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1 PERATURAN PEMARKAHAN PEPERIKSAAN PERCUBAAN SPM 2019 PROGRAM INTERVENSI TERBILANG AKADEMIK SELANGOR (PINTAS) MATEMATIK TAMBAHAN KERTAS 2 NO SOLUTIONS MARKS 1 (a) dx x x V 4 0 2 2 ) 4 ( π[ 16x 3 3 -2x 4 + x 5 5 ] 4 0 (Integrate + limit) [ 16(4) 3 3 − 2(4) 4 + (4) 5 5 ]−0 15 512 π K1 K1K1 N1 4 6 (b) 32 3 3 64 K1 N1 2 2 (a) 8 1 3 n=2 Accept answer without working for K1N1 K1 N1 2 8 b) 3 r (9 s ) 2 = 81 or (3 r ) 2 9 s = 1 3 or equivalent 3 r (3) 4s =3 4 or 3 2r-2s =3 -1 or equivalent r + 4s = 4 or 2r-2s = -1 or equivalent Accept any method of solving simultaneous equation 5 2 r 10 9 s K1 K1 K1 K1 N1 N1 6 SET 2
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### Transcript of SET 2 - trial.spmpaper.me Set 2/2019 Selan… · 1 peraturan pemarkahan peperiksaan percubaan spm...

• 1

PERATURAN PEMARKAHAN PEPERIKSAAN PERCUBAAN SPM 2019

(PINTAS)

MATEMATIK TAMBAHAN KERTAS 2

NO SOLUTIONS MARKS

1 (a) dxxxV

4

0

22 )4(

π [16x3

3-2x4 +

x5

5] 40 (Integrate + limit)

𝜋 [16(4)3

3− 2(4)4 +

(4)5

5] − 0

15

512π

K1

K1K1

N1

4

6

(b)

2 ×32

3

3

64

K1

N1

2

2 (a)

81

3

n = 2 Accept answer without working for K1N1

K1

N1

2

8

b)

3r(9s)2 = 81 or (3r)2

9s=

1

3 or equivalent

3r(3)4s = 34 or 32r-2s = 3-1 or equivalent r + 4s = 4 or 2r-2s = -1 or equivalent Accept any method of solving simultaneous equation

5

2r

10

9s

K1

K1

K1

K1

N1

N1

6

SET 2

cikguJEPcikgujep

• 2

NO SOLUTIONS MARKS

3

a) AD⃗⃗⃗⃗ ⃗ = AB⃗⃗⃗⃗ ⃗ + BC⃗⃗⃗⃗ ⃗ + CD⃗⃗⃗⃗ ⃗

AD⃗⃗⃗⃗ ⃗ = (q – p)a + qb

q – p = -5 or 2

3

pq

p = 13 and q = 8

K1

K1

N1

3

6

(b) 1

2 x DC x 4 = 52

p |a| = 26

2

K1

K1

N1

3

4

5x + y + (x + y) = 24 or 222 )()()5( yxyx

y = 12 – 3x

0)312(12 2 xxx

045 2 xx 0)45( xx or equivalent

5

4x or y =

5

48

48 4

5 5

192

25 or equivalent

P1

K1

K1

K1

N1

K1

N1

7

7

• 3

NO SOLUTIONS MARKS

5

(4.6 × 3) + (5.1 × 5) + (5.6 × 5) + (6.1 × 10) + (6.6 × 12) + (7.1 × 5) or

(4.6 × 2) + (5.1 × 7) + (5.6 × 5) + (6.1 × 12) + (6.6 × 8) + (7.1 × 6)

(4.6×3)+(5.1×5)+(5.6×5)+(6.1×10)+(6.6×12)+(7.1×5)

40 or

(4.6×2)+(5.1×7)+(5.6×5)+(6.1×12)+(6.6×8)+(7.1×6)

40

6.075 or 6.0375

(3 × 4.62) + (5 × 5.12) + (5 × 5.62) + (10 × 6.12) + (12 × 6.62) + (7.12 × 5) or

(2 × 4.62) + (7 × 5.12) + (5 × 5.62) + (12 × 6.12) + (8 × 6.62) + (7.12 × 6)

√(3×4.62)+(5×5.12)+(5×5.62)+(10×6.12)+(12×6.62)+(7.12×5)

40-(6.075)2 or

√(2×4.62)+(7×5.12)+(5×5.62)+(12×6.12)+(8×6.62)+(7.12×6)

