JEE Main 2024 — Differentiation Question with Solution
From: JEE Main 2024 (Online) 6th April Morning Shift
Question
Let 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
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Step-by-step explanation
Let
\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}
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