sulit sept 2008 peperiksaan percubaan spm 2008 · peperiksaan percubaan spm 2008 ... 2. answer all...
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SULIT 1
3472/1 2008 Hak Cipta Zon A Kuching [Lihat Sebelah SULIT
SEKOLAH MENENGAH ZON A KUCHING LEMBAGA PEPERIKSAAN
PEPERIKSAAN PERCUBAAN SPM 2008
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 4 11 3 12 4 13 3 14 3 15 3 16 4 17 3 18 3 19 4 20 4 21 4 22 3 23 3 24 3 25 3 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 Sept 2008 2 Jam
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The following formulae may be helpful in answering the questions. The symbols given are the ones commonly used.
ALGEBRA
1 x =a
acbb2
42
2 am an = a m + n 3 am an = a m n
4 (am) n = a nm 5 log a mn = log a m + log a n
6 log a nm = log a m log a n
7 log a mn = n log a m
8 log a b = ab
c
c
loglog
9 Tn = a + (n 1)d
10 Sn = ])1(2[2
dnan
11 Tn = ar n 1
12 Sn = rra
rra nn
1)1(
1)1( , (r 1)
13 r
aS
1 , r < 1
CALCULUS
1 y = uv , dxduv
dxdvu
dxdy
2 vuy , 2
du dvv udy dx dxdx v
,
3 dxdu
dudy
dxdy
4 Area under a curve
= b
a
y dx or
= b
a
x dy
5 Volume generated
= b
a
y 2 dx or
= b
a
x 2 dy
5 A point dividing a segment of a line
(x, y) = ,21
nmmxnx
nmmyny 21
6. Area of triangle =
1 2 2 3 3 1 2 1 3 2 1 31 ( ) ( )2
x y x y x y x y x y x y
1 Distance = 2 22 1 12( ) ( )x x y y
2 Midpoint
(x , y) =
221 xx ,
221 yy
3 22 yxr
4 2 2
x i yjr
x y
GEOMETRY
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STATISTICS
TRIGONOMETRY
1 Arc length, s = r
2 Area of sector , A = 212
r
3 sin 2A + cos 2A = 1 4 sec2A = 1 + tan2A 5 cosec2 A = 1 + cot2 A
6 sin 2A = 2 sinAcosA 7 cos 2A = cos2A – sin2 A = 2 cos2A 1 = 1 2 sin2A
8 tan2A = A
A2tan1
tan2
9 sin (A B) = sinAcosB cosAsinB
10 cos (A B) = cos AcosB sinAsinB
11 tan (A B) = BABA
tantan1tantan
12 Cc
Bb
Aa
sinsinsin
13 a2 = b2 +c2 2bc cosA
14 Area of triangle = Cabsin21
1 x = N
x
2 x =
ffx
3 = 2( )x x
N =
22x
xN
4 = 2( )f x x
f
=
22fx
xf
5 m = Cf
FNL
m
2
1
6 1000
1 PPI
7 i
i
iw IIw
8 )!(
!rn
nPrn
9 !)!(
!rrn
nCrn
10 P(AB) = P(A) + P(B) P(AB)
11 P(X = r) = rnrr
n qpC , p + q = 1
12 Mean , = np 13 npq
14 z = x
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Answer all questions.
1 Diagram 1 shows the linear function f. DIAGRAM 1
(a) State the value of n.
(b) Using the function notation, express f in terms of x. [ 2 marks ]
Answer : (a) ……………………..
(b) ……………………...
2. Two functions are defined by : 1f x x and 2: 3 1g x x x . Given that 2:gf x x ax b , find the value of a and of b.
[ 3 marks ]
Answer : …………………….....
3
2
For examiner’s
use only
x f(x) f
5 4 n 4
0 1 5 9
2
1
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3 The function of p is defined as p(x) hxx
x
,213 .
Find (a) the value of h, (b) )(1 xp .
[ 3 marks ]
Answer : (a) ……………………..
(b) ……………………... 4 Find the range of values of t if the following quadratic equation has no roots
(t + 2) x2 + 6x + 3 = 0. [ 3 marks ]
Answer : .........…………………
For examiner’s
use only
3
4
3
3
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5 Given that and are the roots of the quadratic equation 22 3 7x x . Form the quadratic equation whose roots are 2 and 2 .
[ 3 marks ]
Answer : .................................
