Add Maths Perfect Score Module Form 4 Marking Scheme Set 1 & Set 2

8/8/2019 Add Maths Perfect Score Module Form 4 Marking Scheme Set 1 & Set 2

1/19

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PERFECT SCORE 2010

FORM 4

ADDITIONAL MATHEMATICS

SET 1 PAPER 1 & 2

SET 2 PAPER 1 & 2


2/19

2

No. Answer

1 (a) k= 14

(b) 2 2f : x x or

22 f ( x ) x

2(a) h = - 3 (b)

11

5

3 (a)p = 9, q = -20 (b) -29

4(a)

2

3x (b)

11

3, x

5 8 12, m

6 k= -2 , p = 4

7 55 2

4, , r q p

8 3x

9 27

10

14 3

2

a b

11r =

3 15

7 2, t

12 2 5(a) (b) 8

5 2t y x

13 2 2 9 6 26 0 x y x y

14 04546210255 22 yxyx

15 (a) 30 (b) 6.21

16 (a) 640 (b) 12

17 a) 0.7855 (b) r = 5

18 Perimeter = 12.0263

19 Area = 2.045

20(a)

2

1(b) 6

21 1x

22

2

1p and 4q

23(a)

2

3,1 (b)

,032

2

dx

ydminimum point

24 (a) 5 (b) 245

5)5(10 2 newV

253

12(a)

(3 2)x (b) -24


3/19

Number Solution and mark scheme Sub Marks Full Marks

1

7,2

6,2

3

0)6)(32(

018152

7)52()52(

52

2

2

x

y

yy

yy

yyyyx

P1

K1

K1

N1,

N1

5

2 (a)

7,2,1

7)2()(

32)2()(

2

22

knm

xxf

xxf

K1

N 0 ,1 ,2

6

(b) )7,2( minimum point N1

(c)

Shape

Minimum point and y-intercept

P1

P1


3 (a)

2

9

0)94)(1(4604

0946

024)3(

2

2

2

22

p

pacb

pxx

xpx

K1

K1

N1 3

y

xo

-3

(-2,-7)


4/19

4


(b)

2

5,3

05622

pqqp

xx

0122

2

1)2)(2(POR,14

2

xx

qpqpSOR

K1

K1k1

N1 4

7

4 (a)

2031.0

8.0lg3lg

8.03

8)33(9

x

x

x

xx

K1

K1

N1 3

7

(b)

2

3221

x

x

K1

N1 2

(c)

2

1

223

2

2)23(log2log

2

22

x

x

x

xx

K1K1

N1 3


5 (a)i) 9

6

54

ii)

5.523

96

5.2

2

2

2

2

x

x

K1 N1

K1

N1 4

8

(b) New mean = kh

95 ,

New standard deviations=h

5.225.1

2

1,2 kh

K1

K1

N1 N1

4


5/19

5


6 (a)

2)2)(

2

1(2

2

1

321

n

m

mx

n = 4

K1K1

N1

N1 4

7(b)

194

)3(4

14

4

1

xy

xy

Gradient

K1

K1

N1 3

7

(a)

4

3

441

8

5

412

1

n

n

m

x

K1

N1

P1 3

10

(b)

x

xxg

2

1)(1

x

xxhg2

141)(1

0,23

)(1

x

x

xxhg

K1

K1

N1 3

(c)

2

9

5

2

1

24

522)21(22

p

pk

k

k

xxpxk

K1

N1

K1

N1 4


6/19

6


8 (a) y=1

P1

10

(b)

)1,3(

3

521

B

x

x

K1

N1 2

(c)

)5,5(

52

152

3122

)3(2

16

2

1

C

xy

xy

xy

xy

m

N1 1

(d)

Area of quadrilateralABCD

16511

1-3531

2

1

2unit15

K1K1

N1 3

9 (a)

50

)1042()1337()1632()727()422( XXXXXMean

= 33.8

K1K1

N1 3

10

(b)

896.5

)8.33(

50

1042()1337()1632()727()422(

2

222222

XXXXX

K1 K1K1

N1 4

(c)

54.38513

27)50(4

3

5.34

x

K1, K1

N1


10 (a)

radCOB

COB

8961.0

16

20tan

K1

N1 2


7/19

7

(b)

0256.86

8961.082

18961.016

2

1 22

xxxxArea

K1K1K1

N1 4

10(c)

9672.65

81620

)8961.0142.3(881620

22

Perimeter

P1

K1

N1

N1 4


11 (a)

