Let $f: R \rightarrow R$ be a function defined by $f(x) = \frac{2e^{2x}}{e^{2x} + e}$. Then $f\left(\frac{1}{100}\right) + f\left(\frac{2}{100}\right) + f\left(\frac{3}{100}\right) + \dots + f\left(\frac{99}{100}\right)$ is equal to

  • A
    $98$
  • B
    $99$
  • C
    $100$
  • D
    $101$

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