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Transcript of Complex Numbers (Package Solutions)
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Solutions of Assignment (Set-2) Complex Numbers and Quadratic Equations (Solutions)
Aakash Educat ional Serv ices Pvt. L td . Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 1 -
Section-A
Q.No. Solution
1. Answer (2)
–9999
9999 2 4999 4999
1 1 1
( ) (–1)i
i i i i =
2
1 –
–
ii
i i
2. Answer (4)1
2
iz
2 22
2
(1 ) 1 2 2
2 22
i i i iz i
Now, z1929 = (z2)964·z
= i 964·z
= z
=1
2
i
3. Answer (2)
–3 –5 3 –1 5 –1
= 3 5i i
= 215i
= – 15
4. Answer (1)
72 5 2 6 2 6 2 2 711 12 13 14 15 ( ) · ( ) ( ) ( )
1 1
i i i i i i i ii i i i i
i i
– (1) ( ) – 1– – – 1– ·
1 1 1 1–
i i i i i i
i i i i
=2
2
– – – 1 –( 1)
1– 2 2
i i i i
i
5Chapter
Complex Numbers and
Quadratic E uations
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Complex Numbers and Quadratic Equations (Solutions) Solutions of Assignment (Set-2)
Aakash Educational Services Pvt . Ltd.
Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 2 -
Q.No. Solution
5. Answer (1)
1 1 1·
1– 1– 1
i i iz
i i i
=2 2
2
(1 ) 1 2
1– 2
i i i
i
= i
z8 = (i)8 = (i2)4
= 1
6. Answer (3)
Let z be the complex number, then
2 5· 1
17
iz
17 17 2 – 5
·
2 5 2 5 2 – 5
iz
i i i
=
2
17 2 – 5 172 – 5
2 – 25 27
ii
i
7. Answer (4)
Additive inverse of 5 + 7i is –5 –7i
8. Answer (2)
2
1 2 1 2 1 (1 2 )(1 )·
1– 1– 1 1–
i i i i i
i i i i
1– 2 (2 1) –1 32 2i i
1 3 –
2 2i
1 3
– ,2 2
lies in the second quadrant.
9. Answer (2)
2 2
2
1 (1 )·(1 ) (1 ) 1 2
1– (1– )(1 ) 1– 2
i i i i i ii
i i i i
and2 2
21– (1– )·(1– ) (1– ) 1 – 2 – 1 (1 )(1– ) 1– 2
i i i i i i ii i i i
3 3
3 31 1– – ( ) – (– )
1– 1
i ii i a ib
i i
– i – i = a + ib
0 – 2i = a + ib
a = 0 and b = –2
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Solutions of Assignment (Set-2) Complex Numbers and Quadratic Equations (Solutions)
Aakash Educat ional Serv ices Pvt. L td . Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 3 -
Q.No. Solution
10. Answer (3)
–2 – 3x i
2 – 3x i
(x + 2)2
=
2
– 3i
x2 + 4 + 4x = 3i2
x2 + 4x + 7 = 0
Now, 2x4 + 5x3 + 7x2 – x + 41
= 2x2(x2 + 4x + 7) –3x(x2 + 4x + 7) + 5(x2 + 4x + 7) + 6
= 0 – 0 + 0 + 6
= 6
11. Answer (4)
9
17
315
1
i i
= 9
82
2 157
1·
( ) ·i i
i i
=9
1 – i
i
=9
2 –
ii
i
= [i + i]9 = (2i)9
= 512(i2)4·i
= 512i
12. Answer (2)
z = 3 – 2i
Re z = 3, Im z = –2
Re z(Im z)2
= 3(–2)2 = 12
13. Answer (3)
z1 – z2 = (4 – 3i) – (3 + 9i)
= (4 – 3) +i(–3–9)
1 – 12i
14. Answer (4)
z1z2 = (2 + 3i)(5 – 3i)
= (10 + 9) + i(15 – 6)
= 19 + i(9)
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Complex Numbers and Quadratic Equations (Solutions) Solutions of Assignment (Set-2)
Aakash Educational Services Pvt . Ltd.
Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 4 -
Q.No. Solution
15. Answer (1)
1 2 1 2 1– 3·
1 3 1 3 1– 3
i i i
i i i
=2
(1 2 )(1– 3 )
1– 9
i i
i
= 1
1 2 1– 310
i i
= 1
(1 6) (2 – 3)10
i
=7 – 7
– 10 10 10
i i
16. Answer (4)
17. Answer (3)
18. Answer (1)
19. Answer (4)
20. Answer (2)
21. Answer (3)
22. Answer (2)
23. Answer (1)
24. Answer (4)
25. Answer (2)
26. Answer (1)
27. Answer (3)
28. Answer (2)
29. Answer (4)
30. Answer (3)
31. Answer (2)
32. Answer (3)
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Solutions of Assignment (Set-2) Complex Numbers and Quadratic Equations (Solutions)
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- 5 -
Q.No. Solution
33. Answer (1)
34. Answer (4)
35. Answer (4)
36. Answer (4)
37. Answer (2)
38. Answer (3)
39. Answer (4)
40. Answer (3)
41. Answer (4)
42. Answer (1)
43. Answer (2)
44. Answer (2)
45. Answer (3)
46. Answer (2)
47. Answer (2)
48. Answer (4)
)32(2)3( iizz
)32(2)3( iizz
|z|2 + 3iz = 4 + 6i
(x2 + y2) + 3i (x + iy) = 4 + 6i
(x2 + y2 – 3y) + 3ix = 4 + 6i
3x = 6 and x2 + y2 – 3y = 4
x = 2 and 4 + y2 – 3y = 4
y = 0, 3
z = x + iy
= 2 + i.0 and 2 + 3i
= 2, 2 + 3i
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Complex Numbers and Quadratic Equations (Solutions) Solutions of Assignment (Set-2)
Aakash Educational Services Pvt . Ltd.
Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 6 -
Q.No. Solution
49. Answer (4)
f (x) = x4 – 8x
3 + 4x2 + 4x + 39
x = 3 + 2i (x – 3)2 = –4
x2 –6x + 13 = 0
Now x4 – 8x
3 + 4x2 + 4x + 39 = (x
2 – 6x + 13) (x2 – 2x – 21) + (–96x + 312)
Now f (3 + 2i) = –96(3 + 2i) + 312
= –288 – 192i + 312
= 24 – 192i
= a + ib
a = 24, b = –192.
Required ratio =8
1
192
24
50. Answer (1)
In regular hexagon OA = AB = BC = CD = ED = EF = FA
Length of perimeter = 6 × |OA|
= 6 1 4
6 5
51. Answer (1)
arg(1 + i) =4
2
arg 1 33
i
5
arg 36
i
arg 36
i
arg2
i
arg 32
i
arg 2 0
arg 1
Required sum11
12
O
A i(1 + 2 )E
F
B
C
D
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Complex Numbers and Quadratic Equations (Solutions) Solutions of Assignment (Set-2)
Aakash Educational Services Pvt . Ltd.
Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Q.No. Solution
56. Answer (4)
If z1 and z2 are two complex number if 1Im( ) 0z and 2Im( ) 0z then z1 > z2 or z1 < z2 does not hold.
57. Answer (3)
2
3
2
1)()( 3231 izzzz
3/3231 )()( iezzzz
Now using concept of rotation.
z1, z2, z3 are vertices of equilateral triangle.
58. Answer (3)
( )ia i
log loge ea i i
2log loge ea i e
log2e a
2a e
Therefore sin(ln ) sin 12
a
Im( ) arg( ) 0 0 0a a
Therefore 1 2 3, ,S S S are correct.
59. Answer (2)
if z2 + z + 1 = 0
(z – ) (z – 2) = 0
z = , 2
if z = , then 21
z
To find the value of2
21
212
3
32
2
22
1......
111
zz
zz
zz
zz
Now 21
,111
,111
3
3
2
2
2
2
z
zz
zz
z
111
,1111
2
2
5
5
4
4
4
4
z
zz
z and
216
6 z
z ...... and so on
z3
z2
z1
/3
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Solutions of Assignment (Set-2) Complex Numbers and Quadratic Equations (Solutions)
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Q.No. Solution
Therefore,
2
21
212
3
32
2
22
1......
111
zz
zz
zz
zz
= {(– 1)2
+ (–1)2
+ (2)2
} + {(–1)2
+ (–1)2
+ (2)2
} × ...... 7 times
= (1 + 1 + 4) + (1 + 1 + 4) × .... 7 times
= 6 + 6 × ..... 7 times
= 6 × 7 = 42
60. Answer (1)
is an imaginary 5th root of unity
5 = 1
1 + + 2
+ 3
+ 4
= 0 … (i) [ sum of nth
roots of unity is zero]
Now,
1
1log 322
1log 4
2 [ from(i) 1 + + 2 + 3 = – 4]
|||2|log|2|log2
log1
log 42
422
5
2
12log2 [ | | = | 2 | = | 3 | = | 4 | = 1 hence nth roots of unity lie on unit circle]
61. Answer (3)
We have
x3n+1 – 1 = (x – 1) (x – 1) (x – 2) ......... (x – 3n)
Thus,)()........)((
)()........()(
321
32
22
12
n
n
)()......()()1(
)()......()()1(.
)1(
1
321
32
22
122
n
n
1.
1.)(.
1
1
1
1)(.
1
13
232
13
132
n
n
n
n
1
1.
1
1 2
[ 3n = 6n = 1]
11
12
2
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Solutions of Assignment (Set-2) Complex Numbers and Quadratic Equations (Solutions)
Aakash Educat ional Serv ices Pvt. L td . Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 11 -
Q.No. Solution
66. Answer (2)
To find area of a whose vertices are represented by complex number 0, z and zei (0 < < )
Area of Abc sin2
1
sin||||2
1zz
sin||2
1 2z
67. Answer (2)
22
z
z
22||
z
z
22
r
r
22
22
r
r
when 22
02
r
r
r 2 – 2r – 2 0
3131 r … (i)
31max r
68. Answer (1)
Let the value of i x iy
2( )i x iy
2 2 2i x y xy
On comparing real and imaginary part
2 2 0x y and 2 1xy
x y and1
2xy
Therefore1 1
,2 2
i ii
| |z
| |z
O(0, 0)
A z( )
B ze( )i
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Complex Numbers and Quadratic Equations (Solutions) Solutions of Assignment (Set-2)
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Q.No. Solution
69. Answer (4)
Roots of the equations
(x – a) (x – b) + (x – b) (x – c) + (x – c) (x – a) = 0
i.e., {x2
– (a + b)x + ab} + {x2
– (b + c)x + bc} + {x2
– (a + c)x + ac} = 0
i.e., 3x2 – 2(a + b + c)x + (ab + bc + ca) = 0
have equal roots, Therefore
B2 – 4 AC = 0
4(a + b + c)2 – 4 × 3 (ab + bc + ca) = 0
4[a2 + b2 + c
2 + 2ab + 2bc + 2ca] – 12ab – 12bc – 12ca = 0
4(a2 + b2 + c
2) – 4ab – 4bc – 4ac = 0
a2 + b
2 + c2 – ab – bc – ca = 0 (Hence (3) is true)
70. Answer (2)
2||2
1||||log
2
3
z
zz
22
)3(||2
1||
z
zz
| z |2 – | z | + 1 < 6 + 3 | z |
| z |2 – 4| z | – 5 < 0
(| z | + 1) (| z | – 5) < 0
but | z | + 1 > 0
| z | – 5 < 0
| z | < 5
71. Answer (3)
arg z =4
–1tan4
yx
1y
x
|y| = |x| x2 – y2 = 0
Re(z2) = 0
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Solutions of Assignment (Set-2) Complex Numbers and Quadratic Equations (Solutions)
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- 13 -
Q.No. Solution
72. Answer (3)
Let z = x + iy
Given equation is,
z2
+ z|z| + |z|2
= 0
(x + iy)2 + (x + iy) + 2 2x y + (x2 + y2) = 0
x2 – y2 + 2ixy + 2 2 2 2 2 2 0x x y iy x y x y
2 2 2 2 22 (2 ) 0x x x y i xy y x y
Now, 2 2 22 0x x x y
2 2(2 ) 0x x x y
x = 0 or x2
+ y2
= 4x2
3x2 – y2 = 0
Al ternat ive
2
21 0
| || |
z z
zz
2
1 0| | | |
z z
z z
2,| |
z
z
z = |z|, z = 2|z|
73. Answer (2)
P
y x=13
3 0y x
Now,
2 2
3·2 3 0
( 3) 1p
6
32
p
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Complex Numbers and Quadratic Equations (Solutions) Solutions of Assignment (Set-2)
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Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 14 -
Q.No. Solution
74. Answer (3)
4 1Re
2 1 2
z
z
4 4
12 1 2 1
z z
z z
4 4
12 1 2 1
z z
z z
2 8 4 2 8 4
1(2 1)(2 1)
zz z z zz z z
z z
4 7 7 8 4 2 2 1 zz z z zz z z
9 9 9 0 z z
1 z z
Hence point z lies on a straight line.
