If $f(x) = \begin{cases} \frac{\log (1 + 2ax) - \log (1 - bx)}{x}, & x \neq 0 \\ k, & x = 0 \end{cases}$ is continuous at $x = 0$,then the value of $k$ is

  • A
    $b + a$
  • B
    $b - 2a$
  • C
    $2a - b$
  • D
    $2a + b$

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