347212347212AT}DITIONALMATTIEIT{ATICSPaper 2Aug / Septz0fi2 % hours
AD DITIONAL MATHEMATICS
Tingkatan 5
Kertas 2Dua jam tiga puluh minit
JANGAIT BTTKA KERTAS SOALAI\I INI SEHINGGA DIBERITAHU
1. Kertas soalan ini adalah dalam dwibahasa.
2. Soalan dalam bahasa Inggeris mendahului soalan yang sepadan dalam
Bahasa Melayu.
3. Calon dikehendaki membaca maklumat di halaman belakang kertas soalan
ini.
Kertas soalan ini mengandungr 20 halaman bercetak.
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The following formulae may be helpful in answering the questions. The symbols given are theones commonly used.Rumus-rumus berilu.t boleh membantu anda meniawab soalan. Simbol-simbol yang diberiadalah yang biasa digunakan.
ALGEBRA
u*rlu'-*_ g log,b=!.og,blx=To-log,a
2 e^Xar:a.^*n 9 Tn=a+(n-L)d,
3 a^ + an:qm-n 10. g = tlrr+(n-Ddl
4 ( a^)n : a^n ll T, = arn-l
o( r^ -l\ o(l- r'\5 logomn=logom+logon 0 S,=f =}/,r*l
6 togo:=logo m-logon ^ 13 g = *, lrl .tn
7 logomn= nlogom
CALCI]LUSKALKULUS
4 Area under a curve
. dy dv . du Luas di bawah lenglamg1 y--tN, +=ttT *v:chc "d)c"dJc b b: I, e or(atau) !- at
du dv
., u dY ' a* - uA 5 Volume of revolution
J=-rT:v ca v' Isi Padu kisaranbb
A dy dy du - lo Y'hc or (atau) = lo x'dY
J dx du dx
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Xx
N
I' fxv-
2f
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STATISTICSSTATISTIK
= zWiI i7 I--2W,
8 '1= +t-' (n- r)!
g nCr= nl
(n - r1t vl3o=
4 oi=
ffi{l"-t
1 Distance liarak
-- J6r-*,Y +(v,-v,Y
10 P(Au B) = P(A) + P(^B) - P(A^ B)
1l P(X =r) ='C,P' Q'-', P * q -l
Meanlmin, P= nP
o = ,lnpq
E x- Pt-
o
GEOMETRYGEOMETRI
4 Area of a triangle I Luas segitiga:
|lt,',r, * xzlt * \!r)-('rY, * xtlz+ x,/,)l
n xi+yiol-ffi
t2
13
6 I=30x100t4
2 Mid point I Titik tengah
G,v)=(ry,ry)3 A point dividing a segment of a line
Titik yang membahagi suatu
tembereng garis
(",r)=( ffi,W)
EG-7\,\/=Nw_=w
1l >,f 1l ,' f
x'+y'
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4
TRIGONOMETRYTRIGONOMETRI
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I Arclength, s:rg 8 sin(r4tB) =sinr4cosB*cosAsinBPanjanglenglmk, s: je sin(,at.B) = sin A ftosBtkos lsin B
2 Areaofasector. A = !,'e' 2 g cos (r4 t .B) = cos .,4 cos B T sin Asin B
Luassektor,L: lft kos(AtB) = ftos AkosBT sin Asin B2'
3 sinzA+cos'A=IsinzA + &os2 A=L
4 sec'A=1+ tan2 A
sekz A -1+ tan2 A
5 cosec'A=l+cotzAkosek2A=l+ kotzA
6 sinZA - 2 sin A cos Asn2A : 2 sin Akos A
to tan(Arr) - ,Yi.n*!|-t' I TtanA tanB
1r tan2.A= ffi1., a b c
sin r4 sin B sin C
13 a' =b2 +c' -zbccosr{A' =b2 +c'-\bc kos A
7 cos 2A
: l4:;T^ u Area ortriangre/ Luas segitiga
: ! obsin C2
kos 2A = kos2 A - sir{ A
: ?yfl;)
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TIIE UPPER TAIL PROBABILITY 0(e) FOR TIm NORIVIAL DISTRIBUTIONIT(0, 1)KEBARANGKALIAN HATUNG ATAS OKI BAGI TABARAN NORMAL I
z 0 32I 654 98734567
Minus /Tolak
II27
0.0
0.1
0.2
0.3
0,4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
r.9
LO
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2,8
2.9
3.0
0.50w
0.4602
0.4207
0,3E21
0.3446
0.3085
0.2743
0,2420
0,21r9
