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:
Correct Option: , 3
$\alpha, \beta$ are roots of $x^{2}+p x+2=0$
$\Rightarrow \alpha^{2}+p \alpha+2=0 \& \beta^{2}+p \beta+2=0$
$\Rightarrow \frac{1}{\alpha}, \frac{1}{\beta}$ are roots of $2 x^{2}+p x+1=0$
But $\frac{1}{\alpha}, \frac{1}{\beta}$ are roots of $2 x^{2}+2 q x+1=0$
$\Rightarrow \mathrm{p}=2 \mathrm{q}$
Also $\alpha+\beta=-\mathrm{p} \quad \alpha \beta=2$
$\left(\alpha-\frac{1}{\alpha}\right)\left(\beta-\frac{1}{\beta}\right)\left(\alpha+\frac{1}{\beta}\right)\left(\beta+\frac{1}{\alpha}\right)$
$=\left(\frac{\alpha^{2}-1}{\alpha}\right)\left(\frac{\beta^{2}-1}{\beta}\right)\left(\frac{\alpha \beta+1}{\beta}\right)\left(\frac{\alpha \beta+1}{\alpha}\right)$
$=\frac{(-\mathrm{p} \alpha-3)(-\mathrm{p} \beta-3)(\alpha \beta+1)^{2}}{(\alpha \beta)^{2}}$
$=\frac{9}{4}(\mathrm{p} \alpha \beta+3 \mathrm{p}(\alpha+\beta)+9)$
$=\frac{9}{4}\left(9-\mathrm{p}^{2}\right)=\frac{9}{4}\left(9-4 \mathrm{q}^{2}\right)$
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