skema add math p2 trial spm zon a 2012
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3472/2MatematikTambahanKertas 22 ½ jamSept 2012
SEKOLAH-SEKOLAH MENENGAH ZON A KUCHING
PEPERIKSAAN PERCUBAANSIJIL PELAJARAN MALAYSIA 2012
MATEMATIK TAMBAHAN
Kertas 2
Dua jam tiga puluh minit
JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU
MARKING SCHEME
Skema Pemarkahan ini mengandungi 13 halaman bercetak
2
ADDITIONAL MATHEMATICS MARKING SCHEME
TRIAL SPM exam Zon A Kuching 2012 – PAPER 2
QUESTIONNO.
SOLUTION MARKS
1 x = 1+ 2 y P1
2 2 Eliminate x or y(1+2y) − 5 y + (1+ 2 y) y − 6 = 0 K1
Solve the quadratic2 K1 equation by quadratic− (5) ± (5) − 4(1)( − 5)
y = formula @ completing the
2(1) square must be shown
y = 0.854, −5.854 N1 Note :
or OW− 1 if the working of solving
x = 2.708, −10.708 N1 quadratic equation is not shown.
5
5
2
(a)
(b)
2 2 K1f(x) = b − ax − 2apx − ap
a =2 N1
2 K1−2ap = −6 or b − ap = 9
N1 3 b = 27/2 N1 p = 2
3 27 − , y 2 2
· N1 (−3/2, 27/2)max and(0, 9)
9N1 The shape of the
function.
x
5
2
7
4
QUESTIONNO.
SOLUTION MARKS
3
(a)
(b)
a = 5, d = 8 K1
77 = 5 + (n −1)(8) K1
n = 10 N1
3
a = 5π , d = 8π K1
10S10 = [2(5π ) + (10 −1)(8π )] K1
2
= 410π N1
∴ the total cost = RM90.20 N1
4
7
4
(a)
(b)(i)
sin 2 x
cos 2x K11 −1
cos 2x
2sin x cos x= 2 K1
1− (1− 2sin x)
= cot x N1
3
y
y = 2 cos 3x2
x0 π π
2
−2
5
QUESTIONNO.
SOLUTION MARKS
(ii)
Shape of cosine curve P1
Amplitude P1
Period P1
2 cos3x = −2k K1
k = −1, 1 N18
5(a)
∑ fx = 30.5(3) + 50.5(7) + 70.5(6) + 90.5(2) +110.5(2) = 1270
K11270 K1
x = 20= 63.5 N1
3
(b)2 2 2 2 2
2 (30.5) (3) + (50.5) (7) + (70.5) (6) + (90.5) (2) + (110.5) (2) σ = 3 + 7 + 6 + 2 + 2 K1
2
− (63.50)91265 K1
= − 4032.2520
= 531 N1
3
6
6
(a)OP = OK + KP K1
= a + h(−a + b ) K1
OP = (1− h)a + hb N1
3
QUESTIONNO.
SOLUTION MARKS
(b) 3 h K1=
4h + 8 1− h
(4h −1)(h + 3) = 0 K1
1h = N1
4
PL : KL = 3 : 4 N1
4
7
0.2 0.3 0.4 0.5
= 1.5 ⇒ n = 3 N1
5
Correct both axes (Uniform scale) K1All points are plotted correctly N1Line of best fit N1
log10 y Q7
(a) Each set of values correct (log10 y must be at least2 decimal places) N1, N1
2.2log 10 y = n
2log 10 x + log 10 (p + 1) K1
where Y = log 10 y, X = log 10 x,
2.0 m = n and c = log 10 (p + 1)(c) (i) X = log 10 6.9 = 0.8388 = 0.84
Y = 1.56 = log 10 y ⇒ y = 36.31 N1
1.8
1.6
1.4
1.2
n (ii) = gradient
2n
2 0.8 − 0
(iii) log 10 (p + 1) = Y-interceptlog 10 (p + 1) = 0.30 K1p = 0.9953 N1
×
×
×
×
.(0.80, 1.50)
×
1.0
0.8×
0.6
0.4
(0, 0.30)
0.2
log10 x
0 0.1 0.6 0.7 0.8 0.9 1.0
log10 x 0.30 0.60 0.70 0.78 0.90 1.00 N1
log10 y 0.75 1.19 1.36 1.46 1.65 61.80 N1
7
8
9
QUESTIONNO.
