Complex Numbers and Quadratic Equations (Solutions)
Solutions of Assignment (Set-2)
Q.No.
5.
Solution
Answer (1) z
1 i 1 i 1 i · 1– i 1– i 1 i (1 i )2 1 i 2 2i 1– i 2 2
=
=i z8 =
(i)8 =
(i2)4
=1 6.
Answer (3) Let z be the complex number, then z·
17
17 17 2 – 5i · 2 5i 2 5i 2 – 5i
z
=
7.
1
2 5i
17
2 – 5i
2 – 25i
2
17
27
2 – 5i
Answer (4) Additive inverse inverse of 5 + 7i is –5 –7i
8.
Answer (2) 1 2 i 1 2 i 1 i (1 2 i )(1 i ) · 1– i 1– i 1 i 1– i 2
1 – 2 i (2 1) –1 3i 2 2 1 2
3 2
– i 1 3 – , lies in the second quadrant. 2 2 9.
Answer (2) 1 i (1 i )·(1 i ) (1 i )2 1 i 2 2i i 1 – i (1 – i )(1 i ) 1– i 2 2 and
1 – i (1 – i )·(1 – i ) (1 – i )2 1 i 2 – 2i – i 1 i (1 i )(1 – i ) 1– i 2 2 3
3
1 i – 1– i (i )3 – (–i )3 a ib 1– i 1 i – i – i = a + ib 0 – 2i = a + ib a = 0 and b = –2 Aakas h Ed uc ati on al Ser vi ces Pvt . Lt d. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
-2-
Complex Numbers and Quadratic Equations (Solutions)
Solutions of Assignment (Set-2)
Q.No.
5.
Solution
Answer (1) z
1 i 1 i 1 i · 1– i 1– i 1 i (1 i )2 1 i 2 2i 1– i 2 2
=
=i z8 =
(i)8 =
(i2)4
=1 6.
Answer (3) Let z be the complex number, then z·
17
17 17 2 – 5i · 2 5i 2 5i 2 – 5i
z
=
7.
1
2 5i
17
2 – 5i
2 – 25i
2
17
27
2 – 5i
Answer (4) Additive inverse inverse of 5 + 7i is –5 –7i
8.
Answer (2) 1 2 i 1 2 i 1 i (1 2 i )(1 i ) · 1– i 1– i 1 i 1– i 2
1 – 2 i (2 1) –1 3i 2 2 1 2
3 2
– i 1 3 – , lies in the second quadrant. 2 2 9.
Answer (2) 1 i (1 i )·(1 i ) (1 i )2 1 i 2 2i i 1 – i (1 – i )(1 i ) 1– i 2 2 and
1 – i (1 – i )·(1 – i ) (1 – i )2 1 i 2 – 2i – i 1 i (1 i )(1 – i ) 1– i 2 2 3
3
1 i – 1– i (i )3 – (–i )3 a ib 1– i 1 i – i – i = a + ib 0 – 2i = a + ib a = 0 and b = –2 Aakas h Ed uc ati on al Ser vi ces Pvt . Lt d. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
-2-
Solutions of Assignment (Set-2)
Complex Numbers and Quadratic Equations (Solutions)
Q.No.
10.
Solution
Answer (3) x –2 – 3i
x 2 – 3i (x + 2)2 = – 3i
2
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)
i 17 1 i 315
9
8 1 = i 2 ·i 2 157 (i ) ·i 1 = i – i
i
9
9
9
= i – 2 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) Aak ash Edu cat io nal Serv ic es Pv t. L td . Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
-3-
Complex Numbers and Quadratic Equations (Solutions) Q.No.
15.
Solutions of Assignment (Set-2) Solution
Answer (1) 1 2i 1 2i 1 – 3i · 1 3i 1 3 i 1 – 3 i =
(1 2i )(1 – 3i ) 1– 9i 2
=
1 1 2i 1 – 3i 10
=
1 (1 6) i (2 – 3) 10
=
i 7–i 7 – 10 10 10
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)
Aakas h Ed uc ati on al Ser vi ces Pvt . Lt d. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
-4-
Solutions of Assignment (Set-2)
Complex Numbers and Quadratic Equations (Solutions)
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)
z( z 3i ) 2(2 3i ) z( z
3i ) 2(2 3i )
|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
Aak ash Edu cat io nal Serv ic es Pv t. L td . Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
-5-
Complex Numbers and Quadratic Equations (Solutions) Q.No.
49.
Solutions of Assignment (Set-2) Solution
Answer (4) 4
3
2
f (x) = x – 8x + 4x + 4x + 39 2
x = 3 + 2i (x – 3) = –4 2
x –6x + 13 = 0
Now x4 – 8x3 + 4x2 + 4x + 39 = (x2 – 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. 24 1 192 8
Required ratio =
50.
F
Answer (1)
A (1 + 2i)
E
In regular hexagon OA = AB = BC = CD = ED = EF = FA Length of perimeter = 6 × |OA| = 6 1 4
B C
6 5 51.
O
D
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
Aakas h Ed uc ati on al Ser vi ces Pvt . Lt d. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
-6-
Solutions of Assignment (Set-2)
Complex Numbers and Quadratic Equations (Solutions)
Q.No.
52.
Solution
Answer (4) z
cos i sin
z
4
1 2
6
1 2
i
2
1 1 2 3 | z | 2 2 2 tan
1 2 2
1 2 1
1 tan 1 2 1 2
arg( z ) tan 1
53.
Answer (2) z
z
1 7i ( 2 i )2 (1 7i ) (1 7i )(3 4i ) 25 25i 1 i 4 1 4i (3 4i ) (3 4i ) 25
|z|
2,
3 4
3 3 z 2 cos i sin . 4 4 54.
Answer (3) From question cos( ) i sin( ) cos( ) i sin( ) cos( ) i sin( ) 2 0i
sin( ) sin( ) sin( ) 0 55.
Answer (3) (i 3 )100 ( i 3 )100 2100 100
= (i )
1 i 3 2
100 100
2
100
( i )
1 i 3 2
100
.2100 2100
= 2100 ((2)100 + ()100 + 1) = 2100 (2 + + 1) {where is complex cube root of unity} =0 Aak ash Edu cat io nal Serv ic es Pv t. L td . Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
-7-
Complex Numbers and Quadratic Equations (Solutions)
Solutions of Assignment (Set-2)
Q.No.
56.
Solution
Answer (4) If z1 and z2 are two complex number if Im(z1 ) 0 and Im(z2 ) 0 then z1 > z2 or z1 < z2 does not hold.
57.
z2
Answer (3)
1 3 i 2 2
(z1 z3 ) ( z2 z3 )
z3
/3
( z1 z3 ) ( z 2 z3 )e i / 3 z1
Now using concept of rotation.
z1, z2, z3 are vertices of equilateral triangle. 58.
Answer (3) a
( i )i
loge a i loge i
loge a i loge e 2 loge a ae
2
2
Therefore sin(ln a) sin 1 2 Im(a ) arg(a ) 0 0 0 Therefore S1,S2 , S3 are correct. 59.
Answer (2) 2
if z + z + 1 = 0
(z – ) (z – 2) = 0 z = , 2 if z = , then
1 z
2
To find the value of z Now z z4
z6
1 z
1 z4
1 z6
1
4
2
1, z 2 1
4
2
2
1 1 1 z 2 2 z 3 3 ...... z 21 21 z z z z 1
1 z
1
2
2
1
2
1, z 5
1, z 3 1 z5
2
1 z3
1
2
2
2 1 and
2 ...... and so on
Aakas h Ed uc ati on al Ser vi ces Pvt . Lt d. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
-8-
Solutions of Assignment (Set-2)
Complex Numbers and Quadratic Equations (Solutions)
Q.No.
Solution
Therefore, 2
2
2
1 1 2 1 3 1 z z 2 z 3 ...... z 21 21 z z z z
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 60.
Answer (1)
is an imaginary 5th root of unity 5 = 1 1 + + 2 + 3 + 4 = 0
… (i)
Now, log 2 1 2 3
log 2 4 log2
1 5
[ from(i) 1 + + 2 + 3 = – 4]
log2 2 1 61.
1
1
log 2
[ sum of nth roots of unity is zero]
2
log 2 | 2 4 | log 2 | 2 | | 4 |
[ | | = | 2 | = | 3 | = | 4 | = 1 hence nth roots of unity lie on unit circle]
Answer (3) We have 3n+1
– 1 = (x – 1) (x – 1) (x – 2) ......... (x – 3n)
x
Thus,
(2 1 ) ( 2 2 )........ ( 2 3n ) ( 1 )( 2 )........ ( 3n )
( 2 1) (2 1 ) (2 2 )...... (2 3n ) . ( 1) ( 1) ( 1 ) ( 2 )...... ( 3n ) 1
1
.
(2 ) 3n 1 1
1 3n 1 1
2 1 1 1 1
.
1
1
.
(2 ) 3n .2 1
3n . 1
[ 3n = 6n = 1]
2 1 1 2 1
Aak ash Edu cat io nal Serv ic es Pv t. L td . Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
-9-
Complex Numbers and Quadratic Equations (Solutions)
Solutions of Assignment (Set-2)
Q.No.
62.
Solution
Answer (1) We have z2 z 1 and z 4 z 3 Therefore z1 z2 z1 z1 | z1 |2 and z3 z 4 z3 z 3 | z3 |2
z1 arg z2 arg z1 z2 z3 z4 z3 z 4
Now arg
2
|z | = arg 1 2 0 | z | 3 63.
( Argument of a positive real number is 0)
Answer (1) |z – (4 – 3i)| 2 | z | | 4 3i | 2
–2 | z | –5 2 3 | z | 7 | z |min = 3, | z |max = 7 64.
Answer (3) 8
|8z2z3 + 27z3z1 + 64z1z2| = | z1 || z 2 || z 3 |
( 2) ( 3 ) ( 4 )
8z1 2
| z1 |
27z2
2
| z2 |
64z3 | z3 |
z1
27 z2
(2) (3) ( 4)
2
64 z3
8z1 27z2 64z3 4 9 16
24 | 2z1 3z2 4z3 | 24 | 2z1 3z2 4 z3 | = 24 |2z1 + 3z2 + 4z3| = 24 (4) = 96 65.
Answer (1) 2
3
n –1
z1 = cos + isin and 1, z1, z1 , z1 ,......, z1
2 n
Now,
, z1 cos
Im(z12 ) Re( z1 ) sin
4 n
2 cos n
2 n
i sin
2 n
, z12 e
i4
are vertices of a regular polygon,
n
cos
4 n
i sin
4 n
5 1 2
2 cos 2 n n 5 1 2 2 cos n
2 sin
5 1 2
5 1 2 sin sin18º sin 4 10 n
2 n
10
n = 20 Aakas h Ed uc ati on al Ser vi ces Pvt . Lt d. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 10 -
Solutions of Assignment (Set-2)
Complex Numbers and Quadratic Equations (Solutions)
Q.No.
66.
Solution
Answer (2) To find area of a whose vertices are represented by complex number 0, z and zei (0 < < ) Area of
67.
