$f(x)=\left\{\begin{array}{ll}\frac{x^{3}}{(1-\cos 2 x)^{2}} \log _{e}\left(\frac{1+2 x e^{-2 x}}{\left(1-x e^{-x}\right)^{2}}\right), & x \neq 0 \\ \,\alpha & , x=0\end{array}\right.$ If $\mathrm{f}$ is continuous at $\mathrm{x}=0$, then $\alpha$ is equal to :
$\lim _{x \rightarrow 0} \frac{x^{3}}{4 \sin ^{4} x}\left(\ln \left(1+2 e^{-2 x}\right)-2 \ln \left(1-x e^{-x}\right)\right)$
$=\alpha$
$\lim _{x \rightarrow 0} \frac{1}{4 x}\left[2 x e^{-2 x}+2 x e^{-x}\right]=\alpha$
$=\frac{1}{4}(4)=\alpha=1$
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$\begin{array}{|l|l|l|l|l|l|} \hline X=x & 0 & 1 & 2 & 3 & 4 \\ \hline P(X=x) & \frac{1}{3} & \frac{1}{2} & 0 & \frac{1}{6} & 0 \\ \hline \end{array}$
, then