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SULIT 3472/1
1
PEPERIKSAAN PERCUBAAN SPM TAHUN 2009 3472/1 ADDITIONAL MATHEMATICS Kertas 1 September 2009 2 jam Dua jam
JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU
1. Kertas soalan ini adalah dalam
dwibahasa. 2. Soalan dalam bahasa Inggeris
mendahului soalan yang sepadan dalam bahasa Malaysia.
3. Calon dibenarkan menjawab keseluruhan
atau sebahagian soalan dalam bahasa Inggeris atau bahasa Malaysia.
4. Calon dikehendaki membaca maklumat di
halaman belakang kertas soalan ini.
Untuk Kegunaan Pemeriksa Kod Pemeriksa:
Soalan Markah Penuh
Markah Diperoleh
1 2 2 4 3 3 4 3 5 3 6 4 7 3 8 4 9 3 10 3 11 2 12 4 13 4 14 2 15 4 16 4 17 3 18 3 19 3 20 3 21 4 22 3 23 3 24 3 25 3
Jumlah 80
Kertas soalan ini mengandungi 20 halaman bercetak.
[Lihat sebelah 3472/1 SULIT
NAMA
ANGKA GILIRAN
SULIT 3472/1
2
INFORMATION FOR CANDIDATES MAKLUMAT UNTUK CALON
1. This question paper consists of 25 questions. Kertas soalan ini mengandungi 25 soalan. 2. Answer all questions. Jawab semua soalan. 3. Give only one answer for each question. Bagi setiap soalan beri satu jawapan sahaja. 4 Write your answers in the spaces provided in this question paper. Jawapan anda hendaklah ditulis pada ruang yang disediakan dalam kertas soalan ini. 5. Show your working. It may help you to get marks. Tunjukkan langkah-langkah penting dalam kerja mengira anda. Ini boleh membantu anda untuk mendapatkan
markah. 6. If you wish to change your answer, cross out the answer that you have done. Then write down the new answer. Jika anda hendak menukar jawapan, batalkan dengan kemas jawapan yang telah dibuat. Kemudian tulis
jawapan yang baru. 7. The diagrams in the questions provided are not drawn to scale unless stated. Rajah yang mengiringi soalan tidak dilukis mengikut skala kecuali dinyatakan. 8. The marks allocated for each question are shown in brackets. Markah yang diperuntukkan bagi setiap soalan ditunjukkan dalam kurungan. 9. A list of formulae is provided on pages 3 to 5. Satu senarai rumus disediakan di halaman 3 hingga 5. 10. A four-figure table for the Normal Distribution N(0, 1) is provided on page 2. Satu jadual empat angka bagi Taburan Normal N(0, 1) disediakan di halaman 2. 11. You may use a non-programmable scientific calculator. Anda dibenarkan menggunakan kalkulator saintifik yang tidak boleh diprogram. 12. Hand in this question paper to the invigilator at the end of the examination. Serahkan kertas soalan ini kepada pengawas peperiksaan pada akhir peperiksaan.
SULIT 3472/1
3
THE UPPER TAIL PROBABILITY Q(z) FOR THE NORMAL DISTRIBUTION N(0, 1) KEBARANGKALIAN HUJUNG ATAS Q(z) BAGI TABURAN NORMAL N(0, 1)
z 0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9
Minus / Tolak
0.0
0.1
0.2
0.3
0.4
0.5000
0.4602
0.4207
0.3821
0.3446
0.4960
0.4562
0.4168
0.3783
0.3409
0.4920
0.4522
0.4129
0.3745
0.3372
0.4880
0.4483
0.4090
0.3707
0.3336
0.4840
0.4443
0.4052
0.3669
0.3300
0.4801
0.4404
0.4013
0.3632
0.3264
0.4761
0.4364
0.3974
0.3594
0.3228
0.4721
0.4325
0.3936
0.3557
0.3192
0.4681
0.4286
0.3897
0.3520
0.3156
0.4641
0.4247
0.3859
0.3483
0.3121
4
4
4
4
4
8
8
8
7
7
12
12
12
11
11
16
16
15
15
15
20
20
19
19
18
24
24
23
22
22
28
28
27
26
25
32
32
31
30
29
36
36
35
34
32
0.5
0.6
0.7
0.8
0.9
0.3085
0.2743
0.2420
0.2119
0.1841
0.3050
0.2709
0.2389
0.2090
0.1814
0.3015
0.2676
0.2358
0.2061
0.1788
0.2981
0.2643
0.2327
0.2033
0.1762
0.2946
0.2611
0.2296
0.2005
0.1736
0.2912
0.2578
0.2266
0.1977
0.1711
0.2877
0.2546
0.2236
0.1949
0.1685
0.2843
0.2514
0.2206
0.1922
0.1660
0.2810
0.2483
0.2177
0.1894
0.1635
0.2776
0.2451
0.2148
0.1867
0.1611
3
3
3
3
3
7
7
6
5
5
10
10
9
8
8
14
13
12
11
10
17
16
15
14
13
20
19
18
16
15
24
23
21
19
18
27
26
24
22
20
31
29
27
25
23
1.0
1.1
1.2
1.3
1.4
0.1587
0.1357
0.1151
0.0968
0.0808
0.1562
0.1335
0.1131
0.0951
0.0793
0.1539
0.1314
0.1112
0.0934
0.0778
0.1515
0.1292
0.1093
0.0918
0.0764
0.1492
0.1271
0.1075
0.0901
0.0749
0.1469
0.1251
0.1056
0.0885
0.0735
0.1446
0.1230
0.1038
0.0869
0.0721
0.1423
0.1210
0.1020
0.0853
0.0708
0.1401
0.1190
0.1003
0.0838
0.0694
0.1379
0.1170
0.0985
0.0823
0.0681
2
2
2
2
1
5
4
4
3
3
7
6
6
5
4
9
8
7
6
6
12
10
9
8
7
14
12
11
10
8
16
14
13
11
10
19
16
15
13
11
21
18
17
14
13
1.5
1.6
1.7
1.8
1.9
0.0668
0.0548
0.0446
0.0359
0.0287
0.0655
0.0537
0.0436
0.0351
0.0281
0.0643
0.0526
0.0427
0.0344
0.0274
0.0630
0.0516
0.0418
0.0336
0.0268
0.0618
0.0505
0.0409
0.0329
0.0262
0.0606
0.0495
0.0401
0.0322
0.0256
0.0594
0.0485
0.0392
0.0314
0.0250
0.0582
0..0475
0.0384
0.0307
0.0244
0.0571
0.0465
0.0375
0.0301
0.0239
0.0559
0.0455
0.0367
0.0294
0.0233
1
1
1
1
1
2
2
2
1
1
4
3
3
2
2
5
4
4
3
2
6
5
4
4
3
7
6
5
4
4
8
7
6
5
4
10
8
7
6
5
11
9
8
6
5
2.0
2.1
2.2
2.3
0.0228
0.0179
0.0139
0.0107
0.0222
0.0174
0.0136
0.0104
0.0217
0.0170
0.0132
0.0102
0.0212
0.0166
0.0129
0.00990
0.0207
0.0162
0.0125
0.00964
0.0202
0.0158
0.0122
0.00939
0.0197
0.0154
0.0119
0.00914
0.0192
0.0150
0.0116
0.00889
0.0188
0.0146
0.0113
0.00866
0.0183
0.0143
0.0110
0.00842
0
0
0
0
3
2
1
1
1
1
5
5
1
1
1
1
8
7
2
2
1
1
10
9
2
2
2
1
13
12
3
2
2
2
15
14
3
3
2
2
18
16
4
3
3
2
20
16
4
4
3
2
23
21
2.4 0.00820 0.00798 0.00776 0.00755 0.00734
0.00714
0.00695
0.00676
0.00657
0.00639
2
2
4
4
6
6
8
7
11
9
13
11
15
13
17
15
19
17
2.5
2.6
2.7
2.8
2.9
0.00621
0.00466
0.00347
0.00256
0.00187
0.00604
0.00453
0.00336
0.00248
0.00181
0.00587
0.00440
0.00326
0.00240
0.00175
0.00570
0.00427
0.00317
0.00233
0.00169
0.00554
0.00415
0.00307
0.00226
0.00164
0.00539
0.00402
0.00298
0.00219
0.00159
0.00523
0.00391
0.00289
0.00212
0.00154
0.00508
0.00379
0.00280
0.00205
0.00149
0.00494
0.00368
0.00272
0.00199
0.00144
0.00480
0.00357
0.00264
0.00193
0.00139
2
1
1
1
0
3
2
2
1
1
5
3
3
2
1
6
5
4
3
2
8
6
5
4
2
9
7
6
4
3
11
9
7
5
3
12
9
8
6
4
14
10
9
6
4
3.0 0.00135 0.00131 0.00126 0.00122 0.00118 0.00114 0.00111 0.00107 0.00104 0.00100 0 1 1 2 2 2 3 3 4
Example / Contoh:
−= 2
2
1exp
2
1)( zzf
π If X ~ N(0, 1), then
Jika X ~ N(0, 1), maka
∫∞
=k
dzzfzQ )()( P(X > k) = Q(k)
P(X > 2.1) = Q(2.1) = 0.0179
Q(z)
z
f (z)
O k
SULIT 3472/1
4
SULIT 3 3472/1 The following formulae may be helpful in answering the questions. The symbols given are the ones commonly used.
