# modul pintas tingkatan 5 peperiksaan percubaan spm 2018 ... set... skema jawapan matematik tambahan...

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• Modul Pintas Tingkatan 5 Peperiksaan Percubaan SPM 2018

Skema Jawapan Matematik Tambahan Kertas 2 3472/2

• 1

NO SOLUTIONS MARKS

1

   

2

22

2 2

4 8 48 4 135

12 12 2

2

12 4 135 4 12 2 12 2 135

2

24 135 0 4 9 0

3 ( 15)( 9) 0

2

Length and breadth: 9 , (15 ( )

3 Height :

2

x y or xy x

x x y OR y

x x x OR y y y

x x OR y

x x OR y

x ignore

y

   

   

        

 

     

   

P1

P1

K1

K1

N1

N1 6 6

2

2 2

2

18 ( ) 6, 18 , 3 3 ,

2

18 18 3

2 2

9

a or seen

or

cm

          

   

P1

N1

N1 3

6

2 ( ) 24, 4 18, 12 24

2

24

2 1

2

81.94

b OR a and r

S

 

P1

K1

N1 3

• 2

NO SOLUTIONS MARKS

3  

  2 2 2 2 2

2

75 70 59 62 68 66.8

5

75 70 59 62 68 66.8 5.706

5

Syafiqah result is more consistent

a    

     

P1

K1N1

N1 4

6

334 ( ) 67.5

6

71

x b

x

 

K1

N1

2

4

 

2

2

2

2

2 2

2 2

sin 1

sin 1

sin

sin

1

2

sec2

x

cos x

x

cos x

cos x x

cos x x

cos x

x

a

 

K1

N1 2

8

(b)

3

( ) 5

x c y or implied

 

Sketch the straight line, correct gradient or y intercept

3 solutions

N1

K1

N1 3

y

1

-1

x

Shape of positive cos graph P1

1 ½ cycles P1

Maximum 1, minimum -1 P1

Use identity :

tan2x=sin2 x/cos2x

atau cos2x+ sin2x = 1

atau cos2x- sin2x = cos2x

• 3

NO SOLUTIONS MARKS

5      

 

 

   

3 2 1 ( )( ) 2 3 2 3

2 3

2 3 3 2

0

m n a i m n i m n i i j OR

m n

m n and m n

m n or equivalent

 

         

     

 

(ii) 2m - n = 0

K1

N1

N1 3

7

( )( ) 2(2 3 ) 3(3 )

(13, 9)

b i OP i j i j

P

   

 

 

( ) ( 13 9 ) ( 2 )

12 11

1 12 11

265

ii PQ i j i j

i j

Vektor unit i j

    

  

  

K1

N1

K1

N1 4

6  

 

2 ( ) 3 2 1

11

6 11 2

11 16

dy a

dx

y x

y x

 

  

 

K1

K1

N1 3

7

2( ) 3 1 11

2,2

( 2, 6)

b x

x

Q

 

 

  

K1

K1

N1 3

(c) Rectangle @ Segi empat tepat P1 1

7 (a) AOB = 6.5/5

K1

N1

N1 3

(b) MN = 5  sin(0.8667 rad.) = 3.811 cm

or ON = 5  cos( 0.8667 rad.) = 3.2367 cm

Length of arc PQ = 6  0.8667 = 5.2002

Perimeter = 1+*MN+ *NQ+*QP

= 12.77 cm

K1

K1

K1

N1 4

• 4

NO SOLUTIONS MARKS

(c ) Area of sector OPQ = ½  62  0.8667

OR Area of triangle OMN=½ (3.811)(3.2367)

Area of shaded region = *15.60  *½ (3.811)(3.2367)

= 9.432

K1

K1

N1 3 10

8

 

1 ( )( ) 2 2(1) 2,

2

1 1 5

2

2 3

dx dy a i y

dy dx

y x

y x

   

  

 

K1

K1

N1 3

10

3

13

4 3

4

:)(

1

0

31

0

2 1

 

