PROGRAM GEMPUR KECEMERLANGAN SIJIL PELAJARAN MALAYSIA 2018 NEGERI PERLIS

  

PROGRAM GEMPUR KECEMERLANGAN

SIJIL PELAJARAN MALAYSIA 2018

NEGERI PERLIS

SIJIL PELAJARAN MALAYSIA 2018 3472/2(PP)

  MATEMATIK TAMBAHAN Kertas 2 Peraturan Pemarkahan Ogos

UNTUK KEGUNAAN PEMERIKSA SAHAJA

  

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   2018 Program Gempur Kecemerlangan SPM Negeri Perlis

  Sub Total No. Solution and Mark Scheme Marks Marks 1(a)

  PR = PO OR or OQ = OP PQ K1 + + (i) PR = − + 3 a 9 b N1

  3 (ii)

  OQ a b

  = +

  3

  6 N1

  OT = OP PT

• (b)

  = + OP k PR = + 3 a k * ( 3 − + a 9 ) b = − + (3* 3 ) k a *9 kb

  Collinear OQ = OT or OQ = TQ or OT = TQ K1

  λ λ λ

  λ (3* 3 ) *9

   

• (3 a 6 ) b = − k a kb + +
• (3 a 6 ) b = + (3* 3 ) − k a k b λ *9 λ
>3 = − k or = k K1 Equate the coefficients of a λ(3* 3 ) *6 *9 λ and b and solve simultan
• 3 *2

  equations for k

  λ = or λ = 3* 3 − k *3 k

   *2   *3  = − = 3 (3* 3 ) k or * 6 *9 k

      −

• 3 k 3* 3 k    

  2

  k = N1

  3

  6

  5

  

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  2

  2 ,

  1

  1

  P1

  = =

  x

  200 100

  2

  h x ax x = + + At maximum point,

  2

  5

  1 ( )

  OR ₂

  = 30 m N1

  2

  4

  −    

  1

  5

  N1

  = 30 m

  2 = − + + K1

  400

  1 (100) (100) 5

  ( ) h x ²

  b x x a a a

  N1 Height of the highest pole,

  − = = −

  a a

  1 100, 4 400

  1

  = − = − = − K1

  K1

  = − +  

  No. Solution and Mark Scheme Sub Marks Total Marks

  1 ( )

             

        = + + − +  

  4 h x a x x a a a

  4

  2

  

5

  1

  1

  1

  h x ax x = + + _{2 2 2}

  2

  5

  1 ( )

  2 ²

  K1 ₂

  1 ( )

  16 400

  1 100, 4 400

  1

  5

  1

  N1 Height of the highest pole

  − = = −

  a a

  1

  5

  x = = P1

  2

  200 100

  At maximum point,

    = + − +    

  16 h x a x a a

  4

  5

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• (8 ) − 34 y y + = or
² 2<
• (8 ) + 34 x x − = K1

  x

  3, 5

  y = or 3, 5 x

  =

  N1 3 cm and 5 cm N1

  7

  7 4(a) _Ahmad

  51.3 48.2 52.0 47.3 45.0 52.4

  6

  6

  − −  − − =

  x

  K1 _Ahmad

  49.37 x = or _Luqman

  49.43 x = N1 _{2 Ahmad} 14666.98 σ *(49.37)

  6 = − or _{2 Luqman} 14698.14 σ *(49.43)

  6 = − K1 _Ahmad

  σ 2.665 = N1 and _{Luqman σ 2.524}

  =

  N1

  a , b, c must correct

  x

  5 K1

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  8 y x = −

   2018 Program Gempur Kecemerlangan SPM Negeri Perlis No. Solution and Mark Scheme Sub Marks Total Marks

  3

  2 π 2π 16π

  x y + = P1 and ^{2 2}

  π π 34π

  x y + = P1

  8 x y = −

  or

  P1 _{2 2}

  ( 8) ( 8) 4(1)(15) 2(1)

  Solve the quadratic equation ₂ ax bx c + + = for b 

  Factorisation

  ( 3)( 5)

  y y − − = or

  ( 3)( 5)

  x x − − = OR Formula ₂

  ( 8) ( 8) 4(1)(15) 2(1)

  y

  − −  − − = or ²

• = or _Luqman 51.3 48.2 52.0 47.3 45.0 52.4
• =

  Sub Total No. Solution and Mark Scheme Marks Marks 4(b) Luqman

  N1

  2 N1

  7 Luqman’s achievement is more consistent 5(a) Use

  cos( A  B ) = cos A cos B sin A sin B

  3

  3

  3

  3

  3

  3  

  cos cos − sin sin or x x K1 cos  

• x x x x

  4

  4

  4 4 

  4 4 

  LHS = RHS N1

  2 (b)(i) y

  2

  1 x

  Shape of positive cosine graph at least 1 cycle P1

  1 1 cycles for   P1

  x

  2 π

  2 Modulus of cosine graph for   P1

  3 x

  2 π

  (Maximum = 2, Minimum = 0) (ii)

