TRIAL ADDMATE SPM 2011 Sarawak Zon a Paper 2 Answer

8/4/2019 TRIAL ADDMATE SPM 2011 Sarawak Zon a Paper 2 Answer

1/13

3472/2

Matematik

TambahanKertas 22 jam

Sept 2011

SEKOLAH-SEKOLAH MENENGAH ZON A KUCHING

PEPERIKSAAN PERCUBAANSIJIL PELAJARAN MALAYSIA 2011

MATEMATIK TAMBAHAN

Kertas 2

Dua jam tiga puluh minit

JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU

Skema Pemarkahan ini mengandungi 13 halaman bercetak

MARKING SCHEME


2/13

2

ADDITIONAL MATHEMATICS MARKING SCHEME

TRIAL SPM exam Zon A Kuching 2011 PAPER 2

QUESTION

NO.SOLUTION MARKS

1

( )

( )( )

2

2 1

(2 1) 2 1 3

2 1 2 0

x y

y y y

y y

= +

+ + =

+ =

1, 2

2

@

2, 3

y y

x x

= =

= =

5

2

(a)

(b)

(c)

2 2 2

2 2

2

( ) 3[ 2( 2) ( 2) ( 2) ] 7

5or 3[( 2) 4] 7 or 3[( 2) ]

3

3( 2) 5

f x x x

x x

x

= + + +

+

=

(2, 5)min

Shape

(2, 5) and (0, 7)

(1, 22) and (4, 7)

2

1

3

Solve the quadratic

equation by using the

factorization @ quadratic

formula @ completing the

square must be shown

Eliminate orx y

Note :

OW 1 if the working of solving

quadratic equation is not shown.

5

6

P1

K1

K1

N1

N1

K1

N1

N1

N1

N1

N1

Ox

(2, 5)

(1, 22)

(4, 7)7

y


3/13

3

QUESTION

NO.SOLUTION MARKS

3

(a)

(b)

(c)

T10 = 100 + 9(4) or 64

Area = 512

2

100 + (n 1)(4) = 4 or100

4@ 100 + (n 1)(4) > 0

n = 25

2

[ ]8640

2(100) ( 1)( 4) or 10802 8

nn+ =

(n 36)(n 15) = 0

n = 15

3

4(a)

(b)xy

5

4=

Number of solutions = 3

4

3

Shape of sine curve

Modulus

Amplitude or period

Translation

P1

P1

P1

P1

y

x0

-1

1

3

1 2siny x=

2

2

2

3

xy5

4=

K1

N1

K1

7N1

K1

N1

K1

K1

N1

7

P1Sketch straight

line correctly


4/13

4

QUESTION

NO.SOLUTION MARKS

5

(a) 43 2 48 4 53 8 58 11 63 9 68 4 73 2 2325fx = + + + + + + =

2325

40x =

58.125=

3

(b)

Q3 = 60.5 +3

(40) 2554

9

= 63.28

OR1

1(40) 6

450.5 58

53

Q

= +

=

Interquartile range = 63.28 53

= 10.28

4

6

(a) BCO = 60

= 1.047 rad

AOB = 0.6982 rad

3

(b) 21 (10) (1.047)2

or 52.35

21 (10) (0.6982)

2

or 34.91

2

(c)Area of segmentBC= 52.35 2

1(10) sin1.047

2

r= 9.237

Area of the shaded region = 25.673

3

N1

7

K1

N1

P1

Lower

boundary OR

3(40) 25

549

N1

7

K1

K1

N1

K1

N1

K1

K1

N1

K1

N1

N1


5/13

5

(0.78, 1.94)

(0, 0.30)

x10log

.0

.2

.8

0.2 0.3 0.4 0.5

.2

0

og10y Q7

x10log 0.30 0.48 0.60 0.78 0.85 0.95

y10log 0.93 1.30 1.57 1.94 2.07 2.31

Correct both axes (Uniform scale) K1

All points are plotted correctly N1

Line of best fit N1

0.1 0.6 0.7 0.8

N1

N1

.4

.6

.8

.0

.2

.4

.6

0.9 1.0

(a) Each set of values correct (log10y must be at least

2 decimal places) N1, N1

log 10y = nlog 10x + log 10 (p + 1) K1

where Y= log 10y, X= log 10x,

m = n and c = log 10 (p + 1)

(c) (i)X= log 10 5.6 = 0.748

Y= 1.88 = log 10yy = 75.86 N1n = gradient

n =1.94 0.30

0.78 0

= 2.103 N1

log 10 (p + 1) = Y-interceptlog 10 (p + 1) = 0.30 K1

p = 0.9953 N1


6/13

6

QUESTION

NO.SOLUTION MARKS

8

(a) (i)

(ii)

(b) (i)

(ii)

2,0

2

82

12

18

2

=>

=

=

=

aa

a

a

aa

3

++=

+

2)2(5,

206

23,

24 yx

D(2, 6)

@

Solving the equationsy = 4x 2 and2

13

4

1+= xy

D(2, 6)

2

0214

5)2()0(

22

22

=++

=++

yyx

yx

Get equation ofBC,y = 4x 19

( )

0544

)264)(17(4136

4

026413617

021)194(4)194(

2

2

2

22

>=

=

=+

=++

acb

xx

xxx

The locus intersects the lineBC.

