Trial Kedah PDF
Trial Kedah PDF
Trial Kedah PDF
SULIT
Name : ....
Form : ..
Dua jam
Soalan
Markah
Penuh
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
3
3
4
3
3
3
3
3
4
4
2
4
3
2
3
4
4
4
3
3
3
3
3
3
3
TOTAL
Markah
Diperolehi
80
SULIT
3472/1
The following formulae may be helpful in answering the questions. The symbols given are
the ones commonly used.
ALGEBRA
1
x=
b b 4ac
2a
log c b
log c a
logab =
am an = a m + n
Tn = a + (n1)d
am an = a m -
10
(am) n = a mn
6
7
n
[2a + (n 1)d ]
2
11 Tn = ar n 1
a (r n 1) a (1 r n )
12 Sn =
, (r 1)
=
r 1
1 r
a
, r <1
13 S =
1 r
Sn =
CALCULUS
1
du
dv
dy
=u +v
dx
dx
dx
y = uv ,
dx or
du
dv
u
u dy
= dx 2 dx ,
y= ,
v
v dx
v
x dy
a
5 Volume generated
3
dy dy du
=
dx du dx
= y 2 dx or
a
dy
GEOMETRY
1 Distance =
( x2 x1 ) 2 + ( y2 y1 ) 2
2 Midpoint
y + y2
x + x2
(x, y) = 1
, 1
2
2
r
3 =
r =
x +y
2
xi + yj
x2 + y 2
m+n
m+n
6 Area of triangle
1
= ( x1 y 2 + x 2 y 3 + x3 y11 ) ( x 2 y1 + x3 y 2 + x1 y 3 )
2
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STATISTICS
x =
x =
fx
f
2
(x x )
N
f (x x)
f
2 N F
m = L+
C
fm
6 =
I
x2
Q1
100
Q0
x2
fx
f
10
11
P (X = r) = nCr p r q n r , p + q = 1
12
Mean = np
13
= npq
14
Z=
x2
Wi I i
Wi
n!
n
Pr =
(n r )!
n!
n
Cr =
(n r )!r!
I=
TRIGONOMETRY
1 Arc length, s = r
2 Area of sector , A =
3 sin 2A + cos 2A = 1
1 2
r
2
4 sec2A = 1 + tan2A
5 cosec2 A = 1 + cot2 A
tan A tan B
1 tan A tan B
12
a
b
c
=
=
sin A sin B sin C
13
a2 = b2 + c2 2bc cosA
14
Area of triangle
1
absin C
2
2 tan A
1 tan 2 A
3472/1
For
examiners
use only
SULIT
3472/1
Set K
5
7
9
Diagram 1
Rajah 1
{ ( 2,9 ) , (1, m ) , ( n , 9 )} .
Hubungan itu ditakrifkan oleh set pasangan tertib { ( 2,9 ) , (1, m ) , ( n , 9 )} .
The relation is defined by the set of ordered pairs
State
Nyatakan
(a) the value of m and of n .
nilai m dan nilai n .
(b) the type of the relation.
jenis hubungan itu .
[3 marks]
[3 markah]
Answer/Jawapan:
(a)
(b)
1
3
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2
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For
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The following information refers to the functions f and g and the composite function
f 1 g .
Maklumat berikut adalah berkaitan dengan fungsi f dan g dan fungsi gubahan f
g.
f : x 5x 3
6
x
f 1 g ( p ) = 3
g: x
2
3
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g ( 3) ,
(b)
k if g 1 f (1) = 8
k jika g 1 f (1) = 8 .
[4 marks]
[4 markah]
Answer/Jawapan:
(a)
(b)
3
4
4
3
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5
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(3 ,q )
Diagram 5
Rajah 5
(a) Find the value of p and of q .
Cari nilai p dan nilai q .
(b) State the equation of the axis of symmetry of the curve.
Nyatakan persamaan paksi simetri bagi lengkung itu.
[3marks]
[3markah]
Answer/Jawapan:
(a)
(b)
5
3
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9 x
.
2
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9 x
.
2
[3 marks]
[3 markah]
Answer/Jawapan:
6
3
3
Given that log 2 h = a and log 2 k = b , express log 8 16 h k in terms of a and b .
[3 marks]
7
3
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8
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7 ( 49 x1 ) =
1
343
[3 marks]
[3 markah]
Answer/Jawapan:
8
3
An arithmetic progression consists of 26 terms. Given the first term is 2 and the sum of
the last 8 terms is 532. Find the 15th term of the progression.
[4marks]
Suatu janjang aritmetik mengandungi 26 sebutan. Diberi sebutan pertama ialah 2 dan
hasil tambah 8 sebutan terakhir ialah 532 .Cari sebutan ke-15 bagi janjang itu.
[4markah]
Answer/Jawapan:
9
4
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10
10
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m3
and m respectively.
9
m3
Sebutan pertama dan kedua suatu janjang geometri masing-masing ialah
dan m.
9
The first term and second term of a geometric progression is
Find
Cari
(a)
the values, other than zero, that are not possible for m.
nilai-nilai yang tidak mungkin bagi m selain daripada sifar.
(b)
Answer/Jawapan:
(a)
(b)
10
4
11
1
Given that 9 + 3 + 1 + + ... is an infinite series of a geometric progression. Find the
3
sum to infinity of the series.
[2 marks]
1
Diberi 9 + 3 + 1 + + ... ialah satu siri takterhingga bagi suatu janjang geometri. Cari
3
hasil tambah hingga sebutan ketakterhinggaan bagi siri itu.
