Mathematics

# If $x$ satisfies the equation $\displaystyle\left(\int_0^1{\frac{dt}{t^2+2t\cos{\alpha}+1}}\right)x^2-\left(\int_{-3}^3{\frac{t^2\sin{2t}}{t^2+1}dt}\right)x-2=0$ for $(0<\alpha<\pi)$then the value of $x$ is?

$\displaystyle\pm2\sqrt{\frac{\sin{\alpha}}{\alpha}}$

##### SOLUTION
$\displaystyle \left( \int _{ 0 }^{ 1 }{ \frac { dt }{ t^{ 2 }+2t\cos { \alpha } +1 } } \right) x^{ 2 }-\left( \int _{ -3 }^{ 3 }{ \frac { t^{ 2 }\sin { 2t } }{ t^{ 2 }+1 } dt } \right) x-2=0$

Here, $\displaystyle \int _{ -3 }^{ 3 }{ \frac { t^{ 2 }\sin { 2t } }{ t^{ 2 }+1 } dt } =0 \left(\because f(t)=\frac { t^{ 2 }\sin { 2t } }{ t^{ 2 }+1 } \text{ is odd}\right)$

So, the equation becomes
$\displaystyle \left( \int _{ 0 }^{ 1 }{ \frac { dt }{ t^{ 2 }+2t\cos { \alpha } +1 } } \right) x^{ 2 }-2=0$   ....(1)

Now,
$\displaystyle \int _{ 0 }^{ 1 }{ \frac { dt }{ t^{ 2 }+2t\cos { \alpha } +1 } } =\int _{ 0 }^{ 1 }{ \frac { dt }{ (t+\cos { \alpha } )^{ 2 }+\sin ^{ 2 }{ \alpha } } }$

$=\displaystyle \frac { 1 }{ \sin { \alpha } } \left[{ \tan ^{ -1 }{ \left(\frac { t+\cos { \alpha} }{ \sin { \alpha } } \right) } }\right]_{ 0 }^{ 1 }$

$\displaystyle =\frac { 1 }{ \sin { \alpha } } \left[\tan ^{ -1 }{\cot {\dfrac{\alpha}2 } -\tan ^{ -1 }{ \cot { \alpha } } }\right]$

$\displaystyle =\frac { 1 }{ \sin { \alpha } } \left[\frac { \pi }{ 2 }-\frac { \alpha }{ 2 } -\frac { \pi }{ 2 } +\alpha\right]$

$\displaystyle =\frac { \alpha }{ 2\sin { \alpha } }$

So, equation (1) becomes
$\displaystyle\frac { \alpha }{ 2\sin { \alpha } } x^{ 2 }-2=0$

$\Rightarrow \displaystyle x=\pm 2 \sqrt {\frac{\sin \alpha}{\alpha}}$

Hence, option D.

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Single Correct Hard Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 111

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