jawapan add math kedah k1

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Additional
Mathematics

August 2017

PROGRAM PEMANTAPAN PRESTASI TINGKATAN 5
SPM 2017

ADDITIONAL MATHEMATICS
Paper 2

( MODULE

1)

.

MARKING SCHEME

1

MARKING SCHEME
ADDITIONAL MATHEMATICS TRIAL EXAMINATION AUGUST 2017
MODULE 1 ( PAPER 2 )

N0.
1

SOLUTION

MARKS

Perimeter = 52 cm
20 + 4x + 4(4y + x) = 52
or

 15 y 
5 x  4 y  x    x2 

  

P1

( x = 0 diabaikan )
x  4  2y

y

or

4 x
2

P1

20 y  5x  15xy  0

20 y  5  4  2 y   15  4  2 y  y  0

K1 Eliminate x/y

 y  1 3 y  2   0

K1 Solve quadratic
equation

3 y2  5 y  2  0

y  1, 

2
3

N1

When y  1, x  4  2 1

=2

N1

Base area of the mold = 5  4 y  x 

= 5  4 1  2 
= 30 cm2

K1
N1

8

2

N0.
2
(a)
(i)

(ii)

56 

SOLUTION

x

K1

10
 x  560

6

2

x


2

N1

 562

K1

10
 x  31720
2

(b)

MARKS

N1

560  2(46)

12
 54  3

Xbaru 
Xbaru

N1

31720  2(462 )
 54  332

12
  6  652

K1

min & sisihan piawai

N1

7
3

(a)

a  2176,

n  15,

T15  120  14 120 

d  12
K1

= 1800
Total = 1800 + 2176
= 3976

K1
N1

3

(b)

a  4096, r 

1
2

P1

ar n 1  1

1
 4096   
2
1
 
2

1
 
2

n 1

n 1

n 1

1


K1

1
4096

1

 
2

12

n  1  12
n  13

 All fish all going to die when n  14 .
On 28 June 2017.

N1

6

4

(a)

RS  RP  PS

 10 2 
2p  p  q
3 
 3
4 2
 p q
3 3

ST  SQ  QT





1
 5 p  q   q  q
3
5 2
  p  q  q
3 3

K1 find (a) triangle
law OR b(ii)
quadrilateral law
(for RS or ST)

N1

N1

4

(b)

RS   ST

4 2
 5 2

 p  q     p  q  q 
3 3
 3 3

4  5
   
3  3
4

5
2 4 2
   
3 5 3
3

2

K1

K1
N1

K1
N1

8

5
(a)

5n  2  10  5n 1   17  5n 

1
= 25  5n   10  5n     17  5n 
5

=

 25  2  17   5n 

= 10  5n 

(b)

K1

log3 q 4 log3 3 3q log 3 p


9
log3 3 log 3 p
log 3 q

N1

K1 for change base,
K1 for power

12q  9
q

3
4

N1

5

5

N0.
6

SOLUTION
(a)

p  0  85

q  0 15

P  X  8

P1

 P  X  9   P  X  10 


10

MARKS

C 9  0  85   0 15  
9

10

C 9  0  85 

 0  5443

P  7  96  X  8.03

10

 0 15 

0

K1

use nCr p r q n r

N1

(b)

8  03  8 
 7  96  8

z

0 1 
 0 1
 P  0  4  z  0  3

 1  P  z  0  4   P  z  0  3
 0  2733

K1

Use score-z
X 
Z=


K1
N1

6

6

N0.
7

SOLUTION

MARKS

(a)

sin x cot 2 x  sin x

 sin x  cot 2 x  1
 sin x cos ec 2 x
 cos ec x

(b)
(i)

y  2 cos

K1
N1

3
x
2

P1
P1

graph cosine curve
amplitude
2

P1

cycle

3
cycle 0 to 2
2

(ii)

P1

 f  x

P1 shifted graph

1  2 cos

3
x
2

K1

y 

N1

equation



x



3
2

(iii)

5
x
 f  x 
 0
2

x
3

y 
2


Number of solutions = 2

y 



x



3
2

N1

10
7

N0.
8
(a)

SOLUTION

sin AOD 

5
10
AOD  0.5236rad

(b)

S AD  10  0.5236

K1
N1

 5.236

K1

1
DC  10
5
2

K1

Perimeter  5.236  2  5  8

 20.24

(c)

MARKS

Luas 

 23.82

1
1
(8  12)5  (10) 2 (0.5236)
2
2

 in rad

Use s  r 

K1
N1

K1 K1 K1
1
A  r 2
2
N1

Use formula

10

8

N0.
9
(a)

