Mocktest 2 Kertas 2

1

Skema Markah Mock Test 2

Matematik Tambahan Kertas 2

Bahagian A

1 12 yx [1]

26)12())(12( 2 yyy [1]

0261442 22 yyyy

02556 2 yy

0)53)(52( yy [1]

2

5y ,

3

5y [1]

4x , 3

13x [1]

3

5,

3

13,

2

5,4 [1]

2 (a) 256)1()8( 2 r [1]

r = 2 cm [1]

(b) 𝑎 = 64, 𝑟 =1

4 [1]

𝑆∞ =64

1−1

4

[1]

= 851

3 [1]

3 (a) 3𝑥 − 2 = −𝑥 + 2 [1]

𝑥 = 1 ,𝑦 = 1

𝑃(1, 1) [1]

Mocktest 2 Kertas 2

2

(b) (i) RQ = RP [1]

𝑥 − 2 2 + (𝑦 − 2)2 = 𝑦2 [1]

𝑥2 − 4𝑥 − 4𝑦+ 8 = 0 [1]

(ii) Show by solving simultaneous equations

𝑥2 − 4𝑥 − 4 −𝑥 + 2 + 8 = 0 [1]

x= 0 ( only one point of intersection) [1]

Therefore the line 𝑦 = −𝑥 + 2 is a tangent.

4 (a) cos 4x = cos [2(2x)] = cos2 2x sin2 2x [1]

= cos2 2x ( 1 𝑠𝑖𝑛 2 2𝑥

𝑐𝑜𝑠2 2𝑥) [1]

= cos2 2x ( 1 tan2 2x) [1]

(b)

- Cosine Shape with correct amplitude = 1 [1] - 2 cyles within the domain [1]

- Draw line y = 1 [1]

- k = 1 [1]

y = k

x

y

Mocktest 2 Kertas 2

3

5 (a) 𝑦 = 𝑎3 [1]

𝑦 1

6 = (𝑎3)1

6 [1]

𝑎 = 𝑦1

6 [1]

(b) 𝑙𝑜𝑔𝑎 𝑦3

𝑎2 = 𝑙𝑜𝑔𝑎𝑦3 − 𝑙𝑜𝑔𝑎𝑎

2 [1]

= 3𝑙𝑜𝑔𝑎𝑦 − 2𝑙𝑜𝑔𝑎𝑎 [1]

= 3 3 − 2 [1]

= 7 [1]

6 (a) 𝑚 = 30.5 + 75 +𝑘

2 − 35

30 × 10 = 35.5 [1][1]

75+𝑘

30− 35 = 15

𝑘 = 25 [1]

(b) 2

2

100

3440

100

135715

[1] [1]

= 173.79 [1]

(c) (i) median = 41.5 g [1]

(ii) 173.79 [1]

Mocktest 2 Kertas 2

4

Bahagian B

7 (a)

[1] *Ignore 𝑝

25

(b)

(c) (i) 𝑦

25= 3.1 [1]

k = 77.5 [1]

(ii) 𝑦

𝑥=

1

𝑎+

𝑏

𝑎𝑥 [1]

a =10 [1]

(iii) 𝑏

𝑎= 0.1214 [1]

b = 1.214 [1]

8 (a) (i) 𝑑𝑦

𝑑𝑥= 2𝑥 − 6

𝑦 = 𝑥2 − 6𝑥 + 𝑐 [1]

0 = (4)2 − 6(4) + 𝑐

c = 8

𝑦 = 𝑥2 − 6𝑥 + 8 [1]

(ii) 3 = 2(h) +1

h = 1 [1]

(iii) Area = 𝑥2 − 6𝑥 + 81

0 𝑑𝑥 − 2𝑥 + 1

1

0 𝑑𝑥 [1]

= 𝑥3

3− 3𝑥2 + 8𝑥

0

1

− 𝑥2 + 𝑥 01 [1]

= 51

3− 2

= 10

3 [1]

x 10 20 25 40 50 60 70 𝑦

𝑥

1.30 2.55 𝑝

25

5.00 6.25 7.30 8.60

𝑦

𝑥

x O

c =0.1

m=0.1214

At least 1 point correctly plotted . [1]

All points correctly plotted. [1]

Best fit line [1]

* Ignore (25, 𝑝

25)

Mocktest 2 Kertas 2

5

(b) 𝜋 4− 𝑦𝑘

0 𝑑𝑦 = 6𝜋 [1]

𝜋 4𝑦 −𝑦2

2

0

𝑘

= 6𝜋 [1][1]

4𝑘 −𝑘2

2= 6

𝑘2 − 8𝑘 + 12 = 0

𝑘 = 6, 𝑘 = 2

k = 2 [1]

9 (a) (i) – u + 4 v [1]

(ii) 𝑀𝑃 = −2v +3u [1]

1

5 (– 2v + 3 u) or

3

5 u −

2

5 v [1]