0.1841
0.1587
0.1357
0.1 151
0.0968
0.0808
0.0668
0.0548
0.M16
0.0359
0.02E7
0,02n
0.0179
0.0139
0,0't07
0.00820
0,00621
0.00466
0,00347
0.m256
0,00r87
0.00135
0.4960 0.4920 0.46E0
0.4562 0.4522 0,4483
0.4168 0.41A 0.4090
0.3783 0.3745 0.3707
0.3409 0.3372 0.3336
0.3050 0.3015 0.2981
0.2709 0.2676 0.2643
0.2389 0.2358 0.2327
0.2090 0.2061 0.2033
0.1814 0.1788 0.1762
0.1562 0.1539 0.151s
0.1335 0.1314 0.1292
0.1131 0.1112 0.1093
0.0951 0.0934 0.0918
0.0793 0.077E 0.0764
0.0655 0.066 0.0630
0.0537 0.0526 0.0516
0.0436 0.0427 0,0418
0.0351 0.0311 0.0336
0.02Er 0.0274 0.0268
0.02n 0.02,17 0.0212
0.017{ 0.0170 0,0166
0.0136 0.0132 0.0129
0.0101 0.0102
0,00776 0.0075s
0.00604 0,00587 0.00570
0.00153 0.00440 0,00427
0.00336 0,00326 0,00317
0.0021E 0.00210 0,00233
0.m181 0.00175 0.00169
q.00122
0.00990
0.4E40 0.4601 0.4761
0.4443 0.4404 0.4364
0.4052 0,4013 0.3974
0.3669 0.3632 0.3594
0.3300 0.3264 0,3228
0,2946 0.2912 0.28n
0,2611 0.2578 0.2546
0.2296 0,2266 0.2236
0,2005 0.1977 0.,1949
0.1736 0.1711 0.1685
0.1492 0.1469 0.1446
0.1271 0.1251 0,1230
0.1075 0.1056 0.103E
0,0901 0.0885 0,0869
0.0749 0.0735 0,0721
0.0618 0.0606 0,0594
0.0505 0.0495 0.0485
0.0409 0.0401 0.0392
0.0329 0.0322 0.0314
0,0262 0,0256 0.0250
0,0207 0,0202 0.0197
0.0162 0.0158 0,0154
0.0125 0.0122 0.0119
0,00964 0,00939
0.00714 0.00695
0,00554 0.00539 0,00523
0.00415 0,00402 0.00391
0.00307 0,00298 0.00289
0.00226 0,00219 0.N212
0.00161 0.00159 0,00154
0,00114 0.001r.|
0.4721 0.4681 0.4641
0.4325 0.4286 0,4247
0,3936 0.3897 0.3859
0.3557 0.3520 0.3483
0.3192 0.3156 0.3121
0.2843 0.2810 0.2776
0.2514 0.2483 0.2451
0.2246 0.2177 0,2148
0j922 0.1894 0.1867
0.1660 0,1635 0.1611
0.1423 0.1401 0,1379
0,1210 0.1190 0.1170
0.1020 0.1003 0.0985
0.0E53 0.0838 0.0E23
0.0708 0.0694 0.0681
0.0582 0.0571 0.0559
0,,0475 0,0465 0.0455
0.0384 0.0375 0.0367
0.0307 0.0301 0.0294
0.0244 0.0239 0.0233
0,0192 0,0188 0.0183
0,0150 0.0146 0,0143
0.01'16 0.0113 0.0110
0.00889
0.00676 0.0065i 0.00639
0,00508 0,00494 0.00480
0,00379 0,00368 0.00357
0.00280 0.00272 0.00264
0,00205 0,00199 0.00193
0.00149 0,00144 0.00139
4
4
4
4
4
3
3
3
3
3
2
2
2
2
1
1
1
I
1
1
III7
7
7
7
b
5
5
5
4
4
3
3
2
2
2
1
1
1
1
1
1
5
5
4
4
3
2
2
1
I
12
12
12
11
11
10
10
III
7
6
0
5
4
4
3
3
2
2
1
I
I
1
I7
6
6
5
3
3
2
I
1
0
0
0
0
3
2
2
2
2
1
1
I
0
0 1
16 20 24
16 20 24
15 19 23
15 19 22
15 18 22
14 17 20
13 16 19
12 15 18
11 14 16
10 13 15
I8
7
b
6
5
4
4
3
2
222212ttr0 13
I 12
12 14
10 12
9 11
81078
7
6
I
4
4
6
5
4
4
3
Io
5
4
2
I7
6
4
3
3
2
2
2
15
14
8
7
11 13
I 11
6
5
4
3
2
2 22
28 32 36
28 32 36
27 31 35
26 30 34
2s 29 32
24 27 31
23 26 29
21 24 27
19 22 25
18 20 23
16 19 21
14 16 18
13 15 17
11 13 14
10 11 t3
81011I7
6
5
34433423322218m2316 16 21
15 17 l9
13 15 1l
11 12 11
I7
5
3
3 4
IE
6
5
7
6
5
4
I t0
I6
4
8
6
4
3
Example / Contoh:
r(z)=#*o(-;r)z)f
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If X-N(0, 1), then
JikaX- lf(0, 1), malw
P(X> k) = Q&)
P(X> 2,r) = QQ.D = 0.0179
QQ) = [n4a,k
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1. Solve the simultaneous equations x - 2y = = 8.