75. Answer (2)
h(x) = xf (x3) + x2g(x6) is divisible by x2 + x + 1,
So, when h(x) will be divided by x – and x – 2 remainder will be 0.
h() = f (1) + 2g(1) = 0 …(i)
h(2) = 2f (1) + g(1) = 0 …(ii)
Now, adding (i) & (ii),
( + 2
)f (1) + ( + 2
)g(1) = 0 – f (1) – g(1) = 0 f (1) = – g(1)
76. Answer (1)
The given expression is (x – 1) (x – 1) ..... (x – 1) ..... till 2n factors.
= (x – 1)2n
77. Answer (2)
101101 101 101cos sin –
6 6
z i i i
101101101 101
cos sin6 6
xi i
1015 5cos sin
6 6i i
3cos sin
6 6 2 2
ii i
3
2 2
i
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Solutions of Assignment (Set-2) Complex Numbers and Quadratic Equations (Solutions)
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Q.No. Solution
Now,
(i101 + z101)103
1033
– 2 2
i
1035 5
cos sin6 6
i
515 515cos sin
6 6i
11 11cos sin
6 6i
cos – sin6 6
i
3
– 2 2
i
78. Answer (2)
1
az
az
|z – a| = |z + a|, let z = x + iy then,
(x – a)2 + (y)2 = (x + a)2 + y2
x2 – 2ax + a
2 + y2 = x
2 + a2 + 2ax + y
2
4ax = 0
x = 0
is y-axis
79. Answer (2)
Let z = x + iy
1)(
122
iyxi
iyxz
ixy
ixy
ixy
yix
)1(
)1(
)1(
212
22
)1(
)12()1()2(2)1()12(
xy
xxyyixyyx
from given condition
2)1(
)12()1()2(22
xy
xxyy
2y – 2y2 – 2x
2 – x = –2(x2 + y2 + 1 – 2y) = 2y – x = –2 + 4y
2y + x – 2 = 0, i.e. a straight line
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Complex Numbers and Quadratic Equations (Solutions) Solutions of Assignment (Set-2)
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- 16 -
Q.No. Solution
80. Answer (4)
z = x + iy
| 2008z – 1 | = 2008 | z – 2 |
|2|2008
1
zz
Put z = x + iy
2222
)2()(2008
1yxyx
xx 442008
12
2008
12
2
2008
14
10044
xx
1004
14
2008
14
2
x
a line parallel to y-axis.
81. Answer (2)
31
1arg
APBz
z
z lies on a circle Al ternatively
put z = x + iy
31
1arg
z
z
3)1(
)1(arg
iyx
iyx
3)1(
)1(
)1(
)1(arg
iyx
iyx
iyx
iyx
3)1(
)2()1(arg
22
22
yx
yiyx
31
2tan
22
1
yx
y
31
222
yx
y 023)(3 22 yyx
013
222 yyx
Which represents a circle having centre at
3
1,0 and radius
3
21
3
1
3
AB
(–1, 0) (1, 0)
P z( )
O
y
x
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Solutions of Assignment (Set-2) Complex Numbers and Quadratic Equations (Solutions)
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- 17 -
Q.No. Solution
82. Answer (1)
,4
exp2
iiz where is parameter put z = x + iy
4sin4cos2 iiiyx
equating real and imaginary parts we get
)1......(4
cos2
x
4sin21y
or )2....(4
sin21
y
squaring and adding (1) and (2), we get
x2
+ (y – 1)2
= 4which represents a circle with centre (0, 1) and radius 2.
83. Answer (4)y
x
iz
A
z
iz
z O
90°
On rotating OA by 90° angle we can find other vertices.
84. Answer (2)
z1, z2, z3 are the vertices of an equilateral triangle such that |z1| = |z2| = |z3|
or |z1 – 0| = |z2 – 0| = |z3 – 0|
origin is the circumcentre of the origin is the centroid of the equilateral
03
321 zzz
z1 + z2 + z3 = 0
85. Answer (4)
|z –(2 + 3i)| + |z – (–2 + 6i)| = 4Let z1 = 2 + 3i, z2 = –2 + 6i
|z1 – z2| = |4 – 3i| = 5 > k
|z – z1| + |z – z2| = 2a, where k < |z1 – z2|
This does not represent any curve
Locus of z is an empty set.
Al ternatively : If we put z = x + iy, then we got an equation in x and y which does not have any solution.
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Complex Numbers and Quadratic Equations (Solutions) Solutions of Assignment (Set-2)
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- 18 -
Q.No. Solution
86. Answer (2)
z1, z2, z3 and u, v, are complex numbers representing the vertices of two triangles such that
z3 = (1 – t) z1 + tz2 and = (1 – t)u + tv, tc
z3 = z1 – tz1 + tz2 and – u = – tu + tv z3 – z1 = t(z2 – z1) and – u = t (v – u)
12
13
zz
zzt
… (1) and
w ut
v u
… (2)
From (1) & (2)
3 1
2 1
z z w u
z z v u
3 1
2 1
arg arg ..........( )*z z w u
z z v u
3 1
2 1
arg 1 arg 1z z w u
z z v u
3 2
2 1
arg argz z w v
z z v u
3 2
1 2
arg arg ....( )**z z w v
z z u v
from (*) & (**) we conclude that two triangles are similar.
87. Answer (3)
max – min = 2 =
5
3sin.2 1
=
5
3cos2 1
88. Answer (1)
11 21
21
zz
zz
|1||| 2121 zzzz 221
221 |1||| zzzz
21212
22
121212
22
1 ||||1|||| zzzzzzzzzzzz
|z1|2 |z2|
2 – |z1|2 – |z2|
2 + 1 = 0 (|z1|2 – 1) (|z2|
2 – 1) = 0
|z1| = 1, |z2| = 1 Both z1 and z2 lie on the circle |z| = 1
min
max
15 (0, 25)
y
xO
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Solutions of Assignment (Set-2) Complex Numbers and Quadratic Equations (Solutions)
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Q.No. Solution
89. Answer (3)
x2 + x + 1 = 0
+ = –1
= –1
(1) 2 + 2 = ( + )2 – 2
= (–1)2 – 2(1)
= 1 – 2 = –1
(2) ( – )2 = ( + )2 – 4
= (–1)2 – 4 1 = –3
(3) 3 + 3 = ( + )(2 + 2 – )
= 2( )(( ) 2 )
= 2( )(( ) 3 )
= (–1)((–1)2 – 31)
= (–1) (1 – 3) = 2
Al ternat ive
x2 + x + 1 = 0
x = , 2 (complex root of unity)
3 + (2)3 = 2
(4) 4 2 = 2 2 2 2 2( ) 2
= (–1)2 – 2 1 = 1 – 2 = –1
90. Answer (2)
l
n
l
n
q
p ,,
Now
p
q
q
p
l
n
l
n
l
n
0 l
n
p
q
q
p
l
n
p
q
q
p
91. Answer (2)
06 0622 xxxx
2,3 023 xxxx
x = ± 2
Two roots are real, with sum 0.
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Q.No. Solution
92. Answer (1)
0)()()( 2 cbaxbacxacb
Put x = 1, 0 cbacbabacacb
1 is the root of the equation.
Roots are rational.
93. Answer (1)
We know
1tansec 22
1)tan)(sectan(sec
142
a
b
a
acb
Squaring both side
422 )4( abacb
04 244 cabba
94. Answer (4)
Let72
71342
2
xx
xxy
0771)342()1(2 yyxyx
For real x, discriminant should be 0
0)771)(1(4)342( 2 yyy 0)771)(1(4)17(4 2 yyy
0)71787()17( 22 yyy 03601128 2 yy
045142 yy 0)5)(9( yy
5 or 9 yy + –
5 9
+
95. Answer (3)
02 cbxax , Given equation is an2 – bx (x –1) + c(x –1)2 = 0
2 – –
1 0 – 1 – 1
x xa b
x x
Now,
Replacing x by1
x
x
0)1()1( 01)1(
22
2
2
xcxbxaxcx
bx
x
ax
1
1x
x
x is the root of the above equation.
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Q.No. Solution
96. Answer (2)
acbDa
c
a
b4,, 2
1
rpqDp
r
p
q4,, 2
2
Let common difference of A.P. be k.
|||| k
p
pr q
a
acb 4422
p
a
D
D
p
a
pr q
acb
2
1
2
2
4
4
2
2
2
1
p
a
D
D
97. Answer (4)
xbcac
bxaxc
abcb
axcxb
caba
cxbxa
))((
))((
))((
))((
))((
))((
is satisfied by x = a, x = b, x = c.
A quadratic equation is satisfied by more than two values of x. So it is an identity. Hence it is satisfied by allvalues of x.
98. Answer (3)
(1) Root will be of the form of a, b, c are rational.
(2) There is no information about b2 – 4ac
Hence statement is false.
(3) As a, b, c are real and one root is i then other root will be i .
(4) If mass are of opposite sign then 0 0 c
a
99. Answer (1)
For an identity
(k2 – 3k + 2) = 0
(k – 1) (k – 2) = 0
k = 1, k = 2
k2 – 5k + 4 = 0 (k – 1)(k – 4) = 0
k = 1, 4
k2 – 6k + 5 = 0 (k – 5)(k – 1) = 0
k = 1, 5
Common value of k = 1.
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Q.No. Solution
100. Answer (1)
(1) Let roots are ,
+ =13
5 … (i)
=5k … (ii)
For reciprocal roots = 1 k = 5.
(2) If roots are cosecutive integer then
| – | = 1
| – |2 = 1
( + )2 – 4 = 1
1 – 4k = 1
k = 0
(3) Let roots are 2,
2 + = 6 … (i)
2 = k … (ii)
By (i), (ii)
= 2, k = 8
(4) In this case
+ = 0
k = 0
101. Answer (2)
bababaxx 4,, 0 22
abababxx 4,, 0 22
Now, abba 44 22
)(4 44 2222 abbaabba
)(4 0)4)(( babababa
102. Answer (1)
Let be common root, 022 r qp …(1)
and 022 qr p …(2)
Now (1) – (2) 2
1 0)(2 qr r q
Common root is ,2
1 substituting in (1)
044 02
12
2
12
pqr r qp
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Q.No. Solution
103. Answer (1)
x2 + x + 1 = 0 … (i)
Discriminant = b2 – 4ac = 1 – 4 1 1 = –3
Hence the roots of x2 + x + 1 = 0 and not real.
