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(A) For $y=|x|+|x-2|$, at $x=2$, the function $y$ is not differentiable because it involves the sum of absolute values where the term $|x-2|$ has a sh

Vedclass · Accepted Answer

$(a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)$

Vedclass · Answer

(A) For $y=|x|+|x-2|$, at $x=2$, the function $y$ is not differentiable because it involves the sum of absolute values where the term $|x-2|$ has a sharp corner at $x=2$. Thus, $\frac{dy}{dx}$ does not exist. (Matches $iv$)$(b)$ For $f(x)=|\cos 2x|$, at $x=\frac{\pi}{4}$, $\cos(2 \cdot \frac{\pi}{4}) = \cos(\frac{\pi}{2}) = 0$. For $x > \frac{\pi}{4}$, $\cos 2x$ is negative, so $f(x) = -\cos 2x$. Then $f^{\prime}(x) = 2\sin 2x$. At $x = \frac{\pi}{4}^{+}$, $f^{\prime}(\frac{\pi}{4}^{+}) = 2\sin(\frac{\pi}{2}) = 2$. (Matches $i$)$(c)$ For $f(x)=\sin(\pi[x])$, for $x$ slightly less than $1$ $(x \in (0, 1))$, $[x]=0$. Thus $f(x) = \sin(0) = 0$. The derivative of a constant function is $0$. (Matches $ii$)$(d)$ For $f(x)=\log|x-1|$, $f^{\prime}(x) = \frac{1}{x-1}$. At $x=\frac{1}{2}$, $f^{\prime}(\frac{1}{2}) = \frac{1}{\frac{1}{2}-1} = \frac{1}{-\frac{1}{2}} = -2$. (Matches $iii$)

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 $f^{\prime}(\frac{\pi}{4}+)=$	$ii$. $0$
$c$. 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

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