Selesailran persamaan serentak x - 2y = = 8.
(x+l) cmxcmE
2x cm
Diagram 2Rajah 2
A wire of length 7.5 m is cut into 10 pieces. Each piece is then bent to form arectangle. The dimensions of the first three smallest rectangles are as shown inDiagram 2.
Segulung dawai yang panjangnya 7.5 m dipotong kepada fi bahagian. Setiap bahagiandibengkolrlrnn untuk membentuk sisiempat tepat. Ulcuran tiga sisiempat tepat yangterkecil ditunjuklmn dalam Rajah 2.
FindCari(a) the value of x,
nilai x,
(b) the area of the largest rectangle.luas sisiempat yang terbesar.
2.
6 3472t2
Section A
Bahagian lt| 40 marksl
[ 40 marlmh ]
Answer all questions.
Jawab semua soalan.
[ 6 marks ]
[ 6 markah ]
(x+2) cm
| 4 marks l| 4 marknh I
[ 3 marks J
| 3 markah l
t2xy
812
Z(x+t) cm 2(x+2) cm
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3. Sketch the graph of y - -3 sin 2x for 0 < x < 2n .
Lalrarkan graf y = -3sin 2x untuk 0 <; <Ztr.
Hence, using the same axes, sketch a suitable straight line to find
of solutions for the equation x + 3n sin 2x = 0 for 0 <x 32n .
State the number of solutions.
the value of k,nilai k,
the gradient of the curve al A,kecerunan lengkung itu di A,
the equation of tangent to the curue at A.persamaan tangen kepada lengkung itu di A,
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14 marksl
| 4 marlmhl
the number
[3 marks ]
l1 mark lI t marlwh]
[ 3 marks J
| 3 markah I
12 marksl[2 marlmhl
Seterusnya, dengan menggunakan palai yang sama, lalcar satu garis lurus yang sesuai
untuk mencari bilangan penyelesaian bagi persamaan x + 3a sin 2x = 0 untuk
0 <r 4r.Ny atalran bil angan p eny ele s aian itu. 13 marlahl
4. The curve ky = ft, where k is a constant, passes through the point A (2, - 6),
Lenghmg y = +, dengan keadaan k ialah pemalar, melalui titik A (2, - 6).4 - 3x' <
FindCari
(a)
(b)
(c)
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5. Table 5 shows the cumulative frequency distribution for the marks of 40 students ina test.
Jadual 5 menuniulckan taburan kekerapan longgokan bagi markah 40 orang murid dalamsatu ujian.
Marks(Markah) <20 <40 <60 <80 <1 00
Number of students(Bilangan murid) 3 13 27 35 40
Table 5Jadual 5
(a) Based on Table 5, copy and complete the table b.1 below:
Berdasarknn Jadual 5, salin dan lengkapkan Jadual 5.1.
Table 5.1Jadual 5.L
(b) Given that the minimum mark to obtain an A* is 80,
find the percentage of students who score A*.
Diberi bahawa markah minimum untuk memperoleh A+ ialah 80,cari peratus murid yang memperoleh A*.
(c) Without drawing an ogive, calculate the median mark.Tanpa menggunakan ogtf, hitungkan markah median.
| 1 markl| | marlwh 7
12 marksl
12 marlwh I
[ 3 marks ][3 marlwh)
Marks(Markah) 0-19 20-39 40-59 60-79 80-99
Number of students(Bilangan murid)
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6.
34722
Diagram 6
Rajah 6
ln Diagram 6, OPQR is a trapezium such that 3RQ = 4OP. S lies on PR such
that PS : SR = 3 : 1, Giventhat G = 6-r and Ofr.= 4y.
Dalam Rajah 6, OPQR ialah se;buah trapezium dengan keadaan 3RQ : 4 OP.
SterletakpadaPRdengankeadaan P,S; ^SR - 3 : l. Diberi OF = 6r dan
oF. - 4y.
(a) Express in terms of x and I of ,
Unglmpkan dalam sebutan 4 dan y bast
(i) m,(ii) B
(b) Show that PQ is parallel to OS.
Tunjuklmn bahawa PQ adalah selari dengan OS.
[ 3 marks ]
13 markahl
12 marksl
12 markahl
[Lihat halaman sebelah
13 marksl
13 markahl
(c) Given that lal = , ,
l1l = 5 , find the area of trapezium OPQR.
Dibert bahawa l1l
= t , ll = 5 , cari tuas trapezium }PQR.
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l1
Solution by scale drawing will not be accepted.
Penyelesaian secara lukisan berslwla tidak akan diterima.
ln Diagram 8, AOC is a straight line and point B lies on the x-axis.Dalam Rajah 8, AOC ialah suatu garis lurus dan titik B terletak pada palai-x.
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8.
