Spm Addmath2 Ans (kedah)

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Question

SubFull mark Mark

Solution and marking scheme

3+ y 2 y = 2x − 3

1.

x=

or P1

Make x or y as the subject

2

⎛ 3+ y ⎞ ⎛ 3+ y ⎞ 2 ⎜ ⎟ −⎜ ⎟ y + y = 10 ⎝ 2 ⎠ ⎝ 2 ⎠ or

x 2 − x ( 2 x − 3 ) + ( 2 x − 3 ) = 10 2

K1

Eliminate x or y

9 + 6 y + y − 6 y − 2 y + 4 y − 40 = 0 2

2

2

or

x 2 − 2 x 2 + 3 x + 4 x 2 − 12 x + 9 − 10 = 0

K1

y = 3.215 , − 3.215 or

Solve quadratic equation N1

x = 3.107 , − 0.107

3 y 2 − 31 = 0 31 y2 = 3 or

3x − 9 x − 1 = 0 2

x = 3.107 / 3.108 , − 0.107 / − 0.108 N1

or y = 3.214 / 3.215 , − 3.214 / − 3.215

Answer must correct to 3 decimal places.

3472/2

Additional Mathematics Paper 2

or using formula or completing the square

5

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Question 2(a)

Submark

Solution and marking scheme

Full Mark

16π ,8π , 4π ,......

a = 16π

r=

1 2

P1

⎡ ⎛ 1 ⎞7 ⎤ 16π ⎢1 − ⎜ ⎟ ⎥ ⎢⎣ ⎝ 2 ⎠ ⎥⎦ S7 = 1 1− 2 3 4

K1

Sn =

Use

a(1 − r n ) 1− r

= 31 π

N1

3

or 31.75 π

(b)

64π ,16π , 4π ,......

a = 64π

r=

1 4

P1

64π S∞ = 1 1− 4 1 = 85 π 3

K1

a 1− r

3

6

f ( x) = x 2 + px + q 2

2

⎛ p⎞ ⎛ p⎞ = x + px + ⎜ ⎟ − ⎜ ⎟ + q ⎝2⎠ ⎝2⎠ 2

2

p⎞ p2 ⎛ = ⎜x+ ⎟ − +q 2⎠ 4 ⎝ p − =3 2 −

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S∞ =

N1

or 85.33 π

3(a)

Use

36 + q = −5 4

p = －6

K1 use x² + bx = ( x + b )² – ( b ) 2 2

2

or N1

use axis of symmetry −

b =3 2a

q=4 N1

Additional Mathematics Paper 2

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Question

Submark

Solution and marking scheme

Full Mark

Alternative solution

x 2 + px + q = ( x − 3) − 5 2

= x2 − 6x + 9 − 5 = x2 − 6x + 4

p = −6

K1

Comparing coefficient of x or constant term

N1

N1

q=4

3

3(b)

x 2 − 6 x + 4 − 31 ≤ 0 x 2 − 6 x − 27 ≤ 0 K1

( x − 9)( x + 3) ≤ 0

−3 ≤ x ≤ 9

Use f ( x) − 31 ≤ 0 and factorization

N1

2

5

4(a)

cos x 1 + sin x sin x cos x + = cos x 1 + sin x sin x(1 + sin x) + cos 2 x = cos x(1 + sin x)

tan x +

sin x + sin 2 x + cos 2 x cos x(1 + sin x) sin x + 1 = cos x(1 + sin x) 1 = cos x = sec x N1

Use tan x =

K1

sin x cos x

=

3472/2

K1

Use identity

sin 2 x + cos 2 x = 1

Additional Mathematics Paper 2

3

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Question

Submark

Solution and marking scheme

Full Mark

4(b)(i)

y=

y

5x

π

−2

3

y = −3sin 2x

π

O

π

3π 2

2

x

2π

–3

P1

Negative sine shape correct. Amplitude = 3

[ Maximum = 3 and Minimum = − 3 ]

Two full cycle in 0 ≤ x ≤ 2π

P1

P1 3

4(b)(ii)

−3sin 2 x =

5x

π

−2

or

y=

5x

π

N1

−2

Draw the straight line

y=

Number of solutions is 3 .

