If \( \alpha \in R \) and \( f(x)=\left(\alpha-\frac{1}{\alpha}-x\right)\left(4-3 x^{2}\right) \) for \( x \in R \). If \( g(\alpha) \) is the difference between the maximum and minimum values of \( f \), then
A) \( g(2)=\frac{125}{18} \)
B) \( g(2)=\frac{125}{8} \)
C) minimum value of \( g(\alpha) \) is \( \frac{32}{9} \)
D) maximum value of \( g(\alpha) \) is \( \frac{256}{18} \)