So roots will be in pair.
Also the roots of ax2 + bx + c = 0 will be non-real.
Clearly both roots of the equations are common.
1 1 1
a b c
a : b : c = 1 : 1 : 1
104. Answer (2)
If 1, 2, 3 are roots of equation then
x3 + ax
2 + bx + c = 0
1 + 2 + 3 = – a a = –6
12 + 23 + 13 = b b = 11
123 = – c c = –6
105. Answer (4)
f (x) = ax2 + bx + c
= 2
b ca x x
a a
=2 2
2
2 24 4
b b b ca x x
a aa a
f (x) =2 2
2
4
2 4
b b aca x
a a
f (x) =2 2 4
2 4
b b aca x
a a
f (x) =2 4
b Da x
a a
Clearly if a > 0 the minimum value of ( )4
Df x
a
Similarly of a < 0 the maximum value4
D
a
If ax2 + bx + c > 0 then a > 0, D < 0 for all x R
Hence option (4) is not true.
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Q.No. Solution
106. Answer (3)
x2 + 2x + 3 = (x + 1)2 + 2 m = 2
– x2 + 4x + 6 = – x
2 + 4x + 4 – 4 + 6
= 6 – (x2 – 4x + 4) + 4
= 10 – (x2
– 4x + 4)= 10 – (x – 2)2
M = 10
m + M = 2 + 10 = 12
107. Answer (3)
Let 1592 mmmxy
We need y > 0
Upward parabola above x-axis.
2 9 5 1 0, .mx mx m x R
0,0 aD
i.e., 0 and 0)15)((481 2 mmmm
61
40 0 and 0)461( mmmm
Also for m = 0,
20 9(0) 0 1 1 0,x x x R
61
4,0m
108. Answer (1)
01)( 2 lxxml
)(3 2
lm
l
lm
l
…(1)
)(2
1
1)2( 2
mlml
…(2)
From (1) and (2)
2
2
)(9)(2
1
ml
l
ml
0992 2 mll
For real l,72
81 09881 mm
8
9m
Greatest value of m is .8
9
y
xO
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Q.No. Solution
109. Answer (3)
0753 2 r qxpx
pr qr pq 8425)7)(3(4)5( 22
222
25342584)(25 r pr ppr r p 025
336
5
175
22
r r p
roots are real and distinct.
110. Answer (2)
x3 – 2x
2 – x + 2 = 0
As x = 1 is the root of the equation
Hence we may write
x3 – 2x
2 – x + 2
= x2 (x – 1) – x(x – 1) – 2(x – 1)
= (x – 1) (x2 – x – 2)
= (x – 1) (x – 2) (x + 1)Roots = 1, –1, 2.
111. Answer (2)
ba ,2
baa 242 222
The other root of equation will be 22
i.e. baa 242 2
Sum of roots, S = –4a
Product of roots, P = bbaa 2)24(422
required equation is 02 PSxx
i.e. 0242 baxx
112. Answer (1)
For roots of opposite sign, product < 0
0)1)(2( 03
232
aaaa
21 a
113. Answer (3)
Here we observe that (a + c)2 < b2
(a – b + c) (a + b + c) < 0
Exactly one real root of the given equation lies in (–1, 1).
02 cbxax
D = 222 )(4)(4 caaccaacb
0
Roots are real.
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Q.No. Solution
114. Answer (3)
p + iq is one root p – iq is other root.
Let be third root.
Now sum = 0 iqpiqp
p2
p2 is root of 03 baxx
2p is root of 0)( 3 baxx
03 baxx
115. Answer (2)
We have, given expression
(a12 + a2
2 + a32 +.....+ an – 1
2)x2 + 2(a1a2 + a2a3 + a3a4 +.....+ an – 1 an)x + (a22 + a3
2 + a42 +......+ an
2) 0
(a1x + a2)2 + (a2x + a3)
2 + (a3x + a4)2 + ....... + (an – 1x + an)
2 0
(a1x + a2)2 + (a2x + a3)
2 + (a3x + a4)2 + ....... + (an – 1x + an)
2 = 0,
as sum of square cann't be negative.
a1x + a2 = 0 = a2x + a3 = a3x + a4 = ....... = an – 1x + an
32 4
1 2 3 1
....... n
n
aa a ax
a a a a
a1, a2, a3, ....... , an – 1, an are in G.P.
116. Answer (2)
,)( 2 cbxaxxf given 0)1( cbaf
Rxxf ,0)( as roots are non-real complex
f (–2) < 0 bcacba 24 024
117. Answer (3)
Given,a
c
a
b ,
Also,22
11
2)())(( 22222 a
c
a
b
a
c
a
b 22
2
2
2
222 2caabbc abbcca 2222
c
b
a
c
b
a2
b
c
a
b
c
a,, are in H.P.
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Q.No. Solution
118. Answer (2)
21)2)(13)(23)(1( xxxx
21)253)(253( 22 xxxx
Put,txx
53
2
25 21)2)(2( 2 ttt 5,5 tt
Now, 553 2 xx and 553 2 xx
0553 2 xx and 0553 2 xx
0553 2 xx has two irrational roots.
whereas roots of 0553 2 xx are imaginary.
119. Answer (4)
S1 : x2 – x – 2 < 0
(x – 2)(x + 1) < 0
–1 < x < 2 … (i)
2sin2x + 3sinx – 2 > 0
2sin2x + 4sinx – sinx – 2 > 0
2sinx(sinx + 2) –1 (sinx + 2) > 0
(sinx + 2)(2sinx – 1) > 0
1
sin2
x
5
,
6 6
x … (ii)
By (i), (ii) , 26
x
S2 : Using A.M. G.M.
22
222
2
tan
tan( )
2
x xx x
x xx x
2
2
2
tan2tanx x
x x
S3 : Using D 0
4(a + b + c)2 – 4(1)(3)(ab + bc + ca) 0
a2 + b
2 + c
2 + 2(ab + bc + ca) – 3(ab + bc + ca) 0
a2 + b
2 + c
2 + (ab + bc + ca) (2 – 3) 0
2 2 2
3 2
a b c
ab bc ca … (i)
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Q.No. Solution
But (a – b)2 c2
(b – c)2 a2
(c – a)2 b2
2 2 2
2
a b c
ab bc ca
3 – 2 2
4
3
S4 :3 23 3 0 x px qx r …(i)
Multiply the second equation by x
3 22 0 x px qx … (ii)
By (i) – (ii)
2
2 0 px qx r
… (iii)But x
2 + 2px + q = 0
px2 + 2p
2x + pq = 0 … (iv)
By (iii) - (iv)
2x(q – p2) + (r – pq) = 0
22( )
r pqx
q p
Putting x in
x2 + 2px + q = 0
We get
4(p2 – q) (q2 – pr ) = (pq – r 2)2
120. Answer (3)
The given quadratic equation is (a – b)x2 – 5(a + b)x – 2(a – b) = 0
The discriminant
D = (– 5(a + b))2 + 8(a – b) (a – b)
= 25(a + b)2 + 8(a – b)2
Hence D > 0 a & b.
So, roots are real and unequal.
121. Answer (4)
Since, , are the roots of the equation ax2 – bx + c = 0
So, ,b c
a a
Now, we have to observe root of the equation
(a + cy)2 = b2y
a2 + 2acy + c2y2 = b2y
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Q.No. Solution
c2y2 + (2ac – b2)y + a2 = 0
2 2
22 2
20
ac b ay y
c c
2 2
2 2 2
20
b a ay ycc c
22 2 2 2
1 1 10y y
Hence the equation (a + cy)2 = b2y has roots2 2
1 1,
122. Answer (1)
Since a, b, c are in G.P.,
So, b2 = ac
4b2
– 4ac = 0D = 0 for the equation ax2 + 2bx + c = 0
Hence, it will have equal roots, and root will be
bx
a
Now, ax2 + 2bx + c and dx2 + 2ex + f = 0 have a common root,
So,b
xa
will satisfy the equation
dx2 + 2ex + f = 0
2
2. 2 . 0b b
d e f
aa
2 2
2
20
db aeb a f
a
db2 – 2aeb + a2f = 0
dac – 2aeb + a2f = 0
dc + af = 2eb
2d f e
a c b
So, , ,d c f
a b c are in A.P.
123. Answer (3)
The given equation is x2 – 2(k + 2)x + 12 + k2 = 0 has distinct real roots when D > 0
4(k + 2)2 – 4(12 + k2) > 0
k2 + 4 + 4k – 12 – k2 > 0
4k – 8 > 0
k > 2
So least integral value of k is 3.
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Q.No. Solution
124. Answer (4)
– 1D
a
2 48 1p
p = ± 7; but p is positive, hence p = 7.
125. Answer (1)
f (x) = (x – a) (x – c) + k(x – b) (x – d)
f (a) = k(a – b) (a – d) which is positive
f (b) = (b – a) (b – c) which is negative
f (c) = k(c – b) (c – d) which is negative
f (d) = (d – a) (d – c) which is positive
So, f (x) = 0 has a root in the interval (a, b) and another in (c, d). So the roots are real and distinct.
126. Answer (4)
Use relation between roots and coefficients
= p ...(i)
= r ...(ii)
22
q ...(iii)
(2 )2
r ...(iv)
(ii) and (iv) are same. (i) and (ii) can be solved to obtain and in terms of p and q, thereby giving r .
127. Answer (4)
Use the idea of rotation to obtain the desired P as
3 4
i
e .ei, where tan =4
3
yielding sin 4
5, cos =
3
5
128. Answer (4)
We have,
2
1 1
11
z
zzz z z
z z
m
1 1
2 I ( )
z z i z
1,
Im
i i R
z z
Thus the locus of21
z
z is y-axis.
