What the function? 2

We are given a function $f$ defined as \left\{ \begin{array}{II} \arccos{(x)} & x \in \mathcal{R} \\ e^{\arcsin{(x)}} & \text{otherwise}. \\ \end{array} \\ \right. , where $\mathcal{R}$ is the set of rational numbers. Within ANY non-singleton interval, there exists an infinite set of rational numbers. Therefore $\mathcal{R}$ is a dense set in $\mathbb{R}$ . As a result, there is an infinite set of points within any interval for which $f$ would be equal to $\arccos{(x)}$ .

We may reason similarly with $f: x \notin \mathcal{R}$ . There are far more irrational numbers than rational numbers within any given interval so if the set of rational numbers is dense in $\mathbb{R}$ , then the set of irrational numbers must also be dense in $\mathbb{R}$ . Therefore, there will also be an infinite set of points within any interval for which $f$ would be equal to $e^{\arcsin{(x)}}$ .

If we were to attempt to view the set of rational numbers in $[-1,1]$ and the set of irrational numbers in $[-1,1]$ as a set of points, they would both be indistinguishable from $-1 \leq x \in \mathbb{R} \leq 1$ . Similarly, both pieces of $f$ appear to occupy the entire domain, although they both exclude dense subsets of the domain of $f$ . In other words, $f$ looks like a plot of $\arccos{(x)}$ superimposed on $e^{\arcsin{(x)}}$ .

Hence, the answer is $B$ .

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It is important to note that $x$ is always either rational or irrational, and is never used in both pieces of $f$ , therefore $f$ is a valid function. Also, neither of the pieces of $f$ are continuous.