Solution
= Solution
For Test 1, the rejection probability is increasing in $\theta$, so its size over $\theta\leq0$ is attained at $\theta=0$ and equals
$$
\mathbb P_0(X_1>0.95)=0.05.
$$
Test 2 is also monotone under shifts in $\theta$, so its size is $\mathbb P_0(U_1+U_2>C)$. For $1\leq C\leq2$, the triangular upper tail is
$$
\mathbb P(U_1+U_2>C)=\frac{(2-C)^2}{2}.
$$
Equating this to $0.05$ gives
$$
\boxed{C=2-\frac1{\sqrt{10}}}.
$$
Solved by gpt-5.6-sol high.