Rumus-rumus berikut boleh membantu anda menjawab soalan. Simbol-simbol yang diberi adalah yang biasa digunakan.
ALGEBRA
1 a
acbbx
2
42 −±−= 8 c
bb
c
ca log
loglog =
2 nmnm aaa +=× 9 dnaTn )1( −+=
3 nmnm aaa −=÷ 10 ])1(2[2
dnan
Sn −+=
4 mnnm aa =)( 11 1−= n
n arT
5 nmmn aaa logloglog += 12 1,1
)1(
1
)1( ≠−−=
−−= r
r
ra
r
raS
nn
n
6 nmn
maaa logloglog −= 13 1,
1<
−=∞ r
r
aS
7 mnm a
na loglog =
CALCULUS / KALKULUS
1 dx
duv
dx
dvu
dx
dyuvy +== , 4 Area under a curve
Luas di bawah lengkung
2
, dx
dy
v
uy =
3
dx
du
du
dy
dx
dy ×=
5 Volume generated
SULIT 3472/1
5
STATISTICS / STATISTIK
1 N
xx ∑= 8
)!(
!
rn
nPr
n
−=
2 ∑∑=
f
fxx 9
!)!(
!
rrn
nCr
n
−=
3 2
22)(x
N
x
N
xx−=
−= ∑∑σ 10 )()()()( BAPBPAPBAP ∪−+=∪
4 2
22)(x
f
fx
f
xxf−=
−=
∑∑
∑∑σ 11 1 , )( =+== − qpqpCrXP rnr
rn
5 Cf
FNLm
m
−+= 2
1
12 Mean / Min , np=µ
6 1000
1 ×=Q
QI 13 npq=σ
7 ∑∑=
i
ii
W
IWI 14
σµ−= X
Z
GEOMETRY / GEOMETRI
1 Distance / Jarak 5 22 yxr +=
212
212 )()( yyxx −+−=
2 Midpoint / Titik tengah 6 22
^^
yx
jyixr
+
+=
++=
2,
2),( 2121 yyxx
yx
3 A point dividing a segment of a line Titik yang membahagi suatu tembereng garis
++
++
=nm
myny
nm
mxnxyx 2121 ,),(
4 Area of triangle / Luas segitiga
( ) ( )312312132212
1yxyxyxyxyxyx ++−++=
SULIT 3472/1
6
5
TRIGONOMETRY / TRIGONOMETRI 1 Arc length, θrs = 8 BABABA sincoscossin)sin( ±=± Panjang lengkok, θjs = BABABA sin kos kos sin)sin( ±=±
2 Area of sector, θ2
2
1rA = 9 BABABA sinsincoscos)cos( ∓=±
Luas sector, θ2
2
1jL = BABABA sinsinkos kos)( kos ∓=±
3 1cossin 22 =+ AA 10 BA
BABA
tantan1
tantan)tan(∓
±=±
1kossin 22 =+ AA
4 AA 22 tan1sec += 11 A
AA
2tan1
tan22tan
−=
AA 22 tan1sek +=
5 AA 22 cot1cosec += 12 C
c
B
b
A
a
sinsinsin==
AA 22 kot1kosek += 6 AAA cossin22sin = 13 Abccba cos2222 −+= AAA kos sin22sin = Abccba kos 2222 −+= 7 AAA 22 sincos2cos −= 14 Area of triangle / Luas segitiga
1cos2 2 −= A cabsin2
1=
A2sin21−= AAA 22 sinkos2kos −= 1kos 2 2 −= A A2sin21−=
SULIT 3472/1
7
For Examiner’s
Use
Answer all questions. Jawab semua soalan.
1 Diagram 1 shows the relation between set P and set Q. Rajah 1 menunjukkan hubungan antara set P dan set Q. State Nyatakan (a) the type of relation between set P and set Q. jenis hubungan antara set P adan set Q.
(b) the value of w if 12
: +→ xxf .
nilai bagi w jika 12
: +→ xxf .
[ 2 marks] [2 markah] Answer / Jawapan: (a) ……………………...………… (b) w =………………………….…
Diagram 1 Rajah 1
x f
12
6
2
10
.
.
.
. w
4
2
6
.
.
.
.
y
Set P Set Q
1
2
SULIT 3472/1
8
2 Given the function )1(3: −→ xxg , find Diberi fungsi )1(3: −→ xxg , cari (a) the value of g 2 (4), [2 marks] nilai bagi g 2 (4), [2 markah] [2 markah] (b) the function of f if gf(x) = 6x. [ 2 marks] fungsi f jika gf(x) = 6x. [2 markah] [2 markah]
Answer / Jawapan: (a) ……………………..……... (b) ……………………………. ______________________________________________________________________ 3 The quadratic equation x2 - (3 – p)x + p - 3 = 0, where p is constant, has two equal
roots. Find the possible values of p. [3 marks]
Persamaan kuadratik x2 - (3 – p)x + p - 3 = 0, dengan keadaan p ialah pemalar, mempunyai dua punca sama. Cari nilai-nilai p yang mungkin.
[3 marks] [3 markah] Answer / Jawapan: p = …………………………...…... [Lihat sebelah
For Examiner’s
Use
2
4
3
3
For Examiner’s
Use
4 Find the range of values of x for which .8)2( 2 xx −≤− [3 marks]
Cari julat nilai x bagi .8)2( 2 xx −≤− [3 markah] Answer / Jawapan: ……………………………...…...….. 5 Given that 26log3log 44 =− xy , express y in terms of x. [3 marks]
Diberi 26log3log 44 =− xy , ungkapkan y dalam sebutan x. [3 markah] Answer / Jawapan: ……………………………...…...…..
5
3
4
4
SULIT 3472/1
10
6 Given that ,log4 px = qy =32log and nqmp
y
x += 2 , find the value of m and n.
[4 marks]
Diberi ,log4 px = qy =32log dan nqmp
y
x += 2 , cari nilai bagi m dan n. [4 markah]
Answer / Jawapan: m = ……….. n = ………… ______________________________________________________________________
7 A piece of string of length 12 m is cut into 20 pieces in such a way that the lengths of the pieces are in arithmetic progressions. If the length of the longest piece is five times of the length of the shortest piece, find the length of the longest piece. [3 marks]
Seutas dawai yang panjangnya 12 m dipotong kepada 20 keratan dengan keadaan ukuran keratan membentuk satu janjang aritmetik. Jika ukuran keratan terpanjang ialah lima kali keratan terpendek, cari ukuran keratan terpanjang. [3 markah]
Answer / Jawapan: ……....……………………………. [Lihat sebelah
For Examiner’s
Use
7
3
6
4
SULIT 3472/1
11
For Examiner’s
Use
8 The fourth and seventh terms of a geometric progression are 18 and 486 respectively. Find the third term. [4 marks] Sebutan keempat dan ketujuh bagi satu janjang geometri masing-masing ialah 18 dan 486. Cari sebutan ketiga. [4 markah] Answer / Jawapan: ..……………………...…...………… ________________________________________________________________________ 9 Point P(h, 7) divides line the segment joining the points E(3, 10) and F(8, k) internally such that EP : PF = 1: 4. Find the values of h and k. [3 marks]
Titik P(h, 7) membahagi dalam tembereng garis yang menyambungkan titik E(3,10) dan F(8, k) dengan keadaan EP : PF =1 : 4. Cari nilai bagi h dan k. [3 markah]
9
3
8
4
SULIT 3472/1
12
Answer / Jawapan: h = ………… k = …………. 10 Solution to this question by scale drawing will not be accepted. Penyelesaian secara lukisan berskala tidak diterima. In Diagram 2, OABC is a quadrilateral. The equation of the straight line AB is
146
=+ yx.
Dalam Rajah 2, OABC adalah sebuah sisiempat. Persamaan bagi garis lurus AB
ialah 146
=+ yx.
Find the area of the quadrilateral OABC. [3 marks] Cari luas bagi sisempat OABC. [3 markah]
Answer / Jawapan: .......………………………………. Lihat sebelah
10
3
y
O
A
B
C .
.
(2, -3)
.
Diagram 2 Rajah 2
(9, -2) x
10
3
SULIT 3472/1
13
For Examiner’s
Use
11 Given ~~~
64 jia −= , ~~~jib −= and
~~~ 21
bac −= , express ~c in the form
~~jyix + .
[2 marks]
Diberi ~~~
64 jia −= , ~~~jib −= dan
~~~ 21
bac −= , ungkapkan ~c dalam bentuk
~~jyix +
. [2 markah] Answer / Jawapan:
~c = …………………..…..
________________________________________________________________________
12 The vector OF has a magnitude of 10 unit and has the same direction as OE .
Given that
−=
4
3OE and
=
y
xOF , find the value of x and y. [3 marks]
Vector OF mempunyai magnitude 10 unit dan mempunyai arah yang sama dengan
OE . Diberi
−=
4
3OE dan
=
y
xOF , cari nilai x dan nilai y. [3 markah]
Answer / Jawapan: x =………… y = …………..
