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  • SKEMA PEMARKAHAN SOALAN KERTAS 2 MATEMATIK TAMBAHAN

    PERCUBAAN 2020

    Soalan Skema pemarkahan Sub

    marka

    h

    Jumlah Markah

    1 2π‘₯ + 4𝑦 = 24

    π‘₯ = 12 βˆ’ 2𝑦

    [(12 βˆ’ 2𝑦) + 𝑦]2 = (12 βˆ’ 2𝑦)2 + (3𝑦)2

    𝑦(𝑦 βˆ’ 2) = 0 π‘œπ‘Ÿ (π‘₯ βˆ’ 8)(π‘₯ βˆ’ 12) = 0

    𝑦 = 0 , 𝑦 = 2

    π‘₯ = 12 , π‘₯ = 8

    π‘Žπ‘Ÿπ‘’π‘Ž = 48π‘π‘š2

    K1

    K1

    K1

    K1

    K1

    K1

    N1

    7

    2 a i)

    100.5 + 150.5

    2

    125.5 (π‘Žπ‘π‘π‘’π‘π‘‘ π‘€π‘–π‘‘β„Žπ‘œπ‘’π‘‘ π‘€π‘œπ‘Ÿπ‘˜π‘–π‘›π‘”)

    ii) 75.5(10)+125.5(40)+175.5(10)+225.5(30)+275.5(20)

    110

    180.05

    K1

    N1

    K1

    N1

    4

    b) 150.5 π‘œπ‘Ÿ 50 π‘œπ‘Ÿ 10

    150.5 + (

    110 2 βˆ’ 50

    10 )50

    175.5

    P1

    K1

    N1

    3

  • 3a) (i)

    ii)

    3b)

    𝐡𝐷⃗⃗⃗⃗⃗⃗ = 𝐡𝐴 βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— + 𝐴𝐷⃗⃗⃗⃗⃗⃗ βƒ—

    𝐡𝐴⃗⃗⃗⃗ βƒ— = βˆ’20π‘₯

    𝐴𝐷⃗⃗ βƒ—βƒ— βƒ— = 4 βˆ— 8𝑦 = 32𝑦

    𝐡𝐷⃗⃗⃗⃗⃗⃗ = 𝐡𝐴 βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— + 𝐴𝐷⃗⃗ βƒ—βƒ— βƒ— = βˆ’20π‘₯ + 32𝑦

    𝐸𝐢⃗⃗⃗⃗ βƒ— = 𝐸𝐷⃗⃗⃗⃗ βƒ— + 𝐷𝐢⃗⃗⃗⃗ βƒ—

    𝐸𝐷⃗⃗⃗⃗ βƒ— = 3𝐴𝐸⃗⃗⃗⃗ βƒ—βƒ— βƒ—βƒ— βƒ— = 24𝑦

    𝐸𝐢⃗⃗⃗⃗ βƒ— = 24𝑦 + (25π‘₯ βˆ’ 24𝑦 ) = 25π‘₯

    N1

    N1

    2

    𝐡𝐷⃗⃗⃗⃗⃗⃗ = βˆ’20π‘₯ + 32𝑦

    𝐸𝐹⃗⃗⃗⃗ βƒ— = 3

    5 𝐸𝐢⃗⃗⃗⃗ βƒ— = 15π‘₯

    𝐹𝐷⃗⃗⃗⃗ βƒ— = 𝐹𝐸⃗⃗⃗⃗ βƒ— + 𝐸𝐷⃗⃗⃗⃗ βƒ—

    = βˆ’πΈπΉβƒ—βƒ—βƒ—βƒ— βƒ— + 24𝑦

    = βˆ’15π‘₯ + 24𝑦

    𝐹𝐷⃗⃗⃗⃗ βƒ— = 3(βˆ’5π‘₯ + 8𝑦)

    𝐡𝐷⃗⃗⃗⃗⃗⃗ = 4(βˆ’5π‘₯ + 8𝑦)

    𝐡𝐷⃗⃗⃗⃗⃗⃗ = 4 ( 𝐹𝐷⃗⃗ βƒ—βƒ— βƒ—

    3 )

