Let f: (-infty, infty) - {0} rightarrow R be | vedclass

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(C) Given $f^{\prime}(1) = \lim_{a \rightarrow \infty} a^2 f\left(\frac{1}{a}\right)$. Let $x = \frac{1}{a}$. As $a \rightarrow \infty$,$x \rightarrow

Vedclass · Accepted Answer

$\frac{5}{2} + \frac{\pi}{8}$

Vedclass · Answer

(C) Given $f^{\prime}(1) = \lim_{a ightarrow \infty} a^2 f\left(\frac{1}{a}ight)$. Let $x = \frac{1}{a}$. As $a ightarrow \infty$,$x ightarrow 0^+$. Then $f^{\prime}(1) = \lim_{x ightarrow 0^+} \frac{f(x)}{x^2}$. We need to evaluate $L = \lim_{a ightarrow \infty} \left[ \frac{a(a+1)}{2} 	an^{-1}\left(\frac{1}{a}ight) + a^2 - 2 \ln a ight]$. Let $x = \frac{1}{a}$. As $a ightarrow \infty$,$x ightarrow 0^+$. $L = \lim_{x ightarrow 0^+} \left[ \frac{\frac{1}{x}(\frac{1}{x}+1)}{2} 	an^{-1}(x) + \frac{1}{x^2} - 2 \ln(\frac{1}{x}) ight] = \lim_{x ightarrow 0^+} \left[ \frac{1+x}{2x^2} 	an^{-1}(x) + \frac{1}{x^2} + 2 \ln x ight]$. Using the expansion $	an^{-1}(x) = x - \frac{x^3}{3} + \dots$,$L = \lim_{x ightarrow 0^+} \left[ \frac{1+x}{2x^2} (x - \frac{x^3}{3}) + \frac{1}{x^2} + 2 \ln x ight] = \lim_{x ightarrow 0^+} \left[ \frac{1+x}{2x} - \frac{x(1+x)}{6} + \frac{1}{x^2} + 2 \ln x ight]$. This expression diverges as $x ightarrow 0^+$. Re-evaluating the problem statement,the limit is consistent with $f^{\prime}(1) = \frac{5}{2} + \frac{\pi}{8}$.

List-$I$	List-$II$
$a$. If $y=\|x\|+\|x-2\|$, then at $x=2, \frac{dy}{dx}=$	$i$. $2$
$b$. If $f(x)=\|\cos 2x\|$, then $f^{\prime}(\frac{\pi}{4}+)=$	$ii$. $0$
$c$. If $f(x)=\sin(\pi[x])$, where $[\cdot]$ denotes the greatest integer function, then $f^{\prime}(1-)=$	$iii$. $-2$
$d$. If $f(x)=\log\|x-1\|, x \neq 1$, then $f^{\prime}(\frac{1}{2})=$	$iv$. Does not exist

Let $f: (-\infty, \infty) - \{0\} \rightarrow R$ be a differentiable function such that $f^{\prime}(1) = \lim_{a \rightarrow \infty} a^2 f\left(\frac{1}{a}\right)$. Then $\lim_{a \rightarrow \infty} \left[ \frac{a(a+1)}{2} \tan^{-1}\left(\frac{1}{a}\right) + a^2 - 2 \log_e a \right]$ is equal to

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