Complex Numbers (Package Solutions)

8/10/2019 Complex Numbers (Package Solutions)

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Solutions of Assignment (Set-2) Complex Numbers and Quadratic Equations (Solutions)

Aakash Educat ional Serv ices Pvt. L td . Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

- 1 -

Section-A

Q.No. Solution

1. Answer (2)

–9999

9999 2 4999 4999

1 1 1

( ) (–1)i

i i i i =

2

1 –

–

ii

i i

2. Answer (4)1

2

iz

2 22

2

(1 ) 1 2 2

2 22

i i i iz i

Now, z1929 = (z2)964·z

= i 964·z

= z

=1

2

i

3. Answer (2)

–3 –5 3 –1 5 –1

= 3 5i i

= 215i

= – 15

4. Answer (1)

72 5 2 6 2 6 2 2 711 12 13 14 15 ( ) · ( ) ( ) ( )

1 1

i i i i i i i ii i i i i

i i

– (1) ( ) – 1– – – 1– ·

1 1 1 1–

i i i i i i

i i i i

=2

2

– – – 1 –( 1)

1– 2 2

i i i i

i

5Chapter

Complex Numbers and

Quadratic E uations



Complex Numbers and Quadratic Equations (Solutions) Solutions of Assignment (Set-2)

Aakash Educational Services Pvt . Ltd.

Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

- 2 -

Q.No. Solution

5. Answer (1)

1 1 1·

1– 1– 1

i i iz

i i i

=2 2

2

(1 ) 1 2

1– 2

i i i

i

= i

z8 = (i)8 = (i2)4

= 1

6. Answer (3)

Let z be the complex number, then

2 5· 1

17

iz

17 17 2 – 5

·

2 5 2 5 2 – 5

iz

i i i

=

2

17 2 – 5 172 – 5

2 – 25 27

ii

i

7. Answer (4)

Additive inverse of 5 + 7i is –5 –7i

8. Answer (2)

2

1 2 1 2 1 (1 2 )(1 )·

1– 1– 1 1–

i i i i i

i i i i

1– 2 (2 1) –1 32 2i i

1 3 –

2 2i

1 3

– ,2 2

lies in the second quadrant.

9. Answer (2)

2 2

2

1 (1 )·(1 ) (1 ) 1 2

1– (1– )(1 ) 1– 2

i i i i i ii

i i i i

and2 2

21– (1– )·(1– ) (1– ) 1 – 2 – 1 (1 )(1– ) 1– 2

i i i i i i ii i i i

3 3

3 31 1– – ( ) – (– )

1– 1

i ii i a ib

i i

– i – i = a + ib

0 – 2i = a + ib

a = 0 and b = –2





- 3 -

Q.No. Solution

10. Answer (3)

–2 – 3x i

2 – 3x i

(x + 2)2

=

2

– 3i

x2 + 4 + 4x = 3i2

x2 + 4x + 7 = 0

Now, 2x4 + 5x3 + 7x2 – x + 41

= 2x2(x2 + 4x + 7) –3x(x2 + 4x + 7) + 5(x2 + 4x + 7) + 6

= 0 – 0 + 0 + 6

= 6

11. Answer (4)

9

17

315

1

i i

= 9

82

2 157

1·

( ) ·i i

i i

=9

1 – i

i

=9

2 –

ii

i

= [i + i]9 = (2i)9

= 512(i2)4·i

= 512i

12. Answer (2)

z = 3 – 2i

Re z = 3, Im z = –2

Re z(Im z)2

= 3(–2)2 = 12

13. Answer (3)

z1 – z2 = (4 – 3i) – (3 + 9i)

= (4 – 3) +i(–3–9)

1 – 12i

14. Answer (4)

z1z2 = (2 + 3i)(5 – 3i)

= (10 + 9) + i(15 – 6)

= 19 + i(9)






- 4 -

Q.No. Solution

15. Answer (1)

1 2 1 2 1– 3·

1 3 1 3 1– 3

i i i

i i i

=2

(1 2 )(1– 3 )

1– 9

i i

i

= 1

1 2 1– 310

i i

= 1

(1 6) (2 – 3)10

i

=7 – 7

– 10 10 10

i i

16. Answer (4)

17. Answer (3)

18. Answer (1)

19. Answer (4)

20. Answer (2)

21. Answer (3)

22. Answer (2)

23. Answer (1)

24. Answer (4)

25. Answer (2)

26. Answer (1)

27. Answer (3)

28. Answer (2)

29. Answer (4)

30. Answer (3)

31. Answer (2)

32. Answer (3)





- 5 -

Q.No. Solution

33. Answer (1)

34. Answer (4)

35. Answer (4)

36. Answer (4)

37. Answer (2)

38. Answer (3)

39. Answer (4)

40. Answer (3)

41. Answer (4)

42. Answer (1)

43. Answer (2)

44. Answer (2)

45. Answer (3)

46. Answer (2)

47. Answer (2)

48. Answer (4)

)32(2)3( iizz

)32(2)3( iizz

|z|2 + 3iz = 4 + 6i

(x2 + y2) + 3i (x + iy) = 4 + 6i

(x2 + y2 – 3y) + 3ix = 4 + 6i

3x = 6 and x2 + y2 – 3y = 4

x = 2 and 4 + y2 – 3y = 4

y = 0, 3

z = x + iy

= 2 + i.0 and 2 + 3i

= 2, 2 + 3i






- 6 -

Q.No. Solution

49. Answer (4)

f (x) = x4 – 8x

3 + 4x2 + 4x + 39

x = 3 + 2i (x – 3)2 = –4

x2 –6x + 13 = 0

Now x4 – 8x

3 + 4x2 + 4x + 39 = (x

2 – 6x + 13) (x2 – 2x – 21) + (–96x + 312)

Now f (3 + 2i) = –96(3 + 2i) + 312

= –288 – 192i + 312

= 24 – 192i

= a + ib

a = 24, b = –192.

Required ratio =8

1

192

24

50. Answer (1)

In regular hexagon OA = AB = BC = CD = ED = EF = FA

Length of perimeter = 6 × |OA|

= 6 1 4

6 5

51. Answer (1)

arg(1 + i) =4

2

arg 1 33

i

5

arg 36

i

arg 36

i

arg2

i

arg 32

i

arg 2 0

arg 1

Required sum11

12

O

A i(1 + 2 )E

F

B

C

D








- 8 -

Q.No. Solution

56. Answer (4)

If z1 and z2 are two complex number if 1Im( ) 0z and 2Im( ) 0z then z1 > z2 or z1 < z2 does not hold.

57. Answer (3)

2

3

2

1)()( 3231 izzzz

3/3231 )()( iezzzz

Now using concept of rotation.

z1, z2, z3 are vertices of equilateral triangle.

58. Answer (3)

( )ia i

log loge ea i i

2log loge ea i e

log2e a

2a e

Therefore sin(ln ) sin 12

a

Im( ) arg( ) 0 0 0a a

Therefore 1 2 3, ,S S S are correct.

59. Answer (2)

if z2 + z + 1 = 0

(z – ) (z – 2) = 0

z = , 2

if z = , then 21

z

To find the value of2

21

212

3

32

2

22

1......

111

zz

zz

zz

zz

Now 21

,111

,111

3

3

2

2

2

2

z

zz

zz

z

111

,1111

2

2

5

5

4

4

4

4

z

zz

z and

216

6 z

z ...... and so on

z3

z2

z1

/3





- 9 -

Q.No. Solution

Therefore,

2

21

212

3

32

2

22

1......

111

zz

zz

zz

zz

= {(– 1)2

+ (–1)2

+ (2)2

} + {(–1)2

+ (–1)2

+ (2)2

} × ...... 7 times

= (1 + 1 + 4) + (1 + 1 + 4) × .... 7 times

= 6 + 6 × ..... 7 times

= 6 × 7 = 42

60. Answer (1)

is an imaginary 5th root of unity

5 = 1

1 + + 2

+ 3

+ 4

= 0 … (i) [ sum of nth

roots of unity is zero]

Now,

1

1log 322

1log 4

2 [ from(i) 1 + + 2 + 3 = – 4]

|||2|log|2|log2

log1

log 42

422

5

2

12log2 [ | | = | 2 | = | 3 | = | 4 | = 1 hence nth roots of unity lie on unit circle]

61. Answer (3)

We have

x3n+1 – 1 = (x – 1) (x – 1) (x – 2) ......... (x – 3n)

Thus,)()........)((

)()........()(

321

32

22

12

n

n

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

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

)1(

1

321

32

22

122

n

n

1.

1.)(.

1

1

1

1)(.

1

13

232

13

132

n

n

n

n

1

1.

1

1 2

[ 3n = 6n = 1]

11

12

2







- 11 -

Q.No. Solution

66. Answer (2)

To find area of a whose vertices are represented by complex number 0, z and zei (0 < < )

Area of Abc sin2

1

sin||||2

1zz

sin||2

1 2z

67. Answer (2)

22

z

z

22||

z

z

22

r

r

22

22

r

r

when 22

02

r

r

r 2 – 2r – 2 0

3131 r … (i)

31max r

68. Answer (1)

Let the value of i x iy

2( )i x iy

2 2 2i x y xy

On comparing real and imaginary part

2 2 0x y and 2 1xy

x y and1

2xy

Therefore1 1

,2 2

i ii

| |z

| |z

O(0, 0)

A z( )

B ze( )i






- 12 -

Q.No. Solution

69. Answer (4)

Roots of the equations

(x – a) (x – b) + (x – b) (x – c) + (x – c) (x – a) = 0

i.e., {x2

– (a + b)x + ab} + {x2

– (b + c)x + bc} + {x2

– (a + c)x + ac} = 0

i.e., 3x2 – 2(a + b + c)x + (ab + bc + ca) = 0

have equal roots, Therefore

B2 – 4 AC = 0

4(a + b + c)2 – 4 × 3 (ab + bc + ca) = 0

4[a2 + b2 + c

2 + 2ab + 2bc + 2ca] – 12ab – 12bc – 12ca = 0

4(a2 + b2 + c

2) – 4ab – 4bc – 4ac = 0

a2 + b

2 + c2 – ab – bc – ca = 0 (Hence (3) is true)

70. Answer (2)

2||2

1||||log

2

3

z

zz

22

)3(||2

1||

z

zz

| z |2 – | z | + 1 < 6 + 3 | z |

| z |2 – 4| z | – 5 < 0

(| z | + 1) (| z | – 5) < 0

but | z | + 1 > 0

| z | – 5 < 0

| z | < 5

71. Answer (3)

arg z =4

–1tan4

yx

1y

x

|y| = |x| x2 – y2 = 0

Re(z2) = 0





- 13 -

Q.No. Solution

72. Answer (3)

Let z = x + iy

Given equation is,

z2

+ z|z| + |z|2

= 0

(x + iy)2 + (x + iy) + 2 2x y + (x2 + y2) = 0

x2 – y2 + 2ixy + 2 2 2 2 2 2 0x x y iy x y x y

2 2 2 2 22 (2 ) 0x x x y i xy y x y

Now, 2 2 22 0x x x y

2 2(2 ) 0x x x y

x = 0 or x2

+ y2

= 4x2

3x2 – y2 = 0

Al ternat ive

2

21 0

| || |

z z

zz

2

1 0| | | |

z z

z z

2,| |

z

z

z = |z|, z = 2|z|

73. Answer (2)

P

y x=13

3 0y x

Now,

2 2

3·2 3 0

( 3) 1p

6

32

p






- 14 -

Q.No. Solution

74. Answer (3)

4 1Re

2 1 2

z

z

4 4

12 1 2 1

z z

z z

4 4

12 1 2 1

z z

z z

2 8 4 2 8 4

1(2 1)(2 1)

zz z z zz z z

z z

4 7 7 8 4 2 2 1 zz z z zz z z

9 9 9 0 z z

1 z z

Hence point z lies on a straight line.