[2 markah]
Answer/Jawapan:
11
2
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P( 2 , 3 )
x2
O
Diagram 12
Rajah 12
b
. Find the value of
y ax =
x
b
Pembolehubah x dan y dihubungkan oleh persamaan y ax =
. Cari nilai bagi
x
(a) a
(b)
b
[4marks]
[4markah]
Answer/Jawapan:
(a)
(b)
12
4
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12
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13
3
14 Given that the points S ( k , 2 ) , T ( 3, 4 ) and U ( 11,8 ) are collinear. Find the value of k.
Diberi titik-titik S ( k , 2 ) , T ( 3, 4 ) dan U ( 11,8 ) adalah segaris. Cari nilai bagi k .
[2 marks]
[2 markah]
Answer/Jawapan:
14
2
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For
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y
B
6
5
4
3
2
1
0
Diagram 15
Rajah 15
(a)
(b)
Answer/Jawapan:
(a)
(b)
15
3
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14
16 Given that OA
= 3i + 8 j , AB
= 3i + 4 j and 3OC = OB , where
OA + AB + mOC =
0 . Find the value of m .
Diberi OA
= 3i + 8 j , AB
= 3i + 4 j dan 3OC = OB , di mana
OA + AB + mOC =
0 . Cari nilai m
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[4 marks]
[4 markah]
Answer/Jawapan:
16
4
17
4
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15
18
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12 cm
Diagram 18
Rajah 18
Diagram 18 shows a sector AOB of a circle with centre O and radius of 12 cm.
Given that point C is the midpoint of OA. Find
(a) AOB , in radians,
(b) the area, in cm, of the shaded region.
[4 marks]
Rajah 18 menunjukkan sebuah sektor AOB bagi sebuah bulatan berpusat O dan
berjejari 12 cm. Diberi bahawa titik C ialah titik tengah bagi garis OA. Cari
(a) AOB , dalam radian ,
(b) luas , dalam cm , kawasan berlorek.
[4 markah]
Answer/Jawapan:
(a)
(b)
18
4
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(b)
19
3
20
3
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21
3
22
Given that y =
( 3x + 1) 2
x 1
and
dy
= h ( x ) , find the value of
dx
1
h ( x ) dx .
5
[3 marks]
Diberi bahawa y =
( 3x + 1) 2
x 1
dan
dy
= h ( x ) , cari nilai bagi
dx
1
h ( x ) dx .
5
[3 markah]
Answer/Jawapan:
22
3
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23 In FIFA World Cup 2014, the probability of Team Y success to enter the final is
3
5
, while the probability that Team Y will win in the final is . Find the probability
5
7
that
(a)
(b)
Team Y will fail to become the winner in FIFA World Cup 2014.
[3 marks]
Dalam Piala Dunia FIFA 2014, kebarangkalian bagi Pasukan Y berjaya memasuki
3
pusingan akhir ialah , manakala kebarangkalian bagi Pasukan Y akan menang
5
5
dalam pusingan akhir ialah . Cari kebarangkalian bahawa
7
(a)
(b)
Pasukan Y akan gagal menjadi juara dalam Piala Dunia FIFA 2014.
[3 markah]
Answer/Jawapan:
(a)
(b)
23
3
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24 A teacher wants to form a team of 8 students to collect the donation from each class.
These 8 students are chosen from 4 monitors, 3 assistant monitors and 5 prefects.
Calculate the number of different ways the team can be form if
(a) there is no restriction,
(b) the team contains only 3 monitors and 2 assistant monitors.
[3 marks]
Seorang guru ingin membentuk satu kumpulan 8 orang murid untuk mengutip derma
dari setiap kelas. Kumpulan 8 orang murid itu mesti dipilih daripada 4 orang ketua
kelas, 3 orang penolong ketua kelas dan 5 orang pengawas. Hitungkan bilangan cara
yang berlainan kumpulan itu boleh dibentuk jika
(a) tiada syarat dikenakan
(b) kumpulan itu hanya terdiri daripada 3 orang ketua kelas dan 2 orang penolong
ketua kelas.
[3 markah]
Answer/Jawapan:
(a)
(b)
24
3
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03835
014
Diagram 25
Rajah 25
The probability represented by the area of the shaded region is 0 3835 . Find the value
of k.
Kebarangkalian yang diwakili oleh luas kawasan berlorek ialah 0 3835 . Cari nilai k.
