3472/2 Matematik Tambahan Kertas 2 2 ½ jam Sept 2010

SEKOLAH-SEKOLAH MENENGAH ZON A KUCHING

PEPERIKSAAN PERCUBAAN SIJIL PELAJARAN MALAYSIA 2010

MATEMATIK TAMBAHAN

Kertas 2

Dua jam tiga puluh minit

JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU

Skema Pemarkahan ini mengandungi 15 halaman bercetak

MARKING SCHEME

2

ADDITIONAL MATHEMATICS MARKING SCHEME

Zon A Kuching 2010 – PAPER 2

QUESTION

NO. SOLUTION MARKS

1

2 2

2

(2 ) 5 (2 ) 2 8 0

( 1)( 2) 0

x y

y y y y

y y

= +

+ − + + − =

+ + =

1, 2

@

1, 0

y y

x x

= − = −

= =

5

2

(a)

(b) (i)

(ii)

2

( ) ( )( 2)

( 2) 2

f x x p x

x p x p

= + −

= + − −

12 [ 2(3)]k= −

2k = −

2 2

2

2

1 12[ 6]

2 2

1 252( )

2 2

1 25 1 1max point , or ,12

2 2 2 2

y x x

x

= − + + − −

= − + +

∴ = − −

2

5

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

7

P1

K1

K1

N1

N1

K1

K1

N1

N1

K1

K1

N1

3

QUESTION NO.

SOLUTION MARKS

3

(a)

(b)

(c)

a = 40 and r = 0.8 |T6 = 40(0.8)5 = 13.1072

3

( )8.01

8.0140 5

5 −−=S

= 134.464

2

8.01

40

−=∞S

= 200

2

4

(a)

2 sincos2 2cos 1 or tan

cos

LHS to RHS or RHS to LHS

xx x x

x= − =

2

N1

K1

N1

K1

N1

P1

K1

7

N1

K1

4

QUESTION NO.

SOLUTION MARKS

(b)

(i)

(ii)

12 −= py

2

1=p

4

2

5 (a)

Refer to the graph Mode = 37

3

(b)

m = 29.5 +

10024

2 (10)30

−

=38.17

3

y ¼ π ½ π ¾ π π x O -1 Shape of sine curve Modulus Amplitude or period Translation

P1

P1

P1

P1

K1

6 N1

N1

8

K1

N1

P1

Lower boundary OR

10024

2 (10)30

−

5

QUESTION NO.

SOLUTION MARKS

6

(a)

(i) 3PR y x= +����

(ii) 4 3QS x y= − +����

2

(b)

3PT my mx= +����

STPSPT += = )34)(1(3 yxny −−+

= ynxn 3)44( +−

Compare coefficient of x and y

4

5n m= =

5

N1

N1

7

N1

K1

K1

N1

N1

6

19.5 29.5 39.5 49.5

5

20

30

0

10

15

marks

Number of students

Correct both axes (Uniform scale) K1 All points are plotted correctly N1

9.5 59.5 69.5

25

7

* y-intercept= 5 must be correct.

QUESTION NO.

SOLUTION MARKS

7

(a)

2500

4dA

rdr r

ππ= −

2500

4 0rr

ππ − =

r = 5 A = 678.58 m2

3

(b) (i)

(ii)

Finding area of trapezium

A1 = )1)(45(*2

1 + or A1 = 1

0

2

25

− x

x

Integrate dxxx )5( 2∫ −

A2 =

−

32

5 32 xx

Using limits ∫1

0 into A2

Area = A1 – A2 = 3

7@

3

12 @ 2.333 unit2

4

Integrate

dxxx 22 )5(∫ −π

+−

54

10

3

25 543 xxxπ

Use the limits ∫5

0into

+−

54

10

3

25 543 xxx

Volume generated = π6

625@ π

6

1104 @ 104.17π unit3

Note: OW − 1 once only for correct answer without showing the process of intergration.

3

N1

K1

K1

K1

K1

K1

N1

K1

K1

N1

10

8

Q8

2 3 4 5

0.2

×

0

×

x − 2

log10 y

×

2−x 3 4 5 7 8 10 log y

0.86 0.94 1.04 1.24 1.33 1.51

(a) Each set of values correct (log10 y must be at least 2 decimal places) N1, N1

Y = mX + c log 10 y = ( 2−x )log 10 b + log 10 a K1 where Y = log 10 y, X = (x − 2), m = log 10 b and c = log 10 a (c) log 10 b = gradient

log 10 b = 1.46 0.57

9.4 0

−−

= 0.09468 K1

b = 1.244 N1 log 10 a = Y-intercept log 10 a = 0.57 K1 a = 3⋅715 N1

Correct both axes (Uniform scale) K1 All points are plotted correctly N1 Line of best fit N1

1 6 7 8

×

×

10

N1

N1

0.4

0.6

0.8

1.0

1.2

1.4

1.6

×

9

QUESTION NO.

