Q53 / 160 · TS EAMCET 2023 (14 May Shift 2) · Mathematics
If $\alpha, \beta$ are the roots of $x^2+a x+2=0$ and $\frac{1}{\alpha}, \frac{1}{\beta}$ are the roots of $x^2-b x+c=0$, then
$\left(\alpha+\frac{1}{\beta}\right)\left(\beta+\frac{1}{\alpha}\right)\left(\alpha-\frac{1}{\alpha}\right)\left(\beta-\frac{1}{\beta}\right)=$
- Correct answerA$\frac{9}{4}\left(9-a^2\right)$
- B$\frac{9}{4}\left(9+a^2\right)$
- C$\frac{9}{4}\left(9-b^2\right)$
- D$\frac{9}{4}\left(9+b^2\right)$
Solution
$x^2+a x+2=0$
$\begin{aligned} & \alpha+\beta=-a, \alpha \beta=2 \\ & x^2-b x+c=0\end{aligned}$
$\begin{aligned} & \frac{1}{\alpha}+\frac{1}{\beta}=b, \frac{1}{\alpha \beta}=c=\frac{1}{2} \\ & \left(\alpha+\frac{1}{\beta}\right)\left(\beta+\frac{1}{\alpha}\right)\left(\alpha-\frac{1}{\alpha}\right)\left(\beta-\frac{1}{\beta}\right) \\ & =\left(\alpha \beta+1+1+\frac{1}{\alpha \beta}\right)\left(\alpha \beta+\frac{1}{\alpha \beta}-\frac{\alpha}{\beta}-\frac{\beta}{\alpha}\right) \\ & =\left(4+\frac{1}{2}\right)\left(2+\frac{1}{2}-\frac{\alpha^2+\beta^2}{\alpha \beta}\right) \\ & =\frac{9}{2}\left(\frac{5}{2}-\frac{(\alpha+\beta)^2-2 \alpha \beta}{\alpha \beta}\right) \\ & =\frac{9}{2}\left(\frac{5}{2}-\frac{a^2-4}{2}\right) \\ & =\frac{9}{2}\left(\frac{9-a^2}{2}\right)=\frac{9}{4}\left(9-a^2\right)\end{aligned}$
Hence, the correct answer is option: 'A'.