jpwpkl trial 2010 p2 mark scheme
DESCRIPTION
KL SPM 2010 Trial AddMaths P2 Marking SchemeTRANSCRIPT
SULIT 3472/2
3472/2 @ 2010 Hak Cipta JPWP SULIT
JABATAN PELAJARAN WILAYAH PERSEKUTUAN KUALA LUMPUR
PEPERIKSAAN PERCUBAAN
SIJIL PELAJARAN MALAYSIA 2010
SKEMA PEMARKAHAN
ADDITIONAL MATHEMATICS
PAPER 2
SULIT 3472/2
[Lihat sebelah
3472/2 @ 2010 Hak Cipta JPWPKL SULIT
1
SECTION A
Question Solution Sub
Mark
Full
Mark
1.
x = 4y – 7
( 4y – 7) 2 – y (4y – 7 ) + y 2 = 9
0404913 2 yy
132
40134)49(49 2 y
196.1,574.2y
x = 3.296 , – 2.216
1
1
1
1
1
[5]
2. (a) 2 2 22 4 30f x x m m m
= 2 22 3 30x m m
(b) (i) 2m = 6
m = 3
(ii)
1
1
1
1
1
1
1
2
5
[7]
3. (a) mtangent = 2
2
6dy h
xdx x
26( 1) 2
1
h
h = 4
24
3( 1)1
k
k = −7
(b) y – (−7) = 2[x – (−1)]
y = 2x − 5
1
1
1
1
1
1
1
5
2
[7]
4
7
xy
0951413
94
7
4
7
2
2
2
xx
xxxx
132
95134)14()14( 2 x
x = 3.295 , −2.218
196.1,574.2y
x
(0, 39) ●
● (8, 7)
y
● (6, 3)
U shape
min point (6, 3)
points: (0, 39) and
(8, 7)
OR
SULIT 2
3472/2 @ 2010 Hak cipta JPWPKL
SULIT
4. (a) List of areas :
2
1xh,
8
1xh,
32
1xh, …
4
1
2
3
1
2 T
T
T
T
This is Geometric Progression and r = 4
1
(b) List of areas of triangles: 20, 5, 4
5, …
a = 20
41
10
41
10
41
4
41
41
])(1[20
1
])(1[20
SorS OR
64
5
4
120
4
5
T
410 SS41
10
41
1
])(1[20
–
41
4
41
1
])(1[20
41
6
41
645
1
1
105
TtoTS
= 26.6667 – 26.5625
= 0.1042
1
1
1
1
1
1
3
3
[6]
5.
(a)
1
1
1
1
1
1
1
4
3
[7]
6. (a) (i) DEADAE
ab 48
(ii) AFBABF
ba 510
(b) (i) AEhAG
= )48( abh or
= ahbh 48
(ii) BABFkorBFkABAG
= bkak 5)1010(
1
1
1
1
1
3
(b) 2(1 )x
y
2
Graph of the straight line
No of solutions = 3
0 x
Sine curve
1 cycle in 0 x
Amplitude = 1
shifted upward 1 unit
or y-intercept = 1
2
1
y
OR
SULIT 3472/2
[Lihat sebelah
3472/2 @ 2010 Hak Cipta JPWPKL SULIT
3
8h = 5k or 4h = −10k +10
kh4
5 1010
4
58
kk
2
1h
5
4k
1
1,1
5
[8]
SECTION B
Question Solution Sub
Mark
Full
Mark
7.
Refer to the attachment
8. (a)
41
)0(4)5)(1(,
41
)10(4)0(1R
41
)0(4)5(1
41
)10(4)0(1
yorx
R( 8, –1)
(b) m PQ = 2
1 or gradient of the perpendicular line = −2
y – (–1) = −2 ( x − 8)
y = −2x + 15
(c) Area of OPR= 0150
0800
2
1
= 20 unit2
(d) PR = 22 510
5
4
OR PR = 22 )]5(1[)08(
= 1255
4 or 8.944 cm or 80
2
1
× perpendicular distance × 125
5
4 = 20
perpendicular distance = 5
10 or 4.472 or
20
1
1
1
1
1
1
1
1
1
1
2
3
2
3
[10]
or
SULIT 4
3472/2 @ 2010 Hak cipta JPWPKL
SULIT
9. (a) x2 – 4 = x + 2
x2 – x – 6 = 0
(x – 3)(x + 2) = 0
x = 3 , x = −2
B(3, 5)
(b)
)5)(5(2
1)(1 ABCA
3
2
2
2 4 dxxA or
2
2
2
3 4 dxxA
3
2
3
43
xx
or
2
2
3
43
xx
)2(4
3
2)3(4
3
3 33
or
)2(4
3
2)2(4
3
2 33
Total Area = A1 – A2 + A3
= 3
32
3
7
2
25
= 6
125 or 6
520 unit
2
(c) V = 3
22
2
4x dx
3
4 2
2
8 16 x x dx
35 3
2
816
5 3
x xx
5 3 5 33 8(3) 2 8(2)
16(3) 16(2)5 3 5 3
7.533 or 15
113 or
15
87 unit
3
1
1
1
1
1
1
1
1
1
1
2
5
3
y
Diagram 9
Rajah 9
Q P
y = x2 – 4
A
B(3, 5)
y = x + 2
x
●
● 2
A3
A2
-2 C
SULIT 3472/2
[Lihat sebelah
3472/2 @ 2010 Hak Cipta JPWPKL SULIT
5
10. (a)
3
2
(b) length of chord PQ = 2 x 3 sin 3
= 5.196 cm
(c) the perimeter of the shaded region = SPRQ + SPQ
= 2
196.5 ( ) + 3(
3
2)
= 598.4 or 14.45 cm