SOLUTION

MARKS
P1

C (1, 5)

 7  2 1  10 
F 
,

3 
 3

K1

  3,3

N1

(b)
P1

D (-1, 1)

5 1
1 7
 1

mBC 

y  1  1 x  1

K1

for using m1m2  1 to
form equation

y  x 2

N1

(c)
Area of quadrilateral
=

=

1 1 7 1 1 1
2 5 1 1 1 5
1
 1  7  1  5  12  1  1  1
2

K1

= 14.5 unit 2

N1
PA  AD

(d)

 x  1   y  1
2

2



 1  1  1  1

x2  y2  2 x  2 y  2  0

2

2

K1
N1

10

9

N0.
10
(a)

(b)

SOLUTION

MARKS

(9  x) 2  9( x  1)

K1

( x  24)( x  3)  0

K1

(3, 6)

N1

Area of trapezium 

 36

1
(3  9)6
2

K1

 y3

y2
Area   (  1)dy    y 
3
2
 27
3

K1

Area of shaded region  36  4

K1

32

N1

6

6

4

(c)

 y2 
v      1 dy
0
 9

4
3 y
2 y2 
  
 1dy
0 81
9



integrate and sub. the
limit correctly

2

3

 y5 2 y3

 

 y
 405 27
0
3

 1.6

K1

K1

integrate and sub. the
limit correctly

N1

10

10

N0.
11
(a)

SOLUTION

1
x
1
y

MARKS

0.33

0.25

0.20

0.17

0.14

0.10

0.98

1.85

2.34

2.55

2.90

3.30

N1

6 correct values

N1

6 correct values

K1

Plot / Correct axes &
uniform scale

N1

6 points plotted
correctly

N1

Line of best-fit

K1

for y-intercept

4.

(b)
(i)

(ii)

(iii)

1
1
 pq 
 p
y
x


p  4.3
q (4.3)  10.06
q  2.340

1
 3.05
y
y  0.3279

N1

K1

finding gradient

N1

N1

10

11

N0.
12
(a)

SOLUTION

MARKS

S   ( 3t 2  4 t  4 )dt

 t 3  2t 2  4 t

K1
K1
N1

 (3)3  2 (3)2  4 (3)
 3
(b)

(c)

( 3t  2 )(t  2 )  0

K1

t2

N1

a 0

6t  4  0
2
t s
3
2
2
v  3( ) 2  4( )  4
3
3
1
 5
3

(d)

S 23  ( 23 )3  2

 23 

2

 4( 23 ) or

K1

sub. t 

2
s into v
3

N1

S4  (4)3  2  4   4(4)
2

K1
Total distance
3

7
 2  16
27

 22

14
27

K1

N1

10

12

N0.
13

SOLUTION

MARKS

(a)

PR  4  8  1 6  6  4

sin PQN
sin 40

64
18  6

sin PQN  0  221174

P1

K1 Use sine rule

PQN  12.778

N1

6  42  18  62  QR2  2 18  6  QR  cos12  778

K1 Use cosine rule

(b)

QR2  36  2787QR  305  0

QR  23  042

 QR  23  042

,

13  2367

(c)
16 1

64 4
18  6
MN 
 4  65
4

NMR  127  222

1
A   4  65 1 6  sin127  222
2
 2  962

K1

Solve quadratic equation

N1

P1

P1

K1 Use A

1
ab sin c
2

N1

10

13

N0.
14
(a)

SOLUTION

MARKS

p
 100  108
2.50

K1

p  2.70

N1

120,100,105,90

P1

(b)

125 120
 150
100
115 100
Q:
 115
100
108 105
R:
 113.40
100
148  90
S:
 133.20
100
P:

I

150  2  115  4  113.40  1  133.20  3
10

 127.30

(c)

1250  2  100  4  105 1  90  3
10
 101 5

I 

p
100  101.5
25
p  25.375

K1
N1

K1

N1

K1

K1

N1

10

14

N0.
15
(a)

SOLUTION

MARKS

x  y  50

N1

y

N1

1
x
2

20 x  15 y  1500
4 x  3 y  300

N1

(b)
y
100

4x

+ 3y =

300
x

=

2y

50

R

x






(c)
(i)
(ii)

+

y=

50

0

50

75

x

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

N1
N1
N1

20
N1

cos t  20(20)  15(30)
 850

K1

P max  1500  850

 RM650

K1
N1

10

END OF MARKING SCHEME
15