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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 shar

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

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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