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  • SKEMA JAWAPAN

    PEPERIKSAAN PERCUBAAN KERTAS 1

    2017

    SOALAN 1

    BIL JAWAPAN SUB MARK MARKAH

    PENUH

    1

    1)32(log:B1

    5)32(:B2

    5

    1

    

    

    xx

    xx

    2

    5 x

    3 3

    2 1

    3

    1

    2 

    x

    x

     

      

     2 log

    x

    x

    3

    B2

    B1

    3

    3

    2527:1

    25133:2

    213

    logloglog

    loglog

    aaa

    aa

    aB

    B

    nm

    

    

    

    3

    4 3 x – 3y

    B2 : y

    x

    3

    2

    1

    2

    3

    )3(

    B1 : m = 9 x atau n = 33

    y

    3 3

    5. 1

    B2 :    

     

    3

    1 3

    log log

    5

    5

    B1 :

    3 3

  • SOALAN 2

    1

    1

    7

    7

    7

    7

    1

    7

    777

    7log

    )753(log

    7

    1 log

    105log :B1

    7log

    7log5log3log :B2

    1

    1 :B3

    

    

    

    

    atau

    kh

    1kh

    4

    4

    2

    kh

    hk

    

    2

    12

    3log2log2

    5log2log3log2

    55

    555

    

    5log2log2log 55 2

    5  or 3log2log 5 2

    5  or

    5log2log3log 1212 2

    12 

    12log

    90log

    5

    5 or 2 log 5 3 or 2 log52

    4

    B3

    B2

    B1

    4

    3 (a)

    p

    1

    (b) 12

    25

    p

    p

    13log

    23log 2

    5

    m

    m

    m

    m

    m

    m

    9log

    243log 2

    1

    3

    B2

    B1

    4

  • 4

    2

    23 xy 

    B3 : 2

    log8loglog2 222 bc 

    B2: 4log

    8loglog

    2

    2

    2

    2 bc 

    B1: bc 8loglog 4 2

    4  @ 4log

    8 log

    2

    2

    2 b

    c

    4 4

    5

    q

    pq

    1

    B3 : log2 3(1 – p) = pq

    B2 : log2 3 = p(log2 4 + log2 3)

    B1 : 12log

    3log

    2

    2 = p

    4 4

    SOALAN 3

    1 15

    B2 : )2)(1()5)(1()42(  nxnx

    B1 : )5)(1()42(  nx atau

    )2)(1(  nx

    3 3

    2 n = 42

    87)2)(1(5  n

    d = 2

    3

    B2

    B1

    3

    3 1256

    )168(2))8(4)200(2( 2

    5 5 S

    )8(42005 T

    3

    B2

    B1

    3

  • 4 a) k = 2h + 1

    k -3h = (h + 2) – k

    b) 9 – 6h

    2

    B1

    1

    3

    5 a) d = 6

    b) 5

    B1 :  

      

     

       )5(283)6(283 )5()6(

    22

    1

    2

    3

    SOALAN 4

    1 7 1 3

    4

    B1 : 282 T atau 7

    28

    2

    2

    6

    1

    3

    1 1

    9

    1

    3

    1 r

    3

    B2

    B1

    3

    3 3

    2

    6 r dan p = 6

    3 2

    6 r atau p = 6

    p

    p

    p

    ppp

    18

    2

    36

    6,0,6

    2

    

    3

    B2

    B1

    3

  • 4 x1 = 3 dan x2 = –1 B2 : (x – 3)(x + 1) = 0

    B1 : 3x 2 – 6x – 9 = 0

    3

    5 -24

    B2 :

     

      

     

    4

    3 1

    6

    B1 : a = - 6 , 4

    3 r

    3 3

    SOALAN 5

    1 (a) 5

    B1 : )2(

    )4(

    )4(

    )2(

     

    x

    x

    x

    x

    2 4

    (b)

    2

    243

    B1 : a = T1 = 81 ...seen atau

    3

    1 1

    81

    2

    2 (a)

    3

    2

    5

    4

    27

    8  or equivalent 

      

     