1 bc sin A 2
i
B(ze )
|z|
1 | z | | z | sin 2
|z|
A(z)
O
1 | z |2 sin 2
(0, 0)
Answer (2) z
2 z
2
2 | z | z 2 r
2
2
r
2
r 2
2
2
r
when 0
r 2
2 r
2
2
r – 2r – 2 0
1 3 r 1 3
1 3
r max
68.
… (i)
Answer (1) Let the value of
i
x iy
i
( x iy )2
i
x 2 y 2 2 xy
On comparing real and imaginary part x2 x
y 2 0 and 2xy 1
y and xy
Therefore
i
1 2
1 i 1 i , 2 2
Aak ash Edu cat io nal Serv ic es Pv t. L td . Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 11 -
Complex Numbers and Quadratic Equations (Solutions)
Solutions of Assignment (Set-2)
Q.No.
69.
Solution
Answer (4) Roots of the equations (x – a) (x – b) + (x – b) (x – c) + (x – c) (x – a) = 0 2
2
2
i.e., {x – (a + b)x + ab} + {x – (b + c)x + bc} + {x – (a + c)x + ac} = 0 2
i.e., 3x – 2(a + b + c)x + (ab + bc + ca) = 0
have equal roots, Therefore 2
B – 4 AC = 0 2
4(a + b + c) – 4 × 3 ( ab + bc + ca) = 0 4[a2 + b2 + c2 + 2ab + 2bc + 2ca] – 12ab – 12bc – 12ca = 0 4(a2 + b2 + c2) – 4ab – 4bc – 4ac = 0
a2 + b2 + c2 – ab – bc – ca = 0 (Hence (3) is true) 70.
Answer (2) log
| z |2 | z | 1 2 | z | 2
3
z 2 | z | 1
2 | z |
( 3 )2
| z |2 – | z | + 1 < 6 + 3 | z |
| z | – 4| z | – 5 < 0
(| z | + 1) (| z | – 5) < 0
2
but | z | + 1 > 0
71.
| z | – 5 < 0
|z| <5
Answer (3) arg z =
4
tan –1
y x
y x
4
1
|y| = |x| x2 – y2 = 0 Re(z2) = 0 Aakas h Ed uc ati on al Ser vi ces Pvt . Lt d. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 12 -
Solutions of Assignment (Set-2)
Complex Numbers and Quadratic Equations (Solutions)
Q.No.
72.
Solution
Answer (3) Let z = x + iy Given equation is, z2 + z|z| + |z|2 = 0
(x + iy)2 + (x + iy) +
x 2 y 2 + (x2 + y2) = 0
x2 – y2 + 2ixy + x x 2 y 2 iy x2 y 2 x2 y 2 0 2x 2 x x 2 y 2 i(2 xy y x2 y 2 ) 0 Now, 2 x 2 x x 2 y 2 0
x (2x x 2 y 2 ) 0 x = 0 or x2 + y2 = 4x2 3x2 – y2 = 0 Al ternat iv e z2 2
|z|
z 1 0 |z|
2
z z 1 0 | z | | z |
z
|z|
, 2
z = |z|, z = 2|z| 73.
Answer (2) P y=
1 x 3
3y x 0 Now, 3·2 3 0 ( 3 )2 12
p
p
6 3 2
Aak ash Edu cat io nal Serv ic es Pv t. L td . Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 13 -
Complex Numbers and Quadratic Equations (Solutions) Q.No.
74.
Solutions of Assignment (Set-2) Solution
Answer (3)
z4 1 2z 1 2
Re
z4 z4 1 2z 1 2z 1
z4 z 4 1 2z 1 2z 1
2zz z 8z 4 2zz 8z z 4 1 (2z 1)(2z 1)
4zz 7 z 7 z 8 4 zz 2 z 2 z 1 9z 9z 9 0
zz
1
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
h(2) =
…(i)
2f (1) + g(1) = 0
…(ii)
Now, adding (i) & (ii), ( + 2)f (1) + ( + 2)g(1) = 0
– f (1) – g(1) = 0 76.
f (1) = – g(1)
Answer (1) The given expression is ( x – 1) (x – 1) ..... (x – 1) ..... till 2n factors. = (x – 1)2n
77.
Answer (2) 101
101
z
i
101
cos i sin – 6 6
i 101
101
101x 101 cos i sin 6 6
cos
i 101
5 5 i sin i 101 6 6
3 i 3 i cos i sin i 2 2 6 6 2 2 Aakas h Ed uc ati on al Ser vi ces Pvt . Lt d. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 14 -
Solutions of Assignment (Set-2)
Complex Numbers and Quadratic Equations (Solutions)
Q.No.
Solution
Now, (i101 + z101)103 103
3 i – 2 2
103
5 5 cos i sin 6 6
cos
515 515 i sin 6 6
cos
11 11 i sin 6 6
cos – i sin 6
78.
6
3 i – 2 2
Answer (2) z a za
1
|z – a| = |z + a|, let z = x + iy then, (x – a)2 + (y)2 = (x + a)2 + y2
x2 – 2ax + a2 + y2 = x2 + a2 + 2ax + y2 4ax = 0 x = 0 is y-axis 79.
Answer (2) Let z = x + iy z
2 x 2iy 1 i ( x iy ) 1
2x 1 i 2y (1 y ) ix (2x 1) (1 y ) 2xy i (2y ) (1 y ) x(2x 1) (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 – 2y2 – 2x2 – x = –2(x2 + y2 + 1 – 2y) = 2y – x = –2 + 4 y
2y + x – 2 = 0, i.e. a straight line
Aak ash Edu cat io nal Serv ic es Pv t. L td . Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 15 -
Complex Numbers and Quadratic Equations (Solutions) Q.No.
80.
Solutions of Assignment (Set-2) Solution
Answer (4) z = x + iy
| 2008z – 1 | = 2008 | z – 2 |
1 | z2| 2008
z
Put z = x + iy 2
1 x ( y ) 2 ( x 2) 2 y 2 2008 2
1 1 4 4x 2 x 2008 2008
1 4x 4 1004 2008 x
1 2008
2
2
4 x
4
1 1004
a line parallel to y-axis. 81.
Answer (2) y
z 1 APB 3 z 1
arg
P(z)
z lies on a circle
3
B
Al ter nativ ely
O
(–1, 0)
A
(1, 0)
x
put z = x + iy
z 1 z 1 3
arg
( x 1) iy arg ( x 1) iy 3
( x 1) iy ( x 1) iy arg ( x 1) iy ( x 1) iy 3
( x 2 y 2 1) i (2y ) arg 2 2 ( x 1) y 3
2y tan 1 2 x y 2 1 3
2y 2
2
x y 1
x2 y 2
3
2 3
y
3 ( x 2 y 2 ) 3 2y 0
1 0
1 and radius Which represents a circle having centre at 0, 3
1 2 1 3 3
Aakas h Ed uc ati on al Ser vi ces Pvt . Lt d. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 16 -
Solutions of Assignment (Set-2)
Complex Numbers and Quadratic Equations (Solutions)
Q.No.
82.
Solution
Answer (1) z
i 2 exp i , where is parameter put z = x + iy 4
x iy
i 2cos i sin 4 4
equating real and imaginary parts we get
or
x
2 cos ......(1) 4
y
1 2 sin 4
y
1 2 sin ....(2) 4
squaring and adding (1) and (2), we get 2
2
x + (y – 1) = 4
which represents a circle with centre (0, 1) and radius 2. 83.
Answer (4) y iz
90° z
O
A z
x
iz
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 z z z 1 2 3 0 3
85.
origin is the centroid of the equilateral z1 + z2 + z3 = 0
Answer (4) |z –(2 + 3i)| + |z – (–2 + 6i)| = 4 Let z1 = 2 + 3 i, 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 ter nativ ely : If we put z = x + iy, then we got an equation in x and y which does not have any solution.
Aak ash Edu cat io nal Serv ic es Pv t. L td . Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 17 -
Complex Numbers and Quadratic Equations (Solutions) Q.No.
86.
Solutions of Assignment (Set-2) Solution
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)
t
z1 … (1) and z2 z1 z3
t
w u v u
… (2)
From (1) & (2) z1 w u z2 z1 v u z3
z z w u arg 3 1 arg v u z2 z1
..........(* )
z z w u arg 3 1 1 arg 1 v u z2 z1 z z w v arg 3 2 arg u v z1 z2
z3 z2 w v arg v u z2 z1
arg
....(**)
from (*) & (**) we conclude that two triangles are similar. 87.
y
Answer (3)
3 max – min = 2 = 2. sin 1 5
(0, 25)
15
3 = 2 cos 1 5
max
min
O
88.
x
Answer (1) z1 z2
1 z1 z2
1 | z1 z2 |2 | 1 z1 z2 |2
| z1 z2 | | 1 z1 z2 |
| z1 |2 | z2 |2 z1 z2 z1z2 1 | z1 |2 | z2 |2 z1z2 z1z2 |z1|2 |z2|2 – |z1|2 – |z2|2 + 1 = 0 |z1| = 1, |z2| = 1
(|z1|2 – 1) (|z2|2 – 1) = 0
Both z1 and z2 lie on the circle |z| = 1
Aakas h Ed uc ati on al Ser vi ces Pvt . Lt d. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 18 -
Solutions of Assignment (Set-2)
Complex Numbers and Quadratic Equations (Solutions)
Q.No.
Solution
89. Answer (3) 2
x + 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 Alter nat ive 2 x + x + 1 = 0
x = ,
2 (complex root of unity)
3 + (2)3 = 2 (4) 4 2 = (2 2 )2 2 22 = (–1)2 – 2 1 = 1 – 2 = –1
90. Answer (2) p q
n n , , l l
Now
p q
q p
n n l l n l
p q
q p
n l
p q
q p
n l
0
91. Answer (2) x2
2
x 6 0 x x 6 0
x 3 x 2 0
x 3, x 2
x = ± 2 Two roots are real, with sum 0. Aak ash Edu cat io nal Serv ic es Pv t. L td . Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 19 -
Complex Numbers and Quadratic Equations (Solutions)
Solutions of Assignment (Set-2)
Q.No.
Solution
92. Answer (1) ( b c a ) x 2 (c a b ) x (a b c ) 0 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. 93. Answer (1) We know sec 2 tan2 1 (sec tan )(sec tan ) 1
b 2 4ac b 1 a a Squaring both side (b 2 4ac )b 2 a 4 a4
94.
b 4 4ab 2c 0
Answer (4) Let y
x2
34 x 71 x 2x 7 2
x 2 (y
1) x (2y 34) 71 7 y 0
For real x, discriminant should be 0
(2y 34)2 4 ( y 1)(71 7 y ) 0
4 ( y 17)2 4 ( y 1)(71 7 y ) 0
( y 17)2 ( 7y 2 78 y 71) 0
8y 2 112y 360 0
y2
( y 9)( y 5) 0
y
14 y 45 0
–
+
9 or y 5
5 95.
+ 9
Answer (3) ax 2
Given equation is an2 – bx (x –1) + c(x –1)2 = 0
bx c 0 , 2
– x – x b a 1 0 x – 1 x – 1 Now, Replacing x by ax 2
( x 1)
2
bx x 1
x x 1
c 0 ax 2 bx ( x 1) c ( x 1)2 0
x x is the root of the above equation. x 1 1
Aakas h Ed uc ati on al Ser vi ces Pvt . Lt d. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 20 -
Solutions of Assignment (Set-2)
Complex Numbers and Quadratic Equations (Solutions)
Q.No.
