Cls Jeead-17-18 Xi Mat Target-2 Set-2 Chapter-5

Chapter 5

Complex Numbers and Quadratic Equations Solutions

SECTION - A

Objective Type Questions (One option is correct) 1.

If

z 

1  i 2

, then the value of z 1929 is

(1) 1 + i

(2) –1

(3)

1  i 2

(4)) (4

1  i 2

Sol. Answer (4) z 

⇒

1  i 2

z

2



(1  i )2

 2

2

1  i 2  2i 2 i    i 2 2

Now, z 1929 = (z (z 2)964·z =i

964·z

= z =

2.

If

z 

1  i 2

 1  i  8  1 – i  , then z equals  

(1) 1

(2) –1

(3) i

(4) 0


1 i 1– i



1  i 1  i � 1 – i 1  i

(1  i )2 1  i 2  2i  = 1 – i 2 2 = i z 8 = (i )8 = (i 2)4 =1 Aakash Educational Services Pvt. Ltd.

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94

3.

Complex Numbers and Quadratic Equations

If the the mult multipl iplica icativ tive e inver inverse se of of a comp complex lex num number ber is

2 – 5i

(1)) (1

(2)) (2

17

2

 5i

29

Solution of Assignment (Set-2) 2

 5i

17

(3)) (3

, then the complex number is

17

 27



2 – 5 i

(4)) (4

17 27



2



 5i

Sol. Answer (3) Let z be be the complex number, then



2

z �

⇒

 1

 5i

17

17

z 

2

17



17

4.

2 – 5i

�

 5i

2





 5i



2 – 5i

2 – 5 i



2

2 – 25 25i 17 27



2 – 5i



The Th e add addit itiv ive e inv inver erse se of 5 + 7 i is is (1) 5 – 7i

(2) –5 +7i

(3) 5 + 7i

(4) –5 –7i

(3) Third quadrant

(4) Fourth quadrant

Sol. Answer (4) Additive inverse of 5 + 7i 7 i is is –5 –7i –7i

5.

The co complex nu number (1) First quadrant

1  2i 1 – i

lies in (2) Second quadrant

Sol. Answer (2) 1  2i 1– i



1  2i 1  i � 1 – i 1  i



(1  2i )(1  i ) 1 – i 2



1 – 2  i (2  1) 2





–1  3i 2

–

 1  – ,  2

1 2



3 2

i

3

 lies in the second quadrant.

2

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Solution of Assignment (Set-2) 3

6.


95

3

 1  i   1 – i  If   –   a  ib, then values of a and b respectively are  1 – i   1  i  (1) 0 and 2

(2) 0 and –2

(3) 2 and 0

(4) 2 and 2

(3) 6

(4) 8

Sol. Answer (2)

1  i (1  i )� )�(1  i )  1 – i (1 – i )(1  i ) (1 i )2  1 – i 2



1 i

2

 2i

2

 i 1 – i (1 – i )�(1 – i )  1 i (1  i )(1 – i )

and





(1– i )2 1 – i 2 1 i

2

– 2i

2

 – i 3

3

 1 i   1 – i  –    ( i )3 – (– i )3  a  ib   1– i   1  i  – i = = a + ib  –i –

 0 – 2i = a + ib  a = 0 and b = –2 7.

If

x

3i , then value of 2 x 4 + 5 x 3 + 7 x 2 – – x x + +

 –2 –

(1) 1

(2) 3

41 is

Sol. Answer (3) x





–2 –

x

2

3i – 3i

 ( x x + + 2)2 =

 – 3i 

2

 x 2 + 4 + 4 x = = 3i 3i 2 + 7 = 0  x 2 + 4 x + Now, 2 x 4 + 5 x 3 + 7 x 2 – – x x + + 41 = 2 x 2( x x 2 + 4 x + + 7) –3 x ( x x 2 + 4 x + + 7) + 5( x x 2 + 4 x + + 7) + 6 =0–0+0+6 =6 Aakash Educational Services Pvt. Ltd.


96


Solution of Assignment (Set-2)

9

8.

1   17  i  i 315  is equal to  

(1) 32 i

(2) –512

(4) 512i

(3) 512

Sol. Answer (4)

 17 i 

9

1   2 8   � i i   = 2 57  315  (i )1 57 �i  i   1

 1 = i –   i   =  i – 

9

9

 2  i  i

9

= [i + i ]9 = (2i (2i )9 = 512( 512(i i 2)4·i = 512i 512i

9.

1  i  2  i

2

(1)) (1

5

3

is equal to

–

6 5

i

(2) 0

(3)

–

1 5



6 5

i

(4)) (4

–

2 5



6 5

i

Sol. Answer (4) (1  i )3 1  i  3i  3 2  2i ( 2  2i ) (2  i ) 2  6 i  2  6         i    5  5 2i 2i 2  i 4 1 5

10.

3   2  3i   2  1 – i  1  i   4  5i  is equal to   

(1)) (1

–

117

–

82

13 82

i

(2)) (2

–

117 82



13 82

i

117

(3)) (3

82

–

13i 82

117

(4)) (4

82



13i 82

Sol. Answer (3)

 2 (1  i )  3 (1  i )  2  3 i   5  i   2  3 i  1 17 13  (1  i ) (1  i )   4  5i    2   4  5 i   8 2  82 i 11.

In the Argan Argand d plane, plane, the conjug conjugate ate of the the complex complex numb number er 3 – 7i 7 i will will lie in (1) First quadrant

(2) Second quadrant

(3) Third quadrant

(4) Fourth quadrant

Sol. Answer (1)

(3  7i )  3  7i (1st quadrant) Aakash Educational Services Pvt. Ltd.



12. Th The e con conju juga gate te of of –13

(1)) (1

10



9 10

(1  2i )2 3 – i


97

is –13

i

(2)) (2

–

10

9 10

i

13

(3)) (3

10



9 10

i

13

(4)) (4

10

–

9 10

i

Sol. Answer (2) 1  4  4i 3  i

 3  4i   3  i    3  i   3  i 

= 

=

=

9  4  12 12i  3i 10

13  9i 10

 13  9   i   10 10

= 

Conjugate is

13 10

 9  10 

 i 

13. If z 1 = 1 + i and and z 2 = – 3 + 2i 2i , then (1) 2

(2) –3

 z1z 2   is  z 1 

Im 

(3) 3

(4) –2

Sol. Answer (2) z z 1 2 z 1

=

(1  i ) ( 3  2i ) (1 i )

=

3  3i  2i  2 1  i

=

5  i (1  i )  1 i (1  i )

=

( 5  1)  i ( 1  5) 2

=

4  6i 2

= – 2 – 3i 3 i In

 z1z 2   z    3 1



98




14.. The mul 14 multip tiplic licati ative ve inv invers erse e of

(1)) (1

1

–

3 5

49

98

i

3

(2)) (2

5i

1





49

2


is

3 5 98

i

(3)) (3

4  6 5i

(4)) (4

4 – 6 5i

(4)) (4

–

Sol. Answer (1)

Multiplicative inverse =

1 (3  i 5 )2 1

=

9  5  i 6 5

1

=

=

=

4  i6 5

4  i 6 5

4  i 6 5 196

1 99

15. If z = = 3 + i + + 9i 9i 2 – 6i 6i 3, then

(1) 2 + i

4  i 6 5





i 3 5

98

  is –1

z

(2)) (2

3

–

79



4 79

i

(3) 1 – i

6 85



7 85

i

Sol. Answer (4) z

 3  i  9  6i

z

 6  7i 1

z



=

=

1

6  7i 6  7i 36  49 6 85

 i

7 85

16. If z 1 = 3 + i and and z 2 = 2 – i , then

(1)) (1

8 5

 z 2

–1

z1 – z2

 i

z1

(2)) (2


8 5

is

8

8

(3)) (3

5

(4)) (4

5




99

Sol. Answer (1)

 3  i ,

z1

z2

 2  i

3i

 2  i  1 3  i  2  i  i 4

=

1  3i 4

=

10

8

=

5

(2  3i )2 17. Th The e mod modul ulus us of is 2  i 13

(1)) (1

147

(2)) (2

5

5

13

(3)) (3

5

(4)) (4

185 5

Sol. Answer (3) ( 2  3i )2 ( 4  9 )2 13   2 2  i 5 2 1 18. Th The e val value ue of (1 + i )(1 )(1 – i 2) (1 + i 4)(1 – i 5) is (1) 2i

(2) 8

(3) –8

(4) 8i

Sol. Answer (2)

(1  i ) (1  i 2 ) (1  i 4 ) (1  i 5 ) 4 (1  i ) (1  i )  4  2  8

19. If

z 

1 (1  i )(1– 2i )

, then |z |z | is

2

(1)) (1

(2)) (2

10

7 10

9

(3)) (3

10

1

(4)) (4

10


1 (1  i ) (1  2i )

| z | 

1  | 1  i | |1 | 1  2 i |

1 2 5




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100 Complex Numbers and Quadratic Equations

1

20. The val alue ue of

–

(1)) (1

2i

–

1 2 – i


is

2

4

(2)) (2

5

25

2

(3)) (3

(4) 0

5

Sol. Answer (3) 1

i

z



1 z

i



2i

5

21. Th The e mod modul ulus us of (1)) (1

 2  i

i

25



2 5

 ( i  2)3 is (2)) (2

47

4 15

(3)) (3

35

(4)) (4

2 37

Sol. Answer (4) i

25

i

 (i  2)3

 ( i )  8  12 12i  6 = 2 + 12i 12i

Now, | 2  12 12i | 

4  14 144  148  2 37

22.. The argu 22 argumen mentt of the the comple complex x number number (1 (1 + i )4 is (1) 135º

(2) 180º

(3) 90º

(4) 45º

(2) cos + i sin sin

(3) –cos – i sin sin

(4)) (4

Sol. Answer (2)

(1  i )4  ( 2 )4 (e i  / 4 )4  e i  Argument =  = 180º 23. Th The e pol polar ar fo form rm of ( i 41)3 is

(1)) (1

cos

 2

 i sin

 2

 –   –   i sin     2   2 

cos 

Sol. Answer (4)



        i sin    2 2

( i 41)3   i  1 cos   

24. If

z 

(1))  (1

–4  2 3i 5  3i

, then the value of arg( z ) is

(2)) (2

 3


(3)) (3

2 3

(4)) (4

 4


Complex Numbers and Quadratic Equations 101


Sol. Answer (3)

4  2 3i

z = z =

5  3i



(5  3i ) (5  3i )

=

20  6  i (10 3  4 3 ) 28

=

14  i (14 3 ) 28

arg (z )    tan1( 3 ) 2

=

25. If

z

3

 cos



 i sin

4



, then

6

arg(z )  (1) |z | = 1, arg(

3

(3)) | z | (3

2

 4

arg(z )  (2) |z | = 1, arg(

5

, ar arg( z ) 

(4)) | z | (4

24

 6

3 1 , arg( z )  tan–1 2 2

Sol. Answer (4) z

 cos

 4

 i sin

 6

1



2

 i

1 2 1

1 1 3   , 2 4 2

| z | 

tan 



2 1



1 2

2



tan

1  1



  2

26.. Th 26 The e squ squar are e roo roott of of –8 –8i is is (1) ± 2(1 – i )

(2) 2(1 + i )

(3) ± (1 – i )

(4) ± (1 + i )

Sol. Answer (1) 0  8i

y  x  iiy

8i  x 2  y 2  2ixy x

x

2



y

2



  y

0

 xy  x

4

(  x )   4

2 x

4

  2

x

 y 

2

  2(1  i ) Aakash Educational Services Pvt. Ltd.




27.. Th 27 The e squ squar are e roo roott of of 3 + 4i 4i is is (1) ± (2 – i )

(2) ± (2 + i )

(3) ± (3 + i )

(4) ± (3 – i )

Sol. Answer (2) 3  4i

3  4i

x

2

iy  x  iy

 x 2  y 2  2ixy

 y 2  3

 2 xy  …………… x  iy   (2  i )

28. If  and  are the roots of 4 x 2 + 3 x + + 7 = 0, then the value of 4

(1)) (1

(2)) (2

7

–

3 7

(3)) (3

1





1



is

3 7

(4)) (4

–

3 4

Sol. Answer (2) 4 x

2

 3 x  7  0

Quadratic equation whose roots are 1

 1





,



1

is

 1





7 x

2

 3 x  4  0

3 7

29. If a and b are the roots of the equation x 2 + + x x + + 1 = 0, then a2 + b2 is equal to (1) 1

(2) 2

(3) –1

(4) 3

Sol. Answer (3) x

2



 x  1  0 a

2

2  b2 = (a  b)  2ab

= ( 1)2  2 =– 1 30.. If the 30 the differen difference ce of the roots roots of the the equation equation x x 2 – – px px + + q = 0 is unity, (1)) p2 + 4q (1 4 q = 1

(2) p2 – 4q = 1

(3)) p2 – 4q2 = (1 + 2q (3 2q)2

(4) 4 p2 + q2 = (1 + 2 p p))2





Sol. Answer (2) x

2

 px  q  0

Let ,  be roots (   )  1

(  )2  1 (   )2  4  1 p

2

 4q  1

31. If  and  are the roots of the equation x 2 – px + + 16 = 0, such that 2 + 2 = 9, then the value of p p is is (1))  (1

(2))  (2

6

41

(3) ± 8

(4) ± 7

(3) 3 + 2i , 2i

(4) 2, 2 + 3i

Sol. Answer (2) x

2

 px  16  0

 2  2  (    )2  2 9

p

 p2  32 2

 41

p 



41

32.. Th 32 The e solu solutio tion n of the the equa equatio tion n (1) 2 + i , 3 – 2i 2 i

z ( z  3i )

 2(2  3i ) is/are

(2) 2 + 2i , 3i

Sol. Answer (4)



z( z

z( z

 3i )  2(2  3i )

 3i )  2(2  3i )

|z |2 + 3iz = = 4 + 6i 6i ( x x 2 + y 2) + 3i 3 i (x (x + iy ) = 4 + 6i 6 i ( x x 2 + y 2 – 3y ) + 3ix = = 4 + 6i 6 i = 6 and and x x 2 + y 2 – 3y = = 4  3 x =

 x = 2 and 4 + y 2 – 3y = = 4 = 0, 3  y = z = z = x x + + iy = 2 + i .0 .0 and 2 + 3i 3 i = 2, 2 + 3i 3 i Aakash Educational Services Pvt. Ltd.




33. If f ( x x ) = x 4 – 8 x 3 + 4 x 2 + 4 x + + 39 and f (3 (3 + 2i 2i ) = a + ib ib,, then a : b equals 1

(1)) (1

1

(2))  (2

8

4

(3)) (3

1

(4))  (4

4

1 8

Sol. Answer (4) f ( x x ) = x 4 – 8 x 3 + 4 x 2 + 4 x + + 39 x = = 3 + 2i 2 i  ( x x – – 3) 2 = –4 x 2 –6 x + + 13 = 0 Now x Now x 4 – 8 x 3 + 4 x 2 + 4 x + + 39 = ( x x 2 – 6 x + + 13) ( x x 2 – 2 x – – 21) + (–96 x + + 312) Now f (3 (3 + 2i 2i ) = –96 –96(3 (3 + 2i 2i ) + 312 = –288 – 192i 192 i + + 312 = 24 – 192i 192i = a + ib

 a = 24, b = –192.

Required ratio =

24

192



1 8

34.. If centre 34 centre of a regular regular hexagon hexagon is at origin origin and and one of the vertex vertex on on Argand Argand diagram diagram is 1 + 2 i , then its perimeter is (1)) (1

6 5

(2)) (2

4 5

(3)) (3

(4)) (4

6 2

2 5

Sol. Answer (1) F A(1 +

E

2i )

O D

B C

In regular hexagon OA OA = = AB AB = = BC = = CD CD = = ED ED = = EF = = FA Length Leng th of of perimet perimeter er = 6 × |OA |OA|| =

6

6

1 4

5

35.. The sum 35 sum of princ principal ipal argum arguments ents of complex complex numb numbers ers 1 + i , –1 

(1)) (1

11 12

(2)) (2

13 12


(3)) (3

12 13

i

3, –

3 – i , 3 – i , i , –3i –3 i , 2, –1 is

(4)) (4

 15




Sol. Answer (1)



arg(1 + i ) =

4





arg 1  i 3 







3  i 

arg  3  i 

arg



arg  i  

2 3

5 6

 6

 2

arg  3i  

 2

arg  2  0 arg 1   Required sum 

11 12

36. If a = cos  + i sin sin, b = cos + i sin sin, c = = cos  + i sin sin and

a b



b c



c a

 2 , then

sin(   )  sin(    )  sin(   ) equals (2) 

(1) 3

3

(3) 0

2

(4)

3 2

Sol. Answer (3) From question cos(  )  i sin(  )  cos(    )  i sin(    )  cos(   )  i sin(   )  2  0 i ⇒

sin(  )  sin(    )  sin(   )  0

37. The val alue ue of (i  3 )100  (i  3 )100  2100  (1) 1

(2) –1

(3) 0

(4) 2

Sol. Answer (3)

(i  3 )100  (i  3 )100  2100 100

100

 ( i )

 1  i 3    2  

100

100

2

10 0

 ( i )

 1  i 3    2  

.2100  2100

= 2 100 ((2)100 + ()100 + 1) = 2 100 ( 2 +  + 1) {where  is complex cube root of unity} =0 Aakash Educational Services Pvt. Ltd.




38.. Whi 38 Which ch of of the the follo followin wing g is not not true true? ? (1)) The numb (1 number er whose whose conju conjugate gate is

1 1  i

is

1 1  i

(2) If sin x + + i cos cos 2 x and and cos x – – i sin2 sin2 x are are conjugate to each other then number of values of x is is zero (3) If x + 1 + iy and and 2 + 3i 3 i are are conjugate of each other then the value of x + y is is –2 (4) 2 + i > > 3 + i Sol. Answer (4) If z 1 and z 2 are two complex number if lm(z lm( z 1)  0 and lm(z lm(z 2)  0 then z 1 > z 2 or z 1 < z 2 does not hold. 39.. Th 39 The e com compl plex ex nu numb mber ers s z 1, z 2 and z 3 satisfying (1) Of area zero

z 1

 z 3

z 2

 z 3



1  i 3 2

are the vertices of a triangle which is

(2) Right angled isosceles (3) Equilateral

(4) Obtuse angle isosceles

Sol. Answer (3) ( z1  z3 )  ( z2

z 2

1 3  z3 )   i  2  2 z 3

 /3

( z1  z3 )  (z2  z3 )e i  /3 z 1

Now using concept of rotation.

 z 1, z 2, z 3 are vertices of equilateral triangle. 40. Let a = i i and consider the following statements S1 :

a

e



 2

S2: The valu value e of sin( sin(ln ln a) = –1 S3: Im(a) + arg(a arg(a) = 0 Now identify the correct combination of the true statements. (1) S1, S2 o on nly

(2) S1, S3 o on nly

(3) S1, S2, S3

(4) S1 only

Sol. Answer (3) a

 (i )i

loge a  i loge i

   log a  i log  e 2    e

e

loge a  

a

e



 2

 2





  Therefore sin(ln a )  sin     1 2 Im(a)  arg(a)  0  0  0 S1 S2 S 3 are

Therefore

,

,

correct. 2

1    2 41. If z 2 + z + + 1 = 0 then the value of  z     z   z  

(1) 21

1 

2

 3    z  2 z  

(2) 42

1 

2

 21 ...   z    3 z  

(3) 0

2

  is equal to 21 z  1

(4) 11

Sol. Answer (2) if z 2 + z + + 1 = 0

 (z – – ) (z ( z – – 2) = 0  z = , 2 if z z = = , then

1

 2

z

2

1   2 To find the value of  z     z    z 

Now,

z

4

z

1





1

1





z

z

6 z



4

1 6 z

 4 



2

1



4

 1, z 2 

1 z



1



2

 1,

 2 

5 z

2

    z3  2  z  1



1



1 5 z

2

3 z

 1,

 2 

2

   ......   z21  3  z  1

1

2



1 3 z

 21   z 1

2

2

 1 and

...... and so on

Therefore, 2

1   2  z     z  z

2

2

1      z3  3   ......   z 21  2   z  z  1

 21   z 1

2

= {(– 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 42. If 1, 1, 2,............. 3n be the roots of equation x 3n + 1 – 1 = 0, and  be an imaginary cube root of unity, then

(2  1 )(2   2 ).....(2   3 ) n

(   1 )(   2 ).....(   3 ) n

(1))  (1

(2) –


(3) 1

(4) 2




Sol. Answer (3) We have x 3n+1 – 1 = ( x x – – 1) ( x x – – 1) ( x x – – 2) ......... ( x x – – 3n)

( 2  1) ( 2   2 )........ ( 2   3 ) Thus, (  1)(    2 )........ (  3 ) n

n





1 (  1) 1)

.

( 2  1) 1) ( 2  1) ( 2   2 )...... ( 2   3n ) (   1) 1) (   1) (    2 )......(    3n )

1 ( 2 )3n 1  1 1 (  2 )3n . 2  1  . .   1 3n 1  1   1 3n .  1

2  1 .  1  1 1



2  1 2  1

[∵ 3n = 6n = 1]

1

43. If z 1, z 2 and z 3, z 4 are two pairs of conjugate complex numbers, then

(1) 0



(2)

(3)) (3

2

 z 1   z    arg 2  is      z 3   z 4 

arg

3 4

(4))  (4

Sol. Answer (1) We have

z 2

Therefore, and

z z 3 4



 z1

z z 1 2

z 4

z z1 1

 | z1 | 2



z z3 3

 z3

and

 | z3 |2

z 

z 

z

1 2 1 Now arg  z   arg  z   arg  z       3

 | z 1 |2  0 = arg  2  | z 3 | 

4

3

z 2 z 4

 

(∵ Argument of a positive real number is 0)

44. If |z – – 4 + 3i 3 i |  2, then the least and the greatest values of | z | are (1) 3, 7

(2) 4, 7

(3) 3, 9

(4) 4, 5

Sol. Answer (1) |z – – (4 – 3i 3i )| )|  2 Aakash Educational Services Pvt. Ltd.




| z |  | 4  3 i |  2

 –2  | z | –5  2  3  | z |  7 | z z ||min = 3, | z z ||max = 7 45. If |z 1| = 2, |z |z 2| = 3, |z |z 3| = 4 and |2z |2 z 1 + 3z 3 z 2 + 4z 4 z 3| = 4, then the expression |8z |8 z 2z 3 + 27 27z z 3z 1 + 64 64z z 1z 2| equals (1) 72

(2) 24

(3) 96

(4) 92

Sol. Answer (3)

|8z |8 z 2z 3 + 27 27z z 3z 1 + 64z 64 z 1z 2| = | z1 || z2 || z 3 |

 (2)(3)(4)

8z1

27z 2



2

| z1 |

2

| z2 |



64z 3 2

| z 3 |

8



z 1

 (2)(3)(4)

27



z 2

64 z 3

8 z1 27 z 2 64z 3   4 9 16

|2z 1 + 3z 2 + 4z 3|  24 | 2z1  3 z2  4 z 3 |  24 | 2z1  3z2  4z 3 | = 24 |2z = 24 (4) = 96 2

46. If z 1 = cos + i sin sin and 1,

z 1, z 12,

z 13,

....,

z 1n –1 are

vertices of a regular polygon such that

I m ( z 1

)

R e ( z 1 )



5 1 2

then the value of n is (1) 20

(2) 10

(3) 18

(4) 15

Sol. Answer (1) z 1 = cos  + i sin sin and 1, z 1, z 12, z 13,......, z 1n–1 are vertices of a regular polygon,



2 n

Now,

, z1  cos

Im( z 12 ) Re( z 1)

sin





2 n

n

,

4 n

 2   n 



 2    n 

sin 

2 z1

e

i 4

 n

 cos

4 n

 i sin

4 n

5 1 2

5

1

2

cos 



 i sin

2

5

1

4

 2   2  cos    n  n    2  cos   n 

2 sin 



 sin18º  sin

 10

5

1

2



2 n



 10

 n = 20 Aakash Educational Services Pvt. Ltd.