75. Answer (2)

h(x) = xf (x3) + x2g(x6) is divisible by x2 + x + 1,

So, when h(x) will be divided by x – and x – 2 remainder will be 0.

h() = f (1) + 2g(1) = 0 …(i)

h(2) = 2f (1) + g(1) = 0 …(ii)

Now, adding (i) & (ii),

( + 2

)f (1) + ( + 2

)g(1) = 0 – f (1) – g(1) = 0 f (1) = – g(1)

76. Answer (1)

The given expression is (x – 1) (x – 1) ..... (x – 1) ..... till 2n factors.

= (x – 1)2n

77. Answer (2)

101101 101 101cos sin –

6 6

z i i i

101101101 101

cos sin6 6

xi i

1015 5cos sin

6 6i i

3cos sin

6 6 2 2

ii i

3

2 2

i





- 15 -

Q.No. Solution

Now,

(i101 + z101)103

1033

– 2 2

i

1035 5

cos sin6 6

i

515 515cos sin

6 6i

11 11cos sin

6 6i

cos – sin6 6

i

3

– 2 2

i

78. Answer (2)

1

az

az

|z – a| = |z + a|, let z = x + iy then,

(x – a)2 + (y)2 = (x + a)2 + y2

x2 – 2ax + a

2 + y2 = x

2 + a2 + 2ax + y

2

4ax = 0

x = 0

is y-axis

79. Answer (2)

Let z = x + iy

1)(

122

iyxi

iyxz

ixy

ixy

ixy

yix

)1(

)1(

)1(

212

22

)1(

)12()1()2(2)1()12(

xy

xxyyixyyx

from given condition

2)1(

)12()1()2(22

xy

xxyy

2y – 2y2 – 2x

2 – x = –2(x2 + y2 + 1 – 2y) = 2y – x = –2 + 4y

2y + x – 2 = 0, i.e. a straight line






- 16 -

Q.No. Solution

80. Answer (4)

z = x + iy

| 2008z – 1 | = 2008 | z – 2 |

|2|2008

1

zz

Put z = x + iy

2222

)2()(2008

1yxyx

xx 442008

12

2008

12

2

2008

14

10044

xx

1004

14

2008

14

2

x

a line parallel to y-axis.

81. Answer (2)

31

1arg

APBz

z

z lies on a circle Al ternatively

put z = x + iy

31

1arg

z

z

3)1(

)1(arg

iyx

iyx

3)1(

)1(

)1(

)1(arg

iyx

iyx

iyx

iyx

3)1(

)2()1(arg

22

22

yx

yiyx

31

2tan

22

1

yx

y

31

222

yx

y 023)(3 22 yyx

013

222 yyx

Which represents a circle having centre at

3

1,0 and radius

3

21

3

1

3

AB

(–1, 0) (1, 0)

P z( )

O

y

x





- 17 -

Q.No. Solution

82. Answer (1)

,4

exp2

iiz where is parameter put z = x + iy

4sin4cos2 iiiyx

equating real and imaginary parts we get

)1......(4

cos2

x

4sin21y

or )2....(4

sin21

y

squaring and adding (1) and (2), we get

x2

+ (y – 1)2

= 4which represents a circle with centre (0, 1) and radius 2.

83. Answer (4)y

x

iz

A

z

iz

z O

90°

On rotating OA by 90° angle we can find other vertices.

84. Answer (2)

z1, z2, z3 are the vertices of an equilateral triangle such that |z1| = |z2| = |z3|

or |z1 – 0| = |z2 – 0| = |z3 – 0|

origin is the circumcentre of the origin is the centroid of the equilateral

03

321 zzz

z1 + z2 + z3 = 0

85. Answer (4)

|z –(2 + 3i)| + |z – (–2 + 6i)| = 4Let z1 = 2 + 3i, z2 = –2 + 6i

|z1 – z2| = |4 – 3i| = 5 > k

|z – z1| + |z – z2| = 2a, where k < |z1 – z2|

This does not represent any curve

Locus of z is an empty set.

Al ternatively : If we put z = x + iy, then we got an equation in x and y which does not have any solution.






- 18 -

Q.No. Solution

86. Answer (2)

z1, z2, z3 and u, v, are complex numbers representing the vertices of two triangles such that

z3 = (1 – t) z1 + tz2 and = (1 – t)u + tv, tc

z3 = z1 – tz1 + tz2 and – u = – tu + tv z3 – z1 = t(z2 – z1) and – u = t (v – u)

12

13

zz

zzt

… (1) and

w ut

v u

… (2)

From (1) & (2)

3 1

2 1

z z w u

z z v u

3 1

2 1

arg arg ..........( )*z z w u

z z v u

3 1

2 1

arg 1 arg 1z z w u

z z v u

3 2

2 1

arg argz z w v

z z v u

3 2

1 2

arg arg ....( )**z z w v

z z u v

from (*) & (**) we conclude that two triangles are similar.

87. Answer (3)

max – min = 2 =

5

3sin.2 1

=

5

3cos2 1

88. Answer (1)

11 21

21

zz

zz

|1||| 2121 zzzz 221

221 |1||| zzzz

21212

22

121212

22

1 ||||1|||| zzzzzzzzzzzz

|z1|2 |z2|

2 – |z1|2 – |z2|

2 + 1 = 0 (|z1|2 – 1) (|z2|

2 – 1) = 0

|z1| = 1, |z2| = 1 Both z1 and z2 lie on the circle |z| = 1

min

max

15 (0, 25)

y

xO





- 19 -

Q.No. Solution

89. Answer (3)

x2 + x + 1 = 0

+ = –1

= –1

(1) 2 + 2 = ( + )2 – 2

= (–1)2 – 2(1)

= 1 – 2 = –1

(2) ( – )2 = ( + )2 – 4

= (–1)2 – 4 1 = –3

(3) 3 + 3 = ( + )(2 + 2 – )

= 2( )(( ) 2 )

= 2( )(( ) 3 )

= (–1)((–1)2 – 31)

= (–1) (1 – 3) = 2

Al ternat ive

x2 + x + 1 = 0

x = , 2 (complex root of unity)

3 + (2)3 = 2

(4) 4 2 = 2 2 2 2 2( ) 2

= (–1)2 – 2 1 = 1 – 2 = –1

90. Answer (2)

l

n

l

n

q

p ,,

Now

p

q

q

p

l

n

l

n

l

n

0 l

n

p

q

q

p

l

n

p

q

q

p

91. Answer (2)

06 0622 xxxx

2,3 023 xxxx

x = ± 2

Two roots are real, with sum 0.






- 20 -

Q.No. Solution

92. Answer (1)

0)()()( 2 cbaxbacxacb

Put x = 1, 0 cbacbabacacb

1 is the root of the equation.

Roots are rational.

93. Answer (1)

We know

1tansec 22

1)tan)(sectan(sec

142

a

b

a

acb

Squaring both side

422 )4( abacb

04 244 cabba

94. Answer (4)

Let72

71342

2

xx

xxy

0771)342()1(2 yyxyx

For real x, discriminant should be 0

0)771)(1(4)342( 2 yyy 0)771)(1(4)17(4 2 yyy

0)71787()17( 22 yyy 03601128 2 yy

045142 yy 0)5)(9( yy

5 or 9 yy + –

5 9

+

95. Answer (3)

02 cbxax , Given equation is an2 – bx (x –1) + c(x –1)2 = 0

2 – –

1 0 – 1 – 1

x xa b

x x

Now,

Replacing x by1

x

x

0)1()1( 01)1(

22

2

2

xcxbxaxcx

bx

x

ax

1

1x

x

x is the root of the above equation.





- 21 -

Q.No. Solution

96. Answer (2)

acbDa

c

a

b4,, 2

1

rpqDp

r

p

q4,, 2

2

Let common difference of A.P. be k.

|||| k

p

pr q

a

acb 4422

p

a

D

D

p

a

pr q

acb

2

1

2

2

4

4

2

2

2

1

p

a

D

D

97. Answer (4)

xbcac

bxaxc

abcb

axcxb

caba

cxbxa

))((

))((

))((

))((

))((

))((

is satisfied by x = a, x = b, x = c.

A quadratic equation is satisfied by more than two values of x. So it is an identity. Hence it is satisfied by allvalues of x.

98. Answer (3)

(1) Root will be of the form of a, b, c are rational.

(2) There is no information about b2 – 4ac

Hence statement is false.

(3) As a, b, c are real and one root is i then other root will be i .

(4) If mass are of opposite sign then 0 0 c

a

99. Answer (1)

For an identity

(k2 – 3k + 2) = 0

(k – 1) (k – 2) = 0

k = 1, k = 2

k2 – 5k + 4 = 0 (k – 1)(k – 4) = 0

k = 1, 4

k2 – 6k + 5 = 0 (k – 5)(k – 1) = 0

k = 1, 5

Common value of k = 1.






- 22 -

Q.No. Solution

100. Answer (1)

(1) Let roots are ,

+ =13

5 … (i)

=5k … (ii)

For reciprocal roots = 1 k = 5.

(2) If roots are cosecutive integer then

| – | = 1

| – |2 = 1

( + )2 – 4 = 1

1 – 4k = 1

k = 0

(3) Let roots are 2,

2 + = 6 … (i)

2 = k … (ii)

By (i), (ii)

= 2, k = 8

(4) In this case

+ = 0

k = 0

101. Answer (2)

bababaxx 4,, 0 22

abababxx 4,, 0 22

Now, abba 44 22

)(4 44 2222 abbaabba

)(4 0)4)(( babababa

102. Answer (1)

Let be common root, 022 r qp …(1)

and 022 qr p …(2)

Now (1) – (2) 2

1 0)(2 qr r q

Common root is ,2

1 substituting in (1)

044 02

12

2

12

pqr r qp





- 23 -

Q.No. Solution

103. Answer (1)

x2 + x + 1 = 0 … (i)

Discriminant = b2 – 4ac = 1 – 4 1 1 = –3

Hence the roots of x2 + x + 1 = 0 and not real.

So roots will be in pair.

Also the roots of ax2 + bx + c = 0 will be non-real.

Clearly both roots of the equations are common.

1 1 1

a b c

a : b : c = 1 : 1 : 1

104. Answer (2)

If 1, 2, 3 are roots of equation then

x3 + ax

2 + bx + c = 0

1 + 2 + 3 = – a a = –6

12 + 23 + 13 = b b = 11

123 = – c c = –6

105. Answer (4)

f (x) = ax2 + bx + c

= 2

b ca x x

a a

=2 2

2

2 24 4

b b b ca x x

a aa a

f (x) =2 2

2

4

2 4

b b aca x

a a

f (x) =2 2 4

2 4

b b aca x

a a

f (x) =2 4

b Da x

a a

Clearly if a > 0 the minimum value of ( )4

Df x

a

Similarly of a < 0 the maximum value4

D

a

If ax2 + bx + c > 0 then a > 0, D < 0 for all x R

Hence option (4) is not true.






