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3472/2
Additional
Mathematics
Sept 2013
PROGRAM PENINGKATAN PRESTASI AKADEMIK SPM 2013
ADDITIONAL MATHEMATICS
Paper 2
(SET A)
.
MARKING SCHEME
SULIT
3472/2
2
MARKING SCHEME
ADDITIONAL MATHEMATICS PAPER 2 2013
N0.
1
SOLUTION
or
2y  x  5
x  2y  5
(2 y  5)2  2 y  7
x2  x  5  7
2 y 2  11y  9  0
(2 y  9)( y  1)  0
x2  x  12  0
( x  4)( x  3)  0
x  4
9
y
2
and
x3
(both)
and
y  1
(both)
MARKS
P1
K1 Eliminate y
K1 Solve quadratic equation
N1
N1
5
2
(a)
3x 2  7 x  6  0
(3x  2)( x  3)  0
K1
2
or x  3
3
2
h  , k  3
3
N1
3x 2  7 x  6  0
K1
x
(b)
x  3
and
N1
x
2
3
N1
5
3
(a)
4
P1 sin shape correct.
2
P1 Amplitude = 4
P1 2 full cycle in 0  x  2
P1 [ Maximum = 4 and
Minimum = 4 ]
(b)
-4
y=
x

x

Number of solutions = 4
draw the straight line y =
N1 For equation
K1 Sketch the straight line
N1
7
3
4
(a)
(i)
BC  BA  AC
BC  4 x  y
K1
N1
(ii)
PC  PA  AC
PC   x  y
N1
(iii)
AQ  AP  PQ
2
1
AQ  x  y
3
3
(b)
AQ  hQR
h 1
A, Q, R are collinear.
K1
N1
K1 find h
N1
N1
8
5
(a)
(b)
P1 for L=24.5 or F=23 or fm=15
1

 2 (48)  23 
median = 25  5  
 (5)
15




= 25  8333
K1 use correct formula
N1
new mean=2(8)+5= 21
K1 N1
new standard devition=2(4)=8
K1 N1
7
4
6
(a)
1
1
1
 p 2 ,  p 2 ,  p 2 ,...
4
16
64
1
1
 p2
 p2
16
 64
,
1
1
2
2
p
p
4
16
r
1
4
K1
K1
N1
(b)
(i)
1
225
900 ( ) n 1 

4
256
n6
K1K1
N1
(ii)
900
S 
1
1
4
1200
K1
N1
8
5
7
(a)
x
1
2
3
4
5
6
N1 6 correct
values of log y
log10 y
- 0.15
0
0.14
0.30
(b)
0.46
0.60
K1
log10 y
Plot log10 y
vs x.
Correct axes &
uniform scale
O
-0.3
(c)
(i)
(iii)
N1 6 points
plotted
correctly
N1 Line of best-fit
log y  kx  log10 h
P1
 log10 h = *y-intercept
K1
N1
h = 2.00
(ii)
x
k = *gradient
= 0.15
y = 1.70
K1
N1
N1
10
6
N0.
8
(a)
(b)
SOLUTION
MARKS
2.095 rad
N1
AB  5.07 cm
K1
S BC  12(30 

180
)
K1 Use s  r
S AC  4(2.095)
or
= 6.28
= 8.38
N1
Perimeter = 5.07 + 6.28 + 8.38
= 19.73
(c)
Area of OBC =
K1
N1
1

1
(12)2 (30 
) or Area of OBC = (4)2 (2.095)
2
180
2
= 37.70 cm2
Area of the shaded region = 37.70 – 16.76 – 13.86
= 7.08 cm2
= 16.76 cm2
K1 Use formula
A
1 2
r 
2
N1
K1
N1
10
7
N0.
9
(a)
SOLUTION
1
2
Area of POR =
0 11  1 0
K1
0 7 2 0
= ½ │-15│
(b)
MARKS
=
N1
7.5
Let Q (x,y)
x=
111  2 1
3
y
,
17   2 2
3
K1 for either x or y
Q ( 3, 1 )
(c)
M PR =
N1
9
3
7   2
=
=
11   1 12 4
y-1 =-
,
m  M PR = -
4
( x -3)
3
3y = -4x + 15
4
3
K1 use gradient
correctly
K1 use forming
quadratic equation
N1
(d)
Let T as (x,y)
TP = 2 TR
x  12   y  22  4( x  112   y  72 )
3x2 +3y2 – 90x – 60y + 675 = 0
x 2  y 2  30 x  20 y  225  0
P1
K1
(use distance
formula)
N1
10
8
N0.
10.
(a) y =3x + 2
SOLUTION
MARKS
, y = x2 + 2
3x=x2
x( x – 3) = 0
x=0,3
When x = 3, y =3(3) + 2 = 11
K1
K1 for solving
quad.equation
N1
P ( 3,11 )
(b)
3
A   (3x  2  ( x 2  2))dx
K1 use
 (y
2
 y1 ) dx
0
 3x 2 x 3
 

