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Nama Pelajar : ………………………………… Tingkatan 5 : …………………….

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Additional

Mathematics

August 2016

PROGRAM PENINGKATAN PRESTASI AKADEMIK

SPM 2016

ADDITIONAL MATHEMATICS

Paper 2

( MODULE 1 )

.

MARKING SCHEME

2

MARKING SCHEME

ADDITIONAL MATHEMATICS TRIAL EXAMINATION AUGUST 2016

MODULE 1 ( PAPER 2 )

N0. SOLUTION MARKS

1 3 2x y or

3

2

xy

3 52

3 2y y

24 17 9 0y y

2( 17) ( 17) 4(4)(9)

2(4)y

3.63y and 0.620y (both)

4.26x and 1.76x

P1

K1 Eliminate x/y

K1 Solve quadratic equation

N1

N1

5

2

(a)

(b)

(i)

(ii)

30x

2 21456030

15

= 70.667

450 232

17

n

47n

2214560 2(32)

3217

9.61

N1

K1

N1

K1

N1

K1

N1

7

3

3

(a)

(b)

13 4 12x x x

1 484

3

xx

x

16x

24 x

1 2x x

1x

3log9 5

x

3log lg9 lg5x

3

lg5log

lg9x

0.7325

0.73253x

2.2361@ 5

K1 (48

3

x

x or 16x

)

N1

K1

K1

N1

5

4

(a)

P1 for cosine curve

P1 for amplitude 0 and 4

P1 for 3

2 cycle 0 to 2

P1 for 3

42

y kos x

4

(b)

2 1 3

3 1

2 2

y k

y k

3 14

2 2k

3k

K1

K1

N1

7

5

(a)

(b)

(c)

15 1 (14)(2)T

29

20 1 (19)(2)T

39

10

3 392

nS or equivalent

210

2

2(1) ( 1)4 4352

4 2 870 0

5 2 29 0

nn

n n

n n

15n 29

2n

Bilangan baris 2 15

30

K1

N1

K1

K1

N1

K1

K1

N1

8

5

6

(a)

(b)

y = 3x 2 + 1

1

h

y

0 x

Shape of quadratic function

Minimum = 1

Complete graph with axis of symmetry x = 0

2 1

3

yx

1

1 9

3 6

h ydy

1

2

21 9

3 6

1 91

2 2 2

2

h

hh

yy

2 2 8 0

4 2 0

h h

h h

4 , 2h h

Panjang Jeli

4 1

3

P1

P1

P1

K1

K1 integrate and sub. the limit

correctly

K1

N1

N1

8

6

N0. SOLUTION MARKS

7

(a)

(b)

(c)

(i)

(ii)

(iii)

x+2

2 3 4 5 6 7

10log y 0.30 0.53 0.78 1.02 1.26 1.50

10log y

10 10 10log ( 2log logy x b a

10log 4.5 = 0.65

3.5x from graf

10log a=*y-intercept

0.6310a

10log b = *gradient

1.754b

N1 6 correct

values of 10log y

K1 Plot 10log y

vs 2x

Correct axes &

uniform scale

N1 6 points plotted

correctly

N1 Line of best-fit

P1

N1

K1 for y-intercept

N1

K1 finding gradient

N1

10

0 -0.2

x+2

7

N0. SOLUTION MARKS

8

(a)

(b)

1.4033 rad 0.3349 rad

6

6

2

BA

2 2 2

2 2 2

6 2 6 2 2 6 cos

24 cos 2 6 6 4

1cos

6

80 4059 @ 1 4033

A

A

A

A rad

2 6 12

2 2 2 1 4033 6.953

Big circle

Small circle s rad

12 6 953

44 652

length

211 2 2 2 1 4033 6.953

2A rad

2 21 12 6 0 3349 6 sin 0 3349 0 1121

2 2A rad rad

6 953 2 0 1121

6 729

Area

K1

N1

K1 Use s r or 2 r

K1

N1

K1 Use formula 21

2A r

K1 in rad

K1

K1

N1

10

8

N0. SOLUTION MARKS

9

(a)

(b)

(i)

(ii)

(c)