Given the area of AOAB is 10 unit2, white the area of AOBC is 20 unif,find the coordinates of B. Hence, show that the coordinates of C is (6, 8).
[ 5 marks ]Diberi luas AOAB ialah l0 unif , manalmla luas AOBC ialah 20 unif , carikoordinat titik B. Seterusnya tunjukkan bahawa koordinat titk C ialah (6, 8).
| 5 marknh I
A point P moves such that its distance from C is always twice the distancefrom A.Satu titik P bergerak dengan keadaan jaralotya dari C adalah sentiasa dua kalijarahtya dari A.
(i) Find the equation of the locus of P,Cari persamaan lolats bagi P,
(ii) Hence, determine the coordinates of the points where this locuscuts the x-axis.Seterusnya, tentukan koordinat titik-titik padamana lola$ ini memotongpalai-x.
[5 marks ]15 markahl
(a)
(b)
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t2
Rajah 9Diagram 9
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12 marksl| 2 markah )
| 4 marks I| 4 markah l
| 4 marks I| 4 marlwhl
9. Diagram 9 shows a sector OPR of a circle, centre O and radius 10 cm. pSRf isa circle, centre Q. oP and oR are tangents to the circle.
Raiah 9 menuniukkan sector bulatan OPR berpusat O dan berjejari 10 cm p^ffif iahhsebuah bulatan berpusat Q. OP dan OR adalah tangen kepada bulatan itu.
[UselGuna n =3.1421
(a) Show that the radius QR = 6.84 cm.Tunjukkan bahawa jejari QR = 6.84 cm
(b) FindCari
(i)
(ii)
the length, in cm, of the major arc pSR,panjang, dalam cm, lengkok major P^tR,
the area, in cm2, of the shaded region.luas , dalam cfrZ , kawasan yang berhrek.
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347212
10 (a) The probability that rain will fall on a certain day in the month of
December is ] . Find the probability that in a certain week in December,4
rain will fall in
Kebarangkalian bahawa hujan alwn turun pada sesuatu hari dalam bulan
Disember nhn ] . Cari kebarangkalian bahawa pada satu minggu tertentu dalam4
bulan Disember, hujan alcan turun
(r) exactly 3 days,tepat 3 hari,
(ii) not more than 5 days.tidak melebihi 5 hari.
[ 5 marks ][ 5 marlwh ]
(b) The height, in cm, of a group of students is found to be normally distributedwith mean 163 cm and standard deviation 16 cm.Tinggi sehtmpulan murid didapati bertaburan secara normal dengan min 163 cmdan sisihan piawai 16 cm
FindCari
(i) the standard score for a height of 175 cm,skor piawai bagi ketinggian 175 crc+
(ii) the height of a student which corresponds to a standard score of -0.6,tinggi seorang murid yang mempunyai skor piawai -0.6,
(iii) the probability that the height of a student chosen at random from thegroup is between 14T cm and 125 cm.ltebaranglwlian bahawa tinggi seorang murid yang dipilih secara rawakdaripada kumpulan itu adalah antara r47 cm dan-l7i cm
[5 marlrs J
[ 5 markah ]
13
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11.
l4 347212
Diagram 1 1 shows part of the graph of the function y = flx) which touches thex-axis at point P and cut the y-axis at point Q. The straight line QR which isparallel to the x-axis is the tangent to the curve at point Q.
Rajah 1l menunjuklwn sebahagian daripada graf fungsi y : f(x) yang menyentuhpalrsi-x pada titik P dan memotong paksi-y di titik Q. Garis lurus QR yang selaridengan palai-x ialah tangen kepada lengkung itu pada titik Q,
Diagram 11
Rajah tl
Given that f '(*) : 3x2 - 12 x.Diberi bahawa f '(x) = 3x2 - 12 x.
(a) FindCari(i) tn:;;,:inates of P,
(ii) (x).
(b) Find ,nli."roinates of R and hence, find the areaof the shaded region.Cari koordinat R dan setentsnya, cari luas rantau berlorek.
[2 marksl12 markah l
[ 3 marks ]| 3 markah j
[ 5 marks ][ 5 markah ]
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12.
l5
Section GBahagian C
[ 20 Marks ]120 Markah l
Answer any two questions from this section.
Jawdb mana-mana dua soalan daripada bahagian ini.
(a) the initial velocity of the particle,halaju awal zarah itu,
(b) the minimum velocity of the particle,halaju minimum zarah itu,
(c) the acceleration of the pailicle at point P,pecutan zarah itu pada titik P,
(d) the range of values of f during which the particle moves to the left.
julat nilai t ketika zarah itu bergerak ke kiri.
A particle moves along a straight line such that its displacement, s fft, is given
by s = f -A( - 15f, where tis thetime, in seconds, after it passes through a
fixed point O. The particle is instantaneously at rest at point P.