96.
Solution
Answer (2) b a
c a
q p
r p
, , D1 b 2 4ac , , D2 q 2 4rp Let common difference of A.P. be k. k
|| | | b2
97.
4ac a
D1 D2
q2
4 pr
p
b2
4ac
q2
4 pr
a p
D1 D2
a p
a2 p2
Answer (4) a ( x b )( x c )
(a b)(a c )
b ( x c )( x a)
(b c )(b a)
c ( x a )( x b)
(c a)(c b)
x
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 all values 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
99.
0
c a
0
Answer (1) For an identity (k2 – 3k + 2) = 0
(k – 1) (k – 2) = 0 k = 1, k = 2 2
k – 5k + 4 = 0
(k – 1)(k – 4) = 0 k = 1, 4
2
k – 6k + 5 = 0
(k – 5)(k – 1) = 0 k = 1, 5
Common value of k = 1. Aak ash Edu cat io nal Serv ic es Pv t. L td . Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 21 -
Complex Numbers and Quadratic Equations (Solutions)
Solutions of Assignment (Set-2)
Q.No.
100.
Solution
Answer (1) (1) Let roots are ,
+ =
=
13 5
… (i)
k
… (ii)
5
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 2 = k
… (i) … (ii)
By (i), (ii)
= 2, k = 8 (4) In this case
+ = 0 101.
k = 0
Answer (2) x2
ax b 0 a, b, a 2 4b
x2
bx a 0 b, a, b 2 4a
Now,
a2
4b b 2 4a
a 2 4b b 2 4a a 2 b 2 4 (b a) (a b)(a b 4) 0 a b 4 (a b) 102.
Answer (1) Let be common root, p 2 2q r 0
…(1)
and p 2 2r q 0
…(2)
Now (1) – (2) 2 (q r ) r q 0 Common root is
1 2
1 , substituting in (1) 2
2
1 1 2q r 0 4r 4q p 0 2 2
p
Aakas h Ed uc ati on al Ser vi ces Pvt . Lt d. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 22 -
Solutions of Assignment (Set-2)
Complex Numbers and Quadratic Equations (Solutions)
Q.No.
103.
Solution
Answer (1) 2
x + 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. 2
Also the roots of ax + 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 : 1 104.
Answer (2) If 1, 2, 3 are roots of equation then 3
2
x + ax + bx + c = 0
1 + 2 + 3 = – a
105.
a = –6
12 + 23 + 13 = b
b = 11
123 = – c
c = –6
Answer (4) 2
f (x) = ax + bx + c
b c x a a
b b2 x a 4a2
= a x2
= a x2
c 4a2 a b2
2 b b2 4ac f (x) = a x 2a 4a2 2
b b2 4ac f (x) = a x 2a 4a
f (x) = a x
b D 2a 4a
Clearly if a > 0 the minimum value of f ( x ) Similarly of a < 0 the maximum value
D 4a
D 4a
If ax2 + bx + c > 0 then a > 0, D < 0 for all x R Hence option (4) is not true.
Aak ash Edu cat io nal Serv ic es Pv t. L td . Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 23 -
Complex Numbers and Quadratic Equations (Solutions)
Solutions of Assignment (Set-2)
Q.No.
106.
Solution
Answer (3) 2
2
x + 2x + 3 = (x + 1) + 2 m = 2
– x2 + 4x + 6 = – x2 + 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 y mx 2 9m 5m 1 We need y > 0
Upward parabola above x-axis. mx 2
y
9mx 5m 1 0, x R .
D 0, a 0 i.e., 81m 2
4 (m )(5m 1) 0 and m 0 O
4 m (61m 4 ) 0 and m 0 0 m 61
x
Also for m = 0, 0 x 2 9(0) x 0 1 1 0, x R
108.
4 61
m 0,
Answer (1) ( l m ) x 2 lx 1 0 l
2
ml
1
(2)
l
m
l 3 (m l )
2
1 2 (l m )
…(1)
…(2)
From (1) and (2) 1 2(l m )
l2
9(l m ) 2
2l 2 9l 9m 0 For real l, 81 8 9 m 0 m
m
81 72
9 8
Greatest value of m is
9 . 8
Aakas h Ed uc ati on al Ser vi ces Pvt . Lt d. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 24 -
Solutions of Assignment (Set-2)
Complex Numbers and Quadratic Equations (Solutions)
Q.No.
109.
Solution
Answer (3) 3 px 2 5qx 7r 0
(5q )2 4 (3 p)(7r ) 25q 2 84 pr 2
17 336 2 25 ( p r ) 2 84 pr 25 p 2 34 pr 25r 2 5 p r r 0 5 25
roots are real and distinct. 110.
Answer (2) 3
2
x – 2x – x + 2 = 0
As x = 1 is the root of the equation Hence we may write 3
2
x – 2x – x + 2
= x2 (x – 1) – x(x – 1) – 2(x – 1) = (x – 1) (x2 – x – 2) = (x – 1) (x – 2) (x + 1) Roots
111.
= 1, –1, 2.
Answer (2)
2a, b 2 2 2a 4a 2 2b The other root of equation will be 2 2 i.e.
2a 4a 2 2b
Sum of roots, S = –4a Product of roots, P = 4a 2 ( 4a 2 2b) 2b
required equation is x 2 Sx P 0 i.e. x
112.
2
4ax 2b 0
Answer (1) For roots of opposite sign, product < 0
a2
3a 2 3
0 (a 2)(a 1) 0
1 a 2 113.
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). ax 2
bx c 0
D = b 2
4ac (a c )2 4ac (a c )2
0 Roots are real. Aak ash Edu cat io nal Serv ic es Pv t. L td . Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 25 -
Complex Numbers and Quadratic Equations (Solutions) Q.No.
114.
Solutions of Assignment (Set-2) Solution
Answer (3)
p – iq is other root.
p + iq is one root
Let be third root. Now sum = p iq p iq 0
2 p 2p is root of x 3 ax b 0 2p is root of ( x )3 ax b 0 115.
x3
ax b 0
Answer (2) We have, given expression (a12 + a22 + a32 +.....+ an – 12)x2 + 2(a1a2 + a2a3 + a3a4 +.....+ an – 1 an)x + (a22 + a32 + a42 +......+ an2) 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 x
a2 a1
a3 a2
a4 a3
.......
an an1
a1, a2, a3, ....... , an – 1, an are in G.P. 116.
Answer (2) f ( x ) ax 2
bx c, given f ( 1) a b c 0
f ( x ) 0,
x R as roots are non-real complex
f (–2) < 0 4a 2b c 0 4a c 2b 117.
Answer (3) b a
Given, , Also,
1
2
c a
1
2
( )( 2 2 ) 2 2 ( )2 2
c 2 b 2 2c b 2 2 a a a a
bc 2 b 2a 2ca 2
2ca 2 bc 2 b 2a
a b c , , are in H.P. c a b
2
a b
c a
b c
Aakas h Ed uc ati on al Ser vi ces Pvt . Lt d. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 26 -
Solutions of Assignment (Set-2)
Complex Numbers and Quadratic Equations (Solutions)
Q.No.
118.
Solution
Answer (2) ( x 1)(3 x 2)(3 x 1)( x 2) 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 t 2 25 t 5, t 5 Now, 3 x 2 5 x 5 and 3 x 2 5 x 5
3 x 2 5 x 5 0 and 3 x 2 5 x 5 0 3 x 2 5 x 5 0 has two irrational roots. whereas roots of 3 x 2
119.
5 x 5 0 are imaginary.
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
sin x
x ,
1 2
5 6 6
… (ii)
By (i), (ii) x , 2 6 S2 : Using A.M. G.M. x2
tan2
x
x2
2
x2
2 x ( x 2 x ) tan x2 x
tan2
x
x2
x
2tan
S3 : Using D 0
4(a + b + c)2 – 4(1)(3)(ab + bc + ca) 0
a + b + c + 2(ab + bc + ca) – 3(ab + bc + ca) 0
a + b + c + (ab + bc + ca) (2 – 3) 0
3 2
2
2
2
2
2
2
a 2 b2 c 2 ab bc ca
… (i)
Aak ash Edu cat io nal Serv ic es Pv t. L td . Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 27 -
Complex Numbers and Quadratic Equations (Solutions) Q.No.
Solutions of Assignment (Set-2) Solution
(a – b)2 c2
But
(b – c)2 a2 (c – a)2 b2
a2 b2 c 2 ab bc ca
3 – 2 2
2
4 3
S4 : x 3 3 px 2 3qx r 0
…(i)
Multiply the second equation by x x3
2 p x 2 qx 0
… (ii)
By (i) – (ii) px 2
2qx r 0
… (iii)
But x2 + 2px + q = 0 2
2
px + 2p x + pq = 0
… (iv)
By (iii) (iii) - (iv) (iv) 2x(q – p2) + (r – pq) = 0 x
pq 2(q p2 ) r
Putting x in 2
x + 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 b a
So, ,
c a
Now, we have to observe root of the equation (a + cy)2 = b2y
a2 + 2acy + c2y2 = b2y Aakas h Ed uc ati on al Ser vi ces Pvt . Lt d. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 28 -
Solutions of Assignment (Set-2)
Complex Numbers and Quadratic Equations (Solutions)
Q.No.
Solution
c2y2 + (2ac – b2)y + a2 = 0 2ac b2 a2 y 2 y 2 0 2 c c b 2 2a a2 y 2 2 y 2 0 c c c 1 1 1 y 2 2 2 y 2 2 0 Hence the equation (a + cy)2 = b2y has roots 122.
1
2
,
1
2
Answer (1) Since a, b, c are in G.P., So, b2 = ac
4b2 – 4ac = 0 D = 0 for the equation ax2 + 2bx + c = 0
Hence, it will have equal roots, and root will be x
b a
Now, ax2 + 2bx + c and dx2 + 2ex + f = = 0 have a common root, b a
So, x will satisfy the equation dx2 + 2ex + f = = 0
d.
b2 a
2
b a
2e. f 0
db2 2aeb a2 f a2
0
= 0 db2 – 2aeb + a2f = = 0 dac – 2aeb + a2f = = 2eb dc + af = d f a c
So, 123.
2e b
d c f , , are in A.P. a b c
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. Aak ash Edu cat io nal Serv ic es Pv t. L td . Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 29 -
Complex Numbers and Quadratic Equations (Solutions) Q.No.
124.
Solutions of Assignment (Set-2) Solution
Answer (4)
–
D a
1
p 2 48
1
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)
2 q
...(iii)
2 (2) r
...(iv)
2
(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 t o obtain the desired P as i
4 3
3 e 4 .ei, where tan = yielding sin 128.
4 3 , cos = 5 5
Answer (4) We have, z
1 z
2
1 1 z
z 1
z
z
1 zz z
z
1 2iIm ( z)
1 i i , R z Im z Thus the locus of
z
1 z2
is y-axis.
Aakas h Ed uc ati on al Ser vi ces Pvt . Lt d. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 30 -
Solutions of Assignment (Set-2)
Complex Numbers and Quadratic Equations (Solutions)
Q.No.
129.
