﻿ If the cube roots of unity are 1, ω, ω2 then the roots of the equation (x – 1)3 + 8 = 0, are from Mathematics Complex Numbers and Quadratic Equations

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# Complex Numbers and Quadratic Equations

#### Multiple Choice Questions

1.

If x = ω – ω2 – 2. Then the value of (x4 + 3x3 + 2x2 – 11x – 6) is

• 0

• -1

• 1

• -3

C.

1

If (x + 2)2 = (ω – ω2 )
2 x2 + 4 + 4x = ω2 + ω4 – 2ω3
x2 + 4 + 4x = ω2 + ω –2 (x2 + 4x + 7) = 0 ...(i)
x4 + 3x3 + 2x2 – 11x – 6
= x2 (x2 + 4x + 7) –x(x2 + 4x + 7) – (x2 + 4x + 7) +1
= x 2 (0) – x(0) – 0 + 1 By (i)
= 1

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2.

Let z, w be complex numbers such that z iw + = 0 and arg zw = π. Then arg z equals

• π/4

• 5π/4

• 3π/4

• π/2

C.

3π/4

Since z + iw = 0 ⇒ z = −iw
⇒ z = iw
⇒ w = -iz
Also arg(zw) = π
⇒ arg (-iz2) = π
⇒ arg (-i) + 2 arg(z) = π

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# 3.If the cube roots of unity are 1, ω, ω2 then the roots of the equation (x – 1)3 + 8 = 0, are -1 , - 1 + 2ω, - 1 - 2ω2 -1 , -1, - 1 -1 , 1 - 2ω, 1 - 2ω2 -1 , 1 + 2ω, 1 + 2ω2

C.

-1 , 1 - 2ω, 1 - 2ω2

(x – 1)3 + 8 = 0
⇒ (x – 1) = (-2) (1)1/3
⇒ x – 1 = -2 or -2ω or -2ω2 or
n = -1 or 1 – 2ω or 1 – 2ω2 .

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4.

The value of α for which the sum of the squares of the roots of the equation x2 – (a – 2)x – a – 1 = 0 assume the least value is

• 1

• 0

• 3

• 2

A.

1

x2 – (a – 2)x – a – 1 = 0
⇒ α + β = a – 2
α β = –(a + 1)
α2 + β2 = (α + β)2 - 2αβ
= a2 – 2a + 6 = (a – 1)2 + 5
⇒ a = 1

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5.

If 2a + 3b + 6c =0, then at least one root of the equation ax+ bx+ c = 0  lies in the interval

• (0,1)

• (1,2)

• (2,3)

• (1,3)

A.

(0,1)

∴ One of the roots of ax2 + bx + c = 0 lies between 0 and 1.
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6.

If (1 – p) is a root of quadratic equation x2 +px + (1-p)=0 , then its roots are

• 0, 1

• -1, 2

• 0, -1

• -1, 1

C.

0, -1

Since (1 - p) is the root of quadratic equation
x2 + px + (1 - p) = 0 ........ (i)
So, (1 - p) satisfied the above equation
∴ (1 - p)2 + p(1 - p) + (1 - p) = 0
(1 - p)[1 - p + p + 1] = 0 (1 - p)(2) = 0
⇒ p = 1 On putting this value of p in equation (i)
x2 + x = 0
⇒ x(x + 1) = 0 ⇒ x = 0, -1

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7.

If roots of the equation x2 – bx + c = 0 be two consectutive integers, then b2 – 4c equals

• – 2

• 3

• 2

• 1

D.

1

Let α, α + 1 be roots
α + α + 1 = b
α(α + 1) = c
∴ b2 – 4c = (2α + 1)2 - 4α(α + 1) = 1.

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8.

The coefficient of xn in expansion of (1+x)(1-x)n is

• (n-1)

• (-1)n(1-n)

• (-1)n-1(n-1)2

• (-1)n-1n

B.

(-1)n(1-n)

The coefficient of xn in expansion of (1+x)(1-x)n is = coefficient of xn + coefficient of xn-1

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9.

If one root of the equation x2+px+12 =0 is 4, while the equation x2 +px +q = 0 has equal roots, then the value of 'q' is

• 49/3

• 4

• 3

• 12

A.

49/3

Since 4 is one of the roots of equation x2 + px + 12 = 0. So it must satisfied the equation.
∴ 16 + 4p + 12 = 0
⇒ 4p = -28
⇒ p = -7
The other equation is x2 - 7x + q = 0 whose roots are equal. Let roots are α and α of above equation

⇒ 2α = 7 ⇒ α = 7/ 2 and product of roots α.α = q ⇒ α2 = q

(7/2)2 = q
q =49/4

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10.

If z = x – i y and z1/3 = p+ iq , then  is equal to

• 1

• -2

• 2

• -1

B.

-2

D.

-1

therefore,(Qz = x − iy)
(x - iy) = (p + iq)3
⇒ (x - iy) = p3 +(iq)3 + 3p2qi + 3pq2i2
⇒ (x - iy) = p3 - iq3 + 3p2qi - 3pq2
⇒ (x - iy) = (p3 - 3pq2 ) + i (3p2 q - q3 ) On comparing both sides, we get
⇒ x = (p3 - 3pq2) and - y = 3p2 q - q3
⇒ x = p(p2 - 3q2 ) and y = q(q2 - 3p2 )
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