[Assume motion to the right is positive]
Satu zarah bergerak di sepanjang suatu garis lurus dengan keadaan sesarannya, .s rn,
diberi oleh s : f - 6/ - I5t, dengan keadaan t ialah masa, dalam saat, selepas melaluisatu titik tetap O. Zarah itu berhenti seketika di titik P.
lAnggapkan gerakan ke arah kanan sebagai positiJl
3472t2
12 marksl12 marknh l
[ 3 marks ]13 marknh l
[ 3 marks ]| 3 markah l
12 marksl12 marlwhl
FindCari
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13.
t6
Solution by scale drawing will not be accepted.
Penyelesaian secara lukisan berslwla tidak alun diterima.
12 cm
Diagram 13Rajah 13
ln Diagram 13 , AB is parallel to DC such that 3AB = 5CD andI ACB is obtuseDalam Rajah 13, AB adalah selari dengan DC dengan keadaan 3AB : 5CD dan
I ACB adalah cakah.
347212
[ 3 marks ][3 markah I
12 marksl
[2 markahl
[ 5 marks ]| 5 markah l
Find the length, in cm, of CD and AC.Cari panjang, dalam cm, bagi AC,(i) €, fr$(ii) Ac.
Find /. ACB .
Cari I ACB .
Find the area, in cm2, of MBC. Hence, findCari luas, dalam cnf , bagi MBC. Seterusnya, cari
(i) the area, in cm2, of MDC,Iuas, dalam crrf , bagi AADC,
(ii) distance between the two parallel lines AB and DC.jarak di antara dua garis selari AB dan DC.
(a)
(b)
(c)
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347212
14. ln an effort to strengthen Bahasa Melayu and to improve the standard of Bahasalnggeris, a training centre offers two courses, BM and Bl. The number ofparticipants for the BM course is x and the number of participants for Bl course isy. The intake of pafticipants is based on the following constraints :
Dalam usaha memperlcasakan Bahasa Melayu dan memperlwkuhlun Bahasa Inggeris,sebuah pusat latihan menawarlcan dua lanrsus, BM dan BI. Bitangan peserta tatrsus nUialah x dan bilangan peserta kursus BI ialah y. Pengambilan peserta adalah berdasarpankekangan berikui
The number of BM participants is more than 20,Bilangan peserta BM adalah melebihi 20,
The number of Bl participants is at least 10,Bilangan peserta BI adalah sehtrang-larangnya 10,
The maximum number of participants is 80,Jumlah maksimam bilangan peserta ialah 80,
The ratio of the number of BM participants to the number of Blparticipants is not more than 4 : 1.Nisbah bilangan peserta BM kepada bilangan peserta BI adalah tidakmelebihi 4 : l.
17
il
ilt
IV
(a) write down four inequalities, other than x >0, y >0, which satisfyall the above constraints.Tulis empat ketaksam),aan, selain x 20, y 20, yang memenuhisetnua kekangan di atas.
| 4 marksl
| 4 markahl
(b) Using a scale of 2 crn to 10 participants on both axes, constructand shadethe region which satisfy all the constraints. [ 3 marks ]Menggunakan skala 2 cm kepada l0 peserta pada kedua-dua palai, bina dan lorekrantau Ryang memenuhi semua kekangan di atas. ll markahl
(c) Using the graph constructed in 14(b), findDengan menggunakan graf yang dibina di L4(b), cari
(i) the range of the number of BM participants if the number of Blparticipants is 20,julat bilangan peserta BM jika bilangan peserta BI ialah 20,
(ii) the maximum total fees that can be collected if the fees for BM and Blcourses are RM 200 and RM 900 respectively.Jumlah maksimum lwtipan yuran yang diperoleh jilw yuran bagi latrsus BMdan BI ialah RM 200 danRNl3}} masing-masing
[ 3 marks ][3 markah]
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15.
18 3472t2
Table 15 shows the price indices of four food items P, Q, R and S in the year
2009, based on the year 2008 and the changes of the price indices from the
year 2009 to the year 2010. The pie chart in Diagram 15 reflects the proportion
of weightage of the 4 items.
Jadual 15 menunjukkan indeks harga bagi empat bahan malwnan P, Q, R, dan S dalamtahun 2009, berasaskan tahun 2008 dan perubahan indeks harga dari tahun 2009 ketahun 2010. Carta pai dalam Rajah 15 menggambarkan kndar pemberat empat bahantersebut.
ItemBahan
Price index inthe year 2009based on the
year 2008
Indeks hargadalam tahun 2009berasaslmn tahun
2008
Changes of priceindex from the year
2009 to the year 2010
Perubahan indeksharga dart tuhun 2009
ke tahun 2010
P 110 lncreased by 10%
Bertambah l0%
o 125 lncreased by 20o/o
Bertambah 20%
R 120 Unchanged
Tidak berubah
s 110 Decreased by 5%
Berhtrang 5o/o
DIAGRAM 15Rajah 15
TABLE 15Jadual 15
(a) Given that the price of item P in the year 2008 is RM 5,find its price in the year 2009.
