Question

If $x$ satisfies the equation $\left(\int\limits_0^1 \frac{d t}{t^2+2 t \cos \alpha+1}\right) x^2-\left(\int\limits_{-3}^3 \frac{t^2 \sin 2 t}{t^2+1} d t\right) x-2=0(0< \alpha< \pi)$, then the value $x$ is

• A. $\pm \sqrt{\frac{\alpha}{2 \sin \alpha}}$
• B. $\pm \sqrt{\frac{2 \sin \alpha}{\alpha}}$
• C. $\pm \sqrt{\frac{\alpha}{\sin \alpha}}$
• D. $\pm 2 \sqrt{\frac{\sin \alpha}{\alpha}}$

Answer 1

Answer: (D)

$\int\limits_{-3}^3 \frac{t^2 \sin 2 t}{t^2+1} d t=0$ as the integrand is an odd function.
also $\int\limits_0^1 \frac{ dt }{ t ^2+2 t \cos \alpha+1}=\left.\frac{1}{\sin \alpha} \tan ^{-1} \frac{ t +\cos \alpha}{\sin \alpha}\right|_0 ^1=\frac{\alpha}{2 \sin \alpha}$
Thus the given equation reduces to
$x ^2 \frac{\alpha}{2 \sin \alpha}-2=0 \Rightarrow x = \pm 2 \sqrt{\frac{\sin \alpha}{\alpha}} $