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(D) For the function $f(x)$ to be defined, both parts must be defined.First, consider $\cos^{-1} \left( \frac{1 + x^2}{2 x} \right)$. This is defined

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$\{-1, 1\}$

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(D) For the function $f(x)$ to be defined, both parts must be defined.First, consider $\cos^{-1} \left( \frac{1 + x^2}{2 x} \right)$. This is defined if $-1 \leq \frac{1 + x^2}{2 x} \leq 1$.This implies $\left| \frac{1 + x^2}{2 x} \right| \leq 1$, which means $|1 + x^2| \leq |2x|$.Since $1 + x^2 \geq 2|x|$ is only true when $|x| = 1$, we have $x = 1$ or $x = -1$.Now, check the first part $\sqrt{\cos(\sin x)}$.For $x = 1$, $\cos(\sin 1) > 0$ since $\sin 1 \approx 0.84$ radians, which is in the first quadrant.For $x = -1$, $\cos(\sin(-1)) = \cos(-\sin 1) = \cos(\sin 1) > 0$.Since both conditions are satisfied only at $x = 1$ and $x = -1$, the domain is $\{-1, 1\}$.

The domain of the real valued function $f(x) = \sqrt{\cos (\sin x)} + \cos^{-1} \left( \frac{1 + x^2}{2 x} \right)$ is

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