  3 y = − 1 x or Implied N1

  5 π Sketch the straight line with *gradient or *y-intercept and straight line

  K1 involves x and y must be correct. N1

  3

  8 No. of solutions = 5

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• = or
• d
• 6 x = K1

  2

  2 (c) ² ₂

  d

  6 d

  y x

  =  K1 (0, 3) is minimum point N1

  2

  7 7(a)

  d

  1

  2 d

  y x x

  (0, 3)

    =  

   

  or

  d d

  y x x

  = K1 6 *2( 2)

  y x

  − = − K1

  2

  2

  y x = + N1

  N1

  x y = =

   2018 Program Gempur Kecemerlangan SPM Negeri Perlis No. Solution and Mark Scheme Sub Marks Total Marks 6(a)

  δ lim *

  Find

  δ δ

  y x ₂

  δ 6 δ 3(δ ) δ δ

  y x x x x x

  δ 6 δ δ

  y x x x

  = −

  K1 Use limit δ x

  → δ

  6 δ ^x

  3

  y x x

  →

  = K1 d 6 d

  y x x

  =

  N1

  3 (b) Solve

  d

  y x

  =

  0,

  3

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•    
•    
• A A −
• 12 *9

  3

   and integrate with respect to y ₂

  2 π

  8

  2

  y y

    −

     

  K1 4 π N1

  3

  10 ^{1 2}

  1

  3 − K1

  Use limit ^{3 6 2} ₄

  (2 8) into*

  2

  3 N1

  

Use limit

^{6 2} ₄

     

    −

  y y

  2

  8

  2 into*

   K1

  2

     

     

     

  −  

     

  y

  4 (c) Use ² π d x y

  8

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   ₂

  No. Solution and Mark Scheme Sub Marks Total Marks 7(b) Find the area of rectangular shape OR Integrate ²

  1 4 d

  2

  x x

   

   ₁ A 6 2 =  OR ³ ₂

  4

  6

  x A x = + K1 Use limit ^{2 3}

  into *

  4

  6

  x x

   

  1

  K1

   _{3 2}

  −      

  y

  2

  2

  3

  (2 8)

  (2 8) d y y −

  9

  2 = 8 x y − P1 Integrate ^{1 2}

  OR

  3 N1

  8

  K1

  3 A =

   K1

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  ( 0.667) ( 2)

  2 (b)(i)

  2 (ii) Find the probability in the correct region

  0.2523 // 0.25239 // 0.2524 N1

  ( * 0.667) P Z  −

  K1 Find the probability in the correct region

  − =

  Z

  140 150 225

  N1

  P Z P Z  −  0.2295 // 0.22964 // 0.2296 K1

  K1 218.4

  σ 960(0.35)(0.65) =

  3 (ii) ²

  0.2616 N1

  P X P X P X = + = + = P1

  K1 Write ( 8) ( 9) ( 10)

  C −

  8 (a)(i) Use ^{10 10} (0.65) (0.35) ^{r r} _r

  K1

• 0.2295(27)

6 N1

  log y

  3 N1

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  (Correct axes and uniform scales) K1 6 *points plotted correctly N1 Line of best fit N1 (Refer graph on page 15)

  log y against x

  1 (b) Plot ₁₀

  0.22 0.15 – 0.09 – 0.16 – 0.31

  0.37

   2018 Program Gempur Kecemerlangan SPM Negeri Perlis No. Solution and Mark Scheme Sub Marks Total Marks

  3

  10

  

9

  6

  5

  3

  10 9(a) x

  12 ₁₀

  Sub Total No. Solution and Mark Scheme Marks Marks

  9 h _{10 ( 10 )} + (c)(i) log y = − log q x log P1 ₁₀

  2 Use m = − log q * ₁₀ K1

  q =

  1.19 N1

  3 h

  (ii) Use c = log K1 * ₁₀

  2

  h

7.96 N1

  2 = (iii)

  1.95 

2.00 N1

  1

  10

  10

  1 p

  (a)(i) =

  4 N1

  Use m m

  1

  (ii)  = − _{JK KL} m = − _KL

  3

  1

  m = − _JK K1

  3

  1

  1

  y − = −

  5 ( x − 3) or y − *4 = − ( x − 6) K1 * *

  3

  3

  1

• y = − x 6 or equivalent N1

  3

  3

  3

  6

  3

  1

  (iii) A =   

  2 5 *4 − 14 5

  1 (*12 84 0) (30 0 42) − + − + − K1

  2

  

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30 N1

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   2018 Program Gempur Kecemerlangan SPM Negeri Perlis

  Sub Total No. Solution and Mark Scheme Marks Marks

  10 (b)(i) Use distance formula for WJ or WK

  WJ = x − y − WK = + x − y − ^{2 2 2 2} OR 2 or Implied K1 WJ = WK _{2 2}

• ( 3) ( 5) or ( 6) ( 4)

