2. Given unit vectors $\boldsymbol{a} 、 \boldsymbol{b}$ satisfy $a \perp b$, vector $\boldsymbol{c}$

satisfies
$$
|c-a|+|c-2 b|=\sqrt{6} .
$$

Then the range of $|\boldsymbol{c}-\boldsymbol{a}|$ is $\qquad$ .
2. $\left[\frac{\sqrt{6}-\sqrt{5}}{2}, \frac{\sqrt{6}+\sqrt{5}}{2}\right]$.

Let $O$ be the origin, and denote
$$
\begin{aligned}
a & =\overrightarrow{O A}=(1,0), 2 b=\overrightarrow{O B}=(0,2), \\
c & =\overrightarrow{O C} .
\end{aligned}
$$

From $|\boldsymbol{c}-\boldsymbol{a}|+|\boldsymbol{c}-2 \boldsymbol{b}|=\sqrt{6}$, we know that the locus of point $C$ is an ellipse with foci at $A$ and $B$, a major axis length of $\sqrt{6}$, and a focal distance of $\sqrt{5}$. $|\boldsymbol{c}-\boldsymbol{a}|$ is one of its focal radii.

Therefore, $|c-a| \in\left[\frac{\sqrt{6}-\sqrt{5}}{2}, \frac{\sqrt{6}+\sqrt{5}}{2}\right]$.