[3 marks]
[3 markah]
Answer/Jawapan:
25
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THE UPPER TAIL PROBABILITY Q(z) FOR THE NORMAL DISTRIBUTION N(0,1)
KEBARANGKALIAN HUJUNG ATAS Q(z) BAGI TABURAN NORMAL N(0, 1)
1
0.4641
12
16
20
24
28
32
36
0.4286
0.4247
12
16
0.3897
0.3859
12
15
20
24
28
32
36
19
23
27
31
35
0.3557
0.3520
0.3483
11
0.3192
0.3156
0.3121
11
15
19
22
26
30
34
15
18
22
25
29
32
0.2877
0.2843
0.2810
0.2776
0.2546
0.2514
0.2483
0.2451
10
14
17
20
24
27
31
10
13
16
19
23
26
29
0.2266
0.2236
0.2206
0.2177
0.2148
0.1977
0.1949
0.1922
0.1894
0.1867
12
15
18
21
24
27
11
14
16
19
22
0.1736
0.1711
0.1685
0.1660
0.1635
0.1611
25
10
13
15
18
20
23
0.1515
0.1492
0.1469
0.1446
0.1423
0.1401
0.1292
0.1271
0.1251
0.1230
0.1210
0.1190
0.1379
12
14
16
19
21
0.1170
10
12
14
16
18
0.1112
0.1093
0.1075
0.1056
0.1038
0.1020
0.0934
0.0918
0.0901
0.0885
0.0869
0.0853
0.1003
0.0985
11
13
15
17
0.0838
0.0823
10
11
13
14
0.0793
0.0778
0.0764
0.0749
0.0735
0.0721
0.0655
0.0643
0.0630
0.0618
0.0606
0.0594
0.0708
0.0694
0.0681
10
11
13
0.0582
0.0571
0.0559
10
11
0.0548
0.0537
0.0526
0.0516
0.0505
0.0495
0.0446
0.0436
0.0427
0.0418
0.0409
0.0401
0.0485
0..0475
0.0465
0.0455
0.0392
0.0384
0.0375
0.0367
1.8
0.0359
0.0351
0.0344
0.0336
0.0329
0.0322
0.0314
0.0307
0.0301
0.0294
1.9
0.0287
0.0281
0.0274
0.0268
2.0
0.0228
0.0222
0.0217
0.0212
0.0262
0.0256
0.0250
0.0244
0.0239
0.0233
0.0207
0.0202
0.0197
0.0192
0.0188
0.0183
2.1
0.0179
0.0174
0.0170
2.2
0.0139
0.0136
0.0132
0.0166
0.0162
0.0158
0.0154
0.0150
0.0146
0.0143
0.0129
0.0125
0.0122
0.0119
0.0116
0.0113
0.0110
2.3
0.0107
0.0104
0.0102
0.00990
0.00964
0.00939
0.00914
10
13
15
18
20
23
12
14
16
16
21
11
13
15
17
19
0.0
0.5000
0.4960
0.4920
0.4880
0.4840
0.4801
0.4761
0.4721
0.4681
0.1
0.4602
0.4562
0.4522
0.4483
0.4443
0.4404
0.4364
0.4325
0.2
0.4207
0.4168
0.4129
0.4090
0.4052
0.4013
0.3974
0.3936
0.3
0.3821
0.3783
0.3745
0.3707
0.3669
0.3632
0.3594
0.4
0.3446
0.3409
0.3372
0.3336
0.3300
0.3264
0.3228
0.5
0.3085
0.3050
0.3015
0.2981
0.2946
0.2912
0.6
0.2743
0.2709
0.2676
0.2643
0.2611
0.2578
0.7
0.2420
0.2389
0.2358
0.2327
0.2296
0.8
0.2119
0.2090
0.2061
0.2033
0.2005
0.9
0.1841
0.1814
0.1788
0.1762
1.0
0.1587
0.1562
0.1539
1.1
0.1357
0.1335
0.1314
1.2
0.1151
0.1131
1.3
0.0968
0.0951
1.4
0.0808
1.5
0.0668
1.6
1.7
Minus / Tolak
0.00889
0.00866
0.00842
2.4
0.00820
0.00798
0.00776
0.00755
0.00734
0.00714
0.00695
0.00676
0.00657
0.00639
11
13
15
17
2.5
0.00621
0.00604
0.00587
0.00570
0.00554
0.00539
0.00523
0.00508
0.00494
0.00480
11
12
14
2.6
0.00466
0.00453
0.00440
0.00427
0.00415
0.00402
0.00391
0.00379
0.00368
0.00357
10
2.7
0.00347
0.00336
0.00326
0.00317
0.00307
0.00298
0.00289
0.00280
0.00272
0.00264
2.8
0.00256
0.00248
0.00240
0.00233
0.00226
0.00219
0.00212
0.00205
0.00199
0.00193
2.9
0.00187
0.00181
0.00175
0.00169
0.00164
0.00159
0.00154
0.00149
0.00144
0.00139
3.0
0.00135
0.00131
0.00126
0.00122
0.00118
0.00114
0.00111
0.00107
0.00104
0.00100
exp z 2
2
2
1
f ( z) =
f (z)
Example / Contoh:
Q(z)
Q( z ) = f ( z ) dz
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3472/2
August 2014
1.
This question paper consists of three sections : Section A, Section B and Section C.
2.
Answer all questions in Section A, four questions from Section B and two questions from
Section C.
3.
4.
5.
The diagrams provided are not drawn according to scale unless stated.
6.
The marks allocated for each question and sub - part of a question are shown in brackets.
7.
8.
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3472/2
August 2014
The following formulae may be helpful in answering the questions. The symbols given are the ones
commonly used.
ALGEBRA
log c b
b b 2 4ac
8. log a b
1. x
log c a
2a
2. am an amn
9. T n a (n 1)d
3. am an amn
10. S n
4. (a m )n amn
11. T n ar n 1
6. log a
n
[ 2a ( n 1 ) d ]
2
a ( r n 1) a (1 r n )
, r1
r 1
1 r
a
13. S
, r <1
1 r
12. S n
m
log a m log a n
n
7. log a m n n log a m
1. y = uv,
CALCULUS
4
Area under a curve
dy
dv
du
u
v
dx
dx
dx
= a y dx
or
= a x dy
2. y =
3.
dy
u
,
v
dx
5.
du
dv
u
dx
dx
2
v
Volume of revolution
b
= a y 2 dx
or
b
= a x 2 dy
dy dy du
dx du dx
GEOMETRY
1. Distance =
( x2 x1 ) 2 ( y 2 y1 ) 2
4. Area of triangle
=
2. Mid point
x x2 y1 y 2
,
(x,y)= 1
2
2
,
( x , y ) = 1
m n
mn
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5.
1
2
( x1 y2 x2 y3 x3 y1 ) ( x2 y1 x3 y2 x1 y3 )
6. r
x2 y2
xi yj
x2 y 2
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August 2014
3
STATISTICS
x
1. x
N
fx
2. x
f
3.