___________________________________________________________________________
6 Diagram 2 shows the graph of a curve y = a(x + p)² + q that passes through the point (0, 5) and has the minimum point (2, 3). Find the values of a, p and q.
[ 3 marks ]
Answer : p = ……........................
q = ……........................
a = ..................................
3
5
3
6
For examiner’s
use only
DIAGRAM 2
(2, 3)
(0, 5)
x
y
O
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7 Find the range of values of x for which x(x − 2) ≤ 15. [3 marks]
Answer : ..................................
8 Solve 279
3 1
x
x
[ 3 marks ]
Answer : ...................................
9 Given that lg 2 0 3 and lg 17 1 23 , find, without using scientific calculator or
mathematical tables, find the value of 34log 2 . [ 3 marks ]
Answer : ......................................
3
7
3
9
3
8
For examiner’s
use only
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10 The thn term of an arithmetic progression is given by .15 nTn Find
(a) the first term and the common difference, (b) the sum of the first 15 terms of the progression. [4 marks]
Answer : (a) ………………………….
(b) ....……………...………..
11 The first three terms of a geometric progression are 219683
, 26561
, 22187
, . . . .
Find the three consecutive terms whose product is 157464. [ 3 marks ]
Answer : ............................................
4
10
For examiner’s
use only
3
11
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12 Diagram 3 shows the straight line obtained by plotting y10log against log 10 x.
The variables x and y are related by the equation ,4kxy where k is a constant. Find the value of
(a) k,
(b) .h [ 4 marks ]
Answer : (a)…...….………..….......
(b) ....................................
___________________________________________________________________________
13 The coordinates of the vertices of a triangle PQR are P(2, h), Q(1, 0) and R(5, h). If the area of the PQR is 9 units 2 , find the values of h.
[ 3 marks ]
Answer : h = …………………….
y10log
x10log 0
(0, 6)
(4, h )
DIAGRAM 3
3
13
4
12
For examiner’s
use only
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14 If the straight line 15
pyx is perpendicular to the straight line
,031210 yx find the value of .p [ 3 marks ]
Answer : .…………………
15 Given the vectors 3a i mj
% % %, 8b i j
% % % and 5 2c i j
% % %. If vector a b
% % is parallel to
vector ~c , find the value of the constant m.
[ 3 marks ] Answer : .………………….
3
15
3
14
For examiner’s
use only
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16 The diagram 4 shows a parallelogram ABCD drawn on a Cartesian plane.
It is given that 3 2AB i j
% %and 4 3BC i j
% %.
Find
(a) BD
,
(b) AC .
[ 4 marks ]
Answer : (a) ……….…….…………...
(b) ………………………….. ___________________________________________________________________________ 17 Solve the equation 2 2sin 5cos 3 cos for 00 3600 . [ 3 marks ]
Answer : …...…………..….......
3
17
For examiner’s
use only
4
16
y
O
A
B
C
D
x
DIAGRAM 4
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18 Given that sin x = 35
and 90 < x < 270, find the value of sec 2x.
[ 3 marks ]
Answer : …...…………..….......
___________________________________________________________________________ 19 The diagram 5 shows a semicircle of centre O and radius r cm. C A O B
The length of the arc AC is 72 cm and the angle of COB is 2692 radians. Calculate (a) the value of r, (b) the area of the shaded region. [Use π = 3.142] [ 4 marks ]
Answer : (a) ……………………..
(b) ……………………..
3
18
For examiner’s
use only
DIAGRAM 5
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20 Find the coordinates of the turning points of the curve y = x3 + 3x2 – 2 . [4 marks]
Answer : …...…………..…....... ___________________________________________________________________________ 21 Given that y = 3m2 and m = 2x + 3.
Find
(a) dxdy in terms of x,
(b) the small change in y when x increases from 3 to 301. [ 4 marks ]
Answer : (a) ……………………..
(b) ……………………..
22 Find dx
x313
[ 3 marks ]
Answer : ……………………..
3
20
4
21
3
22
For examiner’s
use only
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SULIT 14 3472/1
3472/1 2008 Hak Cipta Zon A Kuching SULIT
23 Ben and Shafiq are taking driving test. The probability that Ben and Shafiq pass the test
are 15
and 23
respectively.
Calculate the probability that at least one person passes the test. [ 3 marks ]
Answer : ………………………..
