If α and β are the roots of the quadratic equation x ^ 2 + xsinθ - 2sinθ = 0, θ ∈ (0, (π)/(2)) , then α ^ 12 + β ^ 12 ( α ^ - 12 + β ^ - 12 ) . ( α - β ) ^ 24 is equal to :

Correct answer: C. 2 ^ 12 ( sinθ + 8 ) ^ 12 Let E = α ^ 12 + β ^ 12 ( α ^ - 12 + β ^ - 12 ) (α - β) ^ 24 = α ^ 12 + β ^ 12 α ^ 12 + β ^ 12 α ^ 12 × β ^ 12 (α - β) ^ 24 E = (αβ) ^ 12 (α - β) ^ 24 = (αβ) ^ 12 [ (α + β) ^ 2 - 4αβ] ^ 12 Now α + β = - sinθ αβ = - 2sinθ ∴ E = (- 2 sin θ) ^ 12 [ (- sinθ) ^ 2 - 4(- 2sinθ)] ^ 12 = 2 ^ 12 (8 + sinθ) ^ 12

JEE Main 201910 AprilMorning Shift

Exam: JEE Main
Year: 2019
Date: 10 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Quadratic Equation

JEE Main 2019 · 10 April · Morning Shift · Mathematics · Quadratic Equation

If α and β are the roots of the quadratic equation x ^ 2 + xsinθ -...

\alpha and

\beta

are the roots of the quadratic equation

{x}^{2} + xsin\theta - 2sin\theta = 0, \theta \in

\left(0, \dfrac{\pi}{2}\right)

, then

\dfrac{{\alpha}^{12} + {\beta}^{12}}{( {\alpha}^{- 12} + {\beta}^{- 12} ) {. ( \alpha - \beta )}^{24}}

is equal to :

Options

A $\dfrac{{2}^{6}}{{( sin\theta + 8 )}^{12}}$
B $\dfrac{{2}^{12}}{{( sin\theta - 4 )}^{12}}$
C $\dfrac{{2}^{12}}{{( sin\theta + 8 )}^{12}}$
D $\dfrac{{2}^{12}}{{( sin\theta - 8 )}^{6}}$

Correct answer

C. $\dfrac{{2}^{12}}{{( sin\theta + 8 )}^{12}}$

Step-by-step solution

Let

E = \dfrac{{\alpha}^{12} + {\beta}^{12}}{\left({\alpha}^{- 12} + {\beta}^{- 12}\right){\left(\alpha - \beta\right)}^{24}} = \dfrac{{\alpha}^{12} + {\beta}^{12}}{\dfrac{{\alpha}^{12} + {\beta}^{12}}{{\alpha}^{12} \times {\beta}^{12}}{\left(\alpha - \beta\right)}^{24}}

E = \dfrac{{\left(\alpha\beta\right)}^{12}}{{\left(\alpha - \beta\right)}^{24}} = \dfrac{{\left(\alpha\beta\right)}^{12}}{{\left[{\left(\alpha + \beta\right)}^{2} - 4\alpha\beta\right]}^{12}}

Now

\alpha + \beta = - sin\theta

\alpha\beta = - 2sin\theta

∴ E = \dfrac{{\left(- 2\text{ sin }\theta\right)}^{12}}

{{

\left[{\left(- sin\theta\right)}^{2} - 4\left(- 2sin\theta\right)\right]}^{12}}

= \dfrac{{2}^{12}}{{\left(8 + sin\theta\right)}^{12}}

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