maths t paper 2 2009 johor - marking scheme

8/14/2019 Maths t Paper 2 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 JABATANJABATAN 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 JABATANJABATAN 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 JABATANJABATAN 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

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MATHEMATICS T(MATEMATIK T)

PAPER 2(KERTAS 2)

PERCUBAAN

STPM 2009


2/11

2

Q Steps Marks Notes

1

~a

~b

~~ba

cos =))(( 352

735222 +

0120=

Thus, angle between~a and

~b is 120

0

1

1

1

1

4

Diagram with arrow in

correct direction

using cosine rule

CA

conclusion

1 Alternative method:

=

y

xa~

,

=

n

mb~

x2

+y2

= 25 and m2

+ n2

= 9

=

ny

mxba~~

(x m)2 + (y m)2 = 49x

2+ y

2+m

2+n

2 2mx 2ny = 49

mx + ny =2

15

cos||||.~~~~baba =

n

m

y

x. = 5 x 3cos

xm + yn = 5x3cos

2

15 = 5x3cos

cos =2

1

= 1200Hence, the angle between

~a and

~b is 120

0

1

1

1

1

4

Either one correct

Dot product

CA

conclusion


3/11

3

2 3sinx + 4cos x = r sinx cos + rcos x sin rcos = 3 , rsin = 4

r = 5 or 22 43 + seen

tan =

3

4

= 53.10Hence,

3sinx + 4cos x 5 sin(x + 53.10)

6 sin x + 8 cos x + 5 = 10 sin(x + 53.10

) + 5

-5 10 sin(x + 53.10 ) + 5 15Max. value = 15

Min. value = -5

1

1

1

1

1

1

6

both

rcos = 3rsin = 4 seen,if not, -1

for10 sin(x + 53.10 ) + 5

For his logic

inequality

Both values correct

3(a)

(b)

vQ

vP

Pv

Q

300

1200

QvP= 30

2+ 60

2 2(30)(60)cos 120

0

QvP= 30 7 or 79.37 km

37379

120

30

0

.

sinsin=

= 19.110 / 19.10

Direction ofQvPis S 49.10

W

Shortest distance = 20 sin 40.90

= 13.09 km

1

1

1

1

1

1

1

7

Diagram withcorrect arrows

Cosine rule based on

his diagram

CA

His sine rule

CA

Using his angleCA


4/11

4

4. 2 sin2

x = 1 cos 2x

sin 3x = sin(2x + x)

= sin 2x cos x + cos 2x sin x

= 2 sin x cos2x + ( 1 2sin

2x)sin x

= 2 sin x(1 sin2

x) + sin x 2 sin3

x

= 3 sin x 4 sin3x

8 sin 5 x = (2sin2

x)(4sin3

x)

= (1 cos 2x)(3 sin x sin 3x)

= 3 sin x sin 3x 3cos 2x sin x + sin 3xcos 2x

= 3 sin x sin 3x 3[ sin 3x sin x] + sin 5x

+ sin x

= 5 sin x -2

5sin 3x + sin 5x

Hence, a = 5, b = -2

5, c =

2

1

1

1

1

1

1

1

1

7

CA

Identity(either one)

CA

Using above ans.

Either one of the factor

formula used

CA

conclusion

5(a)

C

D

O

BCO = BAO = 900(radius perpendicular to tangent)BO = BO (common line)

OC = OA (radii of circle)

BAO BCO (RHS)

BC = AB

1

1

1

1

Must provide at least 1

correct reason, if not -1

Must have above 3statements evenwithout RHS

BAO BCO seen


5/11

5

(b)

(c)

A

C

D

E

OB

x0x

0

x0

900-x0

Let AOE be the diameter of the circle,

CAE = 900 x0(tangent to radius)ECA = 900(angle in semicircle)CEA = 1800 (900- x0) - 900

= x0

CDA =CEA = x0(angle in same segment)BAC =ADC

C

D

BAC = ADC = x0(from above)

ACD = x0 (base angle of isosceles ) either oneBCA= x0(base angle of isosceles /alternate segment)ACD =BCADAC = 900 2x0 = ABC

Hence, ACD ADC

1

1

1

1

1

1

1

1

12

Mark accordingly to his

labelling



All the above 3

statements correct




6/11

6

6a

pdt

dp

2

312 = = )( p324

2

1

= dtpdp

2

1

324

ctp +=2

1324

3

1)ln(

c+= )()](ln[ 02

10324

3

1

243

1ln=c

243

1

2

1324

3

1ln)ln( = tp

tp 2

3

324

24=

ln

t

ep

2

3

24

324 =

p = 8 8t

e 23

= 8(1-t

e 23

)

p

t0

8

1

1

1

1

1

1

1

7

Any correct separation

Integrate

Correct subst. for

finding c

Get rid of ln

CA

Shape of curve & 8seen.

