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Preface Page No.

1 . Trigonometric ratio identities & Equations

Exercise 01 - 27

2 . Fundamentals of Mathematics - II

Exercise 28 - 38

3 . Straight Line

Exercise 39 - 70

4 . Circle

Exercise 70 - 92

5 . Mathematical Reasoning, Induction & Statistics

Exercise 92 - 101

6 . Solution of Triangle

Exercise 101 - 125

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MATHEMATICSCLASS : XI

CONTENT

RESONANCE SOLUTIONS (XI) # 1

TRIGONOMETRIC RATIO, IDENTITIES & EQUATIONSEXERCISE # 1

PART - ISection (A) :

A-2. c = 180º

A-4. (a) 3 + 2 + 3 × 31

= 6

(b) 2 × 21

+ 2 × 21

+ 2 × 4 = 10

(c)51

+ 0 = 51

A-6.

2cot)sin(�sin

cos)cos(�

2cot)sin(�sin

cos)cos(�

A-9. tan = � 125

23

< < 2

sin = � 135

and cot = �5

12

LHS =

eccos�eccos�

cot�sin� =

eccos2cotsin

=

513

2�

512

�135

�

= 338181

= RHS

Section (B) :B-4. LHS = cos2 + cos ( + ) { cos cos � sin sin � 2 cos cos }

= cos2 � cos ( + ) . cos ( � ) = cos2 � cos2 + sin2 = sin2 = RHS

B-6. (i)BcosBsinAcosAsin

BsinAsin 22

=

B2sin21

A2sin21

)BAsin()BAsin(

= )BAsin()BAcos(2

)BAsin()BAsin(2

= tan (A + B)

(ii) cot (A + 15º) � tan (A � 15º) = )º15Asin()º15Acos(

� )º15Acos(

)º15�Asin(

= )º15Acos()º15Asin()º15Asin()º15Asin()º15Acos()º15Acos(

=

)º30sinA2(sin21

)º15sinA(sin)º15sinA(cos 2222

=

21

A2sin

A2cos2

= 1A2sin2

A2cos4

B-7 A + B = 45° tan(A + B) = tan(45º)

BtanAtan�1BtanAtan

= 1 tanA + tanB + tanA tanB = 1

(1 + tanA) (1 + tanB) = 2 put A = B = 2º1

22

(1 + tan 2º1

22 )2 = 2 tan 2º1

22 = 1�2


Section (C) :

C-1. LHS =

4cot2

cos

4�

tan1

4�

tan�1�

2

2

sec 2

9

=

4cot

2cos

2�

cos� sec 2

9 =

4sin

4cos2

cos

2sin� sec

29

= 4sin

1

2sin4sin�

2cos4cos sec

29

= 4sin

1 × cos

29

. sec 2

9 = cosec 4 = RHS

C-3. (ii)AsinAcosAsinAcos

�

AsinAcosAsinAcos

=

Asin�Acos

AcosAsin422 =

A2cosA2sin2

= 2 tan 2A

C-9 tan tan(60° + ) tan(60° � ) = tan 3

LHS = tan

tan3�1

tan3

tan31

tan�3

= tan

2

2

tan3�1

tan�3=

3tan

tan3�1

tan�tan32

3

Put = 20° tan 20°

tan 20° tan80° tan 40° = tan60° = 3

Section (D) :

D-1. Let y = cosx .cos

x

3

2cosx

3

2

y =

x2cos

34

cosxcos21

y =

2x2cos21

xcos21

y = xcosxcosx2cos241

y = 41

[cos 3x + cos x � cosx]

y = 41

cos 3x �1 cos 3x 1

ymin = � 41

and ymax = 41

D-3. (i) y = 10 cos2x � 6 sin cosx + 2 sin2x= 5 (1 + cos 2x) � 3 sin 2x + 1 � cos 2x

= 4 cos 2x � 3 sin 2x + 6 � 22 ba a cos + b sin 22 ba

ymax

= 5 + 6 = 11y

min = � 5 + 6 = 1

(ii) y = 1 + 2 sinx + 3 cos2xy = 1 + 2sinx + 3 � 3 sin2 xy = 1 � (3 sin2x � 2 sinx � 3)


y = 1 � 3 (sin2x � 32

sinx + 91

� 91

� 1)

y = 1 � 3

910

�31

�xsin2

= � 3

2

3

1�xsin

+

313

ymax

= 3

13 , y

min = � 3

916

+ 3

13 = � 1

(iii) y = 3 cos

3 + 5 cos + 3

y = 3 cos . 21

� 3 23

sin + 5 cos + 3

y = cos23

� sin233

+ 5 cos + 3

y = cos2

13 �

233

sin + 3

ymax

= 4

274

169 + 3 = 7 + 3 = 10

ymin

= � 4

274

169 + 3 = � 7 + 3 = � 4

Section (E) :

E-2. (i)AsinAcos

AsecAsinecAcosAcos

=

AsinAcosAsin�Acos

)AsinA(cosAcosAsinAsin�Acos 22

(ii) cos

1�

tan�sec1

=

cos)sin�1(sin1�cos

cos1

�sin�1

coscos�1

cos 2

=

cossin

cos)sin�1(sin�sin 2

1sincos

�cos

1

=

cossin

cos)sin1()sin1(sin

cos)sin1(cos�1sin 2

(iii)Asin�AcosAsin�cos

AsinAcosAsinAcos 3333

cos2 A + sin2A � sinA cosA + cos2A + sin2 A + sin A cosA = 2

Section (F) :

F-1. (i) LHS = cos 72

cos 74

cos 76

= � cos 7

cos 72

cos 74

= �

7sin.2

78

sin

3

= 81

= RHS

(ii) LHS = cos 11

cos 112

cos 113

. cos 114

. cos 115

= cos 11

cos 112

cos 114

cos 118

cos 11

16

=

11sin.2

1132

sin

5

=

11sin.32

113sin

= 321

= RHS


F-2. LHS = sin2 + sin2 2 + sin2 3 + ....... + sin2 n

=

2

n2cos1..............

2

4cos1

2

2cos1

= 21

2n [(cos2 + cos4 + cos6 + ........+ cos 2n)]

= 21

2n

2n22

cos.

22

sin

2)2(n

sin

= 21

2n

sin)1ncos(.nsin

= RHS

F-7. cos (S � A) + cos (S � B) + cos (S � C) + cos S

= 2 cos

2BAS2

cos

2AB

+ 2 cos

2CS2

cos

2C

= 2 cos

2

C cos

2

AB + 2 cos

2

BA cos

2C

= 2 cos 2C

2B

cos2A

cos2

Section (G) :

G-4. sin 2 = cos 3 cos

2

2 = cos 3

2

� 2 = 2n ± 3 2

� 2 ± 3 = 2n

= 2n � 2

, 5

n22

= 2n � 2

, 5

n2

21

G-8. tan 2 tan = 1 sin 2 sin = cos 2 cos

0 = cos 3 3 = (2n + 1)2

= (2n + 1)6

.

Section (H) :

H-4. cos2 x + cos2 2 x + cos2 3 x = 1

2

x2cos1+

2x4cos1

+ 2

x6cos1 = 1 cos2x + cos4x + cos6x = � 1

2cos4x cos2x = �2cos22x cos2x = 0 or cos4x + cos2x = 0

2x = (2n +1) 2

or 2cos3x cosx = 0 4

)1n2(x

, (2n + 1) 6

, (2n + 1) 2

Now x = (2n + 1)6

= 3

n +

6

may also be written as

x = (3k + 1)3

+6

,(3k + 2)3

+6

, (3k)3

+6

= 2

k

, 6

5k

,

6k

= (k + 1) 6

, 6

k

( 2

k

is same as (2n + 1) 2

) = 6

m


H-5. sin2n � sin2(n � 1) = sin2

sin(2n � 1) sin = sin2 sin = 0 or sin (2n � 1) = sin

= m , sin(2n � 1) � sin = 0 2

)2n2(sinncos2

0

2

)1p2(n

, )1n( m ,1n

,

21

p n

Section (I) :

I-1. tan2 � (1 + 3 ) tan + 3 = 0 tan = 1, 3 = n + 4

, n + 3

.

I-3. 4 cos � 3 sec = 2 tan 4 cos � cos

3 =

cossin2

4 cos2 � 3 = 2 sin 4 � 4 sin2 � 3 = 2 sin

4 sin2 + 2 sin � 1 = 0

sin = 8

1642 =

8522

= 4

51

sin =

415

, 4

15 = �cos 36º, sin 18º

= �sin 54º, sin 18º = sin

103

, sin 10

= n + (�1)n 10

or n � (�1)n 103

.

Section (J) :

J-1. 3 sin � cos = 2 2

cos

21

sin23

= 2

2 sin

6 = 2 sin

6 = 2

1 = sin

4

� 6

= n + (�1)n 4

.

J-2. 5 sin + 2 cos = 5

29

5 sin +

29

2 cos =

29

5

sin sin + cos cos = 29

5

cos ( � ) = sin = cos

2 � = 2n ±

2

= 2n ± 2

+

= 2n ± 2

, 2n ± 2

+ 2 = 2n + 2

, 2n � 2

+ 2

For = 2n � 2

+ 2,


We have = 2n + 2

4 = 2n + 2

1tan25

tan 11

= 2n + 2 tan�1

25

1

125

= 2n + 2

73

tan 1

= 2n + 2

or 2n + 2 where tan�1 73

=

PART - IISection (A) :A-3. 3{cos4 + sin4} � 2{cos6 + sin6}

= 3{1 � 2 sin2 cos2} � 2 {1 × (cos4 + sin4 � sin4 cos2)}= 3 � 6 sin2 cos2 � 2 { 1 � 3 sin2 cos2}= 3 � 6 sin2 cos2 � 2 + 6 sin2 cos2 = 1

A-6.

10cos1

103

cos1

103

cos�1

10cos�1

=

10cos�1 2

10

3cos�1 2

sin2 10

. sin2 103

=

2

415

·4

1�5

= 2

16

4

= 161

Section (B) :B-2. 3 sin = 5 sin

sinsin

= 35

sin�sinsinsin

= 28

2�

tan

2tan

= 4

B-7. cot (A + B) = cot 225° = 1 BcotAcot

1�BcotAcot

= 1

cot A cot B = 1 + cot A + cot B

NowBcotAcotBcotAcot1

Bcot.Acot

= )BcotAcot1(2

BcotAcot1

=

21

Section (C) :

C-3. tanA = 34

A IIIrd quadrant

5 sin 2A + 3 sinA + 4 cosA= 10 sinA cosA + 3 sinA + 4 cosA= 10 sinA cosA + 3 sinA + 4 cosA= 0


C-6. tan2 = 2 tan2 + 1 ... (i)

cos 2 + sin2 =

2

2

tan1

tan�1 + sin2

= 1tan21

1�tan2�12

2

+ sin2 =

)tan1(2

tan2�2

2

+ sin2

= � sin2 + sin2 = 0which is independent of

C-7*. sin t + cos t = 51

2t

tan1

2t

tan�12t

tan2

2

2

=51

10 tan2 2t

+ 5 � 5 tan2 2t

= 1 + tan2 2t

6 tan2 2t

� 10 tan 2t

� 4 = 0 3 tan2 2t

� 6 tan 2t

+ tan 2t

� 2 = 0

3 tan 2t

2�

2

ttan + 1

2�

2

ttan = 0 tan

2t

= 2 , tan 2t

= � 31

Section (D) :D-1. f() = sin4 + cos2

= sin2 (1 � cos2) + cos2

= sin2 + cos2 � sin2 cos2

f() = 1 � 41

sin22

0 sin22 1f()

max = 1

f()min

= 1 � 41

= 3/4 Range is

1,

43

D-2*. 1 + 4 sin+ 3 cos 4 sin+ 3 cos[� 5, 5]

Max. = 1 + 5 = 6Min. = 1 � 5 = � 4

Section (E) :

E-2. square & adda2 + b2 = 9 + 16 = 25

E-5*. 1 radian ~ 57º (approx.)

sin1 > sin1°

cos1° > cos1


Section (F) :F-3. A = tan 6° tan 42°

B = cot 66° cot 78°

BA

= tan 6° tan 42° tan 66° tan 78°

BA

=

54tan)660(tan)6�60(tan6tan

. tan 78° tan 42°

BA

=

54tan)1860(tan)18�60(tan.18tan

=

54tan54tan

BA

= 1 A = B

F-5*. cos 10

· cos 102

· cos 104

· cos 108

· cos 10

16

=

10sin2

102sin

5

5

= 321

10sin

1032

sin

= 321

10sin

102

3sin

= � 321

·

10sin

10cos

10sin2

= � 161

cos10

= � 641

5210

F-7*. cos2x + cos2y + cos2z � 2 cosx cosy cosz

(Given x + y = z)= 1 + cos (x + y) cos (x � y) + cos2z � 2 cosx cosy cosz

= 1 + cosz [cos (x � y) + cos (x + y)] � 2 cos x cosy cosz

= 1 + cosz . 2cosx cosy � 2 cosx cosy cosz

= 1= cos (x + y � z)

F-9*. tan A + tan B + tan C = 6, tan A tan B = 2In any ABC,tan A + tan B + tan C = tan A tan B tan C 6 = 2 tan C tan C = 3 tan A + tan B + 3 = 6 tan A + tan B = 3 & tan A tan B = 2Now (tan A � tan B)2 = (tan A + tan B)2 � 4tan A tan B

=9 � 8 = 1

tan A � tan B = ± 1

tan A � tan B = 1 or tan A � tan B = � 1

tan A + tan B = 3 tan A + tan B = 3on solving on solving

tan A = 2 tan A = 1tan B =1 tan B = 2

Section (G) :

G-3. tan x + tan

3x + tan

32

x = 3 3 tan 3x = 3

tan 3x = 1 x = 3

n +

12

, n


Section (H) :

H-2. sin 7x + sin 4x + sin x = 0

2 sin 4x cos 3x + sin 4x = 0 sin 4x = 0 or(;k) cos 3x = 21

4x = n or 3x = 2n ± 32

x = 4

n,

92

3n2

= 0, 4

, 2

, 92

, 9

4.

H-4.* 2sin2x = sinx + sin3x 2sin2x = 2sinx2x cosx sin2x = 0 or cosx = 1

2x = n or x = 2m x = 2

n, 2m

options (A), (B), (C), (D) are all a part of 2

nx

.

Section (I) :

I-4. cos 2 + 3 cos = 0 2 cos2 + 3 cos � 1 = 0 cos = 4

893 =

4173

As �1 cos 1 a cos = 4

173 only = 2n ± where cos =

4317

I-5. sin + 7 cos = 5

2t1

t2

+

2

2

t1

t17

= 5 where t = tan

2

2t + 7 � 7t2 = 5 + 5t2 tan 2

is root of 12 t2 � 2t � 2 = 0 or 6t2 � t � 1 = 0.

Section (J) :

J-1. tan = �1 4

7,

43

in [0, 2]

cos = 2

1

47

,4

in [0, 2] common value is x =

47

general solution is 4

7n2

, n I.

J-3.* Let E = sin x � cos2 x � 1 E = sin x � 1 + sin2 x � 1 = sin2 x + sin x � 2

=

2

21

xsin

�

49

assumes least value

when sin x = 21

x = n + (�1)n

6 .

EXERCISE # 2

PART - I

3. sinx + sin y = a .....(1)cosx + cos y = b .....(2)

2yx

cos2

yxcos2

2yx

cos2

yxsin2

= ba


tan

2

yx =

ba

sin

2yx

= 22 ba

a

, cos

2yx

= 22 ba

b

sin (x + y) = 2 sin

2yx

cos

2yx

= 22 ba

ab2

Now for tan

2

yx

(1)2 + (2)2

1 + 1 + 2 cos (x � y) = a2 + b2

cos (x � y) = 2

2ba 22

tan2

2yx

= )yxcos(1)yxcos(1

tan2

2yx

=

22ba

1

22ba

1

22

22

tan

2yx

= ± 22

22

ba

ba4

6. tan = qp

LHS = 21

(p cosec 2 � q sec2) × 22

22

qp

qp

= 21 22

2222qp2sec

qp

q�2eccos

qp

p

22 qp

psin

, 22 qp

qcos

= 22 qp

2cos2sin2sincos�2cossin

21

= 22 qp

4sin)2�sin(

=

22 qp4sin4sin

( = 6)

7. (i) cot 7o

21

= tan 82o

21

=

o

o

21

7sin

21

7cos =

o

o

2

15sin21

7cos2

= )3045sin(

)3045cos(1

=

22

1322

131

= 13

1322

= 13

)13)(1322(

= 6232 = 6432


= )32( )12(

(ii) tan 142o

21

= � cot 52

o

21

= o

21

52tan

1 =

o

21

745tan

1

= � o

o

21

7tan1

21

7tan1

= � oo

oo

21

7sin21

7cos

21

7sin21

7cos

= �

15cos

21

7sin21

7cos

2oo

= �

15cos15sin1

= �

22

1322

131

= � 2

)13)(1322(

= � 2

])13()13(22[ 2

= � 2

)]324()13(22[

= � )]32()13(2[ = � 3226 = 2 + 632

9. (i) tan9° � tan27° � tan 63° + tan81°

= (tan 9º + tan81º) � (tan 27º + tan 63º)

= º27cosº27sin2

2�

º9cosº9sin22

º63cosº27cosº90sin

�º81cosº9cos

º90sin

=

415

2�

41�5

2º54sin

2�

º18sin2

= 44

)15�15(8

(ii) cosec 10° � 3 sec10° = 2

10sin

23

�10cos21

× 22

º10cosº10sin1

= 4

(iii) 22 sin 10°

35sin2�

5sin40cos

25sec

= 22

º10sinº35sin2�

º5sinº40cosº5cosº5sin2

2º5secº5cosº5sin2

= 22 (sin5º + 2cos45º + cos 35º � cos 25º + cos 45º)

= 22 (sin5º + 2cos45º + 2sin 30º sin (� 5º))

= 22 )2( = 4

(iv) cot 70° + 4 cos 70° = º70sin

º70sinº70cos4º70cosº70cos4

70sin70cos

= º70sin

º140sin2º70cos

= º70sin

º140sin)º140sinº20(sinº70sin

º140sin)º140sinº70(cos


= º70sin

º20cosº120sin2º70sin

º140sinº60cosº80sin2

= 2 × 323

(v) tan 10º � tan 50º + tan 70º

= tan 10º � tan (60º � 10º) + tan (60º + 10º)

= tan 10º � º10tan31

º10tan�3

º10tan3�1

º10tan3

= º10tan3�1

º10tan3º�10tan92

3

= 3

º10tan3�1

º10tanº�10tan32

3

= 3 tan 30º

= 3

11.

21AA1

= 31AA

1 +

41AA1

OA1 = OA

2 = OA

3 = OA

4 = r (say)

A1OA

2 =

n2

, AA1OA

3 =

n4

, AA1OA

4 =

n6

nsin

1

=

n2

sin

1

+

n3

sin

1

sin n2

nsin

n3

sin = sin n

3 . sin

n

sin n2

nsin.

n2

cos2 = sin n

3 sin

n

2 sin n2

cos n2

= sin n

3

sin n4

= sin n

3

n4

= � n

3

4 = n � 3 n = 7

13. Pn � P

n�2 = cosn + sinn � cosn�2 � sinn�2

= cosn�2 (cos2 � 1) +sinn�2 (sin2 � 1)

= cosn�2 (�sin2 ) +sinn�2 (� cos2 )= (�sin2 cos2 ) {cosn�4 +sinn�4 }= (�sin2 cos2 ) P

n�4

put n = 4 P

4 � P

2 = (� sin2 cos2) P

0

P4 = P

2 � 2 sin2 cos2

= 1 � 2 sin2 cos2

similarly we can prove the other result also.

15. tan2 + 2 tan . tan 2 = tan2 + 2 tan . tan 2

(tan2 � tan2) + 4 tan tan

22 tan1

1

tan1

1 = 0

(tan2 � tan2) + 4 tan tan )tan1)(tan1(

)tan(tan22

22

= 0


(tan2 � tan2)

)tan1)(tan1(

tantan41

22 = 0

(tan2 � tan2) (1 � tan 2 . tan 2) = 0 tan2 = tan2 or tan 2 . tan 2 = 1

L.H.S. = tan2+ 2 tan .

2tan1

= tan2 +

tan2tan2

. (1 � tan2) = 1

R.H.S. = tan2 + 2 tan . 2tan

1

= tan2 +

tan2tan2

. (1 � tan2) = 1

19. 13 18 tanx = 6 tan x 3 ................(1)

13 � 18 tanx = 36 tan2x + 9 � 36 tanx tanx = 32

, 61

Put in (1) tanx = 32

is correct x = n + tan�1

32

= n + = , + , � + , �2 + in (�2, 2)

22. tan + sin = 23

...(1)

As tan2 + cos2 = 47

2

sin23

+ cos2 =

47

49

+ sin2 � 3 sin + cos2 = 47

23

= 3 sin sin = 21

= n + (�1)n

6

from (1), tan = 23

� sin = 21

23 = 1 = n +

4

.

