Modul Pintas Tingkatan 5Peperiksaan Percubaan SPM 2018

Skema Jawapan Matematik Tambahan Kertas 2 3472/2

1

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1

2

22

2 2

4 8 48 4 135

1212 2

2

124 135 4 12 2 12 2 135

2

24 135 0 4 9 0

3( 15)( 9) 0

2

Length and breadth: 9 , (15 ( )

3Height :

2

x y or xy x

xx y OR y

xx 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

18 183

2 2

9

a or seen

or

cm

P1

N1

N1 3

6

2( ) 24, 4 18, 12 24

2

24

21

2

81.94

b OR a and r

S

P1

K1

N1 3

2

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3

2 2 2 2 2

2

75 70 59 62 6866.8

5

75 70 59 62 6866.8 5.706

5

Syafiqah result is more consistent

a

P1

K1N1

N1 4

6

334( ) 67.5

6

71

xb

x

K1

N1

2

4

2

2

2

2

2 2

2 2

sin1

sin1

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

xc 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

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5

3 2 1( )( ) 2 3 2 3

2 3

2 3 3 2

0

m na 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

112 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

dya

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

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(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

11 5

2

2 3

dx dya i y

dy dx

y x

y x

K1

K1

N1 3

10

3

13

43

4

:)(

1

0

31

0

21

y

ydyyA

CurvetheUnderAreaii

1

22

0

Area of trapezium:

13 1 3 5 4

2A y y or

1 2

Area of the shaded region:

1

3

A A

6

4

62

4

) 4

6 44 4(6) 4(4)

2 2 2

2

b Volume x dx

xx

K1

K1

K1

N1

K1 K1

N1

4

3

5

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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

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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 yor

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

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12 20( ) 100

8

250

a x

K1

N1

2

10

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

2 3 2

5

y yb

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

t s c s kt ht

t s k h

k h k h

k h

K1

K1

K1

K1

N1

5

22 2( )9 3

4 20

9 3

3

2

b v t t

dva t

dt

t s

K1

N1

2

8

NO SOLUTIONS MARKS

1,3

0,5.40

4,6

3

1

27

2)( 23

st

standt

st

ttsb

S = 1+1+4= 6m

P1:

Shape

K1:

Points:

(6,4),

(3,-1),

(0,0)

N1

3

10

14 2 2 2( ) 7 5 6 2(5)(6)

78.46 / /78 28'o o

a kos BAE

BAE

K1

N1

2

\

10

sin sin*78.46( )

8 10

51.61 / / 51 37 '

49.93 / /49 56'

o o

o o

ACDb

ACD

ADC

K1

N1

N1

3

2 2 2( ) 10 8 2(10)(8) * 49.93

10

sin* 49.95 sin*78.46

8

sin* 49.93 sin* 51.61

7.81

o

o o

o o

c AC kos

ACor

ACor

K1

N1

2

1

2

1 2

1( ) 8 10 Sin*49.93

2

16 5 Sin*78.46

2

* *

15.91

o

o

d A Luas ACD

atau A Luas ABE

A A

K1

K1

N1

3

15 (a) (I) 70x y

(II) 2x y

(III) 1

252

x y

NI

N1

N1

3

s

t 6 3 -1

4

0

9

NO SOLUTIONS MARKS

(b) Refer to graph paper

One *straight line drawn correctly

All * straight line drawn correctly

Correct region

K1

N1

N1

3

10

(c) (i) Maximum (40,30)

K = 2.50x + 1.80y

Maximum cost = RM 154

( ) 15 40ii y 10log p m

N1

K1

N1

N1

4

10

10 (b)

11

10 20 30 40 50 60 70

10

0

20

30

40

50

60

70

80

y

80

x

R

(b) Satu *garis lurus dilukis betul K1

Semua *garis lurus dilukis betul N1

Rantau R betul N1

15 (b)