maths t paper 1 2009 johor - marking scheme

8/14/2019 Maths T Paper 1 2009 Johor - Marking Scheme

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JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATANJABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATANJABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN

JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATANJABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN


JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN

JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATANJABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATANJABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN


JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN

954/1

MATHEMATICS T(MATEMATIK T)

PAPER 1(KERTAS 1)

PERCUBAAN

STPM 2009


2/8

No. Marks

1[4]

x+y i + 3 4 i = 3 i22 )4y()3x( ++ = 3

x2 + 6x + 9 + y28y +16 = 9

x2 + y2 + 6x 8y +16 = 0

1

1

1

1

Using his in themodulusEquation correct.

For squaring

CA

2[5]

2)( 3 = xxf

( ) ( ) 233

2)()(

3223

3

+++=

+=+

xxxxxx

xxxxf

( ) ( )

2

322

0

33

0

3

33

lim

)(lim)('

x

x

xxxxx

x

xxxxf

x

x

=

++

=

+=

23)(' aaf =

1

1

1

11

2)()( 3 +=+ xxxxf

correct formulasubstitute with limit

Expand

CA

CA

3[6]

2or12

1

2

1and,2or1

012and023

defined,beto)12(log2and)23(logFor

2

32

3

>>+

+

xx

xxx

xxx

xxx

7

5or0

)12(223

)12(2log)23(log

22

23

23

> 0

2

2

dx

ydlocal minimum point

(c)

1

1

1

1

1

1

1

His = 0

Correct answer incoordinate form

method

nature of both theturning pointscorrect

all parts correct

points

asymptotes andperfect

9[10]

x(8m m2 x ) = 16m

2x

2 8mx + 16 = 0

b2 4ac = 64m2 4 (m2)(16)

= 0

m2 x + y 8m = 0 is a tangent to xy = 16.

Eqn. of OR is y = 2m1

x (1)

m2

x + y 8m = 0 .(2)

(1) (2) m2

x + 2m1

x 8m = 0

x =4

3

m1

m8

+

1

1

1

1

1

quadratic eq.

b2-4ac=0 or other

methodConclusion

Equation of OR

substitution

y

x-1 0 2 3 5

-3/21)-,1( )

9

1,5(


6/8

y =4

m1

m8

+

R (4

3

m1

m8

+,

4m1

m8

+)

m2

=y

x

(m2

x + y )2

= 64m2

(y

x x + y )2

= 64 (y

x )

(x2

+ y2

)2

= 64xy

1

11

1

1

x or y correct

Coordinates Rm or m2 in terms of x

and y.

elimination of m

10[11] A = k 21

13 1

22

121

12

32

= k (61)(42)(2+6)= 7k2 8

= 7k107k10 = 177k = 7

k = 1

BC =

651

432

321

03451

33866

0170

=

15324

9126

6921

ABC =

212

132 111

15324

91266921

=

5100

0510

0051

ABC = 51 I

A1 = 51

1

15324

9126

6921

=

17/517/117/8

17/317/417/2

17/217/317/7

212

132

111

z

y

x=

25

3

1

A

z

y

x=

25

3

1

1

1

*1

1

1

1

1

1

1

CA

( )( ) is seen

CA (if 1 is notobtained here,multiplicationprocess must beseen to obtain *1)

A( *) * his BC

His BC

CA

CA


7/8

dx

z

y

x= 51

1

15324

9126

6921

25

3

1

= 511

408

255

102=

8

5

2

x = 2. y = 5, z = 8

1

1

His inverse

CA

11[11] a)

)4x)(2x(

12x4x52

2

++

++

2x

A

++

4x

CBx2 +

+

5x2

+ 4x + 12 = A(x2

+4) + (Bx+C)(x+2)

x= 2, 20 8+12 = 8AA = 3

coef. of x2, 5 = A + B

B = 2x = 0, 12 = 4A + 2C

C = 0

)4x)(2x(

12x4x52

2

++

++

2x

3

++

4x

x22 +

3

1 )4x)(2x(

12x4x52

2

++

++

= 3

1 2x

3

++

4x

x22 +

dx

= [3 ln(x+2) + ln (x2

+ 4 ) ]1

3

= 3ln (5) + ln(13) 3ln(3) ln(5)

= ln27

325

b) h =4

12 = 0.25

+2

1

2 dx)x1ln( =2

25.0[ 0.6931+1.6094 +

2(0.9410+1.1787+1.4018)]= 1.168

x y

1 0.6931

1.25 0.9410

1.5 1.1787

1.75 1.40182 1.6094

1

1

1

1, 1

1

1

1

1

1

1

CA

Method to find A, B& C

CA

First 1, ln is seen

2nd

1, all correct

his substitution

CA

CA

All correct

Correct formula withhis value

CA


8/8

12[14]

(a) 3)15(lim)(lim11

==

xxfxx

3)86(lim)(lim2

11=+=

++ xxxf

xx

.1atcontinuousis**).1()(lim)(lim

3861)1(

11

=

==

=+=

+

xffxfxf

f

xx

(b)

(c) 18189 =+=y

Range of{ }51: yy

(d)

,3,2,2

862

2)1(5

2

=

+=

=

x

xxx

xx

{ } { } 43:22:

maths t paper 1 2009 johor - marking scheme

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