(iii) 2 u + 4

5 ( –3u + 2v) or – u + 2v +

3

5 u −

2

5 v [1]

−2

5 u +

8

5 v [1]

(b) 𝑇𝑄 = −3

5 u +

12

5 v [1]

𝑅𝑇 = −2

5 u +

8

5 v

= 2 (−1

5 u +

4

5 v )

= 2

3 (−

3

5 u +

12

5 v )

𝑅𝑇 =2

3 𝑇𝑄 [1]

𝑅𝑇

𝑇𝑄=

2

3

RT : TQ = 2 : 3 [1]

(c) Area of PQT

Area of PQM =

1

2 𝑃𝑇 (𝑀𝑄 )

1

2(𝑃𝑀)(𝑀𝑄 )

Area of PQT

25 =

𝑃𝑇

𝑃𝑀=

4

5

Area of PQT = 4

5× 25 [1]

= 20 [1]

Mocktest 2 Kertas 2

6

10 (a) (i) 𝑃 𝑥 ≥ 2 = 1− 𝑃 𝑥 = 1 − 𝑃 𝑥 = 0

= 1 − 𝐶51

1

3

1

2

3

4

− 𝐶50

1

3

0

2

3

5

[1]

= 131

243 atau 0.5391 [1]

(ii) 𝜇 = 19 , [1]

𝜎 = 3.559 [1]

(b) (i) 𝑃 𝑥 > 3.75

= 𝑃 𝑧 >3.75−3.672

0.2704 [1]

= 𝑃 𝑧 > 0.15

= 0.4404 [1]

(ii) 𝑃 3.0 < 𝑥 < 3.75

= 𝑃 3.0−3.672

0.2704< 𝑥 <

3.75−3.672

0.2704 [1]

= 𝑃 −1.292 < 𝑧 < 0.15

= 1 – 0.4404 – 0.0981 [1]

= 0.4615 [1]

= 46.15 % [1]

11 (a) sin(½COD) = 5/8 [1]

½COD = 0.675 rad.

COD = 1.350 rad. [1]

(b) XAD =XBC = /2+0.675 = 2.246 rad. [1]

Length of arc CD or arc CX or arc DX [1]

Perimeter = 13(1.350) + 2 x 5 x 2.246 [1]

= 40.01 cm [1] Accept 40.0 cm

(c) Area of triangle OAB = ½(8)2sin(1.350) or 2 x ½(5)((8

2 - 5

2)) [1]

= 31.223 / 31.225

Area of sector OCD or Area of sector DAX or Area of sector CBX

= ½(13)2(1.350) = ½(5)

2(2.246) [1]

= 114.075 cm2 = 28.075

Area of shaded region = 114.075 – 31.223 – 2(28.075) [1]

= 26.702 cm2 [1]

Mocktest 2 Kertas 2

7

Bahagian C

12 (a)(i) 120 ×30

100= 36 [1][1]

(ii) 140

100×

125

100× 100 = 175 [1][1][1]

(b) (i) 2𝑥+140 4 +120 3 +110 (5)

2+4+3+5= 120 [1][1]

𝑥 = 105 [1]

(ii) 120 ×120

100= 144 [1][1]

13 (a)(i)

]1[39.56

]1[95.87sin

12

sin

10

0

PRT

PRT

(ii)

000 22.67)39.56(2180 RPS

]1[75.7

]1[39.56sin

7

22.67sin 00

cmRSequivalentOR

RS

(iii)

000 73.2022.6795.87 SPT [1]

]1[39.12

]1[)73.20)(sin7)(10(2

1

2

0

cm

PST

(b)

000 17.15522.6795.87 QPT [1]

]1[61.16

]1[17.155)7)(10(2710 0222

cmQT

kosQT

Mocktest 2 Kertas 2

8

14 (a) 𝐼 ∶ 𝑥+ 𝑦 ≤ 7000 [1]

𝐼𝐼 ∶ 𝑦 ≤ 2𝑥 [1]

𝐼𝐼𝐼 ∶ 𝑦 ≥ 1000 [1]

(b)

(c)(i) 5000 [1]

(ii) Maximum profit/ keuntungan maksimum = 50(6000) + 30(1000) [1][1]

= RM 330000 [1]

Mocktest 2 Kertas 2

9

15 (a) a=m+nt

– 10 =m + n(0)

m = – 10 [1]

]1[4

1218600

0,6

]1[122

10

12,12,0

2

2

2

n

n

vt

tn

tv

cvt

ctn

mtv

dtntmv

(b)

]1[2

49

]1[122

52

2

510

2

5

]1[0410

2

v

v

t

t

(c)

]1[3

2262

]1[)6(12)6(5)6(3

2)10(12)10(5)10(

3

2)6(12)6(5)6(

3

2

]1[1253

2125

3

2

]1[1210212102

232323

10

6

236

0

23

10

6

26

0

2

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