( x x )2 =
4.
f ( x x )2 =
f
x2 x 2
Pr
n!
( n r )!
Cr
n!
( n r )!r!
1N F
5. m = L + 2
C
f m
13 npq
Q
6. I 1 100
Q0
14 Z =
TRIGONOMETRY
8. sin ( A B ) = sin A cos B cos A sin B
1. Arc length, s = r
2. Area of sector, A =
fx 2 x 2
f
Wi I i
Wi
1 2
r
2
3. sin A + cos A = 1
9.
10 tan ( A B ) =
4. sec A = 1 + tan A
11 tan 2A =
5. cosec A = 1 + cot A
12
tan A tan B
1 tan A tan B
2 tan A
1 tan 2 A
a
b
c
13 a = b + c 2bc cos A
14 Area of triangle =
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1
ab sin C
2
SULIT
3472/2
August 2014
Section A
Bahagian A
[ 40 marks ]
[ 40 markah ]
Answer all questions.
[5 marks]
[5 markah]
2
7 cm
3 cm
5 cm
Diagram 2 / Rajah 2
Diagram 2 shows part of a structure made up of rectangular blocks. The first column has one
block. For each of the other columns, the number of blocks is doubled the previous column.
(a) Find the number of blocks in the 8 th columns,
[2 marks]
(b) Calculate
(i) the total volume of the blocks if there are 10 columns of blocks.
(ii) the total cost of the 10 columns of blocks if each block cost RM 0 80 .
[5 marks]
Rajah 2 menunjukkan susunan suatu struktur yang terdiri daripada blok yang berbentuk segi
empat tepat. Lajur pertama mempunyai satu blok. Bagi setiap lajur berikutnya, bilangan blok
adalah dua kali ganda daripada lajur sebelumnya.
(a) Carikan bilangan blok bagi lajur ke 8.
[2 markah]
(b) Hitungkan
(i) jumlah isipadu blok jika terdapat 10 lajur bagi struktur itu.
(ii) jumlah kos bagi 10 lajur blok jika setiap blok RM 0 80 .
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[3 marks]
(b) Hence, using the same axes, sketch a suitable straight line to find the number of solutions
for the equation x cos3x 0 for 0 x .
State the number of solutions.
[3 marks]
[3 markah]
(b) Seterusnya, dengan menggunakan paksi yang sama, lakar satu garis lurus yang sesuai
untuk mencari bilangan penyelesaian bagi persamaan x cos3x 0 untuk 0 x .
Nyatakan bilangan penyelesaian itu.
[3 markah]
(a)
Given that a set X has score x1 , x2 , x3 ...............x10 . The mean and standard deviation of
set X are 10 and 4 respectively. Find x and x 2 for set X .
x 3
x1 3 x2 3 x3 3
,
,
................ 10
.
2
2
2
2
(a)
[3 marks]
Diberi bahawa set X mempunyai skor x1 , x2 , x3 ...............x10 . Min dan sisihan piawai
ialah 10 dan 4 masing-masing. Carikan x and x 2 bagi set X .
[4 markah]
x 3
x1 3 x2 3 x3 3
,
,
................ 10
.
2
2
2
2
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[3 marks]
1
8
1
.
2
[5 marks]
[3 markah]
1
8
[5 markah]
9 x2 1
(a) Given that f ( x)
, find f '( x).
3x 1
[2 marks]
(b)
A curve has a gradient function of kx 2 3x , the tangent to the curve at the point
( 2,12) is parallel to the straight line 2 y 4 x 9 , find
9 x2 1
, cari f '( x).
3x 1
[5 marks]
[2 markah]
(b) Fungsi kecerunan suatu lengkung ialah kx 2 3x , tangen pada lengkung di titik ( 2,12)
adalah selari kepada garis lurus 2 y 4 x 9 , cari
(i) nilai bagi k,
(ii) persamaan normal pada lengkung di titik ( 2,12) .
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7
Section B
Bahagian B
[ 40 marks ]
[ 40 markah ]
Answer four questions from this section.
Jawab empat soalan daripada bahagian ini.
3 5
22 0
67 5
152 0
287 5
486 0
Table 7/ Jadual 7
Table 7 shows the values of two variables, x and y, obtained from an experiment. Variables x
p
and y are related by the equation y kx3 x 2 , where k and p are constants.
k
y
(a) Plot 2 against x , using a scale of 2 cm to 1 unit on the x axis and 1 cm to 1 unit
x
y
on the 2 - axis. Hence, draw the line of best fit.
[4 marks]
x
(b) Use the graph in 7 (a) to find the value of
(i) k
(ii) p
(iii)
y when x = 2 5 .
[6 marks]
Jadual 7 menunjukkan nilai-nilai bagi dua pembolehubah, x dan y, yang diperoleh daripada
satu eksperimen. Pembolehubah x dan y dihubungkan oleh persamaan
y kx3
p 2
x ,
k
(a) Plot
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(i)
k,
(ii)
(iii)
p,
y apabila x = 2 5 .
Additional Mathematics Paper 2
http://www.chngtuition.blogspot.com
[6 markah]
[Lihat halaman sebelah
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8
P
M
Q
Diagram 8/ Rajah 8
S
R
1
3
and y , find PS .
2
4
[4 marks]
Rajah 8 menunjukkan sebuah sisiempat PQRT dan segitiga RST. M ialah titik tengah TR.
PQ = 9x , PT = 8y , 2 PQ =3 TS dan PT = 2 QR .
(a) Ungkapkan vektor yang berikut dalam sebutan x dan y
(i) TR
(ii) SR
(iii) MS
Seterusnya, tunjukkan bahawa P, M dan S adalah segaris.
