જો f(x) = tan^{-1}left(frac{sqrt{1+x}-sqrt{1- | vedclass

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(D) આપેલ છે $f(x) = 	an^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}ight)$.ધારો કે $x = \cos 2	heta$,તો $1+x = 2\cos^2	heta$ અને

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$\frac{1}{\sqrt{3}}$

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(D) આપેલ છે $f(x) = 	an^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}ight)$.ધારો કે $x = \cos 2	heta$,તો $1+x = 2\cos^2	heta$ અને $1-x = 2\sin^2	heta$.$f(x) = 	an^{-1}\left(\frac{\sqrt{2}\cos	heta - \sqrt{2}\sin	heta}{\sqrt{2}\cos	heta + \sqrt{2}\sin	heta}ight) = 	an^{-1}\left(\frac{1-	an	heta}{1+	an	heta}ight) = 	an^{-1}\left(	an(\frac{\pi}{4}-	heta)ight) = \frac{\pi}{4}-	heta$.કારણ કે $x = \cos 2	heta$,$	heta = \frac{1}{2}\cos^{-1}x$,તેથી $f(x) = \frac{\pi}{4} - \frac{1}{2}\cos^{-1}x$.લક્ષ $\lim_{x ightarrow \frac{1}{2}} \frac{2[f(x)-f(\frac{1}{2})]}{2x-1} = \lim_{x ightarrow \frac{1}{2}} \frac{f(x)-f(\frac{1}{2})}{x-\frac{1}{2}} = f'(\frac{1}{2})$.$f'(x) = -\frac{1}{2} \left(-\frac{1}{\sqrt{1-x^2}}ight) = \frac{1}{2\sqrt{1-x^2}}$.$f'(\frac{1}{2}) = \frac{1}{2\sqrt{1-(\frac{1}{2})^2}} = \frac{1}{2\sqrt{3/4}} = \frac{1}{2(\frac{\sqrt{3}}{2})} = \frac{1}{\sqrt{3}}$.

List-$I$	List-$II$
$A$. $\sec^{-1} x$	$I$. $\frac{1}{1-x^2}, x \in (-1, 1)$
$B$. $\tanh^{-1} x$	$II$. $\frac{-1}{\|x\| \sqrt{x^2+1}}, x \neq 0$
$C$. $\coth^{-1} x$	$III$. $\frac{1}{\|x\| \sqrt{x^2-1}}, \|x\| > 1$
$D$. $\operatorname{cosech}^{-1} x$	$IV$. $\frac{1}{1-x^2}, x \in R - [-1, 1]$
	$V$. $\frac{-1}{\|x\| \sqrt{1-x^2}}, \|x\| < 1, x \neq 0$

જો $f(x) = \tan^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)$ હોય,તો $\lim_{x \rightarrow \frac{1}{2}} \frac{2[f(x)-f(\frac{1}{2})]}{2x-1} = $

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