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SULIT 1449/1MatematikKertas 1Percubaan SPMSeptember 20131 jam 15minit
PENTAKS IRAN SUMAT IF 3SIJ IL PELAJARAN MALAYS IA 2013
MATEMATIK (KERTAS 1)Satu jam lima belas minit
JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU
1. Kertas soalan in i adalah dalam dwibahasa.2. Soalan dalam bahasa Melayu mendahului soalan yang sepadan dalam bahasa Inggeris.3. Calon dikehendaki membacamaklumat dihalaman belakang kertas soalan ini.
1449/1
Kertassoalan inimengandungi 31 halamanbercetakTERENGGANU NEGERIANJUNG ILMU
Dibiayai oleh:Kerajaan Negeri Terengganu
Dicetak Oleh:Percetakan Yayasan IslamTerengganu Sdn.Bhd.
Tel: 609-666 8611/6652/8601 Faks: 609-666 0611/0063
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SULITMatematikKertas2Percubaan SPMSeptember 20132 jam 30 minit
NAMA
Tingkatan
PENTAKS IRAN SUMAT IF 3SIJ IL PELA JARAN MALAY S I A 2013
MATEMATIK (KERTAS 2)Dua jam tiga puluh minit
J ANGAN BUKA KERTAS SOALAN IN I S E H IN G G A DIBERTTAHU
1. Tuliskan nama dan tingkatan andapada ruangyangdisediakan.2. Kertas soalan
dwibahasa.in i adalah dalam
3. Soalan dalam Bahasa Melayumendahului soalanyang sepadan dalamBahasalnggeris.4. Colon dibenarkan menjawabkeselwuhan atari sebahagian soalansoma ada dalam Bahasa Inggeris atauBahasaMelayu.5. Colon dikehendaki membaca maklumatdi halaman belakangkertassoalanini.
Pemeriksa
Bahagian Soalan MarkahPenuh MarkahDiperoleh
A
1 32 33 44 45 56 47 68 59 610 611 6
B
12 1213 1214 1215 12
Jumlah
Kertas soalan ini mengandungi 36 halaman bercetakTERENGGANU NEGERIANJUNG ILMU
Dibiayai oleh:Kerajaan Negeri Terengganu
Dicetak O leh :Percetakan YayasanIslamTerengganu Sdn. Bhd.Tel: 609-666 8611/6652/860J Faks: 609-666 0611/0063
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1449/1 & 2Matemat ikKertas 1/22013
J A BA T AN P E L AJ A RA N T E R E NG G A NU
P E N I L A I A N S U M A T I F 3 T A H U N 2013
S I J I L P E L A J A R A N M A L A Y S I A
MATEMATIK 1 4 4 9 / 1 / 2
Ker tas 1 & 2
PERATURAN PEMARKAHAN
Ma r k a h = Kertas 1 + Kertas 21 4 0
Peraturan Pemarkahan ini mengandungi 16 halaman bercetak
Disediakan oleh : Guru AKRAM Negeri Terengganu Dibiayai oleh : Kerajaan Negeri Terengganu
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SULIT
No.Soalan
SKEMA PERMARKAHANPENILAIAN SUMATIF 3 TAHUN 2013SIJIL PELAJARAN MALAYSIA
MATEMATIK KERTAS 2BahagianAPeraturan Pemarkahan
2y = x + A
Garis lurus y = 3 dilukis dengan betulRantau di lorek dengan betul.Nota:y = 3 dilukis dengan garis putus-putus , diberi Kl PI
(a) ZGFK@ZKFG
(b) ianZGFK=) 13212'@21.04
y = 3
1449/1/2 Pentaks i ranSumat i f3 20I3
