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SULIT 1 3472/1
3472/1 2009 Hak Cipta Zon A Kuching [ Lihat sebelah SULIT
SEKOLAH-SEKOLAH MENENGAH ZON A KUCHING LEMBAGA PEPERIKSAAN
PEPERIKSAAN PERCUBAAN SPM 2009
Kertas soalan ini mengandungi 15 halaman bercetak
For examiner’s use only
Question Total Marks Marks Obtained
1 2
2 3
3 3
4 3
5 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 3
24 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. Answer all 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 not
drawn to scale unless stated. 8. The marks allocated for each question and sub-part
of 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 scientific calculator. 12 This question paper must be handed in at the end of
the examination .
Name : ………………..…………… Form : ………………………..……
3472/1 Matematik Tambahan Kertas 1 2009 2 Jam
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SULIT 3472/2
3472/1 2009 Hak Cipta Zon A Kuching SULIT
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The following formulae may be helpful in answering the questions. The symbols given are the ones commonly used.
ALGEBRA
1 2 4
2
b b acx
a
− ± −=
2 am × an = a m + n 3 am ÷ an = a m − n
4 (am)n = a mn 5 log a mn = log a m + log a n
6 log a n
m = log a m − log a n
7 log a mn = n log a m
8 log a b = a
b
c
c
log
log
9 Tn = a + (n − 1)d
10 Sn = ])1(2[2
dnan −+
11 Tn = ar n − 1
12 Sn = r
ra
r
ra nn
−−=
−−
1
)1(
1
)1( , (r ≠ 1)
13 r
aS
−=∞ 1
, r <1
CALCULUS
1 y = uv , dx
duv
dx
dvu
dx
dy +=
2 v
uy = ,
2
du dvv udy dx dx
dx v
−= ,
3 dx
du
du
dy
dx
dy ×=
4 Area under a curve
= ∫b
a
y dx or
= ∫b
a
x dy
5 Volume generated
= ∫b
a
y2π dx or
= ∫b
a
x2π dy
5 A point dividing a segment of a line
(x, y) = ,21
++
nm
mxnx
++
nm
myny 21
6 Area of triangle
= 1 2 2 3 3 1 2 1 3 2 1 3
1( ) ( )
2x y x y x y x y x y x y+ + − + +
1 Distance = 221
221 )()( yyxx −+−
2 Midpoint
(x , y) =
+2
21 xx ,
+2
21 yy
3 22 yxr +=
4 2 2
ˆxi yj
rx y
+=+
GEOMETRY
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SULIT 3472/1
3472/1 2009 Hak Cipta Zon A Kuching [ Lihat sebelah SULIT
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STATISTIC
1 Arc length, s = rθ
2 Area of sector , A = 21
2r θ
3 sin 2A + cos 2A = 1 4 sec2A = 1 + tan2A 5 cosec2 A = 1 + cot2 A
6 sin 2A = 2 sinA cosA 7 cos 2A = cos2A – sin2 A = 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
±
12 C
c
B
b
A
a
sinsinsin==
13 a2 = b2 + c2 − 2bc cosA
14 Area of triangle = Cabsin2
1
7 1
11
w
IwI
∑
∑=
8 )!(
!
rn
nPr
n
−=
9 !)!(
!
rrn
nCr
n
−=
10 P(A∪ B) = P(A) + P(B) − P(A∩ B)
11 P(X = r) = rnrr
n qpC − , p + q = 1 12 Mean µ = np
13 npq=σ
14 z = σ
µ−x
1 x = N
x∑
2 x = ∑∑
f
fx
3 σ = 2( )x x
N
−∑ = 2
2xx
N−∑
4 σ = 2( )f x x
f
−∑∑
= 2
2fxx
f−∑
∑
5 m = Cf
FNL
m
−+ 2
1
6 1
0
100Q
IQ
= ×
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SULIT 3472/1
3472/1 2009 Hak Cipta Zon A Kuching SULIT
4
Answer all questions.
1 Diagram 1 shows a graph of the relation between two variables x and y.
State (a) the object of 8,
(b) the type of relation between x and y.
