If $\alpha$ and $\beta$ are the roots of the equation $x^{2}+p x+2=0$ and $\frac{1}{\alpha}$ and $\frac{1}{\beta}$ are the roots of the equation $2 x^{2}+2 q x+1=0$, then $\left(\alpha-\frac{1}{\alpha}\right)\left(\beta-\frac{1}{\beta}\right)\left(\alpha+\frac{1}{\beta}\right)\left(\beta+\frac{1}{\alpha}\right)$ is equal to:

(A) $\frac{9}{4}\left(9-\mathrm{p}^{2}\right)$

(B) $\frac{9}{4}\left(9+q^{2}\right)$

(C) $\frac{9}{4}\left(9-q^{2}\right)$

(D) $\frac{9}{4}\left(9+\mathrm{p}^{2}\right)$

JEE Main · 2020 · 03 Sep 2020 · Shift 1 · Quadratic Equations and Inequalities

α

and

β

are the roots of the equation

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

and

\frac{1}{α}

and

\frac{1}{β}

are the roots of the equation

2 x^{2} + 2 q x + 1 = 0

, then

(α - \frac{1}{α}) (β - \frac{1}{β}) (α + \frac{1}{β}) (β + \frac{1}{α})

is equal to:

Answer: (A)

From

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

α + β = - p

and

α β = 2

. Grouping the factors as

(α - \frac{1}{α}) (β - \frac{1}{β})

and

(α + \frac{1}{β}) (β + \frac{1}{α})

, the first product is

\frac{( α ^{2} - 1 ) ( β ^{2} - 1 )}{α β}

, where

(α^{2} - 1) (β^{2} - 1) = α^{2} β^{2} - (α^{2} + β^{2}) + 1 = 4 - (p^{2} - 4) + 1 = 9 - p^{2}

, so it equals

\frac{9 - p ^{2}}{2}

. The second product is

\frac{( α β + 1 ) ^{2}}{α β} = \frac{9}{2}

. Multiplying, the expression equals

\frac{9}{4} (9 - p^{2})

JEE Main · 2020 · 03 Sep 2020 · Shift 1 · Single correct option