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Complex Numbers and Quadratic Equations

Multiple Choice Questions

251.

$\frac{{(1 + i)}^{2011}}{{(1 - i)}^{2009}} = ?$

252.

$In ∆ PQR, ∠ R = \frac{π}{4}, \tan (\frac{P}{3}), \tan (\frac{Q}{3}) are the roots of the equation {ax}^{2} + bx + c = 0, then$

a +b = c
b + c = 0
a + c = 0
b = c

253.

$The product of real of the equation {|x|}^{\frac{6}{5}} - 26 {|x|}^{\frac{3}{5}} - 27 = 0$

- 3¹⁰
$- 3^{12}$
- 3^12/5
- 3^12/5

254.

If $α, β, γ$ are the roots of the equation x³ + px² + qx + r = 0, then the coefficient of x in the cubic equation whose roots are $α (β + γ), β (γ + α) and γ (α + β) is$

2q
q² + pr
p² - qr
r(pq - r)

255.

If z is complex number such that $|z - \frac{4}{z}| = 2$ , then the greatest value of $|z|$ is

$1 + \sqrt{2}$
$\sqrt{2}$
$\sqrt{3} + 1$
$1 + \sqrt{5}$

256.
$If α is a non - real root of the equation x^{6} - 1 = 0, then \frac{α^{2} + α^{3} + α^{4} + α^{5}}{α + 1} = ?$

$α$

1

0

- 1

- 1

$Since, α is a non - real root of the equation x^{6} - 1 = 0 \Rightarrow α^{6} - 1 = 0 \Rightarrow α^{6} - 1 = 0 \times α - 1 \Rightarrow \frac{α^{6} - 1}{α - 1} = 0 \Rightarrow 1 + α + α^{2} + α^{3} + α^{4} + α^{5} = 0 \Rightarrow (1 + α) + α^{2} (1 + α) + α^{4} (1 + α) = 0 \Rightarrow (1 + α) (1 + α^{2} α^{4}) = 0 . . . (i) ∴ \frac{α^{2} + α^{3} + α^{4} + α^{5}}{α + 1} = \frac{α^{2} (α + 1) + α^{4} (α + 1)}{α + 1} = \frac{(α + 1) (α^{2} + α^{4})}{α + 1} = α^{2} + α^{4} = - 1 [from eq (i)]$