jawapan add math sbp

(1)

2017

BAHAGIAN PENGURUSAN SEKOLAH BERASRAMA PENUH DAN SEKOLAH KECEMERLANGAN

PENTAKSIRAN DIAGNOSTIK AKADEMIK SBP 2017 PERCUBAAN SIJIL PELAJARAN MALAYSIA

ADDITIONAL MATHEMATICS

Kertas 2

PERATURAN PEMARKAHAN

(2)

(3)

(4)

Number Solution and Marking Scheme Sub Marks _MarksFull 5 (a)

0 50 12 5

2 2

 

 x

x

(x – 25)(x – 5) = 0 x = 25 , x = 5

Width = 25 – 5 = 20 unit

K1

N1

(b) ₅ 12

4   x dx dy

OR





15



100



5

2  2 

 x

y

x = 15 OR



15



40 5

2 2

 

 x

y

y = – 40 or max depth = 40

K1

N1

(5)

6 (a) All correct

 2

x y

Sketch graph

(6)

Number Solution and Marking Scheme Sub Marks

Full Marks

7 (a)

(b)

9 12 tan

// 9 12

tanPOQ  POR 1.8548 rad // 1.855 rad

SQR = 9(*1.8548) // 9(*1.855) or sin53.13 9

QM

= 16.6932 // 16.695 Chord QR = 2*7.2 = 14.4

Perimeter = SQR + * Chord QR = 31.09 → 31.098

K1 N1

N1 K1 N1

(c) _{Area of sector OQR or Area of triangle POQ} =

  

9 1.8548



2 1 2

=

  

12 9 2

1

= 75.12 cm2 = 54 cm2

2(Area of triangle *P OQ) − Area of sector *OQR

2(54) − 75.12

32.88 cm2

K1

K1 N1

(7)

8 Refer to graph

9 (a)

(i)

(ii)

(b)

OQ AO

AQ   or OP OAAP

b a AQ

4 3

  

b a OP

4 1 4 3

 

OP m OS

= 

  



 _a_ _b

m

4 1 4 3

   



_ _



 a n a b

OS

4 3

m1n 4

3

or m n

4 3 4 1



3m4n4 and m3n shown

Solve simultaneous linear equations

13 12  m

13 4



n

K1

N1

K1

N1

K1

N1

(8)

Number

Solution and Marking Scheme Sub

(9)

Marks

11 (a) i) 7 3

7 10

) 45 . 0 ( ) 55 . 0 ( )

7

(x C

P  

= 0.1665

K1 N1

(ii) 0 10

0 10

) 55 . 0 ( ) 45 . 0 (

C or 10C₁(0.45)1(0.55)9

) 1 ( ) 0 (

1P X P X equivalent = 0.9767

K1

K1 N1 (b) i)

(ii)

25 . 0

56 . 3 6 . 3 

0.4364

Seen - 0.524 or 0.524

524 . 0 25

. 0

56 . 3

  

k

k = 3.429

K1

N1

P1 K1

N1

(10)

(11)

13 (a)

z or

y or

x     15100

18 100

100 20 175

100 70

x = 40, y = 20, z = 120

K1 N2, 1, 0

(b)

100

46 120 24 100 10 125 12 150 8

175( ) ( ) ( ) ( ) * ( )

I    

123.7

K1

N1

(c)

7 . 123 *

100 880 06

 

P

RM 711.40

K1 N1

(d) Seen I₁₈_/₁₅ 166.25 or165 or 137.5 or 110 or 132

 

100

46 132 24

110 10

5 . 137 12

165 8

25 .

166    



I

97 133.

N1 K1

(12)

Number Solution and Marking Scheme Sub Marks _MarksFull 14 Refer to graph

15 (a) (i)

sin sin 22.18

7.71 3.46

122.73 ADC

ADC

 _ 

  

K1 N1

(ii) 2 2 2

7.71 4.98 4.07 2(4.98)(4.07) cos 116.49

ABC ABC

   

   K1 N1

(iii)

Area of triangle ADC

 

4.98 4.07)



sin116.49

2 1

9.070 cm2 // 9.07 cm2

07 . 9 71 . 7 2

1

  BE

BE = 2.353 cm

K1 N1

(b)

2 2

2.35 2.06 =3.125

BD

cm

  P1

N1 ₁₀