1449/1/2
Markah
K l
P2
PI
K l
N '
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SUL IT
3x+ 2y=S or x =2yor any equivalent4x - 8 or any equivalent
OR
0>>1
(|)(-2)-(l)(l)1
- 2 -1
-i l2)
' - 2 -03
(4 \
Note: (_3)(_2)_(1)(1)-1-
-1 2 y or
'32
'x ^ (4\y j 0) give Kl
^ J
x = 2y= l
Note : Answer '21
3x2+x-4 = 0(3x +4)(x-l) =0x= 1 ,x = - -
give N1
1449/1/2 Pentaksi ran Sumatif 3 2013
K2
1449/1/2
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K l
N lN l
K lK l
N1N1
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(a) (i) Bukan pemyataan / non-statement(ii) Pemyataan / statement
(b) Sebilangan/some(c) 2n3 + 5
n= 1,2 ,3 ,4
1 6 x l 6 x /
227
x7x7x/
2216x l 6x r x7x7x /=8167
(a) mFO=-
2 = -(4) + c2
v=-x-4' 21(b) m0R=--
3 =(6) + c2c = 6
1449/1/2 Pentaksiran Sumat i f 3 2013
1449/1/2
P IP I
P I
K l
N l
K l
K l
K l
N l
P I
K l
N l
P I
K l
N l
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(a) 60(b) (i) 12.40pm @1240
( i i) 20 min
( c ) 160km I60km(60)180min \80jam
1= 53.33 = 53 3
120 _ 22 %.(a) x2x x l4360 790 . 22or x 2 x x 7~" 360 77+Mx2x^xl4+~*-x2x^x7+7+14360 7 360 7
16 8 - atau 68.333
or90 22x x 7 '360 7
90 22 _ ,x 7 "
/1% 120 22 f.a(b) x xl4360 7120 22 1/l2x xl4 -360 7 360 71 6 6 - atau 166.836
NOTE :1.Accept n for K mark.2. Accept correct value from incomplete substitution forKmark.3. Correct answer from incomplete working, award Kk2.
1449/1/2 Pentaks i ranSumat i f3 2013
1449/1/2
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P I
PI
K l
N l
K l
K l
N l
K l
K l
N l
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10
11
(a) (5,5) (5,P) (5,N) (5,T)(P,J) (P,P) (P,N) (P,T)(2, J) (2,P) (2,N) (2,T)(M,J) (M,P) (M,N) (M,T)Nota : Ter ima betul 14 atau 15 untuk PI(b)(i) (P,J) (P,P) (P,N) (P,T)(M,J) (M,P) (M,N) (M,T)
8 * la t au 16 2(ii) (5, J) (5,P) (5,N) (5,T) (P,T)(2,J) (2,P) (2,N) (2,T) (M,T)
10 + 5 a t a u 16 8Nota: Terima tanpa kerja atau tanpa ringkasan
(a) 12(4)-5(-l)' 4 5v-1 2-< 4 5^1 (4 5> a tau 13 13
13 ^ 2, 1v~323 ,
r2 - 5V ^ ri6^(b)
VvJV
Nota : Jika
1 4 >/
12(4)-5(-l)
x= 3
y= - 2
V"5y'4 5Vl6^-1 2 v-5,
K-hsahaja dapat Nl
1449/1/2 PentaksiranSumati f3 2013
1449/1/2
P2
Kl
Nl
PI
Nl
Kl
Nl
Kl
Kl
Nl
Nl
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12 (a) 12.5- 1 6 . 7
(b) Graph:Axes drawn in the correct direction with uniform scales.for - 4 < x < 4 and -25 < y < 25.All 7 points and 2 points* are plotted correctly orthe curvepasses through these points.Note :7@ 8 points are correctly plotted, KlA smooth and continuous curve in range of- 4 < x < 4 with no straight line part andpasses through all 9 correct points using thegiven scales.
