Add Maths Trial Exam P2 Marking Scheme 2013 set B.pdf

7/27/2019 Add Maths Trial Exam P2 Marking Scheme 2013 set B.pdf

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Nama Pelajar : Tingkatan 5 : .

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Additional

Mathematics

Sept 2013

PROGRAM PENINGKATAN PRESTASI AKADEMIK SPM 2013

ADDITIONAL MATHEMATICS

Paper 2

(SET B)

.

MARKING SCHEME

SULIT 3472/2


2/13

3

MARKING SCHEME

ADDITIONAL MATHEMATICS PAPER 2

N0. SOLUTION MARKS

1 3x y

or 3y x

2 2(3 ) (3 ) 6y y y 2(3 ) ( 1) 6x x x 22 7 6 0y y 22 5 3 0x x

(2 9)( 1) 0y y (2 3)( 1) 0x x

3

2x and 1x (both)

3

2y and 2y (both

P1K1 Eliminate x/y

K1 Solve quadratic equation

N1

N1

5

2

(a)

(b)

2

2

[ ( )] [(2 1) 4]

(2 1) 4

3, 12 11

1

f g x f x

m x m n

m n

n

( ) 5

1( )5

3, 5

f x p x y

x p y

p q

K1(find composite function)

N1

N1

K1(find inverse function)

N1N1

7

3

(a)

(b)

y =x

draw the straight line y =x

Number of solutions = 4

P1 cos shape correct.

P1 Amplitude = 6 [ Maximum = 3

and Minimum = -3 ]

P1 2 full cycle in 0 x 2 orP1 reflection the graph

N1 For equation

K1 Sketch the straight line

N1

6

2

3

-3


3/13

4

4

(a)

(i)

(ii)

(iii)

(b)

2( )

3

4 2

OQ OC CR

x y

1( )

2

3 2

BQ BO OC

x y

5

BR BQ QR

x y

.

BR hOC

cannot find h

not parallel

K1

N1

K1

N1

K1

N1

K1 find h

N1

8

5

(a)

(b)

i)6

10

60

x

x

ii)

22 2

2

7 610

850

x

x

new mean 3(6) 5 23

new standard deviation 3(7) 21

P1

K1

N1

K1 N1

K1 N1

7


4/13

5

6

(a)

(b)

2 2 21 1 1, , ,...2 8 32

p p p

2 2

2 2

1 18 321 1

2 8

p p

p p

1

4r

(i)

11 253200( )

4 128

8

n

n

(ii)

3200

114

24266

3

S

K1

K1

N1

K1K1

N1

K1

N1

8


5/13

6

7

(a)

(b)

(c)

(i)

(ii)

(iii)

x 1 2 3 4 5 6

10log y 0.26 0.41 0.53 0.66 0.80 0.94

10 10 10log log logy x h k h

10log h = *gradient

h = 1.37

10logk h = *y-intercept

= 0.88

y = 3.98

N1 6 correct

values of log y

K1 Plot10

log y vs

x.

Correct axes &

uniform scale

N1 6 points plotted

correctly

N1 Line of best-fit

P1

K1

N1

K1

N1

N1

10

10log y

0.12

0x


6/13

7

N0. SOLUTION MARKS

8

(a)

(b)

(c)

21(6) 26.28

2

= 1.46 rad

6(1.46)AD

S = 8.76 cm

Perimeter = 8.76 + 5 + 5 + 8

= 26.76 cm

Area of triangle = 17.89 cm2

Area of rectangle = 40

Area of the shaded region = 17.89 + 4026.28= 31.61 cm

2

K1

N1

K1 Use s r

N1

K1

N1

K1

K1

K1

N1

10


7/13

8

N0. SOLUTION MARKS

9(a)

(b)

c)

2

216

xx

dx

dy

At turning point, (p, 4) ,0

dx

dy

02

162

p

p

16p3

= 2

8

13 p ,2

1p

y = 16x 2x-2 dx

cx

x

2

2

16 2

= cx

x 2

8 2

At point A (2

1, 4 ) ,

4 = 8( )2

+ 4 + c

c = -2

The equation of the curve is , 22

82

xxy

32

24

16xdx

yd

At point A(2

1, 4) , 048)8(416

2

2

dx

yd

Thus,

4,

2

1A is a minimum point

P1

K1

N1

K1 use

dxyy )( 12

K1 integrate

correctly

K1

N1

K1

K1

N1

10


8/13

9

N0. SOLUTION MARKS

10

(a)

