Mark Scheme Additional Mathematics Midyear Paper 1 F4 2017

1

MODUL KECEMERLANAGN NEGERI PAHANG

PERTENGAHAN TAHUN 2017

ADDITIONAL MATHEMATICS

Tingkatan 4

KERTAS 1

PERATURAN PEMARKAHAN

UNTUK KEGUNAAN PEMERIKSA SAHAJA

3472/1(PP)

Tingkatan

Empat

Additional

Mathematics

Kertas 1

Peraturan

Pemarkahan

Mei

2017

Mark Scheme Additional Mathematics Midyear Paper 1 F4 2017

2

ADDITIONAL MATHEMATICS MIDYEAR PARER 1 2017

Question Solutions and marking scheme Sub

Marks

Full

Marks

1(a)

(b)

(c)

3𝑥2 − 3𝑥 − 5 = 0

2 real and different roots

𝑃𝑂𝑅 = −5

3

1

1

1

3

2

𝑥2 + 8𝑥 − 40 = 0

B2 : 4𝛼 + 4𝛽 = 4(−2) and (4𝛼)(4𝛽) = 16 (−5

2)

@SOR= 4(−2) and @POR = 16 (−5

2)

B1 : 𝛼 + 𝛽 = −2 and 𝛼𝛽 = −5

2

@SOR=-2 and @POR=−5

2

3

3

3

𝑝 < 3

B1: 3𝑝 − 9 < 0

2 2

4(a)

(b)

𝑎 = 𝑚 − 3, 𝑏 = 3 𝑎𝑛𝑑 𝑐 = −5𝑚2

B1: (𝑚 − 3)𝑥2 + 3𝑥 − 5𝑚2 = 0

𝑚 = 3

B1: 𝑚 − 3 = 0

2

2

4

5

4𝑚2 < 𝑐2 − 4

B3: (2𝑚𝑐)2 − 4(1 + 𝑚2)(𝑐2 − 4) < 0

B2: (1 + 𝑚2)𝑥2 + 2𝑚𝑐𝑥 + 𝑐2 − 4 = 0

B1: 𝑥2 + (𝑚𝑥 + 𝑐)2 = 4

4

4

6

𝑝 =8

9

B2: 𝑝 = 0 , 𝑝 =8

9 (both)

B1: (3𝑝)2 − 4(2)(𝑝) = 0

3

3

Mark Scheme Additional Mathematics Midyear Paper 1 F4 2017

3

Question Solutions and marking scheme Sub

Marks

Full

Marks

7(a)

(b) {(8,2), (27,3), (125,5), (216,6)}

1

1

2

8(a)

(b)

Many to many

1

1

2

9

ℎ = 2 𝑎𝑛𝑑 𝑘 = 6

B2: ℎ = 2 𝑜𝑟 𝑘 = 6

B1: 𝑓𝑔(𝑥) = ℎ(𝑥2 + 3𝑥 + 5) − 5 or 𝑘 = 3ℎ or

5ℎ − 5 = 5 .

3

3

10

𝑎 = −6 𝑎𝑛𝑑 𝑘 = 3, −3

B3: 𝑎 = −6 𝑜𝑟 𝑘 = 3, −3

B2: 2 = 𝑎 + 8 and 𝑘(𝑘) − 7 = 𝑎 + 8

@ (𝑎+8)+7

𝑘= 𝑘

B1: 2 = 𝑎 + 8 or 𝑘(𝑘) − 7 = 𝑎 + 8 or 𝑓−1(𝑥) =𝑥+7

𝑘

4

4

11

(ii) − 2 ≤ 𝑦 ≤ 7

(a) (i) −1 ≤ 𝑥 ≤ 2

(b) Yes, because it is an one to one relation

1

1

1

3

8

27

125

216

2

3

4

5

6

2

4

7

d

e

f

g

Mark Scheme Additional Mathematics Midyear Paper 1 F4 2017

4

Question Solutions and marking scheme Sub

Marks

Full

Marks

12(a)

(b)

5

6

B1: 𝑔(−4 + 2) = (−4) + 8 or 𝑓(4) = (4) + 2

1

2

3

13(a)

(b)

(c)

Maximum time: 1 hour 36 minutes

2 hours 48 minutes

There were 150 competitors (or half of the total

competitors) who took at most 2 hours 48 minutes to

complete the Marathon race.