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Q.No. Solution
129. Answer (4)
(6, 2)
Z 2Z2
Imaginary axis
Real axis
(7, 6)
Z 0
( 1 , 2 )
90°
1
1
3
5
2 (6 2 cos45 , 5 2 sin45 ) (7, 6) 7 6 Z i
by rotation about (0, 0)
2 2 22 2
2
( )
i iZe Z Z e
Z
2 (7 6 ) cos sin (7 6 )( ) 6 72 2
Z i i i i i
130. Answer (1)
3 3 350 zz zz
2 2 2 2| | ( ) | | ( ) 350 z z z z
2 2 2| | ( ) 350 z z z
(x2 + y2) (x2 – y2 + 2ixy + x2 – y2 – 2ixy) = 350
2(x2 + y2)(x2 – y2) = 350
(x2 + y2)(x2 – y2) = 175 = (32 + 42)(42 – 32)
Which suggests that points (x, y) satisfying the given equation are (4, 3), (–4, –3), (–4, 3), (4, –3)
y
x
A(4, 3)(–4, 3)B
C(–4, 3)
D(4, 3) –
O
Required area = AB × BC
= 8 × 6
= 48 sq. units
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Q.No. Solution
131. Answer (4)
z = cos + isin = ei
Now,15
(2 1)
1
lm
i m
m
e
= sin + sin3 + ..... + sin29
=
15.2sin
2 (15 1) 22 .sin2 2
sin2
=sin15 .sin15
sin
For = 2°, the given expression reduces to
=sin30 .sin30 1
sin2 4sin2
132. Answer (2)
We have + = – p
3 + 3 = q = ( + )3 – 3( + ) = – p3 + 3p()
3
3
p q
p
The quadratic equation with and
as roots is
2 0 x x
2
2 ( ) 21 0
x x
32
2
3
23
1 0
3
p qp
px x
p q
p
(p3 + q)x2 – (p3 – 2q)x + (p3 + q) = 0
133. Answer (3)
We observe that
( + ) (n –1 – n –1) = n – n + (n – 2 – n –2)
6an – 1 = an – 2an – 2
2
1
26
n n
n
a a
a
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Q.No. Solution
2
1
23 , 2
2
n n
n
a an
a
Putting n = 10, we get
10 8
9
23
2
a a
a
134. Answer (2)
Let be a common root between given equations x2 + bx – 1 = 0 and x2 + x + b = 0
2
2
1
1 11
b bb
2
2 1 1and
1 1
b b
b b
22 1 1
1 1
b b
b b
2
2 (1 )1
1
b
bb
b2 – b3 + 1 – b = 1 + 2b + b2
b3 + 3b = 0
0, 3 b b i
3 b i
135. Answer (4)
As, a is real,
So a a gives
z2 + z + 1 = 2 1 z z
( )( 1) 0 z z z z
As, z z
So, 1 z z
1
{where }2
x z x iy
Now,
a = z2 + z + 1
21 1
12 2
iy iy
23
4 y
As, y 0 so,3
4a
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Section-B
Q.No. Solution
1. Answer (1, 2, 4)
If |3z – 1| = 3 |z – 2|
Let z = x + iy
|3x + 3iy – 1| = 3|x + iy – 2|
(3x – 1)2 + (3y)2 = 9 [(x – 2)2 + y2]
9x2 + 1 – 6x + 9y
2 = 9x2 + 9y
2 – 36x + 36
30x = 35
x =6
7
6 Re(z) = 7 (A line parallel to y-axis)
Also mid-point of 0,31 and (2, 0) is 0,
67
2. Answer (1, 2, 3)
We can write11
2cos
11
2sin
ki
k
=
11
2sin
11
2cos
ki
ki
Now
10
111
2cos11
2sink
kikS
10
1
11
2
k
ki
eiS
= ieeei
iii
11.... 11
20
11
4
11
2
(as 1 + + 2 + ..... + 10 = 0)
S = i
iS
0 SS
and 1SS
iiS 212
1 2
)1(2
1iS
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Q.No. Solution
3. Answer (1, 3)
cos cos cos 0 A B C
sin sin sin 0 A B C
Let cos sin iA
a A i A e
cos sin iBb B i B e
cos sin iCc C i C e
0a b c
3 3 3 3a b c abc
2 2 2
3a b c
bc ac ab
2 2 2
3.
iA iB iC
iB iC iA iC iA iB
e e e
e e e e e e
2 2 2 3iA iB iC iB iA iC iC iA iBe e e
cos(2 ) sin(2 ) cos(2 ) sin(2 ) cos(2 ) sin(2 ) 3 A B C i A B C B A C i B A C C A B i C A B
On comparing real and imaginary part
cos(2 ) cos(2 ) cos(2 ) 3 A B C B A C C A B
sin(2 ) sin(2 ) sin(2 ) 0 A B C B A C C A B
4. Answer (2, 4)
x x
2 – 2 cos + 1 = 0
2
4cos4cos2 2 x
= cos isin
= cos + isin, = cos – isin
ninn sincos ; ninn sincos
nS nn cos2 ; P = 1
Equation is 02 PSxx
i.e. 01cos22 nxx
Option (4)
nnnxx 222 cos1coscos2 = 0
0sin)cos( 22 nnx
Option (2)
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Q.No. Solution
5. Answer (2, 3)
a2 + b
2 + c2 = 1
b +ic = (1 + a) z
a
icbz
1
a
icbi
aicbi
iz
iz
11
11
1
1
ibca
ibca
iz
iz
1
1
1
1
)1(
)1(
)1(
)1(
ibca
ibca
ibca
ibca
22
22
)1(
)1(
bca
ciba
c
iba
1
Similarly
ibca
ibca
ibca
ibca
1
1
1
1 =
iba
c
1
6. Answer (3, 4)
2amp
12
14
zz
zz
A = 90°
alsoz1 – z4 = z2 – z3 |z1 – z4| = |z2 – z3|
AD = BC and AD || BC
So AB || CD and AB = CD
ABCD is a rectangle or cyclic quadrilateral
7. Answer (2, 3, 4)
Cube roots of p are1 1 1
23 3 3, ,p p p
i.e. 23
1
3
1
3
1
.,., ppp
Now,222222
222222
zyx
zyx
)( 22222
2222
22422
24222
zyx
zyx
zyx
zyx
= =
Option (4)
We can assign the , , different value we get other options also.
A B
CD
z1 z2
z4 z3
90°
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Q.No. Solution
8. Answer (1, 2)
z + z –1 = 1
11
z
z
z2
– z + 1 = 0
2,2
31
iz
zn + z
– n = (– )n + (– )
–n
Case (1), n = 3m
(– )3m + (– ) –3m = (–1)n + (–1)n = 2(–1)n
whenn = 3m + 1
(– )3m + 1 + (– ) –3m – 1
=13
13
)(
)1()1(
m
nmn
=
1
)1( n
= (–1)n (–1) = (–1)n + 1
9. Answer (2, 3, 4)
|z 1| < |z + 3|
Let z = x + iy
(x 1)2 + y2 < (x + 3)2 + y2
x2 + y
2 2x + 1 < (x2 + y2 + 6x + 9)
8x > –8x > – 1
i = i((2x + i2y) + 3 – i)
= i2x – 2y + 3i + 1
= i(3 + 2x) + (1 – 2y)
as 3 + 2x > 1 option
also, – 1 = 2z + 3 – i – 1
= 2z + 2 – I = 2x + 2iy + 2 – i
= 2(x + 1) + i(2y – 1)
as x > – 1
2(x + 1) > 0 2
)1(arg
option (4)
– 5 = 2(x – 1) + i(2y – 1)
+ 3 = 2(x + 3) + i(2y – 1) as x > –1
| + 3| > | – 5|
Option (2)
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Q.No. Solution
10. Answer (2, 3, 4)
|z + |2 = |z|2 + ||2
Since |z + |2 = |z|2 + ||2 + zz
|z|2 + ||2 = |z|2 + ||2 + zz zz = 0
zz
zz
z
is purely imaginary
Therefore,2
zamp
11. Answer (1, 2, 3)
Clearly |z1 – z2|min = 2 – 1 = 1
|z1 – z2|max = 3
max|2z1 + z2| = |2 + 2| = 4
Now, |z| = 1
|z1| = 1
1. 11 zz
11
1
zz
Now 121
2
1zz
zz
and |||||| 1212 zzzz
3.
12. Answer (3, 4)
12
z
z
1||
2||
zz
02|||| 2 zz
0)1|)(|2|(| zz
21
||
2||1 z
But |z| 0
2||0 z
3|| z and |z| 4
Option (3) and (4)
1
1
2
| | = 1z
| | = 2z
y
xO
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Q.No. Solution
13. Answer (1, 2, 4)
The given equation is |2z – i| = m|z + i|
2(2 )(2 ) ( )( )z i z i m z i z i
24 2 2 1 ( 1)zz iz i z m zz iz i z
2 2 2 2(4 ) (2 ) (2 ) (1 ) 0m zz m iz m i z m
The above equation will does not represent a circle, when,
4m2 = 0 m = 2 ; since m cannot be negative
Hence answer is (1, 2, 4)
14. Answer (1, 2, 3)
(x + 1)3 = (– 4)3
x + 1 = – 4, – 4, – 42
x = – 5, – 1 – 4, – 1 – 42.
Hence, roots are – 5, – 1 – 4, – 1 – 42.
15. Answer (1, 2, 3)
2 21| | 1z a b a2 + b2 = 1 …(i)
2 22| | 1z c d c2 + d2 = 1 …(ii)
1 2Re( )z z = Re[(a + ib)(c – id)]
= Re[ac + bd + i(bc – ad)] = 0
ac + bd = 0 …(iii)
Now, using (i) & (iii) we can prove that b = c, a = d.
Hence, 2 2 2 21| | 1a c a b
Similary we can observe, |2| = 1
1 2Re( ) 0
16. Answer (2, 4)
Given inequality is,
2
1 2
2
| | 2| | 6log 0
2| | 2| | 1
z z
z z
2
2
| | 2| | 61
2| | – 2| | 1
z z
z z
|z|2 + 2|z| + 6 > 2|z|2 – 2|z| + 1
|z|2 – 4|z| – 5 < 0
|z| (– 1, 5), but |z| > 0
0 < |z| < 5
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Q.No. Solution
17. Answer (2, 3, 4)
z1, z2 are the complex numbers satisfying,
1 2
1 2
1 –
z z
z z
|z1 + z2| = |z1 – z2|
1 2 1 2 1 2 1 2( )( ) ( – )( – )z z z z z z z z
1 1 1 2 2 1 2 2 1 1 1 2 2 1 2 2 – – z z z z z z z z z z z z z z z z
1 2 2 12( ) 0z z z z
1 1
2 2
– z z
z z
1 2 2 1 0z z z z
1 2 1 2
0z z z z
2Re( ) 0zz
18. Answer (1, 2, 3, 4)
a
b cossin
a
c cos.sin
1cossin 22
12
2
a
c
a
b 22 2 acab
0222 acba
Also 222)( cbca .
19. Answer (1, 2)
a
b2 ,
a
c
a
acb
2
2||
A
Bhh
2 ,
A
Chh ))((
A
ACB
22||
A
ACB
a
acb
22 22
2
2
2
2
A
a
ACB
acb
Also 2h =a
b
A
B
A
B 22)(
2
A
B
a
bh .
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Q.No. Solution
20. Answer (1, 2)
Let us use the transformation3 1
12
x xy x
So that x + 3 = y + 2 and x – 1 = y – 2
Thus the given inequality reduces to
(y + 2)5 – (y – 2)5 244
(y5 + 5.y4.2 + 10y3.22 + 10y2.23 + 5.y.24 + 25) – (y5 – 5.y4.2 + 10.y3.22 – 10y2.23 + 5.y.24 – 25) 244
2[10y4 + 80y2 + 32] 244
y4 + 8y2 – 9 0
(y2 + 9) (y2 – 1) 0
y2 – 1 0 as y2 + 9 > 0, y.
–ve+ve +ve
–1 1
y 1 or y –1
x + 1 1 or x + 1 –1
x 0 or x –2
Hence the required solution set is
(– , –2] [0, ).
21. Answer (3, 4)
02 cbxax
Let a, b, c is a, ar , ar 2
Now, 022 ar arxax
022 r rxx
231 ir x x = r or 2
r
Roots are imaginary and are in the ratio 1 : or 2 : 1.
22. Answer (3, 4)
0)45)(35( 012525 2 xxxx
01 as
5
3
5
4,
5
3xxxx
cos =
5
4
5
3sin or
5
3
25
24
5
4
5
32cossin22sin
or25
24
5
4
5
32
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Q.No. Solution
23. Answer (1, 2)
Let be common root
r pqr pq ,,02
qpr qpr ,,02
1( ) 0p q r r q
p
Common root is =p
1
Other roots are, qprp and
Equation containing other roots is
0)( 22 rqpxqr px
p
1 is common root 011 2
r
ppq
p
)(12
r qp
Now 0)( 22 qr pxr qpx
0)(2
2
pqr xr qx
p
pp
0)()( 2 pqr xr qxr qp
24. Answer (1, 2, 3)
0)3(2 mxmx
For real distinct roots, 04)3( 2 mm
09102 mm
,9()1,( 0)1)(9( mmm … (1)
For positive roots,
Sum > 0, product > 0
m – 3 > 0 , m > 0 … (2)
From (1) and (2), m (9, )
For negative roots
sum < 0, product > 0
m – 3 < 0, m > 0 … (3)
From (1) and (3), )1,0(m
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Q.No. Solution
25. Answer (1, 2, 3, 4)
012 22 aaxx
1)( 2 ax x = a + 1, a – 1.
