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SKEMA PEMARKAHAN SOALAN KERTAS 2 MATEMATIK TAMBAHAN PERCUBAAN 2020 Soalan Skema pemarkahan Sub marka h Jumlah Markah 1 2 + 4 = 24 = 12 β 2 [(12 β 2) + ] 2 = (12 β 2) 2 + (3) 2 ( β 2) = 0 ( β 8)( β 12) = 0 = 0 , = 2 = 12 , = 8 = 48 2 K1 K1 K1 K1 K1 K1 N1 7 2 a i) 100.5 + 150.5 2 125.5 ( β ) ii) 75.5(10)+125.5(40)+175.5(10)+225.5(30)+275.5(20) 110 180.05 K1 N1 K1 N1 4 b) 150.5 50 10 150.5 + ( 110 2 β 50 10 ) 50 175.5 P1 K1 N1 3

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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 ππ (π₯ β 8)(π₯ β 12) = 0

π¦ = 0 , π¦ = 2

π₯ = 12 , π₯ = 8

ππππ = 48ππ2

K1

K1

K1

K1

K1

K1

N1

7

2 a i)

100.5 + 150.5

2

125.5 (ππππππ‘ π€ππ‘βππ’π‘ π€ππππππ)

ii) 75.5(10)+125.5(40)+175.5(10)+225.5(30)+275.5(20)

110

180.05

K1

N1

K1

N1

4

b) 150.5 ππ 50 ππ 10

150.5 + (

1102 β 50

10)50

175.5

P1

K1

N1

3

• 3a) (i)

ii)

3b)

π΅π·ββββββ = π΅π΄ ββ ββ ββ + π΄π·ββββββ β

π΅π΄ββββ β = β20π₯

π΄π·ββ ββ β = 4 β 8π¦ = 32π¦

π΅π·ββββββ = π΅π΄ ββ ββ ββ + π΄π·ββ ββ β = β20π₯ + 32π¦

πΈπΆββββ β = πΈπ·ββββ β + π·πΆββββ β

πΈπ·ββββ β = 3π΄πΈββββ ββ ββ β = 24π¦

πΈπΆββββ β = 24π¦ + (25π₯ β 24π¦ ) = 25π₯

N1

N1

2

π΅π·ββββββ = β20π₯ + 32π¦

πΈπΉββββ β = 3

5πΈπΆββββ β = 15π₯

πΉπ·ββββ β = πΉπΈββββ β + πΈπ·ββββ β

= βπΈπΉββββ β + 24π¦

= β15π₯ + 24π¦

πΉπ·ββββ β = 3(β5π₯ + 8π¦)

π΅π·ββββββ = 4(β5π₯ + 8π¦)

π΅π·ββββββ = 4 (πΉπ·ββ ββ β

3)

π΅π·ββββββ = 4

3πΉπ·ββββ β

K1

K1

K1

N1

4

• 3c)

|π΅π· β| = β20π₯ + 32π¦

= |β20(2) + 32(3) |

= β(40)2 + (96)2

= 104

K1

N1

2

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

K1

K1

N1

K1

K1

N1

3

3

5a πΏπ’ππ  =

1

2 π2π =

1

2 (8)2(1.956)

= 62.592 π2

K1

N1

2

5b ππΆ + πΆπ + ππ

ππΆ = 6

πΆπ = ππ = 8(1.956) = 15.648 * (* EITHER)

ππ = ππ = 14(3.142 β 1.956) = 16.604*

πππππππ , πππππ π , πππππ π¦πππ πππππππ’πππ

π’ππ‘π’π πππππππ πππ‘ππ  ππ’πππ

= 6 + 15.648 + 16.604

= 38.252 π

P1

K1

K1

N1

4

• 6a (π₯ β 2)2 = π₯ + 4

π₯2 β 5π₯ = 0

π₯ = 0, π₯ = 5

π = 5

K1

N1

2

6b β β« π₯5

0+ 4 ππ₯ β β β« π₯2

5

0β 4π₯ + 4 ππ₯ * 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

K1

K1

K1

N1

4

7.

10

• 8.

• (LIHAT LAMPIRAN )

9. π)

πππ(π₯ β π¦) + π ππ(π₯ + π¦)

2πππ π₯πππ π¦

=(π ππ π₯ πππ  π¦βπππ  π₯ π ππ π¦ ) + (π πππ₯ πππ  π¦+πππ  π₯ π ππ π¦

2πππ π₯πππ π¦

=2π πππ₯πππ π¦

2πππ π₯πππ π¦

= tan x

b)

π₯

πβ π¦ = 1

π¦ =π₯

πβ 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

k

k

kdk

dL

kkLc

kkL

kkABCDtepatsegiempatLuasb

kkB

k

kkykxwhen

k

kOA

xx

xx

xxya

=

β=

=

=β

β=

β=

β=

β=

ββ

β=

βββ=β=

β=

β=

==

=β

=β

β=

10

Equation

draw line correctly

NOS=1

• 11.

10

12. a)

I : 80+ yx

II : 24038 + yx

III : 1202 + yx

Refer to graph

One straight line drawn correctly

10

• 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

• 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 ΞEFG

= Γ 6.7 Γ 4.1 Γ sin 94.17Β°

= 13.7 cm2

Area of ΞEGH

= Γ 3.4 Γ 6.4 Γ sin 79.91Β°

= 10.71 cm2

= 13.7 + 10.71

= 24.41 cm2

b) (i)

ii β E'F'G' = 180Β° β 55.53Β°

= 124.47Β°

10

• 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

N1

N1

K1

N1

K1

N1

K1

K1

N1

N1

10

π‘ = 3

2π

π  = β5(π‘ β 3

2)2

+ 45

4

Tinggi maksimum = 45

4+ 20

= 125

4 m

• LAMPIRAN SOALAN NO 8