jawapan addmath tganu k1

MATEMATIK TAMBAHAN

1

BK3

Q

Marks
(a)

fungsi f memetakan x kepada x2 – 2x + 3
Or function f maps x onto x2 – 2x + 3

B1

f(z) = z2 – 2z + 3

B1

p(x) = 23 – 15x

3

p(x) = 5(6 – 3x) – 7

B2

m-1(x) = 6 – 3x

B1

p = 2 dan q = 7

3

P = 2 atau q = 7

B2

m+n= 

B1

1

2
(b)

2

3

5
9
and mn =
2
2

m =  1 and n = 4
1

3

3

3
4

4
m =  1 atau n = 4

x = 0.8042

(0.6990)  x(0.9031)  (0.3010)  x(1.3979)

B2

3

B1

3
B2

(3x  1)(0.3010)  (2 x  1)(0.6990)
atau

5

log10 5  log10 8x  log10 2  log10 25x

(3x  1) log10 2  (2 x  1) log10 5
atau

3472/1  Percubaan SPM 2017 Skema

3
B1

MATEMATIK TAMBAHAN

(a )
6 (b)

y a
8

2

y

a2

x=4

( x  4)( x  1)  0

5 x  x2   7 x  4   5 x
(a) k = 0.0045

(b)
8

27
11

9
p=

5p =

3
B2

3

B1

4
B2

B1

log3 y  5 p log3 x  log3 q

p=

B1

3

r  0.01

(b)

3

B1

0.45

2
1
1
100

(a)

BK3

B1

log y 1  log

7

2

2
5
2

5

dan q = 81

B1

3
4

atau q = 81

80
62

atau log3 q = 4

3472/1  Percubaan SPM 2017 Skema

B2

B1

MATEMATIK TAMBAHAN

10

3

17.6 cm3

3

176  ( 0.1 )

B2

2 (4)(22)

B1

a  7 and b  4
a  7 or b  4
2a  3  a  10 ( solve the equations )B2
2a  b  10 or 3 - a  b

11

8  32  62 

35

3

4
B3
4
B1

4
B3

 2 x2 
(
)
g
x



2
 2 6
3

6

12

BK3

B2

 g ( x)dx   g ( x)dx   2 x dx
3

6

3

2

3

6

4

B1

(a)
B1

y

• (5,7)
13

3
(b)

35

2
2

5x7

B1
0

3472/1  Percubaan SPM 2017 Skema

x

MATEMATIK TAMBAHAN

4

BK3

(4, 2)

4

x  2  x  6

14

atau

y  x2

B2

mQS  1

 m  4


 5 

UV   OU  OV

15

(b)

  4   m
or     
 7  2 
m=4
4+m=0

B1

2

B1

3

12 + (4 – k)2 = 1

B2

12  (4  k)2

3
B1

k = 12.31, 77.69,192.31, 257.69

4

2k = 24.62, 155.38, 384.62, 515.38

B3

3
2(sin 2k) 
3.6
3
3.6

3472/1  Percubaan SPM 2017 Skema

4

2
B1

k=4

kos y 

4

( persamaan QS )

5 1
 1 or
1 5

(a)

17

B3

y   x  6 ( persamaan PR)

mPR 

16

(menyelesaikan persamaan)

4
B2
B1

MATEMATIK TAMBAHAN
234

L=

18

(a)
19

1
(30)2 (0.52)
2

B2

15.6 = j(0.52)

B1

54

2

(13+5)3

B1

22

3

3

B1

x  20  4 y4

21

BK3
3

(b) 6

2 y2 

20

5

100

5

 x

2
2

, using formula

B1

(a) 5 ! = 120

B1

(b)

2

72

2  4!

B1

10

2

n!
 45
(n  2)!2!

3472/1  Percubaan SPM 2017 Skema

2

3

B1

2

MATEMATIK TAMBAHAN

(a)

23 (b)

24

25

BK3

1
7

B1

3
17

2

15 14

35 34
(a)

6

B1

1
 2k 2
4
4k2 + t +

3

2

1
+t= 1
2

B1

(b)

7
8

B1

(a)

0.02275

B1

(b)

51.78

3

60  x
 1.645
5
1.645 dilihat

3472/1  Percubaan SPM 2017 Skema

3

4
B2
B1