- 24 -

Q.No. Solution

106. Answer (3)

x2 + 2x + 3 = (x + 1)2 + 2 m = 2

– x2 + 4x + 6 = – x

2 + 4x + 4 – 4 + 6

= 6 – (x2 – 4x + 4) + 4

= 10 – (x2

– 4x + 4)= 10 – (x – 2)2

M = 10

m + M = 2 + 10 = 12

107. Answer (3)

Let 1592 mmmxy

We need y > 0

Upward parabola above x-axis.

2 9 5 1 0, .mx mx m x R

0,0 aD

i.e., 0 and 0)15)((481 2 mmmm

61

40 0 and 0)461( mmmm

Also for m = 0,

20 9(0) 0 1 1 0,x x x R

61

4,0m

108. Answer (1)

01)( 2 lxxml

)(3 2

lm

l

lm

l

…(1)

)(2

1

1)2( 2

mlml

…(2)

From (1) and (2)

2

2

)(9)(2

1

ml

l

ml

0992 2 mll

For real l,72

81 09881 mm

8

9m

Greatest value of m is .8

9

y

xO





- 25 -

Q.No. Solution

109. Answer (3)

0753 2 r qxpx

pr qr pq 8425)7)(3(4)5( 22

222

25342584)(25 r pr ppr r p 025

336

5

175

22

r r p

roots are real and distinct.

110. Answer (2)

x3 – 2x

2 – x + 2 = 0

As x = 1 is the root of the equation

Hence we may write

x3 – 2x

2 – x + 2

= x2 (x – 1) – x(x – 1) – 2(x – 1)

= (x – 1) (x2 – x – 2)

= (x – 1) (x – 2) (x + 1)Roots = 1, –1, 2.

111. Answer (2)

ba ,2

baa 242 222

The other root of equation will be 22

i.e. baa 242 2

Sum of roots, S = –4a

Product of roots, P = bbaa 2)24(422

required equation is 02 PSxx

i.e. 0242 baxx

112. Answer (1)

For roots of opposite sign, product < 0

0)1)(2( 03

232

aaaa

21 a

113. Answer (3)

Here we observe that (a + c)2 < b2

(a – b + c) (a + b + c) < 0

Exactly one real root of the given equation lies in (–1, 1).

02 cbxax

D = 222 )(4)(4 caaccaacb

0

Roots are real.






- 26 -

Q.No. Solution

114. Answer (3)

p + iq is one root p – iq is other root.

Let be third root.

Now sum = 0 iqpiqp

p2

p2 is root of 03 baxx

2p is root of 0)( 3 baxx

03 baxx

115. Answer (2)

We have, given expression

(a12 + a2

2 + a32 +.....+ an – 1

2)x2 + 2(a1a2 + a2a3 + a3a4 +.....+ an – 1 an)x + (a22 + a3

2 + a42 +......+ an

2) 0

(a1x + a2)2 + (a2x + a3)

2 + (a3x + a4)2 + ....... + (an – 1x + an)

2 0

(a1x + a2)2 + (a2x + a3)

2 + (a3x + a4)2 + ....... + (an – 1x + an)

2 = 0,

as sum of square cann't be negative.

a1x + a2 = 0 = a2x + a3 = a3x + a4 = ....... = an – 1x + an

32 4

1 2 3 1

....... n

n

aa a ax

a a a a

a1, a2, a3, ....... , an – 1, an are in G.P.

116. Answer (2)

,)( 2 cbxaxxf given 0)1( cbaf

Rxxf ,0)( as roots are non-real complex

f (–2) < 0 bcacba 24 024

117. Answer (3)

Given,a

c

a

b ,

Also,22

11

2)())(( 22222 a

c

a

b

a

c

a

b 22

2

2

2

222 2caabbc abbcca 2222

c

b

a

c

b

a2

b

c

a

b

c

a,, are in H.P.





- 27 -

Q.No. Solution

118. Answer (2)

21)2)(13)(23)(1( xxxx

21)253)(253( 22 xxxx

Put,txx

53

2

25 21)2)(2( 2 ttt 5,5 tt

Now, 553 2 xx and 553 2 xx

0553 2 xx and 0553 2 xx

0553 2 xx has two irrational roots.

whereas roots of 0553 2 xx are imaginary.

119. Answer (4)

S1 : x2 – x – 2 < 0

(x – 2)(x + 1) < 0

–1 < x < 2 … (i)

2sin2x + 3sinx – 2 > 0

2sin2x + 4sinx – sinx – 2 > 0

2sinx(sinx + 2) –1 (sinx + 2) > 0

(sinx + 2)(2sinx – 1) > 0

1

sin2

x

5

,

6 6

x … (ii)

By (i), (ii) , 26

x

S2 : Using A.M. G.M.

22

222

2

tan

tan( )

2

x xx x

x xx x

2

2

2

tan2tanx x

x x

S3 : Using D 0

4(a + b + c)2 – 4(1)(3)(ab + bc + ca) 0

a2 + b

2 + c

2 + 2(ab + bc + ca) – 3(ab + bc + ca) 0

a2 + b

2 + c

2 + (ab + bc + ca) (2 – 3) 0

2 2 2

3 2

a b c

ab bc ca … (i)






- 28 -

Q.No. Solution

But (a – b)2 c2

(b – c)2 a2

(c – a)2 b2

2 2 2

2

a b c

ab bc ca

3 – 2 2

4

3

S4 :3 23 3 0 x px qx r …(i)

Multiply the second equation by x

3 22 0 x px qx … (ii)

By (i) – (ii)

2

2 0 px qx r

… (iii)But x

2 + 2px + q = 0

px2 + 2p

2x + pq = 0 … (iv)

By (iii) - (iv)

2x(q – p2) + (r – pq) = 0

22( )

r pqx

q p

Putting x in

x2 + 2px + q = 0

We get

4(p2 – q) (q2 – pr ) = (pq – r 2)2

120. Answer (3)

The given quadratic equation is (a – b)x2 – 5(a + b)x – 2(a – b) = 0

The discriminant

D = (– 5(a + b))2 + 8(a – b) (a – b)

= 25(a + b)2 + 8(a – b)2

Hence D > 0 a & b.

So, roots are real and unequal.

121. Answer (4)

Since, , are the roots of the equation ax2 – bx + c = 0

So, ,b c

a a

Now, we have to observe root of the equation

(a + cy)2 = b2y

a2 + 2acy + c2y2 = b2y





- 29 -

Q.No. Solution

c2y2 + (2ac – b2)y + a2 = 0

2 2

22 2

20

ac b ay y

c c

2 2

2 2 2

20

b a ay ycc c

22 2 2 2

1 1 10y y

Hence the equation (a + cy)2 = b2y has roots2 2

1 1,

122. Answer (1)

Since a, b, c are in G.P.,

So, b2 = ac

4b2

– 4ac = 0D = 0 for the equation ax2 + 2bx + c = 0

Hence, it will have equal roots, and root will be

bx

a

Now, ax2 + 2bx + c and dx2 + 2ex + f = 0 have a common root,

So,b

xa

will satisfy the equation

dx2 + 2ex + f = 0

2

2. 2 . 0b b

d e f

aa

2 2

2

20

db aeb a f

a

db2 – 2aeb + a2f = 0

dac – 2aeb + a2f = 0

dc + af = 2eb

2d f e

a c b

So, , ,d c f

a b c are in A.P.

123. Answer (3)

The given equation is x2 – 2(k + 2)x + 12 + k2 = 0 has distinct real roots when D > 0

4(k + 2)2 – 4(12 + k2) > 0

k2 + 4 + 4k – 12 – k2 > 0

4k – 8 > 0

k > 2

So least integral value of k is 3.






- 30 -

Q.No. Solution

124. Answer (4)

– 1D

a

2 48 1p

p = ± 7; but p is positive, hence p = 7.

125. Answer (1)

f (x) = (x – a) (x – c) + k(x – b) (x – d)

f (a) = k(a – b) (a – d) which is positive

f (b) = (b – a) (b – c) which is negative

f (c) = k(c – b) (c – d) which is negative

f (d) = (d – a) (d – c) which is positive

So, f (x) = 0 has a root in the interval (a, b) and another in (c, d). So the roots are real and distinct.

126. Answer (4)

Use relation between roots and coefficients

= p ...(i)

= r ...(ii)

22

q ...(iii)

(2 )2

r ...(iv)

(ii) and (iv) are same. (i) and (ii) can be solved to obtain and in terms of p and q, thereby giving r .

127. Answer (4)

Use the idea of rotation to obtain the desired P as

3 4

i

e .ei, where tan =4

3

yielding sin 4

5, cos =

3

5

128. Answer (4)

We have,

2

1 1

11

z

zzz z z

z z

m

1 1

2 I ( )

z z i z

1,

Im

i i R

z z

Thus the locus of21

z

z is y-axis.





- 31 -

Q.No. Solution

129. Answer (4)

(6, 2)

Z 2Z2

Imaginary axis

Real axis

(7, 6)

Z 0

( 1 , 2 )

90°

1

1

3

5

2 (6 2 cos45 , 5 2 sin45 ) (7, 6) 7 6 Z i

by rotation about (0, 0)

2 2 22 2

2

( )

i iZe Z Z e

Z

2 (7 6 ) cos sin (7 6 )( ) 6 72 2

Z i i i i i

130. Answer (1)

3 3 350 zz zz

2 2 2 2| | ( ) | | ( ) 350 z z z z

2 2 2| | ( ) 350 z z z

(x2 + y2) (x2 – y2 + 2ixy + x2 – y2 – 2ixy) = 350

2(x2 + y2)(x2 – y2) = 350

(x2 + y2)(x2 – y2) = 175 = (32 + 42)(42 – 32)

Which suggests that points (x, y) satisfying the given equation are (4, 3), (–4, –3), (–4, 3), (4, –3)

y

x

A(4, 3)(–4, 3)B

C(–4, 3)

D(4, 3) –

O

Required area = AB × BC

= 8 × 6

= 48 sq. units






- 32 -

Q.No. Solution

131. Answer (4)

z = cos + isin = ei

Now,15

(2 1)

1

lm

i m

m

e

= sin + sin3 + ..... + sin29

=

15.2sin

2 (15 1) 22 .sin2 2

sin2

=sin15 .sin15

sin

For = 2°, the given expression reduces to

=sin30 .sin30 1

sin2 4sin2

132. Answer (2)

We have + = – p

3 + 3 = q = ( + )3 – 3( + ) = – p3 + 3p()

3

3

p q

p

The quadratic equation with and

as roots is

2 0 x x

2

2 ( ) 21 0

x x

32

2

3

23

1 0

3

p qp

px x

p q

p

(p3 + q)x2 – (p3 – 2q)x + (p3 + q) = 0

133. Answer (3)

We observe that

( + ) (n –1 – n –1) = n – n + (n – 2 – n –2)