5x

π

−2

K1

N1 3

3472/2

Additional Mathematics Paper 2

9

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Question

Solution and marking scheme

Submark

Full Mark

7

7

5 (a)

K1

∑ x = 12

∑ x = 240

20

Use x =

∑x N

N1

The mean X=

240 + 5 + 8 + 10 + 11 + 14 288 = 25 25

N1

= 11.52

(b)

∑ x2 20

K1

− 122 = 3

∑x

2

= 3060

N1 Use formula

σ=

∑x N

2

− x2

The standard deviation

σ=

3060 + 52 + 82 + 102 + 112 + 142 2 − (11.52 ) 25

=

3566 2 − (11.52 ) 25

=

9.9296

= 3.151

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K1 For the new x 2 and X

∑

N1

Additional Mathematics Paper 2

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Question

Submark

Solution and marking scheme

6(a)

(i)

uuur uuur uuur PR = PO + OR = −6a + 15b % %

Full Mark

K1 N1

Use

uuur uuur uuur PR = PO + OR

uuur uuur uuur (ii) OQ = OP + PQ

uuur oruuur uuur OQ = OP + PQ

3 uuur = 6a + OR % 5 N1

= 6a + 9b % %

3

(b)

uuur uuur (i) OS = hOQ

= h(6a + 9b) % % uuur uuur uuur (ii) OS = OP + PS

N1

uuur = 6a + k PR %

= 6a + k ( −6a + 15b ) % % %

N1

h(6a + 9b) = 6a + k ( −6a + 15b ) % % % % %

6h = 6 − 6k

9h = 15k

h = 1− k

5 h= k 3

K1

5 k = 1− k 3 k=

3 8

N1

5⎛3⎞ 5 h= ⎜ ⎟ = 3⎝8⎠ 8

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Equate coefficient of a or b % and % Eliminate h or k

N1

Additional Mathematics Paper 2

5

8

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Qn. 7(a)

log y =

log x log y

(b)(i)

(ii)

3472/2

Submark

Solution and Marking Scheme 1 1 log x + log 4k n n

P1

0.18

0.30

0.40

0.60

0.74

N1

0.48

0.54

0.59

0.69

0.76

N1

gradient 1 = n

n=2

K1

N1

Full Mark

K1

Correct axes and scale

N1

All points plotted correctly

N1

Line of best-fit

K1

N1

6

intercept 1 = log 4k n

k = 1.51

Additional Mathematics Paper 2

4

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log 10 y Graph of log10 y against log10 x 0.9

0.8 •

0.7

•

0.6

• •

0.5

•

0.4 0.39 0.3

0.2

0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

log 10 x

3472/2

Additional Mathematics Paper 2

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Qn. 8. (a)

Use ∫ ( y 2 − y1 ) dx

∫ (x + 4 −

Full Mark

K1

Solving simultaneous equation

P(– 2, 2)

(b)

Submark

Solution and Marking Scheme

N1

Q(4, 8)

N1

3

K1

x2 ) dx 2

Use correct

K1

Integrate ∫ ( y 2 − y1 ) dx

4

K1

x2 x3 + 4x − 2 6

limit ∫ into −2

2

x x3 + 4x − 2 6

18

N1

Note : If use area of trapezium and

4

∫

ydx , give the marks

accordingly.

(c)

Integrate π ∫ (

x2 2 ) dx 2

K1

2

K1

⎡ x5 ⎤ =π ⎢ ⎥ ⎣ 20 ⎦

limit ∫ into 0

⎡ x5 ⎤ ⎥ ⎣ 20 ⎦

π⎢ 8 π 5

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Use correct

N1

Additional Mathematics Paper 2

3

10

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Qn.

Submark

Solution and Marking Scheme

Full Mark

9(a)

Equation of AD :

Use m = – 2 and find K1 equation of straight line

y – 6 = –2 ( x – 2 )

N1

y = –2x + 10 or equivalent

2 (b)

y = –2x + 10 and x – 2y = 0

D(4, 2)

(c)

p = 3 or C(8, 4)

N1

2

P1

Substitute (8, 4) into y = 3x + q

K1

q = – 20

(d)

Solving simultaneous equations

K1

N1

Area OABC = 4 × Δ OAD

3

K1

Using formula 1 0 4 2 0 Δ OAD = × 2 0 2 6 0

K1

40

Find area of triangle

N1

3 10

Alternative solution :

B(10, 10)

P1

Using formula 0 8 10 2 0 1 Area OABC = × 2 0 4 10 6 0

40

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K1

Find area of parallelogram

N1

Additional Mathematics Paper 2

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Qn. 10(a)

Submark

Solution and Marking Scheme 6⎞ ⎜ ⎟ or equivalent ⎝ 10 ⎠ K1 ∠ POQ = 2 × α

α = sin

−1 ⎛

Full Mark

K1

∠ POQ = 1.287 rad

N1

3

Alternative solution :

122 = 102 + 102 – 2(10)(10) cos ∠ POQ ⎛ 10 2 + 10 2 − 12 2 ∠POQ = cos −1 ⎜⎜ 2(10)(10) ⎝

⎞ ⎟ ⎟ ⎠

Using (2π – 1.287)

N1

K1 K1

Major arc PQ = 10 ( 2π – 1.287 ) = 49.96 cm

(c )

Use formula s = rθ

N1

3

1 2

Lsector = × (10 ) 2 × 1.287

Ltriangle =

Use cosine rule

N1

∠ POQ = 1.287 rad

(b)

