pdf

7/28/2019 2009-PERCUBAAN Matematik Tambahan+Skema [SARAWAK].PDF

1/47

SULIT 1 3472/1

3472/1 2009 Hak CiptaZon AKuching [ Lihat sebelah

SULIT

SEKOLAH-SEKOLAH MENENGAH ZON A KUCHING

LEMBAGA PEPERIKSAANPEPERIKSAAN PERCUBAAN SPM 2009

Kertas soalan ini mengandungi 15 halaman bercetak

For examiners use only

Question Total MarksMarks

Obtained

1 2

2 3

3 3

4 35 3

6 3

7 3

8 3

9 3

10 3

11 3

12 4

13 3

14 3

15 4

16 3

17 4

18 4

19 3

20 3

21 3

22 3

23 324 4

25 4

TOTAL 80

MATEMATIK TAMBAHAN

Kertas 1

Dua jam

JANGAN BUKA KERTAS SOALAN INI

SEHINGGA DIBERITAHU

1 This question paper consists of 25 questions.

2. Answerall questions.

3. Give only one answer for each question.4. Write your answers clearly in the spaces provided in

the question paper.

5. Show your working. It may help you to get marks.6. If you wish to change your answer, cross out the work

that you have done. Then write down the new

answer.

7. The diagrams in the questions provided are notdrawn to scale unless stated.

8. The marks allocated for each question and sub-partof a question are shown in brackets.

9. A list of formulae is provided on pages 2 to 3.10. A booklet of four-figure mathematical tables is

provided.

.

11 You may use a non-programmable scientificcalculator.

12 This question paper must be handed in at the end of

the examination .

Name : ..

Form : ..

3472/1

Matematik Tambahan

Kertas 1

20092 Jam

http://mathsmozac.blogspo

Dapatkan KERTAS SOALAN RAMALAN 2010 di www.maths-catch.com

Bimbingan Matematik percuma di blog www.mathgayapos.blogspot.com


2/47

SULIT 3472/2

3472/1 2009 Hak CiptaZon AKuching SULIT

2

The following formulae may be helpful in answering the questions. The symbols given are

the ones commonly used.

ALGEBRA

12 4

2

b b acx

a

=

2 am an = a m + n

3 am a

n= a

m n

4 (am)

n= a

mn

5 log amn = log am + log an

6 log an

m= log am log an

7 log amn

= n log am

8 log a b =a

b

c

c

log

log

9 Tn = a + (n 1)d

10 Sn = ])1(2[2

dnan

+

11 Tn = arn 1

12 Sn =r

ra

r

ra nn

=

1

)1(

1

)1(, (r 1)

13r

aS

=

1, r


3/47

SULIT 3472/1

3472/1 2009 Hak CiptaZon AKuching [ Lihat sebelah

SULIT

3

STATISTIC

1 Arc length, s = r

2 Area of sector ,A = 212

r

3 sin2A + cos

2A = 1

4 sec2A = 1 + tan2A

5 cosec2A = 1 + cot2A

6 sin 2A = 2 sinA cosA

7 cos 2A = cos2A sin2A

= 2 cos2A 1

= 1 2 sin2A

8 tan 2A =A

A2tan1

tan2

TRIGONOMETRY

9 sin (A B) = sinA cosB cosA sinB

10 cos (A B) = cosA cosB m sinA sinB

11 tan (A B) =BA

BA

tantan1

tantan

m

12C

c

B

b

A

a

sinsinsin==

13 a2

= b2

+ c2

2bc cosA

14 Area of triangle = Cabsin21

71

11

w

IwI

=

8 )!(

!

rn

n

Prn

=

9!)!(

!

rrn

nC

r

n

=

10 P(A B) = P(A) + P(B) P(A B)

11 P(X= r) = rnrrn qpC

, p + q = 1

12 Mean = np

13 npq=

14 z =

x

1 x =N

x

2 x = ffx

3 =

2( )x x

N

=

22x x

N

4 =

2( )f x x

f

=

22fx

xf

5 m = Cf

FN

Lm

+ 2

1

6 1

0

100Q

IQ

=





4/47

SULIT 3472/1


4

Answer all questions.

1 Diagram 1 shows a graph of the relation between two variablesx andy.

State

(a) the object of 8,

(b) the type of relation betweenx andy.

[ 2 marks ]

Answer: (a) ..

(b) ...

2 Given that the function ( ) 2 5f x x + , 2( ) 4g x x= , find

(a) ( )gf x

(b) ( 2)gf [ 3 marks ]

Answer: (a) ..

(b) ...

