modul pintas tingkatan 5 peperiksaan percubaan spm 2018 ... set... skema jawapan matematik tambahan...

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  • Modul Pintas Tingkatan 5 Peperiksaan Percubaan SPM 2018

    Skema Jawapan Matematik Tambahan Kertas 2 3472/2

  • 1

    NO SOLUTIONS MARKS

    1

       

    2

    22

    2 2

    4 8 48 4 135

    12 12 2

    2

    12 4 135 4 12 2 12 2 135

    2

    24 135 0 4 9 0

    3 ( 15)( 9) 0

    2

    Length and breadth: 9 , (15 ( )

    3 Height :

    2

    x y or xy x

    x x y OR y

    x x x OR y y y

    x x OR y

    x x OR y

    x ignore

    y

       

       

            

     

         

       

    P1

    P1

    K1

    K1

    N1

    N1 6 6

    2

    2 2

    2

    18 ( ) 6, 18 , 3 3 ,

    2

    18 18 3

    2 2

    9

    a or seen

    or

    cm

              

       

    P1

    N1

    N1 3

    6

    2 ( ) 24, 4 18, 12 24

    2

    24

    2 1

    2

    81.94

    b OR a and r

    S

     

    P1

    K1

    N1 3

  • 2

    NO SOLUTIONS MARKS

    3  

      2 2 2 2 2

    2

    75 70 59 62 68 66.8

    5

    75 70 59 62 68 66.8 5.706

    5

    Syafiqah result is more consistent

    a    

         

    P1

    K1N1

    N1 4

    6

    334 ( ) 67.5

    6

    71

    x b

    x

     

    K1

    N1

    2

    4

     

    2

    2

    2

    2

    2 2

    2 2

    sin 1

    sin 1

    sin

    sin

    1

    2

    sec2

    x

    cos x

    x

    cos x

    cos x x

    cos x x

    cos x

    x

    a

     

    K1

    N1 2

    8

    (b)

    3

    ( ) 5

    x c y or implied

     

    Sketch the straight line, correct gradient or y intercept

    3 solutions

    N1

    K1

    N1 3

    y

    1

    -1

    x

    Shape of positive cos graph P1

    1 ½ cycles P1

    Maximum 1, minimum -1 P1

    Use identity :

    tan2x=sin2 x/cos2x

    atau cos2x+ sin2x = 1

    atau cos2x- sin2x = cos2x

  • 3

    NO SOLUTIONS MARKS

    5      

     

     

       

    3 2 1 ( )( ) 2 3 2 3

    2 3

    2 3 3 2

    0

    m n a i m n i m n i i j OR

    m n

    m n and m n

    m n or equivalent

     

             

         

     

    (ii) 2m - n = 0

    K1

    N1

    N1 3

    7

    ( )( ) 2(2 3 ) 3(3 )

    (13, 9)

    b i OP i j i j

    P

       

     

     

    ( ) ( 13 9 ) ( 2 )

    12 11

    1 12 11

    265

    ii PQ i j i j

    i j

    Vektor unit i j

        

      

      

    K1

    N1

    K1

    N1 4

    6  

     

    2 ( ) 3 2 1

    11

    6 11 2

    11 16

    dy a

    dx

    y x

    y x

     

      

     

    K1

    K1

    N1 3

    7

    2( ) 3 1 11

    2,2

    ( 2, 6)

    b x

    x

    Q

     

     

      

    K1

    K1

    N1 3

    (c) Rectangle @ Segi empat tepat P1 1

    7 (a) AOB = 6.5/5

    = 1.3 rad.

    POQ = 0.8667 rad.

    K1

    N1

    N1 3

    (b) MN = 5  sin(0.8667 rad.) = 3.811 cm

    or ON = 5  cos( 0.8667 rad.) = 3.2367 cm

    Length of arc PQ = 6  0.8667 = 5.2002

    Perimeter = 1+*MN+ *NQ+*QP

    = 12.77 cm

    K1

    K1

    K1

    N1 4

  • 4

    NO SOLUTIONS MARKS

    (c ) Area of sector OPQ = ½  62  0.8667

    OR Area of triangle OMN=½ (3.811)(3.2367)

    Area of shaded region = *15.60  *½ (3.811)(3.2367)

    = 9.432

    K1

    K1

    N1 3 10

    8

     

    1 ( )( ) 2 2(1) 2,

    2

    1 1 5

    2

    2 3

    dx dy a i y

    dy dx

    y x

    y x

       

      

     

