Q44 / 90 · JEE Main 2024 (06 Apr Shift 1) · Mathematics
Let $f:(-\infty, \infty)-\{0\} \rightarrow \mathbb{R}$ be a differentiable function such that $f^{\prime}(1)=\lim _{a \rightarrow \infty} a^2 f\left(\frac{1}{a}\right)$. Then $\lim _{a \rightarrow \infty} \frac{a(a+1)}{2} \tan ^{-1}\left(\frac{1}{a}\right)+a^2-2 \log _e a$ is equal to
- A$\frac{3}{2}+\frac{\pi}{4}$
- B$\frac{3}{4}+\frac{\pi}{8}$
- C$\frac{3}{8}+\frac{\pi}{4}$
- Correct answerD$\frac{5}{2}+\frac{\pi}{8}$
Solution
$\begin{aligned} & f:(-\infty, \infty)-\{0\} \rightarrow R \\ & f^{\prime}(1)=\lim _{a \rightarrow \infty} a^2 f\left(\frac{1}{a}\right) \\ & \lim _{a \rightarrow \infty} \frac{a(a+1)}{2} \tan ^{-1}\left(\frac{1}{a}\right)+a^2-2 \ln (a) \\ & \lim _{a \rightarrow \infty} a^2\left(\frac{\left(1+\frac{1}{a}\right)}{2} \tan ^{-1}\left(\frac{1}{a}\right)+1-\frac{2}{a^2} \ln (a)\right) \\ & f(x)=\frac{1}{2}(1+x) \tan ^{-1}(x)+1-2 x^2 \ln (x)\end{aligned}$
$\begin{aligned} & \mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{2}\left(\frac{1+\mathrm{x}}{1+\mathrm{x}^2}+\tan ^{-1}(\mathrm{x})+4 \mathrm{x} \ell(\mathrm{x})\right)+2 \mathrm{x} \\ & \mathrm{f}^{\prime}(1)=\frac{1}{2}\left(1+\frac{\pi}{4}\right)+2 \\ & \mathrm{f}^{\prime}(1)=\frac{5}{2}+\frac{\pi}{8} \end{aligned}$
$\begin{aligned} & \mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{2}\left(\frac{1+\mathrm{x}}{1+\mathrm{x}^2}+\tan ^{-1}(\mathrm{x})+4 \mathrm{x} \ell(\mathrm{x})\right)+2 \mathrm{x} \\ & \mathrm{f}^{\prime}(1)=\frac{1}{2}\left(1+\frac{\pi}{4}\right)+2 \\ & \mathrm{f}^{\prime}(1)=\frac{5}{2}+\frac{\pi}{8} \end{aligned}$
Hence, the correct answer is option: 'D'.