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SKEMA PEMARKAHAN SOALAN KERTAS 2 MATEMATIK TAMBAHAN
PERCUBAAN 2020
Soalan Skema pemarkahan Sub
marka
h
Jumlah Markah
1 2π₯ + 4π¦ = 24
π₯ = 12 β 2π¦
[(12 β 2π¦) + π¦]2 = (12 β 2π¦)2 + (3π¦)2
π¦(π¦ β 2) = 0 ππ (π₯ β 8)(π₯ β 12) = 0
π¦ = 0 , π¦ = 2
π₯ = 12 , π₯ = 8
ππππ = 48ππ2
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2 a i)
100.5 + 150.5
2
125.5 (ππππππ‘ π€ππ‘βππ’π‘ π€ππππππ)
ii) 75.5(10)+125.5(40)+175.5(10)+225.5(30)+275.5(20)
110
180.05
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b) 150.5 ππ 50 ππ 10
150.5 + (
1102 β 50
10)50
175.5
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3a) (i)
ii)
3b)
π΅π·ββββββ = π΅π΄ ββ ββ ββ + π΄π·ββββββ β
π΅π΄ββββ β = β20π₯
π΄π·ββ ββ β = 4 β 8π¦ = 32π¦
π΅π·ββββββ = π΅π΄ ββ ββ ββ + π΄π·ββ ββ β = β20π₯ + 32π¦
πΈπΆββββ β = πΈπ·ββββ β + π·πΆββββ β
πΈπ·ββββ β = 3π΄πΈββββ ββ ββ β = 24π¦
πΈπΆββββ β = 24π¦ + (25π₯ β 24π¦ ) = 25π₯
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π΅π·ββββββ = β20π₯ + 32π¦
πΈπΉββββ β = 3
5πΈπΆββββ β = 15π₯
πΉπ·ββββ β = πΉπΈββββ β + πΈπ·ββββ β
= βπΈπΉββββ β + 24π¦
= β15π₯ + 24π¦
πΉπ·ββββ β = 3(β5π₯ + 8π¦)
π΅π·ββββββ = 4(β5π₯ + 8π¦)
π΅π·ββββββ = 4 (πΉπ·ββ ββ β
3)
π΅π·ββββββ = 4
3πΉπ·ββββ β
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3c)
|π΅π· β| = β20π₯ + 32π¦
= |β20(2) + 32(3) |
= β(40)2 + (96)2
= 104
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4a
b)
π₯ log10 3 = π¦ log10 5 *
π¦ log10 5 = π§ log10 3 + π§ log10 5 * *Either
π¦ log10 5 = π§ (π¦
π₯log10 5) + π§ log10 5
5π¦ = 5π§π¦π₯ Γ 5π§
π§ =π₯π¦
π¦ + π₯
log10(2π₯(4π₯ β 1)) = 1
8π₯2 β 2π₯ = 10
π₯ =5
4 , π₯ = 1
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3
5a πΏπ’ππ =
1
2 π2π =
1
2 (8)2(1.956)
= 62.592 π2
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5b π πΆ + πΆπ + π π
π πΆ = 6
πΆπ = ππ = 8(1.956) = 15.648 * (* EITHER)
π π = ππ = 14(3.142 β 1.956) = 16.604*
πππππππ , πππππ π , πππππ π¦πππ πππππππ’πππ
π’ππ‘π’π πππππππ πππ‘ππ ππ’πππ
= 6 + 15.648 + 16.604
= 38.252 π
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6a (π₯ β 2)2 = π₯ + 4
π₯2 β 5π₯ = 0
π₯ = 0, π₯ = 5
π = 5
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6b β β« π₯5
0+ 4 ππ₯ β β β« π₯2
5
0β 4π₯ + 4 ππ₯ * Either
β [π₯2
2+ 4π₯]
0
5
β β [π₯3
3β 2π₯2 + 4π₯]
0
5
* Either
[((5)2
2+ 4(5)) β (0)]
β [((5)3
3β 2(5)2 + 4(5)) β (0)]
= 20.83
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7.
10
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8.
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(LIHAT LAMPIRAN )
9. π)
πππ(π₯ β π¦) + π ππ(π₯ + π¦)
2πππ π₯πππ π¦
=(π ππ π₯ πππ π¦βπππ π₯ π ππ π¦ ) + (π πππ₯ πππ π¦+πππ π₯ π ππ π¦
2πππ π₯πππ π¦
=2π πππ₯πππ π¦
2πππ π₯πππ π¦
= tan x
b)
π₯
πβ π¦ = 1
π¦ =π₯
πβ 1
Ο
Shape
Cycles
Amplitude
Modulus
-
10
10.
2
3
2
2
3
3
2
2
2
2
2672.49
)3094.2(2)3094.2(32
3094.2
0632
632
232)
232
)16(2)
)16,4(
16
)4()4(8,4
4
2
28
8,0
0)8(
0,
)8()
unit
umLuasmaksim
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k
kdk
dL
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kkABCDtepatsegiempatLuasb
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kkykxwhen
k
kOA
xx
xx
yxpaksipada
xxya
answer
=
β=
=
=β
β=
β=
β=
β=
ββ
β=
βββ=β=
β=
β=
==
=β
=β
β=
10
Equation
draw line correctly
NOS=1
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11.
10
12. a)
I : 80+ yx
II : 24038 + yx
III : 1202 + yx
Refer to graph
One straight line drawn correctly
10
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All straight lines drawn correctly
Correct region
i)
Draw line x = 2y
Coconut, x = 53; Rambutan, y = 27
(ii)
700x + 250y = 7000
14x + 5y = 140
From graph, Profitmin = 700(10) + 250(55)
RM20 750.00
13. (a)
(b) Composite index in the year 2007 based on the year
2006
=
= 114.4
(c)
Let the price index of item S in the year 2008 based on the
year 2007 be z.
105 =
10
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1 050 = 745 + 2z
305 = 2z
z = 152.5
Thus, the price of item S increases by 52.5% from the year
2007 to the year 2008
14.
a) i) =
sin β EFG =
= 0.8245
β EFG = 55.53Β°
ii) 6.72 = 3.42 + 6.42 β 2 Γ 3.4 Γ 6.4 Γ cos β EHG
cos β EHG =
= 0.1753
β EHG = 79.91Β°
iii β EGF = 180Β° β 30.3Β° β 55.53Β°
= 94.17Β°
Area of ΞEFG
= Γ 6.7 Γ 4.1 Γ sin 94.17Β°
= 13.7 cm2
Area of ΞEGH
= Γ 3.4 Γ 6.4 Γ sin 79.91Β°
= 10.71 cm2
Area of quadrilateral EFGH
= 13.7 + 10.71
= 24.41 cm2
b) (i)
ii β E'F'G' = 180Β° β 55.53Β°
= 124.47Β°
10
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15 .(a) v = 15 - 10t
v = -5 ms-1
(b) 15 - 10t = 0
(c) 15(4) - 5(4)Β² + h = 0
h = 20
(d)
(e) v = 15 - 10t
a = -10
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π‘ = 3
2π
π = β5(π‘ β 3
2)2
+ 45
4
Tinggi maksimum = 45
4+ 20
= 125
4 m
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LAMPIRAN SOALAN NO 8