12
4
11
2
SULIT 3472/1
14
13 Diagram 3 shows a triangle ABC. Rajah 3 menunjukkan sebuah segitiga ABC.
The point E is the midpoint of AC and D lies on the line BC such that BC = 5DC.
Given ~xAB = and
~5yBC = , express in term of
~x and
~y ,
Titik E ialah titik tengah bagi AC dan D terletak pada garis BC dengan
keadaan BC = 5DC. Diberi ~xAB = dan
~5yBC = , ungkapkan dalam sebutan
~x dan
~y ,
(a) AD ,
(b) DE . [4 marks]
[4 markah]
Answer / Jawapan: (a) .……….………………..….. (b) .……….………………..…..
14 Find the value of 34
3lim 2
2
2 +−−
→ xx
xxx
. [2 marks]
Cari nilai bagi 34
32
2
2 +−−
→ xx
xxhadx
. [2 markah]
Answer / Jawapan: …….………………………..…….. [Lihat sebelah SULIT
For Examiner’s
Use
14
3
Diagram 3 Rajah 3
13
4
A
B C D
E .
.
SULIT 3472/1
15
15 Volume, V cm3, of a solid is given by 32
32
8 rrV ππ += , r is the radius. Find the
approximate change in V if r increases from 3 cm to 3.005 cm. (Give your answers in terms of π ).
Isipadu, V cm3, bagi sebuah pepejal diberi oleh 32
32
8 rrV ππ += , r ialah jejari.
Cari perubahan hampir bagi V jika r bertambah daripada 3 cm kepada 3.005 cm. (Beri jawapan anda dalam sebutan π ). [4 marks] [4 markah] Answer / Jawapan : ……………………………… ________________________________________________________________________ 16 Diagram 4 shows part of a straight line graph drawn to represent linear form of the
equationx
y625= .
Rajah 4 menunjukkan sebahagian daripada graph garis lurus yang dilukis untuk
mewakili bentuk linear bagi persamaan x
y625= .
Find the values of h and k. [4 marks] Cari nilai bagi h dan k. [4 markah] Answer / Jawapan: h = …………… k = ……………..
16
4
Diagram 4 Rajah 4
log5 y
log5 x O
P (0, h)
Q (k, 1)
15
4
SULIT 3472/1
16
17 Find the value of dxx
xx∫−
−+1
1 4
)6)(6(. [3 marks]
Cari nilai dxx
xx∫−
−+1
1 4
)6)(6(. [3 markah]
Answer / Jawapan: …..……………………………….
18 The gradient function of a curve passing through (1, 2) is given by 2)43(
1−x
.
Find the equation of the curve. [3 marks]
Fungsi kecerunan suatu lengkung yang melalui (1, 2) diberi oleh 2)43(
1−x
. Cari
persamaan lengkung itu. [3 markah] Answer / Jawapan: ……………………………...…...….. [Lihat sebelah
For Examiner’s
Use
18
3
17
3
SULIT 3472/1
17
For Examiner’s
Use
19 In Diagram 5, OAC is a right-angled triangle and OAB is a sector of a circle with
centre A. Dalam Rajah 5, OAC ialah sebuah segitiga tegak dan OAB ialah sebuah sektor bulatan berpusat A. Given that OA = 6 cm , OC = 8 cm and radOAB 927.0=∠ , find the area of the shaded region. [3 marks] Diberi OA = 6 cm, OC = 8 cm dan radOAB 927.0=∠ , cari luas rantau berlorek. [3 markah] Answer / Jawapan: …………………………..….. ________________________________________________________________________ 20 Given that h=070cos and k=035sin , express in terms of h and/or k
(a) 0140cos ,
(b) 0105sin [3 marks]
Given that h=070cos and k=035sin , express in terms of h and/or k, (a) cos1400, (b) sin 1050. [3 markah] Answer / Jawapan: (a) ………………………… (b) …………………………
20
3
19
3
Diagram 5 Rajah 5
B
C
A
8 cm O
0.927 rad. 6 cm
SULIT 3472/1
18
21 Solve the equation 02tan4sec3 2 =−− xx for 00 3600 ≤≤ x . [4 marks]
Selesaikan persamaan 02tan43 2 =−− xxsek bagi 00 3600 ≤≤ x [4 markah] Answer / Jawapan: ……….………………………..….. 22 Five boys and four girls are to stand in a line. Calculate the number of possible
arrangements if (a) there is no restriction, (b) no two boys are to stand beside each other. [3 marks]
Lima orang lelaki dan empat orang perempuan berdiri pada satu baris. Kira bilangan susunan yang mungkin jika
(a) tiada syarat yang dikenakan, (b) tiada dua orang lelaki yang berdiri sebelah menyebelah. [3 markah]
Answer / Jawapan: (a) …………………………………. (b) ………………………………… [Lihat sebelah
For Examiner’s
Use
22
3
21
4
SULIT 3472/1
19
23 Lee will play against players E, F and G in a badminton competition. The
probabilities that Lee will beat E, F and G are 43
,65
and 32
respectively. Calculate
the probability that Lee will beat at least two of the three players. [3 marks]
Lee akan berlawan dengan pemain E, F dan G dalam satu pertandingan badminton. Kebarangkalian bahawa Lee akan mengalahkan E, F dan G masing-
masing ialah 43
,65
dan 32
. Hitungkan kebarangkalian bahawa Lee akan
mengalahkan sekurang-kurang dua daripada tiga orang pemain. [3 markah] Answer / Jawapan : ……………………………… ________________________________________________________________________ 24 In an examination, 40 % of the students passed. If a sample of 10 students is randomly selected, find the probability that less than 2 students passed. [3 marks] Dalam satu peperiksaan,didapati 40 % daripada pelajar lulus. Jika satu sampel 10 orang pelajar dipilih secara rawak, cari kebarangkalian bahawa kurang daripada 2 orang pelajar lulus. [3 makah]
Answer / Jawapan: …….……………………...…...…..
24
3
23
3
SULIT 3472/1
20
25 Diagram 7 shows a standardised normal distribution graph. Rajah 7 menunjukkan satu graf taburan normal piawai. Given that the area of the shaded region is 30.5 % of the total area under the curve, find Diberi bahawa luas rantau berlorek ialah 30.5 % daripada keseluruhan luas rantau dibawah lengkung, cari (a) )( kzP < ,
(b) the value of k, cari nilai k.
[3 marks] [3 markah] Answer / Jawapan: (a) ……………………..…….. (b) …………………………….