)12,2int(max

6

)12,2(int

122412

)2(12)2(3

2

0126

123

2

2

2

2

po

dx

yd

poTurning

y

x

x

xxy

K1

K1

N1

N1

4

(b)0.1

dx

dt

6.0

)1.0)(1236(

x

dt

dx

dx

dy

dt

dy

P1

K1

N1 3

(c)

6

6

)52(42

)2(21

x

u

dx

dx

du

dy

dx

dy

1.2)05.0(]5)2(2[42 6 XXx

K1

K1N1 310


8/19

8


12 (a)

9

40

11

SinSinC

02211.128

7789.51180

7789.519

4011

C

ACB

C

SinSinC

K1

N1

K1N1 4

10

(b) CD2=9

2+9

2-2(9)(9)cos76.4422

CD=11.1366 K1

K1

N1 3

(c)

48.492211.88sin9112

1

2211.88

4422.76

7789.11221112840180

xxArea

BAD

CAD

BAC

N1

K1

N1 3


9/19

9


13 (a)

681.8

60sin

8

70sin 00

BC

BC

K1

N1 3

10

(b)

22 681.815 CV

= 17.3309

0764.750sin60sin

8 00

xAB

5854.160764.715 22 AV

.894.38

)5854.16)(3307.17(285854.163307.17cos

0

222

AVC

AVC

24.90894.38sin5854.163309.172

1 0 XXXArea

P1

K1

K1

K1

N1

K1K1N1


10/19

10


14 (a)

(i)

(ii)50.2

10050.3

140

rmx

Xx

121

100

110110

x

Xx

K1

N1

N1 3

10

(b)

120

1.126100

)35()31140()12110()22125(

n

nXXXX

K1

K1N1

3

76.27

1.12610035

x

Xx

K1

K1

K1

N1 4

Number Solution and marking scheme Sub Marks Full Marks

15(a)

(b)

25.1311003242 m

95.12210

)2(25.131)4(118)3(125)1(120

I

K1N1

K1K1

N1

5

10

(c)

(d)27.1053

95.1221001295

x

x

1875.16100

12595.122

Xx

K1, K1

N1

K1N1

5


11/19

11

No. Answer

1(a)a= 6, b=10 (b)

10

6

y(c)

18

20y

2(a) h = - 3 (b)

11

5

3(a)

11

2(b)

1

2

4(a)

3

2h (b)p =12

5 = 16 (a) p = 5 , h = 2, k = 3 (b) y = 5(x 2)23

7 k = 3 , 1 and r =3 , 5

8

5

2x

9 116

16,

10 2x 11 t = 0 , -22

12 (a) R = (9 , 16) ) (b) y = 2x + 34

13 2 2 4 6 8 0 x y x y 14 04546210255 22 yxyx

15 (a)(i) 2640 (ii)175680 (b)Mean = 71.375 , S Deviation= 7.792

16 (a) 24 cm (b) 53.60

17 43.80

18 3p

19 2.044

20 4

4 3 7 21 14 x x x

21 1 6.

22 1 367 4 355. , . 23

(a)1

62

,

(b) minimum point

24(a)

3

2

1

2x

(b) 0.2002

25 4 2x


12/19

12

No. Marking scheme Marks Total1

2

2

2 2

2 2 (2 2 ) 10

4 11 6 0

( 11) 185

8

3.08, 0.33

4.16,2.66

y x

x x x

x x

x

x

y

1

1

1

1

1

5

2 a)

x = 3 and x =2

5

x - 3 = 0 and 5x + 2 = 0

(x - 3)(5x + 2) = 02

5 6 0x x

b)2

2

2 2

4 0

(1 2 ) 4( )( 3) 0

4 4 12 0

16 1

1

16

b ac

k k k

k k k

k

k

1

1

1

1

4

3 (a) (i) babxaxf 22 )(

Compare to2( ) 9 16 f x x

2 9 , 16

3

4

a ab b

a

b

(ii)4(3 4) 3(2 5) 4

1

2

x x

x

(b)Letx

y1

,y

x1

1

1

11

1

1

1

8


13/19

13

2

2

3 14 ( ) ( )

( )1

y yg y

y

2( ) 4 3 1g x x x

1

4 f(x) = 2x2 - 4x + 5

=

2

2

2

2( 2 ) 5

2 ( 1) 1 5

2( 1) 3

x x

x

x

(b) Maximum value = 3

(c)

21

3

-2 1 4

1

1

1

36

5

(a) Class median = 5059

2610

249.5 (10) 54.5

12

m

m = 6

1

1,1

1 9


14/19

14

(b)

Mode = 5 5.0

Mean,

New mean = 3

=

1

1

11

1

6 a) OX: OA = 3 : 4

OX= 3 124

= 9 cm

b)9

XY= sin

3

XY= 7.794 cm

9

OY= cos

3

OY= 4.5 cmYB = 124.5 = 7.5 cm

Length of arcAB, = 12 3

= 12.568cm

Perimeter of the shaded region= 3 + 7.794 + 7.5 + 12.568 =30.762cm

11

1

1

1

10


15/19

15

c) Area of shaded region = Area of sector OBCArea

of triangle OXY

=

2

112

3

-

2

14.5 7.794

= cm

2

2

1

7 a) i) The gradient of QR is 1

Straight lines PQ and QR are perpendicular to each other.Hence, the gradient of PQ is -1.

