JEE Main 2024 — Limits Question with Solution

$\begin{array}{rl}& f:(-\infty ,\infty )-\{0\}\to R\\ & f^{\prime }(1)=\underset{a\to \infty }{\lim }{a}^{2}f\left(\frac{1}{a}\right)\\ & \underset{a\to \infty }{\lim }\frac{a(a+1)}{2}{\tan }^{-1}\left(\frac{1}{a}\right)+a^2-2\ln (a)\\ & \underset{a\to \infty }{\lim }{a}^2\left(\frac{\left(1+\frac{1}{a}\right)}{2}{\tan }^{-1}\left(\frac{1}{a}\right)+1-\frac{2}{a^2}\ln (a)\right)\\ & f(x)=\frac{1}{2}(1+x){\tan }^{-1}(x)+1-2x^2\ln (x)\end{array}$
$\begin{aligned} & \mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{2}\left(\frac{1+\mathrm{x}}{1+\mathrm{x}^2}+\tan ^{-1}(\mathrm{x})+4 \mathrm{x} \ell(\mathrm{x})\right)+2 \mathrm{x} \\ & \mathrm{f}^{\prime}(1)=\frac{1}{2}\left(1+\frac{\pi}{4}\right)+2 \\ & \mathrm{f}^{\prime}(1)=\frac{5}{2}+\frac{\pi}{8} \end{aligned}$