Solution
Answer (4)
Imaginary axis
Z2 (7, 6)
Z2
1
) 2 1 , ( Z 0
90°
1 3 5
Z2
(6, 2)
Real axis
(6 2 cos 45, 5 2 sin 45) (7, 6) 7 6i
by rotation about (0, 0) Z2
i
Z2 Z2
130.
i
e 2 Z2 Z2 (e 2 ) (7 6i ) cos i sin (7 6i )( i ) 6 7 i 2 2
Answer (1) zz 3
zz 3 350
| z |2 (z 2 ) | z |2 (z 2 ) 350 | z |2 (z 2 z 2 ) 35 3 50
(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
(–4, 3)
A(4, 3)
B
x
O D
C
(–4, 3)
(4, – 3)
Required area = AB × BC =8×6 = 48 sq. units Aak ash Edu cat io nal Serv ic es Pv t. L td . Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 31 -
Complex Numbers and Quadratic Equations (Solutions)
Solutions of Assignment (Set-2)
Q.No.
131.
Solution
Answer (4) z = cos + isin = ei 15
lm e
Now,
i (2 m 1)
m 1
= sin + sin3 + ..... + sin29 15.2 2 .sin 2 (15 1) 2 = 2 2 sin 2 sin
=
sin15.sin15 sin
For = 2°, the given expression reduces to = 132.
sin30.sin30 1 sin2 4sin2
Answer (2) We have + = – p
3 + 3 = q = ( + )3 – 3( + ) = – p3 + 3p()
p 3
q
3p
The quadratic equation with
x2
x 0
x2
( )2 2 p2 2
and as roots is
x2
p3 q 3p
p3 q 3p
x 1 0
x 1 0
(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
6
an 2an 2 an 1
Aakas h Ed uc ati on al Ser vi ces Pvt . Lt d. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 32 -
Solutions of Assignment (Set-2)
Complex Numbers and Quadratic Equations (Solutions)
Q.No.
Solution
3
an
2an 2 2an 1
,n 2
Putting n = 10, we get a10 2a8
2a9 134.
3
Answer (2) Let be a common root between given equations x2 + bx – 1 = 0 and x2 + x + b = 0
2 1 2 b 1 1 b 1 b
2
b2 1 1 b 1 b 1 b
b2 1
b2 – b3 + 1 – b = 1 + 2b + b2
b3 + 3b = 0
b
135.
b2
1 1 b and 1 b 1 b 2
(1 b)2 1 b
0, b 3i
b 3i
Answer (4) As, a is real, So a a gives
z2 + z + 1 = z 2 z
( z z )( z z 1) 0
1
As, z z So, z z 1
x
1 {where z x iy } 2
Now, a = z2 + z + 1 2
1 1 iy iy 1 2 2
3 y 2 4
As, y 0 so, a
3 4
Aak ash Edu cat io nal Serv ic es Pv t. L td . Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 33 -
Complex Numbers and Quadratic Equations (Solutions)
Solutions of Assignment (Set-2)
Section-B Q.No.
1.
Solution
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 + 9y2 = 9x2 + 9y2 – 36x + 36 30x = 35 7 6
x =
6 Re(z) = 7 (A line parallel to y-axis) 1 7 Also mid-point of , 0 and (2, 0) is , 0 3 6 2.
Answer (1, 2, 3) We can write sin
2k 2k i cos 11 11
2k 2k i sin = i cos 11 11 Now 10
S
sin 211k i cos 211k k 1
10
S i
e
i
2 k 11
k 1
i 4 i 20 i 2 11 11 11 = i e e .... e 1 1 i (as 1 + + 2 + ..... + 10 = 0)
S = i S i SS 0 and SS 1
S
1 1 i 2 2i 2
S
1 2
(1 i )
Aakas h Ed uc ati on al Ser vi ces Pvt . Lt d. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 34 -
Solutions of Assignment (Set-2)
Complex Numbers and Quadratic Equations (Solutions)
Q.No.
3.
Solution
Answer (1, 3) cos A cos B cos C 0 sin A sin B sin C 0 Let a cos A i sin A eiA b cos B i sin B eiB
cos C i sin C eiC
c
abc
0
a3
b 3 c 3 3abc
a2 bc
b2 ac
e2iA eiB .eiC
e2iA iB iC
c2 ab
3
e2iB eiA eiC
e2iC eiA eiB
3
e2iB iA iC e2iC iA iB 3
cos(2 A B C ) i sin(2 A B C ) cos(2B A C) i sin(2 B 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.
Answer (2, 4) 2
x – 2xcos + 1 = 0
2 cos 4 cos 2 4 x 2
= cos isin
= cos + isin, = cos – isin n cos n i sin n ; n cos n i sin n S
n n 2 cos n ; P = 1
Equation is x 2 Sx P 0 i.e. x 2 2 x cos n 1 0
Option (4) x2
2 x cos n cos 2 n 1 cos 2 n = 0
( x cos n)2 sin 2 n 0 Option (2) Aak ash Edu cat io nal Serv ic es Pv t. L td . Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 35 -
Complex Numbers and Quadratic Equations (Solutions) Q.No.
5.
Solutions of Assignment (Set-2) Solution
Answer (2, 3) 2
2
2
a + b + c = 1 b +ic = (1 + a) z
z
b ic 1 iz 1 a 1 iz b ic 1 i 1 a 1 i
b ic
1 a
1 iz 1 a c ib 1 iz 1 a c ib
(1 a c ib ) (1 a c ib ) (1 a c ib ) (1 a c ib ) (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 c = a ib 1 a c ib 1 a c ib
6.
Answer (3, 4)
z4
D
Cz
3
z 4 z1 z2 z1 2
amp
A
A = 90°
90° z1
B z2
|z1 – z4| = |z2 – z3|
alsoz1 – 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 are i.e. Now,
1 3 p ,
1 3 p .
1 1 3 3 p ,p
,
1 3 ,p
2
1 3 p . 2
x 2 2 y 2 4z 2 x 2 2 y 2 z 2 2 x 2 2 y 2 2z 2 2 x 2 2 y 2 2z 2 2 x 2 4 y 2 z 2 2 ( x 2 2 y 2 z 2 )
= =
Option (4)
We can assign the , , different value we get other options also.
Aakas h Ed uc ati on al Ser vi ces Pvt . Lt d. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 36 -
Solutions of Assignment (Set-2)
Complex Numbers and Quadratic Equations (Solutions)
Q.No.
8.
Solution
Answer (1, 2) –1
z + z = 1 z
1
1
z
2
z – z + 1 = 0 z n
1 i 3 , 2 2 – n
n
–n
z + z = (– ) + (– )
Case (1), n = 3m (– )3m + (– ) –3m = (–1)n + (–1)n = 2(–1)n when n = 3m + 1 (– )3m + 1 + (– ) –3m – 1 = (1)n 3m 1
= (1)n
( 1)n ()3m 1
1
= (–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 2
2
x + y
2x + 1 < (x2 + y2 + 6x + 9)
8x > –8 x > – 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
arg ( 1)
2
option (4) – 5 = 2(x – 1) + i(2y – 1) + 3 = 2(x + 3) + i(2y – 1) as x > –1 | + 3| > | – 5| Option (2) Aak ash Edu cat io nal Serv ic es Pv t. L td . Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 37 -
Complex Numbers and Quadratic Equations (Solutions)
Solutions of Assignment (Set-2)
Q.No.
10.
Solution
Answer (2, 3, 4) |z + |2 = |z|2 + ||2 Since |z + |2 = |z|2 + ||2 + z z
|z|2 + ||2 = |z|2 + ||2 + z z
z z = 0
z z
z
z
z
is purely imaginary
z 2
Therefore, amp
11.
Answer (1, 2, 3) Clearly |z1 – z2|min = 2 – 1 = 1 |z1 – z2|max = 3
y
max|2z1 + z2| = |2 + 2| = 4 Now, |z| = 1 |z1| = 1 z1.z1 z1
1
1 O
1
1
z1
Now z2
x
2 1
z1
|z| = 2
z2 z1 |z| = 1
and | z2 z1 | | z2 | | z1 |
3. 12.
Answer (3, 4) z
2 z
|z|
1 2 |z|
1
| z |2 | z | 2 0 (| z | 2)(| z | 1) 0
|
1
|
2
1 | z | 2 But |z| 0
0|z| 2 | z | 3 and |z| 4 Option (3) and (4) Aakas h Ed uc ati on al Ser vi ces Pvt . Lt d. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 38 -
Solutions of Assignment (Set-2)
Complex Numbers and Quadratic Equations (Solutions)
Q.No.
13.
Solution
Answer (1, 2, 4) The given equation is |2 z – i| = m|z + i|
(2z i )(2z i ) m2 (z i )(z i ) 4zz 2iz 2i z 1 m 2 ( zz iz i z 1) (4 m2 )zz (2 m 2 )iz (2 m 2 )i z (1 m 2 ) 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) 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) | z1 | a 2 b2 1
a2 + b2 = 1 …(i)
| z2 | c 2 d 2 1
c2 + d2 = 1 …(ii)
Re( z1 z2 )
= 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, | 1 | a 2 c 2 a 2 b 2 1 Similary we can observe, |2| = 1 Re( 1 2 ) 0 16.
Answer (2, 4) Given inequality is, log
1 2
| z |2 2| z | 6 0 2| z |2 2| z | 1
| z |2 2| z | 6 1 2| z |2 – 2| z | 1
|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 Aak ash Edu cat io nal Serv ic es Pv t. L td . Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 39 -
Complex Numbers and Quadratic Equations (Solutions)
Solutions of Assignment (Set-2)
Q.No.
17.
Solution
Answer (2, 3, 4) z1, z2 are the complex numbers satisfying, z1 z2
1
z1 – z2
|z1 + z2| = |z1 – z2| ( z1 z 2 )(z1 z 2 ) (z1 – z2 )(z1 – z2 ) z1 z1 z1 z2 z2 z1 z2 z 2 z1z1 – z1z 2 – z 2 z1 z 2 z 2 2( z1 z2 z2 z1) 0 z z 1 – 1 z2 z2 z1 z2 z2 z1 0 z1 z2 z1 z2 0 Re( z z2 ) 0 18.
Answer (1, 2, 3, 4) sin cos
b a
c
sin . cos
a
sin2 cos2 1 2
2c b 1 b 2 2ca a 2 a a
a 2 b 2 2ac 0 Also (a
19.
c )2 b 2 c 2 .
Answer (1, 2)
2b a
,
hh
A
2 b 2 ac
a
Also 2h =
2B
h
b a
2B A
B A
c a
| | 2
, ( h )( h ) 2 B 2 AC A
( )
2B A
C A
b2
ac a
||
2 B 2 AC A
b2
ac a2 B 2 AC A 2 2b a
.
Aakas h Ed uc ati on al Ser vi ces Pvt . Lt d. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 40 -
Solutions of Assignment (Set-2)
Complex Numbers and Quadratic Equations (Solutions)
Q.No.
Solution
20. Answer (1, 2) Let us use the transformation y
x 3 x 1
2
x 1
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 +ve
y2 – 1 0 as y2 + 9 > 0, y.
–ve –1
+ve 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) ax 2
bx c 0
Let a, b, c is a, ar , ar 2 Now, ax 2 arx ar 2 0
1 3i x = r or 2r x 2 rx r 2 0 x r 2 Roots are imaginary and are in the ratio 1 : or 2 : 1. 22.
