Final k2 Skema Mt Pat f4 2013

8/11/2019 Final k2 Skema Mt Pat f4 2013

1/15

PERATURAN PEMARKAHAN DIAGNOSTIK TINGKATAN 4 2013

1

SECTION A

(40 marks)

NO Solution and Mark Scheme Sub

Marks

Total

Mark

1

yx 31 or3

1 xy

Substitution

91031 22 yy Solve

or 93

110

2

2

xx quadratic equation

0)2()4( yy

or

0)13()7( xx

2,4 yy

or

13,7 xx

13,7 xx

or

2,4 yy

5

P1

K1 K1

N1

N1


2/15


2

2 (a)

Usea

acbbx

2

42

)13(2

)16)(13(423232 x

764.1,6978.0 xx

2

(b)

General form

0962 nxmx

Use 042 acb

0)9(46 2 mn

2nm

3

5

K1

K1

N1

K1

N1


3/15


3

3a)

1233

xor

423

xor

42

12

3

x

Equate the base

4

2

12

3123

3

xx

Equate the index

1

6 3 2 82

x x

1

7x

4

(b)

Use maa xxm loglog

4logloglog 3

2

5

22 yxxy

Use mnnm aaa logloglog

and

n

mnm aaa logloglog

4log

5

3

2

x

yxy

Convert log form to index

form

45

3

2

x

yxy

xy 2 4

8

P1

K1

K1

N1

K1

K1

K1

N1


4/15


4

4 (a) 0,4A or 8,0 B

4

3014 **

x or

4

3810 ** y

6,1P 3

(b)

Use 121 mm

2

1m

Use cmxy or

)( 11 xxmyy

c )4(*2

10*

or

)4*(2

10* xy

22

1 xy

3

(c)

)6(*0)0(*1*)4(*0)0(1*)6(*4*)0(*02

1

12

2

8

K1

N1

P1

K1

K1

N1

K1

N1


5/15


5

5 (a) Differentiate

16

543

xy

16

)3(435 4

xdxdy

16

4315 4x

Note: Power reduce by 1

2

(b)

16

1503.0

*

dx

dyorx

xdx

yyUse

)03.0(16

15y

02813.0y 3

(c) 15dx

dy

dt

dx

dt

dy

dt

dyUse

02.015 dt

dy

3.0dt

dy

3

8

N1

K1

P1

N1

K1

P1

N1

K1


6/15


6

6 (a) Refer to the graph 4

(b) 4)35.68(*

286

2

6

K1

N1


7/15


7

SECTION B

(60 marks)

7 (a)

Substitute ( 3 , 1) or ( -1 , -7)

intof(x)= ax + b

ba 7 or ba 31

Solve simultaneous

equation

5,2 ba 4

(b) Use 95*2* x

7x 2

(c)

Find )(1 xf Find )(2 xf

2

5

xy 5*)5*2(*2* x

Equate

154*2

5*

x

x

410

K1

K1

N2,1

K1

N1

K1

N1

K1 K1

5


8/15


8

8 (a)

Completing the square

5)1(3)1(3 22

x

8)1(3 2 x 3

(b)

)8,1(

1x 2

(c)

3

(d) 563)(//8)1(3)( 22 xxxfxxf

2

10

K1

N2,1,0

N1

N1

Maximum shape

Graph passes through,

(i) Maximum point * (-1, 8) or ( 0, 5) or

(-2.633,0) and (0.633,0)N1

And (ii) 2 points : (-3, -4) and (3, - 40)N1

N1

(-1, 8)

x

x

y

x

(-3, -4)

(3, -40)

5

O

N2,1,0


9/15


9

9(a) (i)

2

3k

(ii) Solving simultaneous equation

xx3

213

2

3*

)4,6(A 3

(b) (i) 02

6*

x or 13

2

4*

y

)22,6(

B

(ii) c )6(*3

222* or ))6(*(

3

222* xy

263

2 xy

4

(c) Use distance formula for PA or PQ22 )4*()6*( yx

or 22 )13()0( yx

Use PA=2PQ22 )4*()6*( yx = 2 22 )13()0( yx

020832422 yxyx

3

10

N1

K1

N1

K1

N1

K1

N1

K1

K1

N1


10/15


10

10(a)

180

142.330COD

0.5237 rad.

2

(b)

Use *5CDS Use5

60tan OA

and

)5237.0(*5 Find AC560tan5 AC

OR

UseAB

560cos and

FindAD.

AD

60cos

510

Use

perimeter =

ADACS CD ***

)5237.0(*5 + 560tan5 +

60cos

510

28.11//2785.11 4

K1

N1

K1

K1

K1

N1


11/15


11

(c)

Use 2

52

1

A to find areaof sector OBD

180

142.360*5

2

1 2

Find area of ODB

Use Cabsin21 or other

valid method

60*sin)5(2

1 2

UseArea = *Area of sector

OBD - * Area of ODB

)047.1(*)5(2

1 2 60sin*)5(2

1 2

2.26 2.266 cm2

4

10

K1

K1

K1

N1


12/15


12

1112)(

dx

dvand

dx

dua or imply

2)2(

)1(*)62()2(*)2(

x

xx

dx

dy

10k

3

6

1

10

)0,3())(( Pib

Find mtangent and

use mtangent mnormal= -1to obtain mnormal

2

5

5

2

normal

tangent

m

m

2

15

2

5 xy

)3,0(2

15,*0)( QorRii

unitsRQ2

21

rootnohas

x

OR

xc

02

10*

02

10*)(

2

2

P1

K1

N1

P1

K1 K1 )3*(25*0* xy

or

other valid method

N1

N1

N1

P1


13/15


13

12

0

222

74.95

cos)5)(7(2579

findtorulecosineUse)(

ACB

ACB

ACBa

2.11//20.11

//199.11

56.33*sin

3

7.50sin

findtorulesinee Us

56.3374.95*7.50180)(

00

000

cmBD

CD

CD

CDEii o

5

b) (i)

Note: acuteisAFB

00000

48.1174.95*1802180

7.50)(

CAFor

BACii

2

02

01

90.19

48.11*sin552

1

74.95sin572

1

orofAreafindtosin

2

1Use

cm

Aor

A

ACFABCCab

A

BF

P1

K1

K1

N1

N1

P1

K1

K1

*A1 +* A2

N1

N1


14/15


14

OR

00000

48.1174.95*1802180

7.50)(

CAFor

BACii

018.62048.11*07.50 BAF

018.62sin*592

1Area

19.90 cm2

5

10

K1

P1

K1

P1


15/15


15

13

(a) (i) 227 seen

equivalentor1005.137

2272010/2013 I

= 165.09 3

(ii)

74.445

270100

09.165

equivalentorP2013

2

(b)

5.137

10

)2(16041200

)1(125)3(150

a

a=10

210

10

)2(2041500

)1(200)3(300

b

a (ii)

, Substitute the value of a=*10 into (ii)

to find b

b = 10

P1

K1

N1

K1

N1

K1

K1

(i)

K1

N1

N1

Final k2 Skema Mt Pat f4 2013

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