(c) (i) - 1.9 < x < -1.7 and > The graph isf drawn(ii) - 9.0 < y < - 7.0(d)
Note:
Identify equation y - -4x - 5Straight line y = -4x - 5 correctly drawn and meet thecurve.The values ofx : 1.8 < x< 2.0
-3 .2 < x < -3 .0(i) Allow Nmark if thevalues ofx are shown onthe
graph,(ii) The values of x obtained by calculation, give NO
1449/1/2 Pentaksiran Sumat i f 3 2013
1449/1/2
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P I
K2
N l
P IP I
K l
K l
N lN l
12
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SUL IT
13
to
(a) (i) ( -5 .8)(ii) (3, 16)
Note : (5, 13)marked on diagram, award PI
(b) (i) V- Rotation of 180 about the centre (-1,10)Note :1. Acceptwithoutdirection for P3mark.2. P2 : Rotation of 180 or Rotation about the centre
(-1, 10)3. PI : Rotation
(ii) W- Enlargement centre L// (-6, 7)with scale factor3
Note : 1. P2 : Enlargement centre L // (-6, 7) orEnlargementwith scale factor 32. PI : Enlargement
(c) 26.5 x(3)2
26.5 x(3)2-26.5212
1449/1/2 Pentaksiran Sumat i f 3 2013
1449/1/2
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P2
P3
P3
K l
K l
N l 12
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SULIT
14(a)
(b)
W
id)
II 1449/1/2
J is im b ad an/?oi/v mass
3 6 - 4 04 1 - 4 54 6 - 5 05 1 - 5 55 6 - 6 06 1 - 6 56 6 - 7 071 - 757 6 - 8 0
Sempadan AtasUpperBoundary
40.545.5
50.555.560 .565.570.575.580.5
KekerapanFrequency
581011
94
KekerapanLonggokanCumulativeFrequency
0
152536
45
49
50
Sempadan atas semua betulKekerapan semua betulKekerapan longgokan semua betulNota : 7 atau 8 kekerapan betul beri PI
OgifPaksi dilukis mengikut arah yang betul.Skala seragam digunakan dan paksi x dilabel dengan sempadan atasSemua titik ditanda dengan betul.Nota: 7 atau 8 titik ditanda dengan betul dapat KlOgifdilukisdengan betuldan licin.
(43 x2) +(48x 5)+(53 x8) +(58 x10) +(63 x11) +(68x 9) +(73 x4)+(78 x1)
300550
- 60.1 atau 60110
50
3005Nota: Jika dilunjukkan tanpa kiraan danjavvapan yang betul dapat Kk250
Median = 60.5 0.5
P IP2P I
P I
K2
N l
K2
N l
Kl
1449/1/2 Pentaksi ran Sumatif 3 2013 SUL IT
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15(a)
13
G
B
Bcntuk kelihatan betul, semua garis penuhFG = GK
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SULIT
b(i)
14
w ./>
T Q
H R
F G K
Bentuk kelihatan betul dengan segiempat tepat PQTW, QRST,FGHS dan GHRK.Semua garis penuh
SF = FG = GK= KR, WT= TS= PQ = QR< SF
Ukuran betul sehingga 0.2 cm (sehala) dan sudut di semua bucusegiempat tepat = 90 1
1449/1/2 Pen taks i ranSumat i f3 2013
1449/1 / :
K l
K lBgtKl
N2BgtKlKl
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SUL IT 15 1449/1/2
b(ii) R Q\K J \G H \
B C PBentuk kelihatan betul dengan segiempat tepat BCJK dan trapeziumCPQR.Semua garis penuh K l
Nota : (i) Abaikan garis G clan //disambung dengan garisputus-putusmembentuk segiempat tepatBCGH. Kl Bgt.Kl(ii) QR
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SULIT
16.(a)
(b)
16 1449/1/2
100 B P2Nota: 100 atau 0B dapat PI
(i) (35t7, 100 5) P1P1
(ii) (180-35-35)x60 K l6600 N l
(c) (i) 180x60xfaM35 K2
8846.84 N l
Nota : Penggunaan kos 35 dapat Kl(ii) 70 x 60 K l
8846.84 + 70x60 K l850
15.35 N l12