If $f(x) = \begin{cases} \frac{1 - \cos 4x}{x^2}, & x < 0 \\ a, & x = 0 \\ \frac{\sqrt{x}}{\sqrt{16 + \sqrt{x}} - 4}, & x > 0 \end{cases}$ is continuous at $x = 0$,then the value of $a$ will be

  • A
    $8$
  • B
    $-8$
  • C
    $4$
  • D
    None of these

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