Now 4]1[3 a
and 4]1[3 a
41][3 a and 41][3 a
3][4 a and 5][2 a
3][2 a [a] = –2, –1, 0, 1, 2, 3.
26. Answer (1, 2, 3)
4 , A
+ = 36, = B
Let , , , be a, ar , ar 2, ar
3
a + ar = 4
3632 ar ar 9
1
)1(
12
r r
r
92 r r = ± 3, a = 1.
A a(ar ) = A A = 3
B = B = (ar 2) (ar
3) B = 243 B = 81 A.
27. Answer (1, 2, 3)
4
5loglog
4
32
22 xx
x = 2 .
Taking log with base 2 on both side.
2
12loglog
4
5log)(log
4
3222
22
xxx
Put log2 x = t,2
1
4
5
4
3 2
ttt
02543 23 ttt 0)273)(1( 2 ttt
0)2)(13)(1( ttt
2,3
1,1 t 2,
3
1,1log
2
x x = 2, 23/1 2,2
28. Answer (3, 4)
a
c
a
bcbxax ,,02
a
cab 42
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Q.No. Solution
A
C
A
BCBx Ax ))((,,02
A
CAB 42
Now, A
CAB
a
acb 44 22
2
2
2
2 44
A
CAB
a
acb
2
2
2
4
4
A
a
ACB
acb
)(2 2 A
B
A
B
A
B
a
b
2
1
29. Answer (1, 2, 3)
Let , be the roots of the corresponding equation
x2 + ax + a2 + 6a = 0 …(i)
As the coefficient of x2 = 1 > 0 x2 + ax + a2 + 6x < 0 will be satisfied for all values of x (, ) if , arereal and unequal (let < ).
1 2
Hence the inequality will hold for all real x (1, 2) if the interval (1, 2) is a subject of the interval (, ). Thusfor (1) we should have D > 0 and < 1, > 1 as well as < 2, > 2.
Now, D > 0 a2 – 4(a2 + 6a) > 0
a2 + 8a < 0
a (– 8, 0) …(ii)
< 1, > 1 – 1 < 0, – 1 > 0
( – 1) ( – 1) < 0
a2 + 7a + 1 < 0
–7 – 45 –7 45,
2 2a
…(iii)
< 2, > 2 ( – 2) ( – 2) < 0
a2 + 8a + 4 < 0
(–4 – 2 3, – 4 2 3)a …(iv)
Common values of a satisfying (ii), (iii) and (iv) are
–7 – 45, – 4 2 3
2a
…(v)
Hence answer is (1), (2), (3) those are subject of (v)
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Q.No. Solution
30. Answer (2, 3)
The given equation is
1 1 1
x p x q r
1
( )( )
x q x p
x p x q r
(2x + p + q)r = x2 + (p + q)x + pq
x2 + (p + q – 2r )x + (pq – qr – rp) = 0
According to the question the given equation has roots equal in magnitude but opposite in sign, hence
Coefficient of x = 0
p + q – 2r = 0
r =2
p q
Product of roots
= + [(p + q)r – pq]
=2( ) –
2
p qpq
= 2 21( )
2p q
31. Answer (1, 2, 4)
(a + 2) x2 + 2(a + 1)x + a = 0
Let , be roots
2 1
2
a
a (integer)
2
a
a (integer)
2a
a will integer if
For a = 0, a = –1, a = –3
Also for a = 0, a = –1, a = –3
–2 1
2
a
a
is integer
32. Answer (1, 3)
Since 1 is the repeated roots of ax3 + bx2 + c = 0
So, 1 + 1 + = – b
a
1.1 + + = 0 1
– 2
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Q.No. Solution
1.1. =1 1
– – 2 2
c c
a a
3 –
2
b
a
– 3bc
Now, by the equation,
cx3 + bx + a = 0
3 0b a
x xc c
x3 – 3x + 2 = 0
x3 – x2 + x2 – x – 2x + 2 = 0
x2(x – 1) + x(x – 1) – 2(x – 1) = 0
(
x – 1) (
x2
+x – 2) = 0
(x – 1) (x2 + 2x – x – 2) = 0
(x – 1) (x – 1) (x + 2) = 0
x = + 1, – 2
Hence answer is (1), (3)
33. Answer (3, 4)
Let x2 = y
So the equation ay2 + by + c = 0 should have both roots non-negative in order to all roots of the equationax4 + bx2 + c = 0 are real for this
– 0 0b b
a a …(i)
0c
a …(ii)
From (i) and (ii)
b > 0, a < 0, c < 0
or b < 0, a > 0, c > 0
34. Answer (1, 2, 3)
– 5
( – )2 < 5
( + )2 – 4 < 5
k2 – 4 < 5
k2 < 9 k (– 3, 3)
Hence answers is (1, 2, 3)
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Q.No. Solution
35. Answer (1, 2, 3)
We know that in a triangle sum of two sides of a triangle is greater than third side.
So, a2 + 2a + 2a + 3 > a2 + 3a + 8 4a > 3a + 5 a > 5
a2 + 2a + a2 + 3a + 8 > 2a + 3 2a2 + 3a + 5 > 0 a R
2a + 3 + a2 + 3a + 8 > a2 + 2a 3a > – 11 a >11
– 3
Combining these three,
a (5, )
Hence answer is (1, 2, 3)
36. Answer (1, 3, 4)
1 21
zz z
t t
we havez = (1 –
t)z
1 +tz
2, 0 <t < 1,
2 11
1
t z t zz
t t
2 11 z t z t z
z1, z, z2 are collinear
so Arg(z – z1) = Arg(z – z2) = Arg(z2 – z1)
37. Answer (3, 4)
Note that || = 1
i are possible value of z1
i are possible value of z2
(i = 1,2,3)
3
2 2
i
6
ie
2 3
ie
3 2
ie
24 3
ie
55 6
i
e
So, 1 2z oz can be2 5
,3 6
1
3
2
1
2
3
30°
30°30°30°
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Q.No. Solution
Now let x = cos + isin, then equation (i) becomes,
cos + isin – cos + isin = i
2sin = 1 sin =1
2
5or 6 6
1. Answer (3)
We have to find out the value of x51
51
cos sin6 6
i
17 17cos sin
2 2i
= 0 + i = i
2. Answer (1)
We have to evaluate 2020
1x
x
20 20 20 20cos sin cos – sin
6 6 6 6i i
10 42cos 2cos
3 3
2cos3
– 2cos –13
3. Answer (2)
For finding 20132013
1x
x
2013 2013 2013 2013cos sin – cos sin
6 6 6 6
i i
20132 sin
6i
6712 sin
2i
2 sin2
i i
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Q.No. Solution
Comprehension-III
1. Answer (2)
2. Answer (4)
3. Answer (2)
Solution of Q.1 to Q.3
Given expression is
(1 + x)n = a0 + a1x + a2x2 + a3x
3 + .............+ anxn …(i)
Putting x = ± 1, we get
(1 + 1)n = a0 + a1 + a2 + ...... + an
(1 – 1)n = a0 – a1 + a2 + ...... ± an
Adding these,
2n = 2(a0
+ a2
+ a4
+ ......)
a0 + a2 + a4 + ...... = 2n – 1 …(ii)
Hence, answer of question 1 is (2)
Again, putting x = ± i in (i), we get
(1 + i)n = a0 + a1i – a2 – a3i + a4 + a5i – a6 – a7i + a8 + ......
(1 – i)n = a0 – a1i – a2 + a3i + a4 – a5i – a6 + a7i + a8 + ......
Adding these,
2(a0 – a2 + a4 – a6 + .......)(1 ) (1– )
2
n ni i
/2
/22.2 · cos
4 2 cos2 4
n
n
nn
…(iii)
Hence, answer of question 2 is (4)Now, adding (ii) & (iii), we can get
2(a0 + a4 + a8 + .......) = 1 /22 2 cos4
n n n
a0 + a4 + a8 + a12 + ...... = –1
2 22 2 cos4
nn n
Hence, answer of question 3 is (2)
Comprehension-IV
1. Answer (1)
kxxxf 3)( 3
033)(' 2 xxf x = ±1
For exactly one positive root,
0)1( f and f (1) < 0
–1 + 3 + k < 0 and 1 – 3 + k < 0
2k and k < 2 )2,( k .
1 –1
y
xO
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Q.No. Solution
2. Answer (1)
For exactly one negative root,
0)1( f , 0)1( f
–1 + 3 + k > 0, 1 – 3 + k > 0
k > – 2, k > 2
),2( k
3. Answer (2)
For one negative and two positive root
0)1( f , f (0) > 0, f (1) < 0
–1 + 3 + k > 0, k > 0, 1 – 3 + k < 0
2k , 0k , 2k
20 k i.e., )2,0(k .
Comprehension-V
1. Answer (3)
2 2 2| | (| | 2 1) 1x x k k
)12( 22 kxx = 21 k 01)12( 224 kxkx
All roots are imaginary, if 042 acbD
0)1(4)12( 22 kk
4
5k …. (1)
Also roots are imaginary if 0D , but 2x is negative, i.e. roots of 01))(12()( 2222 kxkx are both
negative.
Sum < 0, and product > 0
012 k and 012 k )1,( k
All roots are imaginary if
,
4
5)1,(k
2. Answer (2)
For exactly two real roots of
01)12( 22 ktkt
0D and one value of 2xt is positive and one is negative.
0)1(4)12( 22 kk
4
5k … (1)
Product = 012 k –1 < k < 1 … (2)
From (1) and (2) )1,1(k .
1 –1
y
xO
1
–1
y
xO
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Q.No. Solution
3. Answer (1)
For repeated roots = 0,4
5k
or P = 0, 012 k , 1k
When product = 0, x = 0 is repeated root.
Comprehension-VI
1. Answer (2)
Let z = x + iy
Set A corresponds to the region y 1 ...(i)
Set B consists of points lying on the circle, centred at (2, 1) and radius 3,
i.e. x2 + y2 – 4x – 2y = 4 ...(ii)
Set C consists of points lying on the x + y = 2 ...(iii)
P
y = 1
x
y
(2, 1)
(0, 2)
( 2,0)
Clearly, there is only one point of intersection of the line 2 x y , and circle x2 + y2 – 4x – 2y = 4
2. Answer (2)2 2
1 – – 5 – z i z i
= (x + 1)2 + (y – 1)2 + (x – 5)2 + (y – 1)2
= 2(x2 + y2 – 4x – 2y) + 28
= 2(4) + 28 2 2 – 4 – 2 4 x y x y
= 36
3. Answer (4)
– 2 3 w i
| | – | 2 |w i < 3
3 5 3 5 w
– 3 – 5 – 3 – 5 w ...(i)
Also, – 2 3 z i
– 3 5 3 5 z ...(ii)
– 3 < |z| – |w| +3 < 9
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Q.No. Solution
OR
Let z = x + iy
iyx
iyx
z
z
)1(
)1(
1
1
iyxiyx
iyxiyx
zz
)1()1(
)1()1(
11
22
22
)1(
)2()1(
1
1
yx
yiyx
z
z
21
1arg
z
z
0122 yx
122 yx and y > 0
Locus of z is semicircle.