' Diberi bahawa harga bahan P pada tahun 2008 ialah RM 5,
cari harga bahan P pada tahun 2009.
12 marksl
12 markahl
1500
0
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19 347212
on the year
12 marksl
(b) The price of item Q in the year 2009 is RM 7.50. Find its correspondingprice(i) in the year 2008,(ii) in the year 2010.
12 marksl
Harga bahan Q pada tahun 2009 ialah RM 7.50. Cari harga yang sepadan
bagi bahan itu
(i) pada tahun 2008,
(ii) Pada tahun 2o1o' t 2 marrrahf
(c) Find the price index ol each item for the year 2010 based2008.
Cari indeks harga bast setiap bahan pada tahun 2010 berasaslmn tahun 2008.
12 markahJ
Calculate the composite index for the four items for the year 2010 basedon the year 2008.
| 4 marks I
Hitunglwn indela gubahan bast empat bahan itu pada tahun 2010 berasasknntahun 2008.
| 4 markah)
END OF QUESTION PAPERKERTAS SOAIAN TAMAT
(d)
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3472/2 Trial SPM (PP)
1 3472/2
PEPERIKSAAN PERCUBAAN SPM 2011
ADDITIONAL MATHEMATICSTingkatan 5
KERTAS 2
PERATURAN PEMARKAHAN
UNTUK KEGUNAAN PEMERIKSA SAHAJA
3472/2(PP)Tingkatan LimaAdditional MathematicsKertas 2PeraturanPemarkahanSeptember2011
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3472/2 Trial SPM (PP)
ADDITIONAL MATHEMATICS PAPER 2 TRIAL SPM 2011 (MARK SCHEME)
No. PERATURAN PEMARKAHAN MARKAH
MARKAH
PENUH
1 x – 2y = 8 ………… (1)
8 12
8x y
− = ………….(2) (any two equations : one linear, one non-linear) P1
x = 2y + 8 or y = 8
2
x − ( x in terms of y or y in terms of x) P1
Substitute x or y into non-linear equation …
8 128
2 8y y− =
+ or
8 128
( 8)2
xx− =−
K1
y2 + 5y +6 = 0 or x2 – 6x + 8 = 0 * (y +2) (y+3) = 0 or (x – 2) (x – 4) = 0 [factorisation or formula] K1
y = -2 , y = -3 ; x = 2, x = 4 N1
x = 4 , x = 2 ; y = -3 , y = -2
[ If answers given without factorisation or using formula : K1 N1 N1 OW-1]
N1
6
No. PERATURAN PEMARKAHAN MARKAH
MARKAH
PENUH
2(a) Listing : 6x, 6x + 6, 6x + 12, … or a = 6x
d = 6
[ ]10
102(6 ) 9(6) 750
2S x= + = [Equating Sn = 750]
x = 8
P1
N1
K1
N1
(b) Area : 2(x)2, 2(x + 1)2 , 2(x + 1)2, … => 2(8)2, 2(9)2 , 2(10)2, …AP : x , x + 1, x + 2, … => 8, 9, 10, …T10 = *8 + (10 – 1)1
= 17Area of largest rectangle = 2[*8 + (10 – 1)1]2
= 578 cm2
ORPerimeter : 6(8), 6(8) + 6, 6(8) + 12, …T10 = 6(8) + (10 – 1)6 = 102 = 6y [any letter except x] K1
2y y = 17
y y Area of largest rectangle = y 2y = 17 2(17) K1
2y = 578 cm2 N1
K1
K1N1
7
2 3472/2
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3472/2 Trial SPM (PP)
No. PERATURAN PEMARKAHAN MARKAH
MARKAH
PENUH
3 Shape of graph of y = sin x , 0 ≤ x ≤ 2π [must begin from origin, O(0, 0)] N1
Shape of graph of y = sin 2x , 0 ≤ x ≤ 2π [exactly 2 complete cycles] N1
Amplitude of graph = 3 N1
Reflection of graph y = 3 sin 2x at x – axis, 0 ≤ x ≤ 2π [all correct] N1
x
yπ
= [Equation of straight line]
Draw any straight line with positive gradient passing through (0, 0)Number of solutions = 5
P1
K1N1
7
4 (a) k = (-6) [4 – 3(2)] = 12 P1
(b) )3()34)(1(12* 2 −−−= −xdx
dyK1
[ ] )3()2(34)1(12* 2 −−−= −
dx
dy[substituting x = 2 in his
dx
dy]
K1
= 9 N1
(c) y – (–6) = 9(x – 2) or equivalent [finding equation of tangent using his dy/dx ]
y = 9x – 24
K1N1
6
3 3472/2
x
y
O 2ππ½ π 3/2 π
+3
-3
2
y = - 3 sin 2x
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3472/2 Trial SPM (PP)
No. PERATURAN PEMARKAHAN MARKAH
MARKAH
PENUH
5 (a)Marks
(Markah) 0-19 20-39 40-59 60-79 80-99
Number of students(Bilangan murid) 3 10 14 8 5 N1
(b) Percentage of students scoring A+ = 10040
5* ×
= 12.5%
K1
N1
(c) L = 39.5
[ ]( )
140 13
239.5 2014
m−
= +
= 49.5
P1
K1
N1
6
6 (a) ~ ~
6 4PR x y= − +uuur
N1
~ ~
3
49
32
PS PR
x y
=
= − +
uuur uuuvK1
N1
(b)
~ ~
33
2
OS OP PS or OR RS
x y
= + +
= +
uuur uuur uuur uuur uuur
N1
~~42 yxPQ += N1
+=
~~4
2
3
3
4yxPQ and PQ is parallel to OS
or
+=~~
424
3yxOS and OS is parallel to PQ .