3

2

2

1

For

examiners

use only

x

0 2 4 6 8 10

48

1216

24

y

DIAGRAM 1

20





5/47

SULIT 3472/1

3472/1 2009 Hak Cipta Zon A Kuching [ Lihat sebelah

SULIT

5

3 Given that the function3 2

:5

xf x

+ , find

(a) 1( )f x

(b) the value ofx such that 1( ) 3f x =

[ 3 marks ]

Answer: (a) ..

(b) ...

4 The quadratic equation 2x25x+ p 3 = 0has two different roots, find the range of

values ofp.[ 3 marks ]

Answer: .........

For

examiners

use only

3

3

3

4





6/47

SULIT 3472/1


6

5 Given and are the roots of the quadratic equation 3x2 + 4x 6 = 0, form the

quadratic equation whose roots are 3 and 3 .

[ 3 marks ]

Answer: .................................

___________________________________________________________________________

6 Diagram 2 shows the graph of the quadratic function y = 2(x 3)2 p which has a

minimum value of5.

Find(a) the value ofp,

(b) the value ofq,

(c) the equation of the axis of symmetry.[ 3 marks ]

Answer: (a) ........................

(b) ........................

(c)..................................

3

5

3

6

For

examiners

use only

DIAGRAM 2

q

x

y

O





7/47

SULIT 3472/1


SULIT

7

7 Find the range of values ofx for which (2x 3)(x + 1) x +1. [ 3 marks ]

Answer: ..................................

8 Solve the equation 279x + 1 =

1

3x.

[ 3 marks ]

Answer: ...................................

9 Solve the equation log 2 (x + 3) = 1 + log 2 (3x 1).[ 3 marks ]

Answer: ......................................

3

7

3

9

3

8

For

examiners

use only





8/47

SULIT 3472/1


8

10 The sum of the first n terms of an arithmetic progression is given by [ ]nnSn 372

3 2 = .

Find

(a) the common difference,

(b) the eleventh term

of the progression[ 3 marks ]

Answer: (a) ..

(b)...

11 Express the recurring decimal 0.21212121... as a fraction in its simplest form.

[ 3 marks ]

Answer:............

3

10

For

examiners

use only

3

11





9/47

SULIT 3472/1


SULIT

9

12 Diagram 3 shows the graph pxxy +=2 .

Based on the graph above a table ofx

yagainstx is obtained as show in table 1

x

y 3 r

x q 1

Calculate the values ofp, q and r.[ 4 marks ]

Answer:p =.............

q= ....................................

r= ....................................

___________________________________________________________________________

13 Find the coordinates of point M which divides line segment joining the points

( 3, 3)A and (7, 8)B such thatAM:AB = 2 : 5. [ 3 marks ]

Answer: ...

3

13

4

12

DIAGRAM 3

For

examiners

use only

3x

y

0

pxxy +=2

TABLE 1





10/47

SULIT 3472/1


10

14 Given the straight lines 3y ax+ = and 4 4y bx+ = are perpendicular to each other.

Express a in terms ofb.[ 3 marks ]

Answer: .

15 In Diagram 4, QR is parallel to PSand Tis the midpoint ofQR.[ 4 marks ]

Given that PS: QR = 3 : 5, PQ

= 3u and PS

= 6v, express, in terms ofu and v, of

(a) RS

,

(b) TS

.

[ 4 marks ]

Answer:(a)............

(b) ....................................

4

15

3

14

For

examiners

use only

Q

S

R

T

DIAGRAM 4

P





11/47

SULIT 3472/1


SULIT

11

16 Given 2 3OP i j= +

% %

and 10 2OQ i j=

% %

. Find, in terms of the unit vectors, i%

and j%

,

(a) PQ

(b) the vector whose magnitude is 2 units and in the direction of PQ .[ 3 marks ]

Answer: (a) PQ

= ....

(b) ..___________________________________________________________________________

17 Solve the equation 2 sinx +xsin

1= 3 for 0x 360. [ 4 marks ]

Answer: ............4

17

For

examiners

use only

3

16





12/47

SULIT 3472/1


12

18 Diagram 5 shows a semicircle with centre O.

The diameter of the circle is 16 cm and POQ = 0.6 radian.Calculate

(a) the length of arc QP,

(b) area of the shaded region.[ 4 marks ]

Answer: (a) ..

(b) ...___________________________________________________________________________

19 Find the coordinates of the minimum point of the curve 23

102

y x x= + .