___________________________________________________________________________ 24 A committee of 5 members is to be selected from 6 boys and 4 girls. Find the number of
ways in which this can be done if
(a) the committee has no girls, (b) the committee has exactly 3 boys.
[ 3 marks ]
Answer : (a) ……………………..
(b) ……………………..
For examiner’s
use only
3
24
3
23
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SULIT 15 3472/1
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25 A random variable X has a normal distribution with mean 50 and variance 2 . Given that P[X > 51] = 0288, find the value of .
[ 3 marks ]
Answer : …...…………..….......
END OF QUESTION PAPER
3
25
For examiner’s
use only
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SULIT 3472/1 Additional Mathematics Paper 1 Sept 2008
SEKOLAH MENENGAH ZON A KUCHING LEMBAGA PEPERIKSAAN
PEPERIKSAAN PERCUBAAN SPM TINGKATAN 5
2008
ADDITIONAL MATHEMATICS
Paper 1
MARKING SCHEME
This marking scheme consists of 6 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)
0 x 5 or f : x x 5 or f(x) = x 5
1 1
2
2
a = 1 and b = 1 gf(x) = x2 + x 1 (x 1)2 + 3(x 1) + 1
3
B2
B1
3
3 (a)
(b)
12
3 1,
2 1 2x xx
y = 31 2x
x
1 2
B1
3
4
t > 1 –12t < 12 or equivalent (6)2 – 12 (t +2)(3) < 0
3
B2
B1
3
5
x2 + 3x + 14 = 0 2(2) = 14 and 2 + 2 = 3
3 7and =2 2
3 B2 B1
3
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3
Number Solution and marking scheme Sub Marks
Full Marks
6
a = 21
a(−2)² + 3 = 5 p = −2 and q = 3
3
B2
B1
3
7
−3 ≤ x ≤ 5 (x − 5)(x + 3) ≤ 0 x² − 2x − 15 ≤ 0
3
B2
B1
3
8
4x
1 2 3x x or equivalent
1 2 33 3x x or 1
233 3
3
x
x
3
B2
B1
3
9
5 1 lg 2 lg17
lg 2
lg34lg 2
3
B2
B1
3
10 (a)
(b)
5d
a @ T1 = 4 or T2 = 9 585
2
B1 1
3
11 18, 54, 162 n = 12 or 54 or 18 or equivalent (Solving)
a = 219683
and r = 3
3
B2
B1
3
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4
Number Solution and marking scheme Sub Marks
Full Marks
12 (a)
(b)
1000000k
xy 1010 log4log + k10log
22h
4046
h
2
B1 2
B1
4
13
2h
1 2 0 5 0 92
h h h h
1 2 0 5 02
h h h h
3
B2
B1
3
14
6p
5 1
5 6p
or equivalent
1 5pm or 2
56
m
3
B2
B1
3
15 m = 1 1 2
5 5m
5 1a b i m j
% % % %
3 B2
B1
3
16 (a)
(b)
5BD i j uuur
% %
or BD BA AD BA BC
uuur uuur uuur uuur uuur
50
7AC i j
uuur
% %
2
B1 2 B1
4
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5
Number Solution and marking scheme Sub Marks
Full Marks
17
66.42 ,293.58
2cos5
1 5cos 3 or equivalent
o o
3
B2
B1
3
18
257
21
31 25
1
cos 2x
3
B2
B1
3
19 (a)
(b)
r = 16 AOC = 0.45 or 7.2 = r (0.45) 344576 or 33458 or 3346
21 (16) (2 692)2
2
B1 2
B1
4
20
(2, 2) and (0, 2) x = 0, 2
dxdy = 0 or 3x(x + 2) = 0
dxdy = 3x2 + 6x
4
B3
B2 B1
4
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6
Number Solution and marking scheme Sub Marks
Full Marks
21 (a)
(b)
24 36 or equivalent
6 and 2
dy xdx
dy dmmdm dx
108
[24(3) 36] 0 01y
2
B1
2
B1
4
22
122(1 3 )x c
12
12
3(1 3 )3x c
123(1 3 )x or
12 3
3
B2
B1
3
23
1115
4 11 or equivalent5 3
4 1or5 3
3
B2
B1
3
24 (a) (b)
6 120 6 4
3 2C C
1 2
B1
3
25
1.789
51 50 0.559
0.559
3 B2 B1
3
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