3
 2

27 27

2
3
=
=
3
K1 integrate correctly
K1 Subtitute the limit
correctly
0
4
1
2
N1
Note : If use area of right angle triangle and
 x dy , give marks
accordingly.
(c)
11

1
3
V = π x 2 dy  r 2 h
2
11
K1 correct limit or
use volume of cone
=  ( y  2)dy   3 9
1
3

2
2
K1 integrate
correctly
11
 y2

 2 y   27
= 
2
2
=  40
1
  27
2
1
2
= 13 
N1
10
9
N0.
SOLUTION
11
(a) X= Students have their breakfast
(i)
p = 0.6
, q = 1- 0.6 = 0.4
,n = 10
MARKS
K1 Use
n
P(X =3) =
10
P ( X=r ) =
Cr p r q nr
c3 0.6 3 0.4 7
=0.0425
N1
(ii)
P (X≥2) = 1 – P(X=0) – P(X = 1)
Or
K1
= P( X  2)  P( X  3)  .........  P( X  10)
10
=1-
c0 0.6 0 0.410 -
10
c1 0.610.4 9
K1 Use
n
P ( X=r ) =
Cr p r q nr
=0.9983
N1
(b)
X= masses of a group of boys, X  N ( 45,5)
(i)
µ= 45
,
σ =5
P(X < 40 ) = P ( Z <
= P(Z < -1)
(ii)
40  45
5
)
K1 Use Z =
X 

= P ( Z > 1)
= 0.1587
P ( X > m) = 0.3
m  45
) = 0.3
5
m  45
From table ,
 0.524
5
P(Z >
N1
K1
use
σ
x-µ
K1 equate with z
score
m- 45 = 2.62
m = 47.62 kg
N1
10
10
N0.
SOLUTION
12
(a)
ainitial = 4 ms-2
(b)
v=  (4  2t)dt
MARKS
N1
K1 for
integrating v
2
= 4t-t +c
t = 0, v = 12, c = 12
v = 4t - t2+12
a = 0, t = 2
Vmax = 4(2)- (2)2 + 12
= 16 m s-1
(c)
v=0
K1
N1
K1
, (t+ 2)(-t+6) = 0
N1
t=6=p
(d)
Total distance
K1 for
6
6 2 2
 8 2 2

(
4
t

t

12
)
dt

   (4t  t  12)dt 
0
 6

=

 2 t3
  2 t3
= 2t   12t    2t   12t 
3
3

 0 
6

6

= 2(6) 2 

= 90
2
3
8



(6) 3
(8) 3
(6) 3
 12(6)  0   2(8) 2 
 12(8)  2(6) 2 
 12(6)
3
3
3



8

or 
0
6
K1 (for
Integration;
either one)
K1
(for use and
summation)
N1
10
11
N0.
SOLUTION
MARKS
13 (a)
1
(8)(7) sin QRS
2
SinQRS = 0.75
QRS = 48.59o
18 
(b)
(c)
QS2
QS
K1
N1
= 82 +62 – 2(8)(6) cos 48.59o
= 6.042 cm
K1
N1
K1
6.5
6.042

o
sin QPS
sin 49
Sin QPS= 0.7015
QPS=44.55O
PQS=180- 49o- 44.55o
= 86.45o
(d)
Area of PQRS = Area of triangle QRS +
K1
N1
1
(6.042)(6.5) sin 86.45 o
2
= 18 + 19.598
K1
K1
= 37.598
N1
10
12
N0.
14
(a)
SOLUTION
MARKS
x + y ≤ 60
60x +120y ≥ 3600
or
x + 2y ≥ 60
N1
≥ 2x
y
N1
N1
y
60
(b)
R
30
60
0

At least one straight line is drawn correctly from inequalities involving
x
(c)(i)
(ii)
x
K1
and y.

All the three straight lines are drawn correctly.
N1

Region is correctly shaded.
N1
N1
12
Minimum point (12, 24)
N1
30x + 90y = k
Minimum profit = 30(12) + 90(24)
= RM 2520
K1
N1
10
13
N0.
SOLUTION
MARKS
15 (a)
p
9
x100
6
atau
140 
3.5
x100
q
K1
N1
p = 150
N1
q = RM 2.50
(b)
720(can be seen)

I
N1
130 x84  150 x96  140 x72  135 x108
84  96  72  108
K1

I 1110  138.83
(c) (i)

I 1210  138.83x
N1
115
100
K1
N1
= 159.65
(ii)
Cost of making bread in the year 2011
p11
x100
50
K1
p11  RM 69.42
N1
138.83 =
10
END OF MARKING SCHEME
14