,



47. The area area of of a triangl triangle e whose whose vertic vertices es are are repres represent ented ed by complex numbers 0, z and and zei (0 <  < ) equals (1)) (1

1 2

| z |2 cos 

(2)) (2

1 2

| z |2 sin 

1

(3)) (3

2

| z |2 sin  cos 

(4)) (4

1 2

| z |2

Sol. Answer (2)  To find area of a  whose vertices are represented by complex number 0, z and and zei  (0 <  < ) i 

1

Area of  

B(ze

bc sin A

2

)

|z | 



1 | z | | z | sin  2



1 | z |2 si s in  2

|z |

A(z )

O

(0, 0)

48.. Th 48 The e max maxim imum um va valu lue e of of |z | z | when z satisfies satisfies the condition

z



2

 2 is

z

3  1

(1)) (1

3  1

(2)) (2

(3)) (3

(4)) (4

3

2 

3

Sol. Answer (2) z 

2 z

2

 2  | z |  z   2  

r



2



2

r

r

2 

2

2



r

when

0



r

2

2

2



2

r

r 2 – 2r – – 2  0 1

3



r max

 1

r

 1

…(i)

3

49. The val alue ue of

(1)) (1

3

i

1  i 1  i , 2 2

is

(2)) (2

1  i 1  i , 2 2


(3)) (3

1  i 1  i ,

2

2

(4)) (4

1  i 1  i ,

2

2




Sol. Answer (1) Let the value of

i  x  iy iy

i  ( x  iy ) 2 i  x 2  y 2  2xy On comparing real and imaginary part x 2  y 2  0 and

x

  y and

Therefore

 1 2 xy 

 xy 

i



1 2

1  i 1  i , 2 2

50.. The 50 The ro root ots s of th the e equa equati tion on ( x x – – a)( x x – – b)+( x x – – b)( x x – – c )+( )+( x x – – a)( x x – – c) = 0 are equal, then which of the following is not true? (1)) a + b + c 2 = 0 (1

(2) a + b2 + c  = 0

(3)) a2 + b2 + c 2 – ab – bc (3 bc – – ca ca = = 0

(4) a + b + c = 0

Sol. Answer (4) Roots of the equations ( x x – – a) ( x – – b) + ( x x – – b) ( x – – c ) + ( x x – – c ) ( x – – a) = 0 i.e., i.e ., { x x 2 – (a ( a + b) x x + + ab ab}} + { x 2 – (b ( b + c ) x x + + bc } + { x x 2 – (a + c ) x x + + ac } = 0 i.e., i.e ., 3 x 2 – 2(a 2( a + b + c ) x + x + (ab + bc bc + + ca ca)) = 0 have equal roots, Therefore B2 – 4 AC = = 0 4(a 4( a + b + c )2 – 4 × 3 (ab ( ab + bc bc + + ca ca)) = 0 4[a 4[ a2 + b2 + c 2 + 2ab + 2bc bc + + 2ca ca]] – 12ab 12ab – 12 12bc bc – – 12 12ca ca = 0 4(a 4( a2 + b2 + c 2) – 4ab 4 ab – 4bc bc – – 4ac 4 ac = =0

 a2 + b2 + c 2 – ab – bc bc – – ca = 0

 | z |2  | z | 1     2 , then the locus of z z is is 3  2 | z |   

51. If log

(1) |z | = 5

(2) |z | < 5

(3) |z | > 5

(4) |z | = 0

Sol. Answer (2)

log



 | z |2  | z | 1 2  2 | z |  

3

z

2

 | z | 1  ( 3 )2 2 | z |





z ||2 – | z | + 1 < 6 + 3 | z z ||  | z

 | z z ||2 – 4| z | | – 5 < 0  (| z | | + 1) (| z | | – 5) < 0 but | z | | + 1 > 0 | – 5 < 0  | z |

 | z z || < 5

52. If arg

z





4

, then

(1) Re(z 2) = 9Im(z 9Im(z 2)

(2) Im I m(z 2) = 0

(3) Re(z 2) = 0

(4) Re Re(z ) = 0

Sol. Answer (3) arg z = =





tan

–1

 4

y x

y x



 4

1

 |y | = x |x |  x 2 – y 2 = 0  Re( Re(z z 2) = 0 53. If z 2 + z |z | + |z |2 = 0, then locus of z z is is (1) Circle

(2) Straight line

(3) A pair of straight line

(4) None of these

Sol. Answer (3) Let z z = = x x + + iy Given equation is, z 2 + z |z | + |z |2 = 0

 ( x x + + iy )2 + ( x x + + iy ) +

x

2

 y 2 + ( x 2 + y 2) = 0

 x 2 – y 2 + 2ixy + + x x 2  y 2  iy x 2  y 2  x 2  y 2  0  2 x 2  x x 2  y 2  i (2xy  y x 2  y 2 )  0 Now,



2 x

2

x (2x 

x x

2

x

2

 y 2  0

 y 2 )  0

 x = = 0 or or x x 2 + y 2 = 4 x 2  3 x 2 – y 2 = 0 Aakash Educational Services Pvt. Ltd.




Alternative z

2



2

| z |

z

 1 0

| z | 2

 z  z         1 0  | z |  | z | z



| z |

 ,  2

 z z = = |z |, |, z z = = 2|z |

54.. Th 54 The e lea least st va valu lue e of of p p for for which the two curves arg

p 

(1)) (1

3

z 

 6

(2)) p (2 p = = 3

and | z – 2 3i |  p intersect is

(3)

p 

1 3

(4)) (4

p 

1 3

Sol. Answer (2)

 x  0

3y

P 3 �2 3  0

Now,



55. If

2

2

y=

 p

1 3

x

( 3 ) 1

p 



6 2

z

3

4 

 2 – 1 z  

Re 

1 2

, then z is is represented by a point lying on

(1) A circle

(2) An ellipse

(3) A straight line

(4) No real locus

Sol. Answer (3)

 z  4  1   2z  1 2

Re 

 z  4   z  4   1    2z  1  2z  1 



4  2z  1 z

4 1 2z  1 z

2zz  z  8z  4  2zz  8 z  z  4 1 (2z  1)(2z  1)







4zz  7 z  7 z  8  4 zz  2 z  2 z  1



9z  9 z  9  0



z

 z  1

Hence point z lies lies on a straight line. 56. I f f ( x x ) and g ( x x ) are two polynomials such that the polynomial h( x x ) = xf ( x x 3 ) + x 2g ( x x 6) is divisible by x 2 + x + + 1, then (1)) f (1) (1 (1) + g (1) = 1

(3) f (1) (1) = g (1) (1)  0

(2) f (1) (1) = – g (1)

(4) f (1) (1) = ± g (1) (1)

Sol. Answer (2) h( x x ) = xf = xf ( x x 3) + x + x 2g ( x x 6) is divisible by by x x 2 + x + + 1,

∵

So, when h( x x ) will be divided by x by x – –  and and x x – – 2 remainder will be 0. h() = f (1) (1) + 2g (1) = 0

…(i)

h(2) = 2f (1) (1) + g (1) = 0

…(ii)

Now, adding (i) & (ii), ( + 2)f (1) (1) + ( + 2)g (1) (1) = 0

 – f (1) (1) – g (1) (1) = 0

 f (1) (1) = – g (1) (1)

57. The va valu lue e of of ( x + x +  + 2) ( x + x + 2 + 4) ( x + x + 4 + 8) .... till 2n 2n factors (1) ( x x – – 1) 2n

(2) ( x x – – 1)2n + 1

(3) ( x x – – 1)2n – 1

(4) ( x x – – 1)2n + 2

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

58. If

z  

3 – i 2

, then ((i i 101 + z 101)103 equals

(1)) iz (1

(2)) z (2

(3)) (3

(4)) z, ( is complex cube root of unity) (4

z

Sol. Answer (2)

101

z

i

101

        cos    i sin –    6  6  

101

 i 101

 101   101 x   cos   i sin      6   6  Aakash Educational Services Pvt. Ltd.

101

 i 101




 cos

5 6

 i sin

5 6

 i 101

      cos  i sin 







6

6

3 2

2

3 2

i

  i



i

2

Now,

(i 101 + z 101)103

  – 

i 

3

103

  2

2

5 5     cos  i sin   6 6 

 cos

 cos

 cos



3 2

515 6 11 6

 6

–

 i sin

 i sin

– i si s in

103

515 6

11 6

 6

i

2

59.. The regio 59 region n of the comp complex lex plane plane for which which (1)) x -axis (1

(2) y -axis

a  1 is (a is real) z  a z

(3) Straight line x = a

(4) The str straig aight ht lin line ey = a

Sol. Answer (2)

a 1 z a z

|z z – – a| = |z | z + + a|, let z z = = x x + + iy then, then, ( x x – – a)2 + (y )2 = ( x x + + a)2 + y 2

 x 2 – 2ax + + a2 + y 2 = x 2 + a2 + 2ax + + y 2 

4ax ax = =0

 x x = =0 is y -axis -axis Aakash Educational Services Pvt. Ltd.



60.. Th 60 The e imag imagin inar ary y par partt of

2z  1 iz  1

(1) A circle


is –2; then the locus of the point representing z in in the Argand plane is

(2) A straight line

(3) A parabola

(4) An ellipse

Sol. Answer (2) Let z z = = x x + + iy z 

2 x  2iy  1 i ( x  iy )  1

(2 x  1) (1  y )  2 xy  i  (2 y ) (1  y )  x(2 x  1) 2 x  1  i 2y (1  y )  ix   (1  y )  ix (1  y )  ix (1  y )2  x 2

from given condition ( 2y ) (1  y )  x (2x  1) (1 y )2  x 2

 2

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

 2y y + + x – – 2 = 0, i.e. i.e.,, a straight line 61. If z is is a complex number satisfying |2008 z – – 1| = 2008|z 2008| z – – 2|, then locus z is is (1)) y (1 - axis

(2) x - axis

(3) Circle

(4) A line parallel to y -axis -axis

Sol. Answer (4) z = z = x x + + iy | 2008z 2008z – – 1 | = 2008 | z – – 2 |

⇒

z



1 2008

 | z  2 |

Put z z = = x x + + iy 2

1   2 2 2  x  2008   ( y )  ( x  2)  y 2

⇒

⇒

x



1  1   2008   2  x  2008  4  4 x

4 x 

x

1004

 1  4  2008  4

 1   4  2008 

2

2

1 1004

a line parallel to y -axis. -axis. Aakash Educational Services Pvt. Ltd.




62.. Th 62 The e locu locus s of of the the poin pointt z satisfying satisfying the condition arg (1) A straight line

(2) Circle

1   is 3 z  1 z

(3) A parabola

(4) Ellipse

Sol. Answer (2)

  z  1   APB    z  1 3

y

arg 

P z  3

 z lies lies on a circle

B (–1, 0)

Alternatively

A x (1, 0)

O

put z z = = x x + + iy

 z  1    z  1 3

arg 

 ( x  1)  iy    arg  3   x iy ( 1 )  

 ( x  1) 1)  iy ( x  1) 1)  iy    arg     ( x  1)  iy ( x  1)  iy  3

 ( x 2  y 2  1)  i (2 y )    arg    ( x  1)2  y 2   3

2y



2

 y  1

2

 y2 

x



x

2



2 3

3



tan

1 

2 y

 2 2  x  y

    1 3

3 ( x 2  y 2 )  3  2y  0

y  1  0

1    Which represents a circle having centre at  0,  and radius 3

63. Th The e lo locu cus s of

    

z  i  2 exp i   

(1) A circle

1 3

1 

2 3

  

  , (where  is parameter) is

4  

(2) An ellipse

(3) A parabola

(4) A hyperbola

Sol. Answer (1)

z

    i  2 exp  i      , where  is parameter put z = = x + + iy 4    



x  iy  i  2  cos    



     i sin      4 4 







equating real and imaginary parts we get

  2 cos    

x

  ......(1) 4

 

y  1  2 si sin   

y  1 

or

  4

 

2 sin   

  4

....(2)

squaring and adding (1) and (2), we get x 2 + (y – – 1)2 = 4 which represents a circle with centre (0, 1) and radius 2. 64.. If one 64 one vertex vertex and cent centre re of a square square are are z and and origin then which of the following cannot be the vertex of the square? (1)) iz (1

(2) –z

(3) –iz

(4) 2z

Sol. Answer (4) y iz 90°

 z

O

A z

x

 iz On rotating OA OA by by 90° angle we can find other vertices. 65. If z 1, z 2, z 3 represent the vertices of an equilateral triangle such that | z 1| = |z 2| = |z 3|, then (1)) z 1 + z 2 = z 3 (1

(2)) z 1 + z 2 + z 3 = 0 (2

(3) z 1z 2 =

1 z 3

(4)) z 1 – z 2 = z 3 – z 2 (4

Sol. Answer (2) z 1, z 2, z 3 are the vertices of an equilateral triangle such that |z 1| = |z2| = |z 3| or |z | z 1 – 0| = |z |z 2 – 0| = |z |z 3 – 0|

 origin is the circumcentre of the  

z 1

 z2  z 3 3

0

 origin is the centroid of the equilateral 

 z 1 + z 2 + z 3 = 0

66. If |z – – 2 – 3i 3i | + |z | z + + 2 – 6i 6 i | = 4, i i = =

 1 , then locus of z is

(1) An ellipse

(2) A point

(3)) Seg (3 Segmen mentt joining joining the the points points (2+3 (2+3i i ) and (–2+6i (–2+6i )

(4) Empty





Sol. Answer (4) |z –(2 –(2 + 3i 3i )| )| + |z |z – – (–2 + 6i 6 i )| )| = 4 Let z 1 = 2 + 3i 3i , z 2 = –2 + 6i 6 i |z 1 – z 2| = |4 – 3i 3 i | = 5 > k |z z – – z 1| + |z | z – – z 2| = 2a, whe here re k < < |z 1 – z 2|  |

 This does not represent any curve  Locus of z is an empty set. Alternatively :: If If we put z = = x + + iy , then we got an equation in x and and y which which does not have any solution. 67. If z 1, z 2, z 3 and u , v , w are complex numbers representing the vertices of two triangles such that z 3 = (1 – t )z 1 + tz 2 and w = = (1 – t ) u + + tv , where t is is a complex number, then the two triangles (1) Have the same area

(2) Are similar

(3) Are congruent

(4) Are equilateral

Sol. Answer (2) z 1, z 2, z 3 and u, v,  are complex numbers representing the vertices of two triangles such that z 3 = (1 – t ) z 1 + tz 2 and  = (1 – t )u u + + tv , t c z 3 = z 1 – tz 1 + tz 2 and  – u u = = – tu tu + + tv z 3 – z 1 = t (z 2 – z 1) and  – u = = t (v ( v – – u )



z3

t 

and

z2

 z 1  z 1

w

t 

v

…(1)

 u  u

…(2)

 From (1) & (2) 



z 3 z 2

 z 1   z1

arg

w v

 u u

 z3  z 1   w  u   z  z   arg  v  u  …(3) 2

1

 z3  z 1   w  u   1  arg  1   v  u   z2  z1 

 arg 



arg

arg

 z3  z 2   w  v   z  z   arg  v  u  2

1

 z3  z 2   w  v   z  z   arg  u  v  …(4) 1

2

From (3) & (4) we conclude that two triangles are similar.

68.. For tw 68 two o com comple plex x num numbe bers rs z 1 and z 2, we have

z 1

 z 2

1  z1z 2

 1, then

(1) Both z 1 and z 2 lie on circle | z z || = 1

 z 1      z 2  3

(2)) arg (2

(3)) At lea (3 least st on one e of of z 1 and z 2 lies on the circle | z z || = 1 (4) | z 1 | = 2| z 2 | Aakash Educational Services Pvt. Ltd.




Sol. Answer (1) z 1

1

 z 2 z z 1 2

1

 | z1  z2 |  | 1  z1

z 2

|

 | z1  z2 |2  | 1  z1

 | z1 |2  | z2 |2  z1 z2  z1z2  1 |

z1

|2|

|z z 1|2 | |z z 2|2 – |z 1|2 – |z 2|2 + 1 = 0  |

 |z 1| = 1, |z | z 2| = 1

z2

z2

|2  z1 z2 

|2 z1z2

(|z z 1|2 – 1) (|z (|z 2|2 – 1) = 0  (|

 Both z 1 and z 2 lie on the circle |z |z | = 1

69. Let  and  are the roots of the equation x 2 + x + + 1 = 0 then (1)) 2 + 2 = 4 (1

(2) ( – )2 = 3

(3) 3 + 3 = 2

(4) 4 + 4 = 1

Sol. Answer (3) x 2 + x + + 1 = 0

 +  = –1  = –1 (1)) 2 + 2 = ( + )2 – 2 (1 = (–1)2 – 2(1) = 1 – 2 = –1 (2) ( – )2 = ( + )2 – 4 = (–1)2 – 4  1 = –3 (3)) 3 + 3 = ( + )(2 + 2 – ) (3 = (   )((    )2  2  ) = (   )((   )2  3) = (–1)((–1)2 – 3 1) = (–1) (1 – 3) = 2 Alternative

x 2

+ x + 1 = 0

x = = , 2 (complex root of unity)

 3 + (2)3 = 2 (4))  4  2 = ( 2  2 )2  2 22 (4 = (–1)2 – 2 1 =1–2 = –1 70.. If the rat 70 ratio io of the roo roots ts of lx 2 – nx + n = 0 is p is p : : q , then

(1)) (1

p q



q p



n l

0

(2)) (2


p q



q p



n l

0

(3)) (3

q p



p q



l n

1

(4)) (4

q p



p q



l n

0




Sol. Answer (2) p



q

n n     l l  ,

,

p

Now,

q

q



  



p

 

n  





l



n

n



l

l

p q

q



p



n

p

⇒

l

q

q



p



n l

0

71. Fo Forr the the eq equa uati tion on x |x 2| + x |x | – 6 = 0, the roots are (1) Real and equal

(2) Real with sum 0

(3) Real with sum 1

(4) Real with product 0

Sol. Answer (2) x

2







x

x

6  0

⇒

 3   x  2  0

x

2



⇒

x

x

60  3,

x

2

 x = = ± 2 Two roots are real, with sum 0. 72. If a + b + c = = 0 and a, b, c are are rational, then the roots of the equation ( b + c – a) a ) x x 2 + (c + a – b) b ) x + x + (a + b – c ) = 0 are (1) Rational

(2) Irrational

(3) Imaginary

(4) Equal

(3) a2 – b2 = 4ac 4 ac

(4)) a2 + b2 = ac (4

Sol. Answer (1)

( b  c  a) x 2  (c  a  b ) x  (a  b  c )  0 Put x Put x = = 1,

b

 c  a  c a b a b c  a  b  c  0

 1 is the root of the equation.  Roots are rational. 73. If sec, tan are roots of ax 2 + bx + + c = = 0, then (1)) a4 – b4 + 4ab2c = (1 = 0

(2) a4 + b4 – 4ab2c = = 0

Sol. Answer (1) We know sec

2

  tan2   1

(sec   tan  )(sec   tan )  1 Aakash Educational Services Pvt. Ltd.



  

b


 4ac   b        1   a  a

2

Squaring both side

(b2  4ac )b 2  a 4 a

4

 b 4  4ab2c  0

74. If x is is real, then the expression

x

2

x

 34 x – 71 2

 2 x – 7

(1) Lies between 4 and 7

(2) Lies between 5 and 9

(3) Has no value between 4 and 7

(4) Has no value between 5 and 9

Sol. Answer (4)

y 

Let



x

2

x

x

2

 34 x  71 2

 2x  7

( y  1)  x (2y  34)  71  7y  0

For real x real x , discriminant should be  0

 (2y  34)2  4 ( y  1)(71  7y )  0  4 ( y  17)2  4 ( y  1)(71  7y )  0  ( y  17)2  ( 7 y 2  78y  71)  0  

y



2

y

8y

2

 112y  360  0

 14y  45  0  ( y  9)( y  5)  0 9

–

+

or y  5

+

5

9

75. If ,  are roots of ax 2 + bx + + c = = 0, then the equation ax 2 – bx ( ( x – – 1) + c ( x x – – 1)2 = 0 has roots



(1)) (1

,



(2)) (2

1–  1– 

1–  1–  ,





(3)) (3

  , 1  1  

(4)) (4

1  1  ,





Sol. Answer (3) ax

2

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

 bx  c  0 , 2

 – x   – x   b 1 0 a   x – 1  x – 1 Now, Replacing x Replacing x by by   

x x

1





ax

2 2

( x  1)

bx



x  1

 x x  1



⇒

c 0

x



⇒

ax

2

 bx ( x  1)  c ( x  1)2  0

 is the root of the above equation. 1 

76. Let ,  be the roots of ax 2 + bx + c = 0,  ,  be the roots of px 2 + qx + r = 0 and D1 and D2 be their respective discriminant. If , ,  ,  are in A.P., then the ratio D1 : D2 is equal to a

(1)) (1

2

b

(2)) (2

2

a p

2

(3)) (3

2

b2

2

(4)) (4

q2

c

2

r

Sol. Answer (2) b

c

a

a

     ,  

    

q p

,  

r p

2

, D1  b  4ac

2

, D2  q  4rp

Let common difference of A.P. be k . k 

|  | |   |





b 2  4ac a

D1 D2





q 2  4 pr p



b 2  4ac q 2  4 pr



a

⇒

p

D1 D2



a p

a2 p2

77. The equ equa atio ion n a( x  b )( x

 c ) b( x c )(x  a) c ( x a )( x  b )    x (a  b)(a  c ) (b  c )( b  a) (c  a)( c  b)

is satisfied by (1)) No va (1 valu lue e of of x x

(2)) Exa (2 Exactl ctly y two valu values es of of x

(3)) Exa (3 Exactl ctly y three three value values s of x of x

(4) Al Alll valu values es of of x

Sol. Answer (4)

 b )( x  c ) b (x  c )( x  a ) c ( x  a )( x  b )    x (a  b)(a  c ) ( b  c )(b  a) ( c  a)(c  b)

a(x

is satisfied by x x = = a, x x = = b, x x = = c . A quadratic equation is satisfied by more than two values of x . So it is an identity. Hence it is satisfied by all values of x of x . Aakash Educational Services Pvt. Ltd.




78.. Co 78 Cons nsid ider er the the equ equat atio ion n ax 2 + bx + c = 0, where a  0, a, b, c  R then then (1) If on one e roo roott is is    , then other root is    (2) If a = 1 and b and c are are integers, then root will be integer (3) If on one e roo roott is is  + i , then other root will be  – i  (4)) If roots (4 roots are are of oppos opposite ite sign, sign, then then b  0 Sol. Answer (3) (1) Roo Roott will will be of of the for form m    of a, b, c are are rational. (2)) Ther (2 There e is no no informa information tion abo about ut b2 – 4ac Hence statement is false. (3) As a, b, c are are real and one root is   i  then other root will be   i  . (4)) If mass (4 mass are are of opposi opposite te sign sign then then   0

c ⇒

a

0

79. If th the e equ equat atio ion n ((k k 2 – 3k k + + 2) x x 2 + (k 2 – 5k k + + 4) x + x + (k 2 – 6k + + 5) = 0 is an identity then the value of k k is is (1) 1

(2) 2

(3) 3

(4) 4

Sol. Answer (1) For an identity (k 2 – 3k + 2) = 0

 (k – – 1) (k (k – – 2) = 0  k = 1, k = 2 k 2 – 5k + + 4 = 0

 (k – – 1)(k 1)(k – 4) = 0

k = 1, 4 k 2 – 6k + 5 = 0  (k – – 5)(k 5)( k – 1) = 0 k = 1, 5 Common value of k = = 1. 80. Th The e va valu lue e of k if if (1) Th The e roo roots ts of 5 x 2 + 13 x + + k = = 0 are reciprocal to each other is 5 (2) Th The e roo roots ts of of x x 2 + x x + + k = = 0 are consecutive integer is 1 (3) Th The e roo roots ts of of x x 2 – 6 x x + + k = = 0 are in the ratio 2 : 1 is 7 (4) The roots roots of of the equat equation ion x x 2 + kx – 1 = 0 are real, equal in magnitude but opposite in sign is 1 Sol. Answer (1) (1) Le Lett root roots s are are , 

+=  

=

13

k 5

5

…(i) …(ii)

For reciprocal roots  = 1  k = 5. Aakash Educational Services Pvt. Ltd.




