Let $f\left( \frac{x + 8y}{9} \right) = \frac{f(x) + 8f(y)}{9}$ for all real $x$ and $y$. If $f'(0)$ exists and is equal to $2$,and $f(0) = -5$,then $f(7)$ is equal to:

  • A
    $3$
  • B
    $7$
  • C
    $5$
  • D
    $9$

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