MODUL KECEMERLANAGN NEGERI PAHANG PERTENGAHAN TAHUN …mpsmpahang.com/Download/Matematik Tambahan T5...
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Mark Scheme Additional Mathematics Midyear Paper 1 F5 2017
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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
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ADDITIONAL MATHEMATICS MIDYEAR PARER 1 2017
Question Solutions and marking scheme Sub
Marks
Full
Marks
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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Question Solutions and marking scheme Sub
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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
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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
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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