5 [cikgujep.com] MS Temerloh P1 2020

Mark Scheme Additional Mathematics Trial Paper 1 2020

3472/1(PP)
Tingkatan
Lima
Additional
Mathematics
Kertas 1
Peraturan
Pemarkahan
Oktober
2020

PEPERIKSAAN PERCUBAAN SPM 2020

ADDITIONAL MATHEMATICS
Tingkatan 5

KERTAS 1
PERATURAN PEMARKAHAN

UNTUK KEGUNAAN PEMERIKSA SAHAJA

1

Mark Scheme Additional Mathematics Trial Paper 1 2020

ADDITIONAL MATHEMATICS TRIAL SPM PAPER 1 2020

Question Solutions and marking scheme Sub Full
Marks Marks

1 (a) 5i + 3j 1 2


(b)  8  1
 
 −4 

2 x= x + 6 2
y 2
B1 : m = 9 − 6
3−0

3 r = −3 , s = − 5 (both) 3 3
2 2
2 2
B2 : r2 = 9 or 1 2
B1 : r(rx − s) − s = 9x − 5 2
1
4 p = − 1 and q = 7 1
33

B1 : p = − 1 or q = 7
33

5 12x2 − 11x + 2 =0
=B1 : SOR 1=1 or POR 2 OR (3x – 2)(4x – 1) = 0
12 12

6(a) x = 2
(b) − 1

3

2

Mark Scheme Additional Mathematics Trial Paper 1 2020

7 n5 4
m
4
B3 : m2n2 ( )or1 5y 5
m3n−3 5x 3
3
B2 : (5x )2 (5y )2 or 52( x+ y)−3(x −y) 2 4
(5x )3 (5y )−3 2
4
B1 : Seen 52(x+ y) or 53(x−y) 3
3
8 x=2 4

B3 : 0.3011x = 0.6021 OR 2x – 2 = x

B2 : x log10 3 + (x −1) log10 4 =x log10 6 OR 22x−2 = 2x

B1: log10 (3x )(4x−1) = log10 6x OR 4 x −1 = 6x or 4 x −1 =  6 x
3x  3 

=9 a =10 and b 3 10
=B2 : a =10 or b 3 10
B1 : a2 + (3a)2 =102

10 B(2,3)

B2 : − 1 x + 4 = 2x −1
2

B1 : m= −1
2

11(a) a = 27
(b)
B1 : 10 [2a + 9(−5)] =45

2

−3
B1 : T7 = 27 + 6(−5)

3

Mark Scheme Additional Mathematics Trial Paper 1 2020

12 39 3
110 3

B2 : 0.3 + 0.054 4
1 − 0.01 4

B1 : a = 0.054=and r 0=.00054 0.01 4
0.054
4
13 5, 5, 6, 7, 12 or 5, 5, 6, 8, 11 or 5, 5, 6, 9, 10
B3 : x + y = 16
B2 : 5 + 5 + 6 + x + y = 7
5
B1 : 5 5 6 x y

14 Eric is not qualifying, because his marks 24 < 24.5.

 30 − 9 
 2 
B3: Median = 19.5 +  12  (10)
 


B2 : L = 19.5 or f = 12 or F = 9 or c = 10

B1 : Median class = 20 – 29
Notes:

Marks Lower Frequency Cumulative
Boundary Frequency

0–9 0 5 5

10 – 19 9.5 4 9

20 – 29 19.5 12 21

30 – 39 29.5 8 28

40 – 49 39.5 2 30

** Not accept without working.

4

Mark Scheme Additional Mathematics Trial Paper 1 2020

15 0.00324π or 0.01018 4

=B3: dV 4(3) π (0.27)2 × 0.1 4

dt 27 3
3
B2: V = 1 π  2 h 2 h 0.4
3  3  1
r 23
B1 : r = h h 0.6 3
0.4 0.6
3
16 (2, 3) is a minimum point. 1
3
B2 : d2 y2= 16 + 4
dx 23 4

B1 : d2 y = 16 − (−2)2x−2−1
dx
2

17(a) 14
(b)
k=2

B1 : 14 +  kx2  5 = 39
 2  0
 

18 x = 90°, 123.69°, 270°, 303.69°

B2 : cos x = 0 and 3cos x + sin x =0 or tan x = − 3
2

B1 : 3cos2 x + 2sin x cos x =0

19(a) 1
q

(b)

B2 : 2  q   − 1− q2 
 1   
1 

B1 : 2sinθ cosθ

5

Mark Scheme Additional Mathematics Trial Paper 1 2020

20 59 cm2 3
3
B2 : 1 (10)2 (1.5) − 1 (8)2 (0.5)
2
22
2
B1 : 1 (10)2 (1.5) or 1 (8)2 (0.5) 4

22

21(a) 1
16

B1 : π (x)2
π (4x)2

(b) Method and Reason
B1 : Method or reason

Method : Added the radius of small circle
Cara : Menambahkan jejari bulatan kecil

Reason : increase the value of probability
Sebab: untuk meningkatkan nilai kebarangkalian.

22(a) 120 1
(b) 3
( )( )Note: 3C1 5C3 4C1
4
315 3

( )( ) ( )( ) ( )( )B2 : 3C1 5C2 4C2 + 3C1 5C3 4C1 + 3C1 5C4 4C0
( )( ) ( )( ) ( )( )B1 : 3C1 5C2 4C2 or 3C1 5C3 4C1 or 3C1 5C4 4C0

23 64 meter 3
B2 : The maximum point (4, 64)

( )B1 : −4  t − 42 − (−4)2 

6

Mark Scheme Additional Mathematics Trial Paper 1 2020

24 k < − 3 3
2 3

B2 : (−2)2 − 4(2)(1+ k) > 0 2

B1 : 2x2 − 2x +1+ k =0 24

25(a) 3
(b) 8

B1 : 1 + 1 + k + 1 + 1 =1
16 4 4 16

5
16

B1 : 1 + 1
4 16

7