Hard MCQ +4 / -1 PYQ · JEE Mains 2024

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

  1. A $\frac{5}{2}+\frac{\pi}{8}$ Correct answer
  2. B $\frac{3}{8}+\frac{\pi}{4}$
  3. C $\frac{3}{4}+\frac{\pi}{8}$
  4. D $\frac{3}{2}+\frac{\pi}{4}$

Solution

<p>Let $f^{\prime}(1)=k$</p> <p>$$\Rightarrow \quad \lim _\limits{x \rightarrow 0} \frac{f(x)}{x^2}=k \quad\left(\frac{0}{0}\right)$$</p> <p>$$\begin{aligned} & \lim _\limits{x \rightarrow 0} \frac{f^{\prime}(x)}{2 x}=\lim _{x \rightarrow 0} \frac{f^{\prime \prime}(x)}{2}=k \\ \Rightarrow & f^{\prime \prime}(0)=2 k \end{aligned}$$</p> <p>Given information is not complete.</p>

About this question

Subject: Mathematics · Chapter: Differentiation · Topic: Derivatives of Standard Functions

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