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: , 4
$\alpha \cdot \beta=2$ and $\alpha+\beta=-p$ also $\frac{1}{\alpha}+\frac{1}{\beta}=-q$
$\Rightarrow p=2 q$
Now $\left(\alpha-\frac{1}{\alpha}\right)\left(\beta-\frac{1}{\beta}\right)\left(\alpha+\frac{1}{\beta}\right)\left(\beta+\frac{1}{\alpha}\right)$
$=\left[\alpha \beta+\frac{1}{\alpha \beta}-\frac{\alpha}{\beta}-\frac{\beta}{\alpha}\right]\left[\alpha \beta+\frac{1}{\alpha \beta}+2\right]$
$=\frac{9}{2}\left[\frac{5}{2}-\frac{\alpha^{2}+\beta^{2}}{2}\right]=\frac{9}{4}\left[5-\left(p^{2}-4\right)\right]$
$=\frac{9}{4}\left(9-p^{2}\right)$ $\left[\because \alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2 \alpha \beta\right]$
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