If a function $f$ defined by $f(x)=\left\{\begin{array}{cc}\frac{1-\cos 4 x}{x^2}, & x < 0 \\ a, & x=0 \\ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}, & x>0\end{array}\right.$ is continuous at $\mathrm{x}=0$, then $\mathrm{a}=$

• Correct answer
  A

  8
• B

  4
• C

  3
• D

  2

Hence, the correct answer is option: 'A'.