Let $f : [0,1] \to R$ be such that $f(xy) = f(x)f(y)$ for all $x, y \in [0,1],$ and $f(0) \ne 0.$ If $y = y(x)$ satisfies the differential equation $\frac{dy}{dx} = f(x)$ with $y(0) = 1,$ then $y\left( \frac{1}{4} \right) + y\left( \frac{3}{4} \right)$ is equal to

  • A
    $4$
  • B
    $3$
  • C
    $5$
  • D
    $2$

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