JEE PYQ: Quadratic Equation Question 38
Question 38 - 2019 (10 Apr Shift 1)
If $\alpha$ and $\beta$ are the roots of the quadratic equation, $x^2 + x\sin\theta - 2\sin\theta = 0$, $\theta \in \left(0, \frac{\pi}{2}\right)$, then $\frac{\alpha^{12} + \beta^{12}}{(\alpha^{-12} + \beta^{-12})(\alpha - \beta)^{24}}$ is equal to:
(1) $\frac{2^{12}}{(\sin\theta - 4)^{12}}$
(2) $\frac{2^{12}}{(\sin\theta + 8)^{12}}$
(3) $\frac{2^{12}}{(\sin\theta - 8)^6}$
(4) $\frac{2^6}{(\sin\theta + 8)^{12}}$
Show Answer
Answer: (2)
Solution
Given equation is $x^2 + x\sin\theta - 2\sin\theta = 0$
$\alpha + \beta = -\sin\theta$ and $\alpha\beta = -2\sin\theta$
$\frac{(\alpha^{12} + \beta^{12})(\alpha\beta)^{12}}{(\alpha^{12} + \beta^{12})(\alpha - \beta)^{24}} = \frac{(\alpha\beta)^{12}}{(\alpha - \beta)^{24}}$
$\therefore |\alpha - \beta| = \sqrt{(\alpha+\beta)^2 - 4\alpha\beta} = \sqrt{\sin^2\theta + 8\sin\theta}$
$\therefore \frac{(\alpha\beta)^{12}}{(\alpha - \beta)^{24}} = \frac{(2\sin\theta)^{12}}{(\sin^2\theta + 8\sin\theta)^{12}} = \frac{2^{12}}{(\sin\theta + 8)^{12}}$
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