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

$ \pm \,\,\sqrt {\frac{{\sin \,\,\alpha }}{\alpha }} $

$ \pm 4\,\,\sqrt {\frac{{\sin \,\,\alpha }}{\alpha }} $

JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

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Analyzing +4 | -1

If x satisfies the equation

{x^2}\,\left( {\int \limits_0^1 \,\,\frac{{dt}}{{{t^2} + \,\,2t\,\,\cos \,\,\alpha \,\, + \,\,1}}} \right)\,\, - \,\,x\left( {\int \limits_{ - 3}^3 \frac{{{t^2}\,\,\sin \,\,2t}}{{{t^2} + \,\,1}}\,\,dt} \right) - 2 = 0

(0 < \,\,\alpha \,\, < \pi )

, then the value of

x

View answer and explanation

Correct option: A

\pm 2\,\,\sqrt {\frac{{\sin \,\,\alpha }}{\alpha }}

Option A (

\pm 2\\,\sqrt {\frac{{\sin \\,\alpha }}{\alpha }}

) is correct because after simplifying the equation using properties of odd functions and evaluating the integral, the value of x can be found.