$f(x)=\left\{\begin{array}{1 1}\large\frac{1-\cos 4x}{x^2}&x < 0\\a&x=0\\\large\frac{\sqrt x}{\sqrt{[16+\sqrt x]-4}}&x > 0\end{array}\right.$If the function be continuous at $x=0$ then $a=$

Question

$(a)\;4\qquad(b)\;6\qquad(c)\;8\qquad(d)\;10$