[ 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 examiner’s
use only
x
• •
•
•
•
0 2 4 6 8 10
4 8 12 16
24
y
DIAGRAM 1
20
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SULIT 3472/1
3472/1 2009 Hak Cipta Zon A Kuching [ Lihat sebelah SULIT
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3 Given that the function 3 2
:5
xf x
+→ , find
(a) 1( )f x−
(b) the value of x such that 1( ) 3f x− = [ 3 marks ]
Answer : (a) ……………………..
(b) ……………………... 4 The quadratic equation 2x2− 5x+ p − 3 = 0 has two different roots, find the range of
values of p. [ 3 marks ]
Answer : .........…………………
For examiner’s
use only
3
3
3
4
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SULIT 3472/1
3472/1 2009 Hak Cipta Zon A Kuching SULIT
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5 Given α and β are the roots of the quadratic equation 3x2 + 4x − 6 = 0, form the quadratic equation whose roots are 3α and 3β .
[ 3 marks ]
Answer : .................................
___________________________________________________________________________
6 Diagram 2 shows the graph of the quadratic function y = 2(x – 3)2− p which has a minimum value of −5.
Find (a) the value of p, (b) the value of q, (c) the equation of the axis of symmetry.
[ 3 marks ]
Answer : (a) ……........................
(b) ……........................
(c)..................................
3
5
3
6
For examiner’s
use only
DIAGRAM 2
q
x
y
O
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SULIT 3472/1
3472/1 2009 Hak Cipta Zon A Kuching [ Lihat sebelah SULIT
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7 Find the range of values of x for which (2x − 3)(x + 1) ≥ x +1. [ 3 marks ]
Answer : ..................................
8 Solve the equation 27×9x + 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 examiner’s
use only
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SULIT 3472/1
3472/1 2009 Hak Cipta Zon A Kuching SULIT
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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 examiner’s
use only
3
11
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SULIT 3472/1
3472/1 2009 Hak Cipta Zon A Kuching [ Lihat sebelah SULIT
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12 Diagram 3 shows the graph pxxy += 2 .
Based on the graph above a table of x
y against x is obtained as show in table 1
x
y −3 r
x q −1
Calculate the values of p, 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 that AM : AB = 2 : 5. [ 3 marks ]
Answer : ………………..…….
3
13
4
12
DIAGRAM 3
For examiner’s
use only
−3
x
y
0
pxxy += 2
TABLE 1
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SULIT 3472/1
3472/1 2009 Hak Cipta Zon A Kuching SULIT
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14 Given the straight lines 3y ax+ = and 4 4y bx+ = are perpendicular to each other. Express a in terms of b.
[ 3 marks ]
Answer : .…………………
15 In Diagram 4, QR is parallel to PS and T is the midpoint of QR. [ 4 marks ]
Given that PS : QR = 3 : 5, PQ→
= 3u and PS→
= 6v, express, in terms of u and v, of
(a) RS→
,
(b) TS→
. [ 4 marks ]
Answer : (a)…...…………..…....... (b) ....................................
4
15
3
14
For examiner’s
use only
Q
S
R
T
DIAGRAM 4
P
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SULIT 3472/1
3472/1 2009 Hak Cipta Zon A Kuching [ Lihat sebelah SULIT
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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 sin x + xsin
1 = –3 for 0° ≤ x ≤ 360°. [ 4 marks ]
Answer : …...…………..….......
4
17
For examiner’s
use only
3
16
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SULIT 3472/1
3472/1 2009 Hak Cipta Zon A Kuching SULIT
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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 2 310
2y x x= − + .
[ 3 marks ]
Answer : ………………………
3
19
4
18
For examiner’s
use only
DIAGRAM 5 P O
Q
0.6 rad
R
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SULIT 13 3472/1
3472/1 2009 Hak Cipta Zon A Kuching SULIT
20 Two variables, x and y, are related by the equation .16
3x
xy += Given that y increases
at a constant rate of 10 unit per second, find the rate of change of x when x = 2. [3 marks]
Answer : …...…………..…....... ___________________________________________________________________________
21 Given that 6
3( ) 2,g x dx=∫ find
(a) 363 ( )g x dx∫ ,
(b) the value of k if 6
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 2.56
respectively. 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 examiner’s
use only
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SULIT 14 3472/1
3472/1 2009 Hak Cipta Zon A Kuching SULIT
23 The probability that Kamal is chosen as a school librarian is 5
2 whereas the
probability that Alisa is chosen as a school librarian is 12
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 3
students.