END OF QUESTION PAPER KERTAS SOALAN TAMAT
For Examiner’s
Use
25
3
Diagram 7 Rajah 7
z k O
f (z)
SULIT 3472/1
21
PEPERIKSAAN PERCUBAAN SPM TAHUN 2009
ADDITIONAL MATHEMATICS
KERTAS 1
PERATURAN PEMARKAHAN
UNTUK KEGUNAAN PEMERIKSA SAHAJA
SULIT 3472/1 Additional Mathematics Kertas 1 Peraturan Pemarkahan August/September 2009
SULIT 3472/1
22
Question Working / Solution Marks Total
1 (a)
1 (b)
One- to- one or 1- to- 1 or 1-1 w = 7
1 1
2
2 (a)
2(b)
24 g (4) = 9
12 +x
xxf 6)1)((3 =−
2 B1
2
B1
4
3 73 and
0)7)(3( =−− pp or equivalent
0)3)(1(4)]3([ 2 =−−−− pp
3 B2 B1
3
4
41 ≤≤− x or Must indicate the range correctly by shading or other ways
0)4)(1( ≤−+ xx or 0)1)(4( ≥++− xx
0432 ≤−− xx or 0432 ≥++− xx
4
B3
B2
B1
3
1− 4x
1− 4 x
SULIT 3472/1
23
5 xy 32=
24
63 =
x
y
263
log4 =x
y
3
B2
B1
3
6 m = 2 and n = -5 m = 2 or n = -5
qp 522 −
p4 or q32
4
B3
B2
B1
4
7 1 m or 100 cm
)]194
(192[220
12a
a += or ]5[220
12 aa += or
)]95
4(192[
220
12−
+= aa or ]
51
[220
12 aa +=
194a
d = or aa 5+ or 95
4ad −= or aa
5
1+
3
B2
B1
3
8
6
2)3(27
18
r =3 and 27
18=a
183 =ar and 4866 =ar
4
B3
B2
B1
4
SULIT 3472/1
24
9 h = 4 and k = -5
h = 4 or k = -5
41)8(1)3(4
++
or 41
)(1)10(4++ k
3
B2
B1
3
10 29.5
)...3(0[21 −× or equivalent
0
0
4
0
2
9
3
2
0
0
21
−− or other correct arrangement
3
B2
B1
3
11 ~~
2 ji −
)()64(21
~~~~jiji −−−
2
B1
2
12 x = 6 and y = -8
−×=
4
3
51
10y
x
−=
4
3
51
OE
OE or
OE
OEOF 10=
3
B2
B1
4
13 (a)
(b)
~~4 yx +
~~ 2
3
2
1yx−−
)5(21
~~~xyyCE −−+=
~~5 xyCA −−= or CADCDE
21+=
1 3
B2
B1
4
14 2
1lim
2 −→ x
xx
2
B1
2
SULIT 3472/1
25
15 π33.0
005.0))3(2)3(16( 2 ×+ ππ
dr
dv 2216 rr ππ +=
rπ16 or 22 rπ or 005.0=rδ
4
B3
B2
B1
4
16 h = 4 and k = 3
h = 4 or 10
14 −=−−
k
10
1 −=−−
k
h or 625log5=h
xy 555 log625loglog −=
4
B3
B2
B1
4
17 -22
−+
−−−
+−)1(
1)1(
1211
112
3
1
1
13
1336
−
−−
−−
−xx
3
B2
B1
3
18
35
)43(31 +−
−=x
y or equivalent
cx
y +−
−=)43(3
1 or equivalent
)1(3)43( 1
−− −x
or equivalent
3
B2
B1
3
SULIT 3472/1
26
19 7.314 cm2
927.0621
8621 2 ××−×× or equivalent
927.0621 2 ×× or equivalent
3
B2
B1
3
20(a)
(b)
12 2 −h
22 11 khhk −+− or equivalent
0000 35sin70cos35cos70sin +
1
2 B1
3
21 18.43o(180 26’) , 45o, 198.43o (198o 26’), 225o 18.43o and 45o 0)1)(tan1tan3( =−− xx
02tan4)1(tan3 2 =−−+ xx
4
B3
B2
B1
4
22 (a)
(b)
362880 2880
!4!5 × or 55
44 PP × or 112233445
1 2
B1
3
23
7261
+××32
43
65 +××
31
43
65 +××
32
41
65
32
43
61 ××
3
2
4
3
6
5 ×× or 31
43
65 ×× or
32
41
65 ×× or
32
43
61 ××
3
B2
B1
3
SULIT 3472/1
27
24
0.04636 10C0 (0.4)0(0.6)10 – 10C1(0.4)1(0.6)9 10C0 (0.4)0(0.6)10 or 10C1 (0.4)1(0.6)9 or P(x=0)-P(x=1)
3
B2
B1
3
25 (a)
25(b)
0.805 0.86 0.195
1
2
B1
3
SULIT 3472/1
28
SULIT 3472/1
29
SULIT 3472/1
30
SULIT 3472/1
31
NAMA DAN LOGO
SEKOLAH
PEPERIKSAAN PERCUBAAN SPM TAHUN 2009
ADDITIONAL MATHEMATICS
Paper 2 Two hours and thirty minutes
JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU
1. Kertas soalan ini adalah dalam dwibahasa.
2. Soalan dalam bahasa Inggeris mendahului soalan yang sepadan dalam Bahasa
Malaysia.
3. Calon dikehendaki membaca maklumat di halaman belakang kertas soalan ini.
4. Calon dikehendaki menceraikan halaman 18 dan ikat sebagai muka hadapan
bersama-sama dengan buku jawapan.
Kertas soalan ini mengandungi 19 halaman bercetak.
3472/2 Form Five Additional Mathematics Paper 2 September 2009 2 ½ hours
CONFIDENTIAL 3472/2
CONFIDENTIAL
2
The following formulae may be helpful in answering the questions. The symbols given are the ones commonly used.
ALGEBRA
1 a
acbbx
2
42 −±−= 8 a
bb
c
ca log
loglog =
2 am x an = a m + n 9 ( 1)nT a n d= + −
3 am ÷ an = a m – n 10. [ ]2 ( 1)2n
nS a n d= + −
4 ( am )n = a m n 11 1nnT a r −=
5 nmmn aaa logloglog += 12 ( ) ( )1 1
, 11 1
n n
n
a r a rS r
r r
− −= = ≠
− −
6 nmn
maaa logloglog −= 13 , 1
1
aS r
r∞ = <−
7 log a mn = n log a m
CALCULUS KALKULUS
1 y = uv , dx
duv
dx
dvu
dx
dy +=
4 Area under a curve Luas di bawah lengkung
= ( )b b
a a
y dx or atau x dy∫ ∫
2 2
,v
dx
dvu
dx
duv
dx
dy
v
uy
−==
5 Volume generated Isipadu janaan
= 2 2( )b b
a a
y dx or atau x dyπ π∫ ∫
3 dx
du
du
dy
dx
dy ×=
CONFIDENTIAL 3472/2
CONFIDENTIAL
3
STATISTICS STATISTIK
1 N
xx
Σ=
7 i
ii
W
IWI
ΣΣ=
2 f
fxx
ΣΣ=
8 !
( )!n
r
nP
n r=
−
3 ( ) 2
22
xN
x
N
xx −Σ=−Σ=σ
9 !
( )! !n
r
nC
n r r=
−
4 ( ) 2
22
xf
fx
f
xxf −Σ
Σ=Σ
−Σ=σ
10 ( ) ( ) ( ) ( )P A B P A P B P A B∪ = + − ∩
5 Cf
FNLm
m
−+= 2
1
11 ( ) , 1n r n rrP X r C p q p q−= = + =
12 / min,Mean npµ =
6 1
0100
QI
Q= ×
13 npqσ =
14 x
Zµ
σ−=
GEOMETRY GEOMETRI
1 Distance/jarak
= ( ) ( )221
221 yyxx −+−
4 Area of a triangle/ Luas segitiga =
( ) ( )3123121332212
1yxyxyxyxyxyx ++−++
2 Mid point / Titik tengah
( )
++=2
,2
, 2121 yyxxyx
5 2 2
~r x y= +
3 A point dividing a segment of a line
Titik yang membahagi suatu tembereng garis
( )
++
++=
nm
myny
nm
mxnxyx 2121 ,,
6 ^
~ ~
2 2~
x i y jr
x y
+=
+
CONFIDENTIAL 3472/2
CONFIDENTIAL
4
TRIGONOMETRY TRIGONOMETRI
1 Arc length, s = rθ Panjang lengkok, s= jθ
8 ( )sin sin cos cos sinA B A B A B± = ±
( )sin sin s sinA B A kos B ko A B± = ±
2 Area of a sector, 21
2A r θ=
Luas sektor, L = 21
2j θ
9 ( )cos cos cos sin sinA B A B A B± = ∓
( )s os os sin sinko A B k A k B A B± = ∓
3 2 2cos 1sin A A+ =
2 2os 1sin A k A+ =
10 ( ) tan tantan
1 tan
A BA B
A tanB
±± =∓
4 2 2sec 1 tanA A= +
2 2se 1 tank A A= +
11 2
2 tantan 2
1 tan
AA
A=
−
5 2 2sec 1 cotco A A= +
2 2se 1 otko k A k A= +
12 sin sin sin
a b c
A B C= =
6 sin 2A = 2 sin A cos A sin 2A = 2 sin A kos A
13 2 2 2 2 cosa b c bc A= + −
2 2 2 2a b c bc kos A= + −
7 cos 2A = cos2 A – sin2 A = 2 cos2A – 1 = 1 – 2 sin2 A kos 2A = kos2 A – sin2 A = 2 kos2A – 1 = 1 – 2 sin2 A
14 Area of triangle/ Luas segitiga
= 1
sin2
ab C
CONFIDENTIAL 3472/2
CONFIDENTIAL
5
THE UPPER TAIL PROBABILITY Q(z) FOR THE NORMAL DIST RIBUTION N(0, 1) KEBARANGKALIAN HUJUNG ATAS Q(z) BAGI TABURAN NORMAL N(0, 1)
z 0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9
Minus / Tolak
0.0
0.1
0.2
0.3
0.4
0.5000
0.4602
0.4207
0.3821
0.3446
0.4960
0.4562
0.4168
0.3783
0.3409
0.4920
0.4522
0.4129
0.3745
0.3372
0.4880
0.4483
0.4090
0.3707
0.3336
0.4840
0.4443
0.4052
0.3669
0.3300
0.4801
0.4404
0.4013
0.3632
0.3264
0.4761
0.4364
0.3974
0.3594
0.3228
0.4721
0.4325
0.3936
0.3557
0.3192
0.4681
0.4286
0.3897
0.3520
0.3156
0.4641
0.4247
0.3859
0.3483
0.3121
4
4
4
4
4
8
8
8
7
7
12
12
12
11
11
16
16
15
15
15
20
20
19
19
18
24
24
23
22
22
28
28
27
26
25
32
32
31
30
29
36
36
35
34
32
0.5
0.6
0.7
0.8
0.9
0.3085
0.2743
0.2420
0.2119
0.1841
0.3050
0.2709
0.2389
0.2090
0.1814
0.3015
0.2676
0.2358
0.2061
0.1788
0.2981
0.2643
0.2327
0.2033
