JEE PYQ: Quadratic Equation Question 17
Question 17 - 2020 (03 Sep Shift 1)
If $\alpha$ and $\beta$ are the roots of the equation $x^2 + px + 2 = 0$ and $\frac{1}{\alpha}$ and $\frac{1}{\beta}$ are the roots of the equation $2x^2 + 2qx + 1 = 0$, 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:
(1) $\frac{9}{4}(9 + q^2)$
(2) $\frac{9}{4}(9 - q^2)$
(3) $\frac{9}{4}(9 + p^2)$
(4) $\frac{9}{4}(9 - p^2)$
Show Answer
Answer: (4)
Solution
$\alpha \cdot \beta = 2$ and $\alpha + \beta = -p$ also $\frac{1}{\alpha} + \frac{1}{\beta} = -q$
$\Rightarrow p = 2q$
Now $\left(\alpha - \frac{1}{\alpha}\right)\left(\beta - \frac{1}{\beta}\right)\left(\alpha + \frac{1}{\beta}\right)\left(\beta + \frac{1}{\alpha}\right)$
$= \left[\alpha\beta + \frac{1}{\alpha\beta} - \frac{\alpha}{\beta} - \frac{\beta}{\alpha}\right]\left[\alpha\beta + \frac{1}{\alpha\beta} + \frac{\alpha}{\beta} + \frac{\beta}{\alpha} + 2\right]$
$= \frac{9}{2} \cdot \frac{5 - \frac{\alpha^2 + \beta^2}{2}}{2} \cdot \frac{9}{4}[5 - (p^2 - 4)]$
$= \frac{9}{4}(9 - p^2)$
BETA
