PEPERIKSAAN PERCUBAAN SPM TAHUN 2018

  Nama Pelajar : ………………………………… Tingkatan 5 : ……………………. 3472/2 Additional Mathematics August 2018

  PEPERIKSAAN PERCUBAAN SPM TAHUN 2018 JABATAN PENDIDIKAN NEGERI KEDAH MAJLIS PENGETUA SEKOLAH MALAYSIA ( KEDAH ) ADDITIONAL MATHEMATICS Paper 2 ( MODULE 1 ) . MARKING SCHEME

  

MARKING SCHEME

ADDITIONAL MATHEMATICS TRIAL EXAMINATION AUGUST 2018

MODULE 1 ( PAPER 2 )

N0. SOLUTION MARKS

1 P1 P1 P1

      (90 45) (90 2 )( ) x y 3300 K1 Eliminate x or y K1 Solve quadratic equation N1

  When = 15

(abaikan)

= 50

  AD

  50  N1 Both sides of the pond

  AB 

  15 correct

  (a)

  20 Any of thesevalue

  3  p   

  13 P1  

  2 19 5 , 13 , 5 , 10  m 19 5 10 23 5

         5  

  K1 formula of m   20  p

  

  15

  2 p 

  10 N1 both p & N N 

  30 (b)

  Students

  10

  9 K1 Plot / Correct axes & uniform scale

  5 K1 Correct histogram

  4 ( 5 bar )

  2 K1 Find mode 4.5 14.5 24.5 34.5 44.5 Marks mode = 33.5

  Mode = 33 5  N1

  7

  (a) P1 any 2 correct

  1, 4,

  10 a  d  n 

  10 S   2 1  9 4  _{10    }  

  2 K1 use S ₁₀ N1

  = 190 (b)

  Yellow tiles = 400

• – 190

  N1 = 210 OR

  10 S   2 3  9 4  _{10    }  

  2  210

  Difference = 210

• – 190

  N1 = 20

  5

  4 (a)

  (i) QT  QP  PT

  K1

  

  6

  5    

       

  

  2

  4    

  

  1  

    

  2  

  N1 QR   

  2 i 4 j

  (ii) SR  SP  PQ QR  K1

  

  1 6 

  2               

  

  5

  2

  4       N1

   3 i  j

  (b)

K1 N1 N1

  

  18 2 , 3 ,18 2 3 3 18

  3

  2

  5 (a) (b) ^{1 1 1 1 1 1 1}

  7

          

  3

      

  k k SR is parallel to PQ

  2 SR k PQ

  1

  2

  1

  6

           

K1 K1 N1 K1 K1 N1

  2 ^{x y q y q x y y q x} k k k k k k k k x y q xy q x y

  3

  1

       

  

  x  

  45 ^{x x k x x}

  3

  9

  2

  9

  5

  log 5, 3

  2

  3 _{1 1 2 2( 1 1 ) 2}

     

  1

  P1 graph cosine curve (a) P1 amplitude 3

   y 3 cos 2 x

  P1 2 cycle 0 to

  2

   (b)

  P1 shifted graph y  f x 

  1

    P1 y  f x 

  1

    x

  K1 line y  

  2

  

  x f x   

  1

  2

  (c)  

  

  x

  2

  x y  

  N1 equation y  

  2

   

  Number of solutions = 2 N1

  8

N0. SOLUTION MARKS

  7 (a) (i) P1 p   0 4 q   0 6 P X

  8   

   P X  9  P X 

  10     _{n r n r}

   K1 use _{10 9 1 10 10 r} C p q

   C 0 4  0 6   C 0 4  0 6  ₉         ₁₀   0 001573 0 0001049  

  N1   0 001678

  (ii) K1 n 0.4 0.6 

  12

     n  600

  N1 (b) (i) 72 55 

  Z  K1

  5 N1   3 4

  (ii) K1 use score-z

  P X  k  0.185  

  

  X  Z = k

  55 

    

  P z    0 185

   

  5  

  k 

  55 0 896  

  K1

  5 N1

  k  59 48 

  (a)

  7 SWT / 0 7 / 40 10 / 40 11        K1 use s  r 

  10 25 

  

     0 4364 rad / 0 4363  rad 180  K1

      0 7 0 4363 1 1363   rad  65 10  

   

    1 136 rad 3 . . d p 

    N1 (b) K1

       ArcQV 8 1 136 9 088

  K1 Perimeter     8 8 9 088

  N1  25 088 

  (c)

  1 ₂ A  8 1 136   36 352  K1 _QPV  

   

  2 K1

  1 ² A  10 0 7  

  35

  2   use formula A  r 

  2

  1 _SWT  

  2 to find A , A _{QPV SWT}

  K1 71 352 63 88

  Area     N1

  7 472  

  (a) (i) m  _QR

  1 K1 for using m m  

  1

  m  _RS

  1 ^{1 2} R  0,6

    S  h ,0

    K1

  6   1  h

  N1 h 

  6 (ii) K1

  18 2 4 

   

  x y , ,  

    

