1. The range of the function $f(x)=|x+1|+|x-1|+\sqrt{4-x^{2}}$ is $\qquad$ .
- 1. $[2+\sqrt{3}, 2 \sqrt{5}]$.

Notice that, $f(x)$ is an even function with the domain $[-2,2]$.
When $0 \leqslant x \leqslant 1$, $f(x)=2+\sqrt{4-x^{2}}$ is monotonically decreasing, and its range is $[2+\sqrt{3}, 4]$;
When $1 \leqslant x \leqslant 2$, let $\theta=\arccos \frac{x}{2}$. Then
$$
\begin{array}{l}
f(x)=2 x+\sqrt{4-x^{2}}=4 \cos \theta+2 \sin \theta \\
=2 \sqrt{5} \cos \left(\theta-\arccos \frac{2}{\sqrt{5}}\right),
\end{array}
$$

its range is $[2+\sqrt{3}, 2 \sqrt{5}]$.
In summary, the range of $f(x)$ is $[2+\sqrt{3}, 2 \sqrt{5}]$.