Question

If $x$ satisfies the equation $\left({\int }_0^1\frac{dt}{t^2+2t\cos \alpha +1}\right)x^2-\left({\int }_{-3}^3\frac{t^2\sin 2t}{t^2+1}dt\right)x-2=0$

for $(0<\alpha <\pi )$

then the value of $x$ 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

The correct option is D $\pm 2\sqrt{\frac{\sin \alpha }{\alpha }}$

$\left({\int }_0^1\frac{dt}{t^2+2t\cos \alpha +1}\right)x^2-\left({\int }_{-3}^3\frac{t^2\sin 2t}{t^2+1}dt\right)x-2=0$
Here, ${\int }_{-3}^3\frac{t^2\sin 2t}{t^2+1}dt=0(\because f(t)=\frac{t^2\sin 2t}{t^2+1}\text{is odd})$
So, the equation becomes
$\left({\int }_0^1\frac{dt}{t^2+2t\cos \alpha +1}\right)x^2-2=0$ ....(1)
Now,

${\int }_0^1\frac{dt}{t^2+2t\cos \alpha +1}={\int }_0^1\frac{dt}{(t+\cos \alpha {)}^2+{\sin }^2\alpha }$
$=\frac{1}{\sin \alpha }{\left[{\tan }^{-1}\left(\frac{t+\cos \alpha }{\sin \alpha }\right)\right]}_0^1$
$=\frac{1}{\sin \alpha }\left[{\tan }^{-1}\cot \frac{\alpha }{2}-{\tan }^{-1}\cot \alpha \right]$
$=\frac{1}{\sin \alpha }[\frac{\pi }{2}-\frac{\alpha }{2}-\frac{\pi }{2}+\alpha ]$
$=\frac{\alpha }{2\sin \alpha }$
So, equation (1) becomes
$\frac{\alpha }{2\sin \alpha }{x}^2-2=0$
$\Rightarrow x=\pm 2\sqrt{\frac{\sin \alpha }{\alpha }}$

Hence, option D.