(2)) If roots (2 roots are cosecu cosecutiv tive e integer integer then then | – | = 1

 |  – |2 = 1 ( + )2 – 4 = 1

 1 – 4k = 1  k = 0 (3) Le Lett root roots s are are 2, 

 2  +  = 6

…(i)

2 = k

…(ii)

By (i), (ii)

 = 2, k k = = 8 (4) In th this is ca case se

+ =0  k = 0 81.. If the 81 the differe difference nce of tthe he roots roots of the the equati equation on x 2 + ax + b = 0 is equal to the difference of the roots of the equation x equation x 2 + bx + a = 0, then (1)) a + b = 4 (1

(2) a + b = – 4

(3) a – b = 4

(4) a – b = – 4

Sol. Answer (2) x

2

 ax  b  0

⇒

    a,   b,    

x

2

 bx  a  0

⇒

    b,   a,    

Now,       



a

2

a

⇒

 4b  b 2  4a

⇒

 (a  b )(a  b  4)  0

a

2

⇒

2

 4b 

b

2

a

b

2

2

 4b

 4a

 4a

 b 2  4 (b  a ) ab

 4 (a  b )

82.. If th 82 the e eq equa uati tion ons s px 2 + 2qx + + r = = 0 and and px px 2 + 2rx + + q = 0 (q ( q  r ) have a common root, then p p + + 4q + 4r equals equals (1) 0

(2) 1

(3) 2

2

…(1)

(4) –2

Sol. Answer (1) Let  be common root, and

p

2

p

 2q  r  0

…(2)

 2r   q  0

Now (1) – (2)

 2 (q  r )  r  q  0

Common root is  

1 2

⇒



1 2

, substituting in (1)

2

 1  1 p    2q    r  0 ⇒  2  2

4r

 4q  p  0





83. If the the equ equat atio ions ns ax 2 + bx + c = = 0 and x 2 + x + + 1 = 0 has one common root then a : b : c is is equal to (1) 1 : 1 : 1

(2) 1 : 2 : 3

(3) 2 : 3 : 1

(4) 3 : 2 : 1

Sol. Answer (1) x 2 + x + + 1 = 0

… (i)

Discriminant = b2 – 4ac 4 ac = = 1 – 4  1  1 = –3 Hence the roots of x 2 + x + 1 = 0 and not real. So roots will be in pair. Also the roots of ax 2 + bx + c = = 0 will be non-real. Clearly both roots of the equations are common.



a 1



b 1



c 1

 a : b : c = 1 : 1 : 1 84.. If 1, 84 1, 2, 3 are are the the roots roots of of the equ equati ation on x x 3 + ax 2 + bx + c = 0, then (1)) a = 1, b = 2, c = 3 (1

(2) a = –6, b = 11, c c = = –6

(3)) a = 6, b = 11, c = (3 = 6

(4) a = 6, b = 6, c c = = 6

Sol. Answer (2) If 1, 2, 3 are roots of equation then x 3 + ax 2 + bx + c = 0

 1 + 2 + 3 = –a – a  a = –6 12 + 23 + 13 = b  b = 11 123 = –c – c

 c c = = –6

85. Co Cons nsid ider er th that at f ( x x ) = ax 2 + bx + c , D = b2 – 4ac , then which of the following is not true? (1) If a > 0, then minimum value of f ( x x ) is

–D 4a

(3) If a > 0, D < 0, then f ( x x ) > 0 for all x  R

(2) If a < 0, then maximum value of f ( x x ) is

–D 4a

(4) If a > 0, D > 0, then f ( x x ) > 0 for all x  R

Sol. Answer (4) f ( x x ) = ax 2 + bx + c

=

 

ax

2



 2 = ax    a f ( x x ) =   x  

b a

b a

x 

x 

2

c 



a

b

4a

   2a  b

2 2

b

2



b

2

4a

2



c



a

 4ac  4a

2

 





 f ( x x ) = a  x  

 

a  x 

f ( x x ) =

2

2 b  4ac    2a  4a

b

  2a  b

D 4a

Clearly if a > 0 the minimum value of f ( x ) 

Similarly of a < 0 the maximum value 

D 4a

D 4a

If ax 2 + bx + c > 0 then a > 0, D < 0 for all x  R Hence option (4) is not true. 86.. If the 86 the min minim imum um va valu lue e of of x 2 + 2 x + + 3 is m and maximum value of – x 2 + 4 x + + 6 is M then then the value of m + M is (1) 10

(2) 11

(3) 12

(4) 13

Sol. Answer (3) x 2 + 2 x + + 3 = ( x + x + 1)2 + 2  m = 2 – x 2 + 4 x x + + 6 = – x 2 + 4 x + + 4 – 4 + 6 = 6 – ( x 2 – 4 x + + 4) + 4 = 10 – ( x 2 – 4 x + + 4) = 10 – ( x x – – 2)2

 M = = 10 m + M = 2 + 10 = 12 87. For all x  R if if mx 2 – 9mx + + 5m + 1 > 0, then m lies in the interval

 61  , 0  4 

 4 61  ,   61 4 

(1))   (1



(2))  (2

 4  , 0  61 

4 

(4))  (4

(3)) 0 , (3   61

Sol. Answer (3) Let

y

y

2

 mx  9m  5m  1

We need y y > >0

 Upward parabola above x above x -axis. -axis. mx



2

 9mx mx  5m  1  0, 

D

 0,

a

x

 R.

m

x

0

i.e.,, 81m2  4 (m)(5m  1)  0 and i.e.



O

(61m  4)  0 and

m

0


⇒

m

0

0m

4 61




Also for m = 0,

0 x 2  9 (0) x  0  1  1  0,  x R



m

 4  0,   61

l + 1 = 0 is double the other and if l is 88.. If one roo 88 roott of of the the equ equati ation on ( l – – m) x x 2 + x is real, then the greatest value

of m is 9

(1)) (1

(2)) (2

8

7

(3)) (3

8

8 9

(4)) (4

5 9

Sol. Answer (1)

(l  m) x 2  lx  1  0 l

  2 

m

  (2 ) 

l

⇒

l

1 m

⇒



l

3 (m  l )

2 

…(1)

1 2( l  m)

…(2)

From (1) and (2) 2

1 2(l  m)



2l

2



l

9(l  m)2

 9l  9 m  0

For real l ,

81  8

 9m  0

⇒



 Greatest value of m is

m

m





81 72

9 8

9 8

89. If p p,, q, r are are real numbers satisfying the condition p + q + r = 0, then the roots of the quadratic equation 3 px 2 + 5qx + + 7r = = 0 are (1) Positive

(2) Negative

(3) Real and distinct

(4) Imaginary

Sol. Answer (3) 3 px

2

 5qx  7r  0





(3 p )( )(7r )   (5q )2  4 (3

 25q 2  84 pr  25 ( p  r )2  84 pr  25 p2  34 pr  25r 2 2

17  336 2    5 p  r  r  0  5  25

 roots are real and distinct. 90.. The roo 90 roots ts of the equ equati ation on x x 3 – 2 x 2 – x + + 2 = 0 are (1) 1, 1, 2, 3

(2) –1 –1, 1, 2

(3) –1 –1, 0, 1

(4) –1 –1, –2, 3

Sol. Answer (2) x 3 – 2 x 2 – x + + 2 = 0 As x As x = = 1 is the root of the equation Hence we may write x 3 – 2 x 2 – – x x + + 2 = x 2 ( x – – 1) – x – x ( x x – – 1) – 2( x x – – 1) = ( x – – 1) ( x x 2 – x – 2) = ( x – – 1) ( x x – – 2) ( x x + + 1) Roots = 1, –1, 2. 91. If a and b are rational and ,  be the roots of x of x 2 + 2ax 2ax + + b = 0, then the equation with rational coefficients one of whose roots is     2  2 i is s (1)) x 2 + 4ax – (1 – 2b = 0

(2) x 2 + 4ax 4 ax + + 2b = 0

(3) x 2 – 4ax + + 2b = 0

(4) x 2 – 4ax – – 2b = 0

Sol. Answer (2)

    2a,   b

    2  2  2a 

4a

2

 2b

2 2 The other root of equation will be       

i.e.,, 2a  i.e.

4a

2

 2b

 Sum of roots, S = = –4 –4a a Product of roots, P P = = 4a2  ( 4a2  2b )  2b

 required equation is i.e.,, i.e.

x

2

x

2

 Sx  P  0

 4ax  2b  0





92. The The va valu lues es of a, for which the quadratic equation 3 3 x x 2 + 2( 2(a a2 + 1) x + x + (a2 – 3a + 2) = 0 possesses roots of opposite sign, are (2) a  (2, )

(1) 1 < a < 2

(3) 1 < a < 3

(4) –1 –1 < a < 0

Sol. Answer (1) For roots of opposite sign, product < 0 a

 

2

 3a  2 0 3

1

a

⇒

(a  2)(a  1)  0

2

93. Let a, b, c  R and and a  0 be such that (a ( a + c )2 < b2, then the quadratic equation ax 2 + bx + + c = = 0 has (1) Im I maginary roots

(2) Re R eal roots

(3) Ex Exac actl tly y one one re real al ro root ot ly lyin ing g in in the the in inte terv rval al (– (–1, 1, 1)

(4) Ex Exac actl tly y two two root roots s in (–1 (–1,, 1)

Sol. Answer (3) Here we observe that (a ( a + c )2 < b2

 (a – b + c ) (a + b + c ) < 0  Exactly one real root of the given equation lies in (–1, 1). ax

2

 bx  c  0

D=

b

2

 

 4ac  (a  c )2  4ac  (a  c )2 0

 Roots are real. 94. If p p + + iq iq be be one of the roots of the equation x 3 + ax + + b = 0, then 2 p p is is one of the roots of the equation (1)) x 3 + ax (1 ax + + b = 0

(2) x 3 – ax – – b = 0

(3) x 3 + ax ax – – b = 0

(4) x 3 + bx bx + +a=0

Sol. Answer (3) p + p + iq iq is is one root  p p – – iq iq is is other root. Let  be third root. Now sum =   p  iq  p  iq  0

   2 p   2 p is root of

x

3

 ax  b  0

 2 p p is is root of (  x )3  ax  b  0 

x

3

 ax  b  0

95. If a1, a2, a3 , a4,......, an – 1, an are distinct non-zero real numbers such that ( a12 + a22 + a32 +......+ an2 – 1 ) x x 2 + 2( 2(a a1a2 + a2a3 + a3a4 +.......+ an – 1 an) x + x + (a22 + a32 + a42 +......+ an2)  0 then a1, a2, a3,......, an , an are in –1 (1) A.P.

(2) G.P.


(3) H.P.

(4) A.G.P.




Sol. Answer (2) We have, given expression (a12 + a22 + a32 +.....+ an –

1

2) x x 2

+ 2( 2(a a1a2 + a2a3 + a3a4 +.....+ an – 1 an) x + x + (a22 + a32 + a42 +......+ an2)  0

(a a1 x + + a2)2 + (a2 x + + a3)2 + (a3 x + + a4)2 + ....... + (a ( an – 1 x + + an)2  0  ( (a a1 x + + a2)2 + (a2 x + + a3)2 + (a3 x + + a4)2 + ....... + (a ( an – 1 x x + + an)2 = 0,  ( as sum of square cann’t be negative.

 a1 x x + + a2 = 0 = a2 x x + + a3 = a3 x x + + a4 = ....... = an – 1 x x + + an   x 

a2 a 1



a 3 a 2



a4 a 3

 ....... 

a

n

a

n

1

 a1, a2, a3, ....... , an – 1 , an are in G.P. 96. Th The e roo roots ts of ax 2 + bx + + c = = 0, whose a  0, b, c  R , are non-real complex and a + c c < < b. Then (1) 4a + c c > > 2b

(2) 4a + c c < < 2b

(3) 4a + c c = = 2b

(4) No None ne of of thes these e

Sol. Answer (2) f (x )

 ax 2  bx  c, given

 f ( x )  0, 

f ( 1)

 a  b  c  0

 R as roots are non-real complex

x

 f (–2) (–2) < 0 

4a

 2b  c  0

⇒

4a

 c  2b

97.. If the 97 the sum sum of the roots roots of of the quad quadratic ratic equa equation tion ax 2 + bx bx + + c c = = 0 (a, b, c  0) is equal to sum of squares a

of their reciprocals, then

c

(1) A.P.

b ,

a

c ,

b

are in

(2) G.P.

(3) H.P.

(4) None of these

Sol. Answer (3) Given,      Also,    

1



2

b a

,



 

c a

1

2

2  b  c   (   )(  )      (   )  2      2   a  a  2

2

2

2

2

 bc 2  b2a  2ca2 

2

a b



c a





b



c

2ca

a c

2

a

2

a

2



2c

a

 bc 2  b2a

b ,

b

c ,

b

are in H.P.

98.. The numb 98 number er of irrati irrational onal root roots s of the equa equation tion ( x x – – 1) ( x x – – 2) (3 x x – – 2) (3 x + + 1) = 21 is (1) 0

(2) 2


(3) 3

(4) 4




Sol. Answer (2) ( x  1)(3 x  2)(3 x  1)( x  2)  21 21

(3 x 2  5 x  2)(3 x 2  5 x  2)  21 Put,

3 x

2

 5 x  t

(t  2)(t  2)  21 Now,

3 x



3 x

2

3 x

2

2

⇒

2

t

 5 x  5 and

 5 x  5  0 and

 25 

3 x

2

3 x

t

 5,

t 

5

 5 x   5

2

 5 x  5  0

 5 x  5  0 has two irrational roots.

whereas roots of

3 x

2

 5 x  5  0 are imaginary.

99.. Con 99 Conside siderr the the foll followin owing g sta statem tements ents S1: If x is is in radian and 2sin 2 x + + 3sin x – – 2 > 0 and and x x 2 – x – – 2 < 0 then

 

S2: If x   0,



and x x  {–1, 0}, then the expression  and 2

x

2

 x 

tan x

2

2



   , 2 . 6

x





is always greater than or equal to 2

 x

tan. S3: If a, b, c are are the sides of a triangle ABC such such that x 2 – 2( 2(a a + b + c ) x + x + 3(ab + bc + ca ca) = 0 has has re real al roots then  

4 3

.

S4: If the the equa equations tions x 3 + 3 px 2 + 3 qx + r = 0 and x 2 + 2 px + q = 0 have a common root then 4( p p2 – q) (q ( q2 – – pr pr ) = ( pq pq – r )2. The correct combination of true statements is (1) S1 o on nly

(2) S2 o on nly

(3) S3 o on nly

(4) Al All S1, S2, S3, S4

Sol. Answer (4) S1 : x x 2 – x – – 2 < 0 x – – 2)( x x + + 1) < 0  ( x < x < 2  –1 < x

… (i)

2sin2 x + + 3sin x – – 2 > 0 + 4sin x – – sin x – – 2 > 0  2sin2 x + 2sin x (sin (sin x + + 2) –1 (sin x + + 2) > 0

 (sin x + + 2)(2sin x – – 1) > 0 

 sin x 

1 2







  5   , 

x

6

By (i), (ii)

…(ii)

6 

x

    , 2   6

S2 :Using A.M.  G.M.

x

2

 x 

tan2  x

2

 x

 ( x 2  x ) 

2

x



2

tan2  x

tan

 x 

x

2

2



2



x

 2tan 

 x

S3 :Using D  0 4(a a + b + c )2 —– 4(1)(3)( )(ab ab + bc + ca)  0  4(

 a2 + b2 + c 2 + 2( 2(ab ab + bc + ca) – 3(ab + bc + ca) ca )  0  a2 + b2 + c 2 + (ab + bc + ca) (2 – 3 )  0  b2  c 2  3  2  ab  bc  ca a

2

… (i)

But (a (a – b)2  c 2 (b – c )2  a2 (c – a)2  b2

 b 2  c 2 2  ab  bc  ca a

2

 3  – 2  2  

S4 :

x

4 3

3

 3px 2  3qx  r  0

…(i)

Multiply the second equation by x x

3

 2px 2  qx  0

…(ii)

By (i) – (ii) px

2

 2qx  r  0

…(iii)

But x But x 2 + 2 px + + q = 0 + pq pq = = 0  px 2 + 2 p2 x +

…(iv)





By (iii) - (iv) 2 x (q – p 2) + (r – pq) pq) = 0 x 

r

 pq

2(q  p 2 )

Putting x Putting x in in x 2 + 2 px + + q = 0 We get 4( p p2 – q) (q ( q2 – – pr pr ) = ( pq pq – r 2)2 100. If a, b are real, then the roots of the quadratic equation ( a – b) x x 2 – 5( 5(a a + b) x x – – 2( 2(a a – b) = 0 are (1) Real and equal

(2) Non-real complex

(3) Real and unequal

(4) None of these

Sol. Answer (3) The given quadratic equation is ( a – b) x x 2 – 5( 5(a a + b) x x – – 2( 2(a a – b) = 0 The discriminant D = (– 5(a 5( a + b))2 + 8( 8(a a – b) (a ( a – b) = 25( 25(a a + b)2 + 8( 8(a a – b)2 Hence D > 0

 a & b.

So, roots are real and unequal. 101. If ,  are the roots of the equation ax 2 – bx + + c = = 0, then equation (a ( a + cy )2 = b2y in in y has has the roots (1)) (1

1 1 ,

(2)) 2, 2 (2

 

(3)) (3

    ,

(4)) (4

1



2

,

1

2

Sol. Answer (4) Since, ,  are the roots of the equation ax 2 – bx + + c = = 0

  So,  

b a

,

 

c a

Now, we have to observe root of the equation (a + cy )2 = b2y + c 2y 2 = b2y  a2 + 2acy +

 c 2y 2 + (2 (2ac ac – – b2)y + + a2 = 0

 2ac  b2  a2 y  0  c 2  c2 



y2  



y2 



y

 b2  c2

2





1

 

2





2a 

a2

c 

c 2

 y 

0

 1 0 y   2 2     1

2

Hence the equation (a ( a + cy )2 = b2y has has roots Aakash Educational Services Pvt. Ltd.

1



2

,

1

2




102. If a, b, c are in G.P., then the equation ax 2 + 2bx bx + + c = 0 and dx 2 + 2ex + f = 0 have a common root if d a

e ,

b

f ,

c

are in

(1) A.P.

(2) G.P.

(3) H.P.

(4) ab = cd

Sol. Answer (1) Since a, b, c are are in G.P., So, b2 = ac

 4b2 – 4ac 4 ac = =0 D = 0 for the equation ax 2 + 2bx + + c = = 0 Hence, it will have equal roots, and root will be x 



b a

Now, ax 2 + 2bx + + c and and dx 2 + 2ex + + f = = 0 have a common root, So,

x 



b

will satisfy the equation

a

dx 2 + 2ex + + f = = 0



d.

b

2

a

2

db



2

b

 2e.  f  0 a

 2aeb  a 2f a

2

0

 db2 – 2aeb + a2f f = =0  dac – – 2aeb aeb + + a2f = = 0  dc dc + + af af = = 2eb d a



So,

f c



d a

2e

b c

,

b

f ,

c

are in A.P.

103. 10 3. The least least integral integral value of k for for which the equation x 2 – 2(k 2( k + + 2) x x + + 12 + k 2 = 0 has two distinct real roots is (1) 0

(2) 2

(3) 3

(4) 4

Sol. Answer (3) The given equation is x 2 – 2( 2(k k + + 2) x x + + 12 + k 2 = 0 has distinct real roots when D > 0 4(k k + + 2)2 – 4(12 + k 2) > 0  4(

 k 2 + 4 + 4k 4k – 12 – k 2 > 0  4k – – 8 > 0  k > 2 So least integral value of k is is 3. Aakash Educational Services Pvt. Ltd.




104. 10 4. The roots roots x x 1 and and x x 2 of the equation x equation x 2 + + px px + + 12 = 0 are such that their difference is 1. Then the positive value of p of p is is (1) 1

(2) 2

(3) 3

(4) 7

Sol. Answer (4) D

–



p

2

a

1

 48  1

 p p = = ± 7; but p p is is positive, hence p = 7. 105. If a < b < c c < < d and and k > > 0, then the quadratic equation ( x x – – a) ( x x – – c ) + k ( x x – – b)( x x – – d) = 0 has (1) All roots real and distinct

(2) All roots real but not necessarily distinct

(3) All root real and negative

(4) May be imaginary

Sol. Answer (1) f ( x x ) = ( x x – – a) ( x – – c ) + k ( x x – – b) ( x – – d ) f (a) = k (a – b) (a ( a – d ) which is positive f (b) = (b – a) (b ( b – c ) which is negative f (c ) = k (c – – b) (c ( c – – d ) which is negative f (d ) = (d – – a) (d ( d – – c ) which is positive So, f ( x x ) = 0 has a root in the interval ( a, b) and another in (c ( c , d ). ). So the roots are real and distinct. 106. If ,  and  be the roots of the equation x 3 – 3 x 2 + 3 x + 7 = 0 and  be non-real cube root of unity, then

   1   1   1   is equal to      1   1   1 

the modulus of the expression  (1) 1

(2) 2

(3) 3

(4) 4

Sol. Answer (3) x 3 – 3 x 2 + 3 x + + 7 = 0 ( x x – – 1)3 + 8 = 0

 ( x – – 1) = –2, –2, –2 2 x = = –1, 1 – 2, 1 – 22

Now,





 1  1  1    1  1  1 2 2 2 2    2   2 2 2 1





1



 2 = |3 2| = 3





107.. If the roots 107 roots of the quadratic equation equation x 2 – ax + + 2b = 0 are prime numbers, then the value of ( a – b) is (1) 0

(2) 2

(3) –2

(4) 4

Sol. Answer (2) x

2

 ax  2b  0 (Here a, b are integers)

Let ,  be roots Now, sum of roots = a product of roots = 2b 2 b (an even number)

 ‘2’ is one root Now, 4 – 2a 2 a + 2b = 0  a – b = 2

108 10 8. Le Lett ,  be the roots of the equation x 2 – px + r = = 0 and and

 2

, 2  be the roots of the equation x 2 – qx + + r = = 0. [IIT-JEE 2007]

Then the value of r is is 2 ( p  q ) ( 2q  p ) 9

(1)) (1

(2)) (2

2 (q  p ) (2 p  q ) 9

(3)) (3

2 (q  2 p ) (2q  p) 9

(4)) (4

2 (2 p  q ) ( 2q  p) 9

Sol. Answer (4) Use relation between roots and coefficients = p p    =

...(i)

  = r

...(ii)

 2

 2 

...(iii)

q

    (2 )  2

r

...(iv)

(ii) and (iv) are same. (i) and (ii) can be solved to obtain  and  in terms of p of p and and q, thereby giving r . 109.. A man walks a distance of 3 units from the origin towards the north-east 109 north-east (N 45° E) direction. From there, he walks a distance of 4 units towards the north-west (N 45° W) direction to reach a point P . Then the position of [IIT-JEE 2007]

P in in the Argand plane is (1) 3ei  / 4 + 4i

(2) (3 – 4i i )e )ei  / 4

(3) (4 + 3i )e )ei  / 4

(4) (3 + 4i i )e )ei  / 4

Sol. Answer (4) Use the idea of rotation to obtain the desired P as as i



3e 4 .e  , where tan = i

yielding sin 

4 5

4 3

 cos  =

3 5




110. If |z | = 1 and z  ± 1, then all the values of


z 2

1  z

[IIT-JEE 2007]

lie on

(1) A line not passing through the origin

(2) | z |  2

(3) The x -axis

(4) The y -axis -axis

Sol. Answer (4) We have, z 2 z

1



1 1

1



zz

z

z



z

 z

z

1 1   z 2iIm ( z)

 1   i  i ,   R  z Im z  Thus the locus of

z

1

2 z

is y -axis. -axis.

