marking scheme maths t 1 - johor 2010
Post on 07-Apr-2015
472 Views
Preview:
TRANSCRIPT
1
Mathematics T Paper 1 Marking Scheme JOHOR 2010
1
[3]
2
13
133
22
k
22
2
k
42 k
42 ,42 kk
6 ,2 kk
1
1(either)
1(both)
2
[4]
xey x sin
xexedx
dy xx sincos yxe x cos
dx
dyxexe
dx
yd xx cossin2
2
dx
dyy
dx
dyy
i.e. .0222
2
ydx
dy
dx
yd
1 (one correct)
1(all correct)
1
1
3
[7]
(a) dx
dyy2 or
dx
dyxy 44
04422 dx
dyxy
dx
dyyx
xyxydx
dy2442
dx
dy = xy
xy
42
24
= xy
xy
2
2
(b) 02 xy
xy 2
03)2(4222 xxxx
1x
2 ,1 ,21 ,
1
1
1
1
1
1
1
2
4
[7]
2
1
2
1
111
1
xx
x
x
...2
!2
2
3
2
1
2
11...
2
!2
2
1
2
1
2
11 xxxx
...
8
3
2
11...
8
1
2
11 22 xxxx
...8
1
2
1
2
1
2
1
8
3
2
11 22
xxxxxx
2
2
11 xx
Substitute24
1x , 23
5
1
25
24
24
23
24
11
24
11
...24
1
2
1
24
1123
5
12
1152
110523
5
1
1152
552523
24
1gives a better approximation because
24
1<
16
7.
1 Use of Bin.
Theorem
1,1
1
1
1
1
3
5
[7]
3212
3212
3212
1
3212
1
rr
rr
rrrr
2
3212
rr
12322
1 rr
3212
1...
53
1
31
1
nn
n
r rr0 3212
1
n
r
rr0
12322
1
13 1
2
5 3
: :
2 3 2 1n n
1322
1
3212
1
0
nrr
n
r
1
1 (for -2)
1
1
1
1
1
6
[8]
021231211223
xxx
Let xu 2 0213111 23 uuu
By trial and error, 1u is a root of the equation
u2 –10u + 21
u–1 )u3–11u
2 +31u –21
u3 – u
2_____
–10u2 +31u
–10u2 +10u
21u –21
21u –21
0731 uuu
u = 1, u = 3, u = 7
2x = 1, 2
x = 3, 2
x = 7
2lg
7lg,
2lg
3lg,0 xxx ,
= 1.585, = 2.807
x = 1.585, 2.807
1
1
1( right
quotient )
1 (long
division)
1
1(using lg/ln)
1
1 (without 0)
2 1 2 1n n
4
7
[8]
2
1
xu
u
xdx
du
2
1
2
12
1
2dx
udu
x = 1, u = 1
x = 4, u = 2
I =
2
2
1
12
4u du
u u
=
2
14
2du
uu
u
B
u
A
uu
4)4(
2
BuuA 42
2
1 ,0 Au
2
1 ,4 Bu
)4u(u
2
=
u2
1+
u42
1
I =
2
14
11
2
1du
uu
= 214lnln2
1uu
= 3ln1ln2ln2ln2
1
= 3ln2
1
1
1
1
1
1 (both A &
B)
1
1
1
5
8
[8]
Domain of f 4: xx
42 xy
422 yx
xxx 4:f 21
Domain of 2:f 1 xx
Range of 4:f 1 yy
1
1
1
1
1
1 for curve f
1 for inverse f
1(all correct)
24 yx
y
x
f
y = x
4
4
0
(2,-4)
(-4,2)
f -1
6
9
[8]
22 y)4x( or 22)4( yx
22 y)4x( – 22 y)4x( = 2 10
2
222
2241024
yxyx
22
41042
1682
402
1682
yxyxxyxx
2241044016 yxx
2
224102104
yxx
222 168101008016 yxxxx
3053 22 yx
3x2 – 5y
2 = 30………(1)
y = x – 2 ………(2)
(2) (1) 3x2 – 5(x–2)
2 = 30
–2x2 +20x –50= 0
x2 –10x +25 = 0
b2 – 4ac = 100 – 4(1)(25)
= 0
y = x – 2 is a tangent to the curve.
1
1
1
1
1
1
1
1
7
10
[12]
(a) Gradient of the tangent at P(2, 9) 3
2
Gradient of the normal at P(2, 9) 2
3
Normal equation at P(2, 9) is 22
39 xy
R(8, 0) Q(0, 12)
Mid point of QR(4, 6)
(b) 212
6
xdx
dy
dx
xy
212
6
)2(121
1
x
cx
y
)2(121
6
When P(2, 9), c = 10
1012
3
xy
(c) 03.0dt
dx
dt
dx
dx
dy
dt
dy
03.03
2
02.0 unit per second
1
1
1
1
1
1
1
1
1
1 (all correct)
1
1
y
13
10
0 0.5 0.65 x
8
11
[13]
(a) M = a ( 1+2) –2(3–0) +0
= 3a – 6
(b) cofactor of a21
(c) cofactor matrix =
624
2
333
aa
aa
Adj M =
63
23
423
aa
aa
(d) 1M =
63
23
423
63
1
aa
aaa
=
63
6
6363
363
2
6363
363
4
63
2
63
3
a
a
a
a
a
a
a
a
a
a
aaa
110
213
023
z
y
x
=
5
9
1
z
y
x
=
333
633
423
3
1
5
9
1
15273
30273
20183
3
1
z
y
x
=
3
23
5
3 ,2 ,3
5 zyx
1
1
1
1(>5 correct)
1(all correct)
1
1
1
1
1
1
1
1
9
12
[15]
02 cbxax
a
acbbx
2
42
a
acbb
a
acbb
2
4 ;
2
4 22
a
b
a
acbb
a
acbb
2
4
2
4 22
a
acbb
a
acbb
2
4
2
4 22
2
22
4
4
a
acbb
a
c
Alternative: 02 a
cx
a
bx 1 1
0 xx
0)(2 xx 1 1
Compare: a
b and .
a
c 1,1
(a) a
ca 4 ………..(1)
a
c23 ………..(2)
c
a
a
acca
2
222 2
3
16 ,
)2(
)1(
01033 22 acca
033 caca
3c ,3
ca
(b) 0)2( 222 axbacxc
2
2
c
acbSOR
2
2
c
aPOR
1
1
1
1
1
1
1
1
1
1
1
10
2
2
22 )(
2)(11
2
2
22
2 2
a
c
a
ac
a
b
2
2 2
c
acb (*)
2
2
222
111
c
a
a
c
(*) *)
22
11
SOR and
22
11
POR,
0)2( 222 axbacxc has roots .1
and 1
22
1
1(both *
correct)
1
1
top related