(i) f(x) is continuous and defined for all re | vedclass

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(B) From the given information:$1$. $f'(-5) = 0$ and $f'(4) = 0$ implies stationary points at $x = -5$ and $x = 4$.$2$. $f'(x) > 0$ for $x < -5$ and $

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(B) From the given information:$1$. $f'(-5) = 0$ and $f'(4) = 0$ implies stationary points at $x = -5$ and $x = 4$.$2$. $f'(x) > 0$ for $x $3$. $f'(2)$ is undefined,indicating a sharp corner or cusp at $x = 2$.$4$. $f'(x) > 0$ for $2  4$ implies a local maximum at $x = 4$.$5$. $f''(x) Comparing these properties with the given options:- Option $A$ shows a local minimum at $x = -5$,which is incorrect.- Option $B$ shows a local maximum at $x = -5$ and a local maximum at $x = 4$,with a cusp at $x = 2$,and the graph is concave down throughout. This matches all conditions.- Option $C$ shows a local maximum at $x = -5$ but a local minimum at $x = 4$,which is incorrect.- Option $D$ shows a local maximum at $x = -5$ but a local minimum at $x = 4$,which is incorrect.Thus,the correct graph is $B$.

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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