[ 3 marks ]

Answer:

3

19

4

18

For

examiners

use only

DIAGRAM 5

PO

Q

0.6 rad

R





13/47

SULIT 13 3472/1


20 Two variables,x andy, are related by the equation .16

3x

xy += Given thaty increases

at a constant rate of 10 unit per second, find the rate of change ofx whenx = 2.[3 marks]

Answer: ............___________________________________________________________________________

21 Given that6

3( ) 2,g x dx = find

(a)3

63 ( )g x dx ,

(b) the value ofkif6

3[ ( ) ] 10g x kx dx+ = .

[ 3 marks ]

Answer: (a) ..

(b) ...

22 Given that the mean and variance of a set of n numbers x1,x2, . . . ,xn are 3 and 2.56respectively. Find the mean and standard deviation of the new set of n numbers

5x1 2, 5x2 2, . . . , 5xn 2.[ 3 marks ]

Answer: Mean = ..

Standard deviation = ...

3

20

3

21

3

22

For

examiners

use only





14/47

SULIT 14 3472/1


23 The probability that Kamal is chosen as a school librarian is5

2whereas the

probability that Alisa is chosen as a school librarian is12

5.

Find the probability that

(a) neither of them is chosen as a school librarian,

(b) only one of them is chosen as a school librarian.[ 3 marks ]

Answer: (a) ..

(b) ...___________________________________________________________________________

24 Mathematics Club of a school has 8 Form 5 students, 10 Form 4 students and 12 Form 3students.

(a) A teacher wants to choose Form 5 students to form a committee consisting apresident, a vice president and a secretary, find the number of ways the committeecan be formed.

(b) A team is to be formed to take part in a Mathematics competition. How manydifferent teams, each comprising 3 Form 5 students, 2 Form 4 students and 1 Form3 student can be formed?

[ 4 marks ]

Answer: (a) ..

(b) ...

For

examiners

use only

4

24

3

23





15/47

SULIT 15 3472/1


25 The mass of a packet of biscuit is normally distributed with a mean of 125 g and avariance of 16 g

2.

(a) Find the probability that a packet of biscuit chosen at random from a sample willhave mass not less than 128 g.

(b) If 30% of the packets chosen at random have mass more than m g, find the valueofm.

[ 4 marks ]

Answer: (a) ..

(b) ...

END OF QUESTION PAPER

4

25

For

examiners

use only





16/47

SULIT 1 3472/2

3472/2 ZON A KUCHING 2009 SULIT

3472/2

Matematik

Tambahan

Kertas 2

2 jam

2009


PEPERIKSAAN PERCUBAAN

SIJIL PELAJARAN MALAYSIA 2009

MATEMATIK TAMBAHAN

Kertas 2

Dua jam tiga puluh minit

JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU

1. This question paper consists of three sections : SectionA, SectionB andSection C.

2. Answerallquestion in SectionA , four questions from SectionBandtwo questions fromSection C.

3. Give only one answer / solution to each question..

4. Show your working. It may help you to get marks.

5. The diagram in the questions provided are not drawn to scale unless stated.

6. The marks allocated for each question and sub-part of a question are shown in brackets..

7. A list of formulae is provided on pages 2 to 3.

8. A booklet of four-figure mathematical tables is provided.

9. You may use a non-programmable scientific calculator.

Kertas soalan ini mengandungi 11 halaman bercetak





17/47

SULIT 3472/2


2

The following formulae may be helpful in answering the questions. The symbols given are the ones

commonly used.

ALGEBRA

1 x =a

acbb

2

42

2 am an = am + n

3 am an = amn

4 (am)n = amn

5 log amn = log a m + log an

6 log an

m= log a m log an

7 log amn = n log am

8 log a b =a

b

c

c

log

log

9 Tn = a + (n 1)d

10 Sn = ])1(2[2

dnan

+

11 Tn = arn 1

12 Sn =r

ra

r

ra nn

=

1

)1(

1

)1(, (r 1)

13r

aS

=

1, r


18/47

SULIT 3472/2


3

STATISTICS

TRIGONOMETRY

71

11

w

IwI

=

8

)!(

!

rn

nPr

n

=

9!)!(

!