All correct


7/11

7

6(b22vgkg

dt

dv=

=gdt

vk

dv22

1

By writing

kv

B

kv

A

kvkv +

+

=

+ 1111

1

))((

1 = A(1 + kv) + B(1 kv)

A =2

1, B =

2

1

=++

gdtdv

kvkv )()( 12

1

12

1

cgtkv

kv

k+=

+

1

1

2

1ln

( ) cgk += )(ln 0121

kgtkv

kv2

1

1=

+ln

kgte

kv

kv 2

1

1=

+

kgtekvkv 211 )( =+

kgt

kgt

kek

ev

2

21

+

=

1

1

1

1

1

1

1

7

Any correct separation

Correct A & B

His correct integration

base on his A&BCorrect subst.

Correct subs. for

finding c

Get rid of ln

CA

7(a)

(b)

!! 21

221

=ee

4 e - 2

e = 0e (4 - ) = 0e 0, = 4

P(X > 3) = 1 [P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)]

= 1 -

+++

!! 3

4

2

441

32

4e

= 0.5666

1

1

1

1

1

5

Forming correctequation

CA

Complement

At least 3 correct terms

with his .

CA


8/11

8

8

(a)

(b)

P(X) =7

2, P(X Y) =

21

1

P(X|Y) =7

3

7

3=

)'(

)'(

YP

YXP

P(X ) P(X Y) =7

3[1 P(Y)]

[ ])(YP= 17

3

21

1

7

2

P(Y) =9

4

P(X Y) =7

2+9

4-21

1

= 63

43

P(X).P(Y) =7

2x9

4=63

8

P(X).P(Y) P(X Y)X and Y are not independent events.

1

1

1

1

1

1

6

Expansion of P(X|Y)correctly.

Seen/applied in the

equation

Correct formula with his

value

CA

Shown

7

2 x9

4 or his

value

Both statements correct

and

63

8 seen

9(a)

(b)

(c)

m

0

5+ m

1

5+ m

2

5+ m

3

5+ m

4

5+ m

5

5= 1

m + 5m + 10m + 5m + m = 1

m =32

1

x 0 1 2 3 4 5

P(X = x)

32

1

32

5

32

10

32

10

32

5

32

1

E(5X 3) =

5[0( 32

1

)+1(32

5

)+2( 32

10

)+3( 32

10

)+4(32

5

)+5( 32

1

)] - 3

=2

19or 9.5

1

1

1

1

11

1

7

At least one of theequation seen

CA

At least 4 prob. correctbased on his m.

All correct with table

Finding E(X) based on his

mFor correct usage of

formula 5E(X) - 3

CA


9/11

9

10

(a)

(b)

(c)

+

=

otherwise,0

3x,2

2x,0

23

2

3

1

x

x

xf )(


10/11

10

(c) F = S1 + S2 + S3 + S4 G = L1 + L2 + L3

Let T = F G

E(T) = 4(0.760) 3(1.010)

= 0.01

Var(T) = 4(0.0082) +3(0.009

2)

= 0.000499

P(T > 0)

= P

>

0004990

0100

.

.Z

= 0.6728

1

1

1

1

1

10

Correct linear

combination(4S-3L isaccepted if his var(T) is

correct)

All correct

All correct

Standardization based onhis values

CA

12

(a)

(b)

120

7650=

x

= 63.75

2

120

7650

120

510020

=

= 13.64

20.5 30.5 40.5 50.5 60.5 70.5 80.5 90.5 1 00.5

0

10

20

30

40

50

60

70

80

90

100

110

120

Numberofstudents

Marks

median = 50564 ..

105

16.3

(i) From the graph, median 50564 ..

(ii) Range for one deviation from mean for marks

= [50.11, 77.39]

Range for one deviation from mean for number of

students = [14, 107]

1

1

1

1

1

1

1

1

1

1

His x 120

CA

510020

Correct formula

CA

Boundaries

Axes & >6 pts correct

All correct with smoothcurve

Shown in graph by dotted

line

Answer in this range


11/11

11

% students in this range = 100120

16105

= 74.2%

1

1

12

Correct formula based on

his value

Accept 74% - 78%

maths t paper 2 2009 johor - marking scheme

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