21. MAlLIS PENGETUA SEKOLAH MENENGAH MALAYSIA
CAW ANGAN NEGERI SEMBILAN DARUL KHUSUS
PROGRAM PENINGKATAN AKADEMIK TINGKATAN 5
SEKOLAH-SEKOLAH MENENGAH NEGERI SEMBILAN 2014
ADDITIONAL
!
MATHEMATICS
22. I
2
Solution and marking scheme
INumber
5+y
X=2
{5~y)2 _y2=14_{5~Y)_y
(y-39XY+l)=O
y=39,y=-1 x=22,x=2
p=2
q=3 7 = a(O -2)2 + 3 a=l
2(a)
Y ... ~
(b)
(0, 7)
y=2x-5
32 -(2x-5f =14-2x-(2x-5)
(x-2Xx-2~=O
x=2,x=22
y=3~y=-1
Full Marks Sub Marks
PI
Kl
KI
NI
Nl
PI
PI
KI
NI
PI
titik (0,7)
PI
Titik (2,3) (2,3)
I
....
y
PI
x
0
Bentuk graf
f(x) = -(x -2)2 -3
Nl
8(c)
dy =2x-3
Kl
3
dx
d 2x
-=2
Kl
dy2
I
Kl
(x 2 -3xp+x(2x-3)+5=O
Kl
(4x-5Xx -1)= 0
5
Nl
5
x =-1
4'
5
2
23. 3
Sub Marks Full Marks Solution and marking scheme Number
1 15
Kl
4(a)
2x =--x --cuba selesaikan
2 2
Nl
E =(-3, -6)
(b)(i)
(0 -7~) =C(-3)+3x 1(-6)+3Y )
'24' 4
KI
F= (1, -8)
Nl
y = 2x + c or -8 =2(1) + c
Kl
(ii)
y=2x-lO
NI
2J(x + 3)2 + (y + 6)2 =JCx -1)2 + (y + 8)2
KI
(c)
3x2 + 3/ + 26x + 32y + 115 = 0
8
NI
KI
27(5) + 32(8) +37x J.. 42(15) + 47(8) +52( 4) =39.5
5+8+x+15+8+4
Sea)
.
KI
1605 + 37 x =39 .5(40 + x)
NI
10
PI
Seen 39.5
--231
KI
39.5+ 2 5
r5015
Nl
6
40.17
3
24. 4
Number 6(a)
Seen
sec
x
Solution and marking scheme or 2sin x cos x
Sub Marks Kl
Full Marks
sin 2x
Nl
1.5
0.75
6
(b)
-1.5
-2.25
4x 3 . 2
---=Sln x
3rr 2
2x 9 . h h
Y = ---or me on t e grap
rr 4
Number of solutions = 3
Shape
of Sine
PI
Amplitud PI
Period
PI
KI
PI
NI
8
4
25. 5
Number
Solution and marking scheme
Sub Marks
Full Marks
~D ii) (c)
'QS = QR + liS or Iff = RQ +QT or Iff = RQ + ~QP 5 -3~ -12~ 36 12 --x+-y5 -5 _ -3k= 36 h or 5 5h=13 k= g13 51 unit -12k=gh -12 5
KI NI Nl KI KI KI Nl NI KI Nl
10
8 (a) (b) (c)(i) (ii)
I ~----~----~--~~--~----~----~-----, ' loglO y I 0.6335 I 0.7076 Ii 0.7924 I 0.8573 !0.9494 !1.0294 1 Sekurang-kurangnya dua tempat perpuluhan Rujuk graf loglO y = 0.9 y = 7.943
Nl PI KI Nl
(iii)
6 loglOP = m or -loglO q = Y-intercept or loglOP =0.02583 or loglO q = -0.47 P = 0.9612 1.1612 q =0.2388 -0.4388
Kl Nl Nl
10
5
26. 6
Number Solution and marking scheme Sub Marks Full Marks Kl Kl
9(a)
Nl
0.7881 K1
Kl
Nl
(b)
K1 Kl
Kl
(c)
Nl
2IT
lO
(a)
(b)
(c)
tan B= 6/2 e= 1.2492 rad
1.571 x 6 == 9.2426 or AD == 1.2492 x 8 = 9.9936 9.2426 + 9.9936 + 1.6754
20.9 116
1. (8)(8)(1.2492) 2 ~ 2-(6)(8)(1.571)
2 ,
1. (8)(8)(1.2492) -~(6)( 6)(1.571) -~(6)(2)
2 2 2 5.6964
Kl
Nl
Kl Kl Kl Nl
Kl
K1
Kl Nl
10
10
6
27. I
7
Sub Marks Full Marks Solution and marking scheme Number
11 a)(i)
(ii)
(b )(i)
(ii)
12 (a)(i) (ii)
(b)
(c )(i)
P(X= 3) = :>C3(OA)"0.6)L
Kl 0.2304
Nl P(X= 0) + P(X= 1) + P(X= 2) + P(X= 3)
Kl or 1 -P(X = 4) -P(X = 5) 5Co(0.4)o(0.6)5 + 5C[(0.4)1(0.6)4 + 5C2(0.4i(0.6)3 +
Kl 5C3(0.4)3(0.6)2 or 1-5C4(0.4)4(0.6)[ + 5Cs(OA)0.6)o
0.9130
Nl Kl
P(Z:S 57.4 -48 ) 16
Nl 0.7216
P(Z> 0.524) = 0.3
Kl
m -48 = -0.524
16
Kl m = 39.62
Nl
PI
-6
Kl Nl
(t-6)(t-l)<0
l<t<6 a = -2t + 7
Kl
v = 7 (7.-) _ 6 _ (?)2
max 2 2
Kl
6.25 Nl
Shape of graph PI Point (0, -6) and (1, 0) or (6, 0) PI
s=-(5)3 + 7(5)2 -6(5)= 95 m
3 2 6 or
(6)3 7(6)2
(ii)
Kl S =--+ ~--6(6) = 18m 3 2
95 13
Distance =18--=-m
6 6
Nl
10
10
7
28. 13
8
Number
(a)
(b)
(c )(i)
(ii)
14 a)
(b) (c) i) ii)
Sub Marks Full Marks Solution and marking scheme
w
120 = -xl00 5
w= RM6.00 x+3
150 = --x100
x x =6, y=9
128 = 35.00 x 100
Q2008
Q2008 = RM27.34
125 or 130
120(3) +125m +150(1) + 130(4)
=128
8+m
.