(b) Diberi bahawa x
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1
3
dan y , cari PS .
2
4
[6 markah]
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9
y
K
y = 3x + 4
y=x2
Diagram 9/Rajah 9
Diagram 9 shows the straight line y = 3x + 4 intersecting the curve y = x2 at the points K.
Find
(a) the coordinates of K,
[3 marks]
(b)
[3 marks]
(c)
the volume generated, in terms of , when the shaded region A is revolved through 360o
about the y-axis.
[4 marks]
koordinat K,
[3 markah]
(b)
[3 markah]
(c)
isipadu janaan, dalam sebutan , apabila rantau berlorek A dikisarkan melalui 360o
pada paksi-y.
[4 markah]
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10
10
B
C
O
Diagram 10 / Rajah 10
Diagram 10 shows a semicircle ABC with centre O and a sector ABO with centre A. The radius
of semicircle ABC and sector ABO is 8 cm.
[ Use = 3 142 ]
Calculate
(a) the value of , in radian,
[2 marks]
[4 marks]
[4 marks]
Rajah 10 menunjukkan sebuah semi bulatan ABC dengan pusat O dan sector ABO dengan
pusat A. Jejari bagi semi bulatan ABC dan sektor bulatan ABO ialah 8 cm.
[ Guna = 3 142 ]
Hitung
(a) nilai , dalam radian,
[2 markah]
[4 markah]
[4 markah]
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(ii)
(b)
The body mass of 500 students in a school follows a normal distribution with a mean of
52 kg and a standard deviation of 10 kg.
(i)
If a student is chosen at random, find the probability that his body mass is between
40 kg and 60 kg.
(ii) Calculate the number of students whose body mass are between 40 kg and 60 kg.
[5 marks]
(a) Dalam suatu peperiksaan, 85 % calon lulus Matematik. Jika 6 calon dipilih secara rawak,
cari kebarangkalian bahawa
(i)
(ii)
(b) Jisim badan 500 pelajar sebuah sekolah adalah mengikut taburan normal dengan min
52 kg dan sisihan piawai 10 kg.
(i) Jika seorang pelajar dipilih secara rawak, carikan kebarangkalian bahawa
jisim badannya berada di antara 40 kg dan 60 kg.
(ii) Hitung bilangan pelajar yang mempunyai jisim badan di antara 40 kg dan 60 kg.
[5 markah]
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12
Section C
Bahagian C
[ 20 marks ]
[ 20 markah ]
Answer any two questions from this section.
Jawab mana-mana dua soalan daripada bahagian ini.
12 A particle moves along a straight line from a fixed point O. Its velocity, v ms-1, is given by
v = kt 3t2 , where k is a constant and t is the time, in seconds, after leaving the point O. The
velocity of the particle is maximum when t = 2 5 s .
[Assume motion to the right is positive.]
Find
(a)
the value of k,
[2 marks]
(b)
[3 marks]
(c)
[2 marks]
(d)
[3 marks]
Suatu zarah bergerak di sepanjang suatu garis lurus dari satu titik tetap O. Halajunya, v ms-1,
diberi oleh v = kt 3t2 , dengan keadaan k ialah pemalar dan t ialah masa, dalam saat,
selepas meninggalkan titik O. Halaju zarah itu adalah maksimum pada
t = 25 s
nilai k,
[2 markah]
(b)
[3 markah]
(c)
[2 markah]
(d) jarak yang dilalui, dalam m, oleh zarah itu dalam tujuh saat pertama.
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13
6 6 cm
115
M
85
3 0 cm
Diagram 13/Rajah 13
Calculate
(a) the length, in cm, of KM,.
[2 marks]
(b)
KMN,
[2 marks]
(c)
LKM,
[3 marks]
[3 marks]
[2 markah]
(b)
KMN,
[2 markah]
(c)
LKM,
[3 markah]
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14
14 Table 14 shows the prices and the price indices of five components P, Q, R, S and T needed to
produce a certain type of digital camera.
2010
2012
110
121
110
180
120
320
400
150
250
3.05
122
200
280
Component
Komponen
Price (RM)
per unit
Harga (RM)
per unit
Table 14 / Jadual 14
P
Q
T
80
40
35
R
160
S
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15
(a)
[4 marks]
(b)
Calculate the composite index for the production cost of the camera in the year 2012
based on the year 2010.
(c)
[3 marks]
The price of each component increases by 10% from the year 2012 to the year 2014.
Given that the production cost of the camera in year 2010 is RM500, calculate the
corresponding cost in year 2014.
[3 marks]
Jadual 14 menunjukkan harga dan indeks harga bagi lima komponen P, Q, R, S dan T yang
diperlukan untuk menghasilkan sejenis kamera digital. Carta pai 14 menunjukkan kuantiti
relatif bagi komponen yang diperlukan dalam penghasilan kamera digital itu.
(a)
[4 markah]
(b)
Hitungkan indeks gubahan bagi kos penghasilan kamera digital itu pada tahun 2012
berasaskan tahun 2010.
(c)
[3 markah]
Harga setiap komponen meningkat 10% dari tahun 2012 ke tahun 2014. Diberi kos
penghasilan kamera digital itu dalam tahun 2010 ialah RM500, hitungkan kosnya yang
sepadan pada tahun 2014.
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16
The production of the pillow per day is based on the following constraints.
I. : The time taken to make pillows of type A is not more than the time taken to make
pillows of type B.
II. : The total number of pillows produced is not more than 500.
III. : The number of pillows of type B must exceed the number of pillows of type A by
at most 200.