23. a cos 2 + b sin 2 = c

2

2

t1

t1a

+

2t1

t2b

= c where t = tan

(c + a)t2 � 2bt + (c � a) = 0 t1 + t

2 =

acb2

, t1t2 =

aca�c

cos2 + cos2 = 2

2cos12cos1 = 1 +

21

[cos 2 + cos 2]

= 1 +

22

22

21

21

t1

t1

t1

t121

simplifying and using values for t1, t

2 we get

cos2 + cos2 = 1 + 22 ba

ac

= 22

22

ba

acba

.


27. RHS = 3x2 + 2x + 3

Minimum value = )3(4

4)3)(3(4 =

38

> 2

whereas LHS 2 no solution.

PART - II

2. For dodecagon

RR

BA

oA 'OB' = 122

= 30°

OA 'B' = OB 'A' = 75°

75sin

R =

30sin1�3

R =

21

22

)13()1�3(

R = 2

For hexagon AOB = 6

2 = 60°

AOB is equilatecal AB = R = 2

6. A + B + C =

2C

sin

2C

Asin

= 1k

2C

sin2C

Asin

2C

sin�2C

Asin

= 1k1�k

2A

cos2

CAsin2

2A

sin2

CAcos2

= 1k1�k

tan

2B

tan 2A

= 1k1�k

8. 4 cos2

2

x�

4 + xcosxsin4xsin4 224

= 4 cos2

2

x�

4 + | 2 sinx | = 4 cos2

2

x�

4 � 2 sinx

= 2

x�

2cos1 � 2 sinx = 2

10. ecCcosecBcosAcos = CsinBsinAcos

+ CsinAsinBcos

+ BsinAsinCcos

= CsinBsinAsin

CsinCcosBsinBcosAsinAcos


= CsinBsinAsin2

C2sinB2sinA2sin =

CsinBsinAsin2CsinBsinAsin4

(using conditional identity)

= 2

13.xcos10x3cos5x5cos

)1x2(cos10)x2cosx4(cos5)x4cosx6(cos

= xcos10x3cos5x5cos

xcos210xcosx3cos25xcosx5cos2 2

= 2 cos x xcos10x3cos5x5cos

xcos10x3cos5x5cos

= 2 cos x

15.

cos3cos

= 4 cos2 � 3 = 2 (1 + cos 2) � 3

= 2 cos2 � 1 = 2 cos ( � ) � 1

(cos2 + sin2 ) + (cos2 + sin2 ) + 2 cos ( � ) = a2 + b2

2 cos ( � ) = a2 + b2 � 2

cos3cos

= a2 + b2 � 3

20. sin 3 = 4 sin sin 2 sin 4 sin 3 = (2 sin ) (2 sin 2 sin 4) 3 sin � 4 sin3 = 2 sin (cos 2 � cos 6) 3 � 4 sin2 = 2(cos 2 � cos 6) or sin = 0

3 � 2(1 � cos 2) = 2 cos 2 � 2 cos 6or sin = 0 1 = �2 cos 6 cos 6 = 21

or sin = 0

sin = 0 or cos 6 = 21

= n or = 6

32

n2

= 93

n

= 0, , 9

, 93

, 93

2 ,

9

So eight solutions.

22. 2 cos x = 2 2 2 sin x

x2sin1xcos2 = xcosxsin )xcosx(sin2

1xcos

cosx =

4xsin

see from graph or we can put values given in options to verify..

25. 2 tan2 x � 5 sec x � 1 = 0 2(sec2 x � 1) � 5 sec x � 1 = 0

2 sec2 x � 5 sec x � 3 = 0 sec x = 26

, 21

= 3, 21


sec x = 3

2

1xsec

cos x = 31

7 solutions in

215

,0 n = 15.

28. 4cos3x � 4cos2x + cos x � 1 = 0

(4 cos2x + 1) (cos x � 1) = 0

cos x = 1x = 2nsolutions in the interval [0, 315] are 0, 2 , 4 , ....., 100

arithmatic mean = 51

100....420 = 50

29. h = 22 )}(sin22sin2{sin)}(cos22cos2{cos

h = [4 cos2 ( + ) (cos ( � ) + 1)2 + 4 sin2 ( + ) (cos ( � ) + 1)2 ]1/2

h = [4{cos ( � ) + 1}2 {cos2 ( + ) + sin2 ( + )}]1/2

h = 2 (1 + cos ( � )) h = 2 × 2 cos2

2

�

h = 4 cos2

2

�

32. y = a cos2 x + 2b sin x cos x + c sin2 x & tan x = ca

b2

z = a sin2 x � 2b sin x cos x + c cos2 x y + z = a + c

and y � z = (a � c) xsinxcos 22 + 4b sin x cos x

= (a � c) cos 2x + 2b sin 2x ( 2b = (a � c) tan x)

= (a � c) [cos 2x + tan x.sin2x] = (a � c)

x2sin

xcos

xsinx2cos2

= xcos

)xx2cos()ca( = (a � c).

33.

n

2BA

sin.2

BAcos2

2BA

cos.2

BAcos2

+

n

2BA

sin.2

BAsin2

2BA

cos.2

BAsin2

= cotn

2BA

+ (�1)n cotn

2BA


34. sin6x + cos6x = a2

(sin2 x + cos2 x) (sin4 x + cos4 x � sin2 x cos2 x) = a2

(sin2 x + cos2 x)2 � 3sin2 x cos2 x = a2 1 � 3 sin2 x cos2 x = a2

1 � 43

sin2 2x = a2 3

)a�1(4 2 = sin2 2x

0 34

(1 � a2) 1

1 � a2 0 and 4 � 4a2 3

a2 1 and41

a2

� 1 a 1 and a 21

or a � 21

a

21

,�1�

1,

21

38. cos 15x = sin 5x

cos 15x = cos

x5

2 or cos

x5

2

3

15x = 2n ±

x5

2 or 15x = 2n ±

x5

23

x = 10n

+ 40

, n , x = 5

n +

203

, n

and x = 5

n �

20

, n and x = 10n

� 403

, n

40. sin2 x + 2 sin x cos x � 3 cos2 x = 0case-I : cos x 0 tan2 x + 2 tan x � 3 = 0

tan x = 3, 1 x = n + tan�1 (�3), n + 4

case-II : cos x = 0 1 + 0 � 0 = 0 not true.

EXERCISE # 33. (A) sin2 + 3 cos = 3 1 � cos2 + 3cos = 3

cos2 � 3cos + 2 = 0 cos = 1, 2 cos = 1 ( cos 2) = 0 in [�, ] No. of solution = 1

(B) sin x . tan 4x = cos x xcosx4cosx4sin

.xsin

sin4x sinx � cos4x cosx = 0 cos5x = 0 5x = (2n + 1) /2 x = (2n + 1) /10

x = 10

, 103

, 105

, 107

, 109

in (0, )y = 2

xy

x

y = x �12

O 3

So there are five solutions.

(C) (1 � tan2 ) sec2 + 2tan2 = 0

(1 � tan4 ) + 2tan2 = 0

(1 � x2) + 2x = 0 where x = tan2


2x = x2 � 1 x = 3

tan2 = 3 3tan 3

in

2,

2

Number of solutions = 2

(D) [sin x] + [ 2 cosx] = � 3

[sin x] = � 1 and [ 2 cosx] = � 2 < x < 2 and 1xcos22�

2

1xcos2�

2

1xcos1�

4

5x

for ]2,0[x

45

x

, ]2,0[x

4

5x

25

x22

0 < sin 2x < 1 [sin2x] = 0

4. (A)

Number of solutions = 6

(B) sin x = 2

82 = 1 ± 2

sin x = 1 � 2As sin x takes at least four valuesin [0, n] n 4(C) 1 + sin4 x = cos2 3xL.H.S. 1 and R.H.S. 1 L.H.S. = R.H.S. = 1 sin4 x = 0 and cos2 3x = 1 x = n and 3x = m

x = n and 3x = m

x = n and x = 3

m

x = n

x = �2, �, 0, , 2 in

2

5,

2

5

Number of solutions = 5.(D) A, B, C are in A.P. B = 60º

As sin (2A + B) = 21

2A + B = 30º or 150º

2A = �30º or 90º 2A = 90º A = 45º

C = 180º � A � B = 75º = 125

p = 12.

Comprehension # 1 (5, 6, 7)

5. Given cos + cos = a 2cos

2 cos

2 = a ..... (i)

and sin + sin = b


2 sin

2 cos

2 = b ..... (ii)

by (i) & (ii)

tan

2 = ab

tan = ab

sin2 + cos2

=

2

2

a

b1

ab

2

+

2

2

2

2

a

b1

a

b1

= 22 ba

ab2

+ 22

22

ba

ba

= 22

222

ba

ab2b2ba

= 1 + 22 ba

)ba(b2

n = 26. sinn A = x

sin2 A = x sin A sin 2A sin 3A sin 4A= sin A (2 sin A cos A) (3 sin A � 4 sin3 A) (4 sin A cos A (1 � 2 sin2 A))= 8 sin4 A (1 � sin2 A) (1 � 2 sin2 A) (3 � 4 sin2 A)If we put sin2 A = x, then given expression is a polynomial of degree 5 in x.

7. If p = 5 sin x + (p � 5), cos x, tan x

sin x, cos x, tan x are in G.P. cos2 x = sin x tan x

cos3 x = sin2 x cos3 x = 1 � cos2 x cos3 x + cos2 x = 1taking cube both sides

cos9 x + cos6 x + 3 cos5 x = 1 cos9 x + cos6 x + 3 cos5 x � 1 = 0

Comprehension # 4

14. sin6x + cos6x < 167

1 � 3sin2x cos2x < 167

sin2x cos2x > 163

sin22x > 43

2

x4cos1 >

43

1 � cos4x > 23

cos4x < 21

Principal is value 4x

34

,32

General value is

3

4n2,

3

2n2x4

n,

32n

,62

nx


15. cos 2x + 5 cos x + 3 0

2cos2x + 5cosx + 2 0 (cosx + 2)(2 cosx + 1) 0

2cosx + 1 0 ( cosx + 2 > 0)

cosx 21

32

,3

2x

16. 2 sin2

4x + 3 cos 2x 0

1 � cos

2x2 + 0x2cos3

1x2sinx2cos3 21

x2sin21

x2cos23

21

3x2sin

2x +

3

67

n2,6

n2

65

n2,2

n2x2

125

n,4

nx

x

127�

,�

125

,4

�

,

43

in ,

19. Statement-1 : cos A cosec B cosec C

= CsinBsin

Acos =

CsinBsinAsinAcosAsin

= CsinBsinAsin2

A2sin

= CsinBsinAsin2CsinBsinAsin4

= 2

Statement-2 : tan A tan B = 1 iff the triangle ABC is right angled Statment is false

20. y =

3tantan

=

2

3

tan31

tantan3

tan y =

2

2

tan3

tan31

tan2 = 3y1y3

0 y

3

1, (3, )

statement-1 and statement-2 both are true and statement-2 explains statement-1

22. cos x . sin y = 1 Either cosx = 1 and siny = 1 or cosx = �1 and siny = �1

(x,y) =

2,0 ,

25

,0 ,

2,2 ,

25

,2 or (x, y) =

23

, ,

23

,3

Number of pairs = 6

24. log2 ]2cos2cos)(cos)([cos 22

= log2 ]2cos2cos)(sin1)([cos 22

= log2 [1 + cos 2.cos2 � cos 2 cos 2]

= log2 1

= 0


30.12cos2

3sin

=

21

2(3 sin � 4 sin3 ) = 2(1 � 2 sin2 ) + 1

8 sin3 � 6 sin � 4 sin2 + 3 = 0 (2 sin � 1)(4 sin2 � 3) = 0

sin = 21

, ±23

For sin = ±23

, 2 cos 2 + 1 = 0 so given equation

becomes undefined sin = 21

only = n +(�1)n 6

, n .

32. sin x . xcos8 2 = 1 sin x.|cos x| = 8

1 2 sin x |cos x| =

2

1.

33.

xcos

1xcos

22 y2tan1 2

(3 + sin 3z) = 4

cos2 x + xcos

12 = 2, 1 + tan2 2y = 1, 3 + sin 3z = 2

cos2 x = 1, tan2 2y = 0, sin 3z = �1 x = n, n I.

36. cos 20° + 2 sin2 55° � 2 sin 65°

= cos 20° + 1 � cos 110° � 2 sin 65°

= 1 + 2 sin 65° sin 45° � 2 sin 65°

= 1 + 2 sin 65° 2

1 � 2 sin 65° = 1

41.

cos13sinsin3

+

sin13coscos3

= 4 2 cos

4

cos1sin4 3

+

sin1cos4 3

= 4(cos � sin )

cos1sin3

+ sin = cos �

sin1cos3

cos1cossinsinsin3

=

sin1cossincoscos 3

cos1cos1sinsin 2

=

sin1cossinsincos 2

cos11cossinsin 2

=

sin1)1sin(cossin

either sin = 0 or

cos11cossin2

= � cos

= n or sin2 + cos + 1 = �cos � cos2 cos = �1

= n = 2n or (2n + 1)But at = (2n + 1) , 1 + cos = 0 (2n + 1) = 2x.

42. xsin7xcos6 2 + cos x = 0 xcos77xcos6 2

+ cos x = 0

1xcosxcos7 2 + cos x = 0 1xcosxcos7 2

= � cos x

(so cos x 0) 7 cos2 x � cos x � 1 = cos2 x 6 cos2 x � cos x � 1 = 0

(2 cos x � 1) (3 cos x + 1) = 0 cos x = 21

, 31


But cos x 0 cos x = 31

cos x = � cos where cos = 31

cos x = cos( � ) x = 2n ± ( � ).

43. x3 + x2 + 4x + 2 sin x = 0 x3 + x2 + 4x = �2 sin x ...(1)when x = 0, 0 = 0 x = 0 is the solutionwhen x [0, ), x3 + x2 + 4x > 0 where as � 2sin x < 0

no solution for x (0, )when x [, 2], x3 + x2 + 4x 3 + 2 + 4 > 2whereas 0 �2 sin x 2 no solution for [, 2]so given equation has only one solution in [0, 2] and that solution is x = 0.

EXERCISE # 4PART - I

1. Clearly = 30º and (60º, 90º)

Hence + lies in (90º, 120º).

2. Let y = 2 sin t

y = 1x2x3

x5x212

2

(3y � 5) x2 � 2x (y � 1) � (y + 1) = 0

x R �

31

,1

D 0 y2 � y � 1 0

y 2

51or y

251

sin t 4

51or sin t

451

range of t is

10,

2

2,

10

3

3. O1BD, DO

BD

1 = cot 30º

BD = 3 similarly EC = 3

BC = AB = AC = 2 + 2 3

area of ABC = 43

2)322( = 43

)3231( 4 = 6 + 34 sq. unit

4.

� = 0, � 2 or 2

� = 0 = cos 2 = e1


This is true for '4' value of '', ''If � = � 2 = � and = and cos ( + ) = 1 (No solution)similarly if � = 2 = and = again no solution results

5.

4,0

tan in

4,0 and 0 < tan < 1

cot in

4,0 and cot > 1

Let tan = 1 � 1 and cot = 1 +

2 where

1 and

2 are very small and positive, then

t1 = 11

1)1( , t

2 = 21

1)1( , t

3 = 11

2 )1( , t

4 = )1(

22)1(

t4 > t

3 > t

1 > t

2

6. 2sin2 � 5sin + 2 > 0 (sin � 2)(2sin �1) > 0

1/2

O

y

sin < 21

[ �1 1sin ]

From graph, we get

2,

65

6,0

7. 2sin2 � cos2 = 0 ............(i)

sin = 21

2cos2 � 3sin = 0 ............(ii)�2sin2 � 3sin + 2 = 0

sin = 21

, �2

So sin = 21

is the only solution

at = 6

5,

6

8.*2

xsin4

+ 3

xcos4

= 51

2

xsin4

+ 3

)xsin1( 22 =

51

2

xsin4

+ 3

xsin2xsin1 24 =

51

5 sin4x � 4 sin2x + 2 = 56

25 sin4x � 20 sin2x + 4 = 0

(5 sin2x � 2)2 = 0 sin2x = 52

, cos2x = 53

tan2x = 3

2and

8

xsin8 +

27

xcos8 = 125

1

9. f() =

22 cos5cossin3sin

1 =

2)2cos1(5

2sin23

22cos1

1

= 2cos42sin36

2

f()max

= 56

2

= 2


10.

nsin

1

�

n3

sin

1

=

n2

sin

1

n3

sinn

sin

nsin

n2

cos2

= n2

sin

1

sin n4

= sin n

3

n4

= (�1)k n

3 + k , k

If k = 2m n

= 2m

n1

= 2m , not possible

If k = 2m + 1 n

7 = (2m + 1) n = 7, m = 0

11. tan = cot 5

cossin

=

5sin5cos

cos 6= 0

6 = (2n + 1) 2

= (2n + 1) 12

; n

= 125

�

, �4

, 12

�

, 12

, 4

, 125

.........(1)

sin2 = cos4 sin2 = 1 � 2 sin2 2 2sin22 + sin2� 1 = 0

sin2 = � 1, 21

2= (4m � 1)2

, p + (�1)p 6

= (4m � 1)4

, 2

p+ (�1)p

12

; m, p I

= �4

, 12

, 125

...........(2)

From (1) & (2)

125

,12

,4

�

Number of solution is 3.

12. P = {: sin � cos = 2 cos }

sin = ( 2 + 1) cos tan = 2 + 1

= n+ 8

3 ; n I

Q = {: sin + cos = 2 sin } cos = ( 2 � 1) sin

tan = 1�2

1 = 2 + 1 = n+

83

; n I

P = Q


13. As tan(2 � ) > 0, � 1 < sin < �23

, [0, 2]

2

3 < <

35

Now 2cos(1 � sin) = sin2( tan /2 + cot /2)cos � 1

2cos(1 � sin) = 2sin cos � 1 2cos + 1 = 2sin( + )

As

35

,2

3 2cos + 1 (1, 2) 1 < 2sin( + ) < 2

21

< sin( + ) < 1

As + [0, 4] +

6

5,

6 or +

6

17,

6

13

6

� < < 6

5 � or

613

� < < 6

17 �

67

,32

32�

,2

3�

35

,2

3

correct option is (A, C, D)

PART - II

1. u = 2222 sinbcosa +

2222 cosbsina

u2 = a2 cos2 + b2 sin2 + a2 sin2 + b2 cos2 + 2 2222 sinbcosa ×

2222 cosbsina

u2 = (a2 + b2) + 2 22222222 cosabasina�ba

u2 = 22 ba + 2 222222224 cossinababaa

u2 = 22 ba + 2

2sin

2

abba 2

22222

.

min(u2) = a2 + b2 + 2ab = 2ba

and max(u2) = a2 + b2 + 22 ba = 2 22 ba

Now, max(u2) � min(u2) = (a � b)2

2. sin + sin = � 6521

and cos + cos = � 6527

squaring and adding, we getsin2 + sin2 + 2 sin sin + cos2 + cos2 + 2 cos . cos

=

2

6521

�

+

2

6527

�

2 + 2 cos ( � ) = 42251170

cos2

2�

= 42254

1170

= 130

9

cos

2

� =

130

3� ( < � < 3

2

<

2

� <

23

)


3. tan 2P

and tan 2Q

are the roots of equation ax2 + bx + c = 0

tan2P

+ tan 2Q

= �ab

and tan2P

tan 2Q

= ac

2P

+ 2Q

+ 2R

= 2

( P + Q + R = )

2

QP =

2

� 2R

2

QP =

4

( R = 2

)

tan

2QP

= 1

2Q

tan.2P

tan�1

2Q

tan2P

tan

= 1 a/c�1a/b�

= 1 c = a + b

4. cos x + sin x = 21

2/xtan1

2/xtan�12

2

+

2/xtan1

2/xtan22

=

21

, Let tan 2x

= t

2

2

t1

t�1

+ 2t1

t2

=

21

3t2 � 4t � 1 = 0 t = 3

72

as 0 < x < 0 < 2x

< 2

tan 2x

is positive

t = tan 2x

= 3

72

Now tan x = 2/xtan�1

2/xtan22 = 2t�1

t2 tan x = 2

372

�1

372

2

=

374

�

5. Given equation is 2 sin2 x + 5 sin x � 3 = 0

(2 sin x � 1) (sin x + 3) = 0

sin x = 21

( sin x �3)

It is clear from figure that the curve intersect the line at four points in the given interval.Hence, number of solutions are 4.

6. Given, cos x + sin x = 21

2x

tan1

2x

tan1

2

2

+

2x

tan1

2x

tan2

2

= 21


Let tan 2x

= t 2

2

t1

t1

+ 2t1

t2

=

21

3t2 � 4t � 1 = 0 t = 3

72

As 0 < x < 0 < 2x

< 2

tan 2x

is positive.

t = tan 2x

= 3

72

Now, tan x =

2x

tan1

2x

tan2

2

= 2t1

t2

tan x = 2

372

1

372

2

tan x = �

721

723

×

721

721

tan x = �

374

.