K1

K1

1 × ( 10 ) 2 × sin 1.287 2

Using formula Lsector = 12 r2θ

K1 Using1 formula LΔ = 2 absin C K1

= 16.35 cm

2

N1

Lsector - LΔ

4

10

= 3.9149 cm

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Additional Mathematics Paper 2

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Qn. 11 (a)(i)

p=

3 ,p+q=1 5

Use P(X = r) = n Cr prqn–r , p+q=1

3 2 = 5C0( )0( )5 5 5

K1

= 0.01024

b)(i)

Full Mark

P1

P( X = 0 )

(ii)

Submark

Solution and Marking Scheme

N1

3

Using P( X ≥ 4 ) = P( X = 4 ) + P( X = 5 ) 3 2 3 = 5C4( )4( )1 + ( )5 5 5 5

K1

= 0.337

N1

2

P ( 30 ≤ X ≤ 60 ) 30 − 35 60 − 35 ≤Z≤ ) =P( 10 10

use

K1

Z=

X −μ

σ

Use P( – 0.5 ≤ Z ≤ 2.5 ) = 1 – P( Z ≥ 0.5 ) – P( Z ≥ 2.5 )

K1

= 1－ 0.30854 – 0.00621 = 0.68525 (ii)

N1

Number of pupils = P( X ≥ 60 ) × 483

3

3472/2

3

K1

N1

Additional Mathematics Paper 2

2

10

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Qn.

Submark

Solution and Marking Scheme

12.

Subst. t = 0 into

(a)

dv dt

K1

2

a= 15 – 6t

N1

15 ms-2

(b) Use

Full mark

dv = 0 and subst. t in v = 15t – 3t2 dt 5 [t = ] 2

K1

2 75 −1 3 ms /18 ms −1 4 4

Integrate s = ∫ v dt =

(c) Use s = 0

N1

15 2 3 t −t 2

K1

15 1 / 7 / 7.5 2 2

(d)

Subst. t = 3 or t = 4 in 15 s = t2 − t3 2

15

3472/2

N1

3

K1

K1

Note : If use

K1

1 2

S4 – S3

N1

4

∫3 vdt , give the marks accordingly. Additional Mathematics Paper 2

3 10

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Qn. 13. (a)

Solution and Marking Scheme

Use I =

P2007 × 100 P2005

x = 48.6 y = 135 z = 80

3

Value of m : 25, m, 80, 30 or equivalent 120 × 25+130m+135 × 80+139 × 30

P1

∑I W Use Iˆ = i i

∑ Wi 120×25+130m+135×80+139 × 30 132.1 = 135+m

150.00 x

100 132.1

K1

3

N1

(ii)

Full mark

K1

N2, 1, 0

(b) (i)

Submark

m =65

K1

2

N1

(iii)

I 08 / 05 = 132.1 + (132.1x0.3)

171.73

3472/2

RM 113.55

K1

N1

Additional Mathematics Paper 2

2 10

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Qn.

Submark

Solution and Marking Scheme

14. (a) (b)

x + y ≤ 80

or equivalent

N1

y ≤ 4x

or equivalent

N1

x + 4y ≥ 120

or equivalent

N1

Full mark

3

y 100

y = 4x 90

80

x + y = 80 x=30

70

(16,64) 60

50

40

30

20

x + 4y = 120 10

10

20

30

40

50

60

80

70

90

100

x

At least one straight line is drawn correctly from inequalities involving x and y K1

All the three straight lines are drawn correctly

Region is correctly shaded (c)

(i) (ii)

minimum = 23 (16,64)

N1

3

N1

N1

N1

Subst. point in the range

K1

in 20x + 40y RM2880

3472/2

N1

Additional Mathematics Paper 2

4

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Qn. 15. (a)

Submark

Solution and Marking Scheme Use area △= ½ ab sin c in △BCD 1 20 = × 9.3 × 6 × sin BCD 2

Full mark

K1

45o48’ / 45.8o N1

0.74545

4555

N1

(c)

Use cosine rule in ΔBCD BD2 = 9.32 + 62 − 2×9.3×6 cos ∠ 45°48’

K1

(b)

K1

Use sine rule in ΔBCD sin ∠CBD sin 45 o 48 ' = 9 .3 6.685 N1

(d)

4

6.685

94o 10’

2

4444 Obtain ∠ ADB by using K1 qqq11111aaaaaaaaaaaaaaaaaa 180o – 85o50’ – ∠14s4 BAD or equivalent 5.555555 5555

z5555555555555555 K1 Use area Δ ADB =

½ × 6.685 × 13 × sin∠ ADB

K1 Sum of area: 20 cm2 + ΔABD 555

102

N1

3472/2

58.82 cm2

Additional Mathematics Paper 2

4

10

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