111. A parti particle cle P starts starts from the point

z 0

 1  2i , where

i



 1 . It moves first horizontally away from origin by 5

units and then vertically vertically away from from origin by 3 units to reach a point z 1. From z1 the particle moves in the direction of the vector i  j and then it moves through an angle ˆ



ˆ

2

(2) –7 + 6i

units

in anticlockwise direction on a circle [IIT-JEE 2008]

with centre at origin, to reach a point z 2. The point z 2 is given by (1) 6 + 7i

2

(3) 7 + 6i

(4) – 6 + 7i

Sol. Answer (4) 

Z2

Imaginary axis

45, 5  2 si sin 45 45 )  (7, 6)  7  6i  (6  2  cos 45

by rotation about (0, 0) Z 2 (7, 6)

Z 2 Z 2 Z 2 

Z2



e

i

 2

⇒

Z2



Z 2 (e

i

 2

1

) , 2 1 (

)

1 3

90° Z

0

5

    (7 ( 7  6i )  cos  i sin   (7  6 i )( i )  6  7 i  2 2

(6, 2)

Real axis

112 12.. Le Lett z = x = x + + iy be be a complex number where x where x and and y are are integers. Then the area of the rectangle whose vertices are the roots of the equation (1) 48

zz

3

 zz3  350 is

(2) 32


[IIT-JEE 2009] (3) 40

(4) 80




Sol. Answer (1) zz

3

 zz3  350

y

(–4, 3) B

| z |2 ( z 2 )  | z |2 ( z2 )  350

A(4, 3)

| z |2 ( z2  z 2 )  350

x

O

 ( x x 2 + y 2) ( x x 2 – y 2 + 2ixy + x + x 2 – y 2 – 2ixy ) = 350

D (4, –3)

C (–4, 3)

x 2 + y 2)( x x 2 – y 2) = 350  2( x

 ( x x 2 + y 2)( x x 2 – y 2) = 175 = (32 + 42)(42 – 3 2) Which suggests that points ( x , y ) satisfying the given equation are (4, 3), (–4, –3), (–4, 3), (4, –3) Required area = AB = AB × BC =8×6 = 48 sq. units 15

113 13.. Le Lett z = = cos + i sin sin . Then the value of

∑ Im(z

2m 1

m

1

1

(1)) (1

) at  = 2° is

(2)) (2

sin2

[IIT-JEE 2009]

1

3sin2

1

(3)) (3

1

(4)) (4

2sin2

4sin2

Sol. Answer (4) z = = cos + i sin sin = ei  15

Now,

∑ lm e m

i ( 2m 1) 

1

= sin + sin3 + ..... + sin29 15.2 1)  2  2 .sin  2  (15  1)    2  2  = sin    2 sin

=

sin15 sin15.sin15 .sin15 sin 

For  = 2°, the given expression reduces to =

sin 30. sin 30 sin 2



1 4 sin 2

114 14.. Le Lett p p and and q be real numbers such that p  0, p3  q and and p p3  –q –q.. If  and  are nonzero complex numbers satisfying  +  = – p and 3 + 3 = q, then a quadratic equation having

 

and

 as its roots is [IIT-JEE 2010] 

(1) ( p p3 + q) x x 2 – ( p p3 + 2q) x x + + ( p p3 + q) = 0

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

(3) ( p p3 – q) x x 2 – (5 p3 – 2q) x x + + ( p p3 – q) = 0

(4) ( p p3 – q) x x 2 – (5 p3 + 2q) x x + + ( p p3 – q) = 0





Sol. Answer (2) We have  +  = – p p(() 3 + 3 = q = (  + )3 – 3( + ) = – p3 + 3 p p

  

3

q

3 p

The quadratic equation with



x



x

      

2



2



x

2



2

2

 p

3

p

3

and

 as roots is 

   0  

 (   )2  2     

p



x

 

x

1 0

q

3 p

q

x  1  0

3 p

 ( p3 + q) x x 2 – ( p p3 – 2q) x x + + ( p p3 + q) = 0

115 15.. Le Lett  and  be the roots of x 2 – 6 x – – 2 = 0, with  > . If an = n – n for n  1, then the value of

a10

 2a8

2a9

[IIT-JEE 2011] (1) 1

(2) 2

(3) 3

(4) 4

Sol. Answer (3) We observe that ( + ) ( n –1 – n –1) = n – n + (n – 2 – n–2)

 6an – 1 = an – 2an – 

6



a

n

 2a

n

a

n

 3

a

n

2

2

1

 2a

n

2an 1

2

,n 2

Putting n = 10, we get a 10

 2a8

2a9

3





116 16.. A value value of b for which the equations x 2 + bx – – 1 = 0 x 2 + x x + + b = 0, [IIT-JEE 2011]

have one root in common is (1))  (1

(2)) i (2

2

3

(3)) (3

i 5

2

(4)) (4

Sol. Answer (2) Let  be a common root between given equations x 2 + bx – – 1 = 0 and and x x 2 + x + + b = 0

2



b

2

1

  

2

b

2

1   1  b 1  b

2

1 1 b

b

2

b



 1  1 b      1  b  1 b



 1

 1 b   1 b

   

and

2

(1 b )2 1 b

 b2 – b3 + 1 – b = 1 + 2b 2 b + b2  b3 + 3b = 0  b  0, b   3i 



b

3i

117 17.. Let Let z be a complex number such that the imaginary part of z is nonzero and a = z 2 + z +1 is real. Then a cannot take cannot [IIT-JEE 2012] take the value (1) –1

(2)

1 3

(3)) (3

1 2

(4)) (4

3 4

Sol. Answer (4) As, a is real, So

a



a

gives

 z 2 + z + 1 =  ( z  z )( z  As,

z



So,

z

 z  1

z

2

 z  1

 1)  0

z

z

 x  

1 { where z  x  iy } 2





Now, a = z 2 + z z + +1 2

 1   1  =    iy      iy   1  2   2 

=

3 4

 y 2

As, y  0 so,

a



3 4

118 18.. The quadratic quadratic equation equation p p(( x x ) = 0 with real coefficients has purely imaginary roots. Then the equation

p(( p( p p( x x )) )) = 0 has

[JEE(Advanced)-2014]

(1) Only purely imaginary roots

(2) All real roots

(3) Two real and two purely imaginary roots

(4) Neither real nor purely imaginary roots

Sol. Answer (4) Let p Let p(( x x ) = x 2 + a ( (a a > 0)

(∵ roots are purely imaginary)

p(( p( p p( x x )) )) = ( x x 2 + a)2 + a (a  R ) x 4 + 2a( x x 2) + a2 + a = 0

 x 2 =

= a 

2a 

4a

2



2

4a



4a

2

ai

x =  a  ai  x1  iy 1

SECTION - B

Objective Type Questions (More than one options are correct) 1.

If |3z – – 1| = 3|z 3| z – – 2|, then z lies lies on (1) 6Re(z ) = 7

 1



 3



(2)) On the perpen (2 perpendicul dicular ar bisector bisector of line line joining joining  , 0  and (2, 0) (3) A line line para parall llel el to to x -axis -axis (4) A line line para parall llel el to to y -axis -axis Sol. Answer (1, 2, 4) If |3z |3z – – 1| = 3 |z |z – – 2| Let z z = = x x + + iy Aakash Educational Services Pvt. Ltd.




|3 x + + 3iy – – 1| = 3| x x + + iy – – 2|

 (3 x – – 1)2 + (3 (3y y )2 = 9 [( x x – – 2)2 + y 2]  9 x 2 + 1 – 6 x + + 9y 2 = 9 x 2 + 9y 2 – 36 x + + 36 = 35  30 x = 7

 x x = =

6

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

1



7



Also mid-point of  , 0  and (2, 0) is  , 0  3  6 

10

2.

If

S 

(1)) (1

2k   2k i  sin cos   , then ∑  11 11  k 1

S  S

(2)) (2

0

S S

1

(3)) (3

S



1 2

(1 i )

(4)) (4

S  S

0

Sol. Answer (1, 2, 3) We can write

 

= i cos

2k 11

sin

2k 11

 i cos

2k 11

2k 

 i sin

11

 

Now No w 10

S





∑  sin

2k 11

k 1

10

S

 i

∑e

i

 i cos

2k  11



2 k 11

k 1

i 4 i 20   i 2   11 11 11  e  ....  e  1  1  i (as 1 +  + 2 + ..... + 10 = 0) = i  e   

 S = i 

S



S  S

and an d

S S



S



S

 i



0

1 1

1  i 2



1 2

2



 2i

(1  i )




3.


Let cos A A + cos B + cos C = = 0 and sin A A + sin B + sin C = 0 then which of the following statement(s) is/ are correct ? (1))  cos( 2 A  B  C )  3 (1

(2))  cos( 2 A  B  C )  0 (2

(3))  sin(2 A  B  C )  0 (3

(4))  sin(2 A  B  C )  3 (4

Sol. Answer (1, 3) co s A  cos B sin A  sin B

Let

a

 cos C  0

 sin C 

0

 cos A  i sin A  e iA

b

 cos B  i sin B  e iB

c

 cos C  i sinC  e iC

a

 b  c  0 a

∵

 b3  c 3  3abc

a



2

e e

b



bc



e

3

2

ac

2



2 iA

iB

.e

iC iC

2iA  iB  iC

c

ab

e



e

3

2 iB

iA iC e



e e

2 iC

iA iB e

3

 e 2 iB  iA iC  e 2 iC  iA iB  3

cos(2 A  B  C )  i sin(2A  B  C )  cos(2B  A  C )  i sin(2B  A  C )  cos(2C  A  B )  i sin(2C  A  B )  3 On comparing real and imaginary part cos(2 A  B  C )  cos(2B  A  C )  cos(2C  A  B )  3 sin(2 A  B  C )  sin(2B  A  C )  sin(2C  A  B )  0 4.

The Th e eq equa uati tion on wh whos ose e ro root ots s ar are e nth power of the roots of the equation, x 2 – 2 x cos cos  + 1 = 0 is given by (1) ( x x + cos n)2 + sin2 n  = 0

(2) ( x x – cos n)2 + sin2 n  = 0

(3)) x 2 + 2 x (3 x cos cosn n + 1 = 0

(4) x 2 – 2 x x cos cos n + 1 = 0

Sol. Answer (2, 4)

 2

x –

2 x cos + 1 = 0



x



2 cos  

4 cos

2

4

2

= cos  i sin sin Aakash Educational Services Pvt. Ltd.




 = cos + i sin sin,  = cos  – i sin sin n

n

   cos n  i sin n ;   cos n  i sin n S

n

n

P = = 1      2cos n ; P

Equation is i.e.,, i.e.

x

2

x

2

 Sx  P  0

 2x cos n  1  0

 Option (4) x

2

 2x cos n  cos2 n  1  cos2 n  0

 ( x  cos n)2  si s in2 n  0  Option (2) 5.

If a, b, c are are real numbers and z is is a complex number such that, a2 + b2 + c 2 = 1 and b + ic = (1 + a)z , then 1  iz 1  iz

equals.

 ic 1  ia

 ib 1  c

b

(1)) (1

a

(2)) (2

(3)) (3

1  c

a

 ib

(4)) (4

1 a

b

 ic

Sol. Answer (2, 3) a2 + b2 + c 2 = 1 b +ic = (1 + a) z





z 

b  ic 1 a

1  iz 1  iz

 b  ic   1  a  1  iz   1  iz  b  ic  1  i   1  a  1  i 



 c  ib 1  a  c  ib



(1  a  c  ib ) (1  a  c  ib )  (1  a  c  ib ) (1  a  c  ib )



1 a

(1 a  ib )2  c 2 2

(1 a  c )  b

2



a  ib 1  c

Similarly, 1 a

 c  ib 1  a  c  ib  = 1  a  c  ib 1  a  c  ib

1  c

a  ib




6.


If z 1, z 2, z 3, z 4 are the four complex numbers represented by the vertices of a quadrilateral taken in order such

 z 4  z 1     , then the quadrilateral is a that z 1 – z 4 = z 2 – z 3 and amp  z 2  z 1   2 (1) Rhombus

(2) Square

Sol. Answer (3, 4) z 4

D

(3) Rectangle C

 z4  z 1     z2  z 1  2

(4) A cyclic quadrilateral

z 3

amp 

90°

B

A z 1

A = = 90°  A

z 2

also z 1 – z 4 = z 2 – z 3  |z 1 – z 4| = |z 2 – z 3|

 AD AD = = BC and AD and AD || || BC So AB So AB || || CD CD and and AB AB = = CD ABCD is is a rectangle or cyclic quadrilateral  ABCD

7.

If ,   are are cube roots of p of p < < 0, then for any x , y , z ,

(1) 1

(2)

 

 2 x 2   2 y 2   2 z 2  2 x 2   2 y 2   2 z 2 (3)) (3

 

is

(4)) (4

 

Sol. Answer (2, 3, 4) Cube roots of p of p are are

i.e,,   p i.e

Now,

==

p

1

1

1

3

3

3

,

p



,

1

1

1

3

3

3

,



p



.

,

 2 x 2  2 y 2   2 z 2 2 x 2   2 y 2   2 z 2

p



p

2

2

.

x

2

 2 y 2  4 z 2

2 x 2  4 y 2  z 2



x

2

 2y 2  z 2

2 (x 2  2y 2  z 2 )

  Option (4) 

We can assign the , ,  different different value we get other options also. 8.

If z is is a complex number satisfying z z + + z –1 = 1, then z n + z – n, n  N , has the value (1) 2(–1)n, when n i is s a multiple of 3

(2) (–1)n – 1, when n is not a multiple of 3

(3) (–1)n + 1, when n i is s a multiple of 3

(4) 0 when n is not a multiple of 3

Sol. Answer (1, 2) z + z + z –1 = 1

z



1 z

1

z 2 – z + + 1 = 0 Aakash Educational Services Pvt. Ltd.




z 

1  i 3 2

 ,  2

z n + z –n = (–)n + (–)–n Case (1), n = 3m –3m m (–)3m + (– )–3 = (–1)n + (–1)n = 2(–1)n

when n = 3m + 1 (–)3m

–3m m – 1 + (– )–3

+1

3 = ( 1)  n

m

1

 

n = ( 1)   

n



( 1) ( )3

m

1

1  

= (–1)n (–1) = (–1) n + 1 9.

If z satisfies satisfies |z | z – – 1| < |z | z + 3|, then  = 2z + + 3 – i satisfies satisfies (1)) |   5  i |  |   3  i | (1

(2)) |   5 ||   3 | (2

(3)) Im (i )  1 (3

(4)) | arg(   1) |  (4

 2

Sol. Answer (2, 3, 4) |z  1| < |z |z + + 3| Let z z = = x x + + iy ( x x  1) 2 + y 2 < ( x x + + 3)2 + y2 x 2 + y 2  2 x + + 1 < ( x x 2 + y 2 + 6 x + + 9)

 8 x > –8 x > > – 1 i = i ((2 ((2 x + + i 2y ) + 3 – i ) = i 2 x x – – 2y 2 y + + 3i + + 1 = i (3 (3 + 2 x ) + (1 – 2y 2 y ) As 3 + 2 x x > > 1  option Also,  – 1 = 2z 2 z + + 3 – i i – –1 = 2z 2 z + + 2 – I = 2 x x + + 2iy 2 iy + + 2 – i = 2( x x + + 1) + i (2y (2y – – 1) As x > > – 1 Aakash Educational Services Pvt. Ltd.



 2( x + + 1) > 0  arg (  1) 


 2

 option (4)  – 5 = 2( x – – 1) + i (2y (2y – – 1) + 3) + i (2y (2y – – 1) as x x > > –1  + 3 = 2( x +

 |  + 3| > |  – 5|  Option (2) 10. If | z   |2  | z |2  |  |2 , where z and and  are complex numbers, then z

(1)) (1



is i s purely real

z

(2)



is pur purel ely y imag imagin inar ary y (3)

 z       2

(4)) amp (4

  z   0

z

Sol. Answer (2, 3, 4) |z + + |2 = |z |2 + ||2 Since |z |z + + |2 = |z |2 + | |2 +

z

|z z |2 + | |2 = |z |2 + | |2 +  |

  z 





z

  z

z





z





z

  z z

  z  0

z



is purely imaginary

 z    2

Therefore, amp  

11.

Let z 1, z 2 be two complex numbers represented by points on the circle | z 1| = 1 and |z | z 2| = 2 respectively then (1) min |z 1 – z 2| = 1

(2) max |2z 1 + z 2| = 4

(3)

z 2



1

3

(4) min | z 1 – z 2 | = 2

z 1

Sol. Answer (1, 2, 3)

y

Clearly |z |z 1 – z 2|min = 2 – 1 = 1 |z 1 – z 2|max = 3 max|2z max|2 z 1 + z 2| = |2 + 2| = 4

1 O

Now, |z |z | = 1 1

|z 1| = 1 z .z 1 1

z 1



x 2

|z | = 2

1 1

|z | = 1

z 1







z 2

Now,

1



z 1

z 2

 z 1

and | z2  z1 |  | z2 |  | z1 |

 3.

12.. Le 12 Lett comp comple lex x numb number er z satisfy satisfy (1) 1

z



2

 1 , then |z |z | can take all values except

z

(2) 2

(3) 3

(4) 4

Sol. Answer (3, 4)

z



2 z

| z | 



1 2

| z |



|

1

|



2

1

| z |2  | z | 2  0 (| z | 2)(| z | 1) 1)  0

 1  | z |  2 Butt |z Bu |z||  0

 0  | z |  2  | z |  3 and |z |z |  4  Option (3) and (4)

13. If z = = x + + iy , then the equation

(1)) (1

m



1 2

2z – i z  1

 m represent a circle when

(2)) m = 1 (2

(3) m = 2

(4) m = 3

Sol. Answer (1, 2, 4) The given equation is |2z |2 z – – i | = m|z + + i |

 (2z  i )(2z  i )  m2 ( z  i )( z  i )  4z z  2iz  2i z  1  m2 (z z  iz  i z  1)  (4  m2 )z z  (2  m2 )iz  (2  m2 )i z  (1  m2 )  0 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) Aakash Educational Services Pvt. Ltd.



14. If

3

–1  – 1 ,


– , – 2, then roots of the equation ( x ( x + + 1)3 + 64 = 0 are (2) – 1 – 42

(1) – 1 – 4

(3) – 5

(4) – 4

Sol. Answer (1, 2, 3) ( x x + + 1)3 = (– 4) 3

 x + + 1 = – 4, – 4 , – 42  x = = – 5, – 1 – 4, – 1 – 42. Hence, roots are – 5, – 1 – 4 , – 1 – 4 2. 15. If z 1 = a + ib ib and z 2 = c + id are complex numbers such that | z 1| = |z 2| = 1 and Re( z1 z 2 )  0 , then the pair of complex numbers 1 = a + ic and and 2 = b + id satisfy satisfy (1) |1| = 1

(2) | 2 | = 1

(3) Re(1 2 )  0

(4) | 1 | = 2| 2 |

Sol. Answer (1, 2, 3) 2

 b2  1  a2 + b2 = 1

…(i)

2

 d 2  1  c 2 + d 2 = 1

…(ii)

| z1 | 

a

| z2 | 

c

Re( z1 z 2 ) = Re[(a Re[(a + ib ib)( )(c c – – id )] )] = Re[ac Re[ac + + bd bd + + i (bc bc – – ad )] )] = 0

 ac ac + + bd = 0

…(iii)

Now, using (i) & (iii) we can prove that b = c , a = d . Hence, | 1 | 

a

2

 c2 

a

2

 b2  1

Similary we can observe, | 2| = 1 Re(1  2 )  0 16.. If a com 16 compl plex ex nu numb mber er z satisfies satisfies log

 | z |2  2 | z |  6  , then locus/region of the point represented by z 1   2 | z |2 – 2 | z |  1   0  2 

is (1) |z | = 5

(2) |z | < 5

(3) |z | > 1

(4) 2 < |z | < 3

Sol. Answer (2, 4) Given inequality is, log

| z |2 2 | z |  6 1 2



2 | z |2  2 | z | 1

| z |2  2 | z |  6 2 | z |2 – 2 | z |  1

0

1

2|z z | + 6 > 2|z 2| z |2 – 2| 2|z z | + 1  |z |2 + 2|

 | |z z |2 – 4| 4|z z | – 5 < 0  |z |  (– 1, 5), but |z | z | > 0 

0 < |z | < 5





17. If z 1, z 2 be two complex numbers satisfying the equation

(1)) (1

 z1  z 1 – z 2  z2 

(2))  (2

 z2 z1  1

z z 1 2

(3)) (3

z 1

 z 2

z 1

–

z 2

z z 1 2

 1 , then

 z2 z1  0

(4)) Re( z1 z 2 )  0 (4

Sol. Answer (2, 3, 4) z 1, z 2 are the complex numbers satisfying, z 1

 z 2

z 1

–

z 2

1

 | |z z 1 + z 2| = |z 1 – z 2|  ( z1  z2 )( z1  z2 )  ( z1 – 

z z 1 1

z 2

)( z1 –

 z1 z2  z2 z1  z2 z2 

z z 1 1

)

z 2

–

z z 1 2

–

z z 2 1



z z 2 2

 2( z1 z2  z2 z1 )  0

 z1  z 1   z   – z  2 2  

z z 1 2

 z2 z1  0

z1 z2

 z1 z2  0

 Re( z z 2 )  0 18. If sin , cos  are the roots of the equation ax 2 + bx + c = = 0 (c ( c  0), then (1)) a2 – b2+ 2ac = (1 = 0

(2) (a (a + c )2 = b2 + c 2

(3)) (3

b a

[ 2 , 2 ]

(4)) (4

c a

 1 1   2 2

  ,

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



sin 



sin sin .cos .cos 



sin

2

 c os   



b a

c a

  cos2   1 2

2c  b 1 ⇒      a a



a

2

b

2

 2ca  a2

 b2  2ac  0

Also (a  c )2  b 2  c 2 . Aakash Educational Services Pvt. Ltd.