7. 2{cos ( � ) + cos ( � ) + cos ( � )} + 3 = 0(cos + cos + cos )2 + (sin + sin + sin )2 = 0 cos = 0 = sin

8. tan 2 = tan (( + ) + ( � )) = )tan()tan(1)tan()tan(

=

125

.43

1

125

43

= 1548

4)59(

=

33414

= 3356

Hence correct option is (1)

9. A = sin2x + cos4x = sin2x + (1 � sin2 x)2

= sin4x � sin2x + 1

= 2

2

21

�xsin

+

43

= 43 A 1

10. 3sin P + 4 cos Q = 6 ...(i)4 sin Q + 3cos P = 1 ...(ii)

Squaring and adding (i) & (ii) we get sin (P + Q) = 21

P + Q = 6

or 6

5 R =

65

or 6

If R = 6

5 then 0 < P, Q <

6

cos Q < 1 and sin P < 21

3 sinP + 4 cosQ < 211

So R = 6


Fundamentals of mathematics - II

OBJECTIVE QUESTIONSSection (A) :

A-5. logp log

p np

1

)p(

=

n

p p

1log

= �log

p pn = �n

independent of p.

A-7*. N = 3log

135log

15

3 � 3log

5log

405

3

= 5log27log 33 15log3 � log3 5.log3 405

= 5log3 3 5log1 3 � log35 log3(81 × 5)

= (3 + log3 5) (1 + log3 5) � log3 5(4 + log3 5)= 3.

A-8*. log23 > 1, log

1210 < 1 log

23 > log

1210

log65 < 1, log

78 > 1 log

65 < log

78

log326 < 3, log

29 > 3 log

326 < log

29

log16

15 < 1, log10

11 > 1 log16

15 < log10

11

Section (B) :

B-4*. (log5x)2 + log

5x x5 = 1

(log5x)2 + log

5x5 � log

5xx = 1 (log

5 x)2 + xlog5log

5log

55

5

� xlog5log

xlog

55

5

= 1

(log5x)2 + xlog1

1

5 � xlog1

xlog

5

5

= 1 Let log

5x = t

t2 + t1

1

� t1

t

= 1 t1

t1)t1(t2

= 1

t3 + t2 + 1 � t = 1 + t

t3 + t2 � 2t = 0

t(t2 + t � 2) = 0

t(t � 1) (t + 2) = 0

t = 0, 1, � 2 log5x = 0, 1, �2

x = 1, 5, 251

B-6*.

5xlog

29

xlog 32

3

x = 3 3 33 xlog �

29

log3 x + 5 = logx 3 3

23 xlog � 29

log3 x + 5 = 23

logx 3 Let log3 x = t

t2 � 29

t + 5 = t2

3 2t3 � 9t2 + 10t � 3 = 0

t = 1 satisfies it2t3 � 9t2 + 10t � 3 = 2t2(t � 1) � 7t(t � 1) + 3(t � 1)

= (t � 1) 3t7t2 2

= (t � 1) (2t � 1) (t � 3)


t = 1 t = 21

t = 3

log3 x = 1 log3 x = 21

log3 x = 3

x = 3 x = 31/2 x = 27.

B-9. Number of digits in integral part = number of digit in 6012 before decimalP = 6012

logP = log 6012 = 12 log 60 = 12[log 6 + 1] = 12 [log 2 + log 3 + 1]= 12 [.3030 + .4771 + 1] = 12 [1.7801] = 21.3612

number of digits in integral part = 22

B-10. log16 x = 43

x = 163/4 x = 8.

Section (C) :

C-3. log1 � x

(x � 2) 1x > 2 ..................(1)(i) When 0 < 1 �x < 1 0 < x < 1

So no common range comes out.(ii) When 1 � x > 1 x < 0 but x > 2

here, also no common range comes out. , hence no solution.Finally, no solution

C-6.

)15(log7

10log

)x2)(8x(

23.0

0

For )x2)(8x( to be defined

(i) (x � 8) (2 � x) 0(x � 2) (x � 8) 0 2 x 8

Now Let say y = log0.3

7

10(log

25 � log

22) = log

0.3

710

(log25/2)

Let y < 0 (assume)

then log0.3

7

10 (log

2 5/2) < 0

7

10 log

2 5/2 > 1 log

2 5/2 >

107

25

> 2(7/10) which is true

So y < 0so denominator is � ve and numerator is +ve, so inequality is not satisfied,

thus )x2)(8x( = 0

x = 2, 8 .....(i)Now 2x � 3 > 31 (x � 3) > log

2 31 x > 3 + log

224.9 (approx) x > 7.9

x = 8

C-8. Domain x2 + 4x � 5 0 x (� , � 5] [1, )Case I :x (� , � 5] [1, 3)� ve < + ve alsways true x (� , � 5] [1, 3) ... (1)Case II :x [3, ) .. (i)


x � 3 < 5�x4x2

x2 � 6x + 9 < x2 + 4x � 5 x > 57

... (ii)

(i) (ii) x [3, ) ... (2)(1) (2) x (� , � 5] [1, ) Ans. (A)

Section (D) :

D-3. z = 4)i1(4

)i1()i(

11= 4)i1(

4

i2

= 2/i

i4

e

e42i

)i1(2

= 2/ie2

|z| = 2 amp (z) = 2

2

2)z(amp

|z|

= 4 (D)

D-6*. |z1 + z

2|2 = |z

1|2 + |z

2|2

0zzzz 2121 2

1

2

1

zz

�zz

0zz

zz

2

1

2

1

2

1

zz

is purely imaginary

so amp

2

1

zz

is may be 2

or � 2

D-8.

D-9. z1/3 = a � ib

z = (a � ib)3

x + iy = (a3 � 3ab2) + i(b3 � 3a2b)

22 b3�a

ax

by

= b2 � 3a2

by

�ax

= 4(a2 � b2) k = 4

EXERCISE # 2

1. (i)

3/1

10

7log

1.01

log

15 5

= (7 + 1)1/3 = 2

(ii) log3/4

log2 2/12/18 = log

3/4 log

2 (2)3/4 = 1


(iii)

27log1

491

=

2log7log2 77)7( = 147log2 )7(

= 27 )14(log7

= 196

1 & 7log 5/15 = 7log55 = 7

7 + 196

1

(iv) 5log37 + 7log53 � 7log35 � 3log57

= 5log37 + 7log53 � 5log37 � 7log53

{using property blogca = alogcb }

= 0

5. (i) log10 (x2 � 12x + 36) = 2

(i) x2 � 12x + 36 > 0 (x � 6)2 > 0 x R � {6}

(ii) x2 � 12x + 36 = 100 x2 � 12x � 64 = 0

(x � 16) (x + 4) = 0 x = 16, �4.

(ii) log4 log3 log2 x = 0 log3 log2 x = 1 log2 x = 3 x = 23 x = 8.

(iii)

x93 9

21

xloglog = 2x log9 x + 21

+ 9x = 32x

log9 x + 21

+ 9x = 9x log9 x = �21

x = 9�1/2 x = 31

(iv) 2 log4 (4 � x) = 4 � log2 (�2 � x)

(i) 4 � x > 0 x < 4(ii) �2 � x > 0 x < �2

(iii) log2 (4 � x) = 4 � log2 (�2 � x) log2 (4 � x) (�2 � x) = 4

(4 � x) (�2 � x) = 16 �8 � 2x + x2 = 16 x2 � 2x � 24 = 0 (x � 6) (x + 4) = 0

x = 6 (not possible) , x = �4.

(v) log10

2 x + log10

x2 = log10

2 2 1

log10

2 x + 2 log10

x + 1 = log10

2 2 (log10

x + 1)2 = log10

2 2log

10x + 1 = ± log

10 2

x = 201

and 51

(vi) log4(log2 x) + log2(log4 x) = 2 21

log2 (log2 x) + log2 (log4 x) = 2

21

log2 (2 log4x) + log2 (log4x) = 2

21

log2 2 + 21

log2 (log4 x) + log2 (log4 x) = 2

23

log2(log4 x) = 23

log2 (log4 x) = 1

log4 x = 2 x = 42 x = 16.

(vii) 35xlog

x

= xlog510

3

5xlog log x = 5 + log x


log2 x + 2 log x � 15 = 0

(log x + 5)(log x � 3) = 0

log x = �5, log x = 3

x = 10�5 , x = 103.(viii) Domain x � 1 > 0 and x + 1 > 0 and y � x > 0

x > 1 x > � 1 x < 7 x (1, 7) .........(i)

� log2 (x � 1) � log2 (x + 1) = 1 + )x7(log21

2

� log2 (x2 � 1) + log2 (7 � x)2 = 1

log2 1x

)x7(2

2

= 1

1x

)x7(2

2

= 2

x2 + 14x � 51 = 0

(x + 17) (x � 3) = 0

x = 3 or x = � 17 (rejected)

x = 3

6. (a) log10 2 = 0.3010 . log10 3 = 0.4771let x = 615

log10 x = 15 log10 6= 15(log10 2 + log10 3)= 15(0.3010 + 0.4771)= 11.6715

characteristic of 615 is 11 number of digits in 615 is 12.(b) let x = 3�100

log10 x = �100 log10 3= � 47.71

number of zeroes immediately after the decimal in 3�100 is 47.

10. (i) 0x

6x4log

51

x6x4

> 0 x ),0(23

,�

....(i)

& 1x

6x4

0

x2x

x (�, � 2] (0, ) ....(ii)

(i) (ii) x

23�

,2�

(ii) log2 (4x � 2.2x + 17) > 54x � 2.2x + 17 > 0(2x)2 � 2.2x + 17 > 0 x R and 4x � 2.2x + 17 > 32

(2x)2 � 2.2x � 15 > 0 (2x + 3) (2x � 5) > 0

2x < � 3 or 2x > 5 x or x > log2 5 x (log2 5, )

(iii) (logx)2 � logx � 2 0x > 0 ....(i)(log x � 2) (log x +1) 0

log x � 1 or log x 2

x 101

or x 100 ....(ii)

(i) (ii) x

,100

101

,0


(iv) log0.5(x + 5)2 > log1/2 (3x � 1)2

(x + 5)2 > 0 x R � {� 5} ........(i)

(3x � 1)2 > 0 x R �

31

........(ii)

(x + 5)2 < (3x � 1)2

8x2 � 16 x � 24 > 0 x2 � 2x � 3 > 0

(x � 3) (x + 1) > 0 x (�, � 1) (3, ) ........(iii)(i) (ii) (iii) gives

(�, �5) (�5, �1) (3, )

(v)21

2log12x3

3x2 + 1 > 1 x2 > 0 x R � {0}

2 < (3x2 + 1)1/2

3x2 + 1 > 4 (x � 1) (x + 1) > 0

x (�, �1) (1, )

(vi) 2xlog (x + 2) < 1 x + 2 > 0 x > � 2

Case-I : when 0 < x2 < 1 x (�1, 0) (0, 1)then x + 2 > x2 x2 � x � 2 < 0

}0{�)1,1(x

Case-II : x2 > 1 |x| > 1x + 2 < x2 x2 � x � 2 > 0

),2()1,2(x

Hence , ),2()1,0()0,1()1,2(x

11. (i)2x

1x2

< 1

Case-I :x � 2 < 0 x < 2 ........(i)2x � 1 (x � 2)2

x (�, 1) (5, ) ........(ii)x (i) (ii) x (�, 1) .......(A)

Case-II : x � 2 > 0 x > 2 ........(iii)2x � 1 (x � 2)2

2x � 1 < x2 � 4x + 4

x2 � 6x + 5 > 0

x (�, 1) (5, ) ........(iv)x (iii) (iv)x (5, ) .......(B)x (A) (B)x (�, 1) (5, )

(ii) x < |x|1

Case-I : x < 0 ......(i)1 � |x| 0 1 + x 0 x � 1 .........(ii)x (i) (ii) x [�1, 0) .......(A)

Case-II : x 0 .....(i)1 � x 0 x 1 .......(ii)x2 < 1 � x

x2 + x < 1 x2 + x + 41

< 45

2

21

x

<

45

251

< x < 2

15 ......(iii)


x (i) (ii) (iii) x

215

,0 .......(B)

x (A) (B) x

215

,1

(iii) 8x6x2 1x

Domain x + 1 0 x �1

x2 � 6x + 8 0 (x � 2) (x � 4) 0 x 2 or x 4 Domain x [�1, 2] [4, )squaring x2 � 6x + 8 x + 1 x2 � 7x + 7 0

2

27

x

�

421

0 x

2217

,2

217

(iv) 2x�x28 > 6 � 3x

(a) 8 + 2x � x2 0 x [�2, 4] .... (i)case - Iwhen (i) 6 � 3x 0 x 2 ... (ii)so 8 + 2x � x2 > 36 + 9x2 � 36 x

10x2 � 38x + 28 < 0

5x2 � 19x + 14 < 0

(5x � 14) (x � 1)< 0

x

514

,1 .... (iii)

by (1) and (2) and (3)x (1, 2]Case -

6 � 3x < 0 x > 2+ ve > �ve

so x > 2 .... (iv)by (1) and (4)x (2, 4]so by case (1) and (2) x (1, 4]

(v) x2 � 7x + 10 0 and 14x � 20 � 2x2 0(x � 2) (x � 5) 0 and (x � 2) (x � 5) 0 ...........(i)so x = 2 or x = 5now check for x = 2

9 log4

4

1 � 9

� 9 � 9

which is true hence x = 2 is a solutionnow check x = 5

29

log

8

5 � 3

log2

85

� 32

(1.6)3 44.096 4which is falseso only solution is x = 2


(vi) Domain x > 0log2

2x + 2 log

2x 0

log2x (log

2x + 2) 0

log2x � 2 or log

2x 0

0 < x 41

or x 1

x

41

,0 [1, ) .......(i)

Case-I 4 � log2x < 0

positive < negative (false)Case-II 4 � log

2x 0 log

2x 4

log2

2x 2 log2x < 2 (4 � log

2x)2

Let log2x = t

t2 + 2t < 2 (4 � t)2

t2 � 18t + 32 > 0

(t � 16) (t � 2) > 0 t < 2 t > 16log

2x < 2 log

2x > 16 (Rejected)

log2x < 2

x < 4 .........(ii)by (i) and (ii)

x

41

,0 [1, 4)

13. Square root of 7 + 2i = ±

2725

i2

725 = ±(4 + 3i)

where |7 + 24 i| = 25

15. (i) z = R3 � (3 + i)+m + 2i = 03 � 3 + m = 0 & � + 2 = 0

= 28 � 6 + m = 0 m = � 2

(ii) If one root is i then other is � i

Let forth root is .

2 = 23

= 43

2a�

= 2 + i + (� i) + 43

= 411

a = 211�

20. (i) z = 1 + 2518

ie

= 259

ie

259

i�259

iee

z =

259

cos2 259

ie

|z| =

259

cos2 Arg z = 259

(ii) z = 6/5i�6/ii e2ee2

|z| = 2 Arg z = 6

5�

.


(iii) |z| = 2

2 1tan1

= sec21Arg z = 2 Arg(tan 1 � i)

= 2

2�1 = 2 �

(iv) z =

5cosi

5sin

5sin2

)1�i(

|z| =

5eccos

2

1

5sin2

2

Arg(z) = � 5

�4

=

2011


1. 1x � 1x = 1x4 .....(i)

squaring both sides

(x + 1) + (x � 1) � 2 1x2 = 4x � 1

(1 � 2x) = 2 1x2 .....(ii)squaring both sides1 + 4x2 � 4x = 4x2 � 4

4x = 5 x = 45

does not satisfy equation (i)

No solution

2. 2 log2 log

2 x + log

1/2 log

2 x22 = 1

log2 (log2 x)2 � log2 log2 x22 = 1 log2

x22log

xlog

2

22

= 1

xlog23

xlog

2

22

= 2 Let log2 x = y

y2 � 2y � 3 = 0 (y � 3) (y + 1) = 0

y = 3, �1 log2 x = 3, �1,

but log2 x > 0 log2 x = � 1 is not possible x = 8

3. (a) z1 = z2 = z3 = 1

11zz = 22zz = 33zz = 1

Given 1 = 321 z1

z1

z1

= 321 zzz = 321 zzz = 1

1 = 321 zzz

(b) � = arg (z) < 0arg (�z) = � arg (�z) � arg (z) = � � (�) = Hence (A)


4. log3/4 log8 (x2 + 7) + log1/2 log1/4 (x2 + 7) 1 = 2

log3/4 31

log2 (x2 + 7) � log2 2

)7x(log 22

= 2

let log2 (x2 + 7) = t

log3/4 3t

� log2 2t

+ 2 = 0 log3/4 3t

+ 1 �

1

2t

log2 = 0

log3/4 4t

= log2 4t

4t

= 1 t = 4

log2 (x2 + 7) = 4this gives x = ± 3.

5.21

log2(x � 1) = log2 (x � 3)

1x = x � 3

(x � 1) = x2 � 6x + 9

x2 � 7x + 10 = 0

(x � 5) (x � 2) = 0 but x 2 x = 5

6. Let 1zz

zz1

21

21

|1 � z

1 2z | < |z2 � z

1|

(1 � z1 2z ) (1 � 21zz ) < (z

2 � z

1) ( 12 zz )

1 + |z1|2 |z

2|2 � |z

1|2 � |z

2|2 < 0 (1 � |z

1|2) + (|z

1|2 � 1) |z

2|2 < 0

(1 � |z1|2) (1 � |z

2|2) < 0

which is true because of |z1| < 1 < |z

2| .

7 . (2x)n2 = (3y)n3

n2 n(2x) = n3 n(3y) = n3 (n3 + ny) ......... (1)also 3nx = 2ny

nx n3 = ny n2 ......... (2)

by (1) n2 n(2x) = n3 (n3 + ny) n 2 . n (2x) = n3

2n

3nnx3n

n22 n2x = n23 (n2 + nx) 3n2n 22 (n2x) = 0

n2x = 0 x = 21

8. Let .......23

1�4

23

1�4 = t

t23

1�4 = t

4 � 23

1t = t2

t2 + 23

1t � 4 = 0 23 t2 + t � 212 = 0

t = 232

21223411�

=

232

171�

t = 26

16,

26

18�


t = 23

8,

2

3� and

2

3� is rejected

so 6 + log3/2

23

8

23

1 = 6 + log

3/2

94

= 6 + log3/2

2

32

= 6 � 2 = 4

PART - II

1. Let z = r1 ei and w = r

2 ei z = r

1 e�i

Given, |z| = 1 i2

i1 er.er = 1

r1r

2 = 1 ...(i)

and arg (z) � arg () = 2

� = 2

Then,

i2

i1 er.erz

= r1r

2 )(ie

From Eqs. (i) and (ii), we get

z = 1. 2/ie

= cos 2

� i sin 2

z = �i.

2.x

i1i1

=

x

)i1()i1()i1()i1(

=

x

2

2

i1

)i1(

=

x

2i211

x

i1i1

= (i)x = 1 (given)

(i)x = (i)4n ,where n is any positive integer. x = 4n.

3. Since , z + i w = 0 z = �i w z = iw w = �iz

Also, arg(zw) = arg(�iz2) = arg(�i) + 2 arg(z) =

�2

+ 2 arg (z) =

2)iarg(

2 arg (z) = 2

3 arg (z) =

43

.

4. Let z = 1i

1

z =

1i1

= 1i

1

= �1i

1

.

5. Let roots be p + iq and p � iq p, q Rroot lie on line Re(z) = 1 p = 1product of roots = p2 + q2 = = 1 + q2

(1, (q 0, roots are distinct) Ans.


STRAIGHT LINE


Section (A) :

A-3._ (i) centroid

312120

,3

1650 (7, 8)

A(0,0)

B(5,12) C(16,12)

13 20

11

(ii) Let coordinates of circumcentre is O (x, y).Therefore OA = OB = OC

x2 + y2 = (x � 5)2 + (y � 12)2 = (x � 16)2 + (y � 12)2

x2 + y2 = (x � 5)2 + (y � 12)2 10x + 24y = 16g(x � 5)2 + (y � 12)2 = (x � 16)2 + (y � 12)2

2x = 21 x = 221

, y = 38

(iii)

112013

12132012110,

112013

1316205110 (7, 9)

(iv) 2 =

11132012131102012

,111320

13111613205 (27, � 21)

A-4. Let coordinates of P(x,y)given PA = PB (x � 3)2 + (y � 4)2 = (x � 5)2 + (y + 2)2

4x � 12y = 4

x � 3y = 1 ...(i)

125

143

1yx

21

= 10

6x + 2y � 26 = ± 20 3x + y � 13 = ± 10

3x + y = 23 ...(ii) 3x + y = 3 ...(iii)Solving (i) and (ii) we get (7, 2)Solving (i) and (iii) we get (1, 0)

Section (B) :B-2. Let equation of line is x + my + n = 0 ...(i)

given

1a3a

,1a

a 23

,

1b3b

,1b

b 23

and

1c3c

,1c

c 23

are collinear

1t3t

,1t

t 23

is general point which satisfies line (i)

1tt3

+ m

1t3t2

+ n = 0

t3 + m t2 + nt � (3m + n) = 0


a + b + c = �

m

ab + bc + ac =

n

abc =

nm3

Now LHS = abc � (ab + bc + ac) + 3 (a + b + c)

=

)nm3( �

n + 3

m = 0

B-5. Let point is P(x, y) and A(ae, 0) and B(�ae,0)

Given |PA � PB| = 2a 2222 y)aex(y)aex( = 2a

Let 22 y)aex( = A, 22 y)aex( = B

Hence A � B = 2a

A2 � B2 = (A + B) (A � B) A + B = �2xe

Hence 2A = 2a �2xe

A = a � ex

(x � ae)2 + y2 = (a � ex)2 2

2

a

x �

)1e(a

y22

2

= 1

Section (C) :

C-3. ObviousC-4. By parametric form 11

2 2

p(4, 1) 3x � y = 0

Q

sin

22

111,cos

22

114

it lies on 3x � y = 0

12 + 22

33 cos � 1 �

22

11 sin = 0

1 + 22

3 cos �

22

sin = 0 3cos � sin = � 22

squaring both sides9cos2 + sin2 � 6 sin cos = 8(sin2 + cos2)cos2 � 6sin cos � 7 sin2 = 07tan2 + 6tan � 1 = 0

tan = � 1, 71

.