(a) A teacher wants to choose Form 5 students to form a committee consisting a president, a vice president and a secretary, find the number of ways the committee can be formed.
(b) A team is to be formed to take part in a Mathematics competition. How many different teams, each comprising 3 Form 5 students, 2 Form 4 students and 1 Form 3 student can be formed? [ 4 marks ]
Answer : (a) ……………………..
(b) .……………..………
For examiner’s
use only
4
24
3
23
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SULIT 15 3472/1
3472/1 2009 Hak Cipta Zon A Kuching SULIT
25 The mass of a packet of biscuit is normally distributed with a mean of 125 g and a variance of 16 g2.
(a) Find the probability that a packet of biscuit chosen at random from a sample will
have mass not less than 128 g. (b) If 30% of the packets chosen at random have mass more than m g, find the value
of m. [ 4 marks ]
Answer : (a) ……………………..
(b) .……………..………
END OF QUESTION PAPER
4
25
For examiner’s
use only
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SULIT 3472/1 Additional Mathematics Paper 1 Sept 2009
SEKOLAH-SEKOLAH MENENGAH ZON A KUCHING LEMBAGA PEPERIKSAAN
PEPERIKSAAN PERCUBAAN SPM TINGKATAN 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 scheme Sub
Marks Full
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
4
p < 498
49 − 8p > 0 (−5)2 − 4(2)(p − 3) > 0
3
B2
B1
3
5
x2 + 4x − 18 = 0 3α + 3β = −4 or 3α(3β) = −18
α + β = − 43
and αβ = −2
3 B2
B1
3
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3
Number Solution and marking scheme Sub
Marks Full
Marks 6
(a)
(b)
(c)
5
13
x = 3
1
1
1
3
7
x ≤ −1, x ≥ 2
2
( 2)( 1) 0
2 0
x x
x x
− + ≥− − ≥
3
B2
B1
3
8
53
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 IE 3 1
x
x
xx
x
=+−
−
3
B2
B1
3
10 (a)
(b)
21
2 23 17(2) 3(2) 2 7(1) 3(1)
2 2d = − − × − 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 scheme Sub
Marks Full
Marks 12
r = 2 q = −6 y
x = x + 3*
p = 3
B1
B1
B1
B1
4
13 ( )1, 5M
3( 3) 2(7) 3(3) 2(8)
,3 2 3 2
M− + +
+ +
AM : BM = 2 : 3
3
B2
B1
3
14 a = − 4
b
14
ba − × − = −
1 2or 4
bm a m= − = −
3
B2
B1
3
15 (a)
(b)
3 4u v− −% %
5
(6 ) 3 63
v u v− − +% % %
3u v− +% %
( )1 56 3 4
2 3v 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 scheme Sub
Marks Full
Marks 17
2
210 ,270 330
1sin and sin 1
2
(2sin 1)(sin 1) 0
2sin 3sin 1 0
x x
x x
θ θ= − = −
+ + =
+ + =
o o o
4
B3
B2
B1
4
18 (a)
(b)
4.8 cm 8 × 0.6 81.33 1
2(8)2(π − 0.6) or π − 0.6
2
B1
2
B1
4
19 3 151
,4 16
2x − 3
2 = 0 or x =
3
4
3
22
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 scheme Sub
Marks Full
Marks 21 (a)
(b)
−6 16
27
102
26
3
2
=
+ kx
1
2
B1
3
22 Mean = 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
38 P 30 240
112
210
38 CCC ××
2
B1
2
B1
4
25 (a)
0.2266
128 125
4P Z
− ≥ OR P[Z ≥ 0.75]
2
B1
4
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7
Number Solution and marking scheme Sub
Marks Full
Marks (b) 127.1
125
0.34
mP Z
− ≥ = OR
1250.524
4
m − =
2
B1
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