  4

  4  

  N1

   4,1

    (iii)

  4  2 

  4

  4

  1 A 

  2

  1 4 

  2

  1

  1 K1 use formula of Area  16 4 4       2 16 8

     

  2

  1  16 26 

  2

  21 

  N1 (b)  y  4  y 

  K1 for using m m   _{1 2} 1 to  

  1    x  2 x 

  6    _{2 2} form equation y  4 y    x 4 x 

  12 _{2 2} K1

  4

  4

  12 x  y  x  y  

  N1

  (a) _y 1 cm 1 cm

  9 9 cm _logam metal _O

  1 _{x 2}

  2

  3

  2

  1

  y  x 

  9

  y  K1 Substitute x

  2 or r

  3  

  y 9 to find  x  2 or r 

  3 Volume of cylinder ₂

  

  3

    K1

  9   

  

  

81 Volume

  ₉ y

  1 

    dy

   ¹ ₂

  2 ₉  y 

     y

   

  2

  2   ¹

   

  81 1     

    9  

  1      

  K1 integrate and use the

  2

  2

  2    

   

  limit correctly

   

  32

   

  2

  16  

  Volume of metal K1

   

   81 

16 N1

  

  

  65

  (b) (i)

  4

  y  ₂ 2 x  3 y  

  6 x 

  2

    h

  , 4  

h , 4

  

 

OR h 

  2

  6 ₂

h 

  2 

  1

  

 

h

  

3 N1

  h 

  3

  (ii) Area of trapezium K1

  1  2 

  4

  3

    

  2 

  9 Area _{5 2}

  

   4 x  2 dx

     ³ ₅

   

  1

  4   

  

  x 

  2 ₃  

  K1 integrate and use the

  1

  1  

    4 

  limit correctly

   

  3

  1  

  2  

    

  4  

  3  

  8 

  3 Area of the shaded region

  8

  9

K1 N1

  3

  35

  2 / 11 / 11 67

  3

  3    

  10

  (a)

1 N1 6 correct values

  1 25  1 00  0 80  0 50  0 40  0 20 

  x

  1 N1 6 correct values      2 78 2 22  1 69 0 96 0 70 0 20 y

  (b) K1 plot / correct axes & uniform scale N1 6 points plotted correctly N1 line of best-fit

  (c)

  1

  1   (i)

  P1  k  h   y x

    (ii)

  N1 for y-intercept h   

  0 3 N1 finding gradient k  

  2 48

  1 (iii)

  K1    1 25

  1.30 y

  N1 y   

  0 8

  0.77

  (a)  ¹ N1

  8ms

  2

      t 2 t

  8

  (b)

K1 K1

  ( t  2)( t   4)   t

  4 N1

  v (c)

  K1

  8 t

  N1 (d) ₃ t ₂ S     t

  8 t

  3 ₃ K1 (4) ₂

  t 4, S (4) 8(4)

      

  K1

  3 ₃ (5) ₂

  t  5, S    (5)  8(5)

  3 Distance=

  2

  2

  1 26  (26  23 ) K1

  3

  3

  3 

  30 N1

  (a) (i) _{2 2 2} PR 

  8   3 71  2 8 3 71 cos112  

     K1 use cosine rule

  PR 

  10 SR 

  5 N1

  (ii) K1 use sine rule

  PT

  5  sin120  sin 30 

  N1 PT   8 66 Perimeter       8 3 71 5 5 8.66

  K1  30 37 

  N1 (i)

  (b) P'

  S' N1

  T' (ii)

  1

  1 K1 use A ab sin c

   Area  8 66 10 sin30  

    

  2

  2 K1

  P’S’ = 10

    21 65

  N1 OR

  1 Area  8 66 5 

    

  2   21 65

  x   y

  80

  (a)

N1 N1

  y 

  4 x

  N1 x 

  2 y 

  60

  (b) ₈₀ y _{y = 4x 30} R _{7x + 10 y = 710 x + y = 80} x + 2y = 60 _{60 80} x

   At least one straight line is drawn correctly from inequalities involving x and y.

  N1  All the three straight lines are drawn correctly

  N1  Region is correctly shaded

  N1 (c)

  35 x  50 y  3550

  K1 or

  7 10 710

  x  y  K1 Draw a straight line in the graph. K1

  X=30 N1

  Maximum .number of cake A = 30

  N0. SOLUTION MARKS (a) N1 x  135

  (b)

  1 50 100    115

  K1 Q

    1 30

  N1 (c) K1

  120 

  I

  140 

  100

  N1 I  116 67 

  (d)

  120   2 115 3 135 1 110      y K1

  (i)

   116 2 3 1    y

  N1 y 

  4

  (ii) p

   100 116 

  20 K1 23 2

  p   K1

  500  21 55  23 2

  

  N1

  max 

  21

  10