Two functions f and g have first and second d | vedclass

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(D) Given: $6f(0) = 3 \implies f(0) = \frac{1}{2}$.Since $f(0) = \frac{2}{g(0)}$,we have $g(0) = \frac{2}{f(0)} = 4$.Given $f'(0) = 4g(0) = 4(4) = 16$

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All of the above

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(D) Given: $6f(0) = 3 \implies f(0) = \frac{1}{2}$.Since $f(0) = \frac{2}{g(0)}$,we have $g(0) = \frac{2}{f(0)} = 4$.Given $f'(0) = 4g(0) = 4(4) = 16$.Given $2g'(0) = 4g(0) = 16 \implies g'(0) = 8$.Check $A$: $h(x) = \frac{f(x)}{g(x)} \implies h'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}$.$h'(0) = \frac{f'(0)g(0) - f(0)g'(0)}{(g(0))^2} = \frac{(16)(4) - (1/2)(8)}{4^2} = \frac{64 - 4}{16} = \frac{60}{16} = \frac{15}{4}$. (Correct)Check $B$: $k(x) = f(x)g(x)\sin x \implies k'(x) = f'(x)g(x)\sin x + f(x)g'(x)\sin x + f(x)g(x)\cos x$.$k'(0) = 0 + 0 + f(0)g(0)\cos(0) = (1/2)(4)(1) = 2$. (Correct)Check $C$: $\lim_{x 	o 0} \frac{g'(x)}{f'(x)} = \frac{g'(0)}{f'(0)} = \frac{8}{16} = \frac{1}{2}$. (Correct)Thus,all options are correct.

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

Two functions $f$ and $g$ have first and second derivatives at $x = 0$ and satisfy the relations: $f(0) = \frac{2}{g(0)}$,$f'(0) = 2g'(0) = 4g(0)$,$g''(0) = 5f''(0) = 6f(0) = 3$. Then:

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