19. Let a , b, c be real numbers in G.P. such that a and c are positive, then the roots of the equation ax 2 + bx + bx + c c = =0 (1)) Are real (1 real and and are in in the rat ratio io b : ac (2) Ar Are e rea reall (3)) Are imagin (3 imaginary ary and and are are in ratio 1 :  where  is a non-real complex cubic root of constant (4)) Are imagina (4 imaginary ry and and are in the ratio 2 : 1 with usual notation Sol. Answer (3, 4) ax

2

 bx  c  0

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

Now,



ax

x

2

 arx  ar 2  0

 rx  r 2  0 

x

 1 

 r 



3i 

2

=  r or or  2 r   x = 

 Roots are imaginary and are in the ratio 1 :  or 2 : 1. 20. Let cos be a root of the equation 25 25 x x 2 + 5 x – 12 = 0, –1 < x < < 0, then the value of sin2  is 20

(1)) (1

(2))  (2

25

12

(3)) (3

25

24 25

(4))  (4

24 25

Sol. Answer (3, 4)

25 x 2  5 x  12  0



x



3 5

,

x



 cos  = 



sin 



3 5

⇒

(5 x  3)(5 x  4)  0

3  as  1   x  5

4 5

2

  0 

4 5

or 

3 5

sin 2  2 sin  cos   2 

or

x

 3   4        5 5

3

4 24      5 5 25

24 25

21.. If the qua 21 quadra dratic tic equ equati ations ons x x 2 + pqx + r = = 0 and x and x 2 + prx + q = q = 0 have a common root then the equation containing their other roots is/are (1)) x 2 – (1 – p p((q + r ) x x + p + p2qr = = 0

(2) x 2 p p((q + r ) + (q + r ) x x – pqr – pqr = = 0

(3)) p (3 p((q + r ) x x 2 – (q + r ) x x + pqr = = 0

(4) x 2 + + p p((q + r ) x x – p – p2qr = = 0

Sol. Answer (1, 2) Let  be common root

 2  pq  r  0,

     pq,  

 2  pr   q  0,

     pr ,   q


r




 p (q  r )  r  q  0

⇒

1 p

1

Common root is  = Other roots are,  



p

  qp

rp and

 Equation containing other roots is x

2

 p (r  q ) x  p 2rq  0 1

∵

p

1 ⇒

p

2

Now,

2

is common root

⇒

 1  1  p   pq  p   r  0

  (q  r )

x

2



⇒

 p  

⇒

p (q



 p (q  r ) x  p 2qr  0 p p

2

x

2



 (q  r ) x  pqr   0



 r ) x 2  (q  r ) x  pqr  0

22.. Th 22 The e quad quadra rati tic c equa equati tion on x x 2 – (m ( m – 3) x + m = m = 0 has (1)) Real disti (1 distinct nct roots roots if and only if m  (–, 1)  (9, ) (2)) Both posit (2 positive ive roots roots ifif and only if m  (9, ) (3)) Both nega (3 negative tive roots roots if and and only if m  (0, 1) (4) No ro roo ots Sol. Answer (1, 2, 3) x

2

 (m  3 )x  m  0

For real distinct roots, ( m  3)2  4m  0



m

2

 10m  9  0

 (m  9)(m  1)  0

⇒

m

 ( , 1)  (9, 

…(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 From (1) and (3),

m

… (3)

 (0, 1)





23.. If both 23 both roo roots ts of of the the equat equation ion x x 2 – 2ax 2 ax + + a2 – 1 = 0 lie between –3 and 4, then [ a] is/are, where [ ] represents represents the greatest integer function (1) 1

(2) –1

(3) 2

(4) 0

Sol. Answer (1, 2, 3, 4) x

2

 2ax  a 2  1  0

x = = a + 1, a – 1. ( x  a)2  1  x Now, 3  [a  1]  4 and 3  [a  1]  4

 3  [a]  1  4 and 3  [a]  1  4  4  [a]  3 and 2  [a]  5  2  [a]  3  [a] = –2, –1, 0, 1, 2, 3. 24. Let ,  be the roots of x of x 2 – 4 x + A = A = 0 and  ,  be the roots of x of x 2 – 36 x + + B = 0. If     forms an increasing G.P. Then (1)) B = 81 A (1

(2)) A (2 A = = 3

(3) B = 243

(4) A + B = 251


    4 ,   A  +  = 36,  = B Let , ,  ,  be a, ar , ar 2, ar 3 a + ar = = 4

ar

 A

2

1  r

 ar 3  36 

r

2

r

2

(1 r )



1 9

= ± 3, a = 1.  9  r =

     a(ar ) = A  A = 3

B =   ( ar 2) (ar 3)  B = 243  B = 81 A A..   B = (ar

25. Fo Forr the the equ equat atio ion n

3 5 (log2 x )2 log2 x  4 4 x



2

, which one of the following is true?

(1) Has at least one real solution

(2) Has exactly three real solutions

(3) Has exactly one irrational solutions

(4) Has non-real complex roots

Sol. Answer (1, 2, 3) 3 5 log2 x 2  log2 x  4 4 x



2

Taking log with base 2 on both side. 5 1 3 2 l og2 x   log2 x  log2 2   4 (log2 x )  lo 4 2 

3

Put log2 x x = = t ,  

4

t

2

5  t   t  4

1 2







3t

3

2  4t 2  5t  2  0  (t  1)(3t  7t  2)  0

 (t  1)(3t  1)(t  2)  0



 1, t 

1 1  ,  2  log2 x   1,  ,  2  x = = 2, 21/ 3 , 2 2 3 3

26. If f ( x x ) = ax 2 + bx + + c , g ( x x ) = –ax 2 + bx + + c, c, where where ac ≠ 0 then f ( x x ).g ).g ( x x ) = 0 has (1) At least three real roots

(2) No real roots

(3) At least two real roots

(4) At most two imaginary roots

Sol. Answer (3, 4) Let D1 and D2 be the respective discriminates. discriminates. then D1 = b2 – 4ac

and D2 = b2 + 4ac

Adding we get D1 + D2 = 2b2 ∵

D1 + D2 is positive

 at least one of D1 or D2 is positive.  at least least 2 real roots. 27. The va valu lue e of of a for which the equation x 2 + ax + + a2 + 6a < 0 is satisfied for all x  (1, 2)

 –7 – 4 5

(1))  (1



2

 –7– 4 5

(3))  (3



2

 –7 – 4 5



(2))  (2

, – 2 

2





 –7  4 5



(4))  (4

, – 42 3 

2










, – 3 





, 4 3 3  

1

 2



Let ,  be the roots of the corresponding equation x 2 + ax + + a2 + 6a = 0

…(i)

As the coefficient of x 2 = 1 > 0 0 x x 2 + ax + + a2 + 6 x < < 0 will be satisfied for all values of x  ( , ) if ,  are real and unequal (let  < ). Hence the inequality will hold for all real x real x  (1, 2) if the interval (1, 2) is a subject of the interval ( , ). Thus for (1) we should have D > 0 and  < 1,  > 1 as well as  < 2,  > 2. Now, D > 0  a2 – 4( 4(a a2 + 6a) > 0

 a2 + 8a < 0 (– – 8, 0)  a  (

…(ii)

 < 1,  > 1   – 1 < 0,  – 1 >

0

( – 1) ( – 1) < 0

 a2 + 7a + 1 < 0 Aakash Educational Services Pvt. Ltd.



a

 –7 – 45 –7  45  ,  2 2  




…(iii)

< 2,  > 2  (  – 2) ( – 2) < 0

α

 a2 + 8a 8 a + 4 < 0 

a

…(iv)

(– 4 – 2 3 , – 4  2 3 )

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

a

 –7 – 45







, – 4  2 3

…(v)



2

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

28.. If the roo 28 roots ts of the equ equati ation on

(1)) p + q = r (1

1

x  p



1

x q



1

r

(2)) p + q = 2r (2

are equal in magnitude but opposite in sign and its product is 

2

(3))   (3

p

2

 q2 2

 p 2  q 2   2  

(4)     (4)

Sol. Answer (2, 3) The given equation is 1

x  p





1

x q



1

r

x  q  x  p

( x  p )( x  q )



1 r

 (2 x + + p p + + q)r = = x 2 + ( p + p + q) x x + + pq  x 2 + ( p + p + q – 2r ) x x + + ( pq – pq – qr – – rp rp)) = 0 According to the question the given equation has roots equal in magnitude but opposite in sign, hence Coefficient of x of x = = 0 p + + q – 2r r = =0  p

r = =  r

p  q 2

Product of roots = + [( p + p + q)r – – pq pq]]

( p  q )2 – pq = 2

=

1 2 ( p  q 2 ) 2





29.. Th 29 The e inte integr gral al val value ues s of a for which the equation (a ( a + 2) x 2 + 2(a 2( a + 1) x + + a = 0 will have both roots integers (1) 0

(2) – 1

(3) – 2

(4) – 3

Sol. Answer (1, 2, 4) (a + 2) 2) x x 2 + 2( 2(a a + 1) x x + + a = 0 Let ,  be roots

2  a  1

 

a

 

a

2

a

a

 2

(integer)

a2

(integer)

will integer if

For a = 0, a = –1, a = –3 Also for a = 0, a = –1, a = –3 –2

 

 a  1

a

is integer

2

30. If ( x x – – 1)2 is a factor of ax 3 + bx 2 + c , then roots of the equation cx 3 + bx + + a = 0 may be (1) 1

(2) – 1

(3) – 2

(4) 0

Sol. Answer (1, 3) Since 1 is the repeated roots of ax 3 + bx 2 + c c = =0 So, 1 + 1 +  = –

b a

1.1 +  +  = 0    –

–

1.1. =

b a



–

b c

c a

–

1 2

c ⇒

a



1 2

1 2

3 2

 –3

Now, by the equation, cx 3 + bx bx + +a=0



x

3



b c

x 

a c

0

 x 3 – 3 x + + 2 = 0 x 3 – x 2 + x 2 – x – – 2 x + + 2 = = 0 Aakash Educational Services Pvt. Ltd.




 x 2( x x – – 1) + x ( x x – – 1) – 2( x – – 1) = 0  ( x x – – 1) ( x 2 + x – – 2) = 0  ( x x – – 1) ( x 2 + 2 x – – x – – 2) = 0 x – – 1) ( x x – – 1) ( x + + 2) = 0  ( x = + 1, – 2  x = Hence answer is (1), (3) 31. If b2  4 4ac ac for for the equation ax 4 + bx 2 + c = = 0 then all roots of the equation will be real if (1)) b > 0, a < 0, c > 0 (1

(2) b > 0, a > 0, c c > >0

(3)) b < 0, a > 0, c > 0 (3

(4) b > 0, a < 0, c c < < 0

Sol. Answer (3, 4) Let x Let x 2 = y So the equation ay 2 + by by + + c = = 0 should have both roots non-negative in order to all roots of the equation 4 2 ax + bx + c = = 0 are real for this

–

 

c a

b a

0

b ⇒

a

0

0

…(i)

…(ii)

From (i) and (ii) b >

0, a < 0, c < 0

or b < 0, a > 0, c c > >0 32.. The diffe 32 difference rence betwe between en the the roots roots of of the equat equation ion x 2 + kx + + 1 = 0 is less than values of k k is is (1) (– (– 3, 0)

(2) (0 (0, 3)

(3) (– (– 3, 3)

5

, then the set of possible

(4) (3 (3, )


– 

5

 ( – )2 < 5  ( + )2 – 4 < 5  k 2 – 4 < 5  k 2 < 9  k  (– 3, 3) Hence answers is (1, 2, 3) 33.. Th 33 The e set set of rea reall valu values es of of a for which a2 + 2a, 2a 2 a + 3 and a2 + 3a + 8 are the sides of a triangle may be

 

(1))  6, (1

13 



2 

(2) (5 (5, 7)

(3) (5 (5, )

(4) (0, 5)

Sol. 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 2 a + 2a 2 a + 3 > a2 + 3a 3 a + 8  4 4a a > 3a 3 a + 5  a > 5 a2 + 2a + a2 + 3a + 8 > 2a 2a + 3  2 2a a2 + 3a + 5 > 0  a  R Aakash Educational Services Pvt. Ltd.




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

–

11 3

Combining these three, a  (5, ) Hence answer is (1, 2, 3) 34. Let z 1 and z 2 be two distinct complex numbers and let z = (1 – t ) z 1 + tz 2 for some real number t with [IIT-JEE 2010] 0 < t < < 1. If arg (w) denotes the principal argument of a non-zero complex number w, then (1) |z – – z 1| + |z – z 2| = |z 1 – z 2|

 z1

z

(3)) (3

z

 z1

z 2

 z1

 z1

z 2

(2) Arg (z – – z 1) = Arg (z (z – – z 2)

0

(4) Arg (z – – z 1) = Arg (z (z 2 – z 1)

Sol. Answer (1, 3, 4) z

z1

t

z 2

 1  t

we have z = = (1 – t )z 1 + tz 2, 0 < t t < < 1,



z 



z

 1  t  z1

t z2 t



t z2

 1  t

 1  t  z1

z 1, z , z 2 are collinear so Arg(z Arg(z – – z 1) = Arg(z Arg(z – – z 2) = Arg(z Arg(z 2 – z 1) 3

35. Let w = =

 i

2

 

and P = = {w n : n = 1, 2, 3, …}. Further H 1 = z   : Re z 

 

H 2 = z   : Re z 

1

 and

2

 1 2

, 

where  is the set of all complex numbers. If z 1  P  H 1, z 2  P  H 2 and O represents the origin, then [JEE(Advanced) 2013] z 1Oz 2 = (1)) (1



(2)) (2

2



(3)) (3

6

2

(4)) (4

3

5 6

Sol. Answer (3, 4) Note that | | = 1

i are possible value of z 1



i are possible value of z 2 (i = = 1,2,3)





1

30° 2

30°





3 2

i

e



30°

1



30°

2

i

2



3



3

 6




2

  3

  4

  5

 

i

e

i

e

 3

 2

2i

e

i

e


 3

5 6

So, z1oz 2 can be

⇒

2 5  , 3 6

36. Let S be the set of all non-zero real numbers  such that the quadratic equation  x 2 – x x + +  = 0 has two distinct real roots x 1 and x 2 satisfying the inequality | x x 1 – x 2| < 1. Which of the following intervals is(are) a [JEE(Advanced) 2015] subset(s) of S ?



1



2

(1))  – (1

1 

,–



1



5

(2))  – (2

 5



,0 





(3))  0, (3



1 

 5

 1 1 ,  2 5 

(4))  (4

Sol. Answer Sol. Answer (1, 4) +  = 0  x 2 – x + |x 1 – x – x x 2| < 1 1 – 4

2

1



1 – 4 2  |  | 1 – 42 < 2

2

–

1 5

0



1 

 1  ,    5  5 

  –  –



Also Al so,, 1 – 42  0

2

1

–

4

  – 

0

1

  2

1

0

2

 1 1   2 2

 – ,

Taking intersection



1



2

 –

, –

1 

 1 1 ,    5   5 2





SECTION - C

Linked Comprehension Type Questions Comprehension I Let z 1, ,  be complex numbers of which  and  are constants and z 1 varies. If z 2 is given in terms of z 1 by one of the following equations, it is required to find z 2 corresponding to z 1, then 1.

The co cons nstru truct ction ion ind indica icate ted d in the the diag diagram ram is give given n by z 2

Y

z 1

 X

O

(1)) (1

z 2

 e i  .z 1

(3)) (3

z 2



(2) z2 = (cos – i i sin sin)z1

cos   i sin 

(4)) (4

z 1

z 2



| z 2 | | z 1 |

(cos   i sin ) z 1

Sol. Answer (4) From rotation formula z2

 0  ( z1  0).

OB

e

B z 2

Y

i 

OA

A

 z 2

2.



.

z 1

z 2 z 1

(cos   sin )

z 1

X

O

In th the gi given fifigure Y z 2



z 1

 X

O

(1)) z 2 = z 1 –  (1

(2)) (2

z 2



1

z 1 

(3) z2 =   z 1

Sol. Answer (4)

D

In  OAB OAB,,

E(z )

 = |  | ei In  OFE Using rotation formula

2

1

F z 1

1

z

(4)) z 2 = z 1 +  (4

β C B

α

θ θ

O


1 A



z 1 = z 1ei .


OE OF

 OFE OFE ~ ~  OAB 

OE OF



OB OA

z 1 = z 1ei . |  | = z 1 

 z 2  z 1  +  3.

The Th e giv given en fi figu gure re il illu lust stra rate tes s Y z 1

X

1

O

z 2

(1)) z 2 = 1 + z 1 (1

(2)) z 2 = 2z 1 (2

(3)) (3

z 2



1 z 1

(4)) (4

z 2



1 2

z 1

Sol. Answer (3) From rotation formula z1



z2



OB OA

.(1 .(1  0). 0).e i 

OC OA

…(i)

.(1 .(1  0). 0).e  i 

…(ii)



z 1z 2 = 1

z 2



1 z 1

Comprehension II If x If x is is the root of the equation x 2 – ix – – 1 = 0, then The value of x 51 is

1.

(1) 1

2.

(2) –1

The value of (1) –1

3.

2013 x

–

(1) –1

20 x



1 20 x

1

(4) – i

(3) i

(4) – i

(3) –2i

(4) – i

may be (2) 1

2013 x

(3) i

may be equal to (2) 1





Solution of Comprehension II Given equation is is x x 2 – ix – – 1 = 0

 x 2 – 1 = ix 

x –

1 x

 i

…(i)

Now let x let x = = cos  + i sin sin, then equation (i) becomes, cos + i sin sin – cos + i sin sin = i

 2sin = 1  sin  =   1.



or

6

1 2

5 6

Answer (3) We have to find out the value of x 51

     cos  i sin   6 6  cos

17

 i sin

2

51

17 2

= 0 + i = i 2.

Answer (1) 20 x

We have to evaluate

 cos

20

 i sin

6

 2cos

10

20 6

 2cos

3



 cos

1 20 x

20  6

– i sin

20  6

4 3

   2cos      3  – 2c 2cos 3.

 3

 –1

Answer (3) For finding

 cos

2013 x

2013

 2i sin  2i sin  2i sin

6



 i sin

1 2013 x

2013 6

– cos

2013  6

 i sin

2013  6

2013 6 671 2

 2

 i





Comprehension III If (1 + + x x )n = a0 + a1 x + + a2 x 2 + ......+ an x n, then 1.

Find Fi nd th the e su sum m of th the e se seri ries es a0 + a2 + a4 + ...... (1) 2n

2.

(4) 2n – 2

(3) 2

Vallue of Va of th the se series a0 – a2 + a4 – a6 + ......... is

(1)) (1

3.

(2) 2n – 1

n

2 cos

n



(2)) (2

4

n

2

–1

sin

n



(3)) (3

4

n

2

–1

cos

n

n



4

(4)) (4

2 2 cos

(4)) (4

2



n

4

The su sum m of th the e seri rie es a0 + a4 + a8 + a12 + .......... is

(1)) (1

n

2

–1

cos

n



4

n

(2)) (2

n

2

–2

2

2

–1

co c os

n

n



(3)) (3

4

n

2

–1

 2 2 sin

n



4

n

–1

sin



n

4

Solution of Comprehension III 1.

Answer (2)

2.

Answer (4)

3.

Answer (2)

Given expression is (1 + + x x )n = a0 + a1 x x + + a2 x 2 + a3 x 3 + .............+ an x n

…(i)

Putting x Putting x = = ± 1, we get (1 + 1) n = a0 + a1 + a2 + ...... + an (1 – 1)n = a0 – a1 + a2 + ...... ± an Adding these, 2n = 2( 2(a a0 + a2 + a4 + ......)

 a0 + a2 + a4 + ...... = 2n – 1

…(ii)

Hence, answer of question 1 is (2) Again, putting x putting x = = ± i in in (i), we get (1 + i )n = a0 + a1i – – a2 – a3i i + + a4 + a5i i – – a6 – a7i i + + a8 + ...... (1 – i )n = a0 – a1i – – a2 + a3i i + + a4 – a5i – – a6 + a7i + + a8 + ...... Adding these, n /2

n

2.2

n

2(a 2( a0 – a2 + a4 – a6 + .......)  (1  i )  (1 – i ) 2



� co cos 2

n



4

n /2

2

cos



n

4

…(iii)

Hence, answer of question 2 is (4) Aakash Educational Services Pvt. Ltd.




Now, adding (ii) & (iii), we can get 2(a 2( a0 + a4 + a8 + .......) =

n

2

1

2

n /2

cos

n

4 n

 a0 + a4 + a8 + a12 + ...... =

n

2



2

 22

–1

co c os

n



4

Hence, answer is (2) Comprehension-IV Let us consider an equation f ( x x ) = x 3 – 3 x + + k = = 0. Then the values of k for for which the equation has 1.

Exac Ex actl tly y one one roo roott whic which h is po posi sitiv tive, e, the then n k belongs belongs to (1) (–, –2)

(2) (2 (2, )

(3) (0, 2)

(4) (–2, 0)

Sol. Answer (1) f (x ) f



x

3

 3x  k

y

x = = ±1 '( x )  3x 2  3  0  x –1

For exactly one positive root 1) f (  1)

1

O

x

(1) < 0  0 and f (1)

 –1 + 3 + k < < 0 and 1 – 3 + k k < < 0  2.

k

k < <2  2 and k

( ,  2) 2) k 

Exac Ex actl tly y one one roo roott whic which h is ne nega gati tive ve,, then then k belongs belongs to (1) (2, )

(2) (0, 2)

(4) (–, –2)

(3) (–2, 0)

Sol. Answer (1)

y

For exa exactly ctly one nega negative tive root root,, 1) f (  1)

 0 , f (1)  0

k > > 0  –1 + 3 + k > 0, 1 – 3 + k

–1

O

1

x

 k > – 2, k > 2  3.

 (2,  ) k 

One On e neg negat ativ ive e and and tw two o pos posit itiv ive e roo roott ifif k belongs belongs to (1) (2, )

(2) (0, 2)

(3) (–2, 0)

(4) (2, 3)

Sol. Answer (2) For one negative and two positive root 1) f (  1)

 0 , f (0) (0) > 0, f (1) (1) < 0

 –1 + 3 + k > 0, k > 0, 1 – 3 + k < 0 

y

k  



2 , 0

 0 k 

,

1 –1

O

x

k   2

i.e.,,  k  2 i.e.

 (0, 2) k 





Comprehension-V The values of ‘k ‘ k ’ for which the equation | x x |2 (| x x |2 – 2k + + 1) = 1 – k 2, has 1.

No real root, when k belongs belongs to

 

–1, (2)  –1,

(1) (– (–1, 1)

5

5  ,  4 

(3)) (– , – 1)   (3

 4

(4)) R (4

Sol. Answer (3)

| x |2 (| x |2 2k  1)  1 k 2 ⇒ x

2

( x 2  2k  1)  1  k 2

All roots are imaginary, if ⇒

(2k  1)2  4(k 2  1)  0

⇒

k 

x

⇒

D

5

4

 (2k  1)x 2  k 2  1  0

 b2  4ac  0

…(1)

4

Also roots are imaginary if

D

 0 , but

2 x

is negative, i.e. roots of ( x 2 )2  (2k  1)( x 2 )  k 2  1  0 are both

negative.

 Sum < 0, and product > 0 

2

 1  0 and k 2k 

 1 0 

 All roots are imaginary if

2.

k  ( ,  1) 1)

5  ,  4 

k  ( ,  1)  

Exac Ex actl tly y two two re real al ro root ots, s, wh when en k belongs belongs to (1) (–, –1)

(2) (–1, 1)

 

(3) 1,

5



4

(4)) R (4

Sol. Answer (2) For exactly two real roots of t

2

1)t  k 2  1  0  (2k  1)

D > 0 and one value of t = x = x 2 is positive and one is negative.

 (2k  1)2  4(k 2  1)  0



k  

5

…(i)

4

Product =

2

k

1 0

From (1) and (2)

 –1 < k < 1

…(ii)

k  ( 1, 1) .





3.

Rep epea eate ted d ro root ots, s, whe hen n k belongs belongs to (1) {1, –1}

(2) {0, 1}

(3) {0, –1}

(4) {2, 3}

Sol. Answer (1) For repeated roots  = 0, or P = 0, 

k

2

 k 

5 4

 1  0, k  1

When product = 0, x = = 0 is repeated root. Comprehension-VI Let A Let A,, B, C be three sets sets of complex numbers as defined below A = {z { z : : Im z  1} B = {z z :: |z – – 2 – i | = 3} C = C = {z : Re((1  i )z )  2 }. 1.

