If x satisfies the equation ( _0^1 dt t^2 + 2t + 1 ) , x^2 - , ( … - Vidyadip

MCQ

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

A
$±$ $\sqrt {\frac{\alpha }{{2\sin \alpha }}} $
B
$±$ $\sqrt {\frac{{2\sin \alpha }}{\alpha }} $
C
$± $ $\sqrt {\frac{\alpha }{{\sin \alpha }}} $
✓
$±$ $2\,\sqrt {\frac{{\sin \alpha }}{\alpha }} $

✓

Answer

Correct option: D.

$±$ $2\,\sqrt {\frac{{\sin \alpha }}{\alpha }} $

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

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