answer for section a

7/31/2019 Answer for Section A

1/17

MRMU

RULI

PREPARED BY MR MURULI1

1.

2.

0

y

5

4

3

2

1

1

2

3

4

x4 3 2 1 1 2 3 4 5

y = x

4

x +y = 2

y = 4

Ox

32 yx 4 xy

y

y = 3

1 ANSWER


2/17

MRMU

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3. (a) PR

(b) ( P Q ) R

4. (a)

F G

(b)

RP Q

F

G HG

F

G HG

RP

QQ


3/17


4/17

MRMU

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12)

3

6622

50205

104

1625

e

e

ed

ed

ed

2,3

2

1210

101210)3(4

into3Substitute

deissolutionThe

d

d

dd

e

5

13)

3

1

9

3

39

96

63

33

1

2

h

h

h

kh

kh

kh

7,3

1

7

61

63

13

into3

1Substitute

khissolutionThe

k

k

k

h

3

+

oPWV

PWV

PWV

PWV

43.63

)2(tan

2tan

5

10tan

1

p

vw

5cm

10cm

15.

OBHC

BHC

BHC

cmHC

58.52

)36.18

24(tan

36.18

24tan

36.18169

1

22

H

18.36 cm

B

C

24cm

4 ANSWER

14.


5/17

MRMU

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17.

0

1

62.22

)125(tan

12

5tan

PUS

PUS

PUS

U

SP

12cm

5cm

18.

0

1

13.53

)34(tan

3

4

6

8tan

AED

AED

AED

A

DE

8cm

6cm

19.

c

c

cxy

cmxy

Fmb

E

ya

10

46

2

)6,2(,2)(

)9,2(

95)2(2)(

5

102

1020

0,

102

102

x

x

x

yaxisxtheAt

xy

isFGlinestraighttheofequationthe

xy

5 ANSWER

16.

OTNU

TNU

TNU

93.61

)8

15(tan

8

15tan

1

T15 cm

N

U

8cm


6/17

MRMU

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20.

c

c

c

cmxy

FmGHEF EF

9

45

)4(15

)5,4(,1//

13

3m(b)

,3)(0G

3y

62y0xaxis,-yAt

6x2y(a)

GH

9int

9

9)0(

0,9

ercepty

y

y

xaxisyAtxy

isEFlinestraighttheofequationThe

21.

3

//

393

93)(

3

3

93

0,

93)(

MN

KL

m

MNKLif

mxy

yxb

xisML

linestraighttheofequtionthe

x

x

yaxisxat

yxa

5int

5

153

1530

0,

153

,

15

123

)4(33

erceptxThe

x

x

x

yaxisxAt

xy

MNlinestraighttheofequationThe

c

c

ccmxy


7/17

MRMU

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24. (a)

(b)

(c)

22.

64

3

6

4

3

8

6

08

612)()(

)0,5()(

xy

isQSlinestarighttheofequationThe

c

mib

Ra

8int

8

324

4

36

64

30

0,int)(

erceptxThe

x

x

x

x

yerceptxAtii

23.

c

c

c

cmxy

Bmb

k

k

k

ka

EF

1

12

)4(4

12

)2,4(,4

1)(

8

35

4

1

12

5

4

1

)6(6

5)(

4

14

1

14

10

0,int)(

14

1

x

x

x

yerceptxAtc

xy

isEFlinestraighttheofequationThe

True

Antecedent = Set P is a subset of set Q

Consequent : P Q = P

Conclusion : Point P(x, y) is not lies on the x-axis.

6 ANSWER


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MRMU

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25. (a) (i)

(ii)

(b)

(c)

26. (a)

(b)

(c)

(d)

27. (a)

(b)

(c)

28. (a)

(b)

(c)

(d)

If the interior angle of polygon is then the corresponding exterior angle

is ( 180 ).

2 5 5 2 >

-6 -11 3 then x > 9 (False)

Sum of 2 and 8 is 10 or product of 8 and 2 is 4.

Triangle PQR is an obtuse angled triangle.

Statement

A regular polygon with n sides has n axes of symmetry. , n = 3, 4, 5, ..

Statement

If A 510 is an number in standard form then 1 A < 10.

Some

Implication 1 : If Q is an empty set then n(Q) = 0.

Implication 2 : If n(Q) = 0 then Q is an empty set.

2ppp or 63 pp

Premise 1 : Ifx is a multiple of 8 then x is a multiple of 4.

False


10/17

MRMU

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32. (a) Perimeter of whole diagram

cm2

155

77

222

360

135714

7

222

360

4521

ABarcoflengthBCCDarcoflengthAOD

(b) The area of the shaded region

2

222

2

1115

4

11977

4

357

5.37

22

2

114

7

22

360

457

7

22

360

135

]sec[sec

cm

semicircleofAreaCODtorofAreaAOBtorofArea

33.

(a) Perimeter of the whole diagram

cm

STarcoflengthRSQRarcoflengthTOQ

3

264

3

177

3

12921

77

222

360

60714

7

222

360

12021

(b) Area of the shaded region

2

22

0

02

0

0

154

773

1205

3

225

77

22

2

114

7

22

360

1207

7

22

360

60

sec[sec

cm

semicircleofareaQORtorofAreaTOStortheofArea

7 ANSWER


11/17

MRMU

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34.