6an – 1 = an – 2an – 2

2

1

26

n n

n

a a

a





- 33 -

Q.No. Solution

2

1

23 , 2

2

n n

n

a an

a

Putting n = 10, we get

10 8

9

23

2

a a

a

134. Answer (2)

Let be a common root between given equations x2 + bx – 1 = 0 and x2 + x + b = 0

2

2

1

1 11

b bb

2

2 1 1and

1 1

b b

b b

22 1 1

1 1

b b

b b

2

2 (1 )1

1

b

bb

b2 – b3 + 1 – b = 1 + 2b + b2

b3 + 3b = 0

0, 3 b b i

3 b i

135. Answer (4)

As, a is real,

So a a gives

z2 + z + 1 = 2 1 z z

( )( 1) 0 z z z z

As, z z

So, 1 z z

1

{where }2

x z x iy

Now,

a = z2 + z + 1

21 1

12 2

iy iy

23

4 y

As, y 0 so,3

4a






- 34 -

Section-B

Q.No. Solution

1. Answer (1, 2, 4)

If |3z – 1| = 3 |z – 2|

Let z = x + iy

|3x + 3iy – 1| = 3|x + iy – 2|

(3x – 1)2 + (3y)2 = 9 [(x – 2)2 + y2]

9x2 + 1 – 6x + 9y

2 = 9x2 + 9y

2 – 36x + 36

30x = 35

x =6

7

6 Re(z) = 7 (A line parallel to y-axis)

Also mid-point of 0,31 and (2, 0) is 0,

67

2. Answer (1, 2, 3)

We can write11

2cos

11

2sin

ki

k

=

11

2sin

11

2cos

ki

ki

Now

10

111

2cos11

2sink

kikS

10

1

11

2

k

ki

eiS

= ieeei

iii

11.... 11

20

11

4

11

2

(as 1 + + 2 + ..... + 10 = 0)

S = i

iS

0 SS

and 1SS

iiS 212

1 2

)1(2

1iS





- 35 -

Q.No. Solution

3. Answer (1, 3)

cos cos cos 0 A B C

sin sin sin 0 A B C

Let cos sin iA

a A i A e

cos sin iBb B i B e

cos sin iCc C i C e

0a b c

3 3 3 3a b c abc

2 2 2

3a b c

bc ac ab

2 2 2

3.

iA iB iC

iB iC iA iC iA iB

e e e

e e e e e e

2 2 2 3iA iB iC iB iA iC iC iA iBe e e

cos(2 ) sin(2 ) cos(2 ) sin(2 ) cos(2 ) sin(2 ) 3 A B C i A B C B A C i B A C C A B i C A B

On comparing real and imaginary part

cos(2 ) cos(2 ) cos(2 ) 3 A B C B A C C A B

sin(2 ) sin(2 ) sin(2 ) 0 A B C B A C C A B

4. Answer (2, 4)

x x

2 – 2 cos + 1 = 0

2

4cos4cos2 2 x

= cos isin

= cos + isin, = cos – isin

ninn sincos ; ninn sincos

nS nn cos2 ; P = 1

Equation is 02 PSxx

i.e. 01cos22 nxx

Option (4)

nnnxx 222 cos1coscos2 = 0

0sin)cos( 22 nnx

Option (2)






- 36 -

Q.No. Solution

5. Answer (2, 3)

a2 + b

2 + c2 = 1

b +ic = (1 + a) z

a

icbz

1

a

icbi

aicbi

iz

iz

11

11

1

1

ibca

ibca

iz

iz

1

1

1

1

)1(

)1(

)1(

)1(

ibca

ibca

ibca

ibca

22

22

)1(

)1(

bca

ciba

c

iba

1

Similarly

ibca

ibca

ibca

ibca

1

1

1

1 =

iba

c

1

6. Answer (3, 4)

2amp

12

14

zz

zz

A = 90°

alsoz1 – z4 = z2 – z3 |z1 – z4| = |z2 – z3|

AD = BC and AD || BC

So AB || CD and AB = CD

ABCD is a rectangle or cyclic quadrilateral

7. Answer (2, 3, 4)

Cube roots of p are1 1 1

23 3 3, ,p p p

i.e. 23

1

3

1

3

1

.,., ppp

Now,222222

222222

zyx

zyx

)( 22222

2222

22422

24222

zyx

zyx

zyx

zyx

= =

Option (4)

We can assign the , , different value we get other options also.

A B

CD

z1 z2

z4 z3

90°





- 37 -

Q.No. Solution

8. Answer (1, 2)

z + z –1 = 1

11

z

z

z2

– z + 1 = 0

2,2

31

iz

zn + z

– n = (– )n + (– )

–n

Case (1), n = 3m

(– )3m + (– ) –3m = (–1)n + (–1)n = 2(–1)n

whenn = 3m + 1

(– )3m + 1 + (– ) –3m – 1

=13

13

)(

)1()1(

m

nmn

=

1

)1( n

= (–1)n (–1) = (–1)n + 1

9. Answer (2, 3, 4)

|z 1| < |z + 3|

Let z = x + iy

(x 1)2 + y2 < (x + 3)2 + y2

x2 + y

2 2x + 1 < (x2 + y2 + 6x + 9)

8x > –8x > – 1

i = i((2x + i2y) + 3 – i)

= i2x – 2y + 3i + 1

= i(3 + 2x) + (1 – 2y)

as 3 + 2x > 1 option

also, – 1 = 2z + 3 – i – 1

= 2z + 2 – I = 2x + 2iy + 2 – i

= 2(x + 1) + i(2y – 1)

as x > – 1

2(x + 1) > 0 2

)1(arg

option (4)

– 5 = 2(x – 1) + i(2y – 1)

+ 3 = 2(x + 3) + i(2y – 1) as x > –1

| + 3| > | – 5|

Option (2)






- 38 -

Q.No. Solution

10. Answer (2, 3, 4)

|z + |2 = |z|2 + ||2

Since |z + |2 = |z|2 + ||2 + zz

|z|2 + ||2 = |z|2 + ||2 + zz zz = 0

zz

zz

z

is purely imaginary

Therefore,2

zamp

11. Answer (1, 2, 3)

Clearly |z1 – z2|min = 2 – 1 = 1

|z1 – z2|max = 3

max|2z1 + z2| = |2 + 2| = 4

Now, |z| = 1

|z1| = 1

1. 11 zz

11

1

zz

Now 121

2

1zz

zz

and |||||| 1212 zzzz

3.

12. Answer (3, 4)

12

z

z

1||

2||

zz

02|||| 2 zz

0)1|)(|2|(| zz

21

||

2||1 z

But |z| 0

2||0 z

3|| z and |z| 4

Option (3) and (4)

1

1

2

| | = 1z

| | = 2z

y

xO





- 39 -

Q.No. Solution

13. Answer (1, 2, 4)

The given equation is |2z – i| = m|z + i|

2(2 )(2 ) ( )( )z i z i m z i z i

24 2 2 1 ( 1)zz iz i z m zz iz i z

2 2 2 2(4 ) (2 ) (2 ) (1 ) 0m zz m iz m i z m

The above equation will does not represent a circle, when,

4m2 = 0 m = 2 ; since m cannot be negative

Hence answer is (1, 2, 4)

14. Answer (1, 2, 3)

(x + 1)3 = (– 4)3

x + 1 = – 4, – 4, – 42

x = – 5, – 1 – 4, – 1 – 42.

Hence, roots are – 5, – 1 – 4, – 1 – 42.

15. Answer (1, 2, 3)

2 21| | 1z a b a2 + b2 = 1 …(i)

2 22| | 1z c d c2 + d2 = 1 …(ii)

1 2Re( )z z = Re[(a + ib)(c – id)]

= Re[ac + bd + i(bc – ad)] = 0

ac + bd = 0 …(iii)

Now, using (i) & (iii) we can prove that b = c, a = d.

Hence, 2 2 2 21| | 1a c a b

Similary we can observe, |2| = 1

1 2Re( ) 0

16. Answer (2, 4)

Given inequality is,

2

1 2

2

| | 2| | 6log 0

2| | 2| | 1

z z

z z

2

2

| | 2| | 61

2| | – 2| | 1

z z

z z

|z|2 + 2|z| + 6 > 2|z|2 – 2|z| + 1

|z|2 – 4|z| – 5 < 0

|z| (– 1, 5), but |z| > 0

0 < |z| < 5






- 40 -

Q.No. Solution

17. Answer (2, 3, 4)

z1, z2 are the complex numbers satisfying,

1 2

1 2

1 –

z z

z z

|z1 + z2| = |z1 – z2|

1 2 1 2 1 2 1 2( )( ) ( – )( – )z z z z z z z z

1 1 1 2 2 1 2 2 1 1 1 2 2 1 2 2 – – z z z z z z z z z z z z z z z z

1 2 2 12( ) 0z z z z

1 1

2 2

– z z

z z

1 2 2 1 0z z z z

1 2 1 2

0z z z z

2Re( ) 0zz

18. Answer (1, 2, 3, 4)

a

b cossin

a

c cos.sin

1cossin 22

12

2

a

c

a

b 22 2 acab

0222 acba

Also 222)( cbca .

19. Answer (1, 2)

a

b2 ,

a

c

a

acb

2

2||

A

Bhh

2 ,

A

Chh ))((

A

ACB

22||

A

ACB

a

acb

22 22

2

2

2

2

A

a

ACB

acb

Also 2h =a

b

A

B

A

B 22)(

2

A

B

a

bh .





- 41 -

Q.No. Solution

20. Answer (1, 2)

Let us use the transformation3 1

12

x xy x

So that x + 3 = y + 2 and x – 1 = y – 2

Thus the given inequality reduces to

(y + 2)5 – (y – 2)5 244

(y5 + 5.y4.2 + 10y3.22 + 10y2.23 + 5.y.24 + 25) – (y5 – 5.y4.2 + 10.y3.22 – 10y2.23 + 5.y.24 – 25) 244

2[10y4 + 80y2 + 32] 244

y4 + 8y2 – 9 0

(y2 + 9) (y2 – 1) 0

y2 – 1 0 as y2 + 9 > 0, y.

–ve+ve +ve

–1 1

y 1 or y –1

x + 1 1 or x + 1 –1

x 0 or x –2

Hence the required solution set is

(– , –2] [0, ).

21. Answer (3, 4)

02 cbxax

Let a, b, c is a, ar , ar 2

Now, 022 ar arxax

022 r rxx

231 ir x x = r or 2

r

Roots are imaginary and are in the ratio 1 : or 2 : 1.

22. Answer (3, 4)

0)45)(35( 012525 2 xxxx

01 as

5

3

5

4,

5

3xxxx

cos =

5

4

5

3sin or

5

3

25

24

5

4

5

32cossin22sin

or25

24

5

4

5

32






- 42 -

Q.No. Solution

23. Answer (1, 2)

Let be common root

r pqr pq ,,02

qpr qpr ,,02

1( ) 0p q r r q

p

Common root is =p

1

Other roots are, qprp and

Equation containing other roots is

0)( 22 rqpxqr px

p

1 is common root 011 2

r

ppq

p

)(12

r qp

Now 0)( 22 qr pxr qpx

0)(2

2

pqr xr qx

p

pp

0)()( 2 pqr xr qxr qp

24. Answer (1, 2, 3)

0)3(2 mxmx

For real distinct roots, 04)3( 2 mm

09102 mm

,9()1,( 0)1)(9( mmm … (1)

For positive roots,

Sum > 0, product > 0

m – 3 > 0 , m > 0 … (2)

From (1) and (2), m (9, )

For negative roots

sum < 0, product > 0

m – 3 < 0, m > 0 … (3)

From (1) and (3), )1,0(m





- 43 -

Q.No. Solution

25. Answer (1, 2, 3, 4)

012 22 aaxx

1)( 2 ax x = a + 1, a – 1.