5 1[ 3]y x

y= -x +2

ii)2...(1)

...(2)

y x

y x

y = 1 and x = 1

The coordinates of point Q are (1, 1)

b)4 3( 3) 4 3(5)

, ( 1, 1)7 7

x y

Compare4 3( 3) 15 4

1@ 17 7

x y

The coordinates of S are (4, -2)

c)

2 2

2 2

2 2

2 2

( 3) 5 4

3 5 16

6 9 10 25 16

6 10 18 0

x y

x y

x x y y

x y x y

1

1

1

1

1

1

1

1

1

1

10

8 (a) '394oQOR

(b)Perimeter

= '0 394cos200200()5.1(10 )

=16.42+14.63

=31.05cm

(c)Area

=2

'022

2.32

394sin102

1)5.1(10

2

1

cm

(d)Shortest distance

1

1,11

1

1,11

1

10


16/19

16

10 a) n = 10

b) Mean =

=

c)

Marks x f fx fx

2

3539 37 5

4044 42 12

4549 47 17

5054 52 10

5559 57 6

f =

50

fx = fx2 =

1

1

1

11

= 10cos '''0 30147

= 6.817cm

1

9 a)

b)

3 27

9

log 25 log 3

log 5

=10 10 10

10 10 10

log 25 log 3 log 9

log 3 log 27 log 5

= 10 10

10 10

2 log 5 log 3 2log3 4

log 3 3log 3 log5 3

i)

2

3

3 3 3

log81

2 log log 4 log 3

2 4

m n

m n

r t

=

ii) 9 27 9 27log log log 3 log 3r tm n

3 3

3 3

3 3

3 3

log 3 log 3

log 9 log 27

log 3 log 3

2log 3 3log 3

r t

r t

=2 3

r t

1

1

1

1

11

1

1

1

1

10


17/19

17

Standard deviation

22

xf

fx

= 25.5450

5.155112

=

d) Median

= cf

FN

Lm

2

1

= 1017

17)50(2

1

5.49

= 54.206

1,1

1

1

1

10

11 a)

2

2 1

4

1

4

3

4(2 1)

4(2 1) 4 (2)(2 1) (2)

(2 1)

4(2 1) [2 1 16 ]

(2 1)

4(1 14 )

(2 1)

d xdx x

x x x

x

x x x

x

x

x

b i) 23 12dy

xdx

20, 3 12 0dy xdx

( 2)( 2) 0x x

2, 2x 3

3

2, ( 2) 12( 2) 5 21

2, (2) 12(2) 5 11

x y

x y

Turning points are : (-2, -21) and (2, 11)

1

1

1

1

1

2

10


18/19

18

ii)2

26

d yx

dx

2

2

2

2

2, 6( 2) 12

2, 6(2) 12

d yx

dx

d yxdx

(2, 11) is a maximum point.

1

1

1

13a)

1.44100

1.20

120

P x

18.00100 150

.00

xq

q

100 1402.00

r x

r

b)

c)

1

1

1

1

11

1

2

1

10

5.137

100

125110

10002

06

06

07

02

07

Q

Q

Q

Q

Q

Q

139

10

1390

1342

)1(120)3(140)4(150)2(125

I


19/19

19

14 a) 2 2 29 12 7.5 (2 12 7.5)cos ACB

ACB = 051.48

b) 000 6051.48180 DEC 049.71

0 0

4

sin 48.51 sin 71.49

DE

DE cm

c) Area of 01

7.5 12 sin 48.512

ABC

cm2

Area of

1

122 ABC t

cm

1

11

1

1

1

1

1

1

1

10

15 a) 2 2 2 0(7) (8) 2(7)(8) cos150AC

14.491AC cm

b)0sin sin 40

14.491 11

014.491sin 40sin

11

c) area =1

14.491 11 sin(180 40 )2

x x x

114.491

2t

t cm

11

1

1

1

1

11

1

1

10

Add Maths Perfect Score Module Form 4 Marking Scheme Set 1 & Set 2

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