Trial Terengganu SPM 2013 MATHS Ques_Scheme All Paper

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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

    [Lihat halaman sebelahhttp://edu joshuatly com

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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

    1449/2

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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

    http://edu.joshuatly.com

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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

    Kl

    K l

    N lN l

    K lK l

    N1N1

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    SULIT

    (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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    SULIT

    (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

    P I

    P I

    PI

    K l

    N l

    K l

    K l

    N l

    K l

    K l

    N l

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    SULIT

    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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    SULIT

    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

    K lK l

    P I

    K2

    N l

    P IP I

    K l

    K l

    N lN l

    12

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    6 r sO u 4k

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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

    P I

    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

    12

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    SULIT

    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