Let $z = \left | 2^{\pi i/x_{1}} \right |$ , then $f(x_{1},x_{2}) = z^{x_{2}}$
Now I will try to prove that $z=1.$ This is a question of theory of Complex Numbers on euler representation of complex numbers.
A small recap, $e^{i\theta} = \cos \theta + i \sin \theta$ and $ \left | e^{ i \theta} \right | = \sqrt{\cos ^{2} \theta+ \sin ^{2} \theta} = 1$ because if $ z_1 = a+ib$ then $|z_1|= \sqrt{a^2 +b ^2}$
Now coming back to the problem,
Since, $e^{ln\;x}=x$, So, $2^{\pi i/x_{1}}= e^{ln\; 2^{\pi i/x_{1}}}= e^{\pi (\ln2) i/x_{1} } $
Now, $z = \left | 2^{\pi i/x_{1}} \right |=\left | e^{\pi (\ln2) i/x_{1} } \right | = \sqrt{\cos ^{2} \theta+ \sin ^{2} \theta} = 1$ where,
$\theta = \pi (\ln2) /x_{1} $ and $x_1 \neq 0$
Substituting back in $f$,
$f(x_{1},x_{2}) = 1^{x_{2}} = 1$
Hence $f$ is a constant function which means it is degenerate on both $x_{1}, x_{2}$
Option $A$ is correct