Answer (3, 4) 25 x 2 5 x 12 0 (5 x 3)(5 x 4) 0 3 5
x , x cos = sin
4 5
3 x as 1 x 0 5
4 5
3 3 or 5 5
sin 2 2 sin cos 2
or
3 5
24 4 25 5
3 4 24 2 5 5 25
Aak ash Edu cat io nal Serv ic es Pv t. L td . Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 41 -
Complex Numbers and Quadratic Equations (Solutions)
Solutions of Assignment (Set-2)
Q.No.
23.
Solution
Answer (1, 2) Let be common root
2 pq r 0,
pq, r
2 pr q 0,
pr , q
p ( q r ) r q 0 Common root is =
1 p
1 p
Other roots are, rp and qp
Equation containing other roots is x2
p (r q ) x p 2 rq 0 2
1 1 is common root pq r 0 p p p 1
1 p2
(q r )
Now x 2 p (q r ) x p 2qr 0
p p 2 x 2 (q r ) x pqr 0 p 24.
p (q r ) x 2
(q r ) x pqr 0
Answer (1, 2, 3) x2
( 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
… (3)
From (1) and (3), m (0, 1)
Aakas h Ed uc ati on al Ser vi ces Pvt . Lt d. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 42 -
Solutions of Assignment (Set-2)
Complex Numbers and Quadratic Equations (Solutions)
Q.No.
25.
Solution
Answer (1, 2, 3, 4) x2
2ax a 2 1 0
( x a)2 1 x = a + 1, a – 1. 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.
26.
Answer (1, 2, 3)
4 , A + = 36, = B Let , , , be a, ar , ar 2, ar 3 a + ar = 4 ar 2
ar 3 36
r 2 9 A
27.
r 2 (1 r )
1 9
r = ± 3, a = 1.
a(ar ) = A
B =
1 r
A = 3
B = (ar 2) (ar 3)
B = 243
B = 81 A.
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 4 (log2 x ) log2 x 4 log2 x log2 2 2 5 1 3 2 t t t 4 2 4
Put log2 x = t,
3t 3 4t 2 5t 2 0 (t 1)(3t 2 7t 2) 0 (t 1)(3t 1)(t 2) 0 t 1, 28.
1 1 , 2 log2 x 1, , 2 x = 2, 21/ 3 , 2 2 3 3
Answer (3, 4) b
c
a
a
ax 2
bx c 0, ,
b2
4ca a
Aak ash Edu cat io nal Serv ic es Pv t. L td . Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 43 -
Complex Numbers and Quadratic Equations (Solutions) Q.No.
Solution Ax 2
Bx C 0,
B2
b2
Now,
4ac
4ac a
2
B2
B A
2
1 2
C A
4CA A
B2
4CA A 2
4ac a 2 B 4 AC A
A
, ( )( )
A
a
b2
B
4CA
b2
29.
Solutions of Assignment (Set-2)
2
B ( ) A
2
b B a A
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 x 2 + ax + a2 + 6x < 0 will be satisfied for all values of x ( , ) if , are real 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 (, ). Thus for (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) < 1, > 1 – 1 < 0, – 1 > 0
…(ii)
( – 1) ( – 1) < 0
a2 + 7a + 1 < 0 –7 – 45 –7 45 , 2 2
a
…(iii)
< 2, > 2 ( – 2) ( – 2) < 0 a2 + 8a + 4 < 0
a (– 4 – 2 3, – 4 2 3)
…(iv)
Common values of a satisfying (ii), (iii) and (iv) are
–7 – 45 , –4 2 3 2
a
…(v)
Hence answer is (1), (2), (3) those are subject of (v) Aakas h Ed uc ati on al Ser vi ces Pvt . Lt d. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 44 -
Solutions of Assignment (Set-2)
Complex Numbers and Quadratic Equations (Solutions)
Q.No.
30.
Solution
Answer (2, 3) The given equation is 1 xp
1 x q
1 r
1 x qxp ( 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 =
pq
2
Product of roots = + [(p + q)r – pq]
31.
=
( p q )2 – pq 2
=
1 2 ( p q2 ) 2
Answer (1, 2, 4) (a + 2) x2 + 2(a + 1)x + a = 0 Let , be roots
a a2
2 a 1 (integer) a2 a
a2
(integer)
will integer if
For a = 0, a = –1, a = –3 Also for a = 0, a = –1, a = –3
32.
–2 a 1 a2
is integer
Answer (1, 3) Since 1 is the repeated roots of ax3 + bx2 + c = 0 b a
So, 1 + 1 + = –
1 1.1 + + = 0 – 2 Aak ash Edu cat io nal Serv ic es Pv t. L td . Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 45 -
Complex Numbers and Quadratic Equations (Solutions) Q.No.
Solutions of Assignment (Set-2) Solution
1 c c 1 1.1. = – – 2 a a 2 3 b – a 2
b – 3 c
Now, by the equation, cx3 + bx + a = 0 b c
a c
x3 x 0 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 equation ax4 + bx2 + c = 0 are real for this b
b
a
a
– 0
c a
0
0
…(i)
…(ii)
From (i) and (ii) b > 0, a < 0, c < 0
or
34.
b < 0, a > 0, c > 0
Answer (1, 2, 3)
– 5 ( – )2 < 5 ( + )2 – 4 < 5 k2 – 4 < 5 k2 < 9 k (– 3, 3) Hence answers is (1, 2, 3) Aakas h Ed uc ati on al Ser vi ces Pvt . Lt d. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 46 -
Solutions of Assignment (Set-2)
Complex Numbers and Quadratic Equations (Solutions)
Q.No.
35.
Solution
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
11 2a + 3 + a2 + 3a + 8 > a2 + 2a 3a > – 11 a > – 3 Combining these three, a (5, )
Hence answer is (1, 2, 3) 36.
Answer (1, 3, 4) z
z1
t
1 t
z2
we have z = (1 – t)z1 + tz2, 0 < t < 1, t z2
1 t z1 t 1 t
z
z
t z2 1 t z1
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) 2
3 i 2 2
e
i
1
1 30° 30°
30° 30°
2 3
3
6
2
e
i
i
3
3 e 2 4 e 5
e So,
2i
i
3
5 6
z1oz2 can be
2 5 , 3 6
Aak ash Edu cat io nal Serv ic es Pv t. L td . Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 47 -
Complex Numbers and Quadratic Equations (Solutions)
Solutions of Assignment (Set-2)
Section-C Q.No.
Solution Comprehension-I
1.
Answer (4)
y
From rotation formula z2
0 ( z1 0).
OB OA
z2
B
e i A
z2
2.
z1.
z2 z1
z1 x
(cos sin )
Answer (4) In OAB,
= | | ei Dz
In OFE
2
1
E(z
Using rotation formula OE
z = z1e . 1
i
OF
C B
OFE ~ OAB OE
F z1
1
OF
)
O
1 A
OB OA
z = z1e . | | = z1 1
i
z2 z1 + 3.
Answer (3) From rotation formula z1
B z1
OB .(1 0).e i OA
… (i) O
OC z2 .(1 0).e i OA
z1z2 = 1
z2
A
1
… (ii) C z2
1 z1
Comprehension-II
Given equation is x2 – ix – 1 = 0
x2 – 1 = ix 1
x – i x
…(i)
Aakas h Ed uc ati on al Ser vi ces Pvt . Lt d. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 48 -
Solutions of Assignment (Set-2)
Complex Numbers and Quadratic Equations (Solutions)
Q.No.
Solution
Now let x = cos + isin, then equation (i) becomes, cos + isin – cos + isin = i
2sin = 1 sin =
1.
6
or
1 2
5 6
Answer (3) We have to find out the value of x51
cos i sin 6 6 cos
51
17 17 i sin 2 2
= 0 + i = i 2.
Answer (1) We have to evaluate x 20
cos
1 x 20
20 20 20 20 – i sin i sin cos 6 6 6 6
2cos
10 4 2cos 3 3
2cos 3 – 2cos –1 3
3.
Answer (2) For finding x 2013
cos
1 x
2013
2013 2013 2013 2013 – cos i sin i sin 6 6 6 6
2i sin
2013 6
2i sin
671 2
2i sin i 2
Aak ash Edu cat io nal Serv ic es Pv t. L td . Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 49 -
Complex Numbers and Quadratic Equations (Solutions)
Solutions of Assignment (Set-2)
Q.No.
Solution Comprehension-III
1.
Answer (2)
2.
Answer (4)
3.
Answer (2) Soluti on of Q.1 to Q.3
Given expression is (1 + x)n = a0 + a1x + a2x2 + a3x3 + .............+ 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, (1 i )n (1– i ) n 2(a0 – a2 + a4 – a6 + .......) 2
2.2n /2 · cos 2
n
4 2n /2 cos n 4
…(iii)
Hence, answer of question 2 is (4) Now, adding (ii) & (iii), we can get 2(a0 + a4 + a8 + .......) = 2n 1 2n /2 cos n
– 1
n
4
a0 + a4 + a8 + a12 + ...... = 2n 2 2 2 cos
n
4
Hence, answer of question 3 is (2) Comprehension-IV
1.
Answer (1) f ( x )
y
3
x 3x k
f ' ( x )
3 x 2 3 0 x = ±1
–1
1 O
For exactly one positive root,
x
f ( 1) 0 and f (1) < 0
–1 + 3 + k < 0 and 1 – 3 + k < 0 k 2 and k < 2 k ( , 2) . Aakas h Ed uc ati on al Ser vi ces Pvt . Lt d. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 50 -
Solutions of Assignment (Set-2)
Complex Numbers and Quadratic Equations (Solutions)
Q.No.
2.
Solution
Answer (1) y
For exactly one negative root, f ( 1) 0 , f (1) 0
–1 + 3 + k > 0, 1 – 3 + k > 0 –1
k > – 2, k > 2
O
1
x
k (2, ) 3.
y
Answer (2) For one negative and two positive root f ( 1) 0 , f (0) > 0, f (1) < 0
1 –1
–1 + 3 + k > 0, k > 0, 1 – 3 + k < 0
O
x
k 2 , k 0 , k 2 0 k 2 i.e., k (0, 2) . Comprehension-V
1.
Answer (3) | x |2 (| x |2 2k 1) 1 k 2
x 2 ( x 2 2k 1) = 1 k 2 x 4 (2k 1)x 2 k 2 1 0 All roots are imaginary, if D b 2 4ac 0
(2k 1)2 4(k 2 1) 0 k
5 4
…. (1)
Also roots are imaginary if D 0 , but x 2 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 2k 1 0 and k 2 1 0 k ( , 1) 5 All roots are imaginary if k ( , 1) , 4 2.
Answer (2) For exactly two real roots of t2
(2k 1)t k 2 1 0
D
0 and one value of t x 2 is positive and one is negative.
(2k 1)2 4(k 2 1) 0 k
5 4
… (1)
Product = k 2 1 0 –1 < k < 1
… (2)
From (1) and (2) k ( 1, 1) . Aak ash Edu cat io nal Serv ic es Pv t. L td . Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 51 -
Complex Numbers and Quadratic Equations (Solutions)
Solutions of Assignment (Set-2)
Q.No.