N1
(c) OP = 12, OR = 20 and QR = 16 [Using
~x = 2,
~y = 5 to find length]
K1
Area of OPQR = 280 unit2 N18
4 3472/2
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3472/2 Trial SPM (PP)
No. PERATURAN PEMARKAHAN MARKAH
MARKAH
PENUH
7
x2 1.00 4.00 9.00 12.25 16.00 30.25
xy 4.50 8.90 16.20 21.53 27.00 48.40
N1
N1
(a) Graph of xy against x2 : Please refer to Appendix 1 (End of Mark Scheme)Correct axes with uniform scale and a pointAll points correctly plottedLine of best fit
K1N1N1
(b) 22k
xy hxh
= + P1
(i) From Graph, gradient, m = 2h ≈ 1.50 h ≈ 0.75 [0.70 – 0.80] N1
(ii) xy-intercept, c = 3.0
3.00.75
2.25 [ 2.1 2.4]
k
k accept
≈
≈ − N1(iii) xy = 30
From graph, x2 = 184.24x∴ ≈ N1
108
(a) 10)0)4(0()000(2
1 =+−+−++ OB or 1
4 102
OB × = K1
OB = 5B(5, 0) N1
Let C be (h, k),From area ratio, OR Equation of OC : OR Area of ΔOBC
AO:OC = 10 : 20 xy3
4= = 20)5(*2
1 =×k K1
021
)3(2)(1 =+
−+h 20)000()05*0(
2
1 =++−++ k OR hk3
4=
or 021
)4(2)(1 =+
−+kK1
C(6, 8) N1
(b) PC = 2 PA2222 )]4([)]3([2)8*()6*( −−+−−=−+− yxyx K1
(x – 6)2 + (y – 8)2 = 4 [(x + 3)2 + (y + 4)2] N1x2 + y2 + 12x + 16y = 0 N1
At x-axis, y = 0 : x2 + 12 x = 0 K1x (x + 12) = 0x = 0, -12(0, 0) , (-12, 0) N1
10
9 (a) 0.6 radian = 34.370
5 3472/2
K1Equating 2h with his gradient ORequating k/
h with his xy-intercept OR
construction of the line xy = 30 on his graph.
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3472/2 Trial SPM (PP)
tan 0.610
QRrad = or 0tan 34.37
10
QR= K1
QR ≈ 6.84 cm N1
(b) (i) °=∠ 55.63or rad 971.0RQO OR °=∠ 111.26or rad 942.1PQR
OR )90(34.37or rad )2
6.0(2
1major °+°+=∠ π
PQR
°=∠ 248.74or rad 342.4majorPQR
Arc PSR = 6.84 4.342 or 84.62360
74.248 ××°
° π
= 29.70 cm
P1
N1
K1
N1
(ii) Area of OPQR = 284.6102
1 ××× K1
Area of sector OPR = 2.1102
1 2 ×× or 210360
74.68 ××°° π K1
Area of shaded region = 68.4 – 60 or 68.4 – 59.99 K1
= 8.4 cm2 or 8.41 cm2 N110
10 (a) p = 0.75 , q = 0.25
(i) P(X = 3) = 7 3 43 (0.75) (0.25)C or
43
37
4
1
4
3
C K1
= 0.05768 or 16384
945N1
(ii) P(X ≤ 5) = 1 – [P(X = 6) + P(X = 7)] K1= [ ]7 6 1 7
61 (0.75) (0.25) (0.75)C− + K1= 0.5551 N1
(b) (i)175 162
16z
−=
= 0.75 N1
(ii)163
0.616
x − = −
x = 153.4 cm N1
(iii) P(147 < X < 175) = P (–1 ≤ Z ≤ 0.75) K1 = 1 – P(Z> 1) – P(Z > 0.75) K1 = 1 – 0.1587 – 0.2266 = 0.6147
N110
11 (a) (i) At P, 2'( ) 3 12 0f x x x= − = K1 3x (x – 4) = 0
6 3472/2
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3472/2 Trial SPM (PP)
x = 0 , x = 4P(4, 0) N1
(ii) 2
3 2
( ) (3 12 )6
f x x x dxx x c
= −= − +
∫K1
c = 32 N1f(x) = x3 – 6x2 + 32 N1
(b) R(6, 32) P1
Area of = 6 x 32 OR *32 – *(x3 – 6x2 + 32)= 192 …………….… (I) = 6x2 – x3
K1
Area under curve from x = 0 to x = 6 Area of shaded region
( )∫ +−=6
0
23 326* dxxx ( )∫ −=6
0
326 dxxx K1
= 64
3
0
2 324
xx x
− +