Hence required line are x + y = 5, x � 7y + 3 = 0

Section (D) :

D-2. foot of perpendicular

32x

= 13y

= � 22 )1(3

)4323(

(x, y)

1029

,1023

image

32x

= 13y

= � 2 22 )1(3

)4323(

(x, y)

514

,5

13


slope of line perpendicular to the line y = 3x � 4 is � 31

hence its equation

y � 3 = � 31

(x � 2) x + 3y � 11 = 0

D-5. L1 : 4x + 3y � 7 = 0

L2 : 24x + 7y � 31 = 0

a1 a

2 + b

1b

2 = 4 × 24 + 3 × 7 > 0

Hence + sign gives obtuse angle bisector and � sign gives acute angle bisector

Now, put origin in both 4 × 0 + 3 × 0 � 7 < 0

24 × 0 + 7(0) � 31 < 0

Hence sign gives that bisector in which origin lies.Hence origin lies in obtuse angle bisector

Now, equation of bisector

57y3x4

= ±

2531y7x24

+ sign x � 2y + 1 = 0

� sign 2x + y � 3 = 0

Section (E) :

E-4. 12x2 � 10xy + 2y2 + 11x � 5y + = 0This represents pair of straight lines if = abc + 2fgh � af2 � bg2 � ch2 = 0

we get = 2Now12x2 � 10xy + 2y2 + 11x � 5y + 2 = (6x � 2y + p) (2x � y + q)

compair both sides 2p + 6q = 11�p � 2q = �5

solving both we get p = 4, q = 21

Hence required lines are 6x � 2y + 4 = 0 3x � y + 2 = 0

2x � y + 21

= 0 4x � 2y + 1 = 0

solving both equations we get point of intersection

2

5,

2

3

Now angle between both lines

tan = 21

21

mm1

mm

= 231

23

=

71

= tan�1 71

Now equation of pair of angle bisector

21225

y23

x22

= 525

y23

x

2x2 + 4xy � 2y2 + 16x � 4y + 7 = 0

E-5. Homogenize x2 + y2 = a2 by y = mx + c

we get x2 + y2 = a2

2

cmxy

This equation represents pair of lines passing through origin.That will be right angle ifcoeff. x2 + coeff. y2 = 0 2c2 = a2 (1 + m2)

Section (F) %

F-1_. (i) (2, 5, 8) (ii) (�5, �4, �3) (iii) (�3, 0, 7) (iv) (8, 2, 5)


F-3_. (i)

(ii)

PART - IISection (A) :

A-1*. AB = 94 = 13

BC = 1636 = 2 13

CD = 94 = 13

AD = 1636 = 2 13

AC = 164 = 65

BD = 4916 = 65its rectangle

A-4. If H is orthocentre of triangle ABC, then orthocentre of triangle BCH is point A

Section (B) :

B-2*. Since A, B, C are coffe. nearSlope of AB = Slope of BC A(k, 2 � 2k) B(1 � k, 2k) C(� k� 4, 6 � 2k)

k1kk2k22

=

4kk1k26k2

1k2k42

=

56k4

10 � 20k = (4k � 6)(2k � 1)

(4k � 6)(2k � 1) + 10(2k � 1) = 0 k = 21

, � 1


B-3. AP = 22 )4�y(x

BP = 22 )4y(x |AP � BP| = 6

AP � BP = ± 6

22 )4�y(x � 22 )4y(x = ± 6

On squaring we get the locus of P9x2 7y2 + 63 = 0

Section (C) :

C-2. x1 + y

1 = 5 ... (i)

x2 = 4 ... (ii)

co - ordinates of G are (4, 1)

3

xx1 21 = 4 ....(iii)

and ¼rFkk½3

2yy 21 = 1 ... (iv)

solving above equations, we get B & C.

C-4.

Let coordinates of point P by parametricP(2 + r cos 45º, 3 + r sin 45º)

It satisfies the line 2x � 3y + 9 = 0

2

2

r2 � 3

2

r3 + 9 = 0 r = 24

Section (D) :

D-1. a2x + a by + 1 = 0origin and (1, 1) lies on same side.a2 + ab + 1 > 0 a RD < 0 b2 � 4 < 0

b (�2, 2)

but b > 0 b (0, 2)

D-4. p = 22 )16(2

56422

=

26091

q(�11, 4)

64x+8y+35 = 0

1

2

pB : 2x � 16y � 5 = 01

q = 22 864

35481164

p < q Hence 2x � 16y � 5 = - is acute angle besectory


Section (E) :

E-2. m1 + m

2= � 10

m1m

2 =

1a

given m1 = 4m

2 m

2 = � 2, m

1 = � 8,

a = 16

E-5. Homogenize given curve with given line

3x + 4xy � 4x + 1 = 02

2x + y = 1

3x2 + 4xy � 4x(2x + y) + 1(2x + y)2 = 03x2 + 4xy � 8x2 � 4xy + 4x2 + y2 + 4xy = 0� x2 + 4xy + y2 =coeff. x2 + coeff. y2 = 0Hence angle is 90º

Section (F) :

F-3_. x2 + y2 + y2 + z2 + z2 + x2 = 362(x2 + y2 + z2) = 36

23zyx 222

F-4_. The two numbers are x and x + 2(a) x > 10(b) x + 2 > 10 x > 8(c) x + x + 2 < 34

2x < 32 X < 16Now x must be between 10 < x < 16

x (11, 13), (13, 15)

F-6_. Let the third PH reading is x

7.4 < 3

x42.848.7 < 8.2

22.2 < 15.90 + x < 24.66.3 < x < 8.7PH range should be in between 6.3 to 8.7

F-8_. Standard result.


3.

A, S, B are collinear

0

112

1xxx

100

121


3x1 = 2x

2.... (1)

B, R, C are collinear

0

103

1xxx

112

122 x1 � 2x

2 + 3 = 0 ... (2)

Solving (i) and (ii) we get x1 =

23

x2 =

49

Hence

0,

49

Q,0,23

P ,

43

,49

S,43

,23

R

6. (i) D is mid point of BC

Hence co-ordinates of D are

2

yy,

2

xx 3232

Therefore, equation of the median AD is

12

yy

2

xx1yx

1yx

3232

11 = 0

Applying R3 2R

3

2yyxx

1yx

1yx

3232

11

= 0

1yx

1yx

1yx

22

11 + 1yx

1yx

1yx

33

11 = 0

(using the addition property of determinats)(ii) Let P(x, y) be any point on the line parallel to BC

Area of ABP = Area of ACP

1yx

1yx

1yx

22

11 = 1yx

1yx

1yx

33

11

1yx

1yx

1yx

22

11 � 1yx

1yx

1yx

33

11 = 0

This gives the equation of line AP.

(iii) Let AD be the internal bisector of angle A,

DCBD

= CABA

= bc

D

bc

bycy,

bc

bxcx 2323


Let P(x,y) be any point on AD then P,A,D are collinear

1

cb

bycy

cb

bxcx1yx

1yx

2323

11

= 0

R3 (b + c) R

3

cbbycybxcx

1yx

1yx

2323

11

= 0

ccycx

1yx

1yx

33

11 + bbybx

1yx

1yx

22

11 = 0 (Addtion property)

c 1yx

1yx

1yx

33

11 + b 1yx

1yx

1yx

22

11 = 0

This is the equation of AD.

9. equation of line L1 is

y � 25

= 2 . (x � 23

)

or 2x � y � 21

= 0

or 4x � 2y � 1 = 0

equation of line L2 is

y � 25

= 1 (x � 23

) or x � y + 1 = 0

Point C is mirror image of point A w.r.t line L1

4)2(��x

= 2�

)3(�y =

20)1�6�8(�2�

C(4, 0)similarly B is mirror image of A in line L

2 = 0

1�)3�y(

1)2(��x =

2)13�2(�2�

B(2, � 1) D(1, 23

) ; E (0, 1)

median through B is

(y + 1) = 1�2/5

(x � 2) 5x + 2y = 8

median through C is

(y � 1) = )0�x(4�

1 x + 4y = 4

11. a2 + b2 = c2 .... (i)Let L is (x

1, y

1)

L is foot of perpendicular from point P(a, b) on line ABequation of AB is bx + ay � ab = 0

2211

ba

)ab�abab(�a

b�yb

a�x


211

c

ab�

ab�y

ba�x

x1 = a � 2

22

2

2

c

)b�c(a

c

ab = a3/c2

a3 = c2x1

.... (ii)similarly

b3 = c2y1

....... (iii)using these relations (ii) & (iii) in equation (i), we getrequired locus.

14. Given pair of lines are a2x2 +2h(a + b)xy + b2y2 = 0 ax2 +2hxy + by2 = 0Equation of pair of bisectors of first pair is

22

22

b�a

y�x = bah

xy

b�ay�x 22

= hxy

Which is also bisector of second pair.Hence both pair are equally inclined.

15. Let equation of chords hx + ky = 1By homogenisation

3x2 � y2 � 2x (hx + ky) + 4y (hx + ky) = 0

it makes 90º. Hence

coeff. x2 + coeff. y2 = 03 � 2h � 1 + 4k = 0 h � 2k = 1

Hence all chords are concurrent at (1, � 2)

Similarly homogenize 3x2 � y2 � 2x + 4y = 0

3x2 + 3y2 � 2x(hx + ky) + 4y (hx + ky) = 0

again coeff. x2 + coeff. y2 = 0

3 + 3 � 2h + 4k = 0 h � 2k = 3 3h

� 3k2

= 1

Hence, all chords passes though

32

,31

.

PART - II

1. here tan = 51

tan 2 = 2

51

�1

51

2

= 125

required line y = 12

x5

4. p = 5

a00 =

5

a

tan 45º = xp

p = x

Hence area = 21

(2x)(p) = px = p2 = a/5


8. Image of A(3, 10) in 2x + y � 6 = 0

23x

= 110y

= � 2

22 12

6106

23x

= 110y

= � 4

A' = (� 5, 6)

Equation of A'B is y � 3 =

4536

(x � 4)

y � 3 = � 31

(x � 4)

3y � 9 = � x + 4

x + 3y � 13 = 0

10. By geometrya2 + b2 = (a + b)2 ....(i)

By section formula

h = ba

=

a)ba(n

k = ba

=

b)ba(k

Put value of and in (i)

2

22

a

)ba(h + 2

22

b

)ba(k = (a + b)2

2

2

a

h + 2

2

b

k = 1

Locus of (opchim) is 2

2

a

x + 2

2

b

y = 1

12. x2 � 2pxy � y2 = 0

pair of angle bisector of this pair )1(1

yx 22

= p

xy

x2 � y2 + p2

xy = 0

compare this bisector pair with x2 � 2qxy � y2 = 0

p2

= �2q pq = �1.

14. 1 = 2x � 3y � 6 = 0

2 = 3x � y + 3 = 0

3 = 3x + 4y � 12 = 0

Hence [� 1, 3]

[� 2, 3]


16. Both A & B are same side of line 2x � 3y � 9 = 0

Now, perimeter of APB will be least when points A, P, B were collinear. Let B' is image of B

Then 2

0x =

34y

= � 2

22 )3(2

9120

B'

1374

,1384

Now equation of AB' is y = 110

74(x + 2)

point of intersection of given line & Q is P

1737

,1721

.

EXERCISE # 3

1. (A) Slope of such line is ± 1

(B) area of OAB = 21

× 3 × 4 = 6 sq. units

y

xO

B

A(0,�3)

(�4,0)

(C) To represent pair of straight lines c33

311

312

= 0 c = 3

(D) Lines represented by given equation are x + y + a = 0 and x + y � 9a = 0

distance between these parallel lines is = 2

a10 = a25

Comprehension # 2 (5, 6, 7)Slopes of the lines

3x + 4y = 5 is m1 = �

43

and 4x � 3y = 15 is m2 =

34

m1 m

2 = � 1

given lines are perpendicular and A = 2

Now required equation of BC is

(y � 2) = )4/tan(m1)4/tan(m

(x � 1) ......(1)

where m = slope of AB = � 43

equation of BC is (on solving (1))x � 7y + 13 = 0 and 7x + y � 9 = 0

L1 x � 7y + 13 = 0

L2 7x + y � 9 = 0

5. c + f = 46. Equation of a straight line

through (2, 3) and inclined at an angle of (/3) with y-axis ((/6) with x-axis) is

)6/cos(2x

= )6/sin(

3y

x � y3 = 2 � 33

Points at a distance c + f = 4 units from point P are

(2 + 4 cos (/6), 3 + 4 sin (/6)) (2 + 32 , 5)


and (2 � 4 cos (/6), 3 � 4 sin (/6)) (2 � 32 , 1)only (A) is true out of given options

7. Let required line be x + y = a

which is at |b � 2a � 1| = |5 � 4 � 34 � 1| = 34 units from origin

required line is x + y � 64 = 0 (since intercepts are on positive axes only)

8. ax3 + bx2y + cxy2 + dy3 = 0since this is homogeneous pair represent there straight lines passing through origin

ax3 + bx2y + cxy2 + dy3 = (y � m1x)(y � m

2x)(y � m

3x)

or put y = mx in given equation we getm3d + cm2 + bm + a = 0

m1 + m

2 + m

3 =

dc

m1m

2 + m

2m

3 + m

3m

1 =

db

m1m

2m

3 =

da

given two lines + hence m1m

2 = � 1 m

3 = a/d

eliminate m3 from remaining equation

10. S2 is standard result.

equation of angle bisectors of lines given in S1 are

52y4x3

= ± 5

2y3x4 x � y = 0 and 7x + 7y � 24 = 0

14._ Let R(5, 1) divides line segment joining P(2,10) and Q(6, � 2) in : 1

5 = 126

= 3

Hence Harmonic conjugate divides in 3 : 1 externaly

Hence required part is

13106

,13218

= (8, -8)

19. Required point is foot of perpendicular from (0, 0) on the given line which is 3

0 =

40

= 25

)1(


1. A(x1,y

1), B(x

2, y

2), C(x

3, y

3)

x1, x

2 , x

3 and y

1 , y

2, y

3 are in G.P. of common ratio r.

x2

= x1r , x

3 = x

1r2 , y

2 = y

1r, y

3 = y

1r2

Area of triangle ABC = 21

111

ryryy

rxrxx2

111

2111

= 0 A, B & C are collinear..

2. Let m be the slope of PQ then

tan 450 = )2(m1)2(m

1 = m212m


± 1 = m212m

m + 2 = 1 � 2m or �1 + 2m = m + 2

m = 31

or m = 3

PR makes 450 with PQ

equation of PQ y �1 = 31

(x � 2)

x + 3y � 5 = 0

equation of PR is y � 1 = 3(x � 2)

3x � y � 5 = 0

combined equation of PQ and PR (x + 3y � 5) (3x � y � 5) = 0

3x2 � 3y2 + 8xy � 20x � 10y + 25 = 0

3. S is the mid point of Q and R

S

213

,2

67 =

1,

213

slope of PS = m = 2/132

12

=

92

equation of line passing through (1, �1) and parallel to PS is

y + 1 = 92

(x � 1) 2x + 9y + 7 = 0

4.

BC = 2AB = 2AC = 2Hence ABC is an equilateral triangle. In equilateral triangle incentre coincides with centroid. Thus

3300

,3

120

3

1,1

5. p (h, k) be a general point in the first quadrent such that d(P, A) = d(p, O) |h � 3| + |k � 2| = |h| + |k| = h + k

[h and k are positive point (h, k) being in quadrent]If h < 3, k < 2, then (h, k) lie in region If h > 3, k < 2, then (h, k) lie in region If h > 3, k > 2, then (h, k) lie in

y = 2 A(3,2)IIIIV

I IIx+y=5/2

(0,0) (5/2,0)x = 3

O

x = 1/2

x

If h < 3, k > 2, then (h, k) lie in IV

In region 3 � h + 2 � k = h + k h + k = 25

In region h � 3 + 2 � k = h + k k = 21

not

possible


In region h � 3 + k � 2 = h + k �5 = 0 Not

possible

In region IV 3 � h + k � 2 = h + k h = 21

Set consist line segment x + y = 25

of finite length

In Ist region and x = 21

in the IV region.

6. c1 ac

1

= a1

cbyaxbcyacxa

bcycbyaxyabxa

acxaybxacabyxa

2

2

2

c1

c

1 + bc

2 + cc

3

= a1

cbyaxcyb)cba(

bcyaxcbyy)cba(

acxbxayx)cba(

222

222

222

= a1

byaxccyb1

cybaxcbyy

acxbxayx

as a2 + b2 + c2 = 1c

2 c

2 � bc

1 , c

3 c

3 � cc

1

= a1

byaxcy1

baxcy

aayx

R1 x R

1

= ax1

byaxcy1

baxcy

axaxyx2

R1 R

1 + yR

2 + R

3

= ax1

byaxcy1

baxcy

001yx 22

= (x2 + y2 + 1) (ax + by + c)Given = 0 ax + by + c = 0 which represent astraight line

7. The x-coordinate of intersection of lines 3x + 4y = 9 and y = mx + 1 is x = m43

5

For x being an integer 3 + 4m should be divisor of 5i.e. 1, �1, 5 or �5

3 + 4m = 1 m = 21

(Not integer)

4m + 3 = � 1 m = � 1 (Interger)

3 + 4m = 5 m = 21

(Not an integer)

3 + 4m = � 5 m = � 2 (integer)

there are two integral value of m


8.

In parallelogram OABCB(0,1) and point A in the point of intersection of y = mx and y = nx + 1

nm

1x

and y =

nmm

Now area of parallelogram = 2 (OAB)

=

nm1

121

2

= |nm|1

9. y = |x| � 1

y = �|x| + 1

Region is clearly square with vertices at thepoint (1,0), (0,1), (�1,0), (0,�1). So,

its area = 22 = 2.

10. Let XOS = and XOT = 2

let p(cos , sin ), then TOP = � 2

let Q be the image of P in OT. Then QOT = � 2

QOX = �

XOQ = �

Q is image of P in the line whose slope is tan 2

11.

The line segment QR makes an angle 60º with the positive direction of x-axis.

hence bisector of angle PQR will make 120º with +ve direction of x-axis.

Its equationy � 0 = tan120º (x � 0)

y = � x3

0y3x


12.

as OPA ~ OQC

OQOP

= OCOA

= 3

4/9 =

43

13. The line y = mx meets the given lines in P

1mm

,1m

1 and Q

1mm3

,1m

3. Hence equation of L

1 is

y � 1m

m

= 2

1m1

x y � 2x � 1 = � 1m

3

.........(i)

and that of L2 is y �

1mm3

= � 3

1m3

x y + 3x � 3 = 1m

6

.........(ii)

Form (i) and (ii) 3x3y1x2y

= �

21

x � 3y + 5 = 0; which is a straight line

14. equation of line y � 2 = m(x � 8) where m < 0

P

0,

m2

8 and Q (0, 2 � 8m)

Now OP + OQ = m

28 + |2 � 8m|

= 10 + )m(�2

+ 8(�m) 10 + 2 )m(8m2

18

15. The number of integral points that lie in the interior of square OABC is 20 × 20. These points are (x, y) where x,

y = 1, 2, ........., 20. Out of these 400 points 20 lie on the line AC. Out of the remaining exactly half lie in ABC.

number of integeral point in the triangle OAC = 21

[20 × 20 � 20] = 190

Alternative SolutionThere are 19 points that lie in the interior of ABC and on the line x = 1, 18 point that lie on the line x = 2 and soon. Thus, the number of desired points is

19 + 18 + 17 + .... + 2 + 1 = 2

1920 = 190.


16. Refer Figure Equation of altitude BD is x = 3.

slope of AB is 4304

= � 4

slope of OE is 1/4Equation of OE is

y = 41

x.