3.
Solution
Answer (1) For repeated roots = 0, k
5 4
or P = 0, k 2 1 0 , k 1 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)
y P
(0, 2) (2, 1)
y=1 x
( 2,0)
Clearly, there is only one point of intersection of the line x y 2 , and circle x2 + y2 – 4x – 2y = 4 2.
Answer (2) z
1– i
2
2
z–5–i
= (x + 1)2 + (y – 1)2 + (x – 5)2 + (y – 1)2 = 2(x2 + y2 – 4x – 2y) + 28 = 2(4) + 28
x
2
y 2 – 4x – 2y 4
= 36 3.
Answer (4) w – 2 i
3
| w | – | 2 i | < 3
3 5 w 3 5
– 3 – 5 – w 3 – 5
...(i)
Also, z – 2 i 3
– 3 5 z 3 5
...(ii)
– 3 < |z| – |w| +3 < 9 Aakas h Ed uc ati on al Ser vi ces Pvt . Lt d. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 52 -
Solutions of Assignment (Set-2)
Complex Numbers and Quadratic Equations (Solutions)
Q.No.
Solution Comprehension-VII
1.
Answer (2) S1 represent circle with centre (0, 0) and radius 4 S1 : |z| < 4
x2 +
y2 <
A
S
16
z 1 3i 0 1 3i
B
S2 : Im
60°
O
y + 3x = 0
[( x 1) ( y 3i )][1 3i ] 0 2
Im S2
C
y 3x 0
S3 Re(z) > 0, i.e., x > 0 S = S1 S2 S3
Area of shaded region is OAB + OBC =
4
60 (4) 2 360
= 4
16 6
= 4
8 3
= 2.
(4)2
20 3
Answer (3) min|z – (1 – 3i)| Minimum distance of z from (1, –3) From question, minimum distance of (1, –3) f rom y 3 x 0 is
3 3 2
3 3 . 2
Section-D Q.No.
1.
Solution
Answer (1) Clearly Statement-1 is true and Statement-2 is its explanation. (standard results)
2.
Answer (4)
y
z 1 arg z 1 2
z
Locus of z is a semicircle
statement-1 is false and statement-2 is true.
–1
O
1
x
Option (4) Aak ash Edu cat io nal Serv ic es Pv t. L td . Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 53 -
Complex Numbers and Quadratic Equations (Solutions) Q.No.
Solutions of Assignment (Set-2) Solution
OR
Let z = x + iy z 1 z 1 z 1 z 1 z 1 z 1
( x 1) iy ( x 1) iy
( x 1) iy ( x 1) iy ( x 1) iy ( x 1) iy
( x 2 1 y 2 ) i (2 y ) ( x 1)2 y 2
z 1 z 1 2
arg
x 2 y 2 1 0 x 2 y 2
1 and y > 0
Locus of z is semicircle. 3.
Answer (1) = ei( A + B + C)
eiA.eiB.eiC
= ei = cos + isin = –1 4.
Answer (1) (1)1/4 = (cos2r + isin2r )1/4 = cos
r 2
i sin
r 2
where r = 0, 1, 2, 3
11/4 = 1, i, – 1, – i
z12 + z22 + z32 + z42 = 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) n
2i 2i (1– i ) n n 1 i 2 (1 i ) in
( 2)n e
4
Now clearly the least integral value for which the given number is a positive integer is 8. Aakas h Ed uc ati on al Ser vi ces Pvt . Lt d. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 54 -
Solutions of Assignment (Set-2)
Complex Numbers and Quadratic Equations (Solutions)
Q.No.
7.
Solution
Answer (1) A(z1)
B(z 2)
D(z4)
C(z3)
z – z ABC arg 1 2
2
z3 – z2
And AB = |z1 – z2| = BC|z3 – z2|
z1 – z2 z3 – z2
1
1 cos – i sin 2 2 z3 – z2 z1 – z2
Hence,
z3 – z2 = (z1 – z2) cos – i sin 2 2 = – 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) z
2
2
2
1 1 r cos2 r – sin2 z r r 1
r 2
1 r 2
r 2
2cos2 1
a2
r 1 a2 4sin2 r
r 2
2 – 4sin2
2
2
1 r a2 4 r 1
r a2 4 r
r 2 – a2 4r 1 0 Aak ash Edu cat io nal Serv ic es Pv t. L td . Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 55 -
Complex Numbers and Quadratic Equations (Solutions)
Solutions of Assignment (Set-2)
Q.No.
Solution
r lies between
a2 4 – a
2
a2 4 a
and
2
This is true for all real a 0. a2 4 a
| z | max
2
, | z |min
a2 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.
Answer (1) Given equation is ( z – 2)n = zn
z – 2 z
z – 2 z
n
1 11/n 1
|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) ax 2
bx c 0
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 by
1 x
in ax 2 bx c 0
a x
2
b x
c 0 cx 2 bx a 0
Statement-2 is correct and is correct explanation of Statement-1 10 x 2 x 5 0 12.
10 x
2
1 x
5 0 5 x 2 x 10 0
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 equation ax2 + bx + c = 0 are rational only when a, b, c are rationals, otherwise roots are irrationals. To this end, let us consider an equation 4 x 2 4 3 x 1 0 Aakas h Ed uc ati on al Ser vi ces Pvt . Lt d. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 56 -
Solutions of Assignment (Set-2)
Complex Numbers and Quadratic Equations (Solutions)
Q.No.
Solution
whose discriminant = 48 + 16 = 64 = 82 = a perfect square but roots are x
4 3 48 16 2 4
3 2 which are not rationals. 2
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, 1 3 7
can
( x 2) x
16.
be
root
of
infinite
equations
with
real
coefficients,
e.g.
( x 1) x
1 7 0, 3
1 7 0... 3
Answer (3) x2 x
2 2008 x 501 0
2 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.
Aak ash Edu cat io nal Serv ic es Pv t. L td . Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 57 -
Complex Numbers and Quadratic Equations (Solutions) Q.No.
17.
Solutions of Assignment (Set-2) Solution
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) x2
2px q 0
2p q
(i) (ii)
ax 2
2bx c 0
1
c a
2b
(iii)
a
(iv)
2 1 2 2 2 2 a ( p q ) ( b ac ) = 2 2
( )2 = 16
1
2
. a2 0 statement-1 is true
a a ( ) 2 2
Now pa = b =
pa
a 2
b
1
1
2 1, { 1, 0, 1}, correct
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. Aakas h Ed uc ati on al Ser vi ces Pvt . Lt d. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 58 -
Solutions of Assignment (Set-2)
Complex Numbers and Quadratic Equations (Solutions)
Section-E Q.No.
1.
Solution
Answer A(s), B(r), C(p), D(q) 2007
2007
2007
i 2k 2k 2k 2k (A) i cos i sin sin = ( i ) cos = (i ) e 9 9 9 9 k 1 k 1 k 1
2 k 9
i 4 i 2( 2007 ) i 2 9 9 = i e e ..... e 9
Which is G.P.
i 2 = i e 9
i 2 ( 2007) e 9 1 i 2 9 (1 1) e i i 2 = 0 i 2 91 e 9 1 e 1
(B) |z1| = 1, |z2| = 1, |z3| = 1 z1z1
1 z1
1, z2 z2 1, z3 z3 1
z1,
Now,
1 z2
1 z1
z2 , 1 z2
1 z3
1 z3
z3 | z1 z2 z3 | = | z1 z2 z3 | 1
(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 1
(D) Let
(1) 5 1, , 2 , 3 , 4
1 + + 2 + 3 + 4 = 0 1 + + 2 + 3 = – 4 and 5 = 1
1
4 and || = 1
Now 4log4 |– 4 – 4| = 4 log 4 | 2 4 | = 4 log 4 | 2 || 4 | 1 = 4 log4 2 4 log2 2 2 2 Aak ash Edu cat io nal Serv ic es Pv t. L td . Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 59 -
Complex Numbers and Quadratic Equations (Solutions)
Solutions of Assignment (Set-2)
Q.No.
2.
Solution
Answer A(s), B(r), C(p), D(q) (A) (1 + i)n = (1 – i)n n
1 i 1 1 i
in
n = 4
1
(B) (a + ib) = (x + iy)3 2
= x3 – iy3 + 3ix2y – 3xy2
2
and b = – y3 + 3x2y
a = x(x – 3y ) a x b y
b y
x 2 3y 2 a x
y 2 3x 2
2( x 2 y 2 )
k = 2
(C) x
1 i 2 x2
(1 i )2 1 1 2i = i 2 2
1 x 2 x 4 x 6 x 8 x 10 ... x 2004 x 2006 x 2008
1[1 x 2010 ] 1 x 3.
2
1 i 2 1 i
2
1 1 1 1 1
Answer A(s), B(r), C(p, q), D(q) (A) | z z1 | | z z2 | = constant = k, where k | z1 z2 | , represents an ellipse. (B) | z z1 | | z z2 | k, where k | z1 z2 | is a hyperbola having foci at z1 and z2.
z z1 2 z z2
(C) arg
p(z)
This represents a circle with z1 and z2 as the vertices of diameter. (D) If lies on | | = 1, then 2007
/2
z1
z2
2007 2007 | |
2007
p
lies on the circle.
Aakas h Ed uc ati on al Ser vi ces Pvt . Lt d. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 60 -
Solutions of Assignment (Set-2)
Complex Numbers and Quadratic Equations (Solutions)
Q.No.
4.
Solution
Answer A(r, s), B(p, q, r, s, t), C(p, q), D(q, r, s, t) 62 82 k k 10
(A) |z – 6i | + |z – 8| = k will represent ellipse if
5 2 122 k 13 k
(B) |z – 12i + 3| – |z – 2|| will represent ellipse if (C) |z – ki| + |z – 4| =
2 10k will represent line segment if k + 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, (t – 0)2 (t – 1– 0)2 3 3 and
4t 2 (4 1)2 3 3
2t2 – 2t – 35 0 and 5t2 + 2t – 35 > 0
Using sign scheme we have,
1 – 71 –1 – 4 11 2 5
–1 + 4 11 5
1 + 71 2
Hence,
1– 1–
71 –1– 4 11 –1 4 11 1 71 , , 2 5 5 2
Hence, 3, 4 lies in above interval. (B) We have to solve for x, y (1 i )x – 2i (2 – 3i )y i i 3i 3 – i
(1 i )(3 – i )x – 2i (3 – i ) (2 – 3i )(3 i )y i (3 1) i 9 – i 2
(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 Aak ash Edu cat io nal Serv ic es Pv t. L td . Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 61 -
Complex Numbers and Quadratic Equations (Solutions)
Solutions of Assignment (Set-2)
Q.No.
