(c) Given $\mathrm{f}(\mathrm{x})$ is differentiable function.
Take, $\lim _{x \rightarrow 0} A(x)=\lim _{x \rightarrow 0}\left[\begin{array}{l}2 f(x) \operatorname{cosec} 4 x+ \\ 4 f(x)\left(\cos ^2 x+1\right)-4 \cos ^2 x\end{array}\right]$
$$
\lim _{x \rightarrow 0} A(x)=2 \lim _{x \rightarrow 0} f(x) \lim _{4 x \rightarrow 0} \frac{1}{\left(\frac{\sin 4 x}{4 x}\right) 4 x}+4 \lim _{x \rightarrow 0} f(x)
$$
$\lim _{x \rightarrow 0}\left(\cos ^2 x+1\right)-4 \lim _{x \rightarrow 0} \cos ^2 x$
$\lim _{x \rightarrow 0} A(x)=\frac{2}{4} \lim _{x \rightarrow 0} \frac{f(x)}{x} \times \frac{1}{1}+4 \lim _{x \rightarrow 0} f(x)\left(\cos ^2 0+1\right)$
$-4 \cos 0$
$$
\lim _{x \rightarrow 0} \mathrm{aA}(\mathrm{x})=\frac{1}{2} \lim _{x \rightarrow 0} \frac{\mathrm{f}(\mathrm{x})}{\mathrm{x}}+4 \lim _{\mathrm{x} \rightarrow 0} \mathrm{f}(\mathrm{x})(2)-4
$$

Apply L'Hospital Rule $\lim A(x)=\frac{1}{2} \frac{f^{\prime}(x)}{1}+8 \lim _{x \rightarrow 0} f(x)-4$.
Apply the limit
$$
\begin{aligned}
& \lim _{x \rightarrow 0} A(x)=\frac{1}{2} f^{\prime}(0)+8 f(0)-4 \\
& \frac{1}{2} \times 20+8 \times 0-4 \\
& =10-4=6
\end{aligned}
$$

Therefore, option (c) is correct