rrn

nC

r

n

=

10 P(A B) = P(A) + P(B) P(A B)

11 P(X= r) = rnrrn qpC , p + q = 1

12 Mean = np

13 npq=

14 z =

x

1 x =N

x

2 x =

f

fx

3 =

2( )x x

N

=

22x x

N

4 =

2( )f x x

f

=

22fx

xf

5 m = Cf

FN

Lm

+ 2

1

6 1

0

100Q

IQ

=

9 sin (A B) = sinA cosB cosA sinB

10 cos (A B) = cosA cosB m sinA sinB

11 tan (A B) =BA

BA

tantan1

tantan

m

12C

c

B

b

A

a

sinsinsin==

13 a2

= b2

+ c2

2bc cosA

14 Area of triangle = Cabsin21

1 Arc length, s = r

2 Area of sector ,A = 212

r

3 sin2A + cos

2A = 1

4 sec2A = 1 + tan

2A

5 cosec2A = 1 + cot

2A

6 sin 2A = 2 sinA cosA

7 cos 2A = cos2A sin2A

= 2 cos2A 1

= 1 2 sin2A

8 tan 2A =A

A2tan1

tan2





19/47

SULIT 4 3472/2


SECTION A

[40 marks]

Answerall questions in this section .

1 Solve the simultaneous equations 2 6 0x y + = and 2 20 0x xy+ = . Give your answer

correct to 3 decimal places.[5 marks]

2 Diagram 1 shows a circle with centre O.

PTQ is a tangent to the circle at Tand PQ = OQ = 20 cm.Calculate

(a) the length of the arc STR, [4 marks]

(b) the area of the shaded region. [4 marks]

3 Table 1 shows the frequency distribution of scores of a group of players in a game.

Score 0 4 95 1410 15 19 20 24 2925 30 34

Number of players 2 3 10 20 w 6 2

TABLE 1

It is given that the median of the distribution is 17.

(a) Calculate the value ofw. [3 marks]

(b) Hence, calculate the variance of the distribution. [4 marks]

4

DIAGRAM 1

300

O

RS

P QT





20/47

SULIT 3472/2


5

4 (a) If the volume of a cube decreases from 125 3cm to 124.4 cm3, find the small changein the sides of the cube.

[3 marks]

(b) Given that2

3 4( )

3

xf x

x

+=

, find the value off(2). [3 marks]

5 (a) Prove that sin 2x = 2 sin2x cotx. [2 marks]

(b) Sketch the graph ofy = x2sin2 for 0 x . Hence, using the same axes,

sketch a suitable straight line to find the number of solutions of the equation x2sin2

=

x2for 0 x. State the number of solutions.

[6 marks]

6 (a)

A piece of wire is cut into 15 parts which are bent to form circles as shown inDiagram 2.The radius of each circle increases by 3 cm consecutively.Calculate

(i) the radius of the last circle, [2 marks]

(ii) the area of the last circle. [1 mark]

(b) Diagram 3 shows a rectangular geometric pattern.

Diagram 2

The first rectangle isABCD and followed by MBNP andso on. The length and widthof the next rectangle is half of the length and width of the previous rectangle. GiventhatAB = 30 cm andBC= 20 cm. Find the perimeter of the seventh rectangle.

[3 marks]

5 cm2 cm 8 cm

A

D

B

C

M

P N

DIAGRAM 3

DIAGRAM 2

. . .





21/47

SULIT 3472/2


6

SECTION B

[40 marks]

Answerfour questions from this section.

7 Use graph paper to answer this question.

Table 2 shows the values of two variablesx andy which are related by 2+= xpqy , wherep

and q are constants.

(a) Convert 2+= xpqy to a linear form of cmXY += . [2 marks]

(b) Plot y10log against )2( +x by using a scale of 2 cm to 1 unit on the Y-axis and 2 cmto 1 unit on theX-axis. Hence, draw the line of best fit. [4 marks]

(c) From the graph in (b), find the value ofp and ofq. [4 marks]

8 Diagram 4 shows a triangle OPQ. The pointR lies on OP and the point Slies on PQ. Thestraight line QR intersects the straight line OSat point T.

Given OP : OR = 4 : 3, PQ : PS= 2 : 1, 12OP x=

%and 9OQ y=

%

.

(a) Express, in terms ofx and / ory,

(i) QR

,

(ii) OS

.

[3 marks]

(b) IfOT hOS =

and QT k QR=

, where h and kare constants, find the values ofh and k.

[5 marks]

(c) Given that 3x =%

units, 5y =%

units and POQ = 90o, find PQ

. [2 marks]

x 1 2 3 4 5 6

y 25.6 125.9 640 3163 15849 63096

TABLE 2

O R P

T

S

Q

DIAGRAM 4





22/47

SULIT 3472/2


7

9 (a) In a certain area, 30% of the trees are rubber trees.

(i) If 8 trees in the area are chosen at random, find the probability that at least twoof the trees are rubber trees.

(ii) If the variance of the rubber trees is 315, find the number of rubber trees in thearea.[5 marks]

(b) The masses of the children in the Primary One in the school have a normaldistribution with mean 33.5 kg and variance 25 kg2. 150 of the children have massesbetween 30 kg and 36.5 kg. Calculate the total number of children in Primary One inthat school.

