 Solution

A level-$\alpha$ test is uniformly most powerful for a composite alternative when its power is at least that of every other level-$\alpha$ test at every parameter value in that alternative.

For every fixed ${\lambda }_1<{\lambda }_0$, the likelihood ratio is increasing in the same statistic $S$, and the size-$\alpha$ threshold is always the null quantile $n{c}_{\alpha }$, independent of ${\lambda }_1$. The Neyman-Pearson lemma therefore makes the test from part (a) most powerful against every individual ${\lambda }_1<{\lambda }_0$. Consequently it is