(b)

(c)

(d)

1

68

yx

Thus,A( 0,6) and B( 8, 0)

LetD (x,y)

CD : DE= 1:3

x =

84

32

4

10321

y =

74

28

4

10321

ThusD 7,8

4

3

8

6

ABm

Thus ,

3

4

CEm

y - 6 = 03

4x

y = 63

4

x

Area ofAOB=

6706

0880

2

1

28562

1

N1N1

K1

K1

N1

K1

N1

K1 (use area formula)

K1N1

10


9/13

10

N0. SOLUTION MARKS

11

(a)

(i)

(ii)

(b)

(i)

(ii)

X = Students have a laptop

p =10

7= 0.7 , n = 5

q = 3.010

3

10

71

P(X= 5) = 5055 7.03.07.05 c

= 0.1681

P( X< 3) = P( X =0) + P( X = 1) + P( X= 2)

= 32

2

41

1

50

0 3.07.053.07.053.07.05 ccc

= 0.1631

X =weights of pencil box

X 20,250N

P(X>265) =

20

250265ZP

=P(Z > 0.75)

= 0.2266

P(X < w ) = 0.3

P(Z > 20

250w= 0.3

From table,

20

250w= -0.524

W= 239.52 g

K1 Use P (X=r ) =

rnr

r

n qpC

N1

K1

K1 Use P (X=r ) =

rnr

r

n qpC

N1

K1 Use Z =

X

N1

K1

K1

N1

10


10/13

11

N0. SOLUTION MARKS

12 (a)

(b)

(c)

(d)

vinitial= 5 m s-1

v= t26t +5

dt

dva

=2t -6

a = 0

t = 3

v = (3)2-6(3) + 5

= -4 ms-1

v < 0

t26t+5 < 0(t-1) (t-5) < 0

The range of values of t

1 < t < 5

5

1

23

1

0

23

5

1

2

1

0

2

1

0

5

1

533

533

)56()56(

ttt

ttt

dtttdttt

vdtvdt

)1(5)1(33

1)5(5)5(3

3

50)1(5)1(3

3

12

3

2

3

2

3

=13

N1

K1

K1

N1

K1

N1

K1 for

1

0

vdt

K1 (for

Integration;

either one)

K1 (for use and

summation)

N1

10


11/13

12

N0. SOLUTION MARKS

13 (a)

i)

ii)

iii)

(b)

i)

(ii)

oSU 65cos)7)(8(2)7()8( 222

SU = 8.10 cm

Area of triangle SVU= O65sin)8)(7(2

1

= 25.38 cm2

SUTo

sin

6

75sin

1.8

Sin SUT= 0.7155

SUT=45.680

TSU=180O-75

O-45.68

O=59.32

O

STS=61.36O

oo SinSin

SS

32.59

6

36.61

'

= 6.123cm

K1

N1

K1

N1

K1

N1

P1

P1

(for 59.32o

OR

61.36o

K1

N1

10

T S

S

U

6 cm


12/13

13

N0. SOLUTION MARKS

14

(a)

(b)

(c)

i)

ii)

x + y 5

10x +20y 150 or x + 2y 15

x 2y or y 2

x

At least one straight line is drawn correctly from inequalities involvingx and y.

All the three straight lines are drawn correctly. Region is correctly shaded.

3 kg

Minimum point (5,0)

Minimum cost = 10(5) + 20(0)

= RM 50

N1

N1

N1

K1

N1

N1

N1

N1

K1

N1

10

R

7.5

5

5 15


13/13

14

N0. SOLUTION MARKS

15 (a)

i)

ii)

(b) (i)

(ii)

1005.162

20.511 xp

P11 = RM 3.20

100140

15012

13 xI

= 107.14

30+10+20+40 ( can be seen )

14540201030

)40135()20()10150()305.162(

XxXXX

X = 136.25

The usage of 136.25

56100

25.13613 xp

= RM 76.3

K1

N1

K1

N1

P1

K1

N1

P1

K1

N1

10

END OF MARKING SCHEME

Add Maths Trial Exam P2 Marking Scheme 2013 set B.pdf

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