Lines BD and OE meets at (3, 3/4)

17. The lines given by x2 � 8x + 12 = 0 are x = 2 and x = 6.

The lines given by y2 � 14y + 45 = 0 are y = 5 and y = 9

Centre of the required circle is the centre of the square. Required centre is

295

,2

62 = (4, 7).

18. x2 � y2 + 2y = 1x = ± (y � 1)

Bisector of above lines are x = 0, y = 1so Area between x = 0, y = 1 and x + y = 3

= 21

× 2 × 2 = 2 squ. units

19. A line passing through P(h, k) and parallel to x-axis is y = k the other lines given are y = x and y + x = 2Let ABC be the formed by the points of intersection of the lines (i) , (ii) and (iii) be A(k, k) , B(1, 1), C (2�k, k)

Area of ABC = 21

1kk�2

111

1kk

= 4h2

C1 C

1 � C

2 21

1kk2�2

110

1k0

= 4h2

21

|(2 � 2k) (k � 1)| = 4h2 (k � 1)2 = 4h2

k � 1 = 2h, k � 1 = � 2h k = 2h + 1 k = � 2h + 1

locus of (h, k) is y = 2x + 1 y = � 2x + 1

20.

R is centroid hence R

34

,3


21.RQPR

= OQOP

OQOP

RQPR

= 5

22

but statement � 2 is false

Ans. (C)

22. P (� sin ( � ) , � cos )Q (cos ( � ), sin )

R (cos ( � + ) , sin ( � )) 0 < , , < 4

xR = cos ( � ) cos � sin ( � ) sin

xR = x

Q . cos + x

P . sin

yR = sin cos � cos sin

yR = y

Q . cos + y

P . sin

For P, Q, R to be collinear sin + cos = 1

sin

4 = 2

1 not possible for the given interval

4,0

non collinear

23. (1 + p) x � py + p (1 + p) = 0 ......(1)(1 + q) x � qy + q(1 + q) = 0 ......(2)on solving (1) and (2), we get C(pq, (1 + p) (1 + q)) equation of altitude CM passing through C and perpendicular to AB is x = pq .......(3)

slope of line (2) is =

q

q1

slope of altitude BN (as shown in figure) is = q1q

equation of BN is y � 0 = q1q

(x + p)

y = )q1(q

(x + p) ....... (4)

Let orthocentre of triangle be H(h, k) which is the point of intersection of (3) and (4) on solving (3) and (4), we getx = pq and y = � pq h = pq and k = �pq

h + k = 0 locus of H(h, k) is x + y = 0

24. Let slope of line L = m

)3(�m1

)3(��m

= tan 60º = 3

m31

3m

= 3

taking positive sign, m + 3 = 3 � 3m

m = 0taking negative sign

m + 3 + 3 � 3m = 0

m = 3


As L cuts x-axis m = 3

so L is y + 2 = 3 (x �3)

PART - II

1. (h � a1)2 + (k � b

1)2 = (h � a

2)2 + (k � b

2)2

2h(a1 � a

2) + 2k(b

1 � b

2) + 2

121

22

22 baba = 0

compare with (a1 � a

2)x + (b

1 � b

2) y + c = 0

c =

2baba 2

121

22

22

.

2. 3h � 1 = a cos t + b sin t

3k = a sin t � b cos t

squaring and add. (Locus)(3x � 1)2 + 9y2 = a2 + b2

3. x2 � 2pxy � y2 = 0

pair of angle bisector of this pair )1(1

yx 22

= p

xy

x2 � y2 + p2

xy = 0

compare this bisector pair with x2 � 2qxy � y2 = 0

p2

= �2q pq = �1.

4. Equation of AC

y � a sin =

sincoscossin

(x � acos )

y(cos + sin ) + x(cos � sin ) = a(sin cos + sin2 � sin cos + cos2 )y(cos + sin ) + x(cos � sin ) = a.

5. G

32k

,3h

3h2

+ (k � 2) = 1 2h + 3k = 9

Locus 2x + 3y = 9.

6. Let equation of line isax

+ by

= 1

it passes through (4, 3)a4

+ b3

= 1


sum of intercepts is �1 a + b = �1 a = �1 � b

b1

4

+ b3

= 1

4b � 3 � 3b = �b � b2

b2 + 2b � 3 = 0

b = �3, 1

b = 1, a = �22

x

+ 1y

= 1

b = �3, a = 22x

+ 3

y

= 1.

7. x2 � 2cxy � 7y2 = 0

sum of the slopes m1 + m

2 =

7c2

Product of slopes m1m

2 =

71

given m1 + m

2 = 4m

1m

2

7c2

=

74

c = 2.

8. Pair 6x2 � xy + 4cy2 = 0 has its one line 3x + 4y = 0

y = 4

x36x2 +

4x3 2

+ 4c 16x9 2

= 0

24x2 + 3x2 + 9cx2 = 0 c = �3.

9. ax + 2by + 3b = 0bx � 2ay � 3a = 0

ab6ab6x

= 22 a3b3

y

= 22 b2a2

1

Hence point of intersection (0, �3/2)

Line parallel to x-axis y = �3/2.

10. a, b, c are in H.P. b2

= a1

+ c1

a1

� b2

+ c1

= 0

given lineax

+ by

+ c1

= 0

Clearly line passes through (1, �2).

11. Centroid is

3

7 ,1


12. Pair of lines ax2 + 2(a + b)xy + by2 = 0

Area of sector A1 = 1

2r21

A2 = 2

2r21

1 +

2 = 180º

given A1 = 3A

2

1 = 3

2

2 = 45º ,

1 = 135º

Angle between lines is = ba

ab)ba(2 2

= 1

4 abba 22 = a2 + b2 + 2ab

3a2 + 3b2 + 2ab = 0.

13. Let equation of line is

by

ax = 1.

By section formula

2a

= 3 a = 6

2b

= 4 b = 8

6x

+ 8y

= 1 4x + 3y = 24.

14. Since (1, 1) and (a, a2) Both lies same side with respect to both linesa � 2a2 < 0 2a2 � a > 0

a(2a � 1) > 0

a (�, 0)

,

2

1

3a � a2 > 0 a2 � 3a < 0 a (0, 3)

Hence after taking intersection a

3,

21

.

15. AB = 22 )1k()1h(

BC = 1

AC = 22 )1k()2h(

AB2 + BC2 = AC2 (h � 1)2 + (k � 1)2 + 1 = (h � 2)2 + (k � 1)2

2h = 2 h = 1

Area of ABC = 22 )1k()1h(21

× 1 = 1

(K � 1)2 = 4 k � 1 = ±2 k = 3, �1.

16.


The line segment QR makes an angle 60º with the positive direction of x-axis.

hence bisector of angle PQR will make 120º with +ve direction of x-axis.

Its equationy � 0 = tan120º (x � 0)

y = � x3

0y3x

17. Bisector of x = 0 and y = 0 is either y = x or y = �x

If y = x is Bisector, thenmx2 + (1 � m2)x2 � mx2 = 0 m + 1 � m2 � m = 0 m2 = 1 m = ±1.

18. Slope of PQ = k1

1

Hence equation of to line PQ

y � 27

= (k � 1)

2

)k1(x

Put x = 0

y = 27

+ 2

)k1()k1( = �4

7 + (1 � k2) = �8 k2 = 16 k = ±4.

Hence possible answer = �4.

19. p(p2 + 1) x � y + q = 0

(p2 + 1)2 x + (p2 + 1) y + 2q = 0 are perpendicularfor a common line lines are parallel slopes are equal

1

)1p(p 2

= � )1p(

)1p(2

22

p = � 1

20. BPAP

=

13

(x + 1)2 + y2 = 9((x � 1)2 + y2)x2 + 2x + 1 + y2 = 9x2 + 9y2 � 18x + 9

8x2 + 8y2 � 20x + 8 = 0

x2 + y2 � 4

10x + 1 = 0

circumcentre

0,

45

.

21.5x

+ by

= 1

513

+ b

32 = 1

b32

= � 58

b = � 20

5x

� 20y

= 1 4x � y = 20

Line K has same slope � c3

= 4

c = � 43

4x � y = � 3

distance = 17

23



22.

AD : DB = 5:22 OD is angle bisectorof angle AOB St : 1 true

St. 2 false (obvious) Ans.

23. x + y = |a|ax � y = 1

if a > 0x + y = aax � y = 1

------------------------------------x(1 + a) = 1 + a as x = 1y = a � 1

It is in the first quadrantso a � 1 0a 1a [1, )If a < 0x + y = � a

ax � y = 1

+-------------------------------x(1 + a) = 1 � a

x = a1a�1

> 0

1a1�a

< 0 .............(1)

y = � a � a1a�1

= a1

a1�a�a� 2

> 0

�

1a1a2

> 0 1a1a2

< 0 .............(2)

from (1) and (2) a {}

24. = 3h � 2 = 3k

= 3k +2third vertex on the line 2x + 3y = 92 + 3 = 92(3h) + 3(3k + 2) = 92h + 3k = 12x + 3y � 1 = 0


25.

C

514

,58

Line 2x + y = k passes C

514

,58

514

582

= k

k = 6

26. (y � 2) = m(x � 1)

OP = 1 � m2

OQ = 2 � m

Area of POQ = 21

(OP)(OQ) = 21

m

21 (2 � m)

= 21

2

m4

m2

=

m4

m421

m = �2

ADVANCE LEVEL PROBLEMSPART - I

1. Condition for concurrency

cc41

bb31

aa21

= 0 b2

= a1

+ c1

So a, b, c are in H.P.

2. x2(sec2� sin2) � 2xy tan + y2sin2= 0

|m1 � m

2| = 21

221 mm4)mm(

2

222

2 sin

sinsec4

sin

tan2 = 2

3. Equation of family of curves passing through intersection of C1 & C

2 is

x2 + 4y2 � 2xy � 9x + 3 + (2x2 + 3y2 � 4xy + 3x � 1)= 0 .............(i)

It will give the joint equation of pair of lines passing through origin,if coefficient of x = 0 & Constant = 0 = 3put = 3 in equation (i), we getx2 + 4y2 � 2xy + 6x2 + 9y2 � 12xy = 0

It will subtend 90º at origin if coeff. of x2 + coeff. of y2 = 0= �19


4.

5

A

(3, 2)

C

5

B

O

For B and C apply Parametric form

cos3x

=

sin2y

= ± 5

Points are (7, 5) & (�1, �1)

5.

From figure it is clear that A is orthocentre of ABC

6. px2 � qxy � y2 = 0m

1 = tan m

2 = tan

m1 + m

2 = � q , m

1 m

2 = � p

tan (+ ) = p1q

7. (2 + ) x + (1 � 2) y + (4 � 3) = 0

distance from point A is = 22 )21()2(

)34()21(32)2(

= 10

= 1Hence, the required line is 3x � y + 1 = 0

8.

To find equations of AB and CD AB and CD are parallel to 3x � 4y = 0 and at a distance of 2 units from (1, 1)

3x � 4y + k = 0

and5

k4�3 = 2 k � 1 = 10

k = 11, � 9

equations of two sides of the square which are parallel to 3x � 4y = 0 are

3x � 4y + 11 = 0 and 3x � 4y � 9 = 0

Now the remaining two sides will be perpendicular to3x � 4y = 0 and at a distance of 2 unit from (1, 1) 4x + 3y + k = 0

and5

k34 = 2 k + 7 = 10

k = 3, � 17 remaining two sides are4x + 3y + 3 = 0 and 4x + 3y � 17 = 0

9. Givenx cos + y sin = a .......(1)x sin � y cos = b .......(2)square (1) and (2) then add them.x2 + y2 = a2 + b2


10. Let point of concurrency of given family of lines is Q and it can be obtained by solving3x + 4y + 6 = 0and x + y + 2 = 0 Q (� 2, 0)

Now required line will pass through Q(� 2, 0) and perpendicular to PQ.

Equation of required line is

y � 0 = 34�

(x + 2) 4x + 3y + 8 = 0

11. (i) After reflection about line y = x position of point will be (1, 4)(ii) After this step (3, 4)

(iii) (h + ki) = (3 + 4i) ei/4 = (3 + 4i)

i

2

1

2

1 h = �

2

1, k =

2

7

Hence the final position will be

2

7,

2

1

(h, k)

(3, 4)

12. Let the point P be (x, y)|x| + |y| = 3 .......(i)

�P(x, y)Case - 1 x > 0, y > 0Equation (i) will become :

x + y = 3

B

O A

Area (OAB) = 29

Similarly for each quadrant , a triangle will be formed. Hence area enclosed will be 18.

13.

P (� 4, � 2)

and Q (� 2, � 6)

Let slopes of PM and QM be m1 and m

2 respectively.

m1 = 3 and m

2 =

21

.

Let �� be the acute angle between PM and QM

tan = 21

21

mm1

m�m

tan = 1 =

4


14. For collinearity of 3 points

1sincos

13

11

102

= 0

3 sin � cos = 2 = 32

15. For ABCa + b > c, b + c > a, c + a > bx2 + 4x + 3 > x2 + 3x + 8 x > 5

x2 + 5x + 11 > x2 + 2x x > 311

2x2 + 5x + 8 > 2x + 3 2x2 + 3x + 5 > 0 x RCommon to all is x > 5.

16.

point of intersection of the two ray is P(0, 2)

Point A is

0,

3

2 or

0,

3

2

and PO is bisector of the angle between two rays required point is (0, 0)

17. Slope BD = �1,

A

B C

D

(x , y )2 2

(x , y )1 1

(x, y)

(a, b)

O

b2

b2

a+ , ,

Equation of BD, x + y = a + bequation of AC x � y = a

On solving, we get O

2

b,

2

ba

B (a + b, 0)m

AC = 1 = tan

OA = OD = 2

b

Apply parametric form for finding A & C

2

12b

ax

=

2

12b

y

= ± 2

b

A & C are (a, 0) and (a + b, b)

18.qpr

prq

rqp

= 0

p3 + q3 + r3 � 3pqr = 0 (p + q + r) (p2 + q2 + r2 � pq � qr � rp) = 0


19. k1u � k

2v = 0 .... (i)

k1u + k

2v = 0 .... (ii)

equations of bisectors of the angles formed by lines (i) and (ii) are

221

221

21

)akbk()bk�ak(

vk�uk

=

221

221

21

)ak�bk()bkak(

)vkuk(

k1u � k

2v = (k

1u + k

2v) .... (iii)

(i) by taking positive sign in (iii), we getk

1u � k

2v = k

1u + k

2v.

2k2v = 0 v = 0

(ii) by taking negative sing in (iii), we getu = 0

PART - II

1.

cos1x1 =

sin2y1 = ±

36

( (1, 2) lie below the line)

x1 = 1 +

36

cos, y1 = 2 +

36

sin

(x1, y

1) lies on x + y = 4

3 + 36

(sin + cos) = 4

sin + cos = 6

3 =

2·3

3

2

1(sin + cos) =

2

3 ·

2

1

sin( + 4

) = 23

= sin 60º or sin 120º

= 12

, 125

rejectedis

36

2.

tan =

43

·34

1

43

34

= 247

257

24

cosec = 725


P1 is from AB to CD, P

2 is from AD to BC

for finding P1 choose arbitrary point (a, a) on AB

P1=

5a2

5

a3a3a4

for P2 choose arbitrary point (a/2, 0) on AD

P2 =

5a

5

aa20

Area = P

1P

2 cosec =

7a2 2

3. Slope of BC is

= )3(1)1(3

=

22

= � 1

Equation of a line parallel to BC isy = � x + c i.e. x + y � c = 0

its distnace from the origin is

21

2

c c =

2

1

Equations of the lines are

x + y ± 2

1 = 0

Since the required line intersects OB and OC, therefore, it is the line whose y intercept is negative. H e n c e

the required line is x + y + 2

1 = 0.

4. AB = 36164 = 425

BC = 36576 = 612

AC = 169256 = 425AB = AC BC, triangle is isoscelesand in isosceles triangle O, H, I, G are collinear

5. D is mid point of AB and lies on the line 3 x + y = 6

3 · 2

12

+ 2

12 = 6

32 � 7 + 2 = 0 ........(1)

= 31

, 2

multiplication of slope of AB & line = �1

1

12

(�3) = �1

2 � � 2 = 0 ........(2) = �1, 2

= 2 satisfies both (1) & (2)

6. AB = AC

A

B C(1, -10)

x+y-3 = 07x-y+3 = 0

m = mBC


The bisectors are 25

3yx7 = ±

2

)3yx(

Their slopes are 31

, � 3

Required lines are y + 10 = 31

(x � 1) and y + 10 = � 3(x � 1)

i.e. x � 3y � 31 = 0 and 3x + y + 7 = 0

7.

AB = 2d OAPB is a cyclic quadrilateral and OP will be diameter of the circumcircle of this quadrilateralLet Q be the centre of the circle in AQT

sin= 2

12

1 yx

d2

..... (i)

tan = ba

ab�h2 2

.... (ii)

from (i) and (ii), we get

221

21

2

d4�yx

d2

ba

ab�h2

Locus of P(x1, y

1) is

(x2 + y2) (h2 � ab) = d2 {(a � b)2 + 4h2}

8. The slopes of the lines AB, BC and CA are �1, �71

and �7 respectively

Let m1 = �

71

, m2 = �1, m

3 = � 7

m1 > m

2 > m

3

tangent of internal angles of the triangle are

A

B

x +

y �

5 =

0

7x + y + 14 = 0

Cx + 7y � 7 = 0

tan A = 43

, tan B = 43

and tan C = � 724

interior angles A and B are acute and interior angle C is obtuse internal bisector of B = acute bisector of B 3x + 6y � 16 = 0

External bisector of C acute bisector of C 8x + 8y + 7 = 0Internal bisector of A acute bisector of A 12x + 6y � 11 = 0

9. Equation of line passing through P(�1, 2) making angle with + ve direction of x-axis is given by

cos1x

=

sin2y

= r1 , r

2 , r

3(parametric form)

where r1, r

2, r

3 are distances of points A, Q, B from point P respectively.

Hence coordinates of A(r1 cos � 1, r

1 sin + 2)


But A lies on x-axis

Hence r1 sin + 2 = 0 sin = �

1r2

coordinates of point B (r3 cos � 1, r

3 sin + 2)

Point B lies on y-axis hence

r3 cos � 1 = 0 cos =

3r1

Coordinates of point Q (r2cos � 1, r

2 sin + 2)

Hence h = r2 cos � 1 cos =

2r1h

and k = r2 sin + 2 sin =

2r2k

Now given that r1 , r

2, r

3 are in H.P.

2r2

= 1r1

+ 3r1

2r2

= � 2

sin + cos

2r2

= � 21

2r2k

+ 2r

)1h( 2 = �

21

(k � 2) + (h + 1)

4 = � k + 2 + 2h + 2 2h = klocus y = 2xAlt : Use P and Q are harmonic conjugates with respect to A and B.

10. Let P(h, k) be a variable point on the lines passing through the origin.

22

11

kh

hykx

=

(kx1 � hy

1)2 = 2 (h2 + k2)

locus of P(h, k) is (x1y � xy

1)2 = 2 (x2 + y2)

solving it, we get(y

12 � 2) x2 � 2x

1y

1 xy + (x

12 � 2) y2 = 0.

11. Let the line (L) through the origin isx = r cosy = r sin

as L intersects L1 at Q and OQ = r

1

r1 sin = m

1r

1 cos + c

1..............(1)

similarly, L intersects L2 at R and OR = r

2

r2 sin = m

2r

2 cos + c

2..............(2)

Let P (h, k) & OP = r r2 = r

1 r

2..............(3)

& h = r cos ..............(4)k = r sin ..............(5)

putting the values of r1 and r

2 from (1) and (2) in (3)

r2 = )cosm(sinc

1

1

. )cosm(sin

c

2

2

..............(6)

putting the value of cos and sinfrom (4) and (5) in (6), we get

r2 =

rh

mrk

rh

mrk

cc

21

21 (k � m

1h) (k � m

2h) = c

1c

2

replacing (h, k) by (x, y) we get the desired locusas (y � m

1x) ( y � m

2x) = c

1c

2


12. take any point on line3x + 2y + 4 = 0

2x + 3y + 1 = 0

3x + 2y + 4 = 0

134

,1320

(0, �2)

(�2,1)

put x = 0, we get y = � 2

Now image of (0, �2)

in line 2x + 3y + 1 = 0

20x

= 3

2y = � 2

94160

= 1310

Hence x = 1320

and y = 1330

� 2 = 134

Point of intersection of 2x + 3y + 1 = 0 and 3x + 2y + 4 = 0 is (�2, 1)

Hence equation of other line y � 134

= 213/20113/4

1320

x

After simplification, we get 9x + 46y = 28

CIRCLE


Section (A) :

A-2. Since BD is diameter of circleHence (x � a) (x � 0) + (y � 0) (y � a) = 0

x2 + y2 = a (x + y)

A-6. x = �3 + 2sin x + 3 = 2 sin y = 4 + 2cos y � 4 = 2 cos

Squarring and add (x + 3)2 + (y � 4)2 = 4

Section (B) :

B-4. S1 (9)2 + (0)2 � 16 = 65 > 0

Since (9, 0) lies outside the circle. Hence two real tangents can be drawn.Now S x2 + y2 � 16

S1 9x � 16

Hence pair of tangents SS1 = T2

(x2 + y2 � 16) (65) = (9x � 16)2

65x2 + 65y2 � 1040 = 81 x2 + 256 � 288 x

16x2 � 65y2 � 288x + 1296 = 0

Angle between these tangents = )ba(abh2 2

= 6516

651602

=

49658

B-5. given 6gf 22 = 2 f3g3gf 22

3g2 + 3f2 + 12g + 12f + 6 = 0 g2 + f2 + 4g + 4f + 2 = 0

Section (C) :

C-4. Area of triangle formed by pair of tangents & chord of contact is = 22

3

LR

RL

Here R = a

L = 222 akh

Hence Area =

22

2/3222

kh

akha


C-7. T = S1

�2x � 3y +3(x �2) +4 (y � 3) + 9

= 4 + 9 � 12 � 24 + 9

x + y + 5 = 0

Section (D) :D-1. S

1 : x2 + y2 � 2x � 6y + 9 = 0 C

1(1, 3), r

1 = 1

S2 : x2 + y2 + 6x � 2y + 1 = 0 C

2(�3, 1), r

2 = 3

C1C

2 = 416 = 20

n + r2 = 4

Hence C1C

2 > r

1 + r

2Both circles are non-intersecting.