  3

  3

  42 22 174 or equivalent N1

  x y − x − y =

₂

  2 (ii) Substitute b −

  4 ac ₂ x = into the locus of *W and use ( 22) − − 4(3)(174) K1

  − 1604  Locus W not intersect the y-axis N1

  2

  10

    θ

  11(a) Use 2 sin r OR other valid method

   

  2  

    40 

  2(7) sin

  K1

     2 

  2 4.79 // 4.788

  N1

     80 3.142   220 3.142 

  (b) = A ₁

  7 OR A *4.79 K1 ₂ =     180  180 

      ₁

• A A
₂ K1 <
• 9.78 18.39 28.16 // 28.17

  N1

  3

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   2018 Program Gempur Kecemerlangan SPM Negeri Perlis

  Sub Total No. Solution and Mark Scheme Marks Marks

  1  220  3.142  ₂ 1  ₂ 40  3.142 

  11(c) = = A (*4.79) OR A (7) K1 ₁

₂

        2 180

  2 180    

  1 ₂ A = (7) sin 40  K1 ₃

  2 A = A − A * * * _{4 2 3} K1

  A − 2(* A ) _{1 4} K1 * 41.30 

41.34 N1

  5

  10 x 

60 N1

  12(a) y 

  50 N1

• x y 

  N1

  30 20 1500

  x  y N1

  4 (b) Draw correctly at least one straight line from K1 the *inequalities involves x and y

  Draw correctly all four *straight lines N1 Note: Accept dotted lines Region shaded correctly N1

  3 (Refer graph on page 16) (c) Minimum point (30, 30) N1 Substitute any points in shaded *region K1

• into 8 000 x 4 000 y

  10 360 000

  N1

  3

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• =

  3 (ii) ^{2 2 2}

  13 PQ 14 2(13)(14) cos 37 = + −  K1

  13 (a)(i)  58 ACB =  P1

  13

  34 sin 58 sin 85

  AP

   

  K1 15.94 // 15.944 N1

8.62 N1

   2018 Program Gempur Kecemerlangan SPM Negeri Perlis No. Solution and Mark Scheme Sub Marks Total Marks

  2 (b) ₁

  2 A = +  OR ₂

  1 (13)(14) sin 37

  2 A =  K1 _{1 2}

• A A −

  K1 241.31 241.35 

  N1

  3 (c)(i)

  Note: ' ' ' A C B  obtuse angle N1

  1 (ii) ' ' ' 122 A C B  =  N1

  1

  10 37°

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  1 (13 *15.94)(34) sin 37

  Sub Total No. Solution and Mark Scheme Marks Marks 14(a) Substitute t = and 5 v =

• 25 m

  5 n = or 5 m n + = K1

• mt nt
² Differentiate w.r.t t d v a =

  d t

  2 mt n + a = K1

  d

  v Substitute t = and 1 a = into *

  3

  d t

  2 m n + =

  3 K1 Solve simultaneous equation to find m and n N1 m = −

  1 n =

  5 ₂ N1

  5 (b)

• ( − + t 5 ) t  K1

  2   t

5 N1

  ₂ (c) Integrate *( − + t 5 ) d t t ₃  ₂ t

  5 t

  s = − + K1

  3

  2 _{2 2} Use s − s OR *( − + * * t 5 ) d t t _{t = 2 t = 1 3 2 1}  _{3 2}  (2) 5(2)   (1) 5(1) 

• − − − + K1    

  3

  2

  3

  2    

  1

  31

  5 N1

  // // 5.167

  3

  10

  6

  6

  

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   2018 Program Gempur Kecemerlangan SPM Negeri Perlis

  10

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  3 (c)(i) 600 : 400 : 300 : 200

  2

  1 172 430 N1

   = K1

  P

  100 *130.27 900 000

  3 (ii) ¹⁸

  N1

  K1 130.27 // 130.267

  I

  W = _18/15 or Implied (seen) P1

  108, 147, 156, 125 N2 (All correct) N1 (Only 3 correct)

   2018 Program Gempur Kecemerlangan SPM Negeri Perlis No. Solution and Mark Scheme Sub Marks Total Marks

   = K1

  I

  I I

  100

  2 (b) Use ^{16/15 18/16} _16/15

  K1 3 000 N1

   =

  P

  3900 100 130

  15(a) ₁₅

• 600(*108) *400(*147) *300(*156) *200(*125)
• 600 *400 *300 *200
• =

  

Graph for Question 9(b)

  0.6

  0.5

  0.4 (3, 0.37) ×

  0.3 × (5, 0.22)

  0.2 × (6, 0.15)

  0.1

  2

  4

  6

  8

  10

  12

  14 ×

  (9,

• –0.09)
• –0.1

  × (10,

• –0.16)
• – 0.2
• – 0.3

  × (12,

• –0.31)

  

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• – 0.4

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Graph for Question 12(b)

y

  (Factory T)

  80

  60

  70 x =

  60 y = 50

  50

  40 30 (30, 30) R

  20

  10

  10

  20

  30

  40

  50

  60 70 x (Factory S)

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PERATURAN PEMARKAHAN TAMAT

   2018 Program Gempur Kecemerlangan SPM Negeri Perlis