The function f(x) = begin{cases} |x - 3| & x | vedclass

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(D) First,we analyze the continuity and differentiability at $x = 1$ and $x = 3$.For $x = 1$:Left-hand limit: $\lim_{x 	o 1^-} f(x) = \frac{1^2}{4} -

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All of the above

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(D) First,we analyze the continuity and differentiability at $x = 1$ and $x = 3$.For $x = 1$:Left-hand limit: $\lim_{x 	o 1^-} f(x) = \frac{1^2}{4} - \frac{3(1)}{2} + \frac{13}{4} = \frac{1 - 6 + 13}{4} = 2$.Right-hand limit: $\lim_{x 	o 1^+} f(x) = |1 - 3| = 2$.Since $f(1) = 2$,the function is continuous at $x = 1$.Now check differentiability at $x = 1$:$f'(1^+) = \frac{d}{dx}(3 - x) = -1$ (for $1 \leqslant x $f'(1^-) = \frac{d}{dx}(\frac{x^2}{4} - \frac{3x}{2} + \frac{13}{4}) = \frac{2x}{4} - \frac{3}{2} = \frac{1}{2} - \frac{3}{2} = -1$.Since $f'(1^+) = f'(1^-)$,the function is differentiable at $x = 1$.For $x = 3$:The function $f(x) = |x - 3|$ is continuous at $x = 3$ because $\lim_{x 	o 3} |x - 3| = 0 = f(3)$.Thus,all statements are correct.

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 $f'(\frac{\pi}{4} +) =$	$II$. $0$
$C$. If $f(x) = \sin(\pi[x])$, where $[x]$ is the greatest integer function, then $f'(1-) =$	$III$. $-2$
$D$. If $f(x) = \log\|x - 1\|$, $x \neq 1$, then $f'(\frac{1}{2}) =$	$IV$. does not exist

List-$I$	List-$II$
$A. \frac{d}{dx}\left(\tan^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right)\right)$	$(i) \log(x+\sqrt{1+x^2})$
$B. \frac{d}{dx}\left(\frac{3+\|x-1\|}{3x+4}\right)$	$(ii) -\frac{4x}{(1+x^2)^2}$
$C. \sinh^{-1} x$	$(iii) \frac{1}{2}$
$D. \frac{d^2}{dx^2}\left(\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right)$	$(iv) \frac{1}{\sqrt{1+x^2}}$
	$(v) \text{not differentiable at } x=1$

The function $f(x) = \begin{cases} |x - 3| & x \geqslant 1 \\ \frac{x^2}{4} - \frac{3x}{2} + \frac{13}{4} & x < 1 \end{cases}$ is :

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