Now 4]1[3 a

and 4]1[3 a

41][3 a and 41][3 a

3][4 a and 5][2 a

3][2 a [a] = –2, –1, 0, 1, 2, 3.

26. Answer (1, 2, 3)

4 , A

+ = 36, = B

Let , , , be a, ar , ar 2, ar

3

a + ar = 4

3632 ar ar 9

1

)1(

12

r r

r

92 r r = ± 3, a = 1.

A a(ar ) = A A = 3

B = B = (ar 2) (ar

3) B = 243 B = 81 A.

27. Answer (1, 2, 3)

4

5loglog

4

32

22 xx

x = 2 .

Taking log with base 2 on both side.

2

12loglog

4

5log)(log

4

3222

22

xxx

Put log2 x = t,2

1

4

5

4

3 2

ttt

02543 23 ttt 0)273)(1( 2 ttt

0)2)(13)(1( ttt

2,3

1,1 t 2,

3

1,1log

2

x x = 2, 23/1 2,2

28. Answer (3, 4)

a

c

a

bcbxax ,,02

a

cab 42






- 44 -

Q.No. Solution

A

C

A

BCBx Ax ))((,,02

A

CAB 42

Now, A

CAB

a

acb 44 22

2

2

2

2 44

A

CAB

a

acb

2

2

2

4

4

A

a

ACB

acb

)(2 2 A

B

A

B

A

B

a

b

2

1

29. Answer (1, 2, 3)

Let , be the roots of the corresponding equation

x2 + ax + a2 + 6a = 0 …(i)

As the coefficient of x2 = 1 > 0 x2 + ax + a2 + 6x < 0 will be satisfied for all values of x (, ) if , arereal and unequal (let < ).

1 2

Hence the inequality will hold for all real x (1, 2) if the interval (1, 2) is a subject of the interval (, ). Thusfor (1) we should have D > 0 and < 1, > 1 as well as < 2, > 2.

Now, D > 0 a2 – 4(a2 + 6a) > 0

a2 + 8a < 0

a (– 8, 0) …(ii)

< 1, > 1 – 1 < 0, – 1 > 0

( – 1) ( – 1) < 0

a2 + 7a + 1 < 0

–7 – 45 –7 45,

2 2a

…(iii)

< 2, > 2 ( – 2) ( – 2) < 0

a2 + 8a + 4 < 0

(–4 – 2 3, – 4 2 3)a …(iv)

Common values of a satisfying (ii), (iii) and (iv) are

–7 – 45, – 4 2 3

2a

…(v)

Hence answer is (1), (2), (3) those are subject of (v)





- 45 -

Q.No. Solution

30. Answer (2, 3)

The given equation is

1 1 1

x p x q r

1

( )( )

x q x p

x p x q r

(2x + p + q)r = x2 + (p + q)x + pq

x2 + (p + q – 2r )x + (pq – qr – rp) = 0

According to the question the given equation has roots equal in magnitude but opposite in sign, hence

Coefficient of x = 0

p + q – 2r = 0

r =2

p q

Product of roots

= + [(p + q)r – pq]

=2( ) –

2

p qpq

= 2 21( )

2p q

31. Answer (1, 2, 4)

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

Let , be roots

2 1

2

a

a (integer)

2

a

a (integer)

2a

a will integer if

For a = 0, a = –1, a = –3

Also for a = 0, a = –1, a = –3

–2 1

2

a

a

is integer

32. Answer (1, 3)

Since 1 is the repeated roots of ax3 + bx2 + c = 0

So, 1 + 1 + = – b

a

1.1 + + = 0 1

– 2






- 46 -

Q.No. Solution

1.1. =1 1

– – 2 2

c c

a a

3 –

2

b

a

– 3bc

Now, by the equation,

cx3 + bx + a = 0

3 0b a

x xc c

x3 – 3x + 2 = 0

x3 – x2 + x2 – x – 2x + 2 = 0

x2(x – 1) + x(x – 1) – 2(x – 1) = 0

(

x – 1) (

x2

+x – 2) = 0

(x – 1) (x2 + 2x – x – 2) = 0

(x – 1) (x – 1) (x + 2) = 0

x = + 1, – 2

Hence answer is (1), (3)

33. Answer (3, 4)

Let x2 = y

So the equation ay2 + by + c = 0 should have both roots non-negative in order to all roots of the equationax4 + bx2 + c = 0 are real for this

– 0 0b b

a a …(i)

0c

a …(ii)

From (i) and (ii)

b > 0, a < 0, c < 0

or b < 0, a > 0, c > 0

34. Answer (1, 2, 3)

– 5

( – )2 < 5

( + )2 – 4 < 5

k2 – 4 < 5

k2 < 9 k (– 3, 3)

Hence answers is (1, 2, 3)





- 47 -

Q.No. Solution

35. Answer (1, 2, 3)

We know that in a triangle sum of two sides of a triangle is greater than third side.

So, a2 + 2a + 2a + 3 > a2 + 3a + 8 4a > 3a + 5 a > 5

a2 + 2a + a2 + 3a + 8 > 2a + 3 2a2 + 3a + 5 > 0 a R

2a + 3 + a2 + 3a + 8 > a2 + 2a 3a > – 11 a >11

– 3

Combining these three,

a (5, )

Hence answer is (1, 2, 3)

36. Answer (1, 3, 4)

1 21

zz z

t t

we havez = (1 –

t)z

1 +tz

2, 0 <t < 1,

2 11

1

t z t zz

t t

2 11 z t z t z

z1, z, z2 are collinear

so Arg(z – z1) = Arg(z – z2) = Arg(z2 – z1)

37. Answer (3, 4)

Note that || = 1

i are possible value of z1

i are possible value of z2

(i = 1,2,3)

3

2 2

i

6

ie

2 3

ie

3 2

ie

24 3

ie

55 6

i

e

So, 1 2z oz can be2 5

,3 6

1

3

2

1

2

3

30°

30°30°30°







- 49 -

Q.No. Solution

Now let x = cos + isin, then equation (i) becomes,

cos + isin – cos + isin = i

2sin = 1 sin =1

2

5or 6 6

1. Answer (3)

We have to find out the value of x51

51

cos sin6 6

i

17 17cos sin

2 2i

= 0 + i = i

2. Answer (1)

We have to evaluate 2020

1x

x

20 20 20 20cos sin cos – sin

6 6 6 6i i

10 42cos 2cos

3 3

2cos3

– 2cos –13

3. Answer (2)

For finding 20132013

1x

x

2013 2013 2013 2013cos sin – cos sin

6 6 6 6

i i

20132 sin

6i

6712 sin

2i

2 sin2

i i






- 50 -

Q.No. Solution

Comprehension-III

1. Answer (2)

2. Answer (4)

3. Answer (2)

Solution of Q.1 to Q.3

Given expression is

(1 + x)n = a0 + a1x + a2x2 + a3x

3 + .............+ anxn …(i)

Putting x = ± 1, we get

(1 + 1)n = a0 + a1 + a2 + ...... + an

(1 – 1)n = a0 – a1 + a2 + ...... ± an

Adding these,

2n = 2(a0

+ a2

+ a4

+ ......)

a0 + a2 + a4 + ...... = 2n – 1 …(ii)

Hence, answer of question 1 is (2)

Again, putting x = ± i in (i), we get

(1 + i)n = a0 + a1i – a2 – a3i + a4 + a5i – a6 – a7i + a8 + ......

(1 – i)n = a0 – a1i – a2 + a3i + a4 – a5i – a6 + a7i + a8 + ......

Adding these,

2(a0 – a2 + a4 – a6 + .......)(1 ) (1– )

2

n ni i

/2

/22.2 · cos

4 2 cos2 4

n

n

nn

…(iii)

Hence, answer of question 2 is (4)Now, adding (ii) & (iii), we can get

2(a0 + a4 + a8 + .......) = 1 /22 2 cos4

n n n

a0 + a4 + a8 + a12 + ...... = –1

2 22 2 cos4

nn n

Hence, answer of question 3 is (2)

Comprehension-IV

1. Answer (1)

kxxxf 3)( 3

033)(' 2 xxf x = ±1

For exactly one positive root,

0)1( f and f (1) < 0

–1 + 3 + k < 0 and 1 – 3 + k < 0

2k and k < 2 )2,( k .

1 –1

y

xO





- 51 -

Q.No. Solution

2. Answer (1)

For exactly one negative root,

0)1( f , 0)1( f

–1 + 3 + k > 0, 1 – 3 + k > 0

k > – 2, k > 2

),2( k

3. Answer (2)

For one negative and two positive root

0)1( f , f (0) > 0, f (1) < 0

–1 + 3 + k > 0, k > 0, 1 – 3 + k < 0

2k , 0k , 2k

20 k i.e., )2,0(k .

Comprehension-V

1. Answer (3)

2 2 2| | (| | 2 1) 1x x k k

)12( 22 kxx = 21 k 01)12( 224 kxkx

All roots are imaginary, if 042 acbD

0)1(4)12( 22 kk

4

5k …. (1)

Also roots are imaginary if 0D , but 2x is negative, i.e. roots of 01))(12()( 2222 kxkx are both

negative.

Sum < 0, and product > 0

012 k and 012 k )1,( k

All roots are imaginary if

,

4

5)1,(k

2. Answer (2)

For exactly two real roots of

01)12( 22 ktkt

0D and one value of 2xt is positive and one is negative.

0)1(4)12( 22 kk

4

5k … (1)

Product = 012 k –1 < k < 1 … (2)

From (1) and (2) )1,1(k .

1 –1

y

xO

1

–1

y

xO






- 52 -

Q.No. Solution

3. Answer (1)

For repeated roots = 0,4

5k

or P = 0, 012 k , 1k

When product = 0, x = 0 is repeated root.

Comprehension-VI

1. Answer (2)

Let z = x + iy

Set A corresponds to the region y 1 ...(i)

Set B consists of points lying on the circle, centred at (2, 1) and radius 3,

i.e. x2 + y2 – 4x – 2y = 4 ...(ii)

Set C consists of points lying on the x + y = 2 ...(iii)

P

y = 1

x

y

(2, 1)

(0, 2)

( 2,0)

Clearly, there is only one point of intersection of the line 2 x y , and circle x2 + y2 – 4x – 2y = 4

2. Answer (2)2 2

1 – – 5 – z i z i

= (x + 1)2 + (y – 1)2 + (x – 5)2 + (y – 1)2

= 2(x2 + y2 – 4x – 2y) + 28

= 2(4) + 28 2 2 – 4 – 2 4 x y x y

= 36

3. Answer (4)

– 2 3 w i

| | – | 2 |w i < 3

3 5 3 5 w

– 3 – 5 – 3 – 5 w ...(i)

Also, – 2 3 z i

– 3 5 3 5 z ...(ii)

– 3 < |z| – |w| +3 < 9








- 54 -

Q.No. Solution

OR

Let z = x + iy

iyx

iyx

z

z

)1(

)1(

1

1

iyxiyx

iyxiyx

zz

)1()1(

)1()1(

11

22

22

)1(

)2()1(

1

1

yx

yiyx

z

z

21

1arg

z

z

0122 yx

122 yx and y > 0

Locus of z is semicircle.