 

  y

y dyyA

CurvetheUnderAreaii

   1

2 2

0

Area of trapezium:

1 3 1 3 5 4

2 A y y or    

  

1 2

1

3

A A

6

4

6 2

4

) 4

6 4 4 4(6) 4(4)

2 2 2

2

b Volume x dx

x x

 

 

                  

       

K1

K1

K1

N1

K1 K1

N1

4

3

• 5

NO SOLUTIONS MARKS

9 (a) p = 0.85, q = 0.15

𝑃(𝑥 ≥ 6) = 𝑃(𝑟 = 6,7,8)

8C6 (0.85) 6(0.15)2 or 8C7 (0.85)

7(0.15)1 or 8C8 (0.85) 8(0.15)0

= 8C6 (0.85) 6(0.15)2 + 8C7 (0.85)

7(0.15)1 + 8C8 (0.85) 8(0.15)0

= 0.8948

P1

K1

K1

N1 4

10

(𝑏)(𝑖) 𝑃(35 < 𝑋 < 66) = 𝑃 ( 35 − 48

6 < 𝑍 <

66 − 48

6 )

= 𝑃(−2.167 < 𝑍 < 3)

= 1 − 𝑃(𝑍 > 2.167) − 𝑃(𝑍 > 3)

= 1 − 0.01512 − 0.00135

= 0.9835

Number of students between 35 to 66 marks = 0.9835 × 180

= 177 students

(ii) 𝑃(𝑥 < 𝑚) = 0.05

𝑃 (𝑍 < 𝑚 − 48

6 ) = 0.05

1.645 𝑠𝑒𝑒𝑛

𝑚 − 48

6 = −1.645

𝑚 = 38.13

K1

N1

N1

P1

K1

N1

3

3

10 (a) All values of (x+2) and log10y correct.

x + 2 1 2 3 4 5 6

log 10 y 0.92 1.00 1.08 1.12 1.24 1.32

Refer graph paper

Plot log10 y against (x+2) with correct axes, uniform scales and

N1

K1

• 6

NO SOLUTIONS MARKS

at least one point.

6 points plotted correctly

Line of best fit

N1

N1 4

10

b) (i) 14.45

 10 10 10

10

10

10

10

( ) log log 2 log

log

log 0.84

0.1445

log

log 0.08

1.202

ii y p x q

Use q c

q

q

Use p m

p

p

  

 

 

N1

P1

K1

N1

K1

N1 6

11

 

1 ( ) , 2

2

2 2 8

2 18

BC AXa m m

y x

y x

  

   

  

P1

K1

N1 3

10

(b) Solve simultaneously : y = -2x +18 and 2y = x + 10

X = (5.2, 7.6)

4(2) 4(6) 5.2 7.6

5 5

x y or

   

C (18 , 14)

K1

N1

K1

N1

4

   

2

1 ( ) 8(6) 2(14) 18(2) 2(2) 18(6) 8(14)

2

2 *

112

c Area of ABC

Area of trapezium Area of ABC

unit

      

  

K1

K1

N1 3

• 7

NO SOLUTIONS MARKS

12 20 ( ) 100

8

250

a x  

K1

N1

2

10

110( ) 80(2 ) *250(3) 150(2) ( ) 120

2 3 2

5

y y b

y y

y

   

  

K1 K1

N1

3

110 80 ( ) : 110 110

100 100

121 88

121*(5) 88*(10) 250(3) 150(2)

121*(5) 88*(10) 250(3) 150(2)

*5 *10 3 2

126.75

c Seen A or B

I

   

 

  

   

  

K1

K1

K1

K1

N1

5

13

(a)(i)

(a)(ii)

2

3 2

3 2

( ) 3, 0 27 6 0

3 2

0, 0, 0

3, 1 1 27 9

Solve simultaneously: 27 6 0, 27 9 1

2 1 ,

27 3

a t v k h

s kt ht dt

kt ht c