Solution
(C)
n 1 i 1 i 2 2i n 1– i 1 1 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) 2
x – (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 …(i) k (– , 1] [9, ] (A) For both roots to be positive, f (0) > 0 and
k 3
2
0
k>0
…(ii)
and k > 3
…(iii)
From (i), (ii) and (iii) k [ 9, ]
(B) For both roots to be negative D 0
k < 3, k (0, 1] (C) For both roots to be real k (– , 1] [9, ) k > 0, (k – 3) / 2 < 0,
(D) f (–1) < 0, f (1) < 0 1 + (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) f ( x ) 0 , x R , f (0) = c > 0 Parabola is upward a > 0. (B) Roots are real and distinct D 0 f (0) < 0
c < 0
One root is positive and one negative ab < 0. (C) Roots are imaginary D 0 Parabola is downward a < 0. (D) Parabola touches x-axis D 0 Parabola is downward a < 0 Both roots are positive, sum =
b a
0 b > 0
Aakas h Ed uc ati on al Ser vi ces Pvt . Lt d. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 62 -
Solutions of Assignment (Set-2)
Complex Numbers and Quadratic Equations (Solutions)
Q.No.
8.
Solution
Answer A(s), B(p), C(r), D(p) (A) For real roots, D 0
( 2a)2 4(a 2 a 3) 0
4[a 2 a 2 a 3) 0 a 3
f (3) > 0
9 – 6a + a2 + a – 3 > 0
a2
a ( , 2) (3, )
b
5a 6 0
3
2a
…. (i)
… (ii)
2a 3 a<3 2
… (iii)
from (i), (ii) and (iii) a ( , 2) (B) D 0 f (3) > 0
a3
… (i)
a ( , 2) (3, )
…(ii)
For greater than 3,
b 2a
3 a>3
… (iii)
From (i), (ii) and (iii), a . (C) D > 0
a<3
… (i)
f (1) f (3) < 0
(1 2a a 2 a 3)(9 6a a 2 a 3) 0
(a 2 a 2)(a 2 5a 6) 0
a ( 1, 3) {2}
(a 2)2 (a 1)(a 3) 0 … (ii)
From (i) and (ii) a ( 1, 2) ( 2, 3)
(D) D 0 f (1) < 0
a3
… (i)
(1 2a a 2 a 3) 0
(a 2 a 2) 0
–1 < a < 2 f (3) < 0
a2
… (ii)
(9 6a a 2 a 3) 0
5a 6 0
2
… (iii)
From (i), (ii) and (iii) a. Aak ash Edu cat io nal Serv ic es Pv t. L td . Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 63 -
Complex Numbers and Quadratic Equations (Solutions) Q.No.
9.
Solutions of Assignment (Set-2) Solution
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
x 2 Now, 2
k
3
k
3
1,
k>3
...(i)
For 1 x < 2 – x = k – 4 x = 4 – k
2
...(ii)
For 2 x < 3 2 k < 3
x = k,
For x 3 3x – 6 = k x
k6
k 6
3 k
3
3
3
2
3
1
2
3
Clearly (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) x2
ax b 0
a , b | | 1 ( )2 4 a 2 4b
a2
4b 1
a2
1 4b
a2
1 2 4b = 2(1 + 2b)
Aakas h Ed uc ati on al Ser vi ces Pvt . Lt d. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 64 -
Solutions of Assignment (Set-2)
Complex Numbers and Quadratic Equations (Solutions)
Q.No.
Solution a
(C) 2 a 2. b
3
2
b
a2
9b 2a 2 9b . 2
2
b a 2 3
(D)
1
2
2
1
2
( )( 2 2 ) 2 2 ( )2 2
a(b)2
a 2 2b
a 2 ab 2 2b b(ab 2) 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( + + ) 1 2
9 – 2· 8 (C) ( –1 + –1 + –1) – ( + + )
– ( ) 1 2
– 3
5 2
1 (D) [ –1 + –1 + –1] = – –1 2 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 semimajor axis 5 and focus at (±4, 0). ae = 4
e =
4 5
Aak ash Edu cat io nal Serv ic es Pv t. L td . Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 65 -
Complex Numbers and Quadratic Equations (Solutions) Q.No.
Solutions of Assignment (Set-2) Solution
(C) | z | |w | (D) | z | |w |
1 w
1 w
5 3 2
2
Re z | z | 2
Section-F Q.No.
1.
Solution
Answer (3) As, arg of P(z) is /4
z = x + iy,
x, y, > 0
y
(0, 2) P (x + iy)
(0, 1) x
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 k 1
2r i sin 2r 0 7 7
cos
r 1
It is possible only when 2002 + (2 k – 1) should be multiple of 7. 3.
Answer (5) We have
z z 3 z z 3 30 z z z 2 z 2 30 (x2 + y2) ((x2 – y2) – 2i xy + x2 – y2 + 2i xy)) = 0 (x2 + y2) (x2 – y2) = 15 = (22 + 1) (22 – 1) Which suggests the possible values of x and y are x = 2, y = 1
or x = –2, y = –1 Aakas h Ed uc ati on al Ser vi ces Pvt . Lt d. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 66 -
Solutions of Assignment (Set-2)
Complex Numbers and Quadratic Equations (Solutions)
Q.No.
Solution
or x = –2, y = 1
or x = 2, y = –1
Centre of rectangle is (0, 0) Now, |z–3| = 2 (0, 0)
(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 + z22 + z1z2 = 0 Hence
5.
z12 z22 4 z1z2 z1 z2
3 z1 z2
3
z1 z2
Answer (2) Let
3 + i = z, hence other two vertices are iz and z + iz
1 So, area of such triangle is | z |2 2 1 2
1 2
· (2)2 4 2 6.
Answer (0) 2
2
x – 6kx + 9(k – 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, ) Also
...(ii)
sum of roots 3 2
3k < 3 k<1
…(iii)
From (i), (ii), (iii) We observe that there does not exist any real value of k. Aak ash Edu cat io nal Serv ic es Pv t. L td . Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 67 -
Complex Numbers and Quadratic Equations (Solutions) Q.No.
7.
Solutions of Assignment (Set-2) Solution
Answer (2) We have
3x 4 x 2x 2 ax 2 3 x 4 As 2 5 x 2x 2 ax 2 3 x 4 5x 2 10 x 10 0 x 2 2x 2 y
ax 2 2
(a – 5)x2 – 7x – 6 < 0
2 (as x + 2x + 2 > 0, x R)
It is satisfied for all x if a – 5 < 0, 49 + 24 (a – 5) < 0 a
71 24
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.
x ( x 1) (a b) x c . 2
Then f (x) = ax2 + bx + c = 2a
Since x is an integer x(x – 1) is an even integer. x( x 1) 2a (a b) x c is an integer as 2a, a + b, c are integers. 2
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. Aakas h Ed uc ati on al Ser vi ces Pvt . Lt d. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 68 -
Solutions of Assignment (Set-2)
Complex Numbers and Quadratic Equations (Solutions)
Q.No.
10.
Solution
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, y
y
1 3 2
4y
4 (as y > 0) 9
4 so, 6 log 3 4 9 2
Section-G Q.No.
1.
Solution
Answer (1) Given, S1 = {z : Im (z) > 1} S2 = {z : |z – 2 – i| = 3}
p
y s,
S3 = {z : Re ((1 – i)2) = 2
–1+i A
S1 : y 1
5 +i
2+i
B
S2 : |z – 2 – i| = 3} S3 : x + y = 2
Clearly ‘p’ is only point satisfying all three condition. x+y= 2
Now, –1 + i, and 5 + i are end points of a diameter
PA2 + PB2 = (6)2 = 36 Also |z – 2 – i| 2 – 2 i
z – 2 i
3
– 3 |z| – |2 + i| 3
…… (i)
Aak ash Edu cat io nal Serv ic es Pv t. L td . Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 69 -
Complex Numbers and Quadratic Equations (Solutions)
Solutions of Assignment (Set-2)
Q.No.
Solution
Similarly –3 < | z1|– |2 + i| < 3
…… (ii)
(i) _____ (ii) –6 < |z| – |z1| < 6 –3 < |z| – |z1| + 3 < 9 2.
Answer (1) f (x) + f (x) + f (2x) + ...... + f (6x)
Statement-1 :
20
7 A0
A x (1 k
k
k
k
...... 6k ) but when k 7
1
and k 14, then 1 + k + 2k + ...... + 6k = 0 f (x) + f (x) + ....... + f (6x) = 7( A0 + A7x7 + A14x14)
Statement-2 : B 8 A
O
Clearly AB = 8 OB = 16
sin
AB 8 1 6 OB 16 2
So, the argument is
3
Statement-3 : It can be seen from the figure that the maximum value of argument is
3.
2 3
Answer (2) Statement-1 : The number of common vertices is equal to the number of common roots of z1982 – 1 = 0 and z2973 – 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) Let y
x2 x 3 so that x2 – (y + 1)x + 3 – 2y = 0 x2
For real values of x (y + 1)2 – 4(3 – 2y) 0
(y + 11) (y – 1) 0 Aakas h Ed uc ati on al Ser vi ces Pvt . Lt d. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 70 -
Solutions of Assignment (Set-2) Q.No.
Complex Numbers and Quadratic Equations (Solutions) 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.
1.
Solution 3
2
x – 3x + 3x + 7 = 0
(x – 1)3 + 8 = 0
(x – 1) = –2, –2 , –22 2 x = –1, 1 – 2, 1 – 2 Now,
1 1 1 1 1 1
2.
2 2 22 2 22 2 1
1
2 = |32| = 3
z0 3
N-W
z
z 5
N (Imaginary axis) y
4
By Coni method
5
z i z0 e z0 0 z0
z0
O
3
3 3 +i 2 2
° 4 5
E (Real axis)
3 5 3 i (cos i sin ) 2 3 2
z
1 5 3 4 1 i i z 3 5 2 35 2 i
(3 4i )e 4 Aak ash Edu cat io nal Serv ic es Pv t. L td . Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 71 -
Complex Numbers and Quadratic Equations (Solutions) Q.No.
3.
Solutions of Assignment (Set-2) Solution
z1 z2
z2 z1
2
1
2
z1 + z2 = z1z2
z12 z22 0 2 z1z2 0z1 0z2 0, z1, z2 form an equilateral triangle Hence, | z1 | | z2 | | z1 z2 | 3
| z1 z2 |2 3 4.
|z – z1| = 5 5
|z – z2| = 2
z1
|z – z1| = r + 5
2 r
r
z2
z
|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 1 zz 1 z z
Now,
z
1 z
2
z
1 z z
z
1 z
z
if z x iy then z x iy z
z 2ixy z
1 z 6.
2
is purely imaginary therefore
z
1 z2
always lies on y –axis.
|z1 + z2|2 = |z1|2 + |z2|2 ………. (Given) 2 2 2 2 2 |z1 + z2| = |z1| + |z2| + 2 Re ( z1 z 2 ) = |z1| + |z2| ………. (Relation)
Re ( z1 z2 ) 0 z1 z 2 2
amp
z 6 amp 1 3 z 2 2
6
Aakas h Ed uc ati on al Ser vi ces Pvt . Lt d. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 72 -
Solutions of Assignment (Set-2)
Complex Numbers and Quadratic Equations (Solutions)
Q.No.
7.
Solution
Let z = 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 y 2 – 4ax 1 0
…(ii)
–4ay + a = 0
…(iii)
(3) a(– 4y + 1) = 0 a = 0 or y But a cannot be zero. Hence y 2 x2
4 x 2 4x 2
1 gives 4
1 4ax – 1 16 1 (4ax – 1)2 16
1 16a 2 x 2 1– 8ax 4
(4 – 16a2)x2 + 8ax –
1 4
3 = 0 4
–8a 16a2 12 8(1– 4a2 ) 4a 2 4a2 3 4(4a2 – 1)
Now, we can observe that 1 , there is no solution 2
if 0 a if a
8.
