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