Skema Add Math P2 Trial Spm Zon A Kuching 2009

3472/2 Matematik Tambahan Kertas 2 2 ½ jam Sept 2009

SEKOLAH-SEKOLAH MENENGAH ZON A KUCHING

PEPERIKSAAN PERCUBAAN SIJIL PELAJARAN MALAYSIA 2009

MATEMATIK TAMBAHAN Kertas 2 Dua jam tiga puluh minit

JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU

MARKING SCHEME

Skema Pemarkahan ini mengandungi 14 halaman bercetak

2

ADDITIONAL MATHEMATICS MARKING SCHEME TRIAL SBP 2009 – PAPER 2 QUESTION NO.

1

SOLUTION

x = 2y − 6

MARKS

P1

(2 y − 6)2 + (2 y − 6) y − 20 = 0

K1

3 y 2 − 15 y + 8 = 0

K1

y = 0.6070, y = 4.393

N1

@ x = −4.786, x = 2.786

N1

5 Eliminate x or y Solve the quadratic equation by using the factorization @ quadratic formula @ completing the square must be shown

Note : OW-1 if the working of solving quadratic equation is not shown.

5 2 (a)

OT = 17.3205 cm

K1 K1

sSTR = 17.3205 ×

sSTR = 18.1380 cm

π 3

π 3

N1

K1

Use the formula s = rθ

4

3

QUESTION NO.

(b)

SOLUTION

MARKS

1 o Area OPQ = × 20 × 20 × sin 60 @ 173.2051cm2 2 1 π Area OSTR = × 17.32052 × 2 3

@

N1

16.1256 cm2

Score (b)

Frequency, f

0-4 5-9 10-14 15-19 20-24 25-29 30-34

∑ Variance =

2 3 10 20 7 6 2 f = 50

17185  865  −  50  50 

= 44.41

N1

Midpoint, x 2 7 12 17 22 27 32

K1

fx 2

fx

4

∑

2

3

Lower boundary OR  43 + w  − 15    2  (5) 20      

N1

P1

8 P1

 43 + w  − 15    (5) K1 17 = 14.5 +  2 20      

w=7

4

@ 157.0795 cm2 K1

1 1 π × 20 × 20 × sin 60o − × 17.32052 × 2 3 2 K1 173.2051 – 157.0795

3 (a)

K1

4 21 120 340 154 162 64 fx = 865

∑

8 147 1440 5780 3388 4374 2048 fx 2 = 17185

OR P1

7

4

QUESTION NO. 4 ( a)

SOLUTION

δV = −0.6

OR

dV = 3x 2 dx

OR

δ x ≈ −0.008

( b)

3

K1

x=5

K1

−0.6 ≈ 3(5)2 × δ x

f ′( x) =

MARKS

N1

(3 − x 2 )(3) − (3 x + 4)(−2 x) (3 − x 2 )2

3

K1

3 − (2)2  (3) − [ (3(2) + 4)(−2(2))] f ′(2) = 2 3 − (2)2 

K1

N1

37

6 5 ( a)

 cos x    sin x 

2 sin2 x 

K1

2 = 2 sin x cos x N1

= sin 2x (b)

y=

2x

π

Sketch straight line correctly

P1

y y = 2 sin 2 x

2

y=

2x

π

Sine curve

P1

Period

P1

Amplitude

P1

Modulus

P1

6

1

O

Number of solutions = 4

π

x

N1

8

5

QUESTION NO.

6 (a)

SOLUTION

MARKS

K1

(i) T15 = 2 + 14(3)

3

N1

= 44 cm

(ii)Area of the fifteenth circle = 1936 π cm 2

N1

(b) 100, 50 , 25 , ……… a= 100 AND r =

T7 = 1.563

K1

1 2

3 K1

N1

6

6

log10 y

5 4

x+2

3

4

5

6

7

8

log 10 y

1.408

2.100

2.806

3.500

4.200

4.800

(a) Each set of values correct (log10 y must be at least 2 decimal places) N1, N1 Y = mX + c log 10 y = (x + 2) log 10 q + log 10 p K1 where Y = log 10 y, X = (x + 2), m = log 10 q and c = log 10 p (c) log 10 q = gradient 4.2 − (− 0.6 ) log 10 q = K1 7−0 q = 4.85 N1 log p = Y-intercept log p = −0.6 p = 0⋅251

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

× ×

K1 N1

10

3

× ×

×

2 × 1

0 −1 −2

1

2

3

4

5

6

7

8

x+1

7

QUESTION NO. 8 (a) (i)

(ii)

(b)

SOLUTION

uuur QR = 9 x − 9 y % %

N1

uuur 1 OS = 9 y + −9 y + 12 x % % 2 % 9 = 6x + y % 2%

(

)

uuur uuur OT = hOS 9 = 6hx + hy % 2 % OR uuur uuur QT = kQR = 9 kx − 9ky % % uuur uuur uuur QT = QO + OT

3

K1 N1

5 K1

9 = −9 y + 6hx + hy % 2 % % 9   = 6hx +  −9 + h  y %  2 % uuur Comparing QT , 2 9k = 6h ⇒ k = h --------------------------(1) 3 OR 9 1 −9k = −9 + h ⇒ k = 1 − h --------------(2) 2 2 (d)

Solving the simultaneous equations 6 4 h = ,k = 7 7

K1

K1

K1

N1

uuur PQ = −12 x + 9 y % % uuur 2 2 PQ = (12(3) ) + ( 9(5) )

= 57.63 units

MARKS

K1

2

N1

10

8

QUESTION NO.

