$f(x) = \begin{cases} \frac{(2x^2 - ax + 1) - (ax^2 + 3bx + 2)}{x + 1} & ; x \neq -1 \\ k & ; x = -1 \end{cases}$ is a real-valued function. If $a, b, k \in R$ and $f$ is continuous on $R$,then $k =$

  • A
    $-\frac{1}{3}$
  • B
    $6$
  • C
    $a - 2$
  • D
    $a - 3$

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