= Solution
The <likelihood ratio> between the two normal means is
$$
\frac{L(\mu_1;x)}{L(\mu_0;x)}
=\exp\left((\mu_1-\mu_0)\sum_{i=1}^nx_i-\frac n2(\mu_1^2-\mu_0^2)\right).
$$
Since $\mu_1>\mu_0$, it is strictly increasing in $\overline x$. The Neyman–Pearson test therefore rejects for large sample means. Under the null, $\sqrt n(\overline X-\mu_0)\sim N(0,1)$, so choosing a right-tail <probability> $\alpha$ gives
$$
\boxed{c_\alpha=\mu_0+\frac{z_\alpha}{\sqrt n},\qquad C=\{x:\overline x\geq c_\alpha\}.}
$$
This has size exactly $\alpha$ because the <sampling distribution> is <continuous>.
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