[5 marks]

10 Solution by scale drawing is not accepted.

In Diagram 5, point Tlies on the perpendicular bisector ofAB.

(a) Find the equation of straight lineAB. [2 marks]

(b) A point P moves such that 2PA AB= . Find the equation of locus ofP.[3 marks]

(c) Locus ofP intersects thex-axis at points YandZ. State the coordinates ofYandZ.[3 marks]

(d) Find thex-intercept ofCD. [2 marks]

B(3, 2)

T

C Ox

y

A (1, 4)

D

DIAGRAM 5





23/47

SULIT 3472/2


8

11 (a) Given that a curve has a gradient function xpx +2 such that p is a constant.

xy 26 = is the equation of tangent to the curve at the point (2, q). Find the value of

p and ofq.[3 marks]

(b) Diagram 6 shows the curve y = (x 3)2 and the straight line y = 2x + 2 intersect atpoint (1, 4).

Calculate

(i) the area of the shaded region, [4 marks]

(ii) the volume of revolution, in terms of , when the region bounded bythe curve, the x-axis, the y-axis and the straight line x = 2 is revolved through

360 about thex-axis. [3 marks]

y

y = (x 3)2

xO

y = 2x + 2

(1, 4)

DIAGRAM 6





24/47

SULIT 3472/2


9

SECTION C

[20 marks]

Answertwo questions from this section.

12 A particle moves along a straight line and passes through a fixed point O. Its velocity, v

ms1, is given by v = t2 6t+ 5, where tis the time, in seconds, after passing through O.

(Assume motion to the right is positive).Find

(a) the initial velocity, in ms1, [1 mark]

(b) the minimum velocity, in ms1, [3 marks]

(c) the range of values oftat which the particles moves to the left, [2 marks]

(d) the total distance, in m, travelled by the particle in the first 5 seconds. [4 marks]

13 In Diagram 7,ABCis a triangle.BMCandAMare straight lines.

DIAGRAM 7

(a) Calculate

(i) AMB,

(ii) the area, in cm2, of triangleABC. [7 marks]

(b) A new triangleABM is formed withAB =AB,BM =BMand BAM = BAM,

find the length ofAM. [3 marks]

A

B CM





25/47

SULIT 3472/2


10

14 Use the graph paper provided to answer this question.

A factory produces two types of school bags MandNusing two types of machines A andB. Given that machine A requires 20 minutes to produce a bag M and 30 minutes toproduce a bagNwhile machineB requires 25 minutes to produce a bag Mand 40 minutes

to produce a bagN. The machines producex units ofMandy units ofNin a particular dayaccording to the following conditions.

I : MachineA is operated for not more than 8 hours.

II : MachineB is operated for at least 4 hours.

III : The number of units of bagMproduced is not more than twice the number ofunits of bagN.

(a) Write the three inequalities for the above conditions. [3 marks]

(b) Using a scale of 2 cm to 2 units for both axes, construct and shade the region Rwhich satisfies all the above conditions. [3 marks]

(c) Use the graph constructed in 14 (b), to find

(i) the maximum number of units of bag M that can be produced if the factoryproduces 12 units of bag N .

(ii) the maximum profit obtained if the profit from one unit of bag M and bag Nare RM 18 and RM 20 respectively.

[4 marks]





26/47

SULIT 3472/2


11

15 Table 3 shows the price indices and percentage of usage of four components, P, Q,R and S,which are the number of parts in the making of an electronic device.

ItemPrice index for the year 2000 based on

the year 1997

Percentage of usage (%)

P 125 20

Q 140 10

R x 30

S 110 40

TABLE 3(a) Calculate

(i) the price ofQ in the year 1997 if its price in the year 2000 is RM 50.40,

(ii) the price index ofP in the year 2000 based on the year 1994 if its price index inthe year 1997 based on the year 1994 is 120.

[5 marks]

(b) The composite index number for the cost of production in the year 2000 based onthe year 1997 is 122. Calculate

(i) the value ofx,

(ii) the price of an electronic device in the year 1997 if the corresponding price inthe year 2000 was RM 288.