0.1762
0.2946
0.2611
0.2296
0.2005
0.1736
0.2912
0.2578
0.2266
0.1977
0.1711
0.2877
0.2546
0.2236
0.1949
0.1685
0.2843
0.2514
0.2206
0.1922
0.1660
0.2810
0.2483
0.2177
0.1894
0.1635
0.2776
0.2451
0.2148
0.1867
0.1611
3
3
3
3
3
7
7
6
5
5
10
10
9
8
8
14
13
12
11
10
17
16
15
14
13
20
19
18
16
15
24
23
21
19
18
27
26
24
22
20
31
29
27
25
23
1.0
1.1
1.2
1.3
1.4
0.1587
0.1357
0.1151
0.0968
0.0808
0.1562
0.1335
0.1131
0.0951
0.0793
0.1539
0.1314
0.1112
0.0934
0.0778
0.1515
0.1292
0.1093
0.0918
0.0764
0.1492
0.1271
0.1075
0.0901
0.0749
0.1469
0.1251
0.1056
0.0885
0.0735
0.1446
0.1230
0.1038
0.0869
0.0721
0.1423
0.1210
0.1020
0.0853
0.0708
0.1401
0.1190
0.1003
0.0838
0.0694
0.1379
0.1170
0.0985
0.0823
0.0681
2
2
2
2
1
5
4
4
3
3
7
6
6
5
4
9
8
7
6
6
12
10
9
8
7
14
12
11
10
8
16
14
13
11
10
19
16
15
13
11
21
18
17
14
13
1.5
1.6
1.7
1.8
1.9
0.0668
0.0548
0.0446
0.0359
0.0287
0.0655
0.0537
0.0436
0.0351
0.0281
0.0643
0.0526
0.0427
0.0344
0.0274
0.0630
0.0516
0.0418
0.0336
0.0268
0.0618
0.0505
0.0409
0.0329
0.0262
0.0606
0.0495
0.0401
0.0322
0.0256
0.0594
0.0485
0.0392
0.0314
0.0250
0.0582
0..0475
0.0384
0.0307
0.0244
0.0571
0.0465
0.0375
0.0301
0.0239
0.0559
0.0455
0.0367
0.0294
0.0233
1
1
1
1
1
2
2
2
1
1
4
3
3
2
2
5
4
4
3
2
6
5
4
4
3
7
6
5
4
4
8
7
6
5
4
10
8
7
6
5
11
9
8
6
5
2.0
2.1
2.2
2.3
0.0228
0.0179
0.0139
0.0107
0.0222
0.0174
0.0136
0.0104
0.0217
0.0170
0.0132
0.0102
0.0212
0.0166
0.0129
0.00990
0.0207
0.0162
0.0125
0.00964
0.0202
0.0158
0.0122
0.00939
0.0197
0.0154
0.0119
0.00914
0.0192
0.0150
0.0116
0.00889
0.0188
0.0146
0.0113
0.00866
0.0183
0.0143
0.0110
0.00842
0
0
0
0
3
2
1
1
1
1
5
5
1
1
1
1
8
7
2
2
1
1
10
9
2
2
2
1
13
12
3
2
2
2
15
14
3
3
2
2
18
16
4
3
3
2
20
16
4
4
3
2
23
21
2.4 0.00820 0.00798 0.00776 0.00755 0.00734
0.00714
0.00695
0.00676
0.00657
0.00639
2
2
4
4
6
6
8
7
11
9
13
11
15
13
17
15
19
17
2.5
2.6
2.7
2.8
2.9
0.00621
0.00466
0.00347
0.00256
0.00187
0.00604
0.00453
0.00336
0.00248
0.00181
0.00587
0.00440
0.00326
0.00240
0.00175
0.00570
0.00427
0.00317
0.00233
0.00169
0.00554
0.00415
0.00307
0.00226
0.00164
0.00539
0.00402
0.00298
0.00219
0.00159
0.00523
0.00391
0.00289
0.00212
0.00154
0.00508
0.00379
0.00280
0.00205
0.00149
0.00494
0.00368
0.00272
0.00199
0.00144
0.00480
0.00357
0.00264
0.00193
0.00139
2
1
1
1
0
3
2
2
1
1
5
3
3
2
1
6
5
4
3
2
8
6
5
4
2
9
7
6
4
3
11
9
7
5
3
12
9
8
6
4
14
10
9
6
4
3.0 0.00135 0.00131 0.00126 0.00122 0.00118 0.00114 0.00111 0.00107 0.00104 0.00100 0 1 1 2 2 2 3 3 4
Example / Contoh:
−= 2
2
1exp
2
1)( zzf
π If X ~ N(0, 1),
then Jika X ~ N(0, 1), maka
∫∞
=k
dzzfzQ )()( P(X > k) = Q(k)
P(X > 2.1) = Q(2.1) = 0.0179
Q(z)
z
f (z)
O k
CONFIDENTIAL 3472/2
CONFIDENTIAL
6
Section A Bahagian A
[40 marks]
[40 markah]
Answer all questions. Jawab semua soalan
1. Solve the following simultaneous equations , give your answers correct to three decimal
places. Selesaikan persamaan serentak berikut dengan memberi jawapan anda tepat kepada
tiga tempat perpuluhan :
5312 =+=+ yxyx
[6 marks]
[6 markah] 2. In Diagram 1, ABCD is a quadrilateral. BFC and DEF are straight lines. Dalam Rajah 1, ABCD ialah sebuah sisi empat. BFC dan DEF adalah garis lurus.
Given that xBA 24= , yBF 10= , yxCD 3030 −= , BCBF4
1= and DFDE5
2= ,
Diberi xBA 24= , yBF 10= , yxCD 3030 −= , BCBF4
1= dan DFDE5
2= ,
(a) express in terms of x and/or y ,
ungkapkan dalam sebutan x dan/atau y ,
(i) AC (ii) DF [3 marks] [3 markah]
(b) show that the points A, E and C are collinear. [3 marks] tunjukkan bahawa titik A, E dan C adalah segaris . [3 markah]
Diagram 1 Rajah 1
A
B C
D
E ••••
F
CONFIDENTIAL 3472/2
CONFIDENTIAL
7
3.(a) Prove that xxx
xxcot
sin2sin
2coscos1 =+++
. [3 marks]
Buktikan bahawa xxx
xxcot
sin2sin
2coscos1 =+++
. [3 markah]
(b)(i) Sketch the graph of the trigonometric function 1cos3 += xy for the domain π20 ≤≤ x . Lakar graf bagi fungsi trigonometri 1cos3 += xy untuk domain π20 ≤≤ x .
(ii) On the same axes, sketch the graph of a suitable straight line that can be used to solve the equation ππ −= xx 3cos3 . State the number of solutions to the equation ππ −= xx 3cos3 for π20 ≤≤ x .
Pada paksi yang sama, lakar graf bagi satu garis lurus yang sesuai digunakan untuk menyelesaikan persamaan ππ −= xx 3cos3 . Nyatakan bilangan penyelesaian bagi persamaan ππ −= xx 3cos3 untuk
π20 ≤≤ x . [5 marks] [5 markah]
4. Table 1 shows the frequency distribution of the Additional Mathematics marks of a
group of students. Jadual 1 menunjukkan taburan kekerapan markah Matematik Tambahan bagi
sekumpulan pelajar.
Marks Number of students 1 – 10 2 11 – 20 3 21 – 30 5 31 – 40 10 41 – 50 K 51 – 60 2
Table 1 Jadual 1
(a) Given that the median mark is 34.5,
Diberi markah median adalah 34.5,
(i) calculate the value of k, hitungkan nilai k,
(ii) find the median mark if the mark of each student is increased by 8.
cari markah median jika markah setiap pelajar ditambahkan sebanyak 8. [4 marks] [4 markah]
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(b) Given that k = 4, draw a histogram to represent the frequency distribution of the
mark by using a scale of 2 cm to 10 marks on the horizontal axis and 2 cm to 1
student on the vertical axis.
Diberi k = 4, lukis sebuah histogram untuk mewakili taburan kekerapan markah
dengan menggunakan skala 2 cm kepada 10 markah pada paksi ufuk dan 2 cm
kepada 1 pelajar pada paksi tegak.
Hence, find the modal mark. [3 marks]
Seterusnya, cari markah mod. [3 markah]
5. (a) Find the equation of the normal to the curve x
xy1
3 += at (1 , 4). [3 marks]
Cari persamaan normal kepada lengkung x
xy1
3 += pada (1 , 4). [3 markah]
(b) Diagram 2 shows a leaking hemispherical container with a radius of r cm.
Rajah 2 menunjukkan sebuah bekas bocor yang berbentuk hemisfera dengan jejari
r cm.
Given that the radius of the water surface is decreasing at the rate of 0.1 cm s–1, find
in terms of π, the rate of change of the volume of water in the container at the instant
the radius of the water surface is 20 cm. [3 marks]
Diberi jejari permukaan air menyusut dengan kadar 0.1 cm s–1, cari dalam sebutan
π, kadar perubahan isipadu air dalam bekas itu pada ketika jejari permukaan air
adalah 20 cm. [3 markah]
Diagram 2 Rajah 2
r cm
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6. Diagram 3 shows a few circles. The first circle is the largest circle with a radius of
R cm. The second circle has a radius of R3
2cm . The third circle has a radius which is
3
2 of the radius of second circle and this process is continued indefinitely.