m=2
x+y$8 x ~ 2y or -2 1
y>-x
750x + 250y ~ 2250 or 3x +y ~ 9 Refer to graph 1
titik (2, 3) 2(40) + 3 (10) 110
K1
N1
K1
Nl N1 K1 N1
PI
Kl
Nl
N1 Nl
N1
PI Kl N1 Nl
10
10
8
29. 9
Number I 5 (a)
Solution and marking scheme ~(l4)(9)sin B = 28 2 LB=153.61°
Sub Marks KI N1
Full Marks
(b)(i)
QS2 =142 +92-2(l4)(9)cosI53.61° QS= 22.42
Kl NI
(ii)
sinLP sin35 =-22.42 15
or
LP =59.020
Kl
PQ - sin85.98°
15 sin35°
KI
PQ= 26.09 em
Nl
(c)
1 -(22.42)(26.09)sin35 2
.
Kl
Area quadrilateral = l(22.42)(26.09)sin 35+28 2 = 195.75crn2
Kl N1
10
9
30. 10
Question No. 14
y
all
3 straight lines correct any 2 straight lines correct correct shaded region x+ y 8 P2
PI Nl
9
...
8
7
6
5
3x + Y 9
y Ylx
4
3
R
2
1
x
40x + lOy k
31. 11
loglO y I Question No.8
1.0
0.9
0.8
0.6
0.4
0.3
0.2
0.1
o
all 6 points plotted correctly PI uniform scale for axis PI Iine of best fit PI
x
32. 1
ANALISIS JADUAL SPESIFIKASI UJIAN MATEMATIK TAMBAHAN 2014
PROGRAM PENINGKATAN AKADEMIK TINGKATAN 5
SEKOLAH-SEKOLAH MENENGAH NEGERI SEMBILAN 2014
A1
Paper 2 Paper 1 CONTEXTS Section A Section B S M S M S M Function IF4) 1 1 2 (R) 2 I 3 (S) I 3 3(R) -' -; oJ ' 4 4 (S) 5 3 (R) 2 8 (R) 6 3 (S) 1 5 (R) .~. - 7 3 (S) 8 4 (R) 9 4 (S) 10 1 3 (R) 11 I 3 (R) 13 3(R) 8 10 (S) .' 12 4 (R) 10 10 (S) 14 4 (R) .1 7 10 (T) 15 ! 2 (R) 1 16 3 (T) 6 8 (T) 17 3(R) 4 6(R) 18 3 (R) 3 6 (S) 19 3 (T) - ~,}:;. ;" -" -.. 20 3 (S) 21 2(S) _~-' , .~ " ~ I' • 22 3 (S) 5 7 (S) 23 4 (R) 24 4 (R) I Section C S M
33. 2
CONTEXTS
S4 Probability Distributions (F5) Binomial Distributions Normal Distributions AST1 Solutions Of Triangles lF4) Motion On the Straight Line AST2 (F5) ASS1 II Index Number (F4) ASS2 Linear Programmin_9 (F5) JUMLAH Paper 1 Pa.Q.er 2 Section A Section B Section C 25 J 25 41R) 80 I 6 , 40 111a) I 5 (S) 15 10 R) 12 10 (S) 13 10 (R) 14 10 (R) 5 50 4 40
Nota :
A -Komponen Algebra
G -Komponen Geometri
S -Komponen Statistik
T -Komponen Trigonometri
K -Komponen Kalkulus
ASS -Aplikasi Sains Sosial
AST -Aplikasi Sains Teknologi
R -Aras Rendah S -Aras Sederhana T -Aras Tinggi S -Soalan M -Markah
BILANGAN SOALAN MENGIKUT ARAS KESUKARAN:
KERTAS
ARAS KESUKARAN
BILANGAN SOALAN
NISBAH
1
RENDAH SEDERHANA TINGGI JUMLAH SOALAN
15 8 2 25
6:3:1
2
RENDAH SEDERHANA TINGGI JUMLAH SOALAN
6 5 4 15
4:3:3