[5 marks]

END OF QUESTION PAPER





27/47

SULIT

3472/1

Additional

Mathematics

Paper 1

Sept2009


LEMBAGA PEPERIKSAAN

PEPERIKSAAN PERCUBAAN SPMTINGKATAN 5

2009

ADDITIONAL MATHEMATICS

Paper 1

MARKING SCHEME

This marking scheme consists of 7 printed pages





28/47

2

PAPER 1 MARKING SCHEME 3472/1

Number Solution and marking schemeSub

MarksFull

Marks

1

(a)

(b)

4

One to many relation

1

1

2

2(a)

(b)

24 20 21x x+ +

2(2 5) 4x +

3

2

B1

1

3

3(a)

(b)

5 2

3

x

3 2

5

xy

+=

11

5

2

B1

1

3

4p 0

(5)2 4(2)(p 3) > 0

3

B2

B1

3

5 x2 + 4x 18 = 0

3 + 3 = 4 or 3(3) = 18

+ = 4

3and = 2

3

B2

B1

3





29/47

3


MarksFull

Marks

6(a)

(b)

(c)

5

13

x = 3

1

1

1

3

7 x1,x 2

2

( 2)( 1) 0

2 0

x x

x x

+

3

B2

B1

3

85

3

x =

2x + 5 = x

2( 1)3 3 2 23 3 3 or 3 3x x x x+ + + = =

3

B2

B1

3

9

2

1

3=2 or IE

3 1

+3

log =1 or IE3 1

x

x

x

x

x

=

+

3

B2

B1

3

10(a)

(b)

21

2 23 17(2) 3(2) 2 7(1) 3(1)2 2

d = or IE

216

2

B1

1

3

11

7

33

0.21

1 0.01

a = 0.21 and r= 0.01

3

B2

B1

3





30/47

4


MarksFull

Marks

12r= 2

q = 6y

x=x + 3*

p = 3

B1

B1

B1

B1

4

13 ( )1, 5M

3( 3) 2(7) 3(3) 2(8),

3 2 3 2M

+ +

+ +

AM:BM= 2 : 3

3

B2

B1

3

14a =

4

b

14

ba

=

1 2or 4

bm a m= =

3

B2

B1

3

15(a)

(b)

3 4u v % %

5(6 ) 3 6

3v u v +% % %

3u v +% %

( )1 5

6 3 42 3

v u v

+ % % %

2

B1

2

B1

4

16 a)

b)

12 5PQ i j= uuur

% %

24 10

13

i j% %

12 5

13

i j% % or

12 52

13

i j % %

1

2

B1

3

or Use TL

or Use TL





31/47

5


MarksFull

Marks

17

2

210 ,270 330

1sin and sin 1

2

(2sin 1)(sin 1) 0

2sin 3sin 1 0

x x

x x

= =

+ + =

+ + =

o o o

4

B3

B2

B1

4

18

(a)

(b)

4.8 cm

8 0.6

81.33

1

2(8)2( 0.6) or 0.6

2

B1

2

B1

4

19

3 151,

16

2x3

2= 0 or x =

3

4

32

2

dyx

dx=

3

B2

B1

3

20

10=

dt

dx

10 =dt

dx).

2

163(

2

2

163

xdx

dy=

3

B2

B1

3





32/47

6


MarksFull

Marks

21(a)

(b)

6

1627

102

2

6

3

2

=

+

kx

1

2

B1

3

22Mean = 13 AND Standard deviation = 8

Mean = 13 AND Variance (new) = 64

Mean = 13 OR Variance (new) = 64

3

B2

B1

3

23

(a)

(b)

7

20

29

60

+

12

5

5

3

12

7

5

2

1

2

B1

3

24(a)

(b)

336

3

8P

30 240

1

12

2

10

3

8 CCC

2

B1

2

B1

4

25(a) 0.2266

128 125

4P Z

OR P[Z 0.75]

2

B14





33/47

7


MarksFull

Marks

(b) 127.1

1250.3

4

mP Z

=

OR125

0.524

4

m =

2

B1





34/47

3472/2

Matematik

Tambahan

Kertas 22 jam

Sept 2009


PEPERIKSAAN PERCUBAAN

SIJIL PELAJARAN MALAYSIA 2009

MATEMATIK TAMBAHAN

Kertas 2

Dua jam tiga puluh minit

JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU

Skema Pemarkahan ini mengandungi 14 halaman bercetak

MARKING SCHEME





35/47

2

ADDITIONAL MATHEMATICS MARKING SCHEME

TRIAL SBP 2009 PAPER 2

QUESTIONNO. SOLUTION MARKS

1

2

2

2 6

(2 6) (2 6) 20 0

3 15 8 0

x y

y y y

y y

=

+ =

+ =

0.6070, 4.393

@

4.786, 2.786

y y

x x

= =

= =

5

2(a) 17.3205cmOT =

17.32053STR

s

=

18.1380 cmSTRs =

4

Solve the quadraticequation by using thefactorization @ quadraticformula @ completing the

s uare must be shown

Eliminate orx y

3

Use the formula

s = r

Note :OW-1 if the working of solving

quadratic equation is not shown.