3. Answer (1)
eiA.eiB.eiC = ei( A + B + C)
= ei = cos + isin
= –1
4. Answer (1)
(1)1/4 = (cos2r + isin2r )1/4
= cos sin
2 2
r r i
where r = 0, 1, 2, 3
11/4 = 1, i, – 1, – i
z12 + z2
2 + z32 + z4
2 = 1 + i2 + 1 + i2
= 2 – 1 – 1 = 0
5. Answer (4)
|z1| = |z2| = |z3| = |z4|
This may not be the case if centre of the circle is not origin.
6. Answer (4)
2 2 (1– )(1 )
1 2
n nni i i
ii
4( 2)
in
n e
Now clearly the least integral value for which the given number is a positive integer is 8.
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Q.No. Solution
7. Answer (1)
B z( )2 A z( )1
C z( )3 D z( )4
1 2
3 2
– arg
2 –
z z ABC
z z
And AB = |z1 – z2| = BC|z3 – z2|
1 2
3 2
– 1
–
z z
z z
Hence, 1 2
3 2
– 1 cos – sin
– 2 2
z zi
z z
z3 – z2 = (z1 – z2) cos – sin2 2
i
= – i(z1 – z2)
= – iz1 + iz2
z3 = – iz1 + (1 + i)z2
z3 = – i(1 – i) + (1 + i)(1 + i) = – i + i2 + 1 + i2 + 2i
= i – 1
8. Answer (2)
Let z = r (cos + isin)
2 2 22 21 1 1
cos – sinz r r z r r
22
12cos2r
r
2 2 22
12 – 4sina r
r
22 21
4sinr a
r
2
214r a
r
214r a
r
2 2 – 4 1 0r a r
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Q.No. Solution
r lies between22 44 –
and2 2
a aa a
This is true for all real a 0.
2 2
max min
4 4 –
| | , | |2 2
a a a a
z z
Hence, for a = 0
|z|min = |z|max
Hence both statement-1 and statement-2 are true. But statement-2 is not the correct explanation ofstatement-1.
9. Answer (1)
Given equation is (z – 2)n = zn
– 2
1n
z
z
1/ – 21 1nz
z
|z – 2| = |z|
Hence z is the locus of a straight line perpendicular bisector of the segment joining the points (2, 0) and (0,0), i.e., x = 1.
10. Answer (1)
02 cbxax
x = 1, a + b + c = 0
If sum of coefficient is 0 then 1 is the root of the equation.(210 – 3) – 211 + 210 + 3 = 0
Both are true and Statement-2 is correct explanation of Statement-1
11. Answer (1)
For reciprocal roots, replacing x byx
1 in 02 cbxax 0
2 c
x
b
x
a 02 abxcx
Statement-2 is correct and is correct explanation of Statement-1
0510 2 xx 05110
2
xx 0105 2 xx
12. Answer (3)The equation in first statement is x2 – 2009x + 2008 = 0 can be written as (x – 2008)(x – 1) = 0
x = 1, 2008 are roots of the equation where are rationals also.
Statement 1 is True.
Statement 2 is not always true. When D = b2 – 4ac = a perfect square than roots of the equationax2 + bx + c = 0 are rational only when a, b, c are rationals, otherwise roots are irrationals.
To this end, let us consider an equation 24 4 3 1 0x x
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Q.No. Solution
whose discriminant = 48 + 16 = 64 = 82 = a perfect square but roots are
4 3 48 16
2 4x
3 2
2
which are not rationals.
Thus statement 2 is false.
Hence option (3) is correct.
13. Answer (1)
We observe that f (1)f (2) = (1 + 5 – 7)(4 + 10 – 7) < 0
Hence these exists a root of x2 + 5x – 7 = 0 in (1, 2).
Clearly option (1) is correct.
14. Answer (4)
We observe that
(a + c)2 > b2
(a + c)2 – b2 > 0
(a – b + c) (a + b + c) > 0
f (–1)f (1) > 0, where f (x) = ax2 + bx + c
f (x) =ax2 + bx + c = 0 has either no root in (–1, 1) or if real roots exist, then both roots lie in (–1, 1)
Statement 1 is not necessarily true
Hence statement 1 is false.
Answer is 4
15. Answer (4)
Statement-1 is wrong,
73
1
can be root of infinite equations with real coefficients, e.g. 07
3
1)1(
xx ,
...073
1)2(
xx
16. Answer (3)
0501200822 xx
2 2008 4 2008 4(501)
2x
2008 1507
Roots are rational.
Statement-2 is wrong as roots are rational only when coefficients are rational and acb 42 is perfect square.
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Q.No. Solution
17. Answer (4)
a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca) = 0,
a + b + c = 0 a, b, c are not equal. The sign of other root depends on sign of ac.
Hence (4) is answer.
18. Answer (4)
( 2 – 1) is a root if coefficients are real then other root can be rational.
19. Answer (1)
20. Answer (2)
2 2 0 x px q
2 (i)(ii)
pq
2 2 0 ax bx c
1 2(iii)
(iv)
b
ac
a
2 2( ) ( ) p q b ac =
2
22
1
2 2
a
=
222( ) 1
. 016
a statement-1 is true
Now pa = ( )2 2
aa
b =1
2
a
211, { 1,0,1}, correctpa b
Similarly
If
c qa a a
1
0
1
0, and 0 { 1, 0,1}
Statement 2 Is true.
Both statement 1 and statement 2 are true, But statement 2 do not explains statement 1.
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Section-E
Q.No. Solution
1. Answer A(s), B(r), C(p), D(q)
(A)
2007
19
2
cos9
2
sink
k
i
k
=
2007
19
2
sin9
2
cos)(k
k
i
k
i =
2007
1
9
2
)(k
ki
ei
=
9
)2007(2
9
4
9
2
.....
iii
eeei
Which is G.P.
=
1
)11(
1
1
19
2
9
2
9
2
9
)2007(2
9
2
i
i
i
i
i
e
ei
e
e
ei = 0
(B) |z1| = 1, |z2| = 1, |z3| = 1
1,1,1 332211 zzzzzz
33
22
11
1,
1,
1z
zz
zz
z
Now, ||111
321321
zzzzzz
= 1|| 321 zzz
(C) |z1| = |z2| = |z3| and z1, z2, z3 are vertices of equilateral triangle
Origin is its centroid
z1 + z2 + z3 = 0
Now, |z1 + z2 + z3| –1 = –1
(D) Let 4325
1
,,,,1)1(
1 + + 2 + 3 + 4 = 0
1 + + 2 + 3 = – 4
and 5 = 1
41
and || = 1
Now 4log4 |– 4 – 4| = |2|log4 44
= |||2|log4 44
= 22log2
142log4 24
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Q.No. Solution
2. Answer A(s), B(r), C(p), D(q)
(A) (1 + i)n = (1 – i)
n
1
1
1
n
i
i
1ni
n = 4
(B) (a + ib) = (x + iy)3 = x3 – iy
3 + 3ix2y – 3xy
2
a = x(x2 – 3y2) and b = – y3 + 3x
2y
22 3yxx
a 22 3xy
y
b
)(2 22 yxx
a
y
b
k = 2
(C)2
1 ix
2
)1( 22 i
x
= ii
2
211
200820062004108642 ...1 xxxxxxxx
111
11
1
1
1
]1[12
2
2
2010
i
i
x
x
3. Answer A(s), B(r), C(p, q), D(q)
(A) |||| 21 zzzz = constant = k,
where || 21 zzk , represents an ellipse.
(B) ,|||| 21 kzzzz where || 21 zzk is a hyperbola having foci at z1 and z2.
(C)2
arg2
1
zz
zz
This represents a circle with z1 and z2 as the vertices ofdiameter.
(D) If lies on || = 1, then
2007||
20072007
2007 lies on the circle.
p z( )
z1 z2
p
/2
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Q.No. Solution
4. Answer A(r, s), B(p, q, r, s, t), C(p, q), D(q, r, s, t)
(A) |z – 6i | + |z – 8| = k will represent ellipse if 2 26 8 10k k
(B) |z – 12i + 3| – |z – 2|| will represent ellipse if 2 25 12 13k k
(C) |z – ki| + |z – 4| = 10k will represent line segment if k2 + 16 = 10k k = 2, 8
(D) To represent circle
k 1 and also k = 2
5. Answer A(r, s), B(r), C(p, q, s, t), D(p, q)
(A) To satisfy all at a time z should lie on the circle |z| = 3.
Inside the circle |z – {(1 + i) – i}| = 3 and outside the circle |z + 2t – (t + 1)i| = 3
For this,
2 2 2 2( – 0) ( – 1– 0) 3 3 and 4 (4 1) 3 3t t t
2t2 – 2t – 35 0 and 5t2 + 2t – 35 > 0
Using sign scheme we have,
1 – 712
–1 – 4 115
–1 + 4 115
1 + 712
Hence,
71 –1– 4 11 –1 4 11 1 711– 1– , ,
2 5 5 2
Hence, 3, 4 lies in above interval.
(B) We have to solve for x, y
(1 ) – 2 (2 – 3 )
3 3 –
i x i i y ii
i i
2
(1 )(3 – ) – 2 (3 – ) (2 – 3 )(3 ) (3 1)
9 –
i i x i i i i y ii
i
(4 + 2i)x – 6i – 2 + (9 – 7 i)y + 3i – 1 = 10i
(4x + 9y – 3) + i(2x – 7y – 3) = 0 + 10 i
Comparing the real and imaginary parts
4x + 9y – 3 = 0 …(i)2x – 7y – 13 = 0 …(ii)
(i) – 2 (ii) gives,
9y + 14y – 3 + 26 = 0
23y = – 23 y = – 1
Putting y = – 1 in (i), we get
4x – 9 – 3 = 0 x = 3
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Q.No. Solution
(C)21 1 2
1– 1 1
n
ni i ii
i
Hence n = 0, 4, 8, 6
(D) Greatest and least absolute values of z + 1 are 1 and 6.
|z + 4| 3
6. Answer A(r, t), B(s), C(q, r, s, t), D(p)
x2 – (k – 3)x + k = 0
For roots to be real
(k – 3)2 – 4k 0
k 2 – 6k + 9 – 4k 0 k
2 – 10k + 9 0 (k – 1) (k – 9) 0
k (– , 1] [9, ] …(i)
(A) For both roots to be positive,
f (0) > 0 and3
02
k
k > 0 …(ii)
and k > 3 …(iii)
From (i), (ii) and (iii)
k [ 9, ]
(B) For both roots to be negative
D 0
k > 0, (k – 3) / 2 < 0, k < 3, k (0, 1]
(C) For both roots to be real k (– , 1] [9, )
(D) f (–1) < 0, f (1) < 01 + (k – 3) + k < 0 also 1 – (k – 3) + k < 0
2k – 2 < 0 k < 1, 4 < 0
No such value is possible
7. Answer A(q), B(r), C(s), D(p)
(A) 0)( xf , Rx , f (0) = c > 0
Parabola is upward a > 0.
(B) Roots are real and distinct 0D
f (0) < 0 c < 0
One root is positive and one negative ab < 0.(C) Roots are imaginary 0D
Parabola is downward a < 0.
(D) Parabola touches x-axis 0D
Parabola is downward a < 0
Both roots are positive, sum = 0a
b b > 0
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Q.No. Solution
8. Answer A(s), B(p), C(r), D(p)
(A) For real roots, 0D
0)3(4)2( 22 aaa
0)3[422
aaa 3a …. (i)
f (3) > 0 9 – 6a + a2 + a – 3 > 0
0652 aa
),3()2,( a … (ii)
32
a
b 3
2
2
a a < 3 … (iii)
from (i), (ii) and (iii) )2,(a
(B) 0D 3a … (i)
f (3) > 0 ),3()2,( a …(ii)
For greater than 3,
32
a
b a > 3 … (iii)
From (i), (ii) and (iii), a .
