Mark Scheme Additional Mathematics Midyear Paper 1 F5 2017

1

MODUL KECEMERLANAGN NEGERI PAHANG

PERTENGAHAN TAHUN 2017

ADDITIONAL MATHEMATICS

Tingkatan 5

KERTAS 1

PERATURAN PEMARKAHAN

UNTUK KEGUNAAN PEMERIKSA SAHAJA

3472/1(PP)

Tingkatan

Lima

Additional

Mathematics

Kertas 1

Peraturan

Pemarkahan

Mei

2017

Mark Scheme Additional Mathematics Midyear Paper 1 F5 2017

2

ADDITIONAL MATHEMATICS MIDYEAR PARER 1 2017

Question Solutions and marking scheme Sub

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1 (-10, 24)

B1 :26 ∙1

13(−5𝑖 + 12𝑗) or √(−5𝛼)2 + (12𝛼)2 = 26

or −10𝑖 + 24𝑗 𝑜𝑟 (−1024

)

2

2

2

1h and 2k

B2 : 1h or 2k

B1 : 4𝑖 − 3𝑗 + (−2𝑖 + ℎ𝑗 ) = 𝑘𝑖 − 4𝑗 or in column

vector form

3

3

3(a)

(b)

65

Option 1, will choose money on chess board because

total money collected =RM1072

B1 : RM 64

2[2(1.00) + (64 − 1)0.50]

1

2

3

4(a)

(b)

(c)

80

27

2

27

𝐵1 ∶ 80

27 -

26

9 or 3 ( 1 −

1

34) − 3( 1 −1

33)

3

1

2

1

4

5(a)

(b)

(c)

2

20

One to one relation

1

1

1

3

6(a)

(b)

𝑚 + 1

2

𝐵1 ∶ 𝑥 + 1

2

7

B1: ∗(5𝑥+4)+1

2 follow through 𝑔−1

2

2

4

Mark Scheme Additional Mathematics Midyear Paper 1 F5 2017

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7(a)

(b)

8

B1 : 10 = 2(m-3)

8

B1 : 1+3+5+7+7+𝑚+11

7= 6 or

3+9+15+21+21+3𝑚+33

7= 18

2

2

4

8

Different, median from calculation = 37.36

B2 : 34.5 +20−12

14(5)

B1 : median class is identified correctly: 34.5 or 12 or

14 seen

3

3

9

𝑝 <1

4

B2 : (1 − 2𝑝)2 − 4(1)(𝑝2) > 0

B1 : 𝑥2 + (1 − 2𝑝)𝑥 + 𝑝2 = 0

3

3

10

m = -4 and n=8

B2 : m = -4 or n=8

B1 : (x-2)(x-4)=0 or -6 = m-2

3

3

11

𝑦 =10

𝑥

𝑙𝑜𝑔10 (𝑥2𝑦)2

B2: (𝑥4𝑦2) =1000𝑥

𝑦

B1: apply any law of logarithms :

𝑙𝑜𝑔10(1000𝑥)𝑜𝑟 𝑙𝑜𝑔10 (𝑥

𝑦) 𝑜𝑟 𝑙𝑜𝑔10 (

1000

𝑦)or

3

3

12

𝑥 = 3, 1

B2 : (2𝑥 − 8)(2𝑥 − 2) = 0 or (𝑦 − 8)(𝑦 − 2) = 0

where y =2𝑥

B1 : (2𝑥)2 − 10(2𝑥) + 16 = 0 or

(𝑦)2 − 10(𝑦) + 16 = 0 where y =2𝑥

3

3

13 x =1, y = -3

B3: Solve simultaneous equation

B2 : y+2x = -1 and 𝑥−𝑦

4 = y+4

B1 : y+2x = -1 or 𝑥−𝑦

4 = y+4

4 4

Mark Scheme Additional Mathematics Midyear Paper 1 F5 2017

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Question Solutions and marking scheme Sub

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14(a)

(b)

( 3

2 ,

9

2 )

B1 : solve simultaneous equation

5 : 4

B1 : −1 =𝑛(

3

2)+𝑚(−3)

𝑚+𝑛 or 2 =

𝑛(9

2)+𝑚(0)

𝑚+𝑛

2

2

4

15 50.22

B2 :

1

2|2(1) + 5(7) + 17(12) + 9(5) − 5(5) − 17(1) − 9(7) − 2(12)|-

(3.142)(32)

B1 : 1

2|2(1) + 5(7) + 17(12) + 9(5) − 5(5) − 17(1) − 9(7) − 2(12)|

3 3

16(a)

(b)

(c)

0

1

-2, 2 (both)

1

1

1

3

17

12 ≤ 𝑥 ≤ 28

B2:−(40)±√(40)2−4(−1)(−320)

2(−1) or

or equivalent

B1: −25𝑥2 + 1000𝑥 − 3000 ≥ 5000

3

3

18

h = 3 , k = 10 (both)

B2: h = 3 or k = 10

B1: xy = 6-2x2 or m = -2, c= 6 (both) or 0 = -2(h)+6 or

k = -2(-2)+6

3

3

19 (a)

(b)

(0,3)

𝑦 =3

4𝑥3 + 3𝑥

B1 : 𝑦

𝑥=

3

4 𝑥2 +∗ 3 or

𝑦

𝑥−∗ 3 =

3

4 (𝑥2 − 0)

1

2

3

20(a)

(b)

−24𝑥(1 − 3𝑥2)3

1

192

B1 : -192

1

2

3

28.94 11.06

Mark Scheme Additional Mathematics Midyear Paper 1 F5 2017

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Question Solutions and marking scheme Sub

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21

-0.012cm2s

-1 ( image become smaller)

B2 :𝑑𝐴

𝑑𝑡= −

24

103(0.5)

B1 : 𝑑𝐴

𝑑𝑟= −

24

𝑟3 or

𝑑𝑟

𝑑𝑡= 0.5

3

3

22

15511

3

B3: 1

3[ ( 13 + 5)5 − ( 03 + 5)5] + [12 − 0]

B2: 1

3[ ( 𝑥3 + 5)5]1

0 + [𝑥2]1

0

B1 : 1

3[ ( 𝑥3 + 5)5]1

0 or [𝑥2]1

0

4

4

23

−11

12

B2 : 5

12− 2

2

3+ 2 (

5

12) + [

12

2− 0]

B1 : 5

12− 2

2

3 or 2 ∫ 𝑓(𝑦)𝑑𝑦

1

0 + ∫ 𝑦 𝑑𝑦

1

0

3

3

24

1

B1 : 𝑟𝜃 = 𝑟

2

2

25

0.8 radian, r = 5 cm (both)

B3 : 0.8 radian or 5 cm

B2: 1

2(

14

2+𝜃)2𝜃 = 10 or (r-5)(r-2)=0

B1 : r + r +r 𝜃= 14 or 1

2𝑟2𝜃 =10

4

4