Hence there are four common tangents.

Transverse common tangents :

coordinate of P

31

91,

31

33

2

5,0

Let slope of these tangents is m

y � 25

= m(x � 0) mx � y + 25

= 0

Now 2m1

25

3m

= 1 21

m = 2m1

m2 + 41

� m = 1 + m2 m = �43

, other tangents is vertical

Equation of tangents x = 0

�43

x � y + 25

= 0 �3x � 4y + 10 = 0 3x + 3y = 10

Direct common tangents

coordinate of Q

3191

,3133

Q(3, 4)

Hence equations y � 4 = m(x � 3) mx � y + (4 � 3m) = 0

2m1

m343m

= 1

|1 � 2m| = 2m1

1 + 4m2 � 4m = 1 + m2 3m2 � 4m = 0 m = 0, 34

Hence equation y � 4 = 0(x � 3) y = 4

y � 4 = 34

(x � 3) 4x � 3y = 0

D-3. Equation of circle passing through origin is x2 + y2 + 2gx + 2fy = 0This circle cuts the circle x2 + y2 � 4x + 6y + 10 = 0 orthogonally

2g(�2) + 2f(3) = 0 + 10

�2g + 3f � 5 = 0 ...(1)& x2 + y2 + 12y + 6 = 0 also

2g(0) + 2f(6) = 6 + 0 f = 21


�2g + 23

�5 = 0 2g = �27

g = �47

Hence circle x2 + y2 + 2

4

7x + 2

2

1y = 0

2x2 + 2y2 � 7x + 2y = 0

Section (E) :E-1. Equation of circumcircle of this triangle

L1L

2 + L

2L

3 + L

3L

1 = 0

(x + 2y � 5)(x + y � 6)+(x + y � 6)(2x + y � 4)+(x + 2y � 5)(2x + y � 4) = 0

coef. of xy = 0 3 + 3 + 5 = 0 3 + 5 + 3 = 0 ...(1)coef. x2 = coef. y2 1 + 2 + 2 = 2 + + 2

= 1 = �56

Hence (x + 2y � 5) (x + y � 6) + (x + y � 6)(2x + y � 4) �56

(x + 2y � 5) (2x + y � 4) = 0

x2 + y2 � 17x � 19y + 50 = 0

E-2. x2 + y2 � 10x + (2x � y) = 0 ....(i)x2 + y2 + 2x ( � 5) � y = 0Centre (� ( � 5) , /2)Using on y = 2x

)5(22

102

5

Putting = 4x2 + y2 � 2x � 4y = 0

PART - II

Section (A) :

A-1.

diameter = 24

r = 22

A-6.* Let equation of required circle isx2 + y2 + 2gx + 2fy + c = 0it passes through (1, �2) & (3, �4)

2g � 4f + c = �5

6g � 8f + c = �25

4g � 8f + 2c = �10

6g � 8f + c = �25

�2g + c = 15

circle touches x-axis g2 = c g2 � 2g � 15 = 0

g = 5, � 3

g = 5, c = 25, f = 10 x2 + y2 + 10x + 20y + 25 = 0g = �3, c = 9, f= 2 x2 + y2 � 6x + 4y + 9 = 0


Section (B) :B-1. Point on the line x + y + 13 = 0 nearest to the circle x2 + y2 + 4x + 6y � 5 = 0 is foot of from centre

12x

= 1

3y = �

22 11

1332 = �4

x = �6 y = �7

B-3. Let slope of required line is my � 3 = m(x � 2)

mx � y +(3 � 2m) = 0

length of from origin = 3 9 + 4m2 � 12m = 9 + 9m2

5m2 + 12m = 0 m = 0, �5

12

Hence lines are y � 3 = 0 y = 3

y � 3 = �5

12(x � 2) 5y � 15 = �12x + 24 12x + 5y = 39.

B-5. Line parallel to given line 4x + 3y + 5 = 0 is 4x + 3y + k = 0This is tangent to x2 + y2 � 6x + 4y � 12 = 0

5k612

= 5

6 + k = ±25 k = 19, �31

Hence required line 4x + 3y � 31 = 0, 4x + 3y + 19 = 0

B-9. As we knowPA.PB = PT2 = (Length of tangent)2

Length of tangent = 916 = 12

B-10. Let any point on the circle x2 + y2 + 2gx + 2fy + p = 0 (, )This point satisfies 2 + 2 + 2g + 2f + p = 0Length of tangent from this point to circle x2 + y2 + 2gx + 2fy + q = 0

length = 1S = qf2g222 = pq

Section (C) :C-2. Required point is foot of

23x

= 51y

= �

254856

= �1

x = 1, y = 4

C-4.* Let point on line be(h, 4 � 2h) (chord of contact)

hx + y (4 � 2h) = 1

h(x � 2y) + 4y � 1 = 0 Point

4

1,

2

1

Section (E) :E-1. Let required circle is x2 + y2 + 2gx + 2fy + c = 0

Hence common chord with x2 + y2 � 4 = 0

is 2gx + 2fy + c + y = 0This is diameter of circle x2 + y2 = 4 hence c = �4.

Now again common chord with other circle2x(g + 1) + 2y(f � 3) + (c � 1) = 0

This is diameter of x2 + y2 � 2x + 6y + 1 = 0

2(g + 1) � 6(f � 3) + 5 = 0

2g � 6f + 15 = 0

locus 2x � 3y � 15 = 0 which is st. line.


E-2. Common chord of given circle6x + 4y + (p + q) = 0This is diameter of x2 + y2 � 2x + 8y � q = 0

centre (1, �4)

6 � 16 + (p + q) = 0

p + q = 10


3. Equation of circle whose diameter's end points are (a, b) and (h, k)(x � a) (x � h) + (y � b) (y � k) = 0

x2 + y2 � x(a + h) � y(b + k) + ah + bk = 0

it touches x-axis.

Hence g2 = c

2

2ha

= ah + bk

(h � a)2 = 4bk Locus of (h, k) is (x � a)2 = 4by.

5. As we know if two lines are m

1m

2 = �1

hk

= �1

2 � k = �2 + hLocus of (, ) is x2 + y2 = xh + yk

6.

2b

,ah2 lies on circle

2(2h � a)(2h � 2a) 2b

(�2b) = 0

4(h �a)(2h � a) + b2 = 08h2 � 12ah + 4a2 + b2 = 0D > 0144a2 � 4 × 8 (4a2 + b2) = 09a2 � 8a2 � 2b2 > 0 a2 > 2b2

9.

Let required equation of circle is x2 + y2 + 2gx + 2fy + c = 0Now common chord of given circle with required circle areCommon chord 2gx + 2fy + (c + 4) = 0 it is also diameter of circle x2 + y2 = 4. Hence c = �4

similarly with x2 + y2 � 6x � 8y + 10 = 0 2x(g + 3) + 2y(f + 4) � 14 = 0

6(g + 3) + 8(f + 4) � 14 = 0


6g + 8f + 36 = 0 3g + 4f + 18 = 0

With circle x2 + y2 + 2x � 4y � 2 = 0 2x (g � 1) + 2y(f + 2) � 2 = 0

�2(g � 1) + 4(f + 2) � 2 = 0

�2g + 4f + 8 = 0

2g � 4f � 8 = 0

after simplification g = �2, f = �3, c = �4

Hence circle x2 + y2 � 4x � 6y � 4 = 0

13. 42 � 5m2 + 6 + 1 = 0

(3 + 1)2 = 5(2 + m2) 22 m

1m.03

= 5

Hence centre (3, 0), radius = 5

15. Equation of circle having centre (x1, y

1) and radius 'd'

(x � x1)2 + (y � y

1)2 = d2

x2 + y2 = a2

Equation of common chord2xx

1 + 2yy

1 � x

12 � y

12 � a2 + d2 = 0

2xx1 + 2yy

1 � 2a2 + d2 = 0

PART - II

1. Point

t1

,t lies on x2 + y2 = 16

t2 + 2t

1 = 16 t4 �16t2 + 1 = 0 ........(i)

If roots are t1, t2, t3, t4 then

t1t2t3t4 = 1 .........(ii)

5.2

c01 = 2 c � 1 = ±2 c = �1, 3

But c = �1 common point is one

c = 3 common point is infiniteHence c = �1 is Answer.

8. Equation of chords of contact from (0, 0) & (g, f)gx + fy + c = 0gx + fy + g(x + g) + f(y + f) + c = 0

gx + fy +

2cfg 22

= 0

Distance between these parallel lines = 22

22

fg2

cfg

11.

(x + g)(x � 2) + (y + f)(y �1) = 0


12. cos /3 = 5

)3k()2h( 22

Locus (x + 2)2 + (y � 3)2 = 6.25

14.

slope of C1C

2 is tan = �

34

By using parametric coordinates C2 (± 3 cos , ± 3 sin )

C2 (± 3 (�3/5) , ± 3 (4/5)

C2 (± 9/5 , 12/5)

20. (x2 + y2 � 6x � 4y � 12) + (4x + 3y � 6) = 0

This is family of circle passing through points of intersection of circlex2 + y2 � 6x � 4y � 12 = 0 and line 4x + 3y � 6 = 0

other family will cut this family at A & B.Hence locus of centre of circle of other family is thiscommon chord 4x + 3y � 6 = 0

22. Let any point P(x1, y

1) to the circle x2 + y2 �

5x16

+ 15

y64 = 0

x1

2 + y1

2 � 5

16x

1 +

1564

y1 = 0

Length of tangent from P(x1, y

1) to the circle are in ration

2

1

S

S =

60y564

x548

yx

15y532

x524

yx

1121

21

1121

21

= 60y

564

x5

48y

1564

x5

16

15y5

32x

524

y1564

x5

16

1111

1111

= 900y128x96

225y32x24

11

11

= )225y32x24(4

225y32x24

11

11

=

21

24. Two fixed pts. are point of intersection ofx2 + y2 �2x �2 = 0 & y = 0


Point x2 � 2x � 2 = 0

(x � 1)2 � 3 = 0

x �1 = 3 , x � 1 = 3

)0,31( )0,31(

25. 512C3C4

= C C = 1, 6

EXERCISE # 3

Match the column :

1. (A) S1 � S

2 = 0 is the required common chord i.e 2x = a

Make homogeneous, we get x2 + y2 � 8.4 2

2

a

x = 0

As pair of lines substending angle of 90° at origin

coefficient of x2 + coefficient of y2 = 0 a = ± 4

(B) y = 22 3 (x � 1) passes through centre (1, 0) of circle

(C) Three lines are parallel

(D) 2(r1 + r

2) = 4

r1 + r

2 = 2

( 3 , 2)

2rr 21 = 1

Comprehension # 2 (6 to 8)6. PQC

1 and PRC

2 are similar

2

1

PRCofAreaPQCofArea

= 2

2

21

r

r =

259

7. Let mid point m(h, k). Now equation of chordT = S

1

hx + ky + 3(x + h) = h2 + k2 + 6hit passes through (1, 0)h + 3(1 + h) = h2 + k2 + 6hlocus x2 + y2 + 2x � 3 = 0

But clear from Geometry it will be arc of BC

8. Common chord of S1 & answer of 7

4x + 3 = 0 x = �3/4


at x = �3/4

2

343

+ y2 = 9 y2 = 9 �

1681

y2 = 1663

y = ± 473

Hence tan = )4/31(

473

=

773

tan = 7

3

10. Statement-1 is true and statement-2 is false as radius = 21

22

11. Statement-1 : There is exactly one circle whose centre is the radical centre and the radius equal to the lengthof tangent drawn from the radical centre to any of the given circles.

Statement-2 is True But does not explain Statement-1.

13. (0, 0) & (8, 6) lie on the director circle of x2 + y2 � 14x + 2y + 25 = 0

so � = 0

16. P = 29

1856 = 29

r2 = p2 + 32 = 38 r = 38

19. x2 + y2 � 8x � 12y + p = 0

Power of (2, 5) is S1 = 4 + 25 � 16 � 60 + P = P � 47 < 0 P < 47

Circle neither touches nor cuts coordinate axesg2 � c < 0 16 � p < 0 p > 16f2 � c < 0 36 � p < 0 p > 36

taking intersection P (36, 47)


1. The lines given by x2 � 8x + 12 = 0 are x = 2 and x = 6.

The lines given by y2 � 14y + 45 = 0 are y = 5 and y = 9

Centre of the required circle is the centre of the square. Required centre is

2

95,

2

62 = (4, 7).

2. Clearly from the figure the radius of bigger circle

r2 = 22 + {(2 � 1)2 + (1 � 3)2}

r2 = 9 or r = 3

3. The equation of circle having tangent 2x + 3y + 1 = 0 at (1, � 1)

(x � 1)2 + (y + 1)2 + (2x + 3y + 1) = 0x2 + y2 + 2x( � 1) + y(3 + 2) + ( + 2) = 0 ... (i)


equation of circle having end points of diameter (0, � 1) and (�2, 3) is

x(x + 2) + (y + 1) (y � 3) = 0

or x2 + y2 + 2x � 2y � 3 = 0 ... (ii)since (i) & (ii) cut orthogonally

2

)23(21.

2)2�2(2

(� 1) = + 2 � 3

2 � 2 � 3 � 2 = � 1

2 = � 3 = � 3/2

from equation (i), equation of required circle is2x2 + 2y2 � 10x� 5y + 1 = 0

4. 22 )1k()0h( = 1 + |k|

or

h2 + k2 � 2k + 1 = 1 + 2|k| + k2

h2 = 2|k| + 2k x2 = 4y if y > 0 & x = 0 if y 0

5. Clearly P is the incentre of triangle ABC.

r = s

= s

)cs)(bs)(as(

Here 2s = 7 + 8 + 9 s = 12

Here r = 12

3.4.5 = 5

6. Statement-1 is true because point (17, 7) lies on the director circle and Statement-2 is equation of director circleof given circle.

7. )r2)(3(21

18 r = 6

Line, y =

r2

(x � 2) is tangent to circle

(x � r)2 + (y � r)2 = r2

2 = 3r and r = 6r = 2

8. (ax2 + by2 + c) (x2 � 5xy + 6y2) = 0 x = 3y or x = 2y or ax2 + by2 + c = 0If a = b and c is of opposite sign, then it will represent a circleHence (B) is correct option.

9*. PS . ST = QS . SR

R

T

Q

P

s

Now HM < GM

ST1

PS1

2

< ST.PS


PS1

+ ST1

> SR.QS

2 B is correct and A is wrong.

Now QR = QS + SRApplying AM > GM

2SRQS

> SR.QS

QR > 2 SR.QS ST.PS

2QR4

ST1

PS1

> ST.PS

2 >

QR4

D is correct and C is wrong

�B� and �D� are correct.

10.

Let G () be the centre of C

= 233

� 1. cos 30 = 3

= 23

� 1 . sin 30 = 1

equation of C is

(x � 3 )2 + (y � 1)2 = 1

11.

FGD = DGE = 120° F = ( 3 , 0) and

GF = GE = GD = 1 E =

23

,23

12. Slope QR = 3 equation of QR is y � 23

= 3

23

x

y = 3 x and slope of RP = 0 equation RP is y = 0

13. The distance between L1 and L

2 is

13

6 < 2

Statement �1� is True because distance between lines is less than radius but L2 need not be a diameter.

Statement �2� is False because if

L1 is diameter then L

2 has to be a chord of circle

Thus �C� is correct


14. For required circle, P(1, 8) and O(3, 2) will be the end points of its diameter.

(x � 1) (x � 3) + (y � 8) (y � 2) = 0 x2 + y2 � 4x � 10y + 19 = 0

15. (r + 1)2 = 2 + 9r2 + 8 = 2

r2 + 2r + 1 = r2 + 8 + 92r = 16r = 8

16. Since distance between parallel chords is greater than radius, therefore both chords lie on opposite side ofcentre.

2 cos k2

+ 2 cos k

= 3 + 1

Let k2

=

2 cos + 2 cos 2 = 3 + 1

2 cos + 2 (2 cos2 � 1) = 3 + 1 4 cos2 + 2 cos � (3 + 3 ) = 0

cos = )4(2

)33(1642 =

)4(23412122

= 4

11212

=

4)132(1

cos k2

= 23

, 2

)13( Rejected

k2

= 6

k = 3 [k] = 3

17. Let equation of circle isx2 + y2 + 2gx + 2 fy + c = 0

as it passes through (-1,0) & (0,2) 1 � 2g + c = 0

and 4 + 4 f+ c =0

also f2 = c f = �2, c= 4 ; g = 25

equation of circle isx2 + y2 + 5x � 4y + 4 =0

which passes through (�4, 0)

18. 2x � 3y = 1, x2 + y2 6

S

)V()()()(

4

1,

8

1,

4

1,

4

1,

4

3,

2

5,

4

3,2

Plot the two curvesI, III, IV will lie inside the circle and point (I, III, IV) will lie on the P regionif (0, 0) and the given point will lie opposite to the line 2x � 3y � 1 = 0

P(0, 0) = negative, P

43

,2 = positive, P

4

1,

4

1 = positive P

4

1,

8

1 = negative


P

4

3,

2

5 = positive , but it will not lie in the given circle

so point

43

,2 and

41

,41

will lie on the opp side of the line

so two point

43

,2 and

41

,41

Further

4

3,2 and

4

1,

4

1 satisfy S

1 < 0

19. Circle x2 + y2 = 9line 4x � 5y = 20

P

520�t4

,t

equation of chord AB whose mid point is M (h, k)T = S

1

hx + ky = h2 + k2 ........(1)equation of chord of contact AB with respect to P.T = 0

tx +

520�t4

y = 9 ........(2)

comparing equation (1) and (2)

9kh

20�t4k5

th 22

on solving45k = 36h � 20h2 � 20k2 Locus is 20(x2 +y2) � 36 x+ 45y = 0

Sol. 20 to 21

20.

B divides C1 C

2 in 2 : 1 externally

B(6, 0)Hence let equation of common tangent isy � 0 = m(x � 6)

mx � y � 6m = 0

length of r dropped from center (0, 0) = radius

2m1

m6

= 2 m = ±

22

1

equation is x + y = 6 or x � y = 6

21. Equation of L is

x � + c = 0

length of perpendicular dropped from centre = radius of circle


= 1 C = �1, �5

x � y = 1 or x � y = 5

PART - II

1. S1 : (x � 1)2 + (y � 3)2 = r2 C

1 (1, 3), r

1 = r

S2 : x2 + y2 � 8x + 2y + 8 = 0 C

2 (4, �1), r

2 = 3

circles intersect |r1 � r

2| < C

1 C

2 < r

1 + r

2

|r � 3| < 5 < r + 3

|r � 3| < 5 �5 < r � 3 < 5 �2 < r < 8

5 < r + 3 r > 2After intersection 2 < r < 8.

2. Point of intersection of 2x � 3y = 5

3x � 4y = 7 is (1, � 1)

Hence centre (1, �1), Area = 154 = r = 7

equation of circle (x � 1)2 + (y + 1)2 = 72

x2 + y2 � 2x + 2y = 47.

3. Let centre of circle is (h, k) and it passes through (a, b)equation of circle is (x � h)2 + (y � k)2 = (h � a)2 + (k � b)2

This circle cuts x2 + y2 � 4 = 0 orthogonally2g

1g

2 + 2f

1f2 = c

1 + c

2

2g1(0) + 2f

1(0) = �(h � a)2 � (k � b)2 + h2 + k2 � 4

2ah + 2kb �

Hence locus of (h, k) is 2ax + 2by � = 0.

4. Equation of circle(x � p) (x � h) + (y � q) (y � k) = 0

x2 + y2 � x(h + p) � y(q + k) + (ph + qk) = 0

This circle touches x-axis g2 = c

= ph + qk

Locus of (h, k) is (x � p)2 = 4qy.