9 (a) (i)

SOLUTION

K1

p = 0.3, q = 0.7, n = 8 1 − (0.7)8 − 8C1(0.3)(0.7)7

(b) (i)

N1

K1

n(0.3)(0.7) =315

n = 1500

N1

36.5 − 33.5  150  30 − 33.5
5

P

K1

OR (ii)

5

K1

= 0.7447

(ii)

MARKS

P [ −0.7 < Z < 0.6] =

150 n

1 − P[Z > 0.7] − P[Z > 0.6] = 1 − 0.2420 − 0.2743 =

150 n

n = 310.11

150 n

K1

K1 K1

Total number = 310

N1

10

9

QUESTION NO.

10 (a)

SOLUTION

3

Get m = −1

K1

y − 4 = −1( x − 1)

K1 N1

y + x−5 = 0

(b)

( x − 1) + ( y − 4 )

( 4 − 2 ) + (1 − 3) x 2 − 2 x + 1 + y 2 − 8 y + 16 = 4 ( 8 ) 2

2

=2

2

x 2 + y 2 − 2 x − 8 y − 15 = 0

(c)

( x − 5)( x + 3) = 0

K1

2

3

K1

Y / Z = ( 5, 0 ) and Z / Y = ( −3, 0 ) (d)

K1

2

N1

x 2 − 2 x − 15 = 0 x = 5, x = −3 Get both

MARKS

Get T = ( 2, 3 ) ,

K1

x − intercept = −1

N1

N1

2 10

10

QUESTION NO. 11 ( a)

SOLUTION px 2 + x = −2

3

K1 K1

p(2)2 + 2 = −2 p = −1, q = 2

(b) (i)

MARKS

N1

Area of the shaded region 1

∫ (x = ∫ (x =

2

0

1

2

0

− 6 x + 9)dx −

7 1

∫ (2 x + 2)dx

K1

0

− 8 x + 7)dx 1

 x3 8 x 2  =  − + 7x 2 3 0 1 8  =  − + 7 − 0 3 2  1 = 3 unit 2 . 3

(ii)

K1 K1 N1

Volume of revolution 2

= π ∫ y 2 dx 0 2

= π ∫ ( x − 3)4 dx

K1

0

2

5  = π  ( x − 3)   5 0

 (1 − 3)5

= π 

5

K1

 (0 − 3)5   −   5  2 5

= 48 π unit 3 .

N1

10

11

QUESTION NO. 12 ( a)

( b)

SOLUTION

v = 5 ms−1

P1

a = 2t − 6 and

a = 0 or

t=3

( d)

dv =0 dt

K1

2

N1

t3 s= − 3t 2 + 5t 3 | s1 − s0 | + | s5 − s1 | 7 25 7 −0 + − − 3 3 3

OR

3

K1

N1

(t − 1)(t − 5) < 0 1
1

K1

vmin = −4 ms−1

( c)

MARKS

K1

OR

1

∫0

v dt +

4 5

∫1 v dt

K1

K1

13  13 53 − 3(1)2 + 5(1) − 0 + − 3(5)2 + 5(5) −  − 3(1)2 + 5(1)  3 3 3 

13 m

N1

10

12

QUESTION NO.

SOLUTION

K1 5 2 + 8 2 − 12 2 2(5)(8) ∠AMB = 133.43˚ @ 133˚26'

13 (a) (i)

cos ∠AMB =

(ii)

sin ∠ACM sin 46.57° = 5 4 ∠ACM = 65.20˚ @ 65˚24'

MARKS

2 N1

5

K1 K1

∠MAC = 180˚ – 46.57˚ - 65.20˚ = 68.23˚ Area of ∆AMB Area of ∆ACM 1 = (5)(8)sin 133.43˚ 1 = (5)(4)sin 68.23˚ OR 2 2 = 14.52 cm2 = 9.287 cm2 Area of ∆ABC = 14.52 + 9.287 K1 = 23.807 cm2

(b)

Get ∠B 'M 'M = 46.57˚ @ ∠B 'MM ' = 46.57˚ @ ∠M 'B 'M = 86.86˚

K1

N1

3 K1

MM ' 8 K1 = sin 86.86° sin 46.57° MM ' = 10.999 cm @ 11 cm A'M '= 5 + 10.999 = 15.999 cm @ 16 cm

N1

10

13

Answer for question 14

(a)

I.

2 x + 3 y ≤ 48 N1

II.

5 x + 8 y ≥ 48 N1

III.

2y≥ x

N1

Refer to the graph,

(b)

K1

1 or 2 graph(s) correct 3 graphs correct

Correct area

(c) i)

16

N1

N1 N1

6

N1

ii) max point (12, 8)

k = 18x + 20y Maximum Profit = RM 18(12) + RM 20(8)

14

= RM 376

K1

N1

10

12

10

8

6

4

2

R

(12, 8)

14

QUESTION NO. 15 (a) (i)

SOLUTION 140 =

RM 50.40 ×100 Q97

Q97 = RM 36

(ii)

I 00,97 =

125 =

I 00,94 I 97,94

I 00,94

120

I00, 94 = 150

(b) (i)

MARKS

5

K1

N1

×100 K1

×100

K1 N1

125 × 20 + 140 ×10 + 30 x + 110 × 40 = 122 100 30x + 8300 = 12200

K1

5

K1 N1

x = 130

(ii) 122 =

RM 288 ×100 Q1997

Q1997 = RM 236.07

K1

N1

10