The document provides information on the marking scheme for a trial SPM 2017 mathematics additional paper 2 examination in Selangor, Malaysia. It lists several multi-step math problems and their solutions, showing the working and marks allocated for each step. It also includes diagrams with geometry problems involving lines, angles and shapes. The last few questions cover calculus topics like integration and maximum/minimum values. Overall, the document outlines the detailed marking rubric for a high school level math exam, exemplifying the types of problems asked and breakdown of marks for partial and full solutions.

1. SET A
1
PERATURAN PEMARKAHAN PEPERIKSAAN PERCUBAAN SPM 2017
PROGRAM INTERVENSI TERBILANG AKADEMIK SELANGOR
(PINTAS)
MATEMATIK TAMBAHAN KERTAS 2
NO SOLUTIONS MARKS
1 xy – 8 = 2 (2x - y) = 3x + 1
x
x
yor,yx
93
21


    113821  yyy





 

x
x
xxyxor
93
211324
018701252 22
 xxoryy
      0920432  xxoryy
X = -2 , 9
4
2
3
,y 
P1
P1
K1
K1
N1
N1 6
2 (a) r = 2
254
12
122
255
12
121






 






  n
or
n
n = 8 or n= 7
Number of rows = 15
K1
N1
N1
3
6
(b) a =1, n =8 or a = 2, n = 7
( 27
) or 2 (27 - 1
)
(Note: or correct listing P1 K1)
128
P1
K1
N1 3

2. SET A
2
NO SOLUTIONS MARKS
3
(a) 14- Markah Ben = 7 or 10
7
ShimaMarkah+62

Markah Ben = 7 dan Markah Shima = 8
K1
N1 2
6
(b) 210
7
21721421228272725








  **
4.071
K1
N1 2
(c) min = 20, varians = 66.29 N1,N1 2
4
  



  1221 Akos
Akos
Asin
a
2 sin A kos A = sin 2A
K1
N1 2
8
(b) P1
(sin
graph)
P1
(2 cycle)
P1
reflect
P1
translasi 4
2
3
212(c)


k
k K1
N1
2
5 (a) A (30,40) dan C(60,20) P1 1
7
(b) D (90, 60)
22
6090 
108.17
P1
N1
K1 3
(c) DA : DC = 2 : 1







3
80
50
3
402(20)
3
302(60)
,
or
P1
K1
N1 3
x
y
1
O
2
2

3. SET A
3
NO SOLUTIONS MARKS
6
(a)
3
1
normalm or xx
dx
dy
623 
Solve 3623  xx
x = 1
P ( 1, -2)
P1
K1
N1 3
7
7
33
1
=-2*(b)


k
k K1
N1 2
(c) y + 2 = - 3( x – 1)
y= - 3x + 1
K1
N1 2
7
(a) methodvakidotherorsin
5
3
2


36.87o
or 0.6436 rad
1.287
(Note: methodvakidotheror K1N1N1)
K1
N1
N1
3
10
(b) 5 (*1.287)
6.435 cm
K1
N1
2
(c) 1.855
   
 
21
855125
2
1
2
855125
2
1
1
AA*
.*sinAtriangleofArea
.*AsektorofArea



11.19 cm2
P1
K1
K1
K1
N1 5
8 (a) x2
+ 4 = 4x
k = 2
K1
N1 2
(b) A1 =  
2
0
3
2
0
2
4
3
4








 x
x
dxx K1
K1

4. SET A
4
NO SOLUTIONS MARKS
A2 =   
2
0
2
2
0 2
4
482
2
1









x
dxxor
𝐴1 − 𝐴2 = ∫(𝑥2
2
0
+ 4 ) 𝑑𝑥 −
1
2
(2)(8)
Area of the shaded region = 2 [∫ (𝑥22
0
+ 4 ) 𝑑𝑥 −
1
2
(2)(8)]
5.333 unit3
K1
K1
N1 5
10
(c) Volume = 𝜋 ∫ (𝑦 − 4)𝑑𝑦
5
4
= [
𝑦2
2
− 4𝑦]
5
4












 16
2
16
20
2
25
0.5 𝜋 // 1.5708 // 1.571
K1
K1
N1 3
9 (a) (i) n = 10, p = 0.7 , q = .03
       