3. Answer (1)

eiA.eiB.eiC = ei( A + B + C)

= ei = cos + isin

= –1

4. Answer (1)

(1)1/4 = (cos2r + isin2r )1/4

= cos sin

2 2

r r i

where r = 0, 1, 2, 3

11/4 = 1, i, – 1, – i

z12 + z2

2 + z32 + z4

2 = 1 + i2 + 1 + i2

= 2 – 1 – 1 = 0

5. Answer (4)

|z1| = |z2| = |z3| = |z4|

This may not be the case if centre of the circle is not origin.

6. Answer (4)

2 2 (1– )(1 )

1 2

n nni i i

ii

4( 2)

in

n e

Now clearly the least integral value for which the given number is a positive integer is 8.





- 55 -

Q.No. Solution

7. Answer (1)

B z( )2 A z( )1

C z( )3 D z( )4

1 2

3 2

– arg

2 –

z z ABC

z z

And AB = |z1 – z2| = BC|z3 – z2|

1 2

3 2

– 1

–

z z

z z

Hence, 1 2

3 2

– 1 cos – sin

– 2 2

z zi

z z

z3 – z2 = (z1 – z2) cos – sin2 2

i

= – i(z1 – z2)

= – iz1 + iz2

z3 = – iz1 + (1 + i)z2

z3 = – i(1 – i) + (1 + i)(1 + i) = – i + i2 + 1 + i2 + 2i

= i – 1

8. Answer (2)

Let z = r (cos + isin)

2 2 22 21 1 1

cos – sinz r r z r r

22

12cos2r

r

2 2 22

12 – 4sina r

r

22 21

4sinr a

r

2

214r a

r

214r a

r

2 2 – 4 1 0r a r






- 56 -

Q.No. Solution

r lies between22 44 –

and2 2

a aa a

This is true for all real a 0.

2 2

max min

4 4 –

| | , | |2 2

a a a a

z z

Hence, for a = 0

|z|min = |z|max

Hence both statement-1 and statement-2 are true. But statement-2 is not the correct explanation ofstatement-1.

9. Answer (1)

Given equation is (z – 2)n = zn

– 2

1n

z

z

1/ – 21 1nz

z

|z – 2| = |z|

Hence z is the locus of a straight line perpendicular bisector of the segment joining the points (2, 0) and (0,0), i.e., x = 1.

10. Answer (1)

02 cbxax

x = 1, a + b + c = 0

If sum of coefficient is 0 then 1 is the root of the equation.(210 – 3) – 211 + 210 + 3 = 0

Both are true and Statement-2 is correct explanation of Statement-1

11. Answer (1)

For reciprocal roots, replacing x byx

1 in 02 cbxax 0

2 c

x

b

x

a 02 abxcx

Statement-2 is correct and is correct explanation of Statement-1

0510 2 xx 05110

2

xx 0105 2 xx

12. Answer (3)The equation in first statement is x2 – 2009x + 2008 = 0 can be written as (x – 2008)(x – 1) = 0

x = 1, 2008 are roots of the equation where are rationals also.

Statement 1 is True.

Statement 2 is not always true. When D = b2 – 4ac = a perfect square than roots of the equationax2 + bx + c = 0 are rational only when a, b, c are rationals, otherwise roots are irrationals.

To this end, let us consider an equation 24 4 3 1 0x x





- 57 -

Q.No. Solution

whose discriminant = 48 + 16 = 64 = 82 = a perfect square but roots are

4 3 48 16

2 4x

3 2

2

which are not rationals.

Thus statement 2 is false.

Hence option (3) is correct.

13. Answer (1)

We observe that f (1)f (2) = (1 + 5 – 7)(4 + 10 – 7) < 0

Hence these exists a root of x2 + 5x – 7 = 0 in (1, 2).

Clearly option (1) is correct.

14. Answer (4)

We observe that

(a + c)2 > b2

(a + c)2 – b2 > 0

(a – b + c) (a + b + c) > 0

f (–1)f (1) > 0, where f (x) = ax2 + bx + c

f (x) =ax2 + bx + c = 0 has either no root in (–1, 1) or if real roots exist, then both roots lie in (–1, 1)

Statement 1 is not necessarily true

Hence statement 1 is false.

Answer is 4

15. Answer (4)

Statement-1 is wrong,

73

1

can be root of infinite equations with real coefficients, e.g. 07

3

1)1(

xx ,

...073

1)2(

xx

16. Answer (3)

0501200822 xx

2 2008 4 2008 4(501)

2x

2008 1507

Roots are rational.

Statement-2 is wrong as roots are rational only when coefficients are rational and acb 42 is perfect square.






- 58 -

Q.No. Solution

17. Answer (4)

a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca) = 0,

a + b + c = 0 a, b, c are not equal. The sign of other root depends on sign of ac.

Hence (4) is answer.

18. Answer (4)

( 2 – 1) is a root if coefficients are real then other root can be rational.

19. Answer (1)

20. Answer (2)

2 2 0 x px q

2 (i)(ii)

pq

2 2 0 ax bx c

1 2(iii)

(iv)

b

ac

a

2 2( ) ( ) p q b ac =

2

22

1

2 2

a

=

222( ) 1

. 016

a statement-1 is true

Now pa = ( )2 2

aa

b =1

2

a

211, { 1,0,1}, correctpa b

Similarly

If

c qa a a

1

0

1

0, and 0 { 1, 0,1}

Statement 2 Is true.

Both statement 1 and statement 2 are true, But statement 2 do not explains statement 1.





- 59 -

Section-E

Q.No. Solution

1. Answer A(s), B(r), C(p), D(q)

(A)

2007

19

2

cos9

2

sink

k

i

k

=

2007

19

2

sin9

2

cos)(k

k

i

k

i =

2007

1

9

2

)(k

ki

ei

=

9

)2007(2

9

4

9

2

.....

iii

eeei

Which is G.P.

=

1

)11(

1

1

19

2

9

2

9

2

9

)2007(2

9

2

i

i

i

i

i

e

ei

e

e

ei = 0

(B) |z1| = 1, |z2| = 1, |z3| = 1

1,1,1 332211 zzzzzz

33

22

11

1,

1,

1z

zz

zz

z

Now, ||111

321321

zzzzzz

= 1|| 321 zzz

(C) |z1| = |z2| = |z3| and z1, z2, z3 are vertices of equilateral triangle

Origin is its centroid

z1 + z2 + z3 = 0

Now, |z1 + z2 + z3| –1 = –1

(D) Let 4325

1

,,,,1)1(

1 + + 2 + 3 + 4 = 0

1 + + 2 + 3 = – 4

and 5 = 1

41

and || = 1

Now 4log4 |– 4 – 4| = |2|log4 44

= |||2|log4 44

= 22log2

142log4 24






- 60 -

Q.No. Solution

2. Answer A(s), B(r), C(p), D(q)

(A) (1 + i)n = (1 – i)

n

1

1

1

n

i

i

1ni

n = 4

(B) (a + ib) = (x + iy)3 = x3 – iy

3 + 3ix2y – 3xy

2

a = x(x2 – 3y2) and b = – y3 + 3x

2y

22 3yxx

a 22 3xy

y

b

)(2 22 yxx

a

y

b

k = 2

(C)2

1 ix

2

)1( 22 i

x

= ii

2

211

200820062004108642 ...1 xxxxxxxx

111

11

1

1

1

]1[12

2

2

2010

i

i

x

x

3. Answer A(s), B(r), C(p, q), D(q)

(A) |||| 21 zzzz = constant = k,

where || 21 zzk , represents an ellipse.

(B) ,|||| 21 kzzzz where || 21 zzk is a hyperbola having foci at z1 and z2.

(C)2

arg2

1

zz

zz

This represents a circle with z1 and z2 as the vertices ofdiameter.

(D) If lies on || = 1, then

2007||

20072007

2007 lies on the circle.

p z( )

z1 z2

p

/2





- 61 -

Q.No. Solution

4. Answer A(r, s), B(p, q, r, s, t), C(p, q), D(q, r, s, t)

(A) |z – 6i | + |z – 8| = k will represent ellipse if 2 26 8 10k k

(B) |z – 12i + 3| – |z – 2|| will represent ellipse if 2 25 12 13k k

(C) |z – ki| + |z – 4| = 10k will represent line segment if k2 + 16 = 10k k = 2, 8

(D) To represent circle

k 1 and also k = 2

5. Answer A(r, s), B(r), C(p, q, s, t), D(p, q)

(A) To satisfy all at a time z should lie on the circle |z| = 3.

Inside the circle |z – {(1 + i) – i}| = 3 and outside the circle |z + 2t – (t + 1)i| = 3

For this,

2 2 2 2( – 0) ( – 1– 0) 3 3 and 4 (4 1) 3 3t t t

2t2 – 2t – 35 0 and 5t2 + 2t – 35 > 0

Using sign scheme we have,

1 – 712

–1 – 4 115

–1 + 4 115

1 + 712

Hence,

71 –1– 4 11 –1 4 11 1 711– 1– , ,

2 5 5 2

Hence, 3, 4 lies in above interval.

(B) We have to solve for x, y

(1 ) – 2 (2 – 3 )

3 3 –

i x i i y ii

i i

2

(1 )(3 – ) – 2 (3 – ) (2 – 3 )(3 ) (3 1)

9 –

i i x i i i i y ii

i

(4 + 2i)x – 6i – 2 + (9 – 7 i)y + 3i – 1 = 10i

(4x + 9y – 3) + i(2x – 7y – 3) = 0 + 10 i

Comparing the real and imaginary parts

4x + 9y – 3 = 0 …(i)2x – 7y – 13 = 0 …(ii)

(i) – 2 (ii) gives,

9y + 14y – 3 + 26 = 0

23y = – 23 y = – 1

Putting y = – 1 in (i), we get

4x – 9 – 3 = 0 x = 3






- 62 -

Q.No. Solution

(C)21 1 2

1– 1 1

n

ni i ii

i

Hence n = 0, 4, 8, 6

(D) Greatest and least absolute values of z + 1 are 1 and 6.

|z + 4| 3

6. Answer A(r, t), B(s), C(q, r, s, t), D(p)

x2 – (k – 3)x + k = 0

For roots to be real

(k – 3)2 – 4k 0

k 2 – 6k + 9 – 4k 0 k

2 – 10k + 9 0 (k – 1) (k – 9) 0

k (– , 1] [9, ] …(i)

(A) For both roots to be positive,

f (0) > 0 and3

02

k

k > 0 …(ii)

and k > 3 …(iii)

From (i), (ii) and (iii)

k [ 9, ]

(B) For both roots to be negative

D 0

k > 0, (k – 3) / 2 < 0, k < 3, k (0, 1]

(C) For both roots to be real k (– , 1] [9, )

(D) f (–1) < 0, f (1) < 01 + (k – 3) + k < 0 also 1 – (k – 3) + k < 0

2k – 2 < 0 k < 1, 4 < 0

No such value is possible

7. Answer A(q), B(r), C(s), D(p)

(A) 0)( xf , Rx , f (0) = c > 0

Parabola is upward a > 0.

(B) Roots are real and distinct 0D

f (0) < 0 c < 0

One root is positive and one negative ab < 0.(C) Roots are imaginary 0D

Parabola is downward a < 0.

