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# Mathematics : Complex Numbers and Quadratic Equations

#### Multiple Choice Questions

101.

The sum of the coefficients of all odd degree terms in the expansion of

• 2

• -1

• 0

• 1

A.

2

Sum of odd degree terms coefficients
= 2(5 + 1 – 10 + 5)
= 2

102.

If , then the least value of

• $\frac{11}{2}$

• $\frac{11}{4}$

• 3

• $\frac{1}{4}$

B.

$\frac{11}{4}$

103.

If $\mathrm{\alpha }$ is an nth root of unity, then equals

• None of these

A.

104.

If x2 + x + 1 = 0, then the value of  is

• 13

• 12

• 9

• 14

B.

12

Given equation is x2 + x + 1 = 0.

On solving equation, we get

x = w and x = w2

Case - I When x = w

Then,

Case -  II When x = w2

105.

If α, β ∈ C are the distinct roots, of the equation x2 -x + 1 = 0, then α101 + β107 is equal to

• 2

• -1

• 0

• 1

D.

1

x2-x + 1 = 0

Roots are -ω, -ω2

Let α = -ω, β = -ω2

α101 + β107 = (-ω)101 + (-ω2)107

= -( ω101214)
= - (ω2 + ω)
= 1

106.

If the complex number z lies on a circle with centre at the origin and radius = $\frac{1}{4}$, then the 4 complex number - 1 + 8z lies on a circle with radius

• 4

• 1

• 3

• 2

D.

2

Hence, z' lies on a circle with centre (-1, 0) and radius 2.

107.

The normal to the curve x2 + 2xy-3y2 =0 at (1,1)

• does not meet the curve again

• meets the curve again in the second quadrant

• meets the curve again in the third quadrant

• meets the curve again in the fourth quadrant

D.

meets the curve again in the fourth quadrant

Given equation of curve is
x2+ 2xy -3y2 = 0    .... (i)
On differentiating w.r.t. we get
2x + 2xy' + 2y-6yy' = 0

At, x = 1, y = 1, y'=1
i.e,
Equation of normal at (1,1) is
y-1 = -
⇒ y-1 = - (x-1)
⇒ x+y = 2 .... (ii)
On solving Eqs. (i) and (ii) sumultaneously we get
x2+ 2x(2-x)-3(2-x)2 = 0
⇒x2+4x-2x2-3(4+x2-4x)=0
⇒-x2 +4x-12-3x2+12x = 0
⇒-4x2 +16x-12 =0
⇒ 4x2-16x+12 = 0
⇒x2-4x+3 = 0
(x-1)(x-3) = 0x= 1,3
Now when x =1, then y=1
and when x=3 theny = -1
therefore, p = (1,1) an Q = (3,-1)
Hence, normal meets the curve again at (3, -1)in fourth quadrant.

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108.

The sum of all real values of x satisfying the equation
is:

• 3

• -4

• 6

• 5

A.

3

Given,

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109.

If, for a positive integer n, the quadratic equation,
x(x + 1) + (x + 1) (x + 2) + .....
+ (x + n -1 ) (x + n) = 10n
has two consecutive integral solutions, then n is equal to :

• 11

• 12

• 9

• 10

A.

11

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