let $M_0(t) = E(e^{tx})$

$\begin{aligned} &= \int_0^\infty e^{tx}.f(x)dx \\ &= \int_0^\infty e^{tx}.2e^{-2x}dx \\ &=2 \int_0^\infty e^{tx}.e^{-2x}dx \\ &= 2\int_0^\infty e^{tx-2x}dx \\ &= 2[\frac{e^{(t-2)x}}{(t-2)}]_0^\infty\\ &=2[0-\frac{1}{t-2}]\\ M_0(t) &=\frac{2}{2-t}\\ &=\frac{2}{2(1-t/2)}\\ &=\frac{1}{1(1-t/2)}\\ &=1-(\frac {t}{2})^{-1}\\ &=1+\frac{t}{2}+\frac{t^2}{2^2}+\frac{t^3}{2^3}+----\\ mean = \mu_1' &= coefficient \ of \ t\\ mean = \mu_1' &=1/2\\ and \ \mu_2' &= coefficient \ of \ \frac{t^2}{2!} =\frac{2}{2^2}\\ \mu_2' &=1/4\\ var(x) &= \mu_2'-(\mu_1')^2\\ &=\frac{2}{4}-\frac{1}{4}\\ var(x)=1/4 \end{aligned}$