Rajah 3 menunjukkan beberapa bulatan. Bulatan pertama adalah bulatan terbesar dan
mempunyai jejari R cm. Bulatan kedua mempunyai jejari R3
2cm. Bulatan ketiga
mempunyai jejari yang merupakan 3
2 daripada jejari bulatan kedua dan proses ini
diteruskan sehingga ketakhinggaan.
(a) Show that the perimeters of the circles form a geometric progression with common
ratio 3
2by using first three circles. [2 marks]
Tunjukkan bahawa perimeter bulatan-bulatan itu membentuk satu janjang geometri
dengan nisbah sepunya 3
2dengan menggunakan tiga bulatan pertama. [2 marks]
(b) Given that the area of the largest circle is 900π cm2, find in terms of π,
Diberi bahawa luas bulatan yang terbesar ialah 900π cm2, cari dalam sebutan π, (i) the circumference of the tenth circle, Ukurlilit bagi bulatan yang kesepuluh, (ii) the sum of the circumference of infinite number of circles formed. [5 marks] jumlah ukurlilit bagi semua bulatan yang dapat dibentuk sehingga
ketakterhinggaan [5 markah]
R cm
Diagram 3 Rajah 3
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Section B Bahagian B
[40 marks]
[ 40 markah]
Answer any four questions from this section. Jawab mana-mana empat soalan daripada bahagian ini.
7. Use graph paper to answer this question. Gunakan kertas graf untuk menjawab soalan ini.
Table 2 shows the values of two variables, x and y , obtained from an experiment.
Variables x and y are related by the equation q
py
x 2+= , where p and qare constants.
One of the values of y is incorrectly recorded. Jadual 2 menunjukkan nilai-nilai bagi dua pembolehubah, x dan y , yang diperoleh daripada satu eksperimen. Pembolehubah x dan y dihubungkan oleh persamaan
q
py
x 2+= , dengan keadaan p dan qadalah pemalar. Satu daripada nilai y telah salah
direkodkan.
Table 2 Jadual 2
(a) Plot y10log against )2( +x , using a scale of 2 cm to 1 unit on the )2( +x - axis
and 2 cm to 0.05 unit on the y10log -axis.
[Start the y10log -axis with the value 0.8].
Hence, draw the line of best fit. [4 marks] Plot y10log melawan )2( +x , dengan menggunakan skala 2 cm kepada 1 unit
pada paksi- )2( +x dan 2 cm kepada 0.05 unit pada paksi- y10log .
[Mulakan paksi- y10log dengan nilai 0.8]
Seterusnya, lukis garis lurus penyuaian terbaik. [4 markah]
(b) Use your graph from 7(a), find Gunakan graf anda di 7(a), cari
(i) the correct value of y that is wrongly recoded. nilai yang betul bagi nilai y yang salah direkodkan. (ii) the values of p and q . [6 marks] nilai p dan nilai q. [6 markah]
x –1 0 1 2 3 4 y 8.4 10.1 12.1 13.2 17.4 20.9
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8. Solutions to this question by scale drawing will not be accepted. Penyelesaian secara lukisan berskala tidak diterima.
Diagram 4 shows the triangle OAB where O is the origin. Point C lies on the straight line AB. Rajah 4 menunjukkan segitiga OAB dengan O ialah titik asalan. Titik C terletak pada garis lurus AB.
(a) Calculate the area, in unit2, of triangle OAB. [2 marks] Hitungkan luas, dalam unit2, bagi segitiga OAB. [2 markah] (b) Find the equation of the perpendicular bisector of line segment AB. [3 marks] Cari persamaan pembahagi dua sama serenjang bagi tembereng garis AB. [3 markah]
(c) Given that the length BC is 5
4 of the distance of point B from the perpendicular
bisector of the line segment AB, find the coordinates of point C. [2 marks]
Diberi panjang BC ialah 5
4daripada jarak titik B dari pembahagi dua sama
serenjang bagi tembereng garis AB, cari koordinat bagi titik C. [2 markah] (d) A point P moves such that its distance from point B is always twice its distance
from point C. Find the equation of the locus of P. [3 markah] Satu titik P bergerak dengan keadaan jaraknya dari titik B adalah sentiasa dua kali jaraknya dari titik C. Cari persamaan lokus bagi P. [3 markah]
x
y
B(6 , -8)
A(-4 , 2)
O
Diagram 4 Rajah 4
C •
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9. Diagram 5 shows the straight line y = 2x which passes through the maximum point of a
quadratic curve ))(( βα −−−= xxy , where βα and are constants.
Rajah 5 menunjukkan garis lurus y = 2x yang melalui titik maksimum suatu garis
lengkung kuadratik, ))(( βα −−−= xxy , dengan keadaan βα and ialah pemalar.
(a) State
Nyatakan
(i) the coordinates of the maximum point,
koordinat titik maksimum itu,
(ii) the equation of the quadratic curve. [2 marks]
persamaan garis lengkung kuadratik itu. [2 markah]
(b) Calculate the area of the shaded region P. [4 marks]
Hitungkan luas rantau berlorek P. [4 markah]
(c) Find the volume of the solid generated, in terms of π , when the
region Q is revolved through 360o about the x-axis. [4 marks]
Cari isipadu pepejal yang dijanakan, dalam sebutan π , apabila rantau
Q dikisarkan melalui 360o pada paksi-x. [4 markah]
x
y y = 2x
O 4
P
Q
Diagram 5 Rajah 5
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10. (a) In a house to house check carried out in Taman Maju, aedes mosquitoes were
found in 2 out of every 5 houses. If 8 houses in Taman Maju are chosen at random,
calculate the probability that
Dalam suatu pemeriksaan dari rumah ke rumah di Taman Maju, nyamuk aedes telah
dijumpai dalam 2 daripada setiap 5 buah rumah. Jika 8 buah rumah di Taman Maju
dipilih secara rawak, hitung kebarangkalian bahawa
(i) exactly 3 houses are infested with aedes mosquitoes,
tepat 3 buah rumah akan dijumpai dengan nyamuk aedes,
(ii) more than 2 houses are infested with aedes mosquitoes.
lebih daripada 2 buah rumah akan dijumpai dengan nyamuk aedes.
[5 marks]
[5 markah]
(b) A study on the body mass of a group of students is conducted and it is found that the
mass of a student is normally distributed with a mean of 50 kg and a variance of
256 kg2.
Satu kajian jisim badan dijalankan ke atas sekumpulan pelajar dan didapati jisim
seorang pelajar adalah mengikut taburan normal dengan min 50 kg dan varians
256 kg2.
(i) If a student is selected randomly, calculate the probability that his mass is more
than 60 kg.
Jika seorang pelajar dipilih secara rawak, hitungkan kebarangkalian bahawa
jisimnya adalah lebih daripada 60 kg.
(ii) Given that 28% of the students weigh less than m kg, calculate the value of m.
Diberi bahawa 28% daripada pelajar itu mempunyai jisim kurang daripada
m kg, cari nilai m.
[5 marks]
[5 markah]
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11. Use 142.3=π in this question. Gunakan 142.3=π dalam soalan ini.
Diagram 6 shows a circular sector OAC with a radius of 5 cm and ∠AOC is 1.2 radian. BC is an arc of the circle with centre A. Rajah 6 menunjukkan satu sektor bulatan OAC dengan jejari 5 cm dan ∠AOC adalah 1.2 radian. BC adalah lengkok bulatan dengan pusat A.
(a) Find Cari
(i) the length, in cm, of the arc AC. [2 marks] panjang, dalam cm, lengkok AC. [2 markah]
(ii) the length, in cm, of radius AB. [2 marks]
panjang, dalam cm, jejari AB. [2 markah]
(b) (i) Show that ∠BAC = 2.171 radian. Tunjukkan ∠BAC = 2.171 radian.
(ii) Hence, calculate the area, in cm2, of the shaded region.
Seterusnya, hitungkan luas, dalam cm2, rantau yang berlorek. [6 marks]
[6 markah]
O
A
B
C
5 cm
1.2 rad Diagram 6
Rajah 6
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Section C
Bahagian C
[20 marks] [20 markah]
Answer two questions from this section.
Jawab dua soalan daripada bahagian ini.
12. Table 3 shows the price indices and the percentage of usage of 5 different ingredients A, B, C, D and E needed to make a cake. The composite index number for the cost of making the cake in the year 2007 based on the year 2005 is 132. Jadual 3 menunjukkan indeks harga dan peratus penggunaan lima jenis bahan A, B, C, D dan E yang diperlukan untuk membuat sejenis kek. Nombor indeks gubahan kos membuat kek itu pada tahun 2007 berasaskan tahun 2005 ialah 132.
Ingredients Jenis bahan
Price index for the year 2007 based on the year 2005 Indeks harga pada tahun 2007 berasaskan tahun 2005
Percentage of ingredient Peratus bahan (%)
A 140 30 B x 20 C 110 15 D 104 10 E 120 25
Table 3 Jadual 3
(a) Calculate
Hitungkan
(i) the price of A in the year 2005 if its price in the year 2007 is RM7. harga A pada tahun 2005 jika harganya pada tahun 2007 ialah RM7.