[IIT-JEE 2008]

The Th e num numbe berr of of ele eleme ment nts s in in the the se sett A  B  C is is (1) 0

(2) 1

(4) 

(3) 2

Sol. Answer (2) Let z z = = x x + + iy Set A Set A corresponds 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., x i.e., x 2 + y 2 – 4 x – – 2y = = 4

...(ii)

Set C consists consists of points lying on the x x + + y y = =

...(iii)

2

y P

(0, 2 ) (2, 1)

y=1 x

( 2, 0)

Clearly, there is only one point of intersection of the line 2.

x

 y 

2

and circle circle x x 2 + y 2 – 4 x – – 2y = = 4

Let z be be any point in A  B  C . The | z + + 1 – i | i |2 + | z – – 5 – i | i |2 lies between (1) 25 and 29

(2) 30 and 34

(3) 35 and 39

(4) 40 and 44

Sol. Answer (3) z

 1– i

2



z –5– i

2

= ( x + + 1)2 + (y – – 1)2 + ( x x – – 5)2 + (y – – 1)2 Aakash Educational Services Pvt. Ltd.




= 2( x 2 + y 2 – 4 x – – 2y ) + 28



= 2(4) + 28

∵ x

2

 y2

– 4 x – 2y



4



= 36 3.

Let z be be any point in A in A  B  C and and let w be be any point satisfying | w – – 2 – i i || < 3. Then, | z z || – | w w || + 3 lies between (1) –6 and 3

(2) –3 and 6

(3) –6 and 6

(4) –3 and 9

Sol. Answer (4)



w – 2





i

3

 | w | – | 2  i | < 3  3  

–3–

–3



–





5





3

w

5

z – 2

Also,



5

w



5

3–

i

3

z

3

5

5

…(i)

…(ii)

 – 3 < |z | z | – |w | +3 < 9 Comprehension-VII Let S = = S 1  S 2  S 3, where

1.

S1

 {z   :| z |  4} ,

S2

 z   : Im 

S3

 { z   : Re z  0} .



z  1





1

3i 







3i

  0  and

[JEE(Advanced) [JEE(Advance d) 2013]

Area of S = = 10 

(1)) (1

3

(2)) (2

20 3

Sol. Answer (2)

(3)) (3

16 

(4)) (4

3

A

3

S

S 1 represent circle with centre (0, 0) and radius 4 S 1 : |z | < 4  x 2 + y 2 < 16 S 2

32

B O

 z  1  3i  : Im  0 1 3 i   

60°

y+

3x = 0

C

 [( x  1)  ( y  3i )][1  3 i ]   0 2  

Im 





S2  y 

3 x  0

S 3 Re( Re(z z ) > 0, i.e. i.e.,, x x > >0 S = S = S 1  S 2  S 3

(4)2 60   (4) 2 Area of shaded region is OAB OAB + + OBC = 4 360 =

4



=

4



=

2.

16  6 8 3

20 3

min | 1  3i  z |  z S

(1)) (1

2

3

(2)) (2

2

2

3

(3)) (3

2

3



3

3

(4)) (4

2



3

2

Sol. Answer (3) min|z min| z – – (1 – 3i 3 i )| )| Minimum distance of z from from (1, –3)

From question, minimum distance of (1, –3) from

y

 3 x  0 is

3  2

3



3

 2

3

.

SECTION - D

Assertion-Reason Type Questions 1.

STATEMENT-1 : Le Let z 1 and z 2 be two complex numbers such that arg( z 1 ) 

   and arg( z 2 )  then arg( z 1z 2 )  . 2 6 3

and STATEMENT-2 : arg( z 1z 2 )  arg( z 1 )  arg( z 2 )  2k , k  {0, 1,  1} Sol. Answer (1) Clearly Statement-1 is true and Statement-2 is its explanation. (standard results) 2.

 z  1     is a circle.  z  1  2

STA ST ATE TEME MENT NT-1 -1 : Th The e loc locus us of z , if arg and STATEMENT-2 :

2   , then the locus of z is is a circle. z  2 2 z





Sol. Answer (4)

y z

  1  arg    z  1 2 z

–1

Locus of z is a semicircle

O

1

x

 statement-1 is false and statement-2 is true.  Option (4) OR Let z z = = x x + + iy z  1 ( x  1)  iy  z  1 ( x  1)  iy z 1 z 1

z 1 z  1





( x  1)  iy ( x  1)  iy  ( x  1)  iy ( x  1)  iy ( x 2  1  y 2 )  i (2 (2y ) ( x  1)2  y 2

 z  1    z  1 2

arg 

 2

x

x

2

 y 2  1  0

> 0  y 2  1 and y >

is semicircle.  Locus of z is 3.

STATEMENT-1 : If e i  = cos + + i i sin sin and the value of e iA.e iB.e iC is equal to –1, where A A,, B, C are are the angles of a triangle. and STATEMENT-2 : In any  ABC , A A + + B + C = = 180°.

Sol. Answer (1) A + B + eiA.eiB.eiC = ei ( A +

C )

= ei  = cos  + i sin sin = –1 4.

STATEMENT-1 : z 12 + z 22 + z 32 + z 42 = 0 where z 1, z 2, z 3 and z 4 are the fourth roots of unity. and 1

1

4

4

STATEMENT-2 : (1)  (cos 0  i sin 0) . Sol. Answer (1) Aakash Educational Services Pvt. Ltd.




(1)1/4 = (cos2r + + i sin2 sin2r )1/4 =

cos

r 2

 i sin

r 2

where r = = 0, 1, 2, 3

 11/4 = 1, i , – 1, – i  z 12 + z 22 + z 32 + z 42 = 1 + i 2 + 1 + i 2 =2–1–1=0 5.

STATEM STA TEMEN ENT T-1 : For For any any four four com comple plex x numb number ers s z 1, z 2, z 3 and z 4, it is given that the four points are concyclic, then | z 1 | = | z 2 | = | z 3 | = | z 4 |. and STATEMENT-2 : Modulus of a complex number represents the distance from origin.

Sol. Answer (4) |z 1| = |z 2| = |z 3| = |z 4| This may not be the case if centre of the circle is not origin.

n

6.

 2i  STA ST ATE TEME MENT NT-1 -1 : The The exp expre ress ssio ion n   is a positive integer for all the values of n.  1  i  and STATEMENT-2 : Here n = 8 is the least positive for which the above expression is a positive integer.

Sol. Answer (4) n

 2i   2i (1 – i )  1  i    2   (1 i ) n

n

in n

 ( 2)

e 4

Now clearly the least integral value for which the given number is a positive integer is 8. 7.

STATE TEM MEN ENT T-1 : If If 1 – i , 1 + i , z 1 and z 2 are the vertices of a square taken in order in the anti-clockwise sense then z 1 is i i – – 1. and STATEMENT-2 : If the vertices of a square are z 1, z 2, z 3, z 4 taken in order in the anti-clockwise sense, then z 3 = – iz 1 + (1 + i )z 2.

Sol. Answer (1)

 ABC 

B(z 2)

A(z 1)

z – z 2   arg 1 z3 – z 2 2

And AB And AB = = |z 1 – z 2| = BC |z 3 – z 2|



z 1

–

z 2

z 3

–

z 2

1


C(z 3)

D(z 4)


172 Complex Numbers and Quadratic Equations z1 – z 2

Hence,

z3 – z 2

 

 1 cos



– i sin

2

   z 3 – z 2 = (z 1 – z 2)  cos 2






2

– i si s in

  2

= – i (z 1 – z 2) = – iz 1 + iz 2

 z 3 = – iz 1 + (1 + i )z 2  z 3 = – i (1 (1 – i ) + (1 + i )(1 )(1 + i ) = – i i + + i 2 + 1 + i 2 + 2i = i i – –1

8.

z

STATEMENT-1 : If

1

 a where z is is a complex number and a is a real number, the least and greatest

z

a

values of |z | z | are



2

4

–a

2

a

and

2

4 a 2

.

and STATEMENT-2 : For a equal to zero the greatest and the least values of | z | are equal. Sol. Answer (2) Let z = = r (cos (cos + i sin sin)

z



1

2

1   r    r

z

 r 2 

1 r



a

2

2

2

cos

2

1     r –   r 

2

sin

2



 2cos2

1

 r 2 

r

2

2–

2

4sin 4sin



2

1  2 2   r    a  4sin   r 

2

1  2  r    a  4  r 



r

2

–

r



1 r

a



2

a

2

4

 4r  1 

 r lies lies between

0

a

2

4

–a

and

2


a

2

4 a 2




This is true for all real a  0. | z | m ax 

a

2

4 a , | z | min  2

a

2

4 –a 2

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 of statement-1. 9.

STA ST ATEM TEMENT ENT-1 -1 : The loc locus us of comp complex lex nu numbe mberr z , satisfying (z ( z – – 2)n = z n is a straight line. and STATEMENT-2 : The equation of the form ax + + by + + c = = 0 in x – – y plane plane is the general equation of straight line.

Sol. Answer (1) Given equation is (z (z – – 2)n = z n z –



2

n

1

z

z –



z

2

 11/  1 n

 | |z z – – 2| = |z | z | Hence z is is the locus of a straight line perpendicular bisector of the segment joining the points (2, 0) and (0, 0), i.e. i.e.,, x x = = 1. 10.. ST 10 STA ATEMENT TEMENT-1 -1 : A root root of of the equati equation on (210 – 3) x x 2 – 211 x + + (210 + 3) = 0 is 1. and STATEMENT-2 : The sum of the coefficients of a quadratic equation is zero, then 1 is a root of the equation. Sol. Answer (1) ax

2

 bx  c  0

x = = 1, a + b + c c = =0 If sum of coefficient is 0 then 1 is the root of the equation. (210 – 3) – 2 11 + 210 + 3 = 0

 Both are true and Statement-2 is correct explanation of Statement-1 11.

STATEMENT-1 STATEM ENT-1 : The equation whose whose roots are reciprocal of the roots roots of the equation 10 x 2 – x – 5 = 0 is 5 x 2 + x – – 10 = 0. and STATEMENT-2 : To obtain a quadratic equation whose roots are reciprocal of the roots of the given equation ax 2 + bx bx + + c = = 0 change the coefficients a, b, c c to to c , b, a. (c  0)

Sol. Answer (1) Aakash Educational Services Pvt. Ltd.



For reciprocal roots, replacing x by by

1 x


a

in

ax

2

 bx  c  0

2

x



b x

 c  0 

cx

2

 bx  a  0

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

10 x

 x  5  0 

10 2

x



1 x

5  0 

5 x

2

 x  10  0

12.. ST 12 STA ATEM TEMENT ENT-1 -1 : The equa equatio tion n x 2 – 2009 x + + 2008 = 0 has rational roots. and STATEMENT-2 : The quadratic equation ax 2 + bx + + c = = 0 has rational roots iff b2 – 4ac is is a perfect square. Sol. Answer (3) The equation in first statement is x 2 – 2009 x + + 2008 = 0 can be written as ( x ( x – – 2008)( x 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 – 4 ac = a perfect square than roots of the equation ax 2 + bx + + c = = 0 are rational only when a, b, c are are rationals, otherwise roots are irrationals. To this end, let us consider an equation

4 x

2

4

3 x 1  0

whose discriminant = 48 + 16 = 64 = 8 2 = a perfect square but roots are

x





4 3

48  16

2 4

3 2 2

which are not rationals.

Thus statement 2 is false. Hence option (3) is correct. 13.. ST 13 STA ATEMEN TEMENTT-1 1 : One One root root of the equat equation ion x x 2 + 5 x – – 7 = 0 lie in the interval (1, 2). and STATEMENT-2 : For a polynomial f ( x x ), ), if f ( p) p)f (q) < 0, then there exists at least one real root of f ( x x ) = 0 in ( p, p, q). Sol. Answer (1) We observe that f (1)f (1)f (2) (2) = (1 + 5 – 7)(4 + 10 – 7) < 0 Hence these exists a root of x 2 + 5 x – – 7 = 0 in (1, 2). Clearly option (1) is correct. 14.. ST 14 STA ATEMEN TEMENTT-1 1 : The quadra quadratic tic equati equation on ax 2 + bx + + c = = 0 has real roots if (a ( a + c )2 > b2,  a, b, c  R . and STATEMENT-2 : The quadratic equation ax 2 + bx + + c = = 0 has real roots if b2 – 4ac  0. Aakash Educational Services Pvt. Ltd.




Sol. 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)f (1) (1) > 0, where f ( x x ) = ax 2 + bx + c  f ( x x ) = ax 2 + 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. ST STAT ATEMENT-1 EMENT-1 : There is just one quadratic quadratic equation with real coefficients, coefficients, one of whose roots is

1 3

7

.

and STATEMENT-2 : In a quadratic equation with rational coefficients the irrational roots occur in pair. Sol. Answer (4) Statement-1 is wrong.

 

1 3

can be root of infinite equations with real coefficients, e.g. ( x  1)  x 

7

 

( x  2)  x 

1   7  0  3

1   7   0...  3

16.. ST 16 STA ATEM TEMENT ENT-1 -1 : The The roots roots of

x

2

2

20 2008 x  501  0

are irrational.

and + c = = 0, a  0 (a, b, c , R ) is a perfect square, STATEMENT-2 : If the discriminant of the equation ax 2 + bx + then the roots are rational. Sol. Answer (3) x

2

x

2





2008 x  501  0

2 20 2008  4  2008  4(501) 2 2008



1507

Roots are rational. STATEMENT-2 is wrong as roots are rational only when coefficients are rational and

b

2

 4ac is perfect

square. Aakash Educational Services Pvt. Ltd.




17.. STA 17 STATE TEME MENT NT-1 -1 : If a, b, c not not all equal and a  0, a3 + b3 + c 3 = 3abc 3 abc , then the equation ax 2 + bx + + c = = 0 has two real roots of opposite sign. and STATEMENT-2 : If roots of a quadratic equation ax 2 + bx + c = 0 are real and of opposite sign then ac < ac < 0. Sol. Answer (4) a3 + b3 + c 3 – 3abc abc = = (a + b + c )(a )(a2 + b2 + c 2 – ab – bc bc – – ca ca)) = 0, a + b + c c = = 0  a, b, c are are not equal. The sign of other root depends on sign of ac . Hence (4) is answer. 18. ST STAT ATEMENT-1 EMENT-1 : There is just one one quadratic equation with real coefficient coefficient one of whose roots roots is

1 2

1

.

and STA ST ATEMENT-2 : In a quadratic equation with rational coefficients the irrational roots are in conjugate pairs. Sol. Answer (4)

( 2 – 1) is a root if coefficients are real then other root can be rational. 19.. ST 19 STA ATEMEN TEMENTT-1 1 : Let a quadrat quadratic ic equation equation has has a root root 3 – 9i 9i then then the sum of roots is 6. and STATEMENT-2 : If one root of ax 2 + bx + + c = = 0, a  0 a, b, c  R is is  + i , ,   R then then the other root must be  – i . Sol. Answer (4) For, ax 2  bx  c  0 , roots are conjugate of each other if a, b, c are real, which is not mentioned in statement-1. 20. Let a, b, c , p p,, q be real numbers. Suppose ,  are the roots of the equation x 2 + 2 px + + q = 0 and , the roots of the equation ax 2 + 2bx + + c = = 0, where 2  {1, 0, 1}

1



are

[IIT-JEE 2008]

STATEMENT-1 : ( p 2  q )(b 2  ac )  0 and STATEMENT-2 : b  pa

or

c  qa

Sol. Answer (2) x

2

 2px  q  0

     2 p

(i)

  q

(ii)

ax 2  2bx  c  0

   

1 2b  a  c a

(iii) (iv)




Solution of Assignment (Set-2) 2   1        2     2 2  ( p  q ) ( b  ac ) =       2       2         

=

(  )2 16

2

2

 1 2      . a  0 statement-1 is true      2 

Now pa Now pa = =    b= 

a

a



a

2

(   )

 1    2   

a

1     ⇒ 2  1,   { 1, 0, 1}, correc correctt 

pa  b ⇒  

Similarly If

c

 qa ⇒ a

  a 

 1       0       0, and  

1  0 ⇒   {1, 0, 1} 

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

SECTION - E

Matrix-Match Type Questions 1.

Match the foll llo owin ing g Column-I

Column-II 2007

(A)) Th (A The e valu value e of



∑   sin k 1

2k  9

 i cos

2k   9

 i is s 

(p) –1

(B) If z 1, z 2 and z 3 a arre unimodular complex numbers such that |z1 + z2 + z3| = 1, then

1 z 1



1 z 2



1 z 3

(q) 2

is equal to

(C)) If the comp (C complex lex numbe numbers rs z 1, z 2 and z 3 represent the

(r) 1

vertices of an equilateral triangle such that |z 1| = |z 2| = |z 3|, then (z ( z 1 + z 2 + z 3) – 1 is equal to (D) If  i is s an imaginary fifth root of unity, then 4 log4

1

  2  3 

1



(s) 0

is





Sol. Answer A(s), B(r), C(p), D(q) 2007

2007

2007 i 2k  2k   2k  2k     ( i ) c o s i s i n  i cos    ( ) i e (A) ∑  sin = ∑ =    ∑ 9 9  9 9  k 1 k 1 k 1

2 k  9

i 4 i 2( 2007 )   i 2  9 9 9 i e e  .....  e   

Which is G.P. (2007)   i 2(2007)  9  1  e   i 2   i 2  9     ( 1 1 ) e    i e 9   i  i 2 0 i 2    e 9 1  1  e 9 1      

(B) |z 1| = 1, |z |z 2| = 1, |z | z 3| = 1 z z 1 1

1 z 1

 1, z2 z2  1,



Now,

1

,

z 1

z 2

1 z 1





z 2

1 z 2

,



z z 3 3

1 z 3

1 z 3



1

z 3

 | z1  z2  z 3 | = | z  z  z | 1 1 2 3

(C) |z 1| = |z 2| = |z 3| and z 1, z 2, z 3 are vertices of equilateral triangle

 Origin is its centroid  z 1 + z 2 + z 3 = 0 Now, |z |z 1 + z 2 + z 3| –1 = –1 1

(D) Le Lett (1)  1, ,  2 , 3 ,  4 5

1 +  + 2 + 3 + 4 = 0 1 +  + 2 + 3 = – 4 and 5 = 1



1



  4 and | | = 1

Now 4log4 |–4 – 4| = 4 log4 | 2 4 | = 4 log4 | 2 || ||  4 | = 4  log4 2  4  Aakash Educational Services Pvt. Ltd.

1 2

log2 2  2




2.


Column-II

(A) The smallest positive integer for which

(p) 1

(1 + i ) n = (1 – i )n is 3

(B) If

a  ib  x  iy and

b



a

y x

 k ( x 2  y 2 ) , then

(q) –3 –3

k is is equal to

(C) If

1  i

x 

2

, then the value of

(r) 2

1 + x 2 + x 4 + x 6 + x 8 + x 10 +.....+ x 2004 + x 2006 + x 2008 (D) If the minimum value of

(s) 4

|z +1+i +1+i | + |z –1–i –1–i | + |2 – z | + |3 – z | is k then then (k (k – – 8) equals Sol. Answer A(s), B(r), C(p), D(q) (A) (1 + i )n = (1 – i )n n

 1  i  1    1  i  

n

1

i

n=4 (B) (a + ib ib)) = ( x x + + iy )3 = x 3 – iy 3 + 3ix 2y – – 3 xy 2 a = = x x ( x x 2 – 3y 2) and b = –y 3 + 3 x 2y a x

 x 2  3 y 2

b   y 2  3 x 2 y

b a   2( x 2  y 2 ) y x k = 2 (C)) (C

x 



1  i 2

x

1  x

2

2

(1 (1  i )2 1  1  2i    i 2 2

 x 4  x 6  x 8  x10  ...  x 2004  x 2006  x 2008

1[1  x 2010 ] 1  x 2



1  i 2 1  i 2



1 1 1 1

1




3.


Let z 1 and z 2 be two given complex numbers. The locus of z such that Column-I

Column-II

(A) |z z – – z 1| + |z z – – z 2| = constant = k , where k  |z 1 – z 2|

(p) Circle with z1 and z2 as the vertices of diameter.

(B) |z z – – z 1| – |z z – – z 2| = k , where k  |z 1 – z 2|

(q) Circle

 z  z 1      2  z  z 2 

(C)) arg (C

(r) Hyperbola

2007

(D) If  lies on || = 1 then



lilies on

(s) El Ellipse

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

) p (z

A) | z  z1 |  | z  z2 | = constant = k , where

k

 /2

 | z1  z 2 | , represents an ellipse. z 1

(B) | z  z1 |  | z  z2 |  k, where

k

z 2

 | z1  z 2 |

is a hyperbola having foci at z 1 and z 2. p

 z  z 1   (C) arg  z  z    2 (C)  2

This represents a circle with z 1 and z 2 as the vertices of diameter. (D) If  lies on || = 1, then 2007 2007   2007  ||



4.

2007



lies on the circle.


Column-II

(A) |z – – 6i | + |z – – 8| = k wi will represent an ellipse for k equals to

(p) 2

(B) ||z – – 12 12ii + 3| – |z | z – – 2|| = k will represent hyperbola if k equals to

(q) 8

(C) |z z – – ki | + |z – – 4| =

(r) 12

10 k

wiill represent line segment if w

k equals to (D)) (D

 k  2k i  z  2  4 i

z

k will

represent circle if k eq equals to

(s) 11 (t) 10





Sol. Answer A(r, s), B(p, q, r, s, t), C(p, q), D(q, r, s, t) (A) |z z – – 6i 6 i || + |z – – 8| = k will will represent ellipse if

6

(B) |z – – 12 12i i + + 3| – |z |z – – 2|| will represent ellipse if (C) |z z – – ki | + |z – – 4| =

2

 82  k

⇒

52  122  k

k  10

⇒

13

 k

will represent line segment if k 2 + 16 = 10k 10 k  k = 2, 8

10k

(D) To represent circle k  1 and also k k = =2 5.

Matc Ma tch h the the entri entries es of of colum column-I n-I with with tho those se of of colum column-I n-III Column-I

Column-II

(A) t  R such such that there is at least one z sa satisfying (p) 6 | z z || = 3, |z |z – – { t (1 (1 + i i )) – i } i } |  3 and | z z + + 2t t – – (t ( t + + 1) 1)i i || > 3. (B) So Solv lve e for for x x : :

(1  i ) x – 2i (2 – 3i )y  i   i 3i 3 – i

(q) 0

n

 1  i  (C) The int intege egerr n for which  is real  is  1 – i 

(r) 3

(D)) The greatest (D greatest and least absolute absolute value value of z + + 1, where |z |z + + 4|  3 are

(s) 4 (t) 8

Sol. Answer A(r, s), B(r), C(p, q, s, t), D(p, q) (A)) To satisf (A satisfy y all at a time z should should lie on the circle | z | = 3. Inside the circle |z | z – – {(1 + i ) – i }| }| = 3 and outside the circle | z z + + 2t 2 t – (t + 1)i 1)i | = 3 For this,

(t – 0) 0)2  ( t – 1 – 0) 0) 2  3  3 and

4 t 2  (4 1) 1) 2  3  3

 2t 2 – 2t 2 t – – 35  0 and 5 5t t 2 + 2t 2 t – – 35 > 0 Using sign scheme we have,

1–

71

–1 – 4 11

–1 + 4 11

5

5

2



71 –1 –1 – 4 11

Hence, 1 – 1 – , 2 

1+

71

2

 –1  4 11 1  71 ,  5 2  

  

5

Hence, 3, 4 lies in above interval. (B)) We have (B have to to solve solve for for x x , y

(1  i )x – 2i (2 – 3i )y  i   i 3i 3 – i



(1  i )(3 – i )x – 2i (3 – i )  (2 – 3i )(3  i )y  i (3  1) 9 – i 2


 i




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

 (4 x + + 9y – – 3) + i (2 (2 x – – 7y – – 3) = 0 + 10i 10i Comparing the real and imaginary parts 4 x + + 9y – – 3 = 0

…(i)

2 x – – 7y – – 13 = 0

…(ii)

(i) – 2  (ii) gives, 9y + + 14 14y y – – 3 + 26 = 0 23y y = = – 23  y y = =– 1  23 Putting y = = – 1 in (i), we get 4 x – – 9 – 3 = 0  x x = =3

 1  i 2  2i   1 i   (C)  (C)  1  1   i  1 – i    n

n

Hence n = 0, 4, 8, 6 (D)) Greatest and least (D least absolute absolute values values of z + + 1 are 1 and 6. |z | z + + 4|  3

∵

6.

