Maths T Paper 2 2009 Johor - Marking Scheme

1 JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN 954/2 PERCUBAAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN STPM JOHOR 2009 JABATAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN MATHEMATICS T(MATEMATIK T) JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PAPER 2(KERTAS 2)JABATAN PELAJARAN JOHOR JABATAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN

2 Q 1

Steps

Marks

Notes

1

Diagram with arrow in correct direction

1

using cosine rule

1

CA

a− b ~

b

~

θ

~

a ~

52 + 32 − 72 2(5 )(3) 0 θ = 120

cosθ =

Thus, angle between a and b is 1200 ~

1

1

~

conclusion

4

Alternative method:  x  m a =   , b =   ~ ~  y n 2 2 x + y = 25 and m2 + n2 = 9  x − m  a − b =  ~ ~  y −n (x – m)2 + (y – m)2 = 49 x2 + y2 +m2 +n2 – 2mx – 2ny = 49 15 mx + ny = − 2 a . b =| a || b | cos θ ~ ~

~

1

Either one correct

1

Dot product

1 1

CA

~

 x   m  .  = 5 x 3cosθ  y  n  xm + yn = 5x3cosθ 15 = 5x3cosθ − 2 1 cosθ = − 2 0 θ = 120 Hence, the angle between a and b is 1200 ~

~

conclusion

4

3 2

3sinx + 4cos x = r sinx cosα + rcos x sin α rcos α = 3 , rsin α = 4 r = 5 or 32 + 42 seen 4 tanα = 3 α = 53.10 Hence, 3sinx + 4cos x ≡ 5 sin(x + 53.10)

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

1 1

1

6 sin x + 8 cos x + 5 = 10 sin(x + 53.1 ) + 5

1

for 10 sin(x + 53.10 ) + 5

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

1

For his logic inequality

0

1

Both values correct

6

3(a)

Ν

θ 300

v P Q

Diagram with correct arrows

1 Ν

vQ

1200 vP QvP=

302 + 602 – 2(30)(60)cos 1200

1

Cosine rule based on his diagram

1

CA

1

His sine rule

Direction of QvP is S 49.10 W

1

CA

Shortest distance = 20 sin 40.90 = 13.09 km

1 1

Using his angle CA

QvP

= 30 7 or 79.37 km

sinθ sin 120 0 = 30 79.373 θ = 19.110 / 19.10

(b)

7

4 4.

2 sin2 x = 1 – cos 2x

1

CA

sin 3x = sin(2x + x) = sin 2x cos x + cos 2x sin x = 2 sin x cos2x + ( 1 – 2sin2x)sin x = 2 sin x(1 – sin2 x) + sin x – 2 sin3 x

1

Identity(either one)

1

CA

1

Using above ans.

1

Either one of the factor formula used

1

CA

= 3 sin x – 4 sin3 x 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[ ½ sin 3x – ½ sin x] + ½ sin 5x + ½ sin x 5 = 5 sin x - sin 3x + ½ sin 5x 2 Hence, a = 5, b = -

1

5 1 ,c= 2 2

conclusion

7

5(a) C

Β

O D Α

∠BCO = ∠BAO = 900(radius perpendicular to tangent) BO = BO (common line) OC = OA (radii of circle) ∴∆BAO ≡ ∆BCO (RHS)

1

∴BC = AB

1

1 1

Must provide at least 1 correct reason, if not -1

Must have above 3 statements even without RHS

∆BAO ≡ ∆BCO seen

5 (b)

C

E x0

900-x0

0

D

Mark accordingly to his labelling

B

O x

x0 A

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

Must provide at least 1 correct reason, if not -1

1 1

1 1

All the above 3 statements correct

C

(c)

Β D Α ∠BAC = ∠ADC = x0(from above) ∠ACD = x0 (base angle of isosceles ∆) ∠BCA= x0(base angle of isosceles ∆/alternate segment) ∴∠ACD = ∠BCA ∠DAC = 900 – 2x0 = ∠ABC Hence, ∆ ACD ≅ ∆ADC

either one

Must provide at least 1 correct reason, if not -1

1 1

1 1 12

6 6a

dp 3 1 = 12 − p = (24 − 3 p ) dt 2 2 1 dp ∫ 24 − 3 p = ∫ 2 dt 1 1 − ln(24 − 3 p ) = t + c 3 2 1 1 − ln[24 − 3(0 )] = (0) + c 3 2 1 c = − ln 24 3 1 1 1 − ln(24 − 3 p ) = t − ln 24 3 2 3  24  3  = t ln  24 − 3 p  2 3

− t 24 − 3 p =e 2 24

p = 8 – 8e

3 − t 2

= 8(1- e

3 − t 2

)

1

Any correct separation

1

Integrate

1

Correct subst. for finding c

1

Get rid of ln

1

CA

1

Shape of curve & 8 seen.

