= Solution
The <likelihood ratio> of the simple alternative to the simple null is
$$
\frac{f(x\mid1)}{f(x\mid0)}
=e^{-1}\left(\frac{1+e^x}{1+e^{x-1}}\right)^2.
$$
This is a strictly increasing function of $x$. By the <Neyman-Pearson lemma>, the most powerful test therefore rejects for $X>c$. The <logistic distribution> has
$$
F_0(c)=\frac{e^c}{1+e^c}.
$$
The size condition $\mathbb P_0(X>c)=\alpha$ gives
$$
\frac{e^c}{1+e^c}=1-\alpha,
\qquad
c=\log\frac{1-\alpha}{\alpha}.
$$
Because the distribution is continuous, no boundary randomization is needed. The most powerful size-$\alpha$ test is therefore
$$
\boxed{\text{reject }H_0\quad\Longleftrightarrow\quad
X>\log\frac{1-\alpha}{\alpha}}.
$$
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