Let α and β be the roots of the equation x ^ 2 + 2x + 2 = 0, then α ^ 15 + β ^ 15 is equal to

Correct answer: D. - 256 Let the two roots of x ^ 2 + 2x + 2 = 0 be α and β ⇒ α, β = - 2 ± 4 - 8 2 = - 1 ± i ⇒ α, β = 2 ( - 1 2 ± i 1 2 ) = 2 (cos(3π)/(4) ± isin(3π)/(4)) ⇒ α ^ 15 + β ^ 15 = [ 2 (cos(3π)/(4) + isin(3π)/(4))] ^ 15 + [ 2 (cos(3π)/(4) - isin(3π)/(4))] ^ 15 ⇒ α ^ 15 + β ^ 15 = 2(15)/(2)[(cos(45π)/(4) + isin(45π)/(4)) + (cos(45π)/(4) - isin(45π)/(4))] ⇒ α ^ 15 + β ^ 15 = 2.2(15)/(2) . cos(45π)/(4) = 2(17)/(2) . cos(5π)/(4) = - 2(17)/(2) . 1 2 ⇒ α ^ 15 + β ^ 15 = - 256

JEE Main 20199 JanuaryMorning Shift

Exam: JEE Main
Year: 2019
Date: 9 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Quadratic Equation

JEE Main 2019 · 9 January · Morning Shift · Mathematics · Quadratic Equation

Let α and β be the roots of the equation x ^ 2 + 2x + 2 = 0, then α...

Let

\alpha

and

\beta

be the roots of the equation

{x}^{2} + 2x + 2 = 0,

then

{\alpha}^{15} + {\beta}^{15}

is equal to

Options

A $- 512$
B $128$
C $512$
D $- 256$

Correct answer

D. $- 256$

Step-by-step solution

Let the two roots of

{x}^{2} + 2x + 2 = 0

\alpha

and

\beta

⇒ \alpha, \beta = \dfrac{- 2 \pm \sqrt{4 - 8}}{2} = - 1 \pm i

⇒ \alpha, \beta = \sqrt{2}\left(\dfrac{- 1}{\sqrt{2}} \pm i\dfrac{1}{\sqrt{2}}\right)

= \sqrt{2}\left(cos\dfrac{3\pi}{4} \pm isin\dfrac{3\pi}{4}\right)

⇒ {\alpha}^{15} + {\beta}^{15} = {\left[\sqrt{2}\left(cos\dfrac{3\pi}{4} + isin\dfrac{3\pi}{4}\right)\right]}^{15} + {\left[\sqrt{2}\left(cos\dfrac{3\pi}{4} - isin\dfrac{3\pi}{4}\right)\right]}^{15}

⇒ {\alpha}^{15} + {\beta}^{15} = 2\dfrac{15}{2}\left[\left(cos\dfrac{45\pi}{4} + isin\dfrac{45\pi}{4}\right) + \left(cos\dfrac{45\pi}{4} - isin\dfrac{45\pi}{4}\right)\right]

⇒ {\alpha}^{15} + {\beta}^{15} = 2.2\dfrac{15}{2} . cos\dfrac{45\pi}{4} = 2\dfrac{17}{2} . cos\dfrac{5\pi}{4} = - 2\dfrac{17}{2} . \dfrac{1}{\sqrt{2}}

⇒ {\alpha}^{15} + {\beta}^{15} = - 256

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