Match the items of List-I with those of List- | vedclass

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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

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