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ADDITIONAL MATHEMATICS 3472/2 OKTOBER 2020
MAJLIS PENGETUA SEKOLAH MALAYSIA (MPSM) CAWANGAN KELANTAN
TINGKATAN 5
2020
ADDITIONAL MATHEMATICS
KERTAS 2
UNTUK KEGUNAAN PEMERIKSA SAHAJA
SKEMA PEMARKAHAN
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1.
(a) (b)
(c)
f (x) = - (x2 + 3x + (32)
2 - (32)2 – 4)
OR equavalent f(x) = - (x + %
&)2 + &'
(
(− %
&, &'(
)
Shape Maximum point X intercept 0 ≤ f(x) ≤ &'
(
3
3
1
7
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2.
(a)
(b)
(c)
T7 = 100π (+&)7-1 100π, 50π, 25π.
&'+,
π
100π (+&)n-1 = &'
,(π
n = 9
100𝜋𝜋
1 −12
200 π m / 628.4 m
3
2
2
7
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N1
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2.
(a)
(b)
(c)
T7 = 100π (+&)7-1 100π, 50π, 25π.
&'+,
π
100π (+&)n-1 = &'
,(π
n = 9
100𝜋𝜋
1 −12
200 π m / 628.4 m
3
2
2
7
N1
N1
N1
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3.
(a) (b)
x + 2( 2x + 6) = 20 or 01,&
+ 2y = 20 (
2', (,')
3(20 −85)
& +(0 −465 )
&
20.57 40 = &:.'<=>?@
80 =
A:.'+(%
41.14
2
4
6
p Qt t=
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4 (a)
(b)
Shape of 𝑘𝑘𝑘𝑘𝑘𝑘2𝑥𝑥 Max and min 𝑦𝑦 = −2𝑘𝑘𝑘𝑘𝑘𝑘2𝑥𝑥 2 cycles for 0≤ 𝑥𝑥 ≤ 2𝜋𝜋 Modulus graph 𝑦𝑦 = I
&J− 1
Graph straight line c=-1 and m positive Number of solutions = 8
4
3
7
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4 (a)
(b)
Shape of 𝑘𝑘𝑘𝑘𝑘𝑘2𝑥𝑥 Max and min 𝑦𝑦 = −2𝑘𝑘𝑘𝑘𝑘𝑘2𝑥𝑥 2 cycles for 0≤ 𝑥𝑥 ≤ 2𝜋𝜋 Modulus graph 𝑦𝑦 = I
&J− 1
Graph straight line c=-1 and m positive Number of solutions = 8
4
3
7
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5 𝑥𝑥 − 3𝑦𝑦 + 4 = 5 𝑦𝑦(𝑥𝑥 + 2𝑦𝑦) = 5 𝑥𝑥 = 1 + 3𝑦𝑦 or 𝑦𝑦 = I1+
%
𝑦𝑦(1 + 3𝑦𝑦 + 2𝑦𝑦) = 5 or I1+%K𝑥𝑥 + 2 LI1+
%MN = 5
Solve the quadratic equation Formulae
𝑦𝑦 = 1(+)±P(+)Q1((')(1')&(')
or
𝑥𝑥 = 1(1<)±P(1<)Q1((')(1(%)&(')
𝑦𝑦 = 0.905, 𝑦𝑦 = −1.105 Or 𝑥𝑥 = 3.715, x= −2.315 𝑥𝑥 = 3.715, x= −2.315 Or 𝑦𝑦 = 0.905, 𝑦𝑦 = −1.105
7
7
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6(a)
(b)
TU
T== :.%+,1:.:'(
&'
+%+
+&':: or 0.01048
TUTV= 4𝜋𝜋𝑟𝑟& or TX
TV= 8𝜋𝜋𝑟𝑟 Substitute 𝑟𝑟 = 1.2
TX
T== TU
T=× TX
TV× TVTU
+%+
+&'::× 8𝜋𝜋𝑟𝑟 × +
(JVQ
+%+
(': or 0.2911
2
4
6
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7 (a)
