If the function defined by $f(x) = \begin{cases} (x^2 + e^{\frac{1}{2-x}})^{-1}, & x \neq 2 \\ k, & x = 2 \end{cases}$ is right continuous at $x = 2$,then $k =$

  • A
    $-\frac{1}{4}$
  • B
    $0$
  • C
    $\frac{1}{4}$
  • D
    $1$

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