(a)
(b)
[3 marks]
By using the scale of 2 cm to 100 pillows on both axes, construct and shade the region
R which satisfies all the above constraints.
(c)
[3 marks]
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17
Masa yang diambil untuk membuat bantal A tidak melebihi masa yang diambil
untuk membuat bantal jenis B.
II:
III:
(a)
atas.
(b)
[3 markah]
Menggunakan skala 2 cm kepada 100 biji bantal pada kedua-dua paksi, bina dan lorek
rantau R yang memenuhi semua kekangan di atas.
(c)
[3 markah]
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18
THE UPPER TAIL PROBABILITY Q(z) FOR THE NORMAL DISTRIBUTION N(0,1)
KEBARANGKALIAN HUJUNG ATAS Q(z) BAGI TABURAN NORMAL N(0, 1)
1
24
28
32
36
24
28
32
36
19
23
27
31
35
19
22
26
30
34
15
18
22
25
29
32
10
14
17
20
24
27
31
10
13
16
19
23
26
29
12
15
18
21
24
27
11
14
16
19
22
25
10
13
15
18
20
23
12
14
16
19
21
0.1170
10
12
14
16
18
0.0985
11
13
15
17
0.0838
0.0823
10
11
13
14
0.0694
0.0681
10
11
13
0.0582
0.0571
0.0559
10
11
0.0485
0..0475
0.0465
0.0455
0.0392
0.0384
0.0375
0.0367
0.0322
0.0314
0.0307
0.0301
0.0294
0.0256
0.0250
0.0244
0.0239
0.0233
0.0207
0.0202
0.0197
0.0192
0.0188
0.0183
0.0162
0.0158
0.0154
0.0150
0.0146
0.0143
0.0125
0.0122
0.0119
0.0116
0.0113
0.0110
10
13
15
18
20
23
12
14
16
16
21
11
13
15
17
19
0.0
0.5000
0.4960
0.4920
0.4880
0.4840
0.4801
0.4761
0.4721
0.4681
0.4641
12
16
20
0.1
0.4602
0.4562
0.4522
0.4483
0.4443
0.4404
0.4364
0.4325
0.4286
0.4247
12
16
20
0.2
0.4207
0.4168
0.4129
0.4090
0.4052
0.4013
0.3974
0.3936
0.3897
0.3859
12
15
0.3
0.3821
0.3783
0.3745
0.3707
0.3669
0.3632
0.3594
0.3557
0.3520
0.3483
11
15
0.4
0.3446
0.3409
0.3372
0.3336
0.3300
0.3264
0.3228
0.3192
0.3156
0.3121
11
0.5
0.3085
0.3050
0.3015
0.2981
0.2946
0.2912
0.2877
0.2843
0.2810
0.2776
0.6
0.2743
0.2709
0.2676
0.2643
0.2611
0.2578
0.2546
0.2514
0.2483
0.2451
0.7
0.2420
0.2389
0.2358
0.2327
0.2296
0.2266
0.2236
0.2206
0.2177
0.2148
0.8
0.2119
0.2090
0.2061
0.2033
0.2005
0.1977
0.1949
0.1922
0.1894
0.1867
0.9
0.1841
0.1814
0.1788
0.1762
0.1736
0.1711
0.1685
0.1660
0.1635
0.1611
1.0
0.1587
0.1562
0.1539
0.1515
0.1492
0.1469
0.1446
0.1423
0.1401
0.1379
1.1
0.1357
0.1335
0.1314
0.1292
0.1271
0.1251
0.1230
0.1210
0.1190
1.2
0.1151
0.1131
0.1112
0.1093
0.1075
0.1056
0.1038
0.1020
0.1003
1.3
0.0968
0.0951
0.0934
0.0918
0.0901
0.0885
0.0869
0.0853
1.4
0.0808
0.0793
0.0778
0.0764
0.0749
0.0735
0.0721
0.0708
1.5
0.0668
0.0655
0.0643
0.0630
0.0618
0.0606
0.0594
1.6
0.0548
0.0537
0.0526
0.0516
0.0505
0.0495
1.7
0.0446
0.0436
0.0427
0.0418
0.0409
0.0401
1.8
0.0359
0.0351
0.0344
0.0336
0.0329
1.9
0.0287
0.0281
0.0274
0.0268
0.0262
2.0
0.0228
0.0222
0.0217
0.0212
2.1
0.0179
0.0174
0.0170
0.0166
2.2
0.0139
0.0136
0.0132
0.0129
2.3
0.0107
0.0104
0.0102
0.00990
0.00964
0.00939
0.00914
0.00889
0.00866
0.00842
Minus / Tolak
2.4
0.00820
0.00798
0.00776
0.00755
0.00734
0.00714
0.00695
0.00676
0.00657
0.00639
11
13
15
17
2.5
0.00621
0.00604
0.00587
0.00570
0.00554
0.00539
0.00523
0.00508
0.00494
0.00480
11
12
14
2.6
0.00466
0.00453
0.00440
0.00427
0.00415
0.00402
0.00391
0.00379
0.00368
0.00357
10
2.7
0.00347
0.00336
0.00326
0.00317
0.00307
0.00298
0.00289
0.00280
0.00272
0.00264
2.8
0.00256
0.00248
0.00240
0.00233
0.00226
0.00219
0.00212
0.00205
0.00199
0.00193
2.9
0.00187
0.00181
0.00175
0.00169
0.00164
0.00159
0.00154
0.00149
0.00144
0.00139