For the For the qu quad adra rati tic c equ equat atio ion n x 2 – (k (k – – 3) x x + + k = = 0, match the condition in column I with the corresponding values of k in in column II. Column-I

Column-II

(A) Both the roots are positive

(p) 

(B) Both the roots are negative

(q) (–, 1)  (9,  )

(C) Both the roots are real

(r) [10, )

(D) One root is less than –1 and other is

(s) (0, 1]

greater than 1

(t) [125, 1250]

Sol. Answer A(r, t), B(s), C(q, r, s, t), D(p) x 2 – (k k – – 3) x + x + k k = =0 For roots to be real (k k – – 3) 2 – 4k  0

 k 2 – 6k + 9 – 4k 4 k  0  k 2 – 10 10k k + + 9  0  (k – – 1) (k ( k – – 9)  0  k  (–, 1]  [9, ]

…(i)

(A)) For both (A both roots roots to be be positive positive,, f (0) (0) > 0 and

 3 k  2

0

k > 0

…(ii)

and k > 3

…(iii)

From (i), (ii) and (iii) k  [ 9, ] Aakash Educational Services Pvt. Ltd.




(B)) For both (B both roots roots to be be negativ negative e D  0 k > 0, (k – 3) / 2 < 0,  k < 3,  k  (0, 1] (C) For both both roots roots to be real real k  (– , 1]  [9, ) (D)) f (–1) (D (–1) < 0, f (1) (1) < 0 1 + (k – 3) + k < 0 also 1 – (k ( k – 3) + k < 0

 2k – 2 < 0  k k < < 1, 1, 4 < 0 No such value is possible 7.

Let y y = = f ( x x ) = ax 2 + bx bx + + c , a  0 be a given quadratic expression. Then for real values of x of x , match the appropriate graph of y = = f ( x x ) indicated in the Column-II from the conditions given in Column-I. Column-I

Column-II

(A) f ( x x ) > 0, a > 0, c > > 0

(p)

(B) D > 0, c < < 0 and ab ab < < 0

(q)

(C)) D < 0, a < 0 (C

(r)

(D)) D = 0, a < 0, b > 0 (D

(s)

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

f ( x )  0

,  x  R , f (0) (0) = c c > >0

Parabola is upward  a > 0. (B)) Roots are real (B real and distin distinct ct  D > 0 f (0) (0) < 0  c c < <0 One root is positive and one negative  ab ab < < 0. Aakash Educational Services Pvt. Ltd.



(C)) Root (C Roots s are are imagina imaginary ry 

D


0

Parabola is downward  a < 0. (D)) Parab (D Parabola ola touche touches s x-axis x-axis 

D

0

Parabola is downward  a < 0 Both roots are positive, sum = 

8.

b a

0  b > 0

Match th the following Column-I

Column-II (p) a does not exist

(A) If the roots of the equation x 2 – 2ax + + a2 + a – 3 = 0 are real and less than 3, then a belongs to

3 2

 

(q)  , 3 

(B) If the roots of the equation x 2 – 2ax + + a2 + a – 3 = 0 are real and greater than 3, then a belongs to (C) If exactly one root of the above equation

(r) (–1, 2)  (2, 3)

lies in the interval (1, 3), then a belongs to (D) If th the ro roo ots of of th the ab above eq equation ar are su such

(s) (– , 2)

that one root is greater than 3 and other smaller than 1, then a belongs to Sol. Answer A(s), B(p), C(r), D(p) (A)) For rea (A reall roots roots,,

D

0

 ( 2a)2  4(a2  a  3)  0  4[a2  a2  a  3)  0



a

 3

…(i)

f (3) (3) > 0  9 – 6a + a2 + a – 3 > 0



a



a



b 2a

2

 5a  6  0

 ( , 2)  (3, ) 2a

3 

2

 3  a < 3

from (i), (ii) and (iii) (B)

D  0



f (3) (3) > 0 

a

 3

a

…(ii)

a

… (iii)

 ( , 2) …(i)

 ( , 2)  (3, )


…(ii) Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456



For greater than 3, b



 3  a > 3 … (iii)

2a

From (i), (ii) and (iii), (C)) D > 0  a < 3 (C

a

 .

… (i)

f (1) (1) f (3) (3) < 0

(1  2a  a2  a  3)(9  6a  a2  a  3)  0

 (a2  a  2)(a2  5a  6)  0 

 ( 1, 3)  {2} {2}

a

 (a  2)2 (a  1)( a  3)  0

…(ii)

From (i) and (ii) (2, 3) ( 1, 2)  (2,

a

(D)) (D

D

0 

a

 3

…(i)

f (1) (1) < 0  (1  2a  a2  a  3)  0

 (a2  a  2)  0  –1 < a < 2 f (3) (3) < 0



a

2

… (ii)

 (9  6a  a2  a  3)  0

 5a  6  0

 2 < a < 3

… (iii)

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

9.

 .

Let f ( x x ) = x |x – – 1| + | x x – – 2| + | x x – – 3|, match the column I for the value of k in in column II. Column I

Column II

(A) f ( x x ) = k ha has no solution

(p) 1

(B) f (x) (x) = k ha has only one solution

(q) 2

(C)) f ( x (C x ) = k has two solution of same sign

(r) 4

(D)) f ( x (D x ) = k has two solution of opposite sign

(s) 5 (t) 8

Sol. Answer A(p), B(q), C(r, s), D(t) We have, f ( x x ) = x |x – – 1| + | x x – – 2| + | x x – – 3| = k Aakash Educational Services Pvt. Ltd.




when x when x < < 1, then f ( x x ) = –3 x + + 6 = k



x   2 

2

Now,

k

3

k

3

1

 k > 3

...(i)

For 1  x < < 2 – x x = = k k – –4 x = = 4 – k

 2 < k  3

...(ii) 3

For 2  x < < 3 x = x = k , 2  k k < < 3

2

For x For x  3 1

2

3

3x – 6 = k x 



k  6

3

k   6

3

3

k  3 Clearly (A) for k < < 2, there is no solution (B) for k = = 2, there is only one solution (C) fo forr 2 < k < < 6, there are two solution of same sign (D) fo forr k > > 6, there are two solution of opposite sign 10.. Fo 10 Forr giv given en eq equa uati tion on x x 2 – ax + + b = 0, match conditions in column I with possible values in column II Column-I

Column-II

(A)) If roots (A roots differ differ by by unity unity then then a2 i is s equal to

(p) b(ab + 2)

(B)) If roots (B roots differ differ by unity unity then then 1 + a2 i is s equal to

(q) 1 + 4b

(C) If one of the root be twice the other

(r) 2(1 + 2 b)

then 2 2a a2 is equal to (D) If the su sum m of of roo roots ts of th the e equ equat atio ion n equ equal al to th the e sum sum of

(s) 9 b

squares of their reciprocal then a2 is equal to Sol. Answer A(q), B(r), C(s), D(p) (A)  (q), (B)  (r) x

2

 ax  b  0





    a ,  

b

|    |  1  (  )2  4  



a

2

 4b  1



a

2

 1  4b



a

2

 1  2  4b 

(C))   2  a    (C

2.



2

 a    3

b 2

1

(D))     (D

2 



b

2





a

2





2 1



2b

a

2

 4b



a

3 b 2

9b 2



2a

2

 9b .

1

2

(   )( 22 )   2   2  (   )2  2

11.

2



a(b)



a

2

 a2  2b

 ab2  2b  b(ab  2)

If , ,  be be the roots of the equation x equation x (1 (1 + x + x 2) + x + x 2(6 + x + x ) + 2 = 0, then match the entries of column-I with those of column-II. Column I

Column II

(A) –1 + –1 +  –1 is is less than or equal to

(p) 8

(B) 2 + 2 +  2 eq equals

(q)

(C) (–1 + –1 +  –1) – ( +  +  ) is less than or equal to

(r) – 1

(D) [–1 + –1 +  –1] equals where [ ] denotes

(s) 3

–

1 2

the greatest integer less than or equal to (t)

5 2

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

, ,  be be the roots of the equation x (1 (1 + + x x 2) + x 2(6 + x + x ) + 2 = 0 + 6 x 2 + x 3 + 2 = 0  x 3 + x + + 2 = 0  2 x 3 + 6 x 2 + x + Aakash Educational Services Pvt. Ltd.




So,  +  +  = – 3

 +  +  =

1 2

 = –1 Now, –1 (A)  –1  –1   –1 

1





1





1





1      –  2

(B) 2 + 2 +  2 = ( +  +  )2 – 2( +  + )

 9 – 2�

1 2

8

(C) (–1 + –1 +  –1) – ( +  +  )



     – (     ) 

–

1 2

3

5 2

 

1

(D) [–1 + –1 +  –1] = –   – 1 2



Column-I with Column-II.. 12.. Ma 12 Matc tch h the the stat statem emen ents ts in in Column-I with those in Column-II

[IIT-JEE 2010]

Note: Here [Note: Here z takes takes values in the complex plane and Im z and and Re z denote, denote, respectively, the imaginary part and the real part of z .] .] Column I

Column II 4

(A)) The set of poin (A points ts z sa satisfying

(p) An ellipse with eccentricity

5

| z – – i i || z z || || = | z + + i i || z z || || is contained in or equal to (B)) The set (B set of of point points s z satisfying satisfying | z + + 4 | + | z – – 4 | = 10

(q) The set of points z satisfying satisfying Im z = = 0

is contained in or equal to (C) If | w w || = 2, then the set of points

z

 w 

1

is i s

(r) The set of points z satisfyi satisfying ng | Im z z ||  1

is i s

(s) The set of points z satisfyin satisfying g | Re z z ||  2

w

contained in or equal to (D) If | w w || = 1, then the set of points

z

 w 

1 w

contained in or equal to (t)) Th (t The e se sett of of poi point nts s z satisfying satisfying | z z ||  3 Sol. Answer A(q, r), B(p), C(p, s, t), D(q, r, s, t) Aakash Educational Services Pvt. Ltd.




(A) z is is equidistant from the points i | z | | and –i –i | z |, |, whose perpendicular bisector is Im ( z ) = 0. (B)) Sum of dist (B distance ance of z from from (4, 0) and (–4, 0) is a constant 10, hence locus of z is is ellipse with semi-major axis 5 and focus at (±4, 0). ae = ae = 4

e=

4 5

(C)) | z |  | w |  (C

(D)) | z |  | w |  (D

1



w

1

5 3 2

2

w

 Re z  | z | |  2

13. Let

zk



 2k    2k    i sin  ; k = 1, 2, ..., 9.   10   10 

cos 

Column I


Column II

(A)) Fo (A Forr eac each h z k there exists a z such that z k . z = 1 j j

(p) Tr True

(B)) The (B There re exis exists ts a k  {1, 2, ..., 9} such that z 1.z = z k

(q) False

has no solution z in in the set of complex numbers

(C)) (C

(D)) (D

|1

z 1

||1

z 2

| ... | 1 

z 9

|

10

1

eq e quals

 2k   eq e quals  9k  1 cos   10 

(r) 1

(s) 2

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

2 2    i sin   e z k =  cos  10 10 

i 2 k

10

(A) z k .z j = 1



2 ( k  j ) 10 e i

1

 2( k  j )  2(k  j )  cos   1 and sin  0 10 10   k + j = 10 10n n and k + j = 5m; so true Aakash Educational Services Pvt. Ltd.



(B)

z



z k




2 ( k 1) 10 e i

z

1

So, if k  {1, 2, .....9}, z has has solution; i .e., false (B)  2 (C)) z 10 – 1 = 0 (C z – – z 1)(  (z )(z z – z 2).....( ).....(z z – – z 9) = 1 + z z + + z 2 + ..... + z 9 So, |1 – z 1||1 – z 2|.......|1 – z 9| = 10 i.e., i.e ., (C)  3

 

1  cos

(D)) (D

2 10

 cos

4 10

 .....  cos

18   10

 

2   4 8  10   1  cos 10  2 2  

SECTION - F

Integer Answer Type Questions 1.

P is a point satisfying arg z = /4, such that sum of its distance from two given points (0, 1) and (0, 2) is minimum, then P must must be

k

3

(1 + i + i ). Then numerical value of k is is __________.

Sol. Answer (2) PA + PB PB will will be minimum

B (0, (0, 2)

where A where A and and A A are mirror image

y = x A(0, 1)

A, P , B are collinear Equation of line line A AB : 2 x + + y = = 2

P

Solve A Solve AB with y = x = x 2

 x x = = P





3



2

, y y = =

2 (1 i ) 3

A(1,

0)

k  2

2002   2 k 1

2.

3

∑

If

 2r    2r   less than 10 may be _____.  i sin   0 , then the non-negative integral values of k less   7   7 

cos 

r  1

Sol. Answer (4) 2002 2 k 1

∑ r



1

 2r    2r    i sin  0   7   7 

cos 

It is possible only when 2002 + (2 k – – 1) should be multiple of 7. Aakash Educational Services Pvt. Ltd.




3.

If z = = x + + iy and and roots of

zz

3



3

zz



30 are

the vertices of a rectangle and z 0 is centre of rectangle. Let d

be distance of z 0 from the point on circle | z – – 3|  2 then maximum value of d is is _________. Sol. Answer (5) We have

 z z3  30



zz



zz z

3



2

 z2   30

x 2 – y 2) – 2i xy + x + x 2 – y 2 + 2i xy )) )) = 0  ( x 2 + y 2) (( x

 ( x 2 + y 2) ( x 2 – y 2) = 15 = (2 2 + 1) (22 – 1) Which suggests the possible values of x and and y are are x = = 2, y = = 1 or x = = –2, y = = –1 or x = = –2, y = = 1 or

|z –3| = 2

x = = 2, y = = –1 rectangle is (0, 0)  Centre of rectangle

(0, 0)

(3, 0)

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

4.

If th the e co com mpl plex ex nu num mbe berr A A((z 1), B(z 2) and origin forms an isosceles triangle such that  AOB  2 z1

 z22  4 z1 z2

2 , then 3

equals _______.

z1 z 2

Sol. Answer (3) If origin z 1, z 2 forms an isosceles triangle then z 12 + z 22 + z 1z 2 = 0 Hence

2 z 1

 z22  4z1z2 z z 1 2

5.



3z1 z2

3

z z 1 2

The Th e area area of the the tri trian angle gle fo forme rmed d by by three three po point ints s

3  i , – 1  3i an and ( 3 – 1) 1)  ( 3  1)i is __________.

Sol. Answer (2) Let

3

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

So, area of such triangle is  6.

1 2 1 1 | z |  � (2)2   4  2 2 2 2

The nu num mbe berr of of va valu lue( e(s s) of of k , for which both the roots of the equation x equation x 2 – 6kx + + 9( 9(k k 2 – – k k + + 1) = 0 are real, distinct and have values atmost 3 is ________.

Sol. Answer (0) x 2 – 6kx kx + + 9( 9(k k 2 – k + + 1) = 0 For real and distinct roots D > 0 Aakash Educational Services Pvt. Ltd.




 k 2 – (k 2 – k + 1) > 0 – 1 > 0  k –

…(i)

f (3) (3)  0 9 – 18k 18k + + 9( 9(k k 2 – k + 1)  0 k + + k 2 – k k + + 1  0  1 – 2k

 k 2 – 3k + 2  0 k  (–, 1]  [2, ) Also

sum of roots 2

...(ii)

3

 3k < 3 k < 1

…(iii)

From (i), (ii), (iii) We observe that there does not exist any real value of k . 7.

The Th e poss possib ible le gre great ates estt inte integr gral al val value ue of of a for which the expression

ax x

2

2

 3x  4

 2x  2

is less than 5 for all real x is is

______ ___ _____. __. Sol. Answer (2) We have

y 

ax x

As

⇒

2

ax x

2

2

2

ax

2

 3 x  4  2x  2  3 x  4  2x  2

5

 3 x  4  5 x 2  10 x  10 x

2

 2x  2

0

x 2 – 7 x – – 6 < 0 (as x 2 + 2 x + + 2 > 0,  x  R )  (a – 5) x It is satisfied for all x if if a – 5 < 0, 49 + 24 (a ( a – 5) < 0 a



71 24

a<3 The possible greatest integral value of a is 2. 8.

Let f ( x x ) = ax 2 + bx + c , where a, b, c are are real numbers. If the numbers 2a 2 a, a + b and c are are all integers, then the number of integral values between 1 and 5 that f ( x x ) can take is ________.

Sol. Answer (5) Firstly, let f ( x x ) = ax 2 + bx + + c ; a, b, c  R be be an integer whenever x is is an integer

 f (0), (0), f (1), (1), f (–1) (–1) are integers  c, a + b + c , a – b + c are integers c – – c , a – b + c c – – c are are integers  c , a + b + c

 c , a + b, a – b are integers Aakash Educational Services Pvt. Ltd.




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

 x( x  1)   (a  b ) x  c 2 

Then f ( ( x ) = ax 2 + bx + + c = = 2a 

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

 x ( x  1)  2a  2 a, a + b, c are are integers.   (a  b )x  c is an integer as 2a 2   f ( ( x ) is an integer for all integer x . 9.

The sm sma all lle est val alu ue of k , for which both the roots of the equation x 2 – 8kx + + 16( 16(k k 2 – k + + 1) = 0 are real, distinct and have values at least 4, is ________.

[IIT-JEE 2009]

Sol. Answer (2) x 2 – 8kx + + 16 ( x 2 – k + + 1) = 0 Roots are real and distinct Let f ( x x ) = x 2 – 8kx 8 kx + + 16( 16(k k 2 – k k + + 1)

 D > 0  x > > 1 Let f (4) (4) > 0  k  (–, 1]  [2, ) can be 2.  Least value of k can 10. If z is is any complex number satisfying | z – – 3 – 2i 2 i ||  2, then the minimum minimum value value of | 2 z – – 6 + 5i 5i || is ________. [IIT-JEE 2011] Sol. Answer (5) From the given condition, | z – – 3 – 2i 2 i ||  2

 | 2z 2 z – – 6 – 4i 4i ||  4  4  | (2 (2z z – – 6 + 5i 5i ) – 9i 9 i | | | – | 2z 2z – – 6 + 5i 5i || ||  || 9i |

 4  9 – | 2z 2z – – 6 + 5i 5 i ||  | 2z 2 z – – 6 + 5i 5 i ||  5 Minimum value of | 2z 2 z – – 6 + 5i 5i | | is = 5

  1 1 1 1  4 4 4 ...  is ________. 11. Th The e va valu lue e of 6  log3  3 2 3 2 3 2 3 2  2  

[IIT-JEE 2012]

Sol. Answer (4) As,



y

y





1 3 2

4 9

4

 y

(as y > > 0)

4 So, 6  log 3    4  9 2





 k   k    i sin    , where  7   7 

12. F or or an an y i nt nt eg eg er er k , let k  cos 

i  

1 . The value of the expression

12

∑ | k 1  k | k 1

is

3


∑ |  4k 1   4k 2 | k 1

Sol. Answer (4) 12

∑ S 

k 1 –  k

k 1 3

∑

 4k –1 –  4k – 2

k 1

r 1 – r = side length of the polygon having 14 sides = a So,

12a

S 

3a



4

SECTION - G

Multiple True-False Type Questions i 2 

1.

STATEMENT-1 : If   e

7

20

and

f (x)  A0

 ∑ A k x k , then f ( x x ) + f ( x ) + f (2 x )+ )+ ...+f ...+f (6 x ) = 5( A A0 + A7 x 7 + A14 x 14). k  1

STATEMENT-2 : The least argument of a complex number z , satisfying | z – – 16 16i i ||  8 is STATEMENT-3 STA TEMENT-3 : Maximum argument of the complex number satisfying | z – – 16 16i i ||  8 is (1) F T T

(2) F T F

(3) F F F

 3

.

2 3

.

(4) T F T

Sol. Answer (1) Statement-1 : f ( x x ) + f ( x ) + f (2 x ) + ...... + f (6 x ) 20

 7 A0  ∑ Ak x k (1   k  ......   6k ) but when k  7 k  1

and k  14, then 1 + k + 2k + ...... + 6k = 0 f ( x x ) + f ( x ) + ....... + f (6 x ) = 7( A A0 + A7 x 7 + A14 x 14) Statement-2 : Clearly AB = 8

sin 

AB OB

8

B

OB = OB = 16

A



8 16



1 ⇒

2

So, the argument is



 6



O

 3

Statement-3 : It can be seen from the figure that the maximum value of argument is Aakash Educational Services Pvt. Ltd.

2 3

.




2.

STATEMENT STA TEMENT-1 -1 : Two Two regular regular polygons polygons are inscribe inscribed d in the same same circle. circle. The first polygon polygon has has 1982 sides sides and and second has 2973 sides. If the polygons have a common vertex, then the number of vertex common to both of them is 991. STATEMENT-2 : The total number of compex numbers z satisfying |z | z – – 1| = |z | z + + 1| = |z | z | is one. STATEMENT-3 : The locus represented by |2011 z + + 1| = 2011|z 2011| z + + 1| is a straight line (1) T T T

(2) T F T

(3) F T T

(4) F F F

Sol. Answer (2) Statement-1 : The number of common vertices is equal to the number of common roots of z 1982 – 1 = 0 and z 2973 – 1 = 0, which is H.C.F. of 1982, 2973, i.e. 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.

3.

STATEMEN ENT T-1 : If x is is real and

y 

x

2

 x  3 , then y  (–, ) – (–11, 1). x  2

STATEMENT-2 : If [ ] represents the greatest integer function and f ( x x ) = x = x – – x [x ] then number of real roots of the

 1 equation f ( x )  f    1 are infinite.  x 

STATEMENT-3 : If the difference of the roots of the equation x 2 + hx + + 7 = 0 is 6, then possible value(s) of h are –8 and 8. (1) T T T

(2) T T F

(3) T F F

(4) F F F

Sol. Answer (1) Let

y



x

2

 x  3 so that x 2 – (y ( y + + 1) x + + 3 – 2y 2y = = 0 x  2

For real values of x of x (y + + 1)2 – 4(3 – 2y 2y )  0

 (y + + 11) (y (y – – 1)  0  y  1 or y  – 11  y lies lies in R – – (–11, 1)  Statement-1 is true The roots of of x x 2 – 4| x x | + 3 = 0 are 1, 3, –1, –3 and hence their sum is zero Statement-3 is also true 4.

|x | + 7 which belongs to the domain of the function STA ST ATEM TEMENT ENT-1 -1 : The sol soluti ution on of of the the equat equation ions s (3 x |x | – 3)2 = x y



x ( x – 3) are

1

given by – , – 2 . 9

STATEMENT-2 : For all x all x , x 2 + 2ax + + 10 – 3a 3 a > 0, then a  [–5, 2]. STATEMENT-3 : If x If x 2 – 2 x + + sin2 = 0 then x then x  [0, 2]. (1) F F T

(2) T T F


(3) T F F

(4) T T T




Sol. Answer (4) Clearly statement 1 and 2 are false We have ( x x – – 1)2 = cos2

 x = = 1 ± cos   x [0, 2]  statement-3 is true SECTION - H

Aakash Challengers Questions 1. Sol.

Let z 1 and z 2 be two given complex numbers such that z 1 z 2



z 2 z 1

z 1



z 2

z 2 z 1

 1 , and |z |z 1| = 3, then |z | z 1 – z 2|2 is equal to

1

z 12 + z 22 = z 1z 2



2 z 1

 z 22  0 2  z 1z 2  0z 1  0z 2

 0, z 1, z 2 form an equilateral triangle Hence, | z1 |  | z2 |  | z1  z2 |  3

 | z1  z 2 |2  9 2.

