Let $f$ and $g$ be twice differentiable even functions on $(-2, 2)$ such that $f(\frac{1}{4}) = 0, f(\frac{1}{2}) = 0, f(1) = 1$ and $g(\frac{3}{4}) = 0, g(1) = 2$. Then,the minimum number of solutions of $f(x)g''(x) + f'(x)g'(x) = 0$ in $(-2, 2)$ is equal to

  • A
    $0$
  • B
    $2$
  • C
    $4$
  • D
    $6$

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