JEE MAIN - Mathematics (2024 - 6th April Morning Shift - No. 13)
Let $$f:(-\infty, \infty)-\{0\} \rightarrow \mathbb{R}$$ be a differentiable function such that $$f^{\prime}(1)=\lim _\limits{a \rightarrow \infty} a^2 f\left(\frac{1}{a}\right)$$. Then $$\lim _\limits{a \rightarrow \infty} \frac{a(a+1)}{2} \tan ^{-1}\left(\frac{1}{a}\right)+a^2-2 \log _e a$$ is equal to
$$\frac{5}{2}+\frac{\pi}{8}$$
$$\frac{3}{8}+\frac{\pi}{4}$$
$$\frac{3}{4}+\frac{\pi}{8}$$
$$\frac{3}{2}+\frac{\pi}{4}$$
Explanation
Let $$f^{\prime}(1)=k$$
$$\Rightarrow \quad \lim _\limits{x \rightarrow 0} \frac{f(x)}{x^2}=k \quad\left(\frac{0}{0}\right)$$
$$\begin{aligned} & \lim _\limits{x \rightarrow 0} \frac{f^{\prime}(x)}{2 x}=\lim _{x \rightarrow 0} \frac{f^{\prime \prime}(x)}{2}=k \\ \Rightarrow & f^{\prime \prime}(0)=2 k \end{aligned}$$
Given information is not complete.
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