JEE Main 2024 — Mathematics Calculus

$$\begin{aligned} & f:(-\infty, \infty)-\{0\} \rightarrow R \\ & f^{\prime}(1)=\lim _{a \rightarrow \infty} a^2 f\left(\frac{1}{a}\right) \\ & \lim _{a \rightarrow \infty} \frac{a(a+1)}{2} \tan ^{-1}\left(\frac{1}{a}\right)+a^2-2 \ln (a) \\ & \lim _{a \rightarrow \infty} a^2\left(\frac{\left(1+\frac{1}{a}\right)}{2} \tan ^{-1}\left(\frac{1}{a}\right)+1-\frac{2}{a^2} \ln (a)\right) \\ & f(x)=\frac{1}{2}(1+x) \tan ^{-1}(x)+1-2 x^2 \ln (x)\end{aligned}$$​f:(−∞,∞)−{0}→Rf′(1)=a→∞lim​a2f(a1​)a→∞lim​2a(a+1)​tan−1(a1​)+a2−2ln(a)a→∞lim​a2(2(1+a1​)​tan−1(a1​)+1−a22​ln(a))f(x)=21​(1+x)tan−1(x)+1−2x2ln(x)​

$\begin{aligned}
& \mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{2}\left(\frac{1+\mathrm{x}}{1+\mathrm{x}^2}+\tan ^{-1}(\mathrm{x})+4 \mathrm{x} \ell(\mathrm{x})\right)+2 \mathrm{x} \
& \mathrm{f}^{\prime}(1)=\frac{1}{2}\left(1+\frac{\pi}{4}\right)+2 \
& \mathrm{f}^{\prime}(1)=\frac{5}{2}+\frac{\pi}{8}
\end{aligned}$