SOLUTION MARKS
8
(a)
(b)
(c)
3 or 3 1 3 or mnormal = K1
1 1 +c K1
+ N1
3
B(4, 0) K1
4 1 1 K1
=2 N1
=1 unit
3
volume of revolution K1 K1
= 1 4 1= K1
= 1 unit 3N1
4
10
QUESTIONNO.
SOLUTION MARKS
9(a) r=5 K1
PR = 8.660 K1
23.66 cm N1
3
(b) π K1s = 5 π − 3
=10.47 N1
2
(c) Area of shaded region of semi circle1 12 2
= (10) (π ) − (5) (π ) K12 2
2
=117.825 / 117.8096 cm
Area of shaded region PQR
1 2 π 1 = (10) − (5)(10) sin 60° K1
2 3 2 K12
=30.716 / 30.7093 cm
Total area of shaded region
=117.825+30.716 K12
=148.541 / 148.5192 cm N1
5
10
QUESTIONNO.
SOLUTION MARKS
10
(a)(i)
(ii)
2( − 4) + 3(0) 2(0) + 3(8) K1or
3 + 2 3 + 2 − 8 24
C = , N1 5 5
8 24 − + x + y
5 5 K1= −2 or or (−2, 0)2 2
12 24 N1A − , − 5 5
4
(b) 1 24 12 24 8 12 24 8 24 = 0(− ) + (− )(0) + (−4)( ) + (− )(0) − 0(− ) − (− )(−4) − (0)(− ) − ( )(0)
2 5 5 5 5 5 5 5 5K1
N1 = 19.2
2
(c) mAB = −3 K1
y = −3x − 12 N1
(d) 2 2( x − (−4)) + ( y − 0) K1
2 2
x + y + 8x = 0 N1
10
QUESTIONNO.
SOLUTION MARKS
11(a)
(i)
(ii)
(b)
(i)
(ii)
(iii)
8 2 3 3 5 K1C3( ) ( )
5 5= 0.2787 N1
8 2 6 3 2 8 2 7 3 2 8 K1 K11 − C6( ) ( ) − C7( ) ( ) − ( )
5 5 5 5 5
0.9052 N1
5
0.75 P1
h − 163 = −0.6 K116
h = 153.4 N1
P(−1 ≤ z ≤ 0.75) or 1 − P(Z > 1) − P(Z > 0.75) or R(−1) − R(0.75)
K1
0.6147 N1
5
10
QUESTIONNO.
SOLUTION MARKS
12(a) −6t + 16 = 0 K1
2 N1t =2 3
2 K1s = −3t + 16t
2 2 2 K1s = − 3( ) + 16( )
3 3
1 N1= 21 m
3
5
(b) t(−3t + 16) = 0 K1
1 N1t =5 3
2
(c) | s83 − s0 | + | s4 − s83 | K1
2 2 K1= | 21 3 − 0 | + |16 − 21 3 |
= 26 23 N1
3
10
11
12
13
Answer for question 15(a) I. x + y ≤ 18 N1
II. 4x + 3 y ≥ 24 N1y
QUESTIONNO.
SOLUTION MARKS
13
(a)
(b)
x = 135 N1
y = 7.20 N1
2
K1
414.75n + 829.5 = 420n + 819 K1
n =2 N1
3
(c) RM 8.80 K1.RM 6.37N1
2
(d) I 11 = 155 K108
155K1 I11 = ×100
10 138.25
112.12 N1
3
10
QUESTIONNO.
SOLUTION MARKS
14
(a)
(b)
(c)
9.7 9= K1
sin ∠ACB sin 60°
∠ACB = 68.97° / 68°58' N1
2
∠ECD =180° − 68.97° =111.03° /111°2′ K1
2 2 2ED = 4.5 + 9 − 2(4.5)(9)cos 111.03° K1
ED = 11.42 cm N1
3
∠BAC = 180° − 60° − 68.97° = 51.03° / 51°2′ P1
1 Area of triangle ABC = ×9.7×9×sin 51.03° K1
2= 33.94
1 K1Area of triangle CDE = ×9×4.5×sin 111.03°
2= 18.90
Area of quadrilateral ABCD =33.94 + 18.90 K1
52.84 N1
5
10
2 4 6 8 10 12 14
III. y ≤ 2xN1
(b) Refer to the graph,
20
18
1 or 2 graph(s) correct3 graphs correct
Correct area
N1(c) (i) x = 8
N1
N1
K1
16 (ii) (6, 12) P1
Maximum Profit = RM [1.50(6) + 2(12)] K1
14
12 (6, 12) ·
= RM 33.00 N1
10
10
8
6
R4
2
x0 16 18
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