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SULIT 3472/1 Additional Mathematics Paper 1 August 2015

BAHAGIAN PENGURUSAN SEKOLAH BERASRAMA PENUH

DAN SEKOLAH KECEMERLANGAN KEMENTERIAN PELAJARAN MALAYSIA

PEPERIKSAAN PERCUBAAN SPM

TINGKATAN 5

2015

ADDITIONAL MATHEMATICS

Paper 1

MARKING SCHEME

This marking scheme consists of 6 printed pages

 

2    

MARKING SCHEME

Question Answers Sub Mark

Mark

1 (a) p =10

(b) h : x → 1x,x ≠ 0 or equivalent

Fungsi kerana hubungan satu dengan satu

1 1 1

3

2

(a) q = 3

B1: 2 = 45−q

(b) h x( ) = 3x + 4x ,x ≠ 0

B1: y =4x −3

or y x −3( )= 4

2 2

4

3 −1.637 or 2.137

B2: −(−1)± (−1)2 − 4(2)(−7)2(2)

B1: 2x 2 − x −7 = 0

3

3

4 p = 8

B1: 3 53!

"#$

%&

2

−53p +5= 0

2

2

5 (a) 3=p (b) 2=q (c) 3=x

1 1 1

3

6 21 <<− x B2: (x − 2)(x +1) < 0 or B1: 022 <−− xx

3

3

   

3    

7 n 3 − n −2 or n 3 − 1

n 2

B2: (23)m − (22 )−m B1: 23 or 22

3 3

8 y = 8x −1

B2: y +1x

= 23 or y +18

= x

B1: 3log2 2 ory +13

or y +18

3 3

9 n = 6 B2: log(1.05)n >log1.3 B1: 300000(1.05)n >390000

3 3

10 (a) 12.6, 25.2, 37.8, 50.4 (b) 12.6

1 1

2

11 (a) 9 (b) 9207

B1: 9(210 −1)2−1

1 2

12 (a)

y 2

x= −2x +10

(b) p = 5 and q = 4 B2: p = 5 or q = 4 B1: 0 = −2 p +10 or q = −2(3)+10

1 3

4

4    

13 −150 = −2(100)+50 Tiang bendera perlu dipindahkan sebab terletak diatas lorong (mesti tunjukkan coordinat diuji) B3: y = −2x +50 Equation of perpendicular bisector Canteen and Block B B2: y −150 = −2* x − (−50)( )

B1: Gradient of canteen and blok B, mCB =225−75

100− (−200)=12

4 4

14 (a) 3

(b) 103

B1: − q2= −53

or m = −53

1 2

3

15 (a) − p

p 2 +1

(b) 1− p1+ p

B1: tan45°− tanθ1+ tan45° tanθ

1 2

3

16 82.22

B2: 126( )2 π2

!

"#

$

%&−12×6×6

B1: 126( )2 π2

!

"#

$

%& or

12×6×6

3 3

17 (a) −3

2

"

#$$

%

&''

(b) 3i - 7j

1 1

2

5    

18 (a) unit vector PQ

! "!!!=

113

2s −3r( )

B1: PQ! "!!

= 2s −3r or 22 + −3( )2

(b) a = 5

b = −8

2 1 1

4

19 −4323125 or −0.1382

B2: f ''(x ) =432

(3x −5)5 or equivalent

B1: −36(3x −5)4

or equivalent

3

3

20 (2.5, 8.75) B3: x = 2.5 B2: −2x +6 =1

B1: dydx

= −2x +6 or gradient, m =1

4 4

21 V = 2t 3 − 3t

2

2+ t −1

B2: 11= 2 2( )3−

3 2( )2

2+ 2+c or c = −1

B1: dAdt

= 6t 2 −3t +1 or A = 6t 2 −3t +1dt∫

3

3

22 k = 4

B2: 42 = 39.5+

24+ k2

−11

12

"

#

$$$$

%

&

''''×10

B1: 39.5 or 11 or 12

6    

23 a) 300 B1 : 6C 2 ×

6C 3 b) 48 B1 : 2! × 4!

2 2

4

24 (a) 8=x

B1: 41

24=

+xx

(b) 24895

B1: 3119

3220

×

2 2

4

25 (a) n = 4 and p = 0.75 B1: np = 3 or npq = 0.75 (b) 0.4219

B1: 4C 3 0.75( )30.25( )

2

2

4

END OF MARKING SCHEME