40-(6.0375)2

0.7242 or 0.7175

ORCHARD B

K1

K1

N1

K1

K1

N1

N1

7

7

6

(a) 1cot

1cot2

2

2

2

2

2

cos1

sin

cos1

sin

2 2

2

2 2

2

cos sin

sin

cos sin

sin

K1

K1

• 4

NO SOLUTIONS MARKS

22 sincos

2cos

N1 3

6

(b)

ekcos

12cos

sin

1

12cos

sinsin21 2

01sinsin2 2

0)1)(sin1sin2(

1sin,2

1sin

270,150,30

K1

K1

N1

3

7

(a) 24012060 oror

3

4189.4 or

K1

N1

2

10

(b) 12OP or 60cos = OP

6

108PS or equivalent

180

142.312012PQRarcofLength or 25.14

92.45

108214.25

K1

K1

K1

K1

N1

5

(c )

120sin)12(2

1

180

142.3120)12(

2

1 22 or

or

12 128 6

2

45.88

35.628.150

K1

K1

N1

3

• 5

NO SOLUTIONS MARKS

8 (a)

2

95,

2

80

( 4, 7)

K1

N1

2

10

(b) )0(1)4(9)8(5)0(0)0(4)1(8)9(0)5(02

1

34

K1

N1

2

(c) m =2

1

midpoint QR ( 6,5)

y – 5 = 2

1 ( x – 6 ) or equivalent

y = 2

1 x +8

K1

K1

K1

N1

4

(d) 2

1

63

14

5

h or 8

3

14

2

1

h

h = 3

17

K1

N1

2

9

(a) p= 0.75 or q = 0.25 or

rnr

rC)25.0()75.0(15

(i) )12( XP

)15()14()13()12( XPXPXPXP

0.4 613

(ii) 75.015

11.25

P1

K1

N1

K1

N1

5

• 6

NO SOLUTIONS MARKS

(b)(i) 370 8

8

370380

8

370355ZP or )25.1875.1( ZP

0.8641 0.86396

(ii) 0.8641 x 3130

2704

P1

K1

N1

K1

N1

5

10

10 (a) All values of

x

1 and

y

1 correct.

x

1

0.67 0.40 0.25 0.20 0.17 0.13 0.10

y

1

0.22 0.91 1.30 1.43 1.52 1.56 1.67

N1, N1

2

10

(b) Refer graph paper

Plot log10 y against x with correct axes, uniform scales and

at least one point.

6 points plotted correctly

Line of best fit

K1

N1

N1

3

(c) hxh

k

y

1)

1(

1

(i) h

192.1

5208.0h

(ii) 5439.25208.0

k

3248.1k

P1

K1

N1

K1

N1

5

• 7

NO SOLUTIONS MARKS

11

(a)r = x – 5 or h = x – 2

V = π (x – 5)2 (x – 2) or equivalent

P1

N1

3

10

(b)

2

5 (1) ( 2)(2)( 5)(1)

3 ( 5)( 3)

dvx x x

dx

x x

3 ( 5)( 3) 0

3 5

x x

or

When x = 3 , V = 4π

K1

N1

K1

N1

N1

4

(c) 0.01x

9π (0.01)

0.09π

P1

K1

N1

3

12 (a) v = 3 N1 1

10

(b) (3t + 1)(t – 3) = 0

t = 3

K1

N1

2

(c) sp = 3t + 4t2 – t3

t = 3 , sp = 18 or t = 4 sp = 12

24

K1

K1

N1

3

(d) 8 – 6t = 0 or t = 4

3

Qv = 3t2 – 6t – 6

Qv = 6

3

46

3

43

2

Qv =` 3

26

K1

K1

K1

N1

4

• 8

NO SOLUTIONS MARKS

13 (a) 80100

80.2

x or 130100

60.2

y or z100

5

85.5

117,2,24.2 zyx

Note : All three values correct N2

Any two values correct N1

One or none values correct N0

K1

N2

3

10

(b) )26117()45130()14120()1580(

100

)26117()45130()14120()1580(

117.72

K1

K1

N1

3

(c) 72.11710035

x

RM 41. 20

K1

N1

2

(d) 100

11472.117

134.2

K1

N1

2

• 9

NO SOLUTIONS MARKS

14 (a) (i) 24)120)(sin)(8(

2

1BC

6.928

(ii) AC2 = 82 + 6.9282 – 2(8)(6.928) kos 120

12.94

(iii) 94.12

120sin

928.6

sin

A

27.62

K1

N1

K1

N1

K1

N1

6

10

(b)

A’

C ‘

B ‘

180 – 27.62 – 60 = 92.38

1(12.94)(6.928)sin 92.38

2

45.12

P1

K1

K1

N1

4

• 10

NO SOLUTIONS MARKS

15 (a) x + y ≥ 10

10x + 8 y ≤ 420

𝑥 ≤ 2𝑦

N1

N1

N1

3

10

(b)Refer to graph paper

One *straight line drawn correctly

All * straight line drawn correctly

Correct region

K1

K1

N1

3

(c) 2113 y

420 – [20(10) + 10(8)]

Seen (20 , 10 )

= 140

K1

N1

N1

N1

4

• 11

• 12