    K1

    K1

    N1 3

    10

    3

    13

    4 3

    4

    :)(

    1

    0

    31

    0

    2 1

     

     

      y

    y dyyA

    CurvetheUnderAreaii

       1

    2 2

    0

    Area of trapezium:

    1 3 1 3 5 4

    2 A y y or    

      

    1 2

    Area of the shaded region:

    1

    3

    A A

    6

    4

    6 2

    4

    ) 4

    6 4 4 4(6) 4(4)

    2 2 2

    2

    b Volume x dx

    x x

     

     

                      

           

    K1

    K1

    K1

    N1

    K1 K1

    N1

    4

    3

  • 5

    NO SOLUTIONS MARKS

    9 (a) p = 0.85, q = 0.15

    𝑃(𝑥 ≥ 6) = 𝑃(𝑟 = 6,7,8)

    8C6 (0.85) 6(0.15)2 or 8C7 (0.85)

    7(0.15)1 or 8C8 (0.85) 8(0.15)0

    = 8C6 (0.85) 6(0.15)2 + 8C7 (0.85)

    7(0.15)1 + 8C8 (0.85) 8(0.15)0

    = 0.8948

    P1

    K1

    K1

    N1 4

    10

    (𝑏)(𝑖) 𝑃(35 < 𝑋 < 66) = 𝑃 ( 35 − 48

    6 < 𝑍 <

    66 − 48

    6 )

    = 𝑃(−2.167 < 𝑍 < 3)

    = 1 − 𝑃(𝑍 > 2.167) − 𝑃(𝑍 > 3)

    = 1 − 0.01512 − 0.00135

    = 0.9835

    Number of students between 35 to 66 marks = 0.9835 × 180

    = 177 students

    (ii) 𝑃(𝑥 < 𝑚) = 0.05

    𝑃 (𝑍 < 𝑚 − 48

    6 ) = 0.05

    1.645 𝑠𝑒𝑒𝑛

    𝑚 − 48

    6 = −1.645

    𝑚 = 38.13

    K1

    N1

    N1

    P1

    K1

    N1

    3

    3

    10 (a) All values of (x+2) and log10y correct.

    x + 2 1 2 3 4 5 6

    log 10 y 0.92 1.00 1.08 1.12 1.24 1.32

    Refer graph paper

    Plot log10 y against (x+2) with correct axes, uniform scales and

    N1

    K1

  • 6

    NO SOLUTIONS MARKS

    at least one point.

    6 points plotted correctly

    Line of best fit

    N1

    N1 4

    10

    b) (i) 14.45

     10 10 10

    10

    10

    10

    10

    ( ) log log 2 log

    log

    log 0.84

    0.1445

    log

    log 0.08

    1.202

    ii y p x q

    Use q c

    q

    q

    Use p m

    p

    p

      

     

     

    N1

    P1

    K1

    N1

    K1

    N1 6

    11

     

    1 ( ) , 2

    2

    2 2 8

    2 18

    BC AXa m m

    y x

    y x

      

       

      

    P1

    K1

    N1 3

    10

    (b) Solve simultaneously : y = -2x +18 and 2y = x + 10

    X = (5.2, 7.6)

    4(2) 4(6) 5.2 7.6

    5 5

    x y or

       

    C (18 , 14)

    K1

    N1

    K1

    N1

    4

       

    2

    1 ( ) 8(6) 2(14) 18(2) 2(2) 18(6) 8(14)

    2

    2 *

    112

    c Area of ABC

    Area of trapezium Area of ABC

    unit

          

      

    K1

    K1

    N1 3

  • 7

    NO SOLUTIONS MARKS

    12 20 ( ) 100

    8

    250

    a x  

    K1

    N1

    2

    10

    110( ) 80(2 ) *250(3) 150(2) ( ) 120

    2 3 2

    5

    y y b

    y y

    y

       

      

    K1 K1

    N1

    3

    110 80 ( ) : 110 110

    100 100

    121 88

    121*(5) 88*(10) 250(3) 150(2)

    121*(5) 88*(10) 250(3) 150(2)

    *5 *10 3 2

    126.75

    c Seen A or B

    I

       

     

      

       

      

    K1

    K1

    K1

    K1

    N1

    5

    13

    (a)(i)

    (a)(ii)

    2

    3 2

    3 2

    ( ) 3, 0 27 6 0

    3 2

    0, 0, 0

    3, 1 1 27 9

    Solve simultaneously: 27 6 0, 27 9 1

    2 1 ,

    27 3

    a t v k h

    s kt ht dt

    kt ht c

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