2. If $x, y \in \mathbf{R}$, and $3 x^{2}+2 y^{2}=6 x$, then the range of the bivariate function $f(x, y)=2 x^{2}+3 y^{2}-4 x-6 y$ is ( ).
(A) $\left[\frac{5}{2}-3 \sqrt{6}, \frac{5}{2}+3 \sqrt{6}\right]$
(B) $[5,10]$
(C) $[2 \sqrt{6}-1,2 \sqrt{6}+1]$
(D) $[7-2 \sqrt{6}, 7+2 \sqrt{6}]$
2.A.

The equation is obtained as $(x-1)^{2}+\left(\frac{y}{\sqrt{\frac{3}{2}}}\right)^{2}=1$.
Let $x-1=\cos \theta, \frac{y}{\sqrt{\frac{3}{2}}}=\sin \theta(-\pi<\theta \leqslant \pi)$, then $x=1+\cos \theta, y=\sