(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-$II$
$(A) \sin ^{-1}\left(\frac{2 x}{1+x^2}\right)$	$(I) \cos x-\sin x$
$(B) \tan ^{-1}\left(\frac{1-x}{1+x}\right)$	$(II) \frac{-1}{1+x^2}$
$(C) e^{\log (\sin x+\cos x)}$	$(III) \frac{2}{1+x^2}$
$(D) \sqrt{1-\sin 2 x} \text{ for } (0 < x < \frac{\pi}{4})$	$(IV) \cos x+\sin x$
	$(V) -\sin x-\cos x$

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