The locu locus s of the the centr centre e of a circ circle le which which touc touches hes the the given given circ circles les |z |z – z 1| = |3 + 4i 4 i | and |z |z – – z 2| = | 1  i 3 | is a hyperbola, then the length of its transverse axis is ......

Sol. |z z – – z 1| = 5 |z – z 2| = 2

5

|z z – – z 1| = r r + +5

z 1

|z z – – z 2| = r r + +2 |z – – z 1| – |z z – – z 2| = 3

2 r

r

z 2

z

which is a constant

 locus is hyperbola and we know PF 1 – PF 2 = 2a = length of transverse axis. length of transverse axis = 3

3.

If |z 1 + z 2|2 = |z 1|2 + |z 2|2, the

6



 z 1   is equal to ....... . z 2  

amp 

Sol. |z 1 + z 2|2 = |z 1|2 + |z 2|2 ………. (Given) |z 1 + z 2|2 = |z 1|2 + |z 2|2 + 2 Re ( z 1

 Re ( z1

z 2

z 2

) = |z 1|2 + |z 2|2 ………. (Relation)

)0

z   amp  1    z2  2  z1  6    6    amp  z       2    3  2 Aakash Educational Services Pvt. Ltd.




4.

For ev every real a  0, find all the complex numbers z that that satisfy the equation 2| z | – 4az + + 1 + ia ia = = 0.

Sol. Let z z = = x x + + iy So, then equation becomes,

2 x 2  y 2 – 4a( x  iy )  1  ia  0

…(i)

Comparing real and imaginary part, 2 x

2

 y2

…(ii)

– 4ax  1  0

–4ay –4 ay + + a = 0

…(iii)

(3)  a(– 4y 4 y + + 1) = 0  a = 0 or

But a cannot be zero. Hence

2

x

2



 

4  x 2 

4 x

2



1 16



 y 

1 4

1 4

gives

 4ax – 1

1 2   ( 4ax – 1) 16

1 4

 16a 2 x 2  1 – 8ax

(4 – 16 16a a2) x x 2 + 8ax – –



y  

3 4

=0

– 8a  16a2  12 8(1– 4a2 ) 4a ∓ 2 4a2  3 4( 4a2 – 1) 1)

Now, we can observe that if 0  a 

if



5.

a



1 2

1 2

, there is no solution

, solution is z z = = x x + + iy

4a  4a 2  3 4( 4a 2 – 1) 1)

 i �

1 4

Let z 1, z 2 be complex numbers with |z | z 1| = |z 2| = 1. Prove that |z | z 1+1| + |z | z 2 + 1| + |z |z 1z 2 + 1|  2.

Sol. We have, |z 1 + 1| + |z |z 2 + 1| + |z | z 1z 2 + 1|  |z 1 + 1| + |z | z 1z 2 + 1 – (z ( z 2 + 1)| = |z | z 1 + 1| + |z | z 1z 2 – z 2|  |z 1 + 1| + |z | z 2|| ||z z 1– 1| = |z 1 + 1| + |z | z 1 – 1|  |z 1 + 1 + z 1 – 1| = 2|z 2| z 1| = 2

 |z 1 + 1| + |z |z 2 + 2| + |z | z 1z 2 + 1|  2 Aakash Educational Services Pvt. Ltd.



6.


If on one e root root of th the e qua quadr drat atic ic eq equa uatio tion n ax 2 + bx + + c = = 0 is equal to the nth power of the other, then show that 1 n

(ac )

n

1

1

 (a

n

c)

n

1

b0

.

Sol. Let  and n be the roots of the given quadratic equation ax 2 + bx + + c = = 0

  + n =  1

 c  n 1

a

 c

    a

   

and  . n =

n

1

a

1

1

1



n

1

n

1

1

c)

n

b  0

1

n

n

n

1

n

c 1 a n1  c n1 a n1  b

1  c n1 a

n

1

1 n

a



1

a

= n +

 c      a

b

c n 1 a

n

c

n



 (a 7.

b

0

1

1  (ac n ) n 1  b  0

Let  and  be the roots of the equation x 2 – px + + q = 0 and V n = n + n. Show that V n + 1 = pV n – qV n – 1 . Find V 5 .

Sol. We have  +  = p p and and  = q. Now, pV Now, pV n – qV n–1 = ( + )(n + n) – (n–1 + n–1) = n+1 + n + n + n+1 – n – n = n+1 + n+1 = V n+1 Also V 5 = 5 + 5 = pV 4 – qV 3 = p p[ pV [ pV 3 – qV 2] – qV 3 = ( p p2 – q)(3 + 3) – pqV – pqV 2 = ( p2 – q)[( + )3 – 3( + )] – pq – pq[( [( + )2 – 2] = ( p p2 – q)[ p p3 – 3 pq pq]] – pq – pq[ p [ p2 – 2q] 8.

If p p,, q are roots of the quadratic equation x 2 – 10 10rx rx – – 11s = 0 and r , s are roots of of x x 2 – 10 px – – 11q 11q = 0, then find the value of p p + + q + r + + s.

Sol. According to the question p + p + q = 10 10rr ; pq pq = = –11s –11s r + r + s = 10 p ; rs rs = = –11q –11q. On subtraction, ( p – p – r ) + (q – s) = 10(r 10( r – – p p))

 (q – s) = 11(r 11(r – – p p))

....(i)

Also, p Also, p is is a root of of x x 2 – 10 10rx rx – – 11 11s s = 0

 p2 – 10 pr – – 11 11s s = 0 Similarly r 2 – 10 pr – – 11 11q q = 0 On subtraction, p2 – r 2 = –11 (q (q – s)

 ( p p – – r ) ( p p + + r ) = –11 × 11 (r ( r – – p p))  p + r = = 121 Now p p + + q + r + + s = 10 ( p + p + r ) = 10 × 121 = 1210. Aakash Educational Services Pvt. Ltd.




9.

+ 1.5 + 0.5 Solve th the eq equation 4x + + 9x + = 10 × 6x .

Sol. The given equation can be written as 22x + 3 + 32x + 1 = 10 × 2 x × 3x . 3  8  3.   2

 3  3   2

2x



 3 3.    2



3. 

2x

2x

 3  10.   2

x

x

 3  10    8  0  2 x

x

 3  3 6  4   8 0  2  2

  3   3   3    2  4    2  0   2   2      2  x

x

x

 3   3      2 3.   4  0  2     2  x

x

x

x

4  3  3  Either    2 or     2  2 3 x

 3 (log3 – log2) = log2  When    2 , Taking logarithm of both sides we get x (log3  2 

x



log2 log3 log3  log2 log2 x

4  3 Also when     2 3

 x(log3 – log2) = log4 – log3 ⇒ x



log4 log4  log3 log3 log3 log3  log2 log2



x

log2 log4 log4  log3 log3 or log3 log3  log2 log2 log3 log3  log2 log2



10.. The twice 10 twice of the the product product of of real roots roots of of the equati equation on (2 x + + 3)2 – 3|2 x + + 3| + 2 = 0 is ________. Sol. |2 x + + 3|2 – 3|2x + 3| + 2 = 0

 (|2 x + + 3| – 2) (|2 x + + 3| –1) = 0  |2 x + + 3| = 2, |2 x + + 3| = 1  2 x + + 3 = ± 2, 2 x + + 3 = ± 1 

x

5 1 ,  2, 2,  1 2 2

 ,

5  5   1 2)( 1)  Product =       ( 2)(  2  2 2 2(Product) = 5 Aakash Educational Services Pvt. Ltd.



11.

Sol.


The eq equa uati tion on ax 2 + bx + + c = = 0 and and x x 3 – 4 x 2 + 8 x – – 8 = 0 have two roots in common. Then 2 b + c is is equal to ______ ___ _____. __. x

3

2  4 x 2  8 x  8  0  ( x  2)( x  2 x  4)  0

 x = = 2, 

x

2

a



1

x

2

 2 x  4  0 will give imaginary roots.

 2 x  4  0 and



b

c



2

4

ax

2

 bx  c  0 will have both roots common.

 k  b = –2 –2k k , c = 4k  2b + c c = = 0

12. If x = = 3 + 3 2/3 + 31/3, then the value of the expression x 3 – 9 x 2 + 18 x – – 10 is equal to ________. Sol.

 

x

33

2/ 3

 31/ 3

  3  31/ 3 (1  31/ 3 )

x

 ( x   3)3  3(1  31/ 3 )3 

x



x

3

3(1  3  3(31/ 3  32/ 3 )]   27 x  9 x 2  27  3(

3

 9 x 2  18 x  12  0 

x

3

n

2

n

n –1

 (a  b

3

3( 4  3( x  3)]  27 x  9 x 2  27  3(

 9 x 2  18 x  10  2

13. Let n be an even positive integer such that

of order n. Prove that

x

2 k

is odd and let 0, 1, ...... n–1 be the complex roots of unity

n

)  ( a  b 2 )2 for any complex numbers a and b. 2

k  0

Sol. If b = 0 result can be shown easily. But if b if b  0. Let n = 2(2m 2(2m + 1) 2 Let the complex number  such that  

a b

and the polynomial.

f ( x x ) = x n – 1 = ( x – – 0)( x x – – 1)( x x – – 2)....... ( x x – – n – 1) We have, n

    1 f      . (  – i  )( – i  ) ........ (  – i  0 1 n – 1)  i   i  n

    1 and f     –  . ( + i 0)( + i 1) ........ (  + i n – 1)  i   i  Hence, f

       f  –  = ( 2 + 02 )(2 – 12 ) ........ ( 2 – 2n – 1) i i n –1

Therefore

 k

n –1

2 k

(a  b )  b

0

n



k  0

a 2    k  b

n –1

b

n

    2 k

k  0





     b f   � f  –   b � (  2 )2  i   i  n

b

b

n

n

 a  2      b

2( 2m  1)

m

1

 a2  

  1 

m

1

b

m

1

 1

2

2

 b2

m

 1

2m  1

2

 

= (a ( an/2 + bn/2) hence proved 14. Let z 1, z 2, z 3 be complex numbers, not all real, such that | z 1| = |z |z 2| = |z |z 3| = 1 and 2(z 2( z 1 + z 2 + z 3) – 3z 3z 1z 2z 3  R . Prove that max(arg z 1, arg z 2, arg z 3) 

 6

. Where 0 < arg(z arg( z 1), arg(z arg(z 2), arg(z arg(z 3) < 2.

Sol. Let z k = cost cost k + i sint sint k : k {1, 2, 3} The condition 2(z 2( z 1 + z 2 + z 3) – 3z 1z 2z 3R

 2(sin 2(sint t 1 + sint sint 2 + sint sint 3) = 3sin(t 3sin(t 1 + t 2 + t 3) Assume by the way contradiction that max( t 1, t 2, t 3) <

t 

t1  t 2

 t 3

3

 6

,



Hence t 1, t 2, t 3 <

Let

…(i)

6

   0,   6

t  t  t 1 (sint1  sin t2  sin t 3 )  s siin 1 2 3 3 3

…(ii)

From (i) & (ii) we have, sin(t1  t2  t3 ) 2

 t  t  t   sin  1 2 3    3

Then, sin3t sin3t  2sin 2sint t

 4sin3t – – sint sint  0 i.e.,, i.e.

 

2

sin t  

t  0 ,

1 4

, hence sint sin t 

1 2

and

t 

 2

which contradicts that

  6

 max( max(t t 1, t 2, t 3) 

 6

Hence proved. Aakash Educational Services Pvt. Ltd.




15. Let z 1, z 2, z 3 be complex numbers such that (i)

| z 1 | = | z 2 | = | z 3 | = 1

(ii) z 1 + z 2 + z 3  0 (iii) z 12 + z 22 + z 32 = 0 Prove that for all integers n  2, |z 1n + z 2n + z 3n|  {0, 1, 2, 3}. Sol. Let s1 = z 1 + z 2 + z 3, s2 = z 1z 2 + z 2z 3 + z 3z 1, s3 = z 1z 2z 3 and take a cubic equation. z 3 – s1z 2 + s2z z – – s3 = 0 with roots z 1, z 2, z 3 It is given that z 12 + z 22 + z 32 = 0 Hence, s12 = 2s2

…(i)

Again we have,



s 2

 s 3 

1

 z1



1 z2



1 z 3



 s3 (z1  z2  z3 )  s3 �s1

…(ii)

Now, from (i) and (ii) 2

s1

 2s 3 � s1 and | s1 |2  2(s3 )(s1) = 2|s 2|s1|

|s s1| = 2  s1 = 2 with || = 1  | Now, again from relation (i) and (ii) it follows that

s2



1 2

2 s1

 2

2

and

s 3



s 2



s 1

2

2

2



Now, the given equation becomes z 3 – 2z 2 + 2 2z – – 3 = 0

 (2 – )( )(z z 2 – z + + 2) = 0 The roots are z = z = ,  = – 2 Now, R n = |z 1n + z 2n + z 3n| = |n + nn + (– 1) nn2n| = | |n |1 + n + (– 1) n2n| R 0 = 3, R 1 = |1 +  – 2| = |– 2 2| = 2 R 2 = |1 + 2 + | = 0 R 3 = |1 + 1 – 1| = 1 R 4 = 0 Aakash Educational Services Pvt. Ltd.




R 5 = 2 R 6 = 3

 R k + 6 = R k for all integers k .  R n  {0, 1, 2, 3} 16.. Fi 16 Find nd all all comp comple lex x numb number ers s z such such that z

– | z  1| 

z

 | z – 1|

Sol. |z z – – |z | z + + 1|| = |z | z + + |z – – 1|| |z z – – |z + + 1|| 2 = |z + + |z – – 1|| 2  |



 z – | z  1|  �  z – | z  1|   z  | z – 1| �  z  | z – 1 |



zz

– | z  1| ( z  z )  | z  1|2  z� z  ( z  z) | z – 1|  | z – 1|2

| z  1|2 – | z – 1|2  ( z  z )(| z  1|  | z – 1 |) ( z  1)( z  1) – ( z – 1)( z – 1)  ( z  z )(| z  1|  | z – 1 |) zz

 z  z  1 – z z  z  z – 1  ( z  z)(| z  1|  | z – 1|) 2( z  z ) – ( z  z ) (| z  1|  | z – 1|)  0

( z  z ) (2 – (| z  1|  | z – 1 |)  0 i.e.,

z

| z + + 1| + |z | z – – 1| = 2  z  0 or |z

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

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

t – 1 t  1

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

The equation

z

z = = bi , b  R .  z  0 has the solutions z

Hence the solutions to the equation are { bi bi :: b  R }  {a  R R :: a  [– bi ]} ]} 17. Suppose p p is is a polynomial with complex coefficients and an even degree. If all the roots of p p are are complex non-real numbers with modulus 1, prove that p p(1) (1)  R iff p iff p(–1) (–1)  R . Sol. In order to prove the result it will be sufficient to prove that

p(1) (1) p(–1)

R

Let x 1, x 2, x 3,........, ,........, x x 2n be roots of p p.. Then p(( x p x ) = ( x x – x – x 1)( x x – x – x 2)( x x – x – x 3) ........ ( x x – x – x 2n) For some   C , and 2n

1 – x k (1 – x1) (1 (1 – x2 ) .. ........ (1 (1 – x 2 )   1  x k ) ........ (– (–1 – x 2 ) p(–1) (–1 – x1 ). k  1

(1) p(1)

n



n

It is given that | x x k | = 1 for all k = = 1, 2, ......., 2n 2 n. Then Aakash Educational Services Pvt. Ltd.



1–

 1 – xk  1 – xk  1  x   1  x   k  k 1  p(1) (1)  Now,  p(–1) 

1 x k

1



x k

–1

x k

1

p(–1)

1–

x k

1

x k

 1 – xk  2n  1 – x k  –  1 x    1  x k    k k 1 k  1





2n

2n

1– x k

 1 x k  1

p(1) (1)

–

x k

2n

  1






k

 

p 1 p

 1

is a real number.

18. The po poin intts A1, A2, ........, A10 are equally distributed on a circle of radius R (taken in order). Prove that A1 A4 – A1 A2 = R . Sol. Without loss of generalization, we take A1= R R + + i 0 Now, in triangle A triangle A1OA4 i6

A4 = Re 10

cos108  

R

2

2

¡

A3 Re

2

 R  ( A1A4 ) 2R 2

30° 30°

( A A1 A4)2 = 4R 4 R 2sin254°

4 10

¡

A2 Re

i2 10

30°

= 2R 2R sin54° sin54°

0

A1 (R + ie)

Similarly in triangle A1OA2

cos36 

R

2

 R2  ( A1A2 )2 2R 2

( A A1 A 2)2 = 2R 2sin218° A1 A 2 = 4R 4 R sin18° sin18° Now A Now A1- A4 – A1 A1 = 2R (sin54° (sin54° – sin18°) = 2R 2 R  2cos36° sin18°



=

4R  



5  1  4

 

5  1 4

 

= R (proved) 19. Let a and b be positive real numbers with a3 + b3 = a – b, and k k = = a2 + 4b 4 b2, then (1)) k < 1 (1

(2) k > 1

(3) k = 1

(4) k > 2

Sol. Answer (1) Given, a3 + b3 = a – b Aakash Educational Services Pvt. Ltd.



Solution of Assignment (Set-2) a

let

 k (a > b)

b

Now a3 + b3 = a – b  b3k 3 + b3 = bk – b



b

2



k  1 3

k

1

Now, a2 + 4b2 = b2k 2 + 4b2 = b2 (k 2 + 4)

k

=

=

=

2

4

  k  1

3

1

k

k

1

2

 4k  5 3

k

1

 k  2

1

3

k

2

1

 as k k > >0

1

 a2 + 4b2 < 1 20. Let k be be a real number such that the inequality

x  3 

6 x

 k has a solution then the maximum value

of k is (1)) (1

(2)) (2

3

6



(3)) (3

3

6

6

(4)) (4

3

Sol. Answer (3) f ( x x ) = Df =

(3) f (3)

x

3 

6

6

 x

[3, 6]

=

3

3

, f (6) (6) =

3 3

from symmetry f (x ) have maxima

at

x



63 2

f ( 4.5 ) 

4.5

6

 4.5

4.5  3  6  4. 4 .5 =

21. Let  and  be the root of

(1) 2

x

2

2 1.5

 px 

(2)

2 2



1 2 p

2

6

 0 , p  R . The minimum value of 4 + 4 is

(3)) (3

2

2

(4)) (4

2

2

Sol. Answer (3) x 2  px 

1 2 p

2

 0, p  R

Let ,  be roots Aakash Educational Services Pvt. Ltd.




1

p,,  =   +  = – p

2 p

2

4 + 4 = ( + )4 – 43 – 43 – 622 ( + )4 – 4(2 + 2) – 622 ( + )4 – 4  ( + )2 – 2  – 622

p

4

p

p

4

 2  p 

2



2

2

 2

p 2

p4 +

4 p

p

2

p



2



4

6 4p

2 p

4

½

4

 1    (A.M. – G.M. inequality)  2

2



4p

1



Now,

4

6

2

4

4

p



4

  

1

 1  2.    2

1 2 p

4

½

 Minimum value of 4 + 4 =

2

2

22.. The numb 22 number er of real solu solution tions s of the the equat equation ion (1) 0

4

(2) 1

97  x

 4 x  5 (3) 2

(4) 4

Sol. Answer (3) 4

97  x

a

m

b



4

x

5 m

 a  b  if  2 

m

2

m (0,

1)

Put a = 97 – x , b = x , m = w/f 4

97  x

 4 x

2



4

97 –

x



 97 

   4

x

2



1 4

 2 × 2.63

Which holds for 4

97 – x

 4 x 

5

Now, Clearly for

4

97 – x

 4 x 

5 has


two solution. Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456



23. If f ( x x ) is a polynomial of degree at least two with integral co-efficients then the remainder when it is divided by ( x – – a) ( x – – b) is, where a  b.

 f (a)  f (b)  af ( b)  bf ( a)  ab   ab  

 f (a)  f (b)  af ( b)  bf ( a)  ba   ab  

(2)) (2

x

 f (b)  f (a)  af ( b)  bf ( a)  ab   ab  

(4)) (4

x

(1)) (1

x

(3)) (3

x

 f (b)  f (a)  bf (a)  af ( a)  ab   ab  

Sol. Answer (2) f ( x x )  polynomial of degree ‘n ‘ n’ = an x n + an –1 x n–1 + …… + ar when this polynomial will be divided by ( x ( x – – a) ( x – – b), remainder will be of form Ax form Ax + + B Now, f (a) = A.a + B f (b) = A A..b + B f (a )  f (n ) a

b

Also B =

= A

 a

f a

…(i)

   f  b

f a

ab

   bf  a  af (a)  af ( b)

af a

ab

   bf  a

af b

ab

 Remainder will be

 f  a   f  b   a f  b   bf  a b  ab 

x 

a 

24. Let p = ( x x 1 – x 2)2 + ( x x 1 – x 3)2 + ........ + ( x x 1 – x 6)2 + ( x x 2 – x 3)2 + ( x x 2 – x 4)2 + ........ + ( x x 1 – x 6)2 + ........ + 6

2

( x x 5 – – x x 6) =

∑

( xi  x j )2 .

1 i  j  6

Then the maximum value of p p if if each x each x i (i = = 1, 2, ......., 6) has the value 0 and 1 is (1) 1

(2) 2

(3) 4

(4) 9

Sol. Answer (4) 6

Let

p

∑  xi  x j 



2

i ,J 1

If we consider any three of x 1, x 2, x 3, x 4, x 5, x 6 is one and another three ‘0’

 p = 9 Aakash Educational Services Pvt. Ltd.




25. If x , y , z are are three real numbers such that x + + y + + z = = 4 and x 2 + y 2 + z 2 = 6, then 2

(1)) (1

3

 x , y , z  2

(2) 0  x , y , z  2

(3) 1  x , y , z  3

(4) 2  x , y , z  3

Sol. Answer (1) x + x + y = = 4 – z x 2 + y 2 = 6 – z 2  2 xy = ( x + y + y )2 – ( x x 2 + y 2) = (4 – z )2 – (6 – z 2) = 2z 2 – 8z z + + 10 The quadratic equation whose roots are x and and y is is 2 t – ( x x + + y ) t + + xy = = 0 t 2 – (4 – z )t t + + z 2 – 4z + + 5 = 0 for real roots, (4 – z )2 – 4( 4(z z 2 – 4z + + 5)  0 – 2) ( z – – 2)  0 3z 2 – 8z z + + 4  0  (3z – 2 3

 z  2

2



Similarly x Similarly x , y   ,2  3  26.. If unit 26 unity y is doub double le repea repeated ted roo roott of of px px 3 + q( x x 2 + + x x ) + r = = 0 then (1)) pr < (1 < 0

(2) pq pq < < 0

(3) pqr > > 0

(4) pqr < < 0

Sol. Answer (2) 1 1

px 3 + qx 2 + qx + + r = = 0

 + 2 =

q p

, 2 + 1 =



q p

,=

q

3 + 3 = 0

 = –1 pq < pq < 0 , pr , pr > > 0

p

 r p

1

r.p>0

q p < . p < 0

 pqr > > 0 or pqr or pqr < < 0 pq < < 0 is suitable option  pq 27.. The numbe 27 numberr of real real soluti solutions ons of the the equation equation 4 x 99 + 5 x 98 + 4 x 97 + 5 x 96 + …… + 4 x + + 5 = 0 is (1) 1

(2) 5

(3) 7

(4) 97

Sol. Answer (1) 4 x 99 + 5 x 98 + 4 x 97 + 5 x 96 x 98 (4 x + + 5) + + x x 96 (4 x + + 5) + + x x 94 (4 x + + 5) + …….. +1 (4 x + + 5) (4 x + + 5) ( x 98 + x 96 + ….. + 1) x =

5 4

as x 98 + x 96 + ….. +1  0

 Only one solution  Aakash Educational Services Pvt. Ltd.






Cls Jeead-17-18 Xi Mat Target-2 Set-2 Chapter-5

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