2

3

K1

K1

K1

N1

N1

N1

K1

N1

K1

K1

N1

K1

10


7/13

7

QUESTION

NO.SOLUTION MARKS

9(a) 1

2OP x y= +

1

(b)

(i)

(ii)

OQ x y = +

( )1

2

11

2

OQ OL LQ

x LP

x LM MP

x y x

x y

= +

= +

= + +

= +

= +

4

x : = 1 1

2

y : =

= 1 1

2

= =2

3

3

(c)Area of triangle OLM=

3

2 24 = 36

Therefore area of parallelogram OLMN= 72

2

P1

10

K1

K1

N1

K1

K1

N1

K1

N1

N1


8/13

8

QUESTION

NO.SOLUTION MARKS

10(a)

3 9 3

r h hr= =

2

9

dV h

dh

=

2

0.89

h dh

dt

=

2

0.89h dh

dt=

4

(b)

(i)

(ii)

h = 1 and k= 1

Area of the shaded region

= ( ) ( )1 0

3 3

0 1

y y dy y y dy

+

=

14 2

04 2

y y

+

0

1

24

24

yy

= ( )1 1

0 04 2

+

( )41 10 0

4 2

=2

1

Note: OW 1 once only for correct answer without showing theprocess of intergration.

6

N1

K1

K1

K1

K1

N1

N1

K1

K1

10

N1


9/13

9

QUESTION

NO.SOLUTION MARKS

11(a)

(i)

(ii)

(b)

(i)

(ii)

0 550( 0) (0.2) (0.8)

0.3277

P X C= =

=

( 2)P X< = 0 5 1 45 50 1(0.2) (0.8) (0.2) (0.8)C C+

= 0.7373

5

( 0.811)P Z> @R(0.811)

= 0.7913 @ 0.79132

P(0.811


10/13

10

QUESTION

NO.SOLUTION MARKS

12

(a) Initial velocity 8=Av 1

(b)

4

12

42

55

2

5

2

5

052

2

=

+

=

=

=+=

Bv

t

tdt

dv

3

(c) ( 2)( 4) 0

( 1)( 4) 0

4

A

B

v t t

v t t

t

= =

= =

=

3

(d)

3

26

)2(8)2(33

)2(

833

83

23

23

2

=

++=

+=

+= tt

t

dtttsA

3

N1

K1

K1

K1

N1

K1

K1

N1

10

K1

N1


11/13

11

QUESTION

NO.SOLUTION MARKS

13

(a)

(b)(i)

(ii)

A : 12510060

08 =P

08 RM75P =

2

(125 4) (120 ) (80 5) 150( 3)120

12 2

n n

n

+ + + +=

+

1440 240 1350 270n n+ = +

n = 3

06

RM30100 120

P =

RM25=

3

(c)

18

)6150()580()3138()4125(

138)15.0120(120

06/09

+++=

=+

I

= 123

3

K1

K1

N1

K1

N1

K1

N1

10

N1

K1

K1


12/13

12

QUESTION

NO.SOLUTION MARKS

14

(a)

(b)

(c)

18 28

sin38 sin QSR=

QSR = 180 7316

= 10644

3

QRS = 3516

2 2 228 18 2(28)(18) cos 35 16

@ Sine Rule

16.88

PS

QS

= +

=

3

2 2 226 11 16.88 2 11 16.88 cos PQS= +

PQS = 13639

Area of triangle PQR = 12

1128sin 17639

= 8.999

2

10

K1

N1

K1

N1

K1

N1

K1

N1

K1

N1


13/13

13

Answer for question 15

2 4 6 8 10 12 140 16

2

4

14

12

10

16

8

6 (15, 6)

R

(a) I. 4832 + yx

II. 4885 + yx

III. xy 3

(b) Refer to the graph,

1 or 2 graph(s) correct3 graphs correct

Correct area

(c) ii) max point (15, 6)

k= RM(500x + 300y)

Maximum Profit = RM 500(15) + RM 300(6)

= RM 7500

(ii) 8 units

10

N1

N1

N1

N1

N1

N1

K1

N1

K1

N1

x

y

TRIAL ADDMATE SPM 2011 Sarawak Zon a Paper 2 Answer

Documents

Transcript of TRIAL ADDMATE SPM 2011 Sarawak Zon a Paper 2 Answer