[2 marks] [2 markah]
(ii) the value of x. [2 marks]
nilai x. [2 markah]
(b) The cost of the cake increased 10% from the year 2007 to the year 2009. Find the price of the cake in the year 2009 if its price in the year 2005 is RM40. Kos penghasilan kek itu bertambah 10 % dari tahun 2007 ke tahun 2009. Cari harga kek itu pada tahun 2009 jika harganya pada tahun 2005 ialah RM40.
[3 marks] [3 markah]
(c) Find the price index of D in the year 2007 based on the year 2003 if its price index
in the year 2005 based on the year 2003 is 125. [3 marks] Carikan indeks harga bagi D pada tahun 2007 berasaskan tahun 2003 jika indeks harganya pada tahun 2005 berasaskan tahun 2003 ialah 125. [3 markah]
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13. Diagram 7 shows a quadrilateral PQRS where the sides PQ and RS are parallel.
Rajah 7 menunjukkan sebuah sisiempat PQRS dengan keadaan sisi PQ dan sisi RS adalah selari.
Given that PQ = 10 cm, ∠RPQ = 1100 , ∠PQR = 500 and SR : PQ = 2 : 5, calculate Diberi PQ = 10 cm, ∠RPQ = 1100 , ∠PQR = 500 dan SR : PQ = 2 : 5, hitungkan (a) the length, in cm, of PR and QR. [3 marks]
panjang, dalam cm, PR dan QR. [3 markah]
(b) the length, in cm, of diagonal QS. [3 marks] panjang, dalam cm, perpenjuru QS. [3 markah]
(c) the area, in cm2, of PQRS. [4 marks]
luas, dalam cm2, PQRS. [4 markah]
14. A particle moves along a straight line and passes through a fixed point O with a velocity of 3 ms-1. Its acceleration, a ms-2, is given by a = 2 – 2t, where t is the time, in seconds, after passing through O. The particle stops momentarily at time, t = k s.
Suatu zarah bergerak di sepanjang suatu garis lurus dan melalui satu titik tetap O dengan halaju 3 ms-1. Pecutannya, a ms-2, diberi oleh a = 2 - 2t, dengan keadaan t ialah masa, dalam saat, selepas melalui O. Zarah itu berhenti seketika pada masa, t = k s.
Find Cari (a) the maximum velocity of the particle, [ 3 marks ]
halaju maksimum zarah itu, [3 markah]
(b) the value of k, [2 marks] nilai k, [2 markah]
(c) the distance travelled in the third second, [2 marks] jarak yang dilalui dalam saat ketiga. [2 markah]
(d) the value of t , correct to two decimal places, when the particle passes O again. [3 marks] nilai t, betul kepada dua tempat perpuluhan, apabila zarah itu melalui titik O semula. [3 markah]
Diagram 7 Rajah 7
10 cm
P
Q R 50o
110o
S
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15. Use the graph paper provided to answer this question Gunakan kertas graf yang disediakan untuk menjawab soalan ini.
A factory produces two types of school bags, type P and type Q. In a day, it can produce x bag of type P and y bag of type Q. The time taken to produce a bag of type P is 40 minutes and a bag of type Q is 50 minutes.
Sebuah kilang menghasilkan dua jenis beg sekolah, jenis P dan jenis Q. Dalam satu hari, kilang itu boleh menghasilkan x beg jenis P dan y beg jenis Q. Masa yang diambil untuk menghasilkan satu beg jenis P ialah 40 minit dan satu beg jenis Q ialah 50 minit.
The production of the bags per day is based on the following constraints : Pengeluaran beg dalam satu hari adalah berdasarkan kepada kekangan berikut :
I : The total number of bags produced is not more than 160.
Jumlah bilangan beg yang dihasilkan tidak melebihi 160.
II : The time taken to make bag P is not more than twice the time taken to make bag Q.
Masa yang diambil untuk membuat beg P tidak melebihi dua kali ganda masa yang diambil untuk membuat beg Q.
III : The number of bag Q exceed the number of bag P by at most 80. Bilangan beg Q melebihi bilangan beg P selebih-lebihnya 80.
(a) Write down three inequalities, other than 0≥x and 0≥y which satisfy all the
above constraints. [3 marks ] Tulis tiga ketaksamaan, selain x ≥ 0 dan y µ 0, yang memenuhi semua kekangan di atas. [3 markah]
(b) By using a scale of 2 cm to 20 bags on both axes, construct and shade the region R
that satisfies all the above constraints. [3 marks] Menggunakan skala 2 cm kepada 20 beg pada kedua-dua paksi, bina dan lorek rantau R yang memenuhi semua kekangan di atas. [3 markah]
(c) Use your graph in 15 (b) to answer the following :
Gunakan graf anda di 15 (b) untuk menjawab yang berikut :
(i) Find the range of the number of bag Q that can be produced if the number of bag P is 50.
Cari julat bilangan beg Q yang boleh dihasilkan jika bilangan beg P ialah 50.
(ii) If the profit of selling bag P is RM20 and bag Q is RM30, find the maximum profit that can be obtained.
Jika untung jualan bagi beg P ialah RM20 dan beg Q ialah RM30, cari keuntungan maksimum yang boleh diperolehi. [4 marks] [4 markah]
END OF QUESTION PAPER KERTAS SOALAN TAMAT
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NAMA:
KELAS :
NO. KAD PENGENALAN: ANGKA GILIRAN Arahan Kepada calon Tulis nama, kelas, nombor kad pengenalan dan angka giliran anda pada ruang yang
disediakan.
Tandakan ( √ ) untuk soalan yang dijawab.
Ceraikan helaian ini dan ikat sebagai muka hadapan bersama-sama dengan kertas jawapan.
Kod Pemeriksa
Bahagian Soalan Soalan Dijawab
Markah Penuh
Markah Diperoleh (Untuk Kegunaan Pemeriksa)
A
1 6
2 6
3 8
4 7
5 6
6 7
B
7 10
8 10
9 10
10 10
11 10
C
12 10
13 10
14 10
15 10
SULIT
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3472/2[PP] Additional Mathematics Paper 2 September 2009
SEKTOR PENGURUSAN AKADEMIK JABATAN PELAJARAN PAHANG
PEPERIKSAAN PERCUBAAN SPM
TAHUN 2009
ADDITIONAL MATHEMATICS
Paper 2
MARKING SCHEME
This marking scheme consists of 12 printed pages
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Question Working/Solution Marks Total 1.
512 =+yx
and 53 =+ yx
or any correct pairs from 5312 =+=+ yxyx
yx 35−= or 3
5 xy
−=
Substitute yx 35−= or 3
5 xy
−= into the non-linear
equation to obtain a quadratic equation in terms of x or y
052615 2 =+− yy or 010245 2 =+− xx Solve quadratic equation using formula or completing the squares
220.0;513.1 == yy or 461.0;339.4 == xx
340.4;461.0 == xx or 513.1;220.0 == yy
P1
P1
K1
K1
N1
N1
6 2(a) Use triangle law correctly to find AC or DF
(i) BCABAC += (ii) CFDCDF +=
yxAC 4024 +−= xDF 30−=
K1
N1 N1
2(b) Find vector CE or EC or AE or EA by correct triangle law.
)53(65
35
3
yx
DFCF
FDCF
FECFCE
−=
−=
+=
+=
CEyx6
153 =−
Find AC in term of CE or vice versa or any equation that can conclude A, C, E are collinear.
CEAC
yxAC
yxAC
3
4
)53(8
4024
−=
−−=
+−=
∴A, C, E are collinear because CEAC3
4−=
K1
K1
N1
6
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Question Working/Solution Marks Total
3(a) Use cos 2x or sin 2x and factorize
xx
x
xx
xx
xx
xx
xxx
xx
xx
xx
cotsin
cos
)1cos2(sin
)1cos2(cos
)1cos2(sin
cos2cos
sincossin2
)1cos2(cos1
sin2sin
2coscos1
2
2
=
=
++=
++=
+−++=
+++
K1
N1
N1
3(b)(i) Correct shape of cosine function with amplitude of 3 units or Correct shape of cosine function and translated 1 unit
OR Correct shape of cosine function with amplitude of 3 units and translated 1 unit.
OR Correct shape of cosine function with amplitude of 3 units
and translated 1 unit and passing through (0 , 4), (2
π , 1),
(π , –2), (2
3π , 1) and (2π , 4).
P1
OR
P2
OR
P3
3(b)(ii) Get the correct linear equation and draw a straight line.
π
πππ
ππ
xx
xx
xx
xx
31cos3
3)1cos3(
3cos3
3cos3
=+
=+=+−=
The equation of straight line : πx
y3=
Straight line drawn correctly and number of solutions = 1
K1
N1
8
4
1
-2 2
π 2
3π
π 2πx
y
O
πx
y3=
1cos3 += xy
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Question Working/Solution Marks Total
4.(a) (i) Use median formula with at least two of the L, N, F, f and c correctly substituted.
