GERAK GEMPUR SPM 2020

b). i). Find ⃗ ⃗ ⃗⃗ ⃗ and use ⃗⃗ ⃗⃗ ⃗ = ⃗⃗ ⃗⃗ ⃗ ,

AC  3 ~x  2 y K1
5m 5m
~

Use triangle law to find ⃗⃗ ⃗⃗ ⃗

⃗ ⃗ ⃗⃗ ⃗ = ⃗ ⃗ ⃗⃗ ⃗ + ⃗⃗ ⃗⃗ ⃗

N1 DC  3 ~x   2 1 y
5m 5m
~

ii). Use triangle law to find ⃗ ⃗ ⃗⃗ ⃗ and use ⃗⃗ ⃗⃗ ⃗ = ⃗ ⃗⃗ ⃗⃗ ⃗
5

⃗ ⃗ ⃗⃗ ⃗ = ⃗ ⃗⃗ ⃗⃗ ⃗ + ⃗⃗ ⃗⃗ ⃗

x  n 1 y N1
~ 5 ~

c). Equate ~x K1

3
5 = 1

3
N1 = 5

Equate coefficient y and substitute =∗ 3
5
~
2 K1
5 ∗ (35) − 1 = 5 − 1
10
N1 = 3

10

9 a).i). Use Binomial formulae and find ( = 1)
5 1(0.6)(0.4)4 K1

ii). Use Binomial formulae N1 0.0768
( = 0) = 5 0(0.4)0(0.6)5 K1
( < 3) = ( = 0) + ( = 1) + ( = 2)
5 0(0.4)0(0.6)5 + 5 1(0.4)1(0.6)4 + 5 2(0.4)2(0.6)3

OR
5 3(0.6)3(0.4)2 + 5 4(0.6)4(0.4)1 + 5 5(0.6)5

K1

N1 0.3174

b). i). Use = − , K1
2 − 1.5 ( < 2.5) = 0.9938 N1

( < 2) = 1 − ( > 0.2 )

ii). Find inverse normal from the area under z-score graph

K1 Equate = − with the value of x-axis.
N1 = 1.423
− 1.5
K1 0.2 =∗ −0.385

10

10 a). Equate the equation of straight line and the curve.

2 K1
3 = − 2 + 8
= −8, = 2

= 3(2)
ℎ = 2, = 6 N1

Substitute = 0 and find the value of x,
2

0=− 2 +8
= ±4

N1 = 4

b). Integrate4 x2  8dx and substitute limit


22

 x3  4 K1
 8x
 6
2

Find the area of triangle and add the area

K1 1 (2 x 6)   x3 4
2   8x
 6
2

N1 38 unit2
3

c). Integrate ∫06 (16 − 2 ) K1
and substitute limit y=0 and y=6.

Find the volume of cone

K1 = 1 (2)2(6)
3

subtract

y16y  26  8 K1
0

N1 52 10

11 11. a

√ . 3.16 4.47 5.48 6.32 7.07 7.75

Table for the values of √ N1

b.
y

10
9
8

7
6

5
4
3
2.1
2

1

0 1 2 3 4 5 6 7 8 √

Correct axes, uniform scale and one point correctly plotted K1
All points correctly plotted
Line of best fit N1
c. i. y = 5a√ + b P1 N1

Find and equate gradient with 5a

K1

N1 a = 0.196

ii. use b = y-intercept K1
b = 2.1 N1

iii. x = 36 N1 10

12 a). Find first equation when = 4, = 0, P1
16 + 4 = 0

Find second equation when = 1, = −2 P1
2 + = −2

Solve simultaneous equation
16 + 4 = 0 and 2 + = −2 K1

N1 = 1, = −4

b). Differentiate V and equate to 0 and find time for turning point.

2 − 4 < 0, K1

0 < < 4 N1

c).

U shape and (2,-4), (6,12) N1
d). Integrate ∫34 2 − 4

K1

Substitute limit t=3 and t=4
K1 43 4(4)2 33 4(3)2
[− ]−[ − ]
32 32

N1 2
13
10
13 a). i). 2014 = 105×75 K1 10
100

N1 2014 = 78.75

ii). 2008 = 183×100 K1
122

N1 2008 = 150

b). Equate

108 = 110 × 3 + 105 × + 120 × + 102 × 2 K1

3 + 4

= 2 N1 P1 Use ratio 3:∗ 2:∗ 2:∗ 4

Find 1̅ 5 ,

+ 08
112 × 3 + 107 ×∗ 2 + 122 ×∗ 2 107 ×∗ 4
1̅ 5 = ∗ 11

08

OR

(∗ 101.82 × 3) + (101.9 ×∗ 2) + (∗ 101.67 ×∗ 2) + (∗ 104.9 ×∗ 4)

∗ 11

Find 15 for P,Q, R and S, (101.82, 101.9,101.67,104.9) K1

14
111.09
1̅ 5 = 108 × 100 K1

14 N1 1̅ 5 = 102.86 ↔ 102.93

14

14 (i)Use cos rule
132 = 122 + 152 – 2(12)(15)COS <DCE K1

< PRS = 56.250 N1

(ii)Find  ,

< BCA = 1800 – 56.250 K1

= 123.750

Use sin rule

K1 = 30

sin 30 sin 123.75

BC = 18.04 cm N1

(iii) < ABC = 1800 – 300 – ( 1800 – 56.250)
= 26.250

Use formulae area of triangle

K1 Area of ∆ ABC = 1(30)(18.04) sin 26.25

2

N1 = 119.68 cm2

(b)
B

18.04 cm 56.25 18.04 cm
cm N
0
C

FindNBC K1
< NBC = 1800 - 2(56.250)
= 67.500

Use sin rule OR cos rule

K1 = 18.0403
sin 67.50 sin 56.25

N1 BC = 20.0453 cm 10

15
10