Spm Addmath2 Ans (kedah)

  • Uploaded by: SimPor
  • 0
  • 0
  • November 2019
  • PDF

This document was uploaded by user and they confirmed that they have the permission to share it. If you are author or own the copyright of this book, please report to us by using this DMCA report form. Report DMCA


Overview

Download & View Spm Addmath2 Ans (kedah) as PDF for free.

More details

  • Words: 2,302
  • Pages: 16
j2k

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

[Lihat sebelah SULIT

j2k

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 −

3472/2

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

[Lihat sebelah SULIT

j2k

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

[Lihat sebelah SULIT

j2k

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



–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

[Lihat sebelah SULIT

j2k

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

3472/2

K1 For the new x 2 and X



N1

Additional Mathematics Paper 2

[Lihat sebelah SULIT

j2k

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

3472/2

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

N1

Additional Mathematics Paper 2

5

8

[Lihat sebelah SULIT

j2k

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

10

[Lihat sebelah SULIT

j2k

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

[Lihat sebelah SULIT

j2k

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

3472/2

Use correct

N1

Additional Mathematics Paper 2

3

10

[Lihat sebelah SULIT

j2k

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

3472/2

K1

Find area of parallelogram

N1

Additional Mathematics Paper 2

[Lihat sebelah SULIT

j2k

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

3472/2

Additional Mathematics Paper 2

[Lihat sebelah SULIT

j2k

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

[Lihat sebelah SULIT

j2k

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

[Lihat sebelah SULIT

j2k

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

[Lihat sebelah SULIT

j2k

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

10

[Lihat sebelah SULIT

j2k

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

[Lihat sebelah SULIT

Related Documents

Spm Addmath2 Ans (kedah)
November 2019 12
Spm Addmath2 (kedah)
November 2019 21
Spm Addmath2 Ans (perak)
November 2019 14
Spm Bi Ans (kedah)
November 2019 9
Spm Mm Ans (kedah)
November 2019 19
Spm Bc2 Ans (kedah)
November 2019 20

More Documents from "SimPor"