)10(10
102
22
5.305.34
−+
+=
k
6=k (ii) median mark = 42.5
K1
N1
N1
P1
4(b) Correct axes and uniform scales with all the lower and upper boundaries correctly labeled and the Height of at least three bars are proportional to the frequency Correct way of finding the value of mode. Modal mark = 35.0
K1
K1
N1
7 5(a)
xxy
13 +=
2
13
xdx
dy −=
Find gradient of normal and use )( 11 xxmyy −=−
at (1 , 4), 2=dx
dy
Gradient of normal = 2
1−
Equation of normal :
)1(2
14 −−=− xy
092 =−+ yx or equivalent
P1
K1
N1
5(b) Get the expression for V and find
dr
dV to determine the
value of dr
dV at r = 20 cm.
3
3
2rV π=
22 rdr
dV π=
Use dt
dr
dr
dV
dt
dV ×=
13
2
80
)1.0()20(2
−−=
−=
scm
dt
dV
π
π
K1
K1
N1
6
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Question Working/Solution Marks Total
6. (a) Perimeters of the first three circles :
Rπ2 , Rπ3
4, Rπ
9
8
Check the ratios 1
2
T
T and
2
3
T
T
Make a correct conclusion :
Since 3
2
2
3
1
2 ==T
T
T
T , the perimeters of the first three
circles form a geometric progression with common
ratio 3
2.
(b)
30
9002
==
R
R ππ
(i) Use the formula 1−= nn arT :
cm561.1
3
2)60(
9
10
π
π
=
=T
(ii) Use the formula r
aS
−=∞ 1
:
cm1803
21
60
π
π
=
−=∞S
K1
N1
P1
K1
N1
K1
N1
7
7(a) All values of y10log correct.
Plot y10log against ( 2+x ) with correct axes, uniform
scales and at least one point plotted correctly. 6 points plotted correctly. Line of best fit, ( passes through as many points as possible and balance in terms of numbers point appear above and below the line, if any .)
2+x 1 2 3 4 5 6 y10log 0.92 1.00 1.08 1.12 1.24 1.32
N1
K1
N1
N1
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7(b) (i) Recognize the wrong recorded value of y and use graph to find the should be value of y.
45..14
16.1log10
==
y
y
(ii) qpxy 101010 loglog)2(log −+=
Use cq =− 10log or Use mp =10log
84.0log10 −=q 08.0log10 =p
p = 0.1445 ; p = 1.202
K1
N1
P1
K1
N1 N1
10 8(a)
Use 1
1
3
3
2
2
1
1
2
1y
x
y
x
y
x
y
x correctly to find the area of triangle
Area = 10 unit2
K1
N1
8(b) Either midpoint of AB or gradient of AB correct. Mid point of AB = (1 , –3)
Gradient of AB = –1 Use )( 11 xxmyy −=− with his midpoint of AB and his gradient of normal.
y + 3 = 1(x -1) y = x – 4
P1
K1
N1
8(c) Use Ratio Theorem follow his midpoint
−+−+=5
)8(1)3(4,
5
)6(1)1(4C
)4,2( −=C
K1
N1
8(d) Use of distance formula correctly for PB or PC 22 )8()6( ++−= yxPB ; 22 )4()2( ++−= yxPC
Use BP = 2 PC
2222 )4()2(2)8()6( ++−=++− yxyx
02016433 22 =−+−+ yxyx
K1
K1 N1
10
9(a) (i) (2 , 4) (ii) )4( −−= xxy or eqivalent
P1 P1
9(b) Correct method of finding area under curve or area of ∆ 4
2
32
32
4
− xx
or
2
0
2
2
2
x or 42
2
1 ××
Find integration value using correct limits
−−
−
3
2
2
)2(4
3
4
2
)4(4 3232
or
−
2
)0(2
2
)2(2 32
Find the sum of two area
3
28 unit2 or equivalent
K1
K1
K1
N1
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25
9(c) Correct method of finding volume of revolution or volume
of cone.
( )dxxxx∫ +−2
0
432 816π or dxx∫2
0
24π or )2()4(3
1 2π
Find integration value using correct limits
−
+− 0
5
2
4
)2(8
3
)2(16 543
π or
− 0
3
)2(4 3
π
Find the difference of two volume
π5
32 cm3 or equivalent
K1
K1
K1
N1
10
10(a) (i) Both
5
2=p and 5
3=q or equivalent
53
38
5
3
5
2
C or equivalent
0.2787 or other more accurate answers.
(ii) 1 - 62
28
71
18
80
08
5
3
5
2
5
3
5
2
5
3
5
2
+
+
CCC
or 08
88
44
48
53
38
5
3
5
2...
5
3
5
2
5
3
5
2
++
+
CCC
All terms must be correct and completed in full 0.6846 or other more accurate answers.
P1
K1
N1
K1
N1
10(b) (i)
−≥16
5060ZP
0.2660
(ii) 28.016
50 =
−< mZP or 28.0
16
50 =
−−≥ mZP
or equivalent
583.016
50 =−− m
40.672 kg
K1 N1
K1
K1 N1
10 11(a) (i) SAC = 5(1.2)
6 cm (ii) Correct method of finding AB
radAC
2
2.1sin
52
1
= or equivalent OR cosine rule
AB = AC = 5.646 cm (5.64578271...)
K1 N1
K1
N1
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26
11(b) (i)
2
2.12.1
++=∠ πBAC or equivalent
2.171 rad or equivalent
(ii) Area of sector BAC = 2
1(5.646)2(2.171)
Area of segment AC = ( )rad2.1sin2.1)5(2
1 2 −
or equivalent Find difference of the two area
)2.1sin2.1()5(2
1)171.2()646.5(
2
1 22 rad−−
Area of shaded region = 31.25 cm2 (31.25261028...)
K1
N1
K1
K1
K1
N1
10 12(a)
(i) 1401007
05
=×A
or equivalent
RM5
(ii) 1322510152030
)25(120)10(104)15(110)20()30(140 =++++
++++ x
165.5
K1 N1
K1 N1
12(b) 132100
4007 =×
Q
Q07 = RM 52.80 Q09 = RM 58.08
K1 N1 N1
12(c) 125100
03
05 =×Q
Q and
100
104
05
07 =Q
Q
Or 10410005
07 =×Q
Q and
100
125
03
05 =Q
Q
10003
0703,07 ×=
Q
QI
10003
05
05
0703,07 ××=
Q
Q
Q
QI or 104
100
125× or 100
104125× or
100
104125×
= 130
P1
K1
N1
10
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27
13(a) Using correct rules to find PR and QR.
oo
PR
20sin
10
50sin= or
oo
QR
20sin
10
110sin=
PR = 22.40 cm (22.39764114...) QR = 27.47 cm (27.47477419...)
K1
N1 N1
13(b) Use cosine rule to find QS. QS2 = 42 + 27.472 – 2(4)(27.47)cos (110 + 20)o QS = 30.20 cm
K1 N1 N1
13(c) Use formula correctly to find area of triangle PQR or PRS.
oPQR 50sin)10)(47.27(2
1Area =∆
or oPRS 110sin)4)(40.22(2
1Area =∆
Use Area PQRS = sum of two area Area PQRS = 147.32 cm2
K1
N1
K1
N1
10 14(a) Find expression for v.
ct
tv +−=2
22
2
Find t for maximum v and substitute in his v. a = 2 - 2t = 0
and 32
)1(22
2
max +−=v
vmax = 4 ms-1
K1
K1
N1
14(b) Try solving v = 0 2k - k2 + 3 = 0 (3 - k)(1 + k) = 0 or equivalent k = 3 s
K1
N1
14(c) Correct method in finding distance traveled. 3
2
32
332
2
+−= t
tts or ttts 3
3
1 32 +−= and 23 ss −
3
5=s m
K1
N1
14(d) Find expression for s.
tt
ts 33
32 +−=
Solve his s = 0
033
32 =+− t
tt
3t2 - t3 + 9t = 0 t2 - 3t -9 = 0 t = 4.854 s
K1
K1
N1
10
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28
15(a) 160≤+ yx note: ss-1 for all answers in terms of P and Q
yx 10040 ≤ or equivalent 80+≤ xy or equivalent
N1 N1 N1
15(b) Axes correct and one correct straight line. All three straight lines are correct. The shaded region of R is correct. Please refer to attached graph.
K1 K1 N1
15(c) (i) Drawing of the straight line, x = 50, completely across the shaded region R. 11020 ≤≤ y (ii) Construction of the line 20x + 30y = k at (40, 120) or any clear indication of point (40, 120) only or 30(120) + 20(40) = k maximum profit = RM4400
K1
N1
K1
N1
10
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29
10
8
1
3
4
5
6
2
7
9
Graph For Question 4(b) Number of students
0.5 10.5 20.5 30.5 40.5 50.5 Modal mark = 35.0
marks 60.5
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30
1 2 3 4 5 6 x + 2
0.85
0.90
0.95
1.00
1.05
1.10
1.15
y10log
x
x
x
x
x
x
No.7(a)
0 7 0.80
1.20
1.25
1.30
(1 , 0.92)
(6 , 1.32)
84.0
08.016
92.032.1
==
−−=
c
m
m
1.16
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31
Qn. 15