2BDm

2 6y x

0 6 6 01

6 4 2 62Area

1

0 12 36 36 24 02

160

2

230 unit

3 2 2 30p q 2 6q p

2 2 10p q

2 2 6 2 10p p

4p , 2q

4, 2B

6 unitOD

PD OD

2 2

0 6 6x y

2 2 12 36 36x y y

2 2 12 0x y y

P1

N1

K1

N1

K1 ( 30ABC )

K1 for substitude q

N1

K1

K1

N1

10

9

N0. SOLUTION MARKS

10

(a)

(i)

(ii)

(b)

(i)

(ii)

Mean, 5.25np

7 5.25p

0.75p

5 5 6 7P X P X P X P X

5 2 6 1 7 07 7 7

5 6 70.75 0.25 0.75 0.25 0.75 0.25C C C

0.31146 0.31146 0.13348

0.7564

1.5 1.3

1.50.5

P X P Z

0.4P Z

0.3446

Percentage = 34.46%

0.9 1.5P X

0.9 1.3 1.5 1.3

0.5 0.5P Z

0.8 0.4P Z

1 0.8 0.4P Z P Z

1 0.2119 0.3446

0.4435

P1

N1

P1

K1 use n r n r

rC p q

N1

K1 Use Z =

X

N1

K1

K1

N1

10

10

N0. SOLUTION MARKS

11

(a)

(i)

(ii)

(b)

(c)

9 x

9 4

19 4

5

PC PH HC

x y

PG x y

13 9 4

5

124 4

5

EG EP PG

x x y

x y

12 2

ER EH HR

x y

212 2

5

12 2

: 2 : 5

5

x yEG

x yER

EG ER

h

12

21

32

2

3

tinggi

tinggi

N1

K1

K1

K1

N1

K1

K1

N1

K1

N1

10

11

N0. SOLUTION MARKS

12

(a)

(b)

(c)

(d)

dva

dt

, a ht k 2

( ) ( ) ( )h k or h k23 3 0 2 3 9

h and k 3 9

tS t

23 9

2

.4 5t s

6(4.5) 9

18

a

( ) ( )( ) ( )S or S

2 23 3

5 3

9 5 9 35 3

2 2

Total distance = 1 1

13 2 122 2

= 1

392

K1

K1

N1

K1

(for integration)

N1

K1

N1

K1

K1

(for summation)

N1

10

12

N0. SOLUTION MARKS

13

(a)

(b)

(c)

(d)

( )( ) cos

.

.

oAC

AC cm

AE cm

2 2 218 26 2 18 26 35

15 27

10 57

sin sin

.

x 35

8 4 7

x = 77.50o

.

.

O OCED

0

180 77 50

102 50

Area = ( )( )sin 0118 26 35

2

= 134.22

.

.

h

h cm

126 134 22

2

10 32

K1

N1

N1

K1

K1

N1

K1

(for using

area= ½absinc)

N1

K1

N1

10

13

N0. SOLUTION MARKS

14

(a)

(b)

(c)

(i)

(ii)

0 35100

0 25

140

p

6 50100 130

5

q

q

4 2 10

4

4

x y

x y

y x

112 5 4 140 130 135 2125

10

720 140 130 4 1250

1240 10 1250

10 10

1

3

x y

x x

x

x

x

AND

y

16

10

125 110

100

137 5

I

137 5 0 60

100

RM0 825 / RM0 83

K1

N1

N1

K1

K1

N1

K1

N1

K1

N1

10

14

N0. SOLUTION MARKS

15

(a)

(b)

(c)

(i)

(ii)

7

50 100 1000

2 20

1

2

x y

x y

x y

y x

R

7

7 80 20

6( 8 , 6 )

x + y = 7x + 2y = 20

x = 2y

x

y

10

At least one straight line is drawn correctly from

inequalities involving x and y.

All the three straight lines are drawn correctly

Region is correctly shaded

8,6

8 6 14

max

15 25

8 , 6

15 8 25 6

RM270

P x y

P

N1

N1

N1

K1

N1

N1

N1

N1

K1

N1

10

END OF MARKING SCHEME