5. Point of intersection of 2x + 3y + 1 = 0 3x � y � 4 = 0 is (1, �1)

and circumference of circle = 2r = 10 r = 5Hence equation of circle (x � 1)2 + (y + 1)2 = 25

x2 + y2 � 2x + 2y � 23 = 0.

6. By family of circle x2 + y2 � 2x + (x � y) = 0

centre of this circle

lies on y = x = = 1

Hence x2 + y2 � x � y = 0.

7. Let S1 : x2 + y2 + 2ax + cy + a = 0

S1 : x2 + y2 � 3ax + dy � 1 = 0

common chord S1 � S

2 = 0 5ax + y(c � d) + (a + 1) = 0

given line is 5x + by � a = 0


compare both = =

a = = �1 �

(i) (ii) (iii)From (i) & (iii) a2 + a + 1 = 0 a = , 2 no real a.

8. Draw a line parallel to x-axis at a distance 2 unit.Now by definition of parabolalocus of a point whose distance from a fixed point (0, 3)is equal to its distance from a fixed line is a parabola.

9. Point of intersectionof lines3x � 4y � 7 = 0

2x � 3y � 5 = 0 is (1, � 1)

Area of circle = r2 = 49 r = 7Hence equation of circle (x � 1)2 + (y + 1)2 = 72 x2 + y2 � 2x + 2y = 47

10. cos = =

Locus of (h, k) is x2 + y2 = .

11. Let equation of circle is (x � h)2 + (y � k)2 = (h + 1)2 + (k � 1)2

it touches x-axis g2 = c

h2 = 2k � 2h � 2 k =

k k .

12.

= � 1 h = �3

= �2 k = �4 Hence Q(�3, �4).

13. S1 + S

2 = 0 should satisfy (1, 1)

(2 + 3 + 7 + 2p � 5) + (1 + 1 + 2 + 2 � p2) = 0

= �

p2 6 p ±

but at p = ± the 2nd circle is

x2 + y2 + 2x + 2y � 6 = 0

satisfies (1, 1) and obviously P and Q

so p = ± is also acceptable


� 1 1 7 + 2p 6 � p2

p2 + 2p + 1 0p � 1

14. r = = 5

< 5

� 25 < m + 10 < 25 � 35 < m < 15


15. x2 + y2 = ax ...........(1)

centre c1 and radius r

1 =

x2 + y2 = c2 .........(2) centre c

2 (0, 0) and radius r

2 = c

both touch each other iff|c

1c

2| = r

1 ± r

2

=

= ± |a| c + c2 |a| = c

16. Circle whose diametric end points are (1, 0) and (0, 1) will be of smallest radius.(x � 1)(x � 0) + (y � 0) (y � 1) = 0

x2 + y2 � x � y = 0

17. Nowh2 = (1 � 2)2 + (h � 3)2

0 = 1 � 6h + 9

6h = 10

h =

Now diameter is 2h =

ADVANCE LEVEL PROBLEMPART - I

1. x2 + y2 � 5x + 2y � 5 = 0 + (y + 1)2 � 5 � � 1 = 0

+ (y + 1)2 =

So the axes are shifted to

New equation of circle must be x2 + y2 =


2.

Equation of circum circle of triangle OAB x2 + y2 � ax � by = 0.

Equation of tangent at origin ax + by = 0.

d1 = and d2 =

d1 + d2 = = diameter

3. Ler r be the radius of new circle

C1C2 = .

So r = 2

Slope of line joining C1 and C2 i.e. tan = 2 Equation of line joining C1 and C2 is

= = 2 + =

x = 2 and y = 5 Centre (2, 5)

4.

Area of ABCD = 4 .

5. Equations of two circles touching both the axes arex2 + y2 � 2c1 x � 2c1y + c1

2 = 0 .....(i)x2 + y2 � 2c2x � 2c2y + c2

2 = 0 .....(ii) (i) & (ii) are orthogonal also 2c1c2 + 2c1c2 = c1

2 + c22

or 6c1 c2 = (c1 + c2)2 ....(iii)Now point P(a, b) lies over the circle

x2 + y2 � 2cx � 2 cy + c2 = 0.so c2 � 2c(a + b) + a2 + b2 = 0 c1 & c2 are roots of this equationso c1 + c2 = 2(a + b) ....(iv)and c1 c2 = a2 + b2 ....(v)from (iii), (iv) & (v), we get

6(a2 + b2) = 4(a + b)2.

6. Let two circles are S = 0 and S = 0having radius r1 and r2 respectively.

=

S r12 = r2

2 S1

S1 � S = 0 Locus of P(h,k)

S � S = 0 which represents the equation of a circle.


7. tan 60º = =

and

sin 60º = =

Let coordinates of any point P on the circle be P (r cos , r sin)

PA2 = + (r sin)2

PB2 = (r cos )2 + (1 � r sin)2

PC2 = (r cos + )2 + (r sin)2

and PD2 = (r cos)2 + (r sin + 1)2

PA2 + PB2 + PC2 + PD2 = 4r2 + 8 = 11 r =

8. = tan�1 tan =

sin = and cos =

A (OA cos , OA sin ) A (3, 2)Similarly B (OB cos , OB sin ) (6, 4)Now it can be checked that circles C1 and C2 touch each other.Let the point of contact be C.

C

required radical axis is a line perpendicular toAB and passing through point C

y � = � (x � 5)

9. Equation of circle (x � 2)2 + (y + 2)2 + (x + y) = 0 ........(i) Centre lies on the x-axis = � 4 put in (i)

equation of circle is x2 + y2 � 8x + 8 = 0

(, ) lies on it2 = � 2 + 8 � 8 0

greatest value of �� is 4 + 2

10. Let �d� be the common difference

the radii of the three circles be 1 � 2d, 1 � d, 1

equation of smallest circle is x2 + y2 = (1 � 2d)2 ........(i) y = x + 1 intersect (i) at real and distinct points x2 + x + 2d � 2d2 = 0 ....(ii) D > 0 8d2 � 8d + 1 > 0

d > or d <

but d can not be greater than

d


11. Let the coordinates of P and Q are (a, 0) and (0, b) respectively equation of PQ is bx + ay � ab = 0 .......(i) a2 + b2 = 4r2 .....(ii) OM PQ equation of OM is ax � by = 0 .......(iii)Let M(h, k) bh + ak � ab = 0 ........(iv) and ah � bk = 0 .......(v)

On solving equations (iv) and (v), we get

a = and b =

put a and b in (ii), we get(h2 + k2)2 (h�2 + k�2) = 4r2

locus of M(h, k) is (x2 + y2)2 (x�2 + y�2) = 4r2

12. Equation of circle passing through (0, 0) and (1, 0) isx2 + y2 � x + 2fy = 0 .......(i)

x2 + y2 = 9 ......(ii)(i) & (ii) touch each other.so equation of Radical axis is x = 2fy + 9 ......(iii)line (iii) is also tangent to the circle (ii) on solving (ii) & (iii), we get(1 + 4f2)y2 + 36fy + 72 = 0 .......(iv)

D = 0 f = ± .

13. a2 � bm2 + 2d + 1 = 0 ......(1)and a + b = d2 .......(2)Put a = d2 � b in equation (1), we get

(d + 1)2 = b(2 + m2)

= ......(3)

From (3) we can say that the line x + my + 1 = 0 touches a fixed circle having centre at (d,0) and radius

=

PART - II1. Let the circumcentre be P(h, k)

Equation of AB is

a =

on solving

= 2(h + k) = + a

locus of circumcentre P(h,k) is

2 (x + y) a =

2. S1 x2 + y2 = a2

S2 x2 + y2 = b2

S3 x2 + y2 = c2

equation of 1 is ax cos + ay sin = b2

1 is tangent to circle S3


c = ca = b2 Hence a,b,c are in G.P.

3. Equation of circle touching y - axis isx2 + y2 + 2gx + 2fy + f2 = 0 it passes through (4, 3) & (2, 5)so 25 + 8g + 6f + f2 = 0

29 + 4g + 10f + f2 = 0solving above two equations, we get(g, f) (�2, � 3) & (� 10, � 11).

So equations of circles are x2 + y2 � 4x � 6y + 9 = 0 and x2 + y2 � 20x � 22y + 121 = 0

for circle x2 + y2 � 4x � 6y + 9 = 0.

tan = =

=

So tan is max at k = 3.at k = 3, tan = 1 = 45°

4. 1 4x + 3y = 10 2 3x � 4y = � 5

Let be the inclination of 2

tan =

equation of 2 in parametric form

= = ± 5

co-ordinates of centres are (5, 5), (�3, �1)

5. centre lies over the line 2x � 2y + 9 = 0

So let coordinate of centre be

Let the radius of circle be 'r'So equation of circle is

(x � h)2 + = r2

x2 + y2 � 2hx � y(2h + 9) + 2h2 + 9h � r2 + = 0

given circle cuts orthogonally to x2 + y2 = 4

so 2h2 + 9h + � r2 = 0 or 2h2 + 9h � r2 = �

so equation of required circle can be written as x2 + y2 � 2hx � y (2h + 9) + 4 = 0

(x2 + y2 � 9y + 4) + h (�2y � 2x) = 0

so this circle always passes through points of intersection of x2 + y2 � 9y + 4 = 0 and x + y = 0

so fixed points are (�4, 4) and

6. Centre of C1 lies over angle bisector of 1 & 2Equations of angle bisectors are

= ±


x = 5 or y = �

Since centre lies in first quadrantso it should be on x = 5.So let centre be (5, )

3 = = 2, � From the figure r =

But � so = 2.

So equation of cirlce C2 is(x � 5)2 + (y � 2)2 = 52

x2 + y2 � 10x � 4y + 4 = 0.

7. OA = a and AQ = QP = QR

OQ =

AQ = = PQ

(OA)2 = (OQ)2 + (AQ)2

a2 = 2 + 2 + (p � )2 + (q � )2

22 + 22 � 2p� 2q+p2+q2 � a2 = 0.Locus of the middle point Q (, ) is2 x

2 + 2y2 2 p x 2 q y + p2 + q2 a2 = 0

8. Let the equation of required straight line be y = mx + c.

= .....(i)

For PCM = tan 2.

PM = 5cot 2 .....(ii)

For PQM = PM sin (90 � )

= cos

on solving, we get = 30°

Equation of tangent at P(� 2, � 2) is

3x + 4y + 14 = 0.

tan 60° =

=

m =

Now on substituting value of 'm' in equation (i), we get

c = or

but c should be (�ve)

So equation of line y = x +


9. Let the centre of the circle be (h, k) and radius equal to �r� h2 + k2 = r2 ......(i)

and = r

2 � h � k = r .....(ii)and h = 1 � r .......(iii)

put h = 1 � r in (ii), we get k = r (1 � ) + 1Now put the values of h and k in (i), we get

(r (1 � ) + 1)2 + (1 � r)2 = r2

r2 (3 � 2 ) � 2 r + 2 = 0

hence radius i.e. r is the root of the equation (3 � 2 ) t2 � 2 t + 2 = 0

10. Let the equation of the circles be x2 + y2 + 2gx + 2fy + d = 0 .......(i) these circles pass through (0, a) and (0, �a)

a2 + 2fa + d = 0 ......(ii)and a2 � 2fa + d = 0 ......(iii)solving (ii) and (iii), we get f = 0, d = � a2

put these value of f and d in (i), we getx2 + y2 + 2gx � a2 = 0 ......(iv)

y = mx + c touch these circles =

g2 + (2cm) g + a2 (1 + m2) � c2 = 0 ......(v)equation (v) is quadratic in 'g' Let g1 and g2 are its two roots g1g2 = a2 (1 + m2) � c2

the two circles represented by (iv) are orthogonal 2g1g2 + 0 = � a2 � a2 g1g2 = �a2

a2 (1 + m2) � c2 = � a2

c2 = a2 (2 + m2) Hence proved

11. Let OAB = and OAB =

+ = and OBA =

length of AB is �a� and length of AB is �b� from the figure

A (b cos , 0) and A(a cos , 0)similarly B(0, a sin ) and B (0, b sin )Let c(h, k) be the centre of circle 2h = a cos + b cos

= �

2h = a cos + b sin ........(i)and 2k = a sin + b sin

= �

2k = a sin + b cos ........(ii)

on solving (i) and (ii), we get cos = and sin =

sin2 + cos2 = 1 locus of C(h, k) is (2ax 2by)² + (2bx 2ay)² = (a² b²)²


12. One circle lies within the other circle C1C2 < |r1 � r2|

<

squaring both sides, we get

� 2gg1 < � 2 � 2c

gg1 > c + .

gg1 � c > . ......(i)

gg1 � c > 0 gg1 > cagain squaring both sides of (i), we get�2cgg1 > � c (g2 + g1

2) c(g � g1)2 > 0 c > 0 and from (i), we can say that gg1 will also be > 0

MATHEMATICAL REASONING, INDUCTION & STATISTICSEXERCISE # 1

PART - I

Section (A) :

A-1. By definition of 'statement'.

A-7. The negation of ��Everyone in Germany speaks German�� is - there is at least one person in Germany who doesnot speak German.

A-10. Statement (A) All prime numbers are even.Statement (B) All prime numbers are odd.Both false

A-12. If it is a holiday as well as sunday than also the office can be closed.

A-13*. Polygon cannot be both concave and convex

A-16. Obvious

A-19. (~ T F) ~T T (F F) F T F F T F T

A-21. p q means(i) p is sufficient for q (ii) q is necessary for q (iii) p implies q (iv) if p then q (v) p only if q

Section (B) :

B-3. Contrapositive of (p q) r is ~ r (p q )

B-4.


Section (C) :

C-2. x1 + x

2 + ...... x

n = nM

(x1 + x

2 ..... + x

n) � x

n + x = nM � x

n + x

so

average = .

C-5. Average speed over the entire distance =

= =

C-7. =

C-10. � , � 3, � , � 2, � , + , + 4, + 5 ( > 0)

C-12. =

=

.(2n�1) =

C-14 Frequency of f = 10C5 which has maximum value

Section (D) :

D-2. 34, 38, 42, 44, 46, 48, 54, 55, 63, 70

median = = 47

= 13 + 9 + 5 + 3 + 1 + 1 + 7 + 8 + 16 + 23 = 86

so mean diviation about median = = 8.6

D-5. new

=

= = odd

D-8. = = 60

= = = 10.4


D-9. = Now =

new

= = .

D-11. = 250

= = 5

coeff. of variation = = 10%

Section (E) :E-4*. Let n = 1 then p(A) = 64

Let p(k) is divisible by 6432k + 2 � 8k � 9 is divisible by 64

Now,P(k + 1) = 32(k + 1) + 2 � 8(k + 1) � 9

= 32k + 2 × 9 � 8 × k × 9 � 9 × 9 � 8 + 72 + 64 k

= 9(32k + 2 � 8k � 9) + 64 (k + 1)

Which is divisible by 64

E-6. Let p(n) = n3 + (n + 1)3 + (n + 2)3 , p(A) = 36, p(B) = 99 both are divisible by 99Let it is true for n = kk3 + (k + 1)3 + (k + 2)3 = 9q ; q adding 9k2 + 27k + 27 both sidesk3 + (k + 1)3 + (k + 2)3 + 9k2 + 27k + 27 = 9q + 9k2 + 27 k + 27(k + 1)3 + (k + 2)3 + (k + 3)3 = 9r ; r

Comprehension # 1 (1 to 3)1. If p then q means p only if q

2. If p then q p is sufficient for q

3. p is false, q is false so p q is true.

Comprehension # 2 (4 to 6)

A.M. = = 12

= = = 3

coeff. of variation = = × 100

MATCH THE COLUMN

1. (A) = 5 + = 15

Md = x

11 + 10 = = 10 + 10 = 20

variance remains unaffected on addition of a constant


(B) = 5 + = 5 + 10 = 15

Md = x

11 = 10

(C) Mean and median get multiplied by 2 and variance by 22

(D) = = 16

Md = x

11 + 11 = 10 + 11 = 21

variance remains unaffected on addition of a constant

EXERCISE # 2

4.

7.

9. (p q) [~ p (p ~ q)]= (p q) [(~ p p) (~ p ~ q)]= (p q) [t (~ p ~ q)]= (p q) (~ p ~ q)= (p q) [ ~ ( p q)] = talso (~p q) t = t

12. (p q) ~p = (p ~ q) (q ~p)= t (q ~p) = q ~p = ~p q

15. p : it rainsq : crops will be goodS

1 : p q , S

2 : ~p S : ~q

Not valid

17. p : it rains tomorrowq : I shall carry my umbrellar : cloth is mendedP : p (r q)Q : p ~rS : ~qP : T, Q : T S : TT S not valid


18. S.D.(xi) = S.D. (x

i � 8) = = = 2

20. n = 200, mean = 40 = = 8000

correct = 8000 � 34 + 43 = 8009 correct mean = = 40.045

Also 2 = � 225 = � 1600

x2 = 36500

correct = 36500 � (34)2 + (43)2 = 365693

correct 2 = � (40.045)2

= 1828.465 � 1603.603 = 14.995

21.

2 =

= = 1.2

so variance of A = 1.2 < 1.25 = variance of Bso more consistent team = A

22. 2 = = = 9

coefficient of variation = = 25

26. (i) Given statement is true for n = 1(ii) Let us assume that the statement is true for n = k

i.e. 1.3 + 2.32 + 3.33 +.......+ k.3k =

(iii) For n = k + 1,L.H.S. = 1.3 + 2.32 + 3.33 +.......+ k.3k + (k + 1) 3k+1

= + (k + 1) 3k+1 = = R.H.S.

so by principle of mathematical induction the statement is true for all n N



MATHEMATICAL REASONING :

1. r : x is a rational number iff y is a transcendental number r = ~p q

Statement-1 is false and Statement-2 is false.

2.

3. Statement-1 :

Statement-2 : False.

4. Negation of Q is ~

It may also be written as ~

5.

6. Let p : I become a teacherq : I will open a schoolNegation of p q is ~ (p q) = p ^ ~ qi.e. I will become a teacher and I will not open a school.

STATISTICS :

7. Let average marks of the girls = x

= 72 x = 65

8. No change median is 5th observation (If observation are in asending order)


9. Correct variance = �

= 222 � 144 = 78.00

10. If we change scale by using x + h then median increases by h. so median is not independent of change of scale.From histtegranm we can see highest frequency so made.

11. = 0

= a2 S.D. = |a| = 2

12.

so median = 22 = , mode = 24

13. 2 0

0 0 n 16

14. Variances remain uneffected by adding some constant to all observationsso V

A = V

Bso V

A/V

B = 1

15. Let no. of student = 100 number of boys = n,

= 50 n = 80

so 80%

16. = 6 a + b = 7 ...(1)

= 6.80 (a � 6)2 + (b � 6)2 = 13

solve a = 3, b = 4

17. Statement-1 : � = �

= (2n + 1 � 3)

Statement-2 : Obvious

18. = = 1 + 50d

Mean deviation = =

= = 225 = 225

.d = 255 d = 10.1


19. x2 = 4 � = 4

� (2)2 = 4 = 40

similarly = 105

2 = � = � = 5.5

20. Median = 25.5 a

Mean deviation about median = 50 = 50

24.5 a + 23.5a + ..... + 0.5a + 0.5a + .... + 24.5a = 2500

a + 3a + 5a + ..... + 49a = 2500 (50a) = 2500 a = 4

21. Correct mean = observed mean + 230 + 2 = 32Correct S.D. = observed S.D. = 2

22. A.M. of 2x1, 2x

2 ..... 2x

n is

= =

So statement-2 is falsevariance (2x

i) = 22 variance (x

i) = 42

so statement-1 is true.

MATHEMATICAL INDUCTION :

23. Put k = 1LHS 1 RHS = 4LHR RHSLet S(k) is truethen 1 + 3 + 5 +....(2k � 1)

= 3 + k2

add (2k + 1) both the side1 + 3 + 5 +.... + (2k � 1) + (2k + 1)

= 3 + k2 + 2k + 1S(k + 1) = 3 + (k + 1)2

then if S(k) is true S(k + 1) is also true.

24. For n 2n2 + n < n2 + n + n + 1n2 + n < (n + 1)2

statement -2 is true

>

> , > . . . . . + +......


ADVANCE LEVEL PROBLEM

1. Statement p q and its contrapositive ~q ~p are logically equivalent and give same meaning.

2.

3.

4.

5. p : Wages will increaseq : there is an inflationr : cost of living will increaseA : p qB : q rC : pS : rA : T, B : T C : T S : T S valid

6. Here = = 8 +12 + 13 + 15 + 22 = 70

& = 64 + 144 + 169 + 225 + 484 = 1086

2 = � = � = � = 21.2


7. If a xi b a b

xi � b � a

(xi � )2 (b � a)2 n(b � a)2

so var(x) (b � a)2

8. Total money per kg. = = so total kg per rupee = = 1.92

11. Let P(n) ; sin + sin2 + .......+ sin n = sin

P(A) is trueLet P(k) is also true

sin + sin2 + ..........+ sin k = sin

add sin(k + 1) both sidessin + sin2 + ............+ sin k + sin(k + 1)

= sin sin cosec + sin(k + 1)

= sin = sin

= . cosec

P(k + 1) is true

SOLUTION OF TRIANGLEEXERCISE # 1

PART - ISection (A) :

A-1. (i) L.H.S. = a sin (B � C) + b sin (C � A) + c sin (A � B)

= k sin A sin (B � C) + k sin B sin (C � A) + k sin C sin (A � B)

= k (sin2 B � sin2 C) + k (sin2C � sin2 A) + k (sin2 A � sin2 B)= 0 = R.H.S.

