Transcript

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Nama Pelajar : ………………………………… Tingkatan 5 : …………………….

3472/2

Additional

Mathematics

August 2014

MODUL PENINGKATAN PRESTASI TINGKATAN 5

TAHUN 2014

ADDITIONAL MATHEMATICS

Paper 2

( MODUL 2 )

MARKING SCHEME

2

SULIT 3472/2

MARKING SCHEME

ADDITIONAL MATHEMATICS PAPER 2 2014

N0. SOLUTION MARKS

1 2 1x y or

1

2

xy

2 2 50x x 2 25x

5 5x x

5x and 5x (both)

2y and 3y (both)

P1

K1 Eliminate x/y

K1 Solve quadratic equation

N1

N1

5

2

(a)

(b)

(i)

(ii)

7

8 (1)(2)

128

T

10

10

(1)(2 1)

2 1

1023

1023 (3)(7)(5)

107415

S

V

1023 0.8

818.4

K1

N1

K1

K1

N1

K1

N1

7

3

(a)

(b)

y = x

draw the straight line y = x

Number of solutions = 3

P1 cos shape correct.

P1 Amplitude = 2 [ Maximum = 1

and Minimum = -1 ]

P1 1

12

cycle in 0 x or

N1 For equation

K1 Sketch the straight line

N1

6

-4

3

4

(a)

(b)

1010

100

x

x

2

2 2

2

4 1010

1160

x

x

10 3

2

6.5

mean

or 2

42

2

4

K1

N1

K1

N1

K1

N1 N1

7

5

(a)

(b)

5 125

55

3

5 5

3

5

3

log log 1

loglog 1

3

log log 3

log 3

125

K V

VK

K V

K

V

KV

i)

1( )2

1 13

2 8 2

4 24

x kf x

m

k

m m

m and k

ii)

1 1( ) 3

8 2

20

p

p

K1

K1

N1

K1

K1

N1

K1

N1

8

4

6

(a)

(b)

(3 1)(3 1)( ) 3 1

3 1

'( ) 3

x xf x x

x

f x

i)

2

2

3

2 (2) 3(2)

2

dykx x

dx

k

k

ii)

1

2

112 (2)

2

111

2

normalm

c

y x

K1

N1

K1

N1

P1

K1

N1

7

5

7

(a)

(b)

(c)

(i)

(ii)

(iii)

x

1 2 3 4 5 6

2

y

x 3.5 5.5 7.5 9.5 11.5 13.5

2

y

x

2

y

x= kx+

p

k

k = *gradient

k = 2.0

p

k= *y-intercept

p = 3.0

y = 40

N1 6 correct

values of 2

y

x

K1 Plot 2

y

x vs x.

Correct axes &

uniform scale

N1 6 points plotted

correctly

N1 Line of best-fit

P1

K1

N1

K1

N1

N1

10

1.5

0 x

6

N0. SOLUTION MARKS

8(a)

i)

ii)

iii)

b)

2(9 )

3TS x

2

PTQR

TR TP PQ QR

PS PT TS

MS MR RS

PS kMS

PS PT TS

6 8x y =

k =4

4PS MS and S is a common point or equivalent

2 21 3

6( ) 8( )2 4

PS

= 45

K1 (TS or QR )

N1

K1

N1

K1

N1

K1

N1

K1

N1

8 9 4

9 4

y x y

x y

3 42

9 43 4

2

32

2

TRx y

x yx y

xy

6 8x y

3( 2 )2

k x y

= 6x

= 4y

6 8x y

7

9

a)

b)

c)

2

2

3 4

3 4 0

( 1)( 4) 0

4, 16

(4,16)

x x

x x

x x

x y

K

4

2

y

x

2

2

0

23

0

3

2

(4)(2)

83

28

3

16

3

x dx

x

cm

16

2

4

162

4

2 2

1(4) (12)

3

642

16 464

2 2

256 1664

2 2

56

ydy

y

K1 for solving

quad.eqn.

N1

N1

K1 use area of

rectangle - ( )y dx

K1 integrate

correctly

and Sub.

the limit

correctly

N1

K1

K1 correct limit

K1 integrate

correctly

N1

10

Area B

Volume A

8

N0. SOLUTION MARKS

10

(a)

(b)

(c)

60o

1.047 rad

8(1.047)OBS or 8(2.095)BCS OR 8(3.142)ACS

= 8.38 = 16.76 = 25.14

Perimeter = 8.38+16.76+8 or Perimeter = 25.14 + 8

= 33.14 = 33.14

Area of OAB = 21(8) (1.047)

2

= 33.50 cm2

Area of triangle OAB = 21(8) sin 60

2

= 27.71

Area of the shaded region = 33.50 – 27.71

= 5.79 cm2

P1

N1

K1 Use s r

N1

K1

N1

K1 Use formula

21

2A r

K1

K1

N1

10

9

N0. SOLUTION MARKS

11

(a)

(i)

(ii)

(b)

(i)

(ii)

X= Students passed Mathematics

p = 0.85 , q = 1- 0.85 = 0.15 , n = 6

P(X =6) = 6 6 0

6(0.85) (0.15)c

=0.3772

P (Y≥2) = 1 – P(Y=0) – P(Y = 1)

Or = ( 2) ( 3) ......... ( 6)P Y P Y P Y

= 1 - 6 1 5

1(0.15) (0.85)c - 6 0 6

0(0.15) (0.85)c

=0.2235

µ= 52 , σ =10

P( 40 < X < 60 ) = P (40 52

10

< Z <

60 52

10

)

= P( -1.2 < Z < 0.8)

= 0.6731

n = 0.6731 x 500

n = 337

P1

K1 Use P ( X=r ) =

rnr

r

n qpC

N1

K1

N1

K1 Use Z =

X

K1

N1

K1

N1

10

10

N0. SOLUTION MARKS

12

(a)

(b)

(c)

(d)

6a k t

6(2.5) 0k

k = 15

2 315

2s t t

2 3150

2t t

t = 7.5 s

215 3 0t t

t = 5

Total distance

=

5 72 2

3 3

0 5

15 15

2 2

t tt t

d = 62.5 + 38

= 100.5

K1

N1

K1

K1

N1

K1

N1

K1 (for

Integration;

either one

and

substitute

the limit

5 7

0 5

or )

K1

(for use and

summation)

N1

10

11

N0. SOLUTION MARKS

13

(a)

(b)

(i)

(ii)

(iii)

(i)

KI (PetuaKosinus)

N1

K1 (Petua Sinus)

N1

K1 (GunaPetua Sinus)

K1

N1

K1

K1

N1

10

12

N0. SOLUTION MARKS

14

(a)

(b)

(c)

(i)

Lihat 45º

I2012/2010 =

= 124.64

I2014/2010 = 124.64

= 137.10

Q2014 =

= RM685.50

K1

N1

K1

N1

K1

K1

N1

K1

K1

N1

10

i)

13

N0. SOLUTION MARKS

15

(a)

(b)

(c)

i)

ii)

iii)

Sekurang-kurangnya 1 garislurusdilukis dengan betul yang

melibatanx dan y.

Semuagarislurusdilukisbetul.

Kawasandilorek dengan betul

i) BilanganmaksimumbantalA = 220

ii) Titikmaksimum (300, 200)

Keuntunganmaksimum;

k = RM8 400

N1

N1

N1

K1

K1

N1

N1

P1

K1

N1

10

END OF MARKING SCHEME

650

600

550

500

450

400

350

300

250

200

150

100

50

-50

-100

-400 -300 -200 -100 100 200 300 400 500 600 700 800 900 1000 1100 1200

220

(300, 200)

Type B

Type A


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