(b)(i)
(ii)
(c)
Write triangle law 𝐵𝐵𝐵𝐵\\\\\\⃗ − 10𝑥𝑥 + 10𝑦𝑦 𝐵𝐵𝐵𝐵\\\\\⃗ − 2𝑥𝑥 + 5𝑦𝑦 𝐴𝐴𝐵𝐵\\\\\⃗ = 8𝑥𝑥 + 5𝑦𝑦 𝐴𝐴𝐴𝐴\\\\\⃗ = 𝑘𝑘
1+𝑘𝑘(12𝑥𝑥 − 5𝑦𝑦)
𝐸𝐸𝐵𝐵\\\\\⃗ = −4𝑥𝑥 + 10𝑦𝑦 𝐴𝐴𝐴𝐴\\\\\⃗ = (4 − 4ℎ)𝑥𝑥 + 10ℎ𝑦𝑦 Equate and solve coefficient of x or y
2c+dc
= 4 − 4ℎ or 'c+dc
= 10ℎ 𝑘𝑘 = &
%
ℎ = +
'
3
4
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8 (a)(i)
(ii)
(b)(i)
(II)
mean, µ = 500
P(Z > ) = 0.2
= 0.842
m = 584.2 Ahmad qualify/ layak 585 > 584.2
or
0.80541
n =
0.2(1554) 310 / 311
1
4
2
3
10
100500-m
100500-m
÷øö
çèæ -
££-
100500680
100500400 zP ( )8.11 ££- zP
8054.01252 K1
N1
K1
N1
K1
K1
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9 (a)
(b)(i)
(ii)
120: × J
+2:e
&J
%
𝑆𝑆𝑆𝑆𝑆𝑆60: = I
i
Use 2𝑗𝑗𝑠𝑠𝑆𝑆𝑆𝑆 k
&
2 l√%&𝑗𝑗n
√3𝑗𝑗
𝑠𝑠𝑆𝑆𝑆𝑆60: = √%&
or 𝜃𝜃 = J%
or 𝜃𝜃 = &J%
𝐴𝐴+ = 𝜋𝜋𝑗𝑗& or 𝐴𝐴& =
+&𝑗𝑗& L&J
%M
𝐴𝐴% =+&𝑗𝑗& J
%− +
&𝑗𝑗& √%
&
𝐴𝐴+ − 𝐴𝐴& − 2𝐴𝐴%
JiQ
%+ √%i
&
2
3
5
10
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9 (a)
(b)(i)
(ii)
120: × J
+2:e
&J
%
𝑆𝑆𝑆𝑆𝑆𝑆60: = I
i
Use 2𝑗𝑗𝑠𝑠𝑆𝑆𝑆𝑆 k
&
2 l√%&𝑗𝑗n
√3𝑗𝑗
𝑠𝑠𝑆𝑆𝑆𝑆60: = √%&
or 𝜃𝜃 = J%
or 𝜃𝜃 = &J%
𝐴𝐴+ = 𝜋𝜋𝑗𝑗& or 𝐴𝐴& =
+&𝑗𝑗& L&J
%M
𝐴𝐴% =+&𝑗𝑗& J
%− +
&𝑗𝑗& √%
&
𝐴𝐴+ − 𝐴𝐴& − 2𝐴𝐴%
JiQ
%+ √%i
&
2
3
5
10
N1
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10 (a)
(b)
(c)(i)
1𝑢𝑢
0.80 0.50 0.33 0.25 0.21 0.17
1𝑣𝑣
0.35 1.60 2.25 2.55 2.75 2.90
Plot +
r against +
s
(Correct axes and uniform scales) 6 *points plotted correctly Line of best fit (At least *5 points plotted) 𝑢𝑢 = 2.33 ± 0.1 +
r= &c
ts+ u
t
Use 𝑐𝑐 = u
t or 𝑚𝑚 = &c
t
ℎ = 2.54 𝑘𝑘 = −5.14
2
3
5
10
N1
N1
K1
K1
N1
N1
If table not shown, all the points are correctly plotted award N1
N1
N1
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S
TOTAL MARK
S 10
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S
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S 10
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11 (a)
(b)
(c)
3 = 4 − IQ
(
𝑥𝑥 = 2, 𝑥𝑥 = −2 𝐵𝐵𝐵𝐵 = 4
∫ l4 − IQ
(n&
1& 𝑑𝑑𝑥𝑥
l4𝑥𝑥 − Iz
+&n &1&
l4(2) − &z
+&n − l4(−2) − (1&)z
+&n
14.67 or ((
%
4(1) ((
%+ 4(1)
18.67 or ',
%
3
4
3
10
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12
(a)
(b)
(c)
r = 6 substitute t=3 in equation V p(3)2 + q (3) + 6 = 42 TrT== 2𝑝𝑝𝑝𝑝 + 𝑞𝑞 9p + 3q = 36 OR
