JEE Main 2024 — 27 January, Shift 2 — Question 27

$f(x)=\int_{0}^{x} g(t) \ln \left(\frac{1-t}{1+t}\right) d t$

$f(-x)=\int_{0}^{-x} g(t) \ln \left(\frac{1-t}{1+t}\right) d t$

$f(-x)=-\int_{0}^{x} g(-y) \ln \left(\frac{1+y}{1-y}\right) d y$

$=-\int_{0}^{x} g(y) \ln \left(\frac{1-y}{1+y}\right) d y$ ( $g$ is odd $)$ $f(-x)=-f(x) \Rightarrow f$ is also odd

Now, $I=\int_{-\pi / 2}^{\pi / 2}\left(f(x)+\frac{x^{2} \cos x}{1+e^{x}}\right) d x$

$I=\int_{-\pi / 2}^{\pi / 2}\left(f(-x)+\frac{x^{2} e^{x} \cos x}{1+e^{x}}\right) d x$

$2 I=\int_{-\pi / 2}^{\pi / 2} x^{2} \cos x d x=2 \int_{0}^{\pi / 2} x^{2} \cos x d x$

$I=\left(x^{2} \sin x\right)_{0}^{\pi / 2}-\int_{0}^{\pi / 2} 2 x \sin x d x$

$=\frac{\pi^{2}}{4}-2\left(-x \cos x+\int \cos x d x\right)_{0}^{\pi / 2}$

$=\frac{\pi^{2}}{4}-2(0+1)=\frac{\pi^{2}}{4}-2 \Rightarrow\left(\frac{\pi}{2}\right)^{2}-2$

$\therefore \alpha=2$