(D) Parabola touches x-axis 0D

Parabola is downward a < 0

Both roots are positive, sum = 0a

b b > 0





- 63 -

Q.No. Solution

8. Answer A(s), B(p), C(r), D(p)

(A) For real roots, 0D

0)3(4)2( 22 aaa

0)3[422

aaa 3a …. (i)

f (3) > 0 9 – 6a + a2 + a – 3 > 0

0652 aa

),3()2,( a … (ii)

32

a

b 3

2

2

a a < 3 … (iii)

from (i), (ii) and (iii) )2,(a

(B) 0D 3a … (i)

f (3) > 0 ),3()2,( a …(ii)

For greater than 3,

32

a

b a > 3 … (iii)

From (i), (ii) and (iii), a .

(C) D > 0 a < 3 … (i)

f (1) f (3) < 0

0)369)(321( 22 aaaaaa

0)65)(2( 22 aaaa 0)3)(1()2( 2 aaa

}2{)3,1( a … (ii)

From (i) and (ii)

)3,2()2,1( a

(D) 0D 3a … (i)

f (1) < 0 0)321( 2 aaa

0)2( 2 aa

–1 < a < 2 … (ii)

f (3) < 0 0)369( 2 aaa

0652 aa

2 < a < 3 … (iii)

From (i), (ii) and (iii)

a .






- 64 -

Q.No. Solution

9. Answer A(p), B(q), C(r, s), D(t)

We have

f (x) = |x – 1| + |x – 2| + |x – 3| = k

when x < 1, then

f (x) = –3x + 6 = k

23

kx

Now, 2 13

k ,

k > 3 ...(i)

For 1 x < 2

– x = k – 4

x = 4 – k

2 < k 3 ...(ii)

For 2 x < 3

x = k, 2 k < 3

For x 3

3x – 6 = k

6

3

kx

6

33

k

k 3Clearly

(A) for k < 2, there is no solution

(B) for k = 2, there is only one solution

(C) for 2 < k < 6, there are two solution of same sign

(D) for k > 6, there are two solution of opposite sign

10. Answer A(q), B(r), C(s), D(p)

(A) (q), (B) (r)

02 baxx

a , b

ba 44)(1|| 22

142 ba

ba 412

ba 4212 = 2(1 + 2b)

2

3

321





- 65 -

Q.No. Solution

(C) a 2 3

a

b.2 2

2 b

23

2ba

2

92 ba ba 92

2

.

(D)22

11

2)())(( 22222

baba 2)( 22

)2(222 abbbaba

11. Answer A(p, q, s, t), B(p), C(p, s, t), D(r)

, , be the roots of the equation

x(1 + x2) + x2(6 + x) + 2 = 0 x3 + x + 6x2 + x3 + 2 = 0

2x3 + 6x2 + x + 2 = 0

So, + + = – 3

+ + =1

2

= –1

Now,

(A) –1 –1 –1 1 1 1 1 –

2

(B) 2 + 2 + 2 = ( + + )2 – 2( + + )

19 – 2· 8

2

(C) ( –1 + –1 + –1) – ( + + )

– ( )

1 5 – 3

2 2

(D) [ –1 + –1 + –1] =1

– –12

12. Answer A(q, r), B(p), C(p, s, t), D(q, r, s, t)

(A) z is equidistant from the points i| z | and – i| z |, whose perpendicular bisector is Im (z) = 0.

(B) Sum of distance of z from (4, 0) and (–4, 0) is a constant 10, hence locus of z is ellipse with semi-major axis 5 and focus at (±4, 0).

ae = 4

e =4

5






- 66 -

Q.No. Solution

(C)1 5

| | | | 32

z ww

(D)1

| | | | 2 z ww

Re z | z | 2

Section-F

Q.No. Solution

1. Answer (3)

As, arg of P(z) is /4

z = x + iy, x, y, > 0

(0, 2)

(0, 1)P x iy( + )

x

y

For sum of distance to be minimum P will lie on perpendicular bisector of (0, 2) and (0, 1) hence

x = y = 2/3

k = 2

2. Answer (4)

2002 2 1

1

2 2cos sin 0

7 7

k

r

r r i

It is possible only when 2002 + (2k – 1) should be multiple of 7.

3. Answer (5)

We have

33 30z z z z

2 2 30z z z z

(x2 + y2) ((x2 – y

2) – 2i xy + x2 – y

2 + 2i xy)) = 0

(x2 + y2) (x2 – y

2) = 15 = (22 + 1) (22 – 1)

Which suggests the possible values of x and y are

x = 2, y = 1

or

x = –2, y = –1





- 67 -

Q.No. Solution

or

x = –2, y = 1

or

x = 2, y = –1

Centre of rectangle is (0, 0)

Now,

(0, 0)

| –3| = 2z

(3, 0)

Maximum distance of (0, 0) from the circle is 5 and minimum is 1.

4. Answer (3)

If origin z1, z2 forms an isosceles triangle then z12 + z2

2 + z1z2 = 0

Hence2 21 2 1 2 1 2

1 2 1 2

4 33

z z z z z z

z z z z

5. Answer (2)

Let 3 + i = z, hence other two vertices are iz and z + iz

So, area of such triangle is 21| |

2z

21 1·(2) 4 2

2 2

6. Answer (0)x

2 – 6kx + 9(k2 – k + 1) = 0

For real and distinct roots

D > 0

k2 – (k2 – k + 1) > 0

k – 1 > 0 …(i)

f (3) 0

9 – 18k + 9(k2 – k + 1) 0

1 – 2k + k2 – k + 1 0

k2 – 3k + 2 0

k (– , 1] [2, ) ...(ii)

Alsosum of roots

32

3k < 3

k < 1 …(iii)

From (i), (ii), (iii)

We observe that there does not exist any real value of k.






- 68 -

Q.No. Solution

7. Answer (2)

We have

2

2

2

2

2 2

2

3 4

2 2

3 4 As 52 2

3 4 5 10 100

2 2

ax xy

x x

ax x

x x

ax x x x

x x

(a – 5)x2 – 7x – 6 < 0 (as x

2 + 2x + 2 > 0, x R)

It is satisfied for all x if

a – 5 < 0, 49 + 24 (a – 5) < 0

71

24a

a < 3

The possible greatest integral value of a is 2.

8. Answer (5)

Firstly, let f (x) = ax2 + bx + c; a, b, c R be an integer whenever x is an integer.

f (0), f (1), f (–1) are integers

c, a + b + c, a – b + c are integers

c, a + b + c – c, a – b + c – c are integers

c, a + b, a – b are integers

c, a + b, a + b + a – b are integers

c, a + b, 2a are integers

Secondly let 2a, a + b and c be integers. Let x be an integer.

Then f (x) = ax2 + bx + c =( 1)

2 ( )2

x xa a b x c

.

Since x is an integer x(x – 1) is an even integer.

( 1)

2 ( )2

x xa a b x c

is an integer as 2a, a + b, c are integers.

f (x) is an integer for all integer x.

9. Answer (2)

x2 – 8kx + 16 (x2 – k + 1) = 0

Roots are real and distinct

Let f (x) = x2 – 8kx + 16(k2 – k + 1)

D > 0 x > 1

Let f (4) > 0 k (– , 1] [2, )

Least value of k can be 2.





- 69 -

Q.No. Solution

10. Answer (5)

From the given condition,

| z – 3 – 2i | 2

| 2z – 6 – 4i | 4

4 | (2z – 6 + 5i) – 9i |

|| 9i | – | 2z – 6 + 5i ||

4 9 – | 2z – 6 + 5i |

| 2z – 6 + 5i | 5

Minimum value of | 2z – 6 + 5i | is = 5

11. Answer (4)

As,1

43 2

y y

4

9y (as y > 0)

so, 3

2

46 log 4

9

Section-G

Q.No. Solution

1. Answer (1)

Given,

S1 = {z : Im (z) > 1}

S2 = {z : |z – 2 – i| = 3}

S3 = {z : Re ((1 – i)2) = 2

S1 : y 1

S2 : |z – 2 – i| = 3}

S3 : x + y = 2

Clearly ‘p’ is only point satisfying all threecondition.

Now, –1 + i, and 5 + i are end points of a

diameter

PA2 + PB

2 = (6)2 = 36

Also

|z – 2 – i| 2 – 2 i

– 2 3 z i

– 3 |z| – |2 + i| 3 …… (i)

p

–1+i 2 + i5 + i

y s,

B A

x y+ = 2






- 70 -

Q.No. Solution

Similarly –3 < |z1|– |2 + i| < 3 …… (ii)

(i) _____ (ii)

–6 < |z| – |z1| < 6

–3 < |z| – |z1| + 3 < 9

2. Answer (1)

Statement-1 : f (x) + f (x) + f (2x) + ...... + f (6x)

206

0

1

7 (1 ...... )k k kk

k

A A x

but when k 7

and k 14, then 1 + k + 2k + ...... + 6k = 0

f (x) + f (x) + ....... + f (6x) = 7( A0 + A7x7 + A14x

14)

Statement-2 :

B 8 A

O

Clearly AB = 8

OB = 16

8 1sin

16 2 6

AB

OB

So, the argument is3

Statement-3 :

It can be seen from the figure that the maximum value of argument is2

3

3. Answer (2)

Statement-1 : The number of common vertices is equal to the number of common roots of z1982 – 1 = 0 andz2973 – 1 = 0, which is H.C.F. of 1982, 2973, i.e., 991. Here consider both vertex has (1, 0) as one vertex.

Statement-2 : Since two parallel lines never meet so no solution

Statement-3 : Clearly the locus is a straight line.

4. Answer (1)

Let2 3

2

x xy

x

so that x

2 – (y + 1)x + 3 – 2y = 0

For real values of x

(y + 1)2 – 4(3 – 2y) 0

(y + 11) (y – 1) 0





- 71 -

y

E (Real axis)

4 5 °

z0

z

O

N-W N (Imaginary axis)

5

4

3

32

32

+i

Q.No. Solution

y 1 or y – 11

y lies in R – (–11, 1)

Statement-1 is true

The roots of x2 – 4|x| + 3 = 0 are

1, 3, –1, –3 and hence their sum is zero

Statement-3 is also true

5. Answer (4)

Clearly statement 1 and 2 are false

We have (x – 1)2 = cos2

x = 1 ± cos

x [0, 2]

statement-3 is true

Section-H

Q.No. Solution

1. x3 – 3x

2 + 3x + 7 = 0

(x – 1)3 + 8 = 0

(x – 1) = –2, –2, –22

x = –1, 1 – 2, 1 – 22

Now,1

1

1

1

1

1

22

22

22

2

2

211

= |32| = 3

2.0 3z

5z

By Coni method

0 0

0

0

izze

z z

3 3 5(cos sin )

32 2z i i

1 1 5 3 4

33 5 52 2

z i i

4(3 4 )i

i e






- 72 -

Q.No. Solution

3.1

1

2

2

1 z

z

z

z

z12 + z2

2 = z1z2

2121

22

2

2

1000 zzzzzz

21,,0 zz form an equilateral triangle

Hence, 3|||||| 2121 zzzz

3|| 221 zz

4. |z – z1| = 5

|z – z2| = 2

|z – z1| = r + 5

|z –

z2| =

r + 2

|z – z1| – |z – z2| = 3 which is a constant

locus is hyperbola and we know

PF1 – PF2 = 2a = length of transverse axis.

length of transverse axis = 3

5. |z| = 1 or |z|2 = 1

11zz z

z

Now, 2

1

11

z z

z zzz z

z

if z x iy then z x iy

2z z ixy

21

z

z is purely imaginary therefore

21

z

z always lies on y –axis.