5

P1

K1

K1

N1

N1

K1

K1

N1

K1





36/47

3

QUESTIONNO.

SOLUTION MARKS

(b)Area OPQ = @ 173.2051cm2

Area OSTR = 21

17.32052 3

@ 157.0795 cm2

120 20 sin 60

2 o 2

117.3205

2 3

@

173.2051 157.0795

16.1256 cm2

4

17 = 14.5 + )5(20

152

43

+ w

w = 7

33(a)

(b)

Score Frequency,

f

Mid-

point,

x

fx fx 2

0-4 2 2 4 8

5-9 3 7 21 147

10-14 10 12 120 1440

15-19 20 17 340 5780

20-24 7 22 154 3388

25-29 6 27 162 4374

30-34 2 32 64 2048

= 50f = 865fx 171852 =fx

Variance =

2

50

865

50

17185

= 44.41

4

120 20 sin 60

2

o

P1

Lower boundary OR

)5(20

152

43

+ w

7

8

K1

K1

K1

N1

K1

N1

K1

P1

N1

OR

P1





37/47

4

QUESTIONNO.

SOLUTION MARKS

V= 0.6 ORdV

dx= 3x2 OR x = 5

20.6 3(5) x

0.008x

34(a)

(b)

[ ]

2

2 2

2

22

(3 )(3) (3 4)( 2 )( )

(3 )

3 (2) (3) (3(2) 4)( 2(2))(2)

3 (2)

37

x x xf x

x

f

+ =

+

=

3

5(a) 2 sin

2xcos

sin

x

x

= 2 sinx cosx

= sin 2x

2

(b)

xy

2=

Number of solutions = 4

6

K1

N1

K1

N1

K1

6

K1

N1

P1

8

xy 2sin2=

x

xy

2=

y

O

2

1

K1

Sine curve

Period

Amplitude

Modulus

P1

P1

P1

P1

Sketch straight

line correctly

N1





38/47

5

QUESTIONNO.

SOLUTION MARKS

6 (a) (i) 15T = 2 + 14(3)

= 44 cm

(ii)Area of the fifteenth circle = 1936 2cm

3

(b)

100, 50 , 25 ,

a= 100 AND r =

2

1

7T = 1.563

3

K1

N1

N1

K1

K1

N1

6





39/47

6

2 3 4 5

2

0

2

x +

og10y

x + 2 3 4 5 6 7 8

log 10y 1.408 2.100 2.806 3.500 4.200 4.800

(a) Each set of values correct (log10y must be at least2 decimal places) N1, N1Y= mX+ clog 10y = (x + 2) log 10 q + log 10p K1where Y= log 10y, X= (x + 2),m = log 10 q and c = log 10p

(c) log 10 q = gradient

log 10 q =( )

07

6.02.4

K1

q = 4.85 N1

logp = Y-intercept

logp = 0.6 K1

p = 0251 N1

Correct both axes (Uniform scale) K1All points are plotted correctly N1

Line of best fit N1

1 6 7 8

10





40/47

7

QUESTIONNO.

SOLUTION MARKS

8(a)

(i)

(ii)

9 9QR x y= uuur

% %

( )1

9 9 122

OS y y x= + +uuur

%% %

96

2x y= +% %

3

OT hOS =uuur uuur

96

2hx hy= +

% % OR

QT kQR=

uuur uuur

9 9kx ky=

% %

QT QO OT = +uuur uuur uuur

99 6

2y hx hy= + +

%% %

96 9

2hx h y

= + +

% %

Comparing QTuuur

,

9 6k h= 23

k h= --------------------------(1)

OR

99 9

2k h = +

11

2k h= --------------(2)

Solving the simultaneous equations

6 4,

7 7h k= =

5

(b)

(d)

12 9PQ x y= +uuur

% %

( ) ( )2 2

12(3) 9(5)PQ = +uuur

= 57.63 units

2

10

K1

K1

K1

K1

N1

N1

N1

N1

K1

K1





41/47

8

QUESTIONNO.

SOLUTION MARKS

9(a)

(i)

(ii)

p = 0.3, q = 0.7, n = 8

1 (0.7)

8

8

C1(0.3)(0.7)

7

= 0.7447

n(0.3)(0.7) =315

n = 1500

5

(b)(i)

(ii)

P30 33.5 36.5 33.5 150

5 5Z

n

< < =

OR

P [ ]150

0.7 0.6Zn

< < =

1 P[Z> 0.7] P[Z> 0.6] =150

n

1 0.2420 0.2743 = 150n

n = 310.11

Total number = 310

5

N1

K1

K1

N1

K1

N1

10

K1

K1

K1

K1





42/47

9

QUESTIONNO.

