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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
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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
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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
=
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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4
24
3
23
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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
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SULIT 1 3472/2
3472/2 ZON A KUCHING 2009 SULIT
3472/2
Matematik
Tambahan
Kertas 2
2 jam
2009
SEKOLAH-SEKOLAH MENENGAH ZON A KUCHING
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
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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
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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
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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
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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
. . .
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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
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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
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SULIT 3472/2
3472/2 ZON A KUCHING 2009 SULIT
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
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SULIT 3472/2
3472/2 ZON A KUCHING 2009 SULIT
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
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SULIT 3472/2
3472/2 ZON A KUCHING 2009 SULIT
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]
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SULIT 3472/2
3472/2 ZON A KUCHING 2009 SULIT
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
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SULIT
3472/1
Additional
Mathematics
Paper 1
Sept2009
SEKOLAH-SEKOLAH MENENGAH ZON A KUCHING
LEMBAGA PEPERIKSAAN
PEPERIKSAAN PERCUBAAN SPMTINGKATAN 5
2009
ADDITIONAL MATHEMATICS
Paper 1
MARKING SCHEME
This marking scheme consists of 7 printed pages
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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
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3
Number Solution and marking schemeSub
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
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4
Number Solution and marking schemeSub
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
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5
Number Solution and marking schemeSub
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
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6
Number Solution and marking schemeSub
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
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7
Number Solution and marking schemeSub
MarksFull
Marks
(b) 127.1
1250.3
4
mP Z
=
OR125
0.524
4
m =
2
B1
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3472/2
Matematik
Tambahan
Kertas 22 jam
Sept 2009
SEKOLAH-SEKOLAH MENENGAH ZON A KUCHING
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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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7/28/2019 2009-PERCUBAAN Matematik Tambahan+Skema [SARAWAK].PDF
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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
http://mathsmozac.blogspo
Dapatkan KERTAS SOALAN RAMALAN 2010 di www.maths-catch.com
Bimbingan Matematik percuma di blog www.mathgayapos.blogspot.com
7/28/2019 2009-PERCUBAAN Matematik Tambahan+Skema [SARAWAK].PDF
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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Dapatkan KERTAS SOALAN RAMALAN 2010 di www.maths-catch.com
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