1

1

1

3

14

16.6976

B2: 2(21.5)2+5(25.5)2+13(29.5)2+23(33.5)2+6(37.5)2+1(41.5)2

50−

(2(21.5)+5(25.5)+13(29.5)+23(33.5)+6(37.5)+1(41.5)

50)

2

B1: �̅� =2(21.5)+5(25.5)+13(29.5)+23(33.5)+6(37.5)+1(41.5)

50

3 3

15

𝑚 = 5 𝑎𝑛𝑑 𝑛 = 9

B3: 𝑚 = 5 𝑜𝑟 𝑛 = 9

B2: (14 − 𝑛) +2 𝑛2 − 106 = 0 𝑜𝑟

𝑚2 + (14 − 𝑚)2 − 106 = 0

B1: 12+12+72+22+12+32+72+𝑚2+𝑛2

9− 42 = (

√76

3)

2

or

1+1+7+2+1+3+7+𝑚+𝑛

9= 4

4

4

16

4 Beta, smallest standard deviation (Both)

B1: 4 Beta

2

2

17 (a)

(b)

𝑚𝑒𝑎𝑛 = 20.75 , 𝑚𝑜𝑑𝑒 = 4.00 𝑎𝑛𝑑 𝑚𝑒𝑑𝑖𝑎𝑛 = 5.25

B1: 𝑚𝑒𝑎𝑛 = 20.75 , 𝑚𝑜𝑑𝑒 = 4.00 , 𝑚𝑒𝑑𝑖𝑎𝑛 = 5.25

(* any two values are correct)

Median, because there is an extreme value (RM 100) in

the data.

B1: Median

2

2

4

Mark Scheme Additional Mathematics Midyear Paper 1 F4 2017

5

Question Solutions and marking scheme Sub

Marks

Full

Marks

18(a)

(b)

New mean= 22

B1: 8(4) − 10

New 𝜎 = 6

B1 : (√2.25) × 4 or √2.25 × 42

2

2

4

19(a)i)

ii)

(b)

𝐴 = (−2, −9)

𝐵(−5,0)

B1: (𝑥 + 2)2 − 9 = 0

−9

1

2

1

4

20

𝑚 <1

3 , 𝑚 > 3 (both)

B2 : (3𝑚 − 1)(𝑚 − 3) > 0 OR

B1 :3𝑚2 − 10𝑚 + 3 > 0

3

4

21(a)

(b)

𝑝 = −3

2 𝑎𝑛𝑑 𝑞 =

5

2

B2: 𝑝 = −3

2 𝑜𝑟 𝑞 =

5

2

B1: 2 (𝑥 −3

2)

2

+5

2

(3

2,5

2)

3

1

4

22(a)

(b)

45

B1: 𝑝 + 30 = 75

100 cm

2

1

3

23(a)

(b)

𝑝 = 1 𝑎𝑛𝑑 𝑞 = −9 ; 𝑝 = −1 𝑎𝑛𝑑 𝑞 = 9 (both)

B1: |𝑝(4 − 1)2 + 𝑞| = 0 𝑜𝑟 |𝑝(1 − 1)2 + 𝑞| = 9

0 ≤ 𝑦 ≤ 16

B1: 𝑦 = |1(6 − 1)2 − 9| or 𝑦 = |1(−2 − 1)2 − 9|

2

2

4

3 1

3

Mark Scheme Additional Mathematics Midyear Paper 1 F4 2017

6

Question Solutions and marking scheme Sub

Marks

Full

Marks

24

B2: Label (-3,0), (7,0), (2,25) and 𝑐 = 21 correctly.

B1: −(𝑥 − 2)2 + 25

3

3

25

𝑎 = −2, 𝑏 = 6 𝑎𝑛𝑑 𝑐 = 8

B2: 0 = 𝑎(−1)2 + 𝑏(−1) + 8 and 0 = 𝑎(4)2 + 𝑏(4) + 8

OR −2(𝑥 + 1)(𝑥 − 4) = 0 OR equivalent

B1: 0 = 𝑎(−1)2 + 𝑏(−1) + 8 or 0 = 𝑎(4)2 + 𝑏(4) + 8

OR 𝑎(𝑥 + 1)(𝑥 − 4) = 0 or 𝑐 = 8

3

(-3,0)

(7,0)

(8,-11)

(2,25)

𝑦

21

𝑥