A single observation $X$ is taken from a $BETA(a, b)$ distribution.
Given the hypothesis
$$ H_0 : a=b=1 \space vs\space H_1 : a=b=\frac{1}{2} $$
I want to find the Most Powerful test.
Well, we know by the Neyman-Pearson lemma that: $$ \frac{f(\widetilde{x}/\theta_1)}{f(\widetilde{x}/\theta_0)} > k $$ to some $$ k > 0 $$ we will have a UMP with size $$ \alpha=0.05 $$ such that $$ P(X \space \epsilon \space R | \theta = \theta_0) = \alpha $$
Then,
$$ \frac{f(\widetilde{x}/\theta_1)}{f(\widetilde{x}/\theta_0)}= \frac{1}{\pi} x^{-\frac{1}{2}} (1-x)^{-\frac{1}{2}} > k$$
If I were to graph that I would get a global minimum at $x=\frac{1}{2}$ and that the support $x \in [0,1]$ but I am not sure where to go from here. Just to mention that is a self-study thing! Thank you in advance.
Symmetry tells us our rejection region is of the form $$ |x-\frac{1}{2}|> f(k).$$ (If you do the algebra, it is $f(k)=\sqrt{\frac{1}{4} - \frac{1}{\pi^2 k^2}}$.) So it's the region outside of some interval centered inside $(0,1)$.
Since our null hypothesis is a uniform distribution, we don't even need to go through anything more than mental math: this immediately means that for $\alpha=0.05$ we reject if $0<x<0.025$ or if $.975<x<1.$
If you want to get a $k$, you just solve for $f(k) = 0.475,$ but it seems almost counterproductive to do so. You might want to compute the power by integrating the $beta(1/2,1/2)$ distribution over the rejection region.