6p +q = 0 p = −4 and q = 24
t =−(−24)±~(24)2−4(−4)(6)
2(−4)
-4t2 + 24t + 6 = 0 t = 3 + +
2 √672 s = (
%𝑝𝑝% +&(
&𝑝𝑝& + 6𝑝𝑝
substitute t = 6 or t = 5 in equation S s = (
%(6)% +&(
&(6)& + 6(6) or
s = (
%(5)% +&(
&(5)& + 6(5)
<<2%
5
2
3
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12
(a)
(b)
(c)
r = 6 substitute t=3 in equation V p(3)2 + q (3) + 6 = 42 TrT== 2𝑝𝑝𝑝𝑝 + 𝑞𝑞 9p + 3q = 36 OR
6p +q = 0 p = −4 and q = 24
t =−(−24)±~(24)2−4(−4)(6)
2(−4)
-4t2 + 24t + 6 = 0 t = 3 + +
2 √672 s = (
%𝑝𝑝% +&(
&𝑝𝑝& + 6𝑝𝑝
substitute t = 6 or t = 5 in equation S s = (
%(6)% +&(
&(6)& + 6(6) or
s = (
%(5)% +&(
&(5)& + 6(5)
<<2%
5
2
3
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13. a) (b) (c)
(i) x ≥ +
%𝑦𝑦 or
(ii) 12x + 8y ≤ 6000 or 3x + 2y ≤ 1500
(iii) 8x + 4y ≥ 2400 or 2x + y ≤ 600
graph- attachment
- draw correctly at least one straight line from
* inequalities involves x and y
- draw correctly ALL the
* straight line
Note: Accept dotted line
Region shaded correctly.
(i) 120
(ii) (200, 450) seen 450
8(200) + 4 (450)
8x + 4y ( profit equation)
RM 3400.00
Noted: inequality for which no symbol "=" is accepted
3
3
4
10
3y x£ N1
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x
y
x = 200
2x + y = 600 3x + 2y = 1 500
x = 𝟏𝟏𝟑𝟑 y
R
100 200 O 300 400 500 600
100
200
300
400
500
600
700
800
450
120
x
y
x = 200
2x + y = 600 3x + 2y = 1 500
x = 𝟏𝟏𝟑𝟑 y
R
100 200 O 300 400 500 600
100
200
300
400
500
600
700
800
450
120
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14. (a) (b)
(i) 135 = &+:ÅQeÇÉ
𝑋𝑋100
RM 155.56
(ii) +&:+::X+%:
+::𝑋𝑋100
156
Increase 56 %
(i) p = 15
(+%'Ö':)d(0Ö&')d(+&:Ö+')d(<:Ö+:)':d&'dÜd(Ü1')
= 119.50
108
(ii) 108 = ÅQeQe+%:
𝑋𝑋100
RM 140.40
2 3 3 2
10
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15. (a) (b) (c)
(i) BD2 = 3.742 + 82 – 2(3.74)(8)(Cos 450)
5.973 km
(ii) á>à∠Xäã
2= á>à('
e
<.2'
46.110
Noted: ∠𝐴𝐴𝐴𝐴𝐴𝐴 = ∠𝐵𝐵𝐴𝐴𝐴𝐴
(iii) 88.890
A = +&x8x7.85xSin88.890
31.39 km2
The farthest hotel from hotel A is hotel C because
the distance between them faces the largest angle
+&x8x(shortestdistance) = 31.39
7.848 km
2 2 3 1 2
10
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N1
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K111