(d) 2
12
2
196.5
( ) or
2
1(3) 2 (
3
2)
2
1(3) 2
sin3
2 or
3cos3
3sin)3)(2(
2
1
Area of the shaded region
= Area semicircle PRQ – [Area sector POQ – Area POQ]
=2
1(
2
196.5) 2 ( ) – [
2
1(3) 2 (
3
2) −
2
1(3) 2
sin3
2]
= 10.6023 – 5.5277
= 5.075 cm 2
1
1
1
1,1
1
1
1
1
1
1
2
3
4
[10]
11. (a)(i) P(x = 2) = 10 2 8
2(0.2) (0.8)C
= 0.3020
(ii) 2 1 0 1P x P x P x
91
1
10100
0
10 )8.0()2.0()8.0()2.0(1 CC
= 0.6242
(b) P ( x ≤ 20 ) = P ( z ≤10
3020)
= P ( z ≤ – 1 )
= 0.1587
P ( x ≥ m) = 2
1587.01
P ( z ≥ 10
30m) = 0.4207
10
30m = 0.2
m = 32
1
1
1
1
1
1
1
1
1
1
5
5
[10]
SULIT 6
3472/2 @ 2010 Hak cipta JPWPKL
SULIT
SECTION C
Question Solution Sub
Mark
Full
Mark
12. (a) v = 5t (t − 4) = 5t2 – 20t
a = 10t − 20 = 0
t = 2
v = 5(2)2 – 20(2)
= − 20 ms-1
(b) s = 25 20 t t dt
= 3
2510
3
tt
32
3 2
510 0
3
5 30 0
tt
t t
5t2(t – 6) = 0
t = 6
(c) 5t2 – 20t = 0
5t (t – 4) = 0
t = 4
3 32 2
4 5
5(4) 5(5)10(4) 10(5)
3 3t ts or s
= 160
3 =
125
3
Total distance travelled = 160 160 125
3 3 3
= 65
1
1
1
1
1
1
1
1
1
1
3
3
4
[10]
OR
5t2 – 20t = 0
5t (t – 4) = 0
t = 4
Total distance travelled
=
45
2 2
40
4 5
3 2 3 2
0 4
45
2 2
40
5 20 (5 20 )
5 20 5 20
3 2 3 2
5(64) 10(16) 0
3
5 5(64)(125) 10(25) 10(16)
3 3
5 20 (5 20 )
65
OR
OR
+
t t dt t t dt
t t t t
or
t t dt t t dt
1
1
1
1
SULIT 3472/2
[Lihat sebelah
3472/2 @ 2010 Hak Cipta JPWPKL SULIT
7
13. (a)
1cos 60
2 oPSR or PSR
2 2 26.5 5.2 2 6.5 5.2 cosPR PSR
2 2 2 16.5 5.2 2 6.5 5.2
2PR
PR = 5.957 cm
(b) '329/04.29
50sin
4.9
sin
oo
o
PQR
PQR
PR
0 0 0 0180 50 29.04 100.96 PRQ
096.100sin957.54.92
1)( PQRAc
060sin2.55.6
2
1PRSA
PRSPQRPQRS AAA
= 42.12 cm 2
1
1
1
1
1
1
1
1
1
1
3
3
4
[10]
14. Refer to the attachment.
15.
(a) (i) y = 110
100100
= 110
04
04
148(ii) 125 100
118.40
P
P RM
(b) 15
)2(110)4(105)3(125)6(130 I
I = 119.67
(c) 67.119100684
07 P
54.81807 RMP
(d) 67.119100
12004/09 I
= 143.6
1
1
1
1,1
1
1
1
1
1
3
3
2
2
[10]
SULIT 8
3472/2 @ 2010 Hak cipta JPWPKL
SULIT
Question 7
1.2
1.0
0.8
0.4
0.2
1.4
0.5 0.4 0.3 0.2 0.1 0.6 0.7
0.6
1.6
1.8
Correct axes and uniform scale .........K1
Plot all the points correctly …………N1
Line of best fit ………N1
5
x10log 0.18 0.30 0.48 0.60 0.70 0.78
y10log 1.60 1.40 1.12 0.88 0.80 0.60
10
0
x
x
x
x
x
x
c = 1.88 ............... K1
6.13.08.0
4.16.0
m ............... K1
xpky 101010 logloglog ............. P1
6.1
p
pm
86.75
88.1log10
k
kc
............................. 1
0.8
y10log
x10log
……......… N1
……… N1
N1
N1
5
1.9
SULIT 3472/2
[Lihat sebelah
3472/2 @ 2010 Hak Cipta JPWPKL SULIT
9
Question 14
x 0 100 400 300 200 700 600 500 800
y
400
100
350
300
250
200
150
450
500
50
R
5x + 7y = 3500
x + 8y = 800
x = 4y
500
(a) 5x + 7y 3500 or equivalent
x + 8y ≥ 800 or equivalent
x 4y or equivalent
(b) Draw correctly any straight line
All three straight lines are correct
Shaded region R
(c) (i) x = 518 [accept integer 510 – 520]
(ii) Use 3x + 2y where (x, y)
for point in the shaded region
Maximum point = (518,130) or
130129,520510 yx ,
x and y = integers
Maximum profit = RM1780 to RM1820
N1
N1
N1
3
K1
K1
N1
3 N1
K1
N1
N1
4
10
Question 14
x