1 , solution is z = x + iy 2
4a 4a 2 3 2
4(4a – 1)
i·
1 4
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. Aak ash Edu cat io nal Serv ic es Pv t. L td . Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 73 -
Complex Numbers and Quadratic Equations (Solutions)
Solutions of Assignment (Set-2)
Q.No.
9.
Solution
Let and n be the roots of the given quadratic equation ax2 + bx + c = 0 1
c n1 + n = and . n = = n + 1 a a a b
1
n
c n 1
c n 1 b a a
a 1
1
1
c n 1 a
1 n 1 n n c a 1
n 1
(a 10.
c ) n1
n n 1 1 n n c a 1
1 n 1 n n c a 1
1 n
c
b0
b0
1
(ac n )n 1
b 0
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.
Aakas h Ed uc ati on al Ser vi ces Pvt . Lt d. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 74 -
Solutions of Assignment (Set-2)
Complex Numbers and Quadratic Equations (Solutions)
Q.No.
12.
Solution
The given equation can be written as 22x + 3 + 32x + 1 = 10 × 2x × 3x. 2x
3 3 8 3. 10. 2 2 2x
x
x
3 3 3 10 8 0 2 2 2x
x
x
3 3 3 3. 6 4 8 0 2 2 2 3 3. 2
x
3 x 3 x 2 4 2 0 2 2
3 x 2 2
3 x 3. 4 0 2 x
x
3 3 4 Either 2 or 2 2 3 x
3 When 2 , Taking logarithm of both sides we get x(log3 – log2) = log2 2
x
log2 log3 log2 x
3 4 Also when 2 3
x(log3 – log2) = log4 – log3 x
13.
log4 log3 log2 log 4 log3 or x log3 log2 log3 log2 log3 log2
The given equation is 2x2 + (2 – )x + 3 = 0
and are roots of the equation, so But
( 2) 2 2
3
2 4 3
2 2 4 ( )2 2 3 2
2 4 2 3
3
2
Aak ash Edu cat io nal Serv ic es Pv t. L td . Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 75 -
Complex Numbers and Quadratic Equations (Solutions)
Solutions of Assignment (Set-2)
Q.No.
Solution
2 – 4 + 4 = 10 2 – 4 – 6 = 0 whose roots are 1 & 2 1 + 2 = 4 & 12 = –6 Now
12 22 13 32 ( 1 2 )3 312 ( 1 2 ) 2 1 1 2 1 2
64 72 136 68 6 6 3 2
12 22 1 2 Also 2 2 2 2 1 2 1 2
2
2 2 ( )2 21 2 1 2 2 1 2 2 1 2 1 2 2
16 12 2 196 2 178 6 9 9 14.
|2x + 3|2 – 3|2x + 3| + 2 = 0
(|2x + 3| – 2) (|2x + 3| –1) = 0
|2x + 3| = 2, |2x + 3| = 1
2x + 3 = ± 2, 2x + 3 = ± 1
x
5 1 , 2, 1 2 2
,
5 5 1 ( 2)( 1) 2 2 2
Product =
2(Product) = 5. 15.
x3
4 x 2 8 x 8 0 ( x 2)( x 2 2 x 4) 0
x = 2, x 2 2x 4 0 will give imaginary roots.
16.
x2
a
x
1
2x 4 0 and ax 2 bx c 0 will have both roots common.
b
2
c
4
k b = –2k, c = 4k 2b + c = 0
3 3 2 / 3 31/ 3
x 3 31/ 3 (1 31/ 3 )
( x 3)3 3(1 31/ 3 )3 x 3 27 x 9 x 2 27 3(1 3 3(31/ 3 3 2 / 3 )]
x3
27 x 9 x 2 27 3( 4 3( x 3)]
x 3 9 x 2 18 x 12 0
x3
9 x 2 18 x 10 2
Aakas h Ed uc ati on al Ser vi ces Pvt . Lt d. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 76 -
Solutions of Assignment (Set-2)
Complex Numbers and Quadratic Equations (Solutions)
Q.No.
17.
Solution
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 and the polynomial. b
f (x) = xn – 1 = (x – 0)(x – 1)(x – 2)....... (x – n – 1)
We have, n
1 f . ( – i0)( – i1) ........ ( – in – 1) i i n
1 and f – . ( + i0)( + i1) ........ ( + in – 1) i i
Hence,
f – = (2 + 2)(2 – 2) ........ (2 – 2 ) 0 1 n – 1 i i
f
Therefore n – 1
n –1
(a b ) b ab 2
n
k
k
0
2 k
k 0
n – 1
bn
2 k
k 0
2 bn f · f – bn · (2 )2m 1 1 i i
a 2m 1 b 1 b
2
n
b
2(2m 1)
a 2m 1 b2 m 1 b 2m 1
2
= (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) < Hence t1, t2, t3 < Let
t
6
,
6
t1 t 2
t3 0, 3 6
t t t 1 (sin t1 sin t2 sin t3 ) sin 1 2 3 3 3
…(ii)
Aak ash Edu cat io nal Serv ic es Pv t. L td . Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 77 -
Complex Numbers and Quadratic Equations (Solutions)
Solutions of Assignment (Set-2)
Q.No.
Solution
From (i) & (ii) we have, sin(t1 t2 t3 ) t t t sin 1 2 3 2 3 Then, sin3t 2sint
4sin3t – sint 0 i.e., sin2 t
1 1 , hence sint then t , which contradicts that t 0 , 4 2 2 6
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, s2
1 1 1 s3 z1 z2 z3 s3 ( z1 z2 z3 ) s3 · s1
…(ii)
Now, from (i) and (ii) s12
2s3 · s1 and | s1 |2 2(s3 )(s1) = 2|s1|
|s1| = 2 s1 = 2 with || = 1
Now, again from relation (i) and (ii) it follows that s2 2 2 1 2 2 s2 s1 2 and s3 2 2 s1
Now, the given equation becomes z3 – 2z2 + 22z – 3 = 0
(2 – )(z2 – z + 2) = 0
The roots are z = , = –
2
Now, Rn = |z1n + z2n + z3n| = |n + nn + (– 1)nn2n|
= ||n |1 + n + (– 1)n2n| R0 = 3, R1 = |1 +
– 2|
= |– 22| = 2
Aakas h Ed uc ati on al Ser vi ces Pvt . Lt d. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 78 -
Solutions of Assignment (Set-2)
Complex Numbers and Quadratic Equations (Solutions)
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
z z – | z 1|( z z ) | z 1|2 z· z
z – | z 1| · z – | z 1| z | z – 1| · z | z – 1| ( 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 0 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 z
t – 1 t
1
, so, z is any real number with – 1 z 1
The equation z z 0 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
p(1) R . p(–1)
Let x1, x2, x3,........, x2n be roots of p. Then p(x) = (x – x1)(x – x2)(x – x3) ........ (x – x2n)
For some C, and 2n
1– xk p(1) (1– x1) (1– x2 ) ........ (1– x2n ) p(–1) 1 xk (–1– x1) ........ (–1– x2n ) k 1
It is given that |xk| = 1 for all k = 1, 2, ......., 2n. Then 1–
1
1– xk 1– xk xk x –1 1– x k k – 1 xk 1 xk 1 1 x k 1 1 x k xk
Aak ash Edu cat io nal Serv ic es Pv t. L td . Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 79 -
Complex Numbers and Quadratic Equations (Solutions)
Solutions of Assignment (Set-2)
Q.No.
Solution
p(1) 2n 1– xk 2n 1– xk – p(–1) k 1 1 xk k 1 1 xk
Now,
2n
2n
(–1)
x 1– 1 x k
k
1
p(1) p(–1)
p(1) is a real number. p(–1)
22.
k
Without loss of generalization, we take A1= R + i0 Now, in triangle A1OA4 cos108
i6
A4 = Re 10
¡ 4
A3 Re
R 2 R 2 ( A1A4 )2
2R
30° 30°
2
( A1 A4)2 = 4R2sin254°
10 i ¡ 2
A2 Re
10
30°
= 2Rsin54°
0
Similarly in triangle
A1( R + ie)
A1OA2
cos36
R 2 R 2 ( A1A2 )2
2R 2
( A1 A2)2 = 2R2sin218° A1 A2 = 4Rsin18°
Now A1 A4 – A1 A1 = 2R(sin54° – sin18°) = 2R 2cos36° sin18°
5 1 5 1 4 4
= 4R
= R (proved) 23.
Answer (1) Given, a3 + b3 = a – b let
a b
k (a > b)
Now a3 + b3 = a – b b3k3 + b3 = bk – b
b2
1 k 1 k
3
2
Now, a2 + 4b2 = b2k2 + 4b2 = b2 (k + 4) k 2 4 k 1 = k3
=1
= 1
k2
1
4k 5 k3 1
2 k 2 1
k3
1
as k > 0
a2 + 4b2 < 1 Aakas h Ed uc ati on al Ser vi ces Pvt . Lt d. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 80 -
Solutions of Assignment (Set-2)
Complex Numbers and Quadratic Equations (Solutions)
Q.No.
24.
Solution
Answer (3) x 3
f (x) =
6x
Df = [3, 6]
6
3 , f (6) =
f (3) =
3 3
from symmetry f (x) have maxima at x
63 4.5 2
f (4.5)
25.
3
4.5
6
4.5 3 6 4.5 = 2 1.5 6
Answer (3) x2
1
px
2 p2
0, p R
Let , be roots
+ = – p, =
1 2 p2
4 + 4 = ( + )4 – 43 – 43 – 622 ( + )4 – 4(2 + 2) – 622 ( + )4 – 4 ( + )2 – 2 – 622 2
1 6 2 4 p 2 2 p p 4p
p4
p4
2 2
4
p +
4 p4 p4
2 p
4
6 4 p4
2 1
½
Now,
2p 4 1 (A.M. – G.M. inequality) 2 2
p4
1 2. 4 2p 2 1
½
2 Minimum value of 4 + 4 = 2 + 2 26.
Answer (2) x2
ax 2b 0 (Here a, b are integers)
Let , be roots Now, sum of roots = a product of roots = 2b (an even number)
‘2’ is one root Now, 4 – 2a + 2b = 0 a – b = 2
Aak ash Edu cat io nal Serv ic es Pv t. L td . Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 81 -
Complex Numbers and Quadratic Equations (Solutions) Q.No.
27.
Solutions of Assignment (Set-2) Solution
Answer (3) 4
97 x + 4 x = 5
am
bm 2
m
ab if m (0, 1) 2
Put a = 97 – x, b = x, m = w/f 4
1
97 4 97 x 4 x 2 2
4 97 – x 4 x 2 × 2.63 Which holds for 4
97 – x 4 x 5
Now,
1 97 4 2 2 5 4
97 97
Clearly for 28.
4
97 – x 4 x 5 has two solution.
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 .,…(i) ab
Also B = f a a
f a f b ab
af a bf a af ( a) af ( b) ab af b bf a ab
Remainder will be
f a f b af b bf a a b ab
x
Aakas h Ed uc ati on al Ser vi ces Pvt . Lt d. Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
- 82 -
