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(A) . For $y = |x| + |x - 2|$, at $x = 2$, the function has a corner point. Thus, $\frac{dy}{dx}$ does not exist. So, $A \rightarrow IV$.$B$. For $f(x

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$A-IV, B-I, C-II, D-III$

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(A) . For $y = |x| + |x - 2|$, at $x = 2$, the function has a corner point. Thus, $\frac{dy}{dx}$ does not exist. So, $A \rightarrow IV$.$B$. For $f(x) = |\cos 2x|$, for $x$ slightly greater than $\frac{\pi}{4}$, $\cos 2x$ is negative, so $f(x) = -\cos 2x$. Then $f'(x) = 2 \sin 2x$. Thus, $f'(\frac{\pi}{4} +) = 2 \sin(\frac{\pi}{2}) = 2$. So, $B \rightarrow I$.$C$. For $f(x) = \sin(\pi[x])$, for $x$ slightly less than $1$, $[x] = 0$. So $f(x) = \sin(0) = 0$. Thus, $f'(1-) = 0$. So, $C \rightarrow II$.$D$. For $f(x) = \log|x - 1|$, $f'(x) = \frac{1}{x - 1}$. Then $f'(\frac{1}{2}) = \frac{1}{1/2 - 1} = \frac{1}{-1/2} = -2$. So, $D \rightarrow III$.Therefore, the correct matching is $A-IV, B-I, C-II, D-III$.

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

Column $I$	Column $II$
$(A)$ $f(x) = x\|x\|$	$(p)$ continuous in $(-1, 1)$
$(B)$ $f(x) = \sqrt{\|x\|}$	$(q)$ differentiable in $(-1, 1)$
$(C)$ $f(x) = x + [x]$	$(r)$ strictly increasing in $(-1, 1)$
$(D)$ $f(x) = \|x - 1\| + \|x + 1\|$	$(s)$ not differentiable at least at one point in $(-1, 1)$

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