    𝐡𝐷⃗⃗⃗⃗⃗⃗ = 4

    3 𝐹𝐷⃗⃗⃗⃗ βƒ—

    K1

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    4

  • 3c)

    |𝐡𝐷 β†’| = βˆ’20π‘₯ + 32𝑦

    = |βˆ’20(2) + 32(3) |

    = √(40)2 + (96)2

    = 104

    K1

    N1

    2

    4a

    b)

    π‘₯ log10 3 = 𝑦 log10 5 *

    𝑦 log10 5 = 𝑧 log10 3 + 𝑧 log10 5 * *Either

    𝑦 log10 5 = 𝑧 ( 𝑦

    π‘₯ log10 5) + 𝑧 log10 5

    5𝑦 = 5 𝑧𝑦 π‘₯ Γ— 5𝑧

    𝑧 = π‘₯𝑦

    𝑦 + π‘₯

    log10(2π‘₯(4π‘₯ βˆ’ 1)) = 1

    8π‘₯2 βˆ’ 2π‘₯ = 10

    π‘₯ = 5

    4 , π‘₯ = 1

    K1

    K1

    N1

    K1

    K1

    N1

    3

    3

    5a πΏπ‘’π‘Žπ‘  =

    1

    2 π‘Ÿ2πœƒ =

    1

    2 (8)2(1.956)

    = 62.592 π‘š2

    K1

    N1

    2

    5b 𝑅𝐢 + 𝐢𝑄 + 𝑅𝑄

    𝑅𝐢 = 6

    𝐢𝑄 = π‘Ÿπœƒ = 8(1.956) = 15.648 * (* EITHER)

    𝑅𝑄 = π‘Ÿπœƒ = 14(3.142 βˆ’ 1.956) = 16.604*

    π‘π‘Žπ‘›π‘—π‘Žπ‘›π‘” , π‘‘π‘Žπ‘™π‘Žπ‘š π‘š , π‘π‘Žπ‘”π‘Žπ‘Ÿ π‘¦π‘Žπ‘›π‘” π‘‘π‘–π‘π‘’π‘Ÿπ‘™π‘’π‘˜π‘Žπ‘›

    π‘’π‘›π‘‘π‘’π‘˜ π‘šπ‘’π‘šπ‘Žπ‘”π‘Žπ‘Ÿ π‘π‘Žπ‘‘π‘Žπ‘  π‘π‘’π‘›π‘”π‘Ž

    = 6 + 15.648 + 16.604

    = 38.252 π‘š

    P1

    K1

    K1

    N1

    4

  • 6a (π‘₯ βˆ’ 2)2 = π‘₯ + 4

    π‘₯2 βˆ’ 5π‘₯ = 0

    π‘₯ = 0, π‘₯ = 5

    π‘˜ = 5

    K1

    N1

    2

    6b βˆ— ∫ π‘₯ 5

    0 + 4 𝑑π‘₯ βˆ’ βˆ— ∫ π‘₯2

    5

    0 βˆ’ 4π‘₯ + 4 𝑑π‘₯ * Either

    βˆ— [ π‘₯2

    2 + 4π‘₯]

    0

    5

    βˆ’ βˆ— [ π‘₯3

    3 βˆ’ 2π‘₯2 + 4π‘₯]

    0

    5

    * Either

    [( (5)2

    2 + 4(5)) βˆ’ (0)]

    βˆ’ [( (5)3

    3 βˆ’ 2(5)2 + 4(5)) βˆ’ (0)]

    = 20.83

    K1

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    N1

    4

    7.

    10

  • 8.

  • (LIHAT LAMPIRAN )

    9. π‘Ž)

    𝑆𝑖𝑛(π‘₯ βˆ’ 𝑦) + 𝑠𝑖𝑛(π‘₯ + 𝑦)

    2π‘˜π‘œπ‘ π‘₯π‘˜π‘œπ‘ π‘¦

    = (𝑠𝑖𝑛 π‘₯ π‘˜π‘œπ‘  π‘¦βˆ’π‘˜π‘œπ‘  π‘₯ 𝑠𝑖𝑛 𝑦 ) + (𝑠𝑖𝑛π‘₯ π‘π‘œπ‘  𝑦+π‘π‘œπ‘  π‘₯ 𝑠𝑖𝑛 𝑦