6
0
43
43
6
−= xx
K1
= 84………………………………. (II)
Area of shaded region = (I) – (II) = 192 – 84 04
)6(
3
)6(6 43
−
−=
= 108 cm2 N110
SECTION C (20 marks)12 S = t 3 – 6t 2 – 15t
(a) 23 12 15ds
v t tdt
= = − − K1
t = 0, v = –15 ms-1 N1
(b) For minimum velocity, 0126 =−= tdt
dvK1
t = 2 N1vmin = 3(2)2 – 12(2) – 15
= – 27 ms-1 N1
(c) At P, v = 0 : 3(t + 1) (t – 5) = 0 K1 t = 5 N1
a = 6 (5) – 12 = 18 ms-2 N1
(d) v < 0 3 (t + 1) (t – 5) < 0 K1 0 < t < 5 N1
10
13 (a) (i) AB = 20 cm
(ii) 2 2 2 020 8.9 2(20)(8.9) cos 54AC = + −
P1
K1AC = 16.43 cm N1
7 3472/2
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3472/2 Trial SPM (PP)
(b) ACB∠−+= cos)43.16)(*9.8(243.16*9.820 222 K11739.0cos −=∠ACB
∠ ACB = 100 o [accept 100.01o ] N1OR
Using sine rule,
43.16*
54sin
9.8
sin °=∠CABOR
43.16*
54sin
20
sin °=∠ACB K1
°=∠ 99.25CAB°−°−°=∠ 99.2554180ACB
= 100.01o °=∠ 100ACB N1
(c) Area of ΔABC = ½ (8.9) (20) sin 540 = 72.00 cm2
K1N1
(i) Area of ΔADC = ×5
3 Area of ΔABC OR °99.25sin)43.16)(12(
2
1
= 43.2 cm2 = 43.20 cm2
OR 2.772202
1 ==>=×× hh
Area of ΔADC = 2.7122
1 ×× K1
= 43.2 cm2 N1
K1N1
(ii) Let h cm be the distance between AB and DC
72202
1 =×× h
h = 7.2 cm N110
14 (a) x > 20 y ≥ 10x + y ≤ 80x ≤ 4y (or y ≥ ¼ x )
(b) Axes correct and one *straight line correct
Draw correctly all 4 straight linesRegion R is correctly shaded and labelled.(Please refer to appendix 2)
(c) (i) 20 < x ≤ 60
(ii) 200 (21) + 300 (59)
The maximum fees collected is RM 21 900
N1N1N1N1
K1N1N1
N1
K1
N1
10
15(a) Item P : 110100
509 =×
PK1
8 3472/2
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3472/2 Trial SPM (PP)
P09 = RM 5.50 N1
(b) Item Q : (i) Q2008 = RM 6.00 N1
(ii) Q2010 = RM 9.00 N1
(c)Item 2008I2010
P 121Q 150R 120S 104.5
N1
N1
(d) Item 2008I2010 W WI
P 121 80 9680Q 150 150 22500R 120 90 10800S 104.5 40 4180
∑W = 360 ∑WI = 47160
Weightage for item S = 40o or 4 [as seen in his formula ]
ii IW∑ = 80(121) + 150(150) + 90(120) + 40(104.5)or 8(121) + 15(150) + 9(120) + 4(104.5)
360
)5.104(40)120(90)150(150)121(80 +++=I
or 36
)5.104(4)120(9)150(15)121(8 +++
= 131
P1
K1
K1
N110
Skema Pemarkahan Tamat
Appendix 1Graph for Question 7
9 3472/2
One mark each for every TWO correct answers.
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10 200 30 40 50 60 70 80
80
50
40
30
70
60
20
10
xy
R
x
100
90
(21, 59)
x+ y = 80
X
x = 4y
x = 20
y = 10
3472/2 Trial SPM (PP)
Appendix 2Graph for Question 14 [LINEAR PROGRAMMING]
10 3472/2
5 100 15 20 25 30 35 40
40
25
20
15
35
30
10
5
xy
X
X
x
x
x2
50
45
(0, 3.0)
x
18
(28, 45)x
Gradient ,
45 3
28 0
1.50
m−=−
=
x
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3472/2 Trial SPM (PP)
END OF MARKING SCHEMESKEMA PEMARKAHAN TAMAT
11 3472/2
y
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