If the 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 $\mathrm{a}=$

• A

  $1$
• B

  $2$
• C

  $4$
• Correct answer
  D

  $8$

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