Jee Main 2024 27 01 2024 Shift 2 - Question17

Answer: (2)

Solution:

$\therefore \quad f(x)=\int _0^{x} g(t) \ln \left(\frac{1-t}{1+t}\right) d t$

$$ \therefore \quad f^{\prime}(x)=g(x) \cdot \ln \left(\frac{1-x}{1+x}\right) $$

Here $f(x)$ is even since $g(x)$ and

$$ \ln \left(\frac{1-x}{1+x}\right) \text { both are odd } $$

Hence $f(x)$ is odd function $\Rightarrow f(x)+f(-x)=0$

Now $\int _{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(f(x)+\frac{x^{2} \cos x}{\left(1+e^{x}\right)}\right) d x$

$$ \begin{aligned} & =\int _0^{\frac{\pi}{2}}{f(x)+f(-x)+x^{2} \cos x\left(\frac{1}{1+e^{x}}+\frac{1}{1+e^{-x}}\right) } d x \\ & =\int _0^{\frac{\pi}{2}} x^{2} \cos x d x \end{aligned} $$

$=\left[x^{2} \sin x\right] _0^{\frac{\pi}{2}}-\int _0^{\frac{\pi}{2}} 2 x \sin x d x$

$=\frac{\pi}{4}-2{[-x \cos x] _0^{\frac{\pi}{2}}-\int _0^{\frac{\pi}{2}}-\cos x d x }$

$=\frac{\pi^{2}}{4}-2$

Given that $\frac{\pi^{2}}{\alpha^{2}}-\alpha=\frac{\pi^{2}}{4}-2$

$\therefore \quad \alpha=2$