1

All correct

p 8

0

t

7

7 6(b

dv = g − gk 2 v 2 dt dv ∫ 1 − k 2 v 2 = ∫ gdt

1 A B By writing = + (1 − kv )(1 + kv) 1 − kv 1 + kv 1 = A(1 + kv) + B(1 – kv) 1 1 A= ,B= 2 2 ⇒∫

1 1 + dv = ∫ gdt 2(1 − kv) 2(1 + kv) 1  1 + kv  ln  = gt + c 2k  1 − kv  1 ln(1) = g (0 ) + c 2k  1 + kv  ln  = 2 kgt  1 − kv   1 + kv  2 kgt  =e  1 − kv  1 + kv = (1 − kv )e 2 kgt

v=

e 2 kgt − 1 k + ke 2 kgt

1 Any correct separation

1

1

Correct A & B

1

His correct integration base on his A&B Correct subst.

1

Correct subs. for finding c

1

Get rid of ln

1

CA

7

7(a)

(b)

2e − λ λ1 e − λ λ2 = 1! 2! −λ −λ 2 4e λ - e λ = 0 e − λ (4 - λ) = 0 e − λ ≠ 0, λ = 4

1

Forming correct equation

1

CA

P(X > 3) = 1 – [P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)]  42 43   = 1 - e − 4  1 + 4 + + 2! 3!  

1

Complement

1

At least 3 correct terms with his λ.

= 0.5666

1

CA

5

8 8

2 1 , P(X ∩Y) = 7 21 3 P(X|Y’) = 7 P( X ∩ Y ' ) 3 = P(Y ' ) 7 3 P(X ) –P(X ∩ Y) = [1 – P(Y)] 7 2 1 3 − = [1 − P(Y )] 7 21 7 4 P(Y) = 9 P(X) =

(a)

2 4 1 + 7 9 21 43 = 63

P(X ∪ Y) =

(b)

P(X).P(Y) =

2 4 8 x = 7 9 63

1

Expansion of P(X|Y’) correctly.

1

Seen/applied in the equation

1

Correct formula with his value

1

CA

1

Shown 2 7

x

4 or his 9

value

P(X).P(Y) ≠ P(X ∩ Y) ∴X and Y are not independent events. 9(a)

(b)

(c)

1 6

5 5 5 5 5 5 m   + m   + m   + m   + m   + m   = 1 0  1 2 3 4 5 m + 5m + 10m + 5m + m = 1 1 m= 32

x P(X = x)

0 1 32

1 5 32

2 10 32

3 10 32

4 5 32

19 or 9.5 2

63

1

At least one of the equation seen

1

CA

5

1

1 32

At least 4 prob. correct based on his m.

1

All correct with table

1 1

Finding E(X) based on his m For correct usage of formula 5E(X) - 3

E(5X – 3) = 1 5 10 10 5 1 5[0( )+1( )+2( )+3( )+4( )+5( )] - 3 32 32 32 32 32 32 =

Both statements correct and 8 seen

1

CA

7

9 10 (a)

(b)

,0 ≤ x ≤ 2 ,2 ≤ x ≤ 3 , otherwise

 31 x  f ( x ) = − 23 x + 2  0 

1

,x <0  0  x2 ,0 ≤ x < 2  F ( x) =  6 - 1 ( x 2 − 6 x + 6 ), 2≤ x <3  3  1 ,x ≥3

1

For 3

x

seen

− 23 x + 2

1 1

For

1

At least one quadratic function correct

1

All correct

1

Curves correct shape

1

All correct

1

App. of formula/integration

1

Correct subst.

1

CA

seen

All correct

F(x) 1 2 3

0 (c)

x 1

2

3

P(0.9 < X ≤ 2.1) = F(2.1) – F(0.9) (0.9)2 1 2 = − [(2.1) – 6(2.1) + 6] 3 6 = 0.73 – 0.135 = 0.595

10

11 (a)

(b)

0.750 − 0.760   P(S < 0.750) = P  z <  0.008   = P(Z < -1.25) = 0.1057

1

Standardization

1

CA

X~B(10, 0.1057)

1

Binomial distrb/implied

P(X ≥ 3) = 1 – P(X = 0) – P(X = 1) – P(X = 2) = 1- [90.8943]10 + (10(0.1057)(0.8943)9 + 45(0.1057)2(0.8943)8]

1

Correct subst of his binomial

1

CA

= 0.0803

10

(c)

F = S1 + S2 + S3 + S4 Let T = F – G

G = L1 + L2 + L3 Correct linear combination(4S-3L is accepted if his var(T) is correct)

1

E(T) = 4(0.760) – 3(1.010) = 0.01

1

Var(T) = 4(0.0082) +3(0.0092) = 0.000499

1

P(T > 0)  0 − 0.01   = P  Z > 0.000499  

All correct

All correct

1

= 0.6728

Standardization based on his values

1

CA

10

12 (a)

−

∑ x ÷120

1

His

1

CA

1 1

510020

1

CA

1

Boundaries

1

Axes & >6 pts correct

1

All correct with smooth curve

(i) From the graph, median 64.5 ± 0.5

1

Shown in graph by dotted line

(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

Answer in this range

7650 120 = 63.75

x=

510020  7650  − σ =  120  120  = 13.64

2

Correct formula

(b) 120 110

105

Number of students

100 90 80 70 60

median = 64.5 ± 0.5

50 40 30 20

16.3

10 0 20.5

30.5

40.5

50.5

60.5

70.5

80.5

90.5

100.5

Marks

11

105 − 16 × 100 120 = 74.2%

% students in this range =

1

Correct formula based on his value

1

Accept 74% - 78%

12