(ii) L.H.S. =

first term = =

= k2 sin (B + C) sin (B � C)

= k2 (sin2 B � sin2 C)

Similarly = k2 (sin2 C � sin2 A)

and = k2 (sin2 A � sin2 B)

L.H.S. = k2 (sin2 B � sin2C + sin2C � sin2A + sin2 A � sin2 B)= 0 = R.H.S.


(iii) L.H.S. = 2bc cos A + 2ca cos B + 2ab cos C= b2 + c2 � a2 + a2 + c2 � b2 + a2 + b2 � c2

= a2 + b2 + c2

= R.H.S

(iv) L.H.S. = a2 � 2ab

= a2 + b2 � 2ab cos C

= a2 + b2 � (a2 + b2 � c2)= c2 = R.H.S.

(v) L.H.S. = b2 sin 2C + c2 sin 2B= 2b2 sin C cos C + 2c2 sin B cos B= 2k2 sin2 B cos C sin C + 2k2 sin2 C sin B cos B (b = ksin B, c = ksin C)= 2k2 sin B sin C [sin B cos C + cos B sin C]= 2(k sin B) (k sin C) sin (B + C)= 2bc sin A

(vi) R.H.S = c = a cos B + b cos A,

b = c cos A + a cos C

= =

= = L.H.S.

A�4. =

sin(B + C) sin(B � C) = sin(A + B) sin(A � B)

sin2 B � sin2 C = sin2 A � sin2 B 2 sin2 B = sin2 A + sin2 C 2b2 = a2 + c2 a2, b2, c2 are in A.P.

A�7. x3 � Px2 + Qx � R = 0

a2 + b2 + c2 = Pa2b2 + b2c2 + c2a2 = Q

a2b2c2 = R abc = + + = [a2 + b2 + c2] =

Section (B) :

B�1. (i) L.H.S. = 2a sin2 + 2 c sin2

= a(1 � cos c) + c(1 � cos A)

= a + c � (a cos C + c cos A)

= a + c � b

= R.H.S.

(ii) L.H.S. = + +

= . + . + .

= = .


(iii) L.H.S. = 2bc(1 + cos A) + 2ca(1 + cos B) + 2ab(1 + cos C)= 2bc + 2ca + 2ab + 2bc cos A + 2ca cos B + 2 ab cos C

= 2 + a2 + b2 + c2 = (a + b + c)2

= R.H.S.

(iv) L.H.S. = (b � c) + (c � a) + (a � b)

(b � c) cot = k(sin B � sin C)

= 2k cos sin

= 2k sin sin

= k [cos C � cos B]

similarly (c � a) cot = k[cos A � cos C]

and (a � b) cot = k[cos B � cos A]

L.H.S. = k[cos C � cos B + cos A � cos C + cos B � cos A]

= 0= R.H.S.

(v) L.H.S. = 4 (cot A + cot B + cot C)

= 4

= 2bc cos A + 2 ca cos B + 2ab cos C= a2 + b2 + c2 = R.H.S.

(vi) L.H.S. = cos . cos . cos

=

= = = R.H.S.

B�3.

Let ADB =

we have to prove that tan =

if we aply m � n rule, then

(1 + 1) cot= 1.cot C � 1.cotA.

= � = �


=

= [2(a2 � c2)]

2cot = tan =

Section (C) :

C�2. (i) r. r1 .r

2 .r

3 = = 2

(ii) r1 + r

2 � r

3 + r = 4R cosC

L.H.S. =

=

=

=

=

= =

= c =

cos C =

L.H.S. =

= = = 4RcosC

(iii) L.H.S. =

= [s2 + (s � a)2 + (s � b)2 + (s � c)2]

= [4s2 � 2s(a + b + c) +a2]

= = R.H.S.

(iv) L.H.S. =


= (s + s � a + s � b + s � c)2 = 4 =

R.H.S. =

= · (s � a + s � b + s � c) = =

(v) =

=

= =

= =

= = = r

similarly we can show that = = r

C�4. = 24 sq. cm .... (i)2s = 24 s = 12 .... (ii)

r1, r

2, r

3 are in H.P.

are in A.P..

are in A.P..

a, b, c are in A.P. 2b = a + c 2s = 24 a + b + c = 24

3b = 24 b = 8 a + c = 16

But = = 24 × 24 = 12 × (12 � a) × 4 × (12 � c) 2 × 6 = 144 � 12 (a + c) + ac

12 = 144 � 192 + ac

ac = 60 and a + c = 16 a= 10, c = 6 or a = 6, c = 10 and b = 8

Section (D) :

D�1. (i) = ,= , = =

R.H.S. =


=

= L.H.S. = R.H.S.

(ii) = = =

R.H.S. = =

=

L.H.S.= R.H.S.PART - II

Section (A) :A�4. (a + b + c) (b + c � a) = kbc (b + c)2 � a2 = kbc

b2 + c2 � a2 = (k � 2) bc = = cos AA

In a ABC �1 < cos A < 1 �1 < < 1

0 < k < 4.

Section (B) :

B�2. b cos2 + a cos2 = c. b + a = c.

[ s � a + s � b] = c × c = c

= a + b = 2c

a, c, b are in A.P.

B�5. = (a + b � c) (a � b + c)

= 4(s � c) (s � b) =

tan = tan A = tan A =

B�6*. (A) tan = cot .........(i)

tan2 = = =

tan = a = 5 and b = 4

from equation (i), we get

= cot = cot cot =


cos C = = = =

cos C = c2 = a2 + b2 � 2ab cos C c = 6

(B), (C) Area = ab sinC cosC = sinC = =

Area = × 5 × 4 ×

Area = sq. unit. From Sine rule

= = sinA = =

sinA =

Section (C) :

C�3. =

=

= 4

= .

C�5*. (A) + +

= + +

=

(B) + +

= + + =

(C) = = cot A = cot B = cot C

A = B = C

true for equilateral triangle only

(D) = =


= =

cot A = cot B = cot C A = B = C true for equilateral triangle only

Section (D) :

D�1. = 2

D�4*. a = cos

(A) correct(B) incorrect

(C) = = = cos

(D) cosec = . = . = cos


3.

If we apply Sine-Rule in ABD , we get

= AB = = ...(i)

sin = and cos =

from equation (i), we get

AB = AB =

7.

required distance = inradius of ABC


2s = a + b + b + c + c + a

= 2 (a + b + c)

s = a + b + c

=

=

required distance

= = =

=

8. (i) L.H.S. = (r3 + r

1) (r

3 + r

2) sin C

= sin C

= sin C

= sin C

=

= = 2 sr3

R.H.S. = 2r3

= 2r3

= 2sr

3

L.H.S. = R.H.S.

(ii) L.H.S. = �

= �

= = R.H.S.

(iii) First term = (r + r1) tan

= cot

= . .

= b � c

similarly second term = c � a & third term = a � b

L.H.S. = b � c + c � a + a � b = 0 = R.H.S.


(iv) r1 + r

2 + r

3 � r = 4R

(r1 + r

2 + r

3 � r)2 = r

12 + r

22 + r

32 + r2 � 2r (r

1 + r

2 + r

3) + 2(r

1r

2 + r

2r

3 + r

3 r

1) ........(i)

r(r1 + r

2 + r

3) = ab + bc + ca � s2

and r1r

2 + r

2r

3 + r

3r

1 = s2

from equation (i)16R2 = r2 + r

12 + r

22 + r

32 � 2 (ab + bc + ca � s2) + 2s2

r2 + r1

2 + r2

2 + r3

2 = 16 R2 � 4 s2 + 2 (ab + bc + ca)= 16R2

� (a + b + c)2 + 2 (ab + bc + ca)

= 16R2 � a2 � b2 � c2

11. (i) EFA is a cyclic quadrilateral

= A

A = r cosec A/2 EF = r cosec A/2.sin A

= 2 r cos A/2similarly DF = 2 r cos B/2and DE = 2r cos C/2.(ii) ECD is a cyclic quadrilateral

CE = DE =

similarly DF = BF =

FDE = =

= �

(iii) area of DEF = FD . DE sin FDE

=

= 2r2 = 2r2

= = =

= =

= = .

PART - II

3. ED = � c cos B

= � c

= �


=

=

5. f = RcosA , g = R cos B, h = R cosC.

+ + = + +

= 2

= 8 + + =

= .8 =

9. MNA is a cyclic quadrilatral

= AA MN = r cosec sin A = 2r cos

M = N = r

x = = , =

similarly y = and z =

xyz = = = r2 R

12. r1 + r

2 =

(r1 + r

2) = =

= = = 4Rs2

= 4

14. A, C1 , G and B

1 are cyclic

BC1 . BA = BG . BB

1

. c =

= (2c2 + 2a2 � b2)

c2 + b2 = 2a2


16. a = 1 2s = 6

2s = 2

R = 1 = 2R sin A =

A =

18. sin C = 1 cos (A � B) 1

cos (A � B) = 1 A � B = 0 A = B

sin C = = 1 C = 90º

20. if we apply m-n Rule in ABE, we get

(1+1) cot = 1.cot B � 1.cot

2 cot = cot B � cot

3 cot = cot B

tan = 3 tan B ..........(1)

Similarly, if we apply m-n Rule in ACD, we get

(1+1) cot (�) = 1.cot � 1.cotC.

cotC = 3 cot tan = 3 tanC .......(2)

form (1) and (2) we can say that

tan B = tan C B=C

A + B + C =

A = � (B + C)

= � 2B B = C

tan A = � tan2B

= � = �

tan A =


22. r1 � r = � = = a tan

(r1 � r) = abc tan tan tan

= abc tan

= abc

= =

= = = 4Rr2

EXERCISE # 32. Match the column

(A) AA1 and BB

1 are perpendicular

a2 + b2 = 5c2

c2 = = 5 c = (

cos C = = =

sin C = = ab sin C =

2 = 11

(B) G.M. H.M.

(r1 r

2 r

3)1/3 (r

1 r

2 r

3)1/3 3r 27

(C) tan2 = a = 5, b = 4 2s = 9 + c

= = = c2 = 36 c = 6

(D) 2a2 + 4b2 + c2 = 4ab + 2ac. (a � 2b)2 + (a � c)2 = 0 a = 2b = c

cos B = =

8 cos B = 7


COMPREHENSION # 2 (Q. No. 7 to 10)7. Clearly

8. Let 3

1

2 =

Then angle of pedal trinagle = � 2 = A

=

9. Side of pedal triangle = I2I

3cos = BC

I2I

3 =

I2I

3 = 4Rcos

10. 1 = 4 R sin

I2I3 = 4 R cos

12 +

2

32 = 16R2

12. 1

2 = 4R cos if we apply Sine-Rule in

1

2

3 , then

2 Rex

= =

=

2Rex

= 4R Rex

= 2R ABC is pedal triangle of I

1 I

2 I

3

statement - 1 and statement - 2 both are correct and statement -2 also explains Statement - 1

14. sin = =

similarly sin =

3 sin � 4 sin3 =

� = r2 = r = a. cm.

19. ax2 + bx + c = 0 ...(1)

x2 + x + 1 = 0 ...(2)

roots of (2) are imaginary and a, b, c are real

= = = k cos C = = = C =



1. We have a2 a2 � (b � c)2 = (a + b � c) (a � b + c)

a2 (2s � 2c) (2s � 2b) = 4(s � b) (s � c)

similarly b2 4 (s � c) (s � a)

and c2 4 (s � a) (s � b).

Multiplying the above inequalities, we geta2b2c2 64 (s � a)2 (s � b)2 (s � c)2

(a + b + c) abc 16 s (s � a) (s � b) (s � c) = 162

Equality occurs if and only if(b � c)2 = 0(c � a)2 = 0

and (a � b)2 = 0i.e if and only if a = b = c.

2. (A) a, sin A, sin B are given one can determine

b = c = So the three sides are unique. So option (a) is incorrect option

(B) The three sides can uniquely determine a triangle.So option (b) is incorrect option.

(C) a , sin B, R are given one can determine b = 2R sin B,

sin A = . So sin C can be determined. Hence side c can also be uniquely determined

(D) for a, sin A, R

= 2R

But this could not determine the exact values of b and c

3. n = 2n × area of OA

1

1

n = 2n × × AA

1

1 × O

1

n = n × sin × cos

n = sin . .........(1)

On = 2n × area of OB

1O

1

On = 2n × × B

1O

1 × O

1O = n × tan × 1 = n tan

On = n tan ......(2)

Now R.H.S. = =

= × 2 cos2 = On. cos2

= n tan .cos2 = sin = n = L.H.S


4. Let angle of the triangle be 4x, x and x .Then 4x + x + x = 180° x = 30°

Longest side is opposite to the largest angle.Using the law of sines

= 2R

a = R, b = R, c = 2S = =

5. Clearly the triangle is right angled. Hence angles are 30º, 60º and 90º are in ratio 1 : 2 : 3

6. Consider =

= = =

7. Clearly P is the incentre of triangle ABC.

r = =

Here 2s = 7 + 8 + 9 s = 12

Here r = =

8. = . b . b . sin 120º = b2 .........(1)

Also a = .........(2)

and = and s = (a + 2b)

= (a + 2b) ..........(3)

From (1), (2) and (3), we get =

9.* We have ABC = ABD + ACD

bc sin A = c AD sin + b × AD sin

AD =

Again AE = AD sec


= AE is HM of b and c.

EF = ED + DF = 2DE = 2 × AD tan = × cos × tan an

= sin

As and DE = DF and AD is bisector AEF is isosceles.

Hence A, B, C and D are correct answers.

10. In ABC , by sine rule

= = C = 45º, C = 135º

When C = 45º A = 180º � (45º + 30º) = 105º

When C = 135º A = 180º � (135º + 30º) = 15º

Area of ABC = AB . AC.sin BAC = × 4 × sin (15º) = × = 2

Area of ABC = AB . AC .sinA = × 4 × sin (105º) = 2

Absolute difference of areas of triangles = | 2 � 2 | = 4

Aliter

AD = 2 , DC = 2 Difference of Areas of triangle ABC and ABC = Area of triangle ACC

= AD × CC = × 2 × 4 = 4

12. cos B + cos C = 4 sin2 2 cos cos = 4 sin2

2 sin = 0

cos � 2 cos = 0 as sin 0

� cos cos + 3 sin sin = 0

tan tan =

=

= 2s = 3a b + c = 2a

Locus of A is an ellipse


11. sin 2C + sin 2A = (a cos C + c cos A) = = 2 sin B = 2 sin 60º =

12. cos =

=

=

= (x2 + x + 1) = 2x2 + 2x � 1

( � 2) x2 + ( � 2) x + ( + 1) = 0

on solving

x2 + x � = 0 we get

x = + 1, �

At x = � , Side c becomes negative. x =

13. Area of triangle = ab sin C = 15

. 6 . 10 sin C = 15 sin C =

C = (C is obtuse angle )

Now cos C =

� = c = 14

r = = = r2 = 3

14. a = 2 = QR

b = = PR

c = = PQ

s = = = 4

= = = = tan2

= = = =


PART - II1. Let a = 3x + 4y, b = 4x + 3y and c = 5x + 5y

as x, y > 0, c = 5x + 5y is the largest side C is the largest angle. Now

cos C =

= < 0

C is obtuse angle ABC is obtuse angled.

2. r1 > r

2 > r

3 > >

s � a < s � b < s � c �a < �b < �c; a > b > c

3. tan = ; sin =

r + R = r + R = .cot

4. a =

= a + b + c = 3b.

a + c = 2b a, b, c are in A.P.

5. AD = 4

AG = × 4 =

Area of ABG = × AB × AG sin 30º

= × × × = Sin 60º = AB = =

Area of ABC = 3(Area of ABG) =

6. cos = = � = 120º

7. C = /2

r = (s � c) tan C = 90º

r = s � 2R

2r + 2R = 2 (s � 2R) + 2R.

= 2s � 2R

= (a + b + c) � C = 90º

= a + b + c � c

= a + b


8. are in H.P.

are in A.P. a,b,c are in A.P.

9. = cos

Let cos = for some n 3, n N

As cos cos cos

3 n < 4, which is not possibleso option (2) is the false statementso it will be the right choiceHence correct option is (2)

ADVANCE LEVEL PROBLEMSPART - I

1. From figure, AD = c sin B

Hence number of triangle is 0 if b < c sin B

one triangle for b = c sin B

two triangles for b > c sin B

2. C = 60°

Hence c2 = a2 + b2 � ab

= = 2 cos

3. Using properties of pedal triangle,

we have MLN = 180° � 2A

LMN = 180° � 2B

MNL = 180° � 2C

Hence the required sum = sin2A + sin2B + sin2C

= 4sinA sinB sinC

4.

From figure, we can observe that OGD is directly similar to PGA

5. BD = s � b, CE = s � c and AF = s � a

Hence BD + CE + AF = s

6.


, as cos = cos

A = B, in either case

7. ,

Using cosine rule in ABO, we get

h =

8. In ABD,

Comprehension # 1

9. + + = b sin B + c sin C + a sin A =

k = 2R

10. cot A + cot B + cot C = (b2 + c2 � a2 + c2 + a2 � b2 + a2 + b2 � c2)

= (b2 + c2 + a2) =

=

= . = k =

11. = = 6

Comprehension # 2 (12 to 14)

12. PG = AD

=

= .ab sin C or

= b sin C ( = ac sin B)


PG = ac sin B

= c sin B

13. Area of GPL = (PL) (PG)

and Area of ALD = (DL) (AD) PL = DL and PG = ADAD

= = =

14. Area of PQR = Area of PGQ + Area of QGR + Area of RGP ...(1)

Area of PGQ = PG.GQ.sin(PGQ)

= × AD × BE sin ( � C)

= × × sin C

= × bc sin A × ac sin B × sin C

= sin A.sin B.sin C

Similarly Area of QGR = sin A.sin B .sin C and Area of RGP = sin A.sin B.sin C

From equation (1), we get

Area of PQR = (a2 + b2 + c2) sin A.sin B.sin C

15. In CDB , =

Also from same triangle = BD =

16. cosAcosB + sinAsinBsinC = 1

(cosA � cosB)2 + (sinA � sinB)2 + 2sinAsinB(1 � sinC) = 0

A = B & C = 90°

a : b : c = 1 : 1 :

17. We have

a : b : c = 5 : 4 : 3


18. from figure, OO = ON � ON = R �

ZO = ZM +

= RcosA +

from OZO, using Pythagorous theorem,

we get (R � )2 = (RcosA + )2 +

=

PART - II

1. from ABC , =

AB = 2Rsin(A + )

from ACB, =

AC� = 2Rsin( � A)

BC = 2R(sin (A + ) � sin( � A))

= 4RcossinA = 2acos

similarly CA = 2bcos area ABC =

=

= 4cos2.

2. c2 � 2bc cosA + (b2 � a2) = 0

c1 & c2 are roots of this quadratic equation

Hence (c1 � c2)2 + (c1 + c2)

2tan2A = 4a2

3. Area =

=

=

= 2Rs

=


4. We know that OA = R, HA = 2RcosA and

applying Appoloneous theorem to AOH, we get

2.(AQ)2 + 2(OQ)2 = OA2 + (HA)2

2.(AQ)2 = R2 + 4R2cos2A �

5. = +

using sine rule, diameter of required circle

= = = 20

radius = 10

6. L.H.S. = (a2 (b + c � a) + b2 (c + a � b) + c2 (a + b � c))

=

=

= abc

= 4R

7. from the parellelogram ABAC, AA = 21 ,

from AAC, AA < b + c

21 < b + c ...(1)

similarly 22 < c + a ...(2)

and 23 < a + b ...(3)

(1) + (2) + (3) gives 1 + 2 + 3 < 2s

8. ZXY =

and

Area of


= 2

Ccos

2

Bcos

2

AcosR2 2

Area of CsinBsinAsinR2Csinab2

1ABC 2

Area of XYZ = 2R2 cos 2A

cos2B

cos 2C

= r2R

9. Drop a perpendicular from the apex P to the base ABC.

The foot of perpendicular is at circum centre O of ABC

Using given data, we get 52

21RBO

from, right angle POB, we get

POh = 22 OBPB

= 8.83 m

10. from cyclic quadrileteral CQFP, we get

BCFPCQP

from cyclic quadriletral AQMF, we get

FQM = FAM = 90º � B

AQM = 90º + 90º � B = 180º � B

180CQPAQM

P, Q, M are collinear

A

CBP D

N

MQ

EF

similarly P, Q, N are collinear

hence, P, Q, M, N are collinear

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