    2π‘˜π‘œπ‘ π‘₯π‘˜π‘œπ‘ π‘¦

    = 2𝑠𝑖𝑛π‘₯π‘˜π‘œπ‘ π‘¦

    2π‘˜π‘œπ‘ π‘₯π‘˜π‘œπ‘ π‘¦

    = tan x

    b)

    π‘₯

    πœ‹ βˆ’ 𝑦 = 1

    𝑦 = π‘₯

    πœ‹ βˆ’ 1

    Ο€

    Shape

    Cycles

    Amplitude

    Modulus

  • 10

    10.

    2

    3

    2

    2

    3

    3

    2

    2

    2

    2

    2672.49

    )3094.2(2)3094.2(32

    3094.2

    0632

    632

    232)

    232

    )16(2)

    )16,4(

    16

    )4()4(8,4

    4

    2

    28

    8,0

    0)8(

    0,

    )8()

    unit

    umLuasmaksim

    k

    k

    k dk

    dL

    kkLc

    kkL

    kkABCDtepatsegiempatLuasb

    kkB

    k

    kkykxwhen

    k

    k OA

    xx

    xx

    yxpaksipada

    xxya

    answer

    =

    βˆ’=

    =

    =βˆ’

    βˆ’=

    βˆ’=

    βˆ’=

    βˆ’=

    βˆ’βˆ’

    βˆ’=

    βˆ’βˆ’βˆ’=βˆ’=

    βˆ’=

    βˆ’ =

    ==

    =βˆ’

    =βˆ’

    βˆ’=

    10

    Equation

    draw line correctly

    NOS=1

  • 11.

    10

    12. a)

    I : 80ο‚£+ yx

    II : 24038 ο‚³+ yx

    III : 1202 ο‚£+ yx

    Refer to graph

    One straight line drawn correctly

    10

  • All straight lines drawn correctly

    Correct region

    i)

    Draw line x = 2y

    Coconut, x = 53; Rambutan, y = 27

    (ii)

    700x + 250y = 7000

    14x + 5y = 140

    From graph, Profitmin = 700(10) + 250(55)

    RM20 750.00

    13. (a)

    (b) Composite index in the year 2007 based on the year

    2006

    =

    = 114.4

    (c)

    Let the price index of item S in the year 2008 based on the

    year 2007 be z.

    105 =

    10

  • 1 050 = 745 + 2z

    305 = 2z

    z = 152.5

    Thus, the price of item S increases by 52.5% from the year

    2007 to the year 2008

    14.

    a) i) =

    sin ∠EFG =

    = 0.8245

    ∠EFG = 55.53°

    ii) 6.72 = 3.42 + 6.42 βˆ’ 2 Γ— 3.4 Γ— 6.4 Γ— cos ∠EHG

    cos ∠EHG =

    = 0.1753

    ∠EHG = 79.91°

    iii ∠EGF = 180Β° βˆ’ 30.3Β° βˆ’ 55.53Β°

    = 94.17Β°

    Area of Ξ”EFG

    = Γ— 6.7 Γ— 4.1 Γ— sin 94.17Β°

    = 13.7 cm2

    Area of Ξ”EGH

    = Γ— 3.4 Γ— 6.4 Γ— sin 79.91Β°

    = 10.71 cm2

    Area of quadrilateral EFGH

    = 13.7 + 10.71

    = 24.41 cm2

    b) (i)

    ii ∠E'F'G' = 180Β° βˆ’ 55.53Β°

    = 124.47Β°

    10

  • 15 .(a) v = 15 - 10t

    v = -5 ms-1

    (b) 15 - 10t = 0

    (c) 15(4) - 5(4)Β² + h = 0

    h = 20

    (d)

    (e) v = 15 - 10t

    a = -10

    N1

    N1

    K1

    N1

    K1

    N1

    K1

    K1

    N1

    N1

    10

    𝑑 = 3

    2 𝑠

    𝑠 = βˆ’5(𝑑 βˆ’ 3

    2 ) 2

    + 45

    4

    Tinggi maksimum = 45

    4 + 20

    = 125

    4 m

  • LAMPIRAN SOALAN NO 8