(C) D > 0 a < 3 … (i)
f (1) f (3) < 0
0)369)(321( 22 aaaaaa
0)65)(2( 22 aaaa 0)3)(1()2( 2 aaa
}2{)3,1( a … (ii)
From (i) and (ii)
)3,2()2,1( a
(D) 0D 3a … (i)
f (1) < 0 0)321( 2 aaa
0)2( 2 aa
–1 < a < 2 … (ii)
f (3) < 0 0)369( 2 aaa
0652 aa
2 < a < 3 … (iii)
From (i), (ii) and (iii)
a .
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Q.No. Solution
9. Answer A(p), B(q), C(r, s), D(t)
We have
f (x) = |x – 1| + |x – 2| + |x – 3| = k
when x < 1, then
f (x) = –3x + 6 = k
23
kx
Now, 2 13
k ,
k > 3 ...(i)
For 1 x < 2
– x = k – 4
x = 4 – k
2 < k 3 ...(ii)
For 2 x < 3
x = k, 2 k < 3
For x 3
3x – 6 = k
6
3
kx
6
33
k
k 3Clearly
(A) for k < 2, there is no solution
(B) for k = 2, there is only one solution
(C) for 2 < k < 6, there are two solution of same sign
(D) for k > 6, there are two solution of opposite sign
10. Answer A(q), B(r), C(s), D(p)
(A) (q), (B) (r)
02 baxx
a , b
ba 44)(1|| 22
142 ba
ba 412
ba 4212 = 2(1 + 2b)
2
3
321
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Q.No. Solution
(C) a 2 3
a
b.2 2
2 b
23
2ba
2
92 ba ba 92
2
.
(D)22
11
2)())(( 22222
baba 2)( 22
)2(222 abbbaba
11. Answer A(p, q, s, t), B(p), C(p, s, t), D(r)
, , be the roots of the equation
x(1 + x2) + x2(6 + x) + 2 = 0 x3 + x + 6x2 + x3 + 2 = 0
2x3 + 6x2 + x + 2 = 0
So, + + = – 3
+ + =1
2
= –1
Now,
(A) –1 –1 –1 1 1 1 1 –
2
(B) 2 + 2 + 2 = ( + + )2 – 2( + + )
19 – 2· 8
2
(C) ( –1 + –1 + –1) – ( + + )
– ( )
1 5 – 3
2 2
(D) [ –1 + –1 + –1] =1
– –12
12. Answer A(q, r), B(p), C(p, s, t), D(q, r, s, t)
(A) z is equidistant from the points i| z | and – i| z |, whose perpendicular bisector is Im (z) = 0.
(B) Sum of distance of z from (4, 0) and (–4, 0) is a constant 10, hence locus of z is ellipse with semi-major axis 5 and focus at (±4, 0).
ae = 4
e =4
5
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Complex Numbers and Quadratic Equations (Solutions) Solutions of Assignment (Set-2)
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Q.No. Solution
(C)1 5
| | | | 32
z ww
(D)1
| | | | 2 z ww
Re z | z | 2
Section-F
Q.No. Solution
1. Answer (3)
As, arg of P(z) is /4
z = x + iy, x, y, > 0
(0, 2)
(0, 1)P x iy( + )
x
y
For sum of distance to be minimum P will lie on perpendicular bisector of (0, 2) and (0, 1) hence
x = y = 2/3
k = 2
2. Answer (4)
2002 2 1
1
2 2cos sin 0
7 7
k
r
r r i
It is possible only when 2002 + (2k – 1) should be multiple of 7.
3. Answer (5)
We have
33 30z z z z
2 2 30z z z z
(x2 + y2) ((x2 – y
2) – 2i xy + x2 – y
2 + 2i xy)) = 0
(x2 + y2) (x2 – y
2) = 15 = (22 + 1) (22 – 1)
Which suggests the possible values of x and y are
x = 2, y = 1
or
x = –2, y = –1
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Solutions of Assignment (Set-2) Complex Numbers and Quadratic Equations (Solutions)
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Q.No. Solution
or
x = –2, y = 1
or
x = 2, y = –1
Centre of rectangle is (0, 0)
Now,
(0, 0)
| –3| = 2z
(3, 0)
Maximum distance of (0, 0) from the circle is 5 and minimum is 1.
4. Answer (3)
If origin z1, z2 forms an isosceles triangle then z12 + z2
2 + z1z2 = 0
Hence2 21 2 1 2 1 2
1 2 1 2
4 33
z z z z z z
z z z z
5. Answer (2)
Let 3 + i = z, hence other two vertices are iz and z + iz
So, area of such triangle is 21| |
2z
21 1·(2) 4 2
2 2
6. Answer (0)x
2 – 6kx + 9(k2 – k + 1) = 0
For real and distinct roots
D > 0
k2 – (k2 – k + 1) > 0
k – 1 > 0 …(i)
f (3) 0
9 – 18k + 9(k2 – k + 1) 0
1 – 2k + k2 – k + 1 0
k2 – 3k + 2 0
k (– , 1] [2, ) ...(ii)
Alsosum of roots
32
3k < 3
k < 1 …(iii)
From (i), (ii), (iii)
We observe that there does not exist any real value of k.
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Complex Numbers and Quadratic Equations (Solutions) Solutions of Assignment (Set-2)
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Q.No. Solution
7. Answer (2)
We have
2
2
2
2
2 2
2
3 4
2 2
3 4 As 52 2
3 4 5 10 100
2 2
ax xy
x x
ax x
x x
ax x x x
x x
(a – 5)x2 – 7x – 6 < 0 (as x
2 + 2x + 2 > 0, x R)
It is satisfied for all x if
a – 5 < 0, 49 + 24 (a – 5) < 0
71
24a
a < 3
The possible greatest integral value of a is 2.
8. Answer (5)
Firstly, let f (x) = ax2 + bx + c; a, b, c R be an integer whenever x is an integer.
f (0), f (1), f (–1) are integers
c, a + b + c, a – b + c are integers
c, a + b + c – c, a – b + c – c are integers
c, a + b, a – b are integers
c, a + b, a + b + a – b are integers
c, a + b, 2a are integers
Secondly let 2a, a + b and c be integers. Let x be an integer.
Then f (x) = ax2 + bx + c =( 1)
2 ( )2
x xa a b x c
.
Since x is an integer x(x – 1) is an even integer.
( 1)
2 ( )2
x xa a b x c
is an integer as 2a, a + b, c are integers.
f (x) is an integer for all integer x.
9. Answer (2)
x2 – 8kx + 16 (x2 – k + 1) = 0
Roots are real and distinct
Let f (x) = x2 – 8kx + 16(k2 – k + 1)
D > 0 x > 1
Let f (4) > 0 k (– , 1] [2, )
Least value of k can be 2.
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Solutions of Assignment (Set-2) Complex Numbers and Quadratic Equations (Solutions)
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Q.No. Solution
10. Answer (5)
From the given condition,
| z – 3 – 2i | 2
| 2z – 6 – 4i | 4
4 | (2z – 6 + 5i) – 9i |
|| 9i | – | 2z – 6 + 5i ||
4 9 – | 2z – 6 + 5i |
| 2z – 6 + 5i | 5
Minimum value of | 2z – 6 + 5i | is = 5
11. Answer (4)
As,1
43 2
y y
4
9y (as y > 0)
so, 3
2
46 log 4
9
Section-G
Q.No. Solution
1. Answer (1)
Given,
S1 = {z : Im (z) > 1}
S2 = {z : |z – 2 – i| = 3}
S3 = {z : Re ((1 – i)2) = 2
S1 : y 1
S2 : |z – 2 – i| = 3}
S3 : x + y = 2
Clearly ‘p’ is only point satisfying all threecondition.
Now, –1 + i, and 5 + i are end points of a
diameter
PA2 + PB
2 = (6)2 = 36
Also
|z – 2 – i| 2 – 2 i
– 2 3 z i
– 3 |z| – |2 + i| 3 …… (i)
p
–1+i 2 + i5 + i
y s,
B A
x y+ = 2
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Complex Numbers and Quadratic Equations (Solutions) Solutions of Assignment (Set-2)
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Q.No. Solution
Similarly –3 < |z1|– |2 + i| < 3 …… (ii)
(i) _____ (ii)
–6 < |z| – |z1| < 6
–3 < |z| – |z1| + 3 < 9
2. Answer (1)
Statement-1 : f (x) + f (x) + f (2x) + ...... + f (6x)
206
0
1
7 (1 ...... )k k kk
k
A A x
but when k 7
and k 14, then 1 + k + 2k + ...... + 6k = 0
f (x) + f (x) + ....... + f (6x) = 7( A0 + A7x7 + A14x
14)
Statement-2 :
B 8 A
O
Clearly AB = 8
OB = 16
8 1sin
16 2 6
AB
OB
So, the argument is3
Statement-3 :
It can be seen from the figure that the maximum value of argument is2
3
3. Answer (2)
Statement-1 : The number of common vertices is equal to the number of common roots of z1982 – 1 = 0 andz2973 – 1 = 0, which is H.C.F. of 1982, 2973, i.e., 991. Here consider both vertex has (1, 0) as one vertex.
Statement-2 : Since two parallel lines never meet so no solution
Statement-3 : Clearly the locus is a straight line.
4. Answer (1)
Let2 3
2
x xy
x
so that x
2 – (y + 1)x + 3 – 2y = 0
For real values of x
(y + 1)2 – 4(3 – 2y) 0
(y + 11) (y – 1) 0
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Solutions of Assignment (Set-2) Complex Numbers and Quadratic Equations (Solutions)
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y
E (Real axis)
4 5 °
z0
z
O
N-W N (Imaginary axis)
5
4
3
32
32
+i
Q.No. Solution
y 1 or y – 11
y lies in R – (–11, 1)
Statement-1 is true
The roots of x2 – 4|x| + 3 = 0 are
1, 3, –1, –3 and hence their sum is zero
Statement-3 is also true
5. Answer (4)
Clearly statement 1 and 2 are false
We have (x – 1)2 = cos2
x = 1 ± cos
x [0, 2]
statement-3 is true
Section-H
Q.No. Solution
1. x3 – 3x
2 + 3x + 7 = 0
(x – 1)3 + 8 = 0
(x – 1) = –2, –2, –22
x = –1, 1 – 2, 1 – 22
Now,1
1
1
1
1
1
22
22
22
2
2
211
= |32| = 3
2.0 3z
5z
By Coni method
0 0
0
0
izze
z z
3 3 5(cos sin )
32 2z i i
1 1 5 3 4
33 5 52 2
z i i
4(3 4 )i
i e
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Q.No. Solution
3.1
1
2
2
1 z
z
z
z
z12 + z2
2 = z1z2
2121
22
2
2
1000 zzzzzz
21,,0 zz form an equilateral triangle
Hence, 3|||||| 2121 zzzz
3|| 221 zz
4. |z – z1| = 5
|z – z2| = 2
|z – z1| = r + 5
|z –
z2| =
r + 2
|z – z1| – |z – z2| = 3 which is a constant
locus is hyperbola and we know
PF1 – PF2 = 2a = length of transverse axis.
length of transverse axis = 3
5. |z| = 1 or |z|2 = 1
11zz z
z
Now, 2
1
11
z z
z zzz z
z
if z x iy then z x iy
2z z ixy
21
z
z is purely imaginary therefore
21
z
z always lies on y –axis.