SOLUTION MARKS

9 (a) (i)

(ii)

3−r (r − 3)2 + 42 = r2 Using Pythagoras theorem on ∆BMO

r =6

25 (or 4.1667)

3

(b)

4.167 3 1.167

,

1 4tan

2 1.1671

1.287 rad.2

2.574 rad.

OM cm

let AOBθ

θ

θ

θ

= − == ∠

=

=

=

3

Area of sector OAB =21

(6

25)2 (2.574) = 22.34 cm2

Area OAB∆ =21

(8)( )36

25 − = 4.667 cm2

Area of shaded region =22.34 − 4.667 = 17.673 cm2

4

K1

P1

N1

K1

P1

N1

10

K1

K1

N1

K1

10

QUESTION NO.

SOLUTION MARKS

10 (a)

(b)

(c)

1

3

12

3

CD ABm m

y x

= =

= −

m = 6

3

3

1 3(3)

3 8

ADm

c

y x

= −

− = − +

= − +

Solving the equations y = −3x + 8 and 3 4 0x y− + = A(2, 2)

5

2 2

( 1) 41

3 8

11 3 20 0

y y

x x

x y x y

− − −× = −− −

+ − − + =

2

10

N1

K1

K1

K1

N1

N1

N1

K1

K1

N1

11

QUESTION

NO. SOLUTION MARKS

11

(a) (i)

(ii)

3

, 15

p p q= + =

52

( 0)5

P X = =

or IE

0.01024= ( 4) ( 4) ( 5)P X P X P X≥ = = + =

4 1 5

54

3 2 3

5 5 5C

= +

= 0.337

3

(b) (i)

(ii)

30 35 60 35

or 10 10

− −

( )0.5 2.5 1 ( 0.5) ( 2.5)P Z P Z P Z− ≤ ≤ = − ≥ − ≥

or 1 − 0.3085 − 0.00621 or R(−0.5) − R(2.5) = 0.6853 or 0.68525 Number of pupils = P( 60) 483X ≥ × = 3

7

N1

K1

K1

10

K1

N1

K1

K1

K1

N1

N1

12

QUESTION NO.

SOLUTION MARKS

12 (a)

(b)

(c)

(d)

v = 15 ms−1

1

a = 2t − 8 and a = 0 or dv

dt = 0

t = 4 vmin = −1 ms−1

3

(t − 3)(t − 5) > 0 0 ≤ t < 3, t > 5

2

324 15

3

ts t t= − +

| s3 − s2 | OR 3

2v dt∫

2

18 163

−

OR 3 3

2 23 24(3) 15(3) 4(2) 15(2)

3 3

− + − − +

11

3 m

4

10

K1

K1

K1

K1

N1

N1

P1

K1

N1

K1

13

QUESTION NO.

SOLUTION MARKS

13

(a) (i)

(ii)

(b)

(c) (i)

(ii)

Electricity : I = 2.1

1001.8

× = 116.7

Telephone and internet : I = 100

2

116.7(4) 133.3(2) 100(3) 120(6)

15

116.9

I+ + +=

=

3

110

210100

231

RM

RM

×

=

110 110(116.7 )(4) 133.3(2) 100(3) (120 )(6)

100 10015

124.8

× + + + ×

=

5

10

P2, 1

N1

K1

K1

K1

K1 K1

N1

N1

Any one of the price indices correct.

Any one of the price indices formula correct.

14

QUESTION

NO. SOLUTION MARKS

14 (a)

(b)

(c)

(d)

1

28 (14) (10) sin2

QRS= ∠

23 35'oQRS∴ ∠ =

2

2 2 2( ) 10 14 2(10)(14)cos23 35'

6.276

oQS

QS

= + −

=

2

6.276 7.5

sin sin30QPS=

∠ �

∠QPS = 24°44′ ∠PQS = 125°16′

3

Area of quadrilateral

128 (7.5)(6.276)sin125 15'

2

= 47.22

o

PQRS

= +

3

10

K1

N1

K1

N1

N1

K1

N1

K1 K1

N1

15

y

Answer for question 15

1 2 3 4 5 6 7 0 8

1

2

7

6

5

8

4

3

(3, 4) •

(a) I. 2y x≤

II. 7x y+ ≥ III. 3y x≥ −

(b) Refer to the graph, 1 or 2 graph(s) correct 3 graphs correct Correct area (c) i) xmin = 3 ii) max point (3, 4) k = 100x + 80y Maximum Profit = RM 100(3) + RM 80(4) = RM 620

10

N1

N1

N1

N1

N1

N1

K1

N1

K1

N1

R