    2

    3

    1

    2

    T

    T

    T

    T

    (b) 3

    4 or equivalent

  • 3

    2 1

    9

    4

    3 a) x = 6

    B1 : 12

    2412 

    x

    b) 6096

    B1 :    

    12

    16

    12

    16 22 310

     

    1

    1

    4

    4 atau 5.5

    B3 :  

      

     

    2

    1 11

    2T

    B2 : a = 11

    B1 : 22

    2

    1 1

    a

    4 4

    5 a) a = 8 , - 2

    B1 : 4

    224

     

    a

    a

    a

    a

    b) 16

    2187 atau 136.6875

    B1 :  

      

     

    2

    3 8

    9 3

    16 T

    1

    1

    4

    SOALAN 6

    1 y = x3

    3

    3

  • B2 : 4 = 4 + c atau 12 = 12 + c atau c = 0

    B1 : m = 1

    2 p = 2 and 1q

    p = 2 or 1q

    04

    )5(3

     p or 55 q

    qpxxy 52 2 

    3

    B2

    B1

    3

    3 s = 3 and t = 5

    7 = 2s + 1 or t = 2(2) + 1

    3

    B1

    3

    4 r = 1, p = 0.001 (BOTH)

    r = 1 or p = 0.001 (either one)

    =-3 or r = or

    3

    5 m =

    2

    3 dan n = 3

    B1 : –2 = m

    3 

    y = –2x

    2 + 3x

    2

    1

    3

    SOALAN 7

    BIL JAWAPAN SUB MARK MARKAH PENUH

    1 2 rad

    B3 : 0)2)(2(  

    4

    4

  • B2 : 100 2

    40

    2

    1 2

     

      

     

    B1 : 40 rrr

    2 (a) 842.1

    5.65 

    (b) 23.025

    )842.1()5(

    2

    1 2 * (candidate’s  from a)

    2

    B1

    2

    B1

    4

    3 (a) 10.8 cm

    (b) 19.44 cm2

    ) 142.3

    180 9.0tan(612

    2

    1 9.012

    2

    1 2 

    6

    9.0tan h

    1

    3

    B2

    B1

    4

    4 (a) θ = 0.842 rad

    kos θ = 15

    10

    (b) 126.708 cm 2

    Area = 88)18.11)(10(

    2

    1 )841.0()15(

    2

    1 2 

    2

    B1

    2

    B1

    4

    5 (a) 1.75 rad

    (b) 46. 5 rad

    SOALAN 8

  • BIL JAWAPAN SUB MARK MARKAH PENUH

    1 (a)

    15

    17

    (b) 221

    21

     

      

      

      

     

      

     

      

     

    13

    5

    17

    15

    13

    12

    17

    8 B1

    1

    2

    3

    2 (a) tan A =

    12

    5 

    (b) sin (A – B) = 65

    56

     

      

      

      

     

      

      

      

    5

    3

    13

    12

    5

    4

    13

    5

    1

    2

    B1

    3

    3 θ = 45o, 63.43o, 206.57o, 225o

    θ = 63.43o, 206.57o OR θ = 45o, 225o

    (tan θ - 2)(tan θ - 1) = 0

    3

    B2

    B1

    3

    4

    7

    17 atau 2.4286

    B2 :

    )1)( 5

    12 (1

    1 5

    12

    

    

    atau tan (112.62 – 45)

    B1 : tan  = 5

    12  atau tan 45 = 1

    3 3

    5 (a)

    p

    1 

    1

    2

    3

  • (b) p

    p21 

    21 p seen

    B1

    SOALAN 9

    BIL JAWAPAN SUB MARK MARKAH PENUH

    1 x = 0 o , 30

    o , 150

    o , 180

    o , 360

    o (all correct)

    B3 : 0 o and 30

    o (both)

    B2 : sin x (-2sinx +1 )= 0

    B1 : (1-2sin 2 x) + sin x - 1 = 0

    4 4

    2 0, 120, 180, 240, 360

    B3 : 0, 180, 360 atau 120, 240

    B2 : sin x(2 kos x + 1) = 0

    B1 : 2 sin x kos x + sin x = 0

    4 4