(a) perimeter of the whole diagram

3

157

3

2147

3

21421

77

222

360

120714

7

222

360

6021

SUarcoflengthRSPRarcoflengthUOTP

0

0

(b)

The area of the shaded region

2

2

0

02

0

02

0

0

45.81

217.213

151

3

2102

3

151

062.672

17

7

22

360

12014

7

22

360

607

7

22

360

120

secsecsec

cm

SOTofAreaPTStorofAreaPORtorofAreaUOStorofArea

O

S

T

h cm

060

7cm

hcm

h

h

062.6

60sin7

760sin

0

0

35.

(a) Perimeter of the whole diagram

cm

TUarcoflengthRTQRoflengthUOPQ

79.50

79.1191119

57

222

360

135914

7

222

360

4519

0

0

0

0

(b) Area of the shaded region

2

2

0

02

0

0

96.81

5.247746.29

772

114

7

22

360

455

7

22

360

135

secsec

cm

OPSofAreaQORtorofAreaUOTtorofArea


12/17

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36.

Volume of the solid

3

2

7

51265

4807

5785

20862

1205

7

22

2

1

cm

prismtriangleofvolumecylinderhalfVolume

37.Volume of remaining solid

3

22

43.423

57.10081.523

827

22205

7

22

3

1

cm

cylinderofvolumeconetheofvolume

8 ANSWER

38.

Volume of the solid = volume of prism + volume of half cylinder

2

2

735

231504

125.37

22

2

112775

2

1

cm

39.

Volume of half cylinder + volume of triangle prism = 801

cmp

p

pp

pp

pp

21

5607267

560716899

801247

99

801682

1

37

22

2

1 2


13/17

MRMU

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41.

42.

(a)

43.

44.

40.

Volume of the solid

3

2

6699

16175082

2177

22

2

1)21(113014

2

1

cm

cylinderhalfofvolumeprismtheofvolume

a) P( WW ) =14

5

15

6

=7

1

b) P( WW ) + P( CC ) + P( TT )

=

14

4

15

5

14

3

15

4

14

5

15

6

=105

31

P( ST ) =10

1

9

3

10

3

b) P( EE) + P(EE) =

132

43

33

4

44

9

12

2

11

8

12

9

11

3

b) P( II) + P( SS ) + P( TT )

=

9

2

10

3

9

2

10

3

9

1

10

2

=45

7

(a) P( EE ) =44

3

12

3

11

3

(a) P( NN ) =14

5

8

5

7

4 b) P( NA ) + P( AN ) =

6

4

7

3

8

3

7

4

=21

72

143

9 ANSWER


14/17

MRMU

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45 (a)

46. a)

25

13M

1MN

35

12

)5(6

1

35

12

11

1

By

pqNcomparison

5

121:

3p , 11q

P( EE ) =

5

1

5

2

6

3

b) P( HH ) + P( EE ) + P( MM )

105

3814

3

70

1

15

220

9

21

10

20

2

21

3

20

7

21

8

10 ANSWER

(b) 133 yx

425 yx

4

13

25

13

y

x

4

13

35

12

11

1

y

x

7722

111

yx

7

2

y

x

:solutionThe

7,2 yx


15/17

MRMU

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48 (a) 3k +16 = 43k = 12 k = 4

(b) 1M =

68

44

4

1

=

2

32

11

c)

24

23

q

p=

2

1

q

p=

68

44

4

1

2

1

q

p=

4

4

4

1

q

p=

1

1

The solution : p = 1 , q = 1

2

32

11

34

22

2

1

34

22

86

11M

Or

47 (a) Inverse matrix =

42

53

)10(12

1

=

42

53

2

1

=

212

5

2

3

By comparison,

h =2

3 2

2

k

4k

(b)

32

54

y

x=

43

y

x=

2

1

42

53

4

3

y

x=

10

11

2

1

y

x=

5

2

11

The solution: x =2

11 , y = 5


16/17

MRMU

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49 (a) 12 0)( k

12 + k = 0

k = 12

(b) P =

31

84

(i)1

P =)8(12

1

41

83

=

41

83

20

1

=

5

1

20

15

2

20

3

ii)

31

84

6

4

y

x

6

4

41

83

20

1

y

x

20

60

20

1

y

x

1

3

y

x

The Solution : x = 3, y = 1

11 ANSWER

50.

2

1-

3

10

3

1828)(

18msu

724u(a)

ms

speedofchangeofRateb

ondT

T

T

T

Tc

sec13

18214

153981469

153)7(1469

1532872

132818

2

1)(

51.

2

2

5

4

10

04

1424)(

ms

speedofchangeofRatea

ondu

u

uu

uu

uub

sec18

1448

1865342

932

5

2

321

9352

1)3(14

2

1)(


17/17

MRMU

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52.

(a) m80810

(b)

2

24

8

ms

3

113

3

40

403

4047

8032475616

80882

1148

2

184

2

1)(

u

u

u

uu

uv

vvc

53.

(a) 6 second

(b)

2

2

13

4

317

ms

speedofchangeofRate

st

t

t

t

tc

14

410

94102

47

236)10(2

4710240

236)10(30172

11764173

2

1)(

answer for section a

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