If $\alpha$ and $\beta$ are the roots of the quadratic equation, $x^{2}+x \sin \theta-2 \sin \theta=0, \theta \in\left(0, \frac{\pi}{2}\right)$ then $\frac{\alpha^{12}+\beta^{12}}{\left(\alpha^{-12}+\beta^{-12}\right)(\alpha-\beta)^{24}}$ is equal to:

(A) $\frac{2^{12}}{(\sin \theta+8)^{12}}$

(B) $\frac{2^{12}}{(\sin \theta-4)^{12}}$

(C) $\frac{2^{12}}{(\sin \theta-8)^{6}}$

(D) $\frac{2^{6}}{(\sin \theta+8)^{12}}$

JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Quadratic Equations and Inequalities

α

and

β

are the roots of the quadratic equation,

x^{2} + x sin θ - 2 sin θ = 0, θ \in (0, \frac{π}{2})

then

\frac{α ^{12} + β ^{12}}{( α ^{- 12} + β ^{- 12} ) ( α - β ) ^{24}}

is equal to:

Answer: (A)

By Vieta's formulas for

x^{2} + x sin θ - 2 sin θ = 0

α + β = - sin θ

and

α β = - 2 sin θ

. The expression

\frac{α ^{12} + β ^{12}}{( α ^{- 12} + β ^{- 12} ) ( α - β ) ^{24}}

simplifies, since

α^{- 12} + β^{- 12} = \frac{α ^{12} + β ^{12}}{( α β ) ^{12}}

, to

\frac{( α β ) ^{12}}{( α - β ) ^{24}}

. Now

(α - β)^{2} = (α + β)^{2} - 4 α β = sin^{2} θ + 8 sin θ

, so the value is

\frac{( - 2 s i n θ ) ^{12}}{( s i n ^{2} θ + 8 s i n θ ) ^{12}} = \frac{2 ^{12} s i n ^{12} θ}{s i n ^{12} θ ( s i n θ + 8 ) ^{12}} = \frac{2 ^{12}}{( s i n θ + 8 ) ^{12}}

JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Single correct option