SOLUTION MARKS

( )

1

4 1 1

5 0

Get m

y x

y x

=

=

+ =

3

( ) ( ) ( ) ( )

( )

2 2 2 2

2 2

2 2

1 4 2 4 2 1 3

2 1 8 16 4 8

2 8 15 0

x y

x x y y

x y x y

+ = +

+ + + =

+ =

2

( )( )

( ) ( )

2 2 15 0

5 3 0

5, 3

Get both

/ 5,0 and / 3,0

x x

x x

x x

Y Z Z Y

=

+ =

= =

= =

3

10(a)

(b)

(c)

(d)( )Get 2, 3 ,

intercept 1

T

x

=

=

2

10

N1

K1

K1

K1

K1

K1

N1

N1

K1

N1





43/47

10

QUESTIONNO.

SOLUTION MARKS

11(a)

2

2

2

(2) 2 2

1, 2

px x

p

p q

+ =

+ =

= =

3

(b)(i)

(ii)

Area of the shaded region

=1 1

2

0 0( 6 9) (2 2)x x dx x dx + +

=1

2

0

( 8 7)x x dx +

=

13 2

0

87

3 2

x xx

+

=1 8

7 03 2

+

=1

33

unit 2 .

Volume of revolution

= 2 20y dx

=2 4

0( 3)x dx

=5 2

0

( 3)

5

x

=5 5(1 3) (0 3)

5 5

=2

485

unit 3 .

7

N1

K1

K1

10

K1

N1

K1

K1

K1

K1

N1





44/47

11

QUESTIONNO.

SOLUTION MARKS

v = 5 ms1 1

a = 2t 6 and a = 0 ordv

dt= 0

t= 3

vmin = 4 ms1

3

(t 1)(t 5) < 0

1 < t< 52

12(a)

(b)

(c)

(d)

323 5

3

ts t t= +

| s1s0 | + | s5s1 | OR1 5

0 1v dt v dt +

7 25 70

3 3 3 +

OR

3 3 32 2 21 5 13(1) 5(1) 0 3(5) 5(5) 3(1) 5(1)3 3 3

+ + + +

13 m

4

10

K1

K1

K1

K1

N1

N1

P1

K1

N1

K1





45/47

12

QUESTIONNO.

SOLUTION MARKS

)8)(5(2

1285

cos

222+

=AMB

AMB = 133.43 @ 13326'

2

4

57.46sin

5

sin =

ACM

ACM= 65.20 @ 6524'

MAC= 180 46.57 - 65.20= 68.23

Area ofACM

=2

1(5)(4)sin 68.23 OR

= 9.287 cm2

Area ofABC= 14.52 + 9.287

= 23.807 cm2

5

13

(a)(i)

(ii)

(b)Get B 'M'M = 46.57@

B 'MM'= 46.57@

M'B 'M = 86.86

=

57.46sin

8

86.86sin

'MM

MM' = 10.999 cm @ 11 cmA'M'= 5 + 10.999

= 15.999 cm @ 16 cm

3

10

K1

K1

K1

N1

K1

N1

K1

Area ofAMB

=2

1(5)(8)sin 133.43

= 14.52 cm2

K1

K1

N1





46/47

13

Answer for question 14

2

4

14

12

10

16

8

6

(12, 8)

(a) I. 2 3 48x y+

II. 5 8 48x y+

III. 2y x

(b) Refer to the graph,

1 or 2 graph(s) correct3 graphs correct

Correct area

(c) i) 6

ii) max point (12, 8)

k= 18x + 20yMaximum Profit = RM 18(12) + RM 20(8)

= RM 37610

N1

N1

N1

N1

N1

N1

K1

N1

K1

N1

R





47/47

14

QUESTIONNO.

SOLUTION MARKS

97

RM 50.40140 100

Q

=

Q97 = RM 36

00,9400,97

97,94

100I

II

=

00,94125 100

120

I=

I00, 94 = 150

515

(a)(i)

(ii)

(b)(i)

(ii)

125 20 140 10 30 110 40122

100

x + + + =

30x + 8300 = 12200

x = 130

1997

RM 288122 100Q

=

1997 RM 236.07Q =

5

10

N1

N1

N1

K1

K1

K1

N1

K1

K1

K1



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