6. |z1 + z2|2 = |z1|

2 + |z2|2 ………. (Given)

|z1 + z2|2 = |z1|

2 + |z2|2 + 2 Re )(

21 zz = |z1|

2 + |z2|2 ………. (Relation)

0)(Re 21 zz

2z

zamp

2

1

32

6

z

zamp

6

2

1

5 2z2z1

r r

z





- 73 -

Q.No. Solution

7. Let z = x + iy

So, then equation becomes,

2 22 – 4 ( ) 1 0x y a x iy ia …(i)

Comparing real and imaginary part,

2 22 – 4 1 0x y ax …(ii)

–4ay + a = 0 …(iii)

(3) a(– 4y + 1) = 0 a = 0 or1

4y

But a cannot be zero. Hence1

4y gives

2 12 4 – 1

16x ax

2 214 (4 – 1)16x ax

2 2 214 16 1– 8

4x a x ax

(4 – 16a2)x2 + 8ax –3

4 = 0

2

2

–8 16 12

8(1– 4 )

a a

a

2

2

4 2 4 3

4(4 – 1)

a a

a

Now, we can observe that

if 0 a 1

2 , there is no solution

if1

2a , solution is z = x + iy

2

2

4 4 3 1·44(4 – 1)

a ai

a

8. We have,

|z1 + 1| + |z2 + 1| + |z1z2 + 1| |z1 + 1| + |z1z2 + 1 – (z2 + 1)|

= |z1 + 1| + |z1z2 – z2|

|z1 + 1| + |z2||z1 – 1|

= |z1 + 1| + |z1 – 1|

|z1 + 1 + z1 – 1| = 2|z1| = 2

|z1 + 1| + |z2 + 2| + |z1z2 + 1| 2.






- 74 -

Q.No. Solution

9. Let and n be the roots of the given quadratic equation ax2 + bx + c = 0

+ n = b

a and . n =

c

a = n + 1

1

1nc

a

11 1

nn nc c b

a a a

1 1

1 11 1 1 1 0

n n

n n n nc a c a b

1 1

1 1 1 1 0n n

n n n nc a c a b

1 1

1 1( ) ( ) 0n nn na c ac b

10. We have + = p and = q.

Now pVn – qVn –1 = ( + )(n + n) – (n –1 + n –1)

= n+1 + n + n + n+1 – n – n

= n+1 + n+1 = Vn+1

Also V5 = 5 + 5 = pV4 – qV3 = p[pV3 – qV2] – qV3

= (p2 – q)(3 + 3) – pqV2

= (p2 – q)[( + )3 – 3( + )] – pq[( + )2 – 2]

= (p2 – q)[p3 – 3pq] – pq[p2 – 2q]

11. According to the question

p + q = 10r ; pq = –11s

r + s = 10p ; rs = –11q.

On subtraction, (p – r ) + (q – s) = 10(r – p)

(q – s) = 11(r – p) ....(i)

Also p is a root of x2 – 10rx – 11s = 0

p2 – 10pr – 11s = 0

Similarly r 2 – 10pr – 11q = 0

On subtraction,

p2 – r 2 = –11 (q – s)

(p – r ) (p + r ) = –11 × 11 (r – p)

p + r = 121

Now p + q + r + s = 10 (p + r ) = 10 × 121 = 1210.





- 75 -

Q.No. Solution

12. The given equation can be written as

22x + 3 + 32x + 1 = 10 × 2x × 3x.

23 3

8 3. 10.2 2

x x

23 3

3 10 8 02 2

x x

23 3 3

3. 6 4 8 02 2 2

x x x

3 3 33. 2 4 2 0

2 2 2

x x x

3 32 3. 4 0

2 2

x x

Either

32

2

x or

3 4

2 3

x

When

32

2

x

, Taking logarithm of both sides we get x(log3 – log2) = log2

log2

log3 log2x

Also when

3 4

2 3

x

x(log3 – log2) = log4 – log3

log4 log3log3 log2

x

log2log3 log2

x or

log4 log3log3 log2

13. The given equation is 2x2 + (2 – )x + 3 = 0

and are roots of the equation, so

2

( 2) 2

2

3

But

4

3

2 2 24 ( ) 2

3

2

2

2

42

3 3







- 77 -

Q.No. Solution

17. If b = 0 result can be shown easily.

But if b 0.

Let n = 2(2m + 1)

Let the complex number such that 2 a

b

and the polynomial.

f (x) = xn – 1 = (x – 0)(x – 1)(x – 2)....... (x – n – 1)

We have,

1 n

f i i

. ( – i0)( – i1) ........ ( – in – 1)

and1

– n

f i i

. ( + i0)( + i1) ........ ( + in – 1)

Hence,

– f f i i

= (2 + 02)(2 – 1

2) ........ (2 – 2n – 1)

Therefore

– 1 – 12 2

0 0

( )n n

nk k

k k

aa b b

b

– 1

2

0

nn

k

k

b

22 2 1· – · ( ) 1 n n mb f f b

i i

22 1

1

mn a

bb

22 1 2 12(2 1)

2 1

m mm

m

a bb

b

= (an/2 + bn/2) hence proved

18. Let zk = costk + isintk : k{1, 2, 3}

The condition 2(z1 + z2 + z3) – 3z1z2z3R

2(sint1 + sint2 + sint3) = 3sin(t1 + t2 + t3) …(i)

Assume by the way contradiction that max(t1, t2, t3) <

6

,

Hence t1, t2, t3 < 6

Let 1 2 3 0,3 6

t t tt

1 2 31 2 3

1(sin sin sin ) sin

3 3

t t tt t t

…(ii)






- 78 -

Q.No. Solution

From (i) & (ii) we have,

1 2 3 1 2 3sin( )sin

2 3

t t t t t t

Then, sin3t 2sint

4sin3t – sint 0

i.e., 2 1sin

4t , hence sint

1

2 then

2t

, which contradicts that 0 ,

6t

max(t1, t2, t3) 6

Hence proved.

19. Let s1 = z1 + z2 + z3, s2 = z1z2 + z2z3 + z3z1, s3 = z1z2z3 and take a cubic equation.

z3 – s1z2 + s2z – s3 = 0 with roots z1, z2, z3

It is given that z12

+ z22

+ z32

= 0

Hence, s12 = 2s2 …(i)

Again we have,

2 31 2 3

1 1 1s s

z z z

3 1 2 3 3 1( ) ·s z z z s s …(ii)

Now, from (i) and (ii)

2 21 3 1 1 3 12 · and | | 2( )( )s s s s s s

= 2|s1| |s1| = 2 s1 = 2 with || = 1

Now, again from relation (i) and (ii) it follows that

22 2 2

2 1 31

1 22 and

2 2

ss s s

s

Now, the given equation becomes

z3 – 2z2 + 22z – 3 = 0

(2 – )(z2 – z + 2) = 0

The roots are

z = , = – 2

Now,

Rn = |z1n + z2

n + z3n| = |n + nn + (– 1)nn2n|

= ||n |1 + n + (– 1)n2n|

R0 = 3, R1 = |1 + – 2|

= |– 22| = 2





- 79 -

Q.No. Solution

R2 = |1 + 2 + | = 0

R3 = |1 + 1 – 1| = 1

R4 = 0

R5 = 2

R6 = 3

Rk + 6 = Rk for all integers k.

Rn {0, 1, 2, 3}

20. |z – |z + 1|| = |z + |z – 1||

|z – |z + 1||2 = |z + |z – 1||2

– | 1| · – | 1| | – 1| · | – 1|z z z z z z z z

2 2 – | 1|( ) | 1| · ( ) | – 1| | – 1|zz z z z z z z z z z z

2 2

| 1| – | – 1| ( )(| 1| | – 1|)z z z z z z ( 1)( 1) – ( – 1)( – 1) ( )(| 1| | – 1|)z z z z z z z z

1– – 1 ( )(| 1| | – 1|)zz z z zz z z z z z z

2( ) – ( ) (| 1| | – 1|) 0z z z z z z

( ) (2 – (| 1| | – 1|) 0z z z z

i.e., 0z z or |z + 1| + |z – 1| = 2

2 = |(z + 1) – (z – 1)| |z + 1| + |z – 1|

Solution of the equation |z + 1| + |z – 1| = 2 satisfy z + 1 = t(1 – z) where t R, t 0

– 1

1

tz

t

, so, z is any real number with – 1 z 1

The equation 0z z has the solutions z = bi, b R.

Hence the solutions to the equation are {bi : b R} {a R : a [– bi]}

21.In order to prove the result it will be sufficient to prove that

(1)

(–1)

pR

p .

Let x1, x2, x3,........, x2n be roots of p. Then

p(x) = (x – x1)(x – x2)(x – x3) ........ (x – x2n)

For some C, and

1 2 2

1 2

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

n

n

x x xpp x x

2

1

1– 1

n

k

kk

xx

It is given that |xk| = 1 for all k = 1, 2, ......., 2n. Then

11–

1– 1– –1 1– –

11 1 1 11

k k k k k

k k k k

k

x x x x x

x x x x

x










- 82 -

97

5

4 97

972

1

42

Q.No. Solution

27. Answer (3)

4 97 x + 4 x = 5

2 2

mm ma b a b if m (0, 1)

Put a = 97 – x, b = x, m = w/f

44 97

2

x x

1

497

2

44 97 – x x 2 × 2.63

Which holds for

44 97 – 5 x x

Now,

Clearly for 44 97 – 5x x has two solution.

28. Answer (2)

f (x) polynomial of degree ‘n’

= anxn + an –1 xn –1 + …… + ar

when this polynomial will be divided by (x – a) (x – b), remainder will be of form Ax + B

Now, f (a) = A.a + B

f (b) = A.b + B

( ) ( )f a f n

a b

= A .,…(i)

Also B = f a f b

f a aa b

( ) ( )

af a bf a af a af b

a b

af b bf a

a b

Remainder will be

f a f b af b bf ax

a b a b




11

Q.No. Solution

29. Answer (4)

Let 6

2

, 1

i j

i J

p x x

If we consider any three of x1, x2, x3, x4, x5, x6 is one and another three ‘0’

p = 9

30. Answer (1)

x + y = 4 – z

x2 + y

2 = 6 – z2

2xy = (x + y)2 – (x2 + y2)

= (4 – z)2 – (6 – z2)

= 2z2 – 8z + 10

The quadratic equation whose roots are x and y is

t2 – (x + y) t + xy = 0

t2 – (4 – z)t + z

2 – 4z + 5 = 0

for real roots,

(4 – z)2 – 4(z2 – 4z + 5) 0

3z2 – 8z + 4 0 (3z – 2) (z – 2) 0

2

23

z

Similarly x, y 2

,23

31. Answer (2)

px3 + qx

2 + qx + r = 0

+ 2 =q

p, 2 + 1 =

q

p, =

r

p

3 + 3 = 0 1

q

p r . p > 0

= –1 q.p < 0

pq < 0 , pr > 0

pqr > 0 or pqr < 0

pq < 0 is suitable option

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Complex Numbers (Package Solutions)