   1091
8108


xPxPor
xP......xPxPxP
    rr
r ..CUse 1010
7030
0.85065 // 0.8507
(ii) 6.72 = n(0.7)(0.3)
n = 32
K1
K1
N1
K1
N1 5
10
(b) X ~ N ( 56 , 32 )
   
 
students.
.
.ZP
ZPXPi
15450030850
30850
50
32
5640
40








 

ii) P(X > m) = 0.12
marksm
m
m
ZP
6.93
175.1
32
56
12.0
32
56








 

K1
N1
N1
K1
N1
5

5. SET A
5
NO SOLUTIONS MARKS
10 (a)
lg(x+1) 0.3 0.48 0.60 0.70 0.78 0.85
lg y 0.85 0.76 0.70 0.65 0.61 0.58
N1
N1 2
10
(b) Refer graph paper
Plot log10 y against log10 (x+1)
(at least one point)
6 points plotted correctly
Line of best fit N1
K1
N1
N1
5
(c)   blogxlogaylog 101010 1  P1
K1
N1
K1
N1
5
11 (a)(i)
1
3
OP OA AB 
= 10 2x y
(ii)
3
5
AQ AB BC 
= 6 6y x
K1
N1
N1
3
10
1(ii)
2
1
2
1
(i)
10




b
blog
a
a
i)

6. SET A
6
NO SOLUTIONS MARKS
b) (6 6 )AR h y x 
= 6 6hx hy 
AR AO OR AO kOP   
= 10 (10 2 )x k x y  
(10 10) 2k x k y  
6 h = 2 k
– 6 h = 10 k – 10
h =
5
18
, k =
5
6
K1
K1
K1
N1, N1
10
(c) OS OA AS 
10 10 (6 6 )y x m y x  
5
3
m 
K1
N1
12
(a) 125100
168

x
or 100
90
99
y
210RMx 
y = 110
K1
N1
N1
3
10
(b) m + n = 35
  115
100
4011010512025125

 nm
8m + 7n = 265
8(35 – n) + 7n = 265
m = 20
n = 15
K1
K1
K1
N1
N1
5
(c) 115
100
120
I
= 138
K1
N1
2

7. SET A
7
NO SOLUTIONS MARKS
13 (a) 200 500 x
700 yx
2002  yx
N1
N1
N1
3
(b) Graf (satu garis betul)
(semua garis betul)
(rantau betul)
K1
N1
N1
3
10
( c )(i) Maximum number of arts students= 450
(ii) (500, 200)
k = 800x + 600y
= 800(500)+600(200)
= RM 520 000
N1
N1
K1
N1
4
14 (a) (2t + 3)(t – 2) = 0
t = 2
a = 4 t – 1
= 4(2) – 1
= 7 m s – 2
K1
K1
N1 3
(b) a = 4t – 1 = 0
111
2
//
8
49
//
8
1
6
6)
4
1
()
4
1
(2
4
1




ms6.125msmsv
v
st
K1
K1
N1 3
(c)   dttts )62( 2
t
tt
6
23
2 23

S2 =
3
2
8122
3
16

S3 =
2
1
418
2
9
18 
K1
K1
K1 4

8. SET A
8
NO SOLUTIONS MARKS
d = m2.83mm 1//
6
77
//
6
5
12)]
2
1
4(
3
2
8[
3
2
8 
OR
N1
15 (a) AE2
= 202
+ AE2
- 2(AE)(20)cos 300
AE = 11.55 cm
K1
N1 2
10
(b) (i) ½ (20)(11.55) sin 300
= 2 ( ½) ( CE)(10) sin 600
6.668 cm
(ii) CD2
= 6.668*2
+ 102
- 2(6.668*)(10) cos 60o
8.819 cm
K1, K1
N1
K1
N1
5
0
0
9040
8198
60
6.668*
CDEsin
(c)(i)
.
.*
sin


(i)
K1
N1
N1
3
C
E
F
D
dtvvdt  
3
2
2
0
K1
Correct integration K1
Correct limit K1
Correct answer N1

9. SET A
9
Question 10
0.1 0.2 0.3 0.4 0.5 0.6 0.7
lg(x+1)
0.1
0.2
0.3
0.4
0.5
0.6
1.0
0.9
0.7
0.8
log10 y
x
x
x
x
x
x
Graph log10 y against log10(x+1)
0.8

10. SET A
10
n
100
1
200
2
300
3
400
4
y
500
4
600
4
700 8000
100
1
200
1
300
1
400
1
500
1
600
1
700
1
800
1
R
500,200
x
Question 15

11. SET A
11