6. |z1 + z2|2 = |z1|
2 + |z2|2 ………. (Given)
|z1 + z2|2 = |z1|
2 + |z2|2 + 2 Re )(
21 zz = |z1|
2 + |z2|2 ………. (Relation)
0)(Re 21 zz
2z
zamp
2
1
32
6
z
zamp
6
2
1
5 2z2z1
r r
z
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Solutions of Assignment (Set-2) Complex Numbers and Quadratic Equations (Solutions)
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Q.No. Solution
7. Let z = x + iy
So, then equation becomes,
2 22 – 4 ( ) 1 0x y a x iy ia …(i)
Comparing real and imaginary part,
2 22 – 4 1 0x y ax …(ii)
–4ay + a = 0 …(iii)
(3) a(– 4y + 1) = 0 a = 0 or1
4y
But a cannot be zero. Hence1
4y gives
2 12 4 – 1
16x ax
2 214 (4 – 1)16x ax
2 2 214 16 1– 8
4x a x ax
(4 – 16a2)x2 + 8ax –3
4 = 0
2
2
–8 16 12
8(1– 4 )
a a
a
2
2
4 2 4 3
4(4 – 1)
a a
a
Now, we can observe that
if 0 a 1
2 , there is no solution
if1
2a , solution is z = x + iy
2
2
4 4 3 1·44(4 – 1)
a ai
a
8. We have,
|z1 + 1| + |z2 + 1| + |z1z2 + 1| |z1 + 1| + |z1z2 + 1 – (z2 + 1)|
= |z1 + 1| + |z1z2 – z2|
|z1 + 1| + |z2||z1 – 1|
= |z1 + 1| + |z1 – 1|
|z1 + 1 + z1 – 1| = 2|z1| = 2
|z1 + 1| + |z2 + 2| + |z1z2 + 1| 2.
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Complex Numbers and Quadratic Equations (Solutions) Solutions of Assignment (Set-2)
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Q.No. Solution
9. Let and n be the roots of the given quadratic equation ax2 + bx + c = 0
+ n = b
a and . n =
c
a = n + 1
1
1nc
a
11 1
nn nc c b
a a a
1 1
1 11 1 1 1 0
n n
n n n nc a c a b
1 1
1 1 1 1 0n n
n n n nc a c a b
1 1
1 1( ) ( ) 0n nn na c ac b
10. We have + = p and = q.
Now pVn – qVn –1 = ( + )(n + n) – (n –1 + n –1)
= n+1 + n + n + n+1 – n – n
= n+1 + n+1 = Vn+1
Also V5 = 5 + 5 = pV4 – qV3 = p[pV3 – qV2] – qV3
= (p2 – q)(3 + 3) – pqV2
= (p2 – q)[( + )3 – 3( + )] – pq[( + )2 – 2]
= (p2 – q)[p3 – 3pq] – pq[p2 – 2q]
11. According to the question
p + q = 10r ; pq = –11s
r + s = 10p ; rs = –11q.
On subtraction, (p – r ) + (q – s) = 10(r – p)
(q – s) = 11(r – p) ....(i)
Also p is a root of x2 – 10rx – 11s = 0
p2 – 10pr – 11s = 0
Similarly r 2 – 10pr – 11q = 0
On subtraction,
p2 – r 2 = –11 (q – s)
(p – r ) (p + r ) = –11 × 11 (r – p)
p + r = 121
Now p + q + r + s = 10 (p + r ) = 10 × 121 = 1210.
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Solutions of Assignment (Set-2) Complex Numbers and Quadratic Equations (Solutions)
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Q.No. Solution
12. The given equation can be written as
22x + 3 + 32x + 1 = 10 × 2x × 3x.
23 3
8 3. 10.2 2
x x
23 3
3 10 8 02 2
x x
23 3 3
3. 6 4 8 02 2 2
x x x
3 3 33. 2 4 2 0
2 2 2
x x x
3 32 3. 4 0
2 2
x x
Either
32
2
x or
3 4
2 3
x
When
32
2
x
, Taking logarithm of both sides we get x(log3 – log2) = log2
log2
log3 log2x
Also when
3 4
2 3
x
x(log3 – log2) = log4 – log3
log4 log3log3 log2
x
log2log3 log2
x or
log4 log3log3 log2
13. The given equation is 2x2 + (2 – )x + 3 = 0
and are roots of the equation, so
2
( 2) 2
2
3
But
4
3
2 2 24 ( ) 2
3
2
2
2
42
3 3
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Solutions of Assignment (Set-2) Complex Numbers and Quadratic Equations (Solutions)
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Q.No. Solution
17. If b = 0 result can be shown easily.
But if b 0.
Let n = 2(2m + 1)
Let the complex number such that 2 a
b
and the polynomial.
f (x) = xn – 1 = (x – 0)(x – 1)(x – 2)....... (x – n – 1)
We have,
1 n
f i i
. ( – i0)( – i1) ........ ( – in – 1)
and1
– n
f i i
. ( + i0)( + i1) ........ ( + in – 1)
Hence,
– f f i i
= (2 + 02)(2 – 1
2) ........ (2 – 2n – 1)
Therefore
– 1 – 12 2
0 0
( )n n
nk k
k k
aa b b
b
– 1
2
0
nn
k
k
b
22 2 1· – · ( ) 1 n n mb f f b
i i
22 1
1
mn a
bb
22 1 2 12(2 1)
2 1
m mm
m
a bb
b
= (an/2 + bn/2) hence proved
18. Let zk = costk + isintk : k{1, 2, 3}
The condition 2(z1 + z2 + z3) – 3z1z2z3R
2(sint1 + sint2 + sint3) = 3sin(t1 + t2 + t3) …(i)
Assume by the way contradiction that max(t1, t2, t3) <
6
,
Hence t1, t2, t3 < 6
Let 1 2 3 0,3 6
t t tt
1 2 31 2 3
1(sin sin sin ) sin
3 3
t t tt t t
…(ii)
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Q.No. Solution
From (i) & (ii) we have,
1 2 3 1 2 3sin( )sin
2 3
t t t t t t
Then, sin3t 2sint
4sin3t – sint 0
i.e., 2 1sin
4t , hence sint
1
2 then
2t
, which contradicts that 0 ,
6t
max(t1, t2, t3) 6
Hence proved.
19. Let s1 = z1 + z2 + z3, s2 = z1z2 + z2z3 + z3z1, s3 = z1z2z3 and take a cubic equation.
z3 – s1z2 + s2z – s3 = 0 with roots z1, z2, z3
It is given that z12
+ z22
+ z32
= 0
Hence, s12 = 2s2 …(i)
Again we have,
2 31 2 3
1 1 1s s
z z z
3 1 2 3 3 1( ) ·s z z z s s …(ii)
Now, from (i) and (ii)
2 21 3 1 1 3 12 · and | | 2( )( )s s s s s s
= 2|s1| |s1| = 2 s1 = 2 with || = 1
Now, again from relation (i) and (ii) it follows that
22 2 2
2 1 31
1 22 and
2 2
ss s s
s
Now, the given equation becomes
z3 – 2z2 + 22z – 3 = 0
(2 – )(z2 – z + 2) = 0
The roots are
z = , = – 2
Now,
Rn = |z1n + z2
n + z3n| = |n + nn + (– 1)nn2n|
= ||n |1 + n + (– 1)n2n|
R0 = 3, R1 = |1 + – 2|
= |– 22| = 2
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Q.No. Solution
R2 = |1 + 2 + | = 0
R3 = |1 + 1 – 1| = 1
R4 = 0
R5 = 2
R6 = 3
Rk + 6 = Rk for all integers k.
Rn {0, 1, 2, 3}
20. |z – |z + 1|| = |z + |z – 1||
|z – |z + 1||2 = |z + |z – 1||2
– | 1| · – | 1| | – 1| · | – 1|z z z z z z z z
2 2 – | 1|( ) | 1| · ( ) | – 1| | – 1|zz z z z z z z z z z z
2 2
| 1| – | – 1| ( )(| 1| | – 1|)z z z z z z ( 1)( 1) – ( – 1)( – 1) ( )(| 1| | – 1|)z z z z z z z z
1– – 1 ( )(| 1| | – 1|)zz z z zz z z z z z z
2( ) – ( ) (| 1| | – 1|) 0z z z z z z
( ) (2 – (| 1| | – 1|) 0z z z z
i.e., 0z z or |z + 1| + |z – 1| = 2
2 = |(z + 1) – (z – 1)| |z + 1| + |z – 1|
Solution of the equation |z + 1| + |z – 1| = 2 satisfy z + 1 = t(1 – z) where t R, t 0
– 1
1
tz
t
, so, z is any real number with – 1 z 1
The equation 0z z has the solutions z = bi, b R.
Hence the solutions to the equation are {bi : b R} {a R : a [– bi]}
21.In order to prove the result it will be sufficient to prove that
(1)
(–1)
pR
p .
Let x1, x2, x3,........, x2n be roots of p. Then
p(x) = (x – x1)(x – x2)(x – x3) ........ (x – x2n)
For some C, and
1 2 2
1 2
(1– ) (1– )........ (1– )(1)(–1) (–1– )........ (–1– )
n
n
x x xpp x x
2
1
1– 1
n
k
kk
xx
It is given that |xk| = 1 for all k = 1, 2, ......., 2n. Then
11–
1– 1– –1 1– –
11 1 1 11
k k k k k
k k k k
k
x x x x x
x x x x
x
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Complex Numbers and Quadratic Equations (Solutions) Solutions of Assignment (Set-2)
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97
5
4 97
972
1
42
Q.No. Solution
27. Answer (3)
4 97 x + 4 x = 5
2 2
mm ma b a b if m (0, 1)
Put a = 97 – x, b = x, m = w/f
44 97
2
x x
1
497
2
44 97 – x x 2 × 2.63
Which holds for
44 97 – 5 x x
Now,
Clearly for 44 97 – 5x x has two solution.
28. Answer (2)
f (x) polynomial of degree ‘n’
= anxn + an –1 xn –1 + …… + ar
when this polynomial will be divided by (x – a) (x – b), remainder will be of form Ax + B
Now, f (a) = A.a + B
f (b) = A.b + B
( ) ( )f a f n
a b
= A .,…(i)
Also B = f a f b
f a aa b
( ) ( )
af a bf a af a af b
a b
af b bf a
a b
Remainder will be
f a f b af b bf ax
a b a b
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11
Q.No. Solution
29. Answer (4)
Let 6
2
, 1
i j
i J
p x x
If we consider any three of x1, x2, x3, x4, x5, x6 is one and another three ‘0’
p = 9
30. Answer (1)
x + y = 4 – z
x2 + y
2 = 6 – z2
2xy = (x + y)2 – (x2 + y2)
= (4 – z)2 – (6 – z2)
= 2z2 – 8z + 10
The quadratic equation whose roots are x and y is
t2 – (x + y) t + xy = 0
t2 – (4 – z)t + z
2 – 4z + 5 = 0
for real roots,
(4 – z)2 – 4(z2 – 4z + 5) 0
3z2 – 8z + 4 0 (3z – 2) (z – 2) 0
2
23
z
Similarly x, y 2
,23
31. Answer (2)
px3 + qx
2 + qx + r = 0
+ 2 =q
p, 2 + 1 =
q
p, =
r
p
3 + 3 = 0 1
q
p r . p > 0
= –1 q.p < 0
pq < 0 , pr > 0
pqr > 0 or pqr < 0
pq < 0 is suitable option