3. If the real number $x$ satisfies $\arcsin x>\arccos x$, then the range of the expression
$$
f(x)=\sqrt{2 x^{2}-x+3}+2^{\sqrt{x^{2}-x}}
$$

is $\qquad$
3. $\{3\}$.

If $-1 \leqslant x \leqslant 0$, then $\arcsin x \in\left[-\frac{\pi}{2}, 0\right], \arccos x \in\left[\frac{\pi}{2}, \pi\right]$. At this time, $\arcsin x\sin (\arccos x) \\
\Rightarrow x>\sqrt{1-x^{2}} \Rightarrow|x|>\frac{\sqrt{2}}{2} \text {. } \\
\text { Also } 0<x \leqslant 1 \Rightarrow \frac{\sqrt{2}}{2}<x \leqslant 1 \text {. } \\
\text { From }\left\{\begin{array}{l}
2 x^{2}-x+3 \geqslant 0, \\
x^{2}-x \geqslant 0
\end{array}\right. \text {, } \\
\Rightarrow x \in(-\infty, 0] \cup[1,+\infty) \text {. } \\
\end{array}
$$

Therefore, $x \in\{1\}$.
Thus, the value of the function $f(x)$ is $\{3\}$.