3.0
0.00135
0.00131
0.00126
0.00122
0.00118
0.00114
0.00111
0.00107
0.00104
0.00100
exp z 2
2
2
1
f ( z)
f (z)
Example / Contoh:
Q(z)
Q( z ) f ( z ) dz
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HALAMAN KOSONG
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Additional
Mathematics
Paper 1
Ogos, 2014
2014
ANJURAN
MAJLIS PENGETUA SEKOLAH MALAYSIA (KEDAH)
ADDITIONAL MATHEMATICS
MARKING SCHEME
Paper 1
MODUL 2
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Answer
(a) B1
: m 3=
=
or n 2
(a)
m=3
(b)
many to one
AND
Marks
n=2
2
6
B2 :
=
p
B1 : f
6
=
f ( 3) or
12
p
p=
6
=3
p
1
2
B1 : 2 3 y =
3
(a)
g ( 3) =
(b)
B1 : 2 3( 3 k ) =
8
(b)
k =5
=
p 1=
AND
p 2
0 or
( p 1)( p 2 ) =
B2 :
=
p 1=
or p 2
B1 :
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1
3
(a)
(2 p)
4(1)( 3 p 2 )
(a) B1 : p =
3 or q =
5
(a)
p = 3 AND q = 5
(b)
x = 3
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Question
6
3472/1
Answer
Marks
B2 :
or
x=
5
2
5
2
5
2
4 + 3a + b
3
3 x
, x = 3
B1: (2 x 5)( x + 3)
log 2 k
or
or log 2 8
B2: 1 + 2 x 2 = 6
B1 : 7(7) 2( x 1) or ( 7 3 )
5
2
44
B3 : 2 + 14 ( 3)
B2 :
26
18
2 ( 2 ) + 25d
2 ( 2 ) + 17 d =
532
2
2
or
d =3
B1 :
26
18
2 ( 2 ) + 25d or
2 ( 2 ) + 17 d
2
2
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Question
10
3472/1
Answer
(a) m = 3 AND
m
=1
m3
9
(a) B1:
1 5
81 1
9
(b) B1: S5 =
1
1
9
Marks
m = 3
(b) 91.123
11
2
27
2
1
3
B1: r =
12
(a) B1: =
xy ax 2 + b
(a) 2
(b) b = 1
3 x 2 + 3 y 2 24 x 8 y + 36 =
0
k = 1
13 7* ( 2 ) + b
=
(b) B1 :
or
3 2*( 2 ) + b
=
13
B2 :
2
2
2
4 ( x 3 ) + ( y 2 ) = x 2 + ( y 4 )
B1 :
( x 3)
+ ( y 2)
or
( x 0)
+ ( y 4)
14
B1:
1
(11 4 ) + ( 3 2 ) + ( k 8 ) ( 8 3) + ( k 4 ) + (11 2 ) =
0
2
OR
84
42
=
11 3 3 k
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Question
15
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Answer
(a) 3i + 4 j
(b)
32 + 4 2
B1 : 5 or
(b)
16
B3 : =
6 + 2m 0 OR
3i + 4 j
Marks
1
12
=
+ 4m 0
3 3 m 6 0
B2 : + + =
8 4 3 12 0
m = 3
=
x 41 81
AND
6
B1 : OB =
12
17
B3 : x =
41 81 or
x=
138 19
6sin 2 x 8sin x + 8 =
0 or
B2 :
2
sin x =
2 or sin x =
3
=
x 138 19
B1 : 3 1 2sin 2 x =
8sin x 5
18
(a) 1.047rad
(b) B2:
B1:
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1
1
2
(12 ) * (1.047 ) ( 6 )(10.392 )
2
2
(b) 44.21
1
1
2
(12 ) * (1.047 ) or ( 6 )(10.392 )
2
2
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Question
19
3472/1
Answer
(a) 70
(b) B1:
20
70 + k
8
(b) k = 18
= 11
Marks
= 24 8
B2 : h (1)
16
k =3
10
B1 : h ( x ) = 12 x 8 x + 1
2
21
B2 : 2k 8 = 2 or
x =3
B1 : 2 x 8
22
2
1 ( 3 x + 1)
1
B2 :
49 ( 1)
or
5 x 1 0
5
2
B1 :
h ( x ) dx =
( 3x + 1) 2
x 1
23
(b) B1:
2 3 2
3 5
+ or 1
5 5 7
5 7
(a)
2
5
(b)
4
7
24
(b) B1 :
C 3 3C 2 5C 3
(a)
495
(b)
120
25
B2 : p( z>k ) = 0.0608
k = 1.548
SULIT
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Nama Pelajar :
Tingkatan 5 : .
3472/2
Additional
Mathematics
August 2014
ADDITIONAL MATHEMATICS
Paper 2
( MODUL 2 )
MARKING SCHEME
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SULIT
3472/2
MARKING SCHEME
ADDITIONAL MATHEMATICS PAPER 2 2014
N0.
1
SOLUTION
x 2y 1
or
MARKS
x 1
2
P1
x2 x2 50
x 2 25
x 5 x 5
K1 Eliminate x/y
K1 Solve quadratic equation
x5
and
x 5
(both)
y2
and
y 3
(both)
N1
N1
5
2
(a)
(b)
(i)
T8 (1)(2)7
K1
128
N1
(1)(210 1)
S10
2 1
1023
K1
V 1023 (3)(7)(5)
K1
N1
107415
(ii)
1023 0.8
K1
N1
818.4
7
3
(a)
P1 cos
shape correct.
P1 Amplitude = 2 [ Maximum = 1
and Minimum = -1 ]
1
P1 1
cycle in 0 x or
2
(b)
-4
y=
N1 For equation
= 3
6
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4
(a)
10
K1
10
x 100
N1
(b)
K1
10
10
x 2 1160
4
10 3
2
6.5
mean
N1
4
2
2
2 4
or
K1
N1
N1
log 5 K log125 V 1
5
(a)
K1
log 5 V
1
3
log 5 K 3 log 5 V 3
log 5 K
K1
K3
log 5
3
V
K3
V
125
(b)
f 1 ( x)
i)
N1
xk
2m
K1
1
1
k
3
2m 8
2m
m 4 and k 24
K1
N1
ii)
1
1
( p) 3
8
2
p 20
K1
N1
8
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6
(a)
(b)
(3x 1)(3x 1)
3x 1
3x 1
f '( x) 3
f ( x)
K1
N1
i)
dy
kx 2 3x
dx
2 k (2) 2 3(2)
k 2
K1
N1
ii)
mnormal
1
2
1
12 (2) c
2
1
y x 11
2
P1
K1
N1
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7
(a)
x
y
x2
(b)
3.5
5.5
7.5
9.5
11.5
13.5
N1 6 correct
values of
y
x2
K1 Plot
y
x2
y
vs x.
x2
N1 6 points plotted
correctly
1.5
N1 Line of best-fit
x
0
(c)
(i)
(ii)
(iii)
P1
y
p
= kx+
2
x
k
k = *gradient
k = 2.0
K1
N1
p
= *y-intercept
k
p = 3.0
K1
N1
y = 40
N1
10
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N0.
SOLUTION
MARKS
8(a)
TS
2
(9 x)
3
= 6x
K1 ( TS or QR )
QR
i)
PT
2
= 4y
TR TP PQ QR
8 y 9 x 4 y
9x 4 y
ii)
N1
PS PT TS
K1
N1
6x 8 y
MS MR RS
iii)
TR
3 x 4 y
2
9x 4 y
3x 4 y
2
3x
2y
2
K1
N1
PS k MS
PS PT TS
b)
6x 8 y
6x 8 y
3
2
= k ( x 2 y)
k =4
K1
1 3
PS 6( ) 8( )
2 4
N1
K1
N1
45
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9
a)
x 2 3x 4
x 2 3x 4 0
( x 1)( x 4) 0
K1
for
quad.eqn.
N1
N1
x 4, y 16
K (4,16)
solving
b)
y4
x2
2
Area B (4)(2) x dx
2
K1 use area of
rectangle -
x
8
3 0
3
K1
23
8
3
16 2
cm
3
( y) dx
integrate
correctly
and Sub.
the limit
correctly
N1
c)
Volume A
16
16
y2
64
2 4
16 4
64
2
2
256 16
64
2
2
56
2
K1
K1 correct limit
K1 integrate
correctly
N1
10
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N0.
10
(a) 60o
1.047 rad
SOLUTION
MARKS
P1
N1
(b)
SOB 8(1.047)
SBC 8(2.095)
or
= 8.38
S AC 8(3.142)
= 16.76
Perimeter = 8.38+16.76+8
or
= 25.14
Perimeter = 25.14 + 8
= 33.14
(c)
OR
= 33.14
1 2
(8) (1.047)
2
= 33.50 cm2
Area of OAB =
1 2
(8) sin 60
2
= 27.71
= 5.79 cm2
K1 Use s r
N1
K1
N1
K1 Use formula
A
1 2
r
2
K1
K1
N1
10
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N0.
SOLUTION
11
(a) X= Students passed Mathematics
p = 0.85 , q = 1- 0.85 = 0.15
,n=6
(i)
P(X =6) =
MARKS
P1
K1 Use
c6 (0.85)6 (0.15)0
=0.3772
(ii)
P ( X=r ) =
Cr p r q nr
N1
K1
c1 (0.15)1 (0.85)5
c0 (0.15)0 (0.85)6
=0.2235
N1
(b)
(i)
= 52
, =10
P( 40 < X < 60 ) = P (
40 52
< Z <
10
60 52
10
K1 Use Z =
K1
= 0.6731
N1
(ii)
n = 0.6731 x 500
K1
n = 337
N1
10
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N0.
12
(a)
SOLUTION
MARKS
a k 6t
K1
k 6(2.5) 0
k = 15
(b)
N1
15 2 3
t t
2
K1
15 2 3
t t 0
2
K1
t = 7.5 s
N1
15t 3t 2 0
K1
t=5
N1
(c)
(d)
Total distance
5
7
15t 2 3 15t 2 3
t
t
=
2
0 2
5
K1 (for
Integration;
either one
and
substitute
the limit
5
or )
d = 62.5 + 38
K1
(for use and
summation)
= 100.5
N1
10
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10
N0.
13
(a)
SOLUTION
(i)
MARKS
KI (PetuaKosinus)
N1
(ii)
K1 (Petua Sinus)
N1
(iii)
K1 (GunaPetua Sinus)
K1
N1
(b)
(i)
K1
K1
N1
10
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11
N0.
14
(a)
SOLUTION
MARKS
K1
(i)
N1
i)
K1
N1
(b)
Lihat 45
K1
K1
I2012/2010 =
= 124.64
(c)
N1
I2014/2010 = 124.64
K1
= 137.10
Q2014
=
= RM685.50
K1
N1
10
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12
N0.
15
(a)
SOLUTION
MARKS
i)
ii)
iii)
(b)
N1
N1
N1
Type B
650
600
550
500
450
400
350
300
250
(300, 200)
200
150
100
50
-400
-300
-200
-100
100
-50
200
300
400
500
220
600
700
800
900
1000
1100
1200
Type A
-100
i) BilanganmaksimumbantalA = 220
N1
ii)
P1
(c)
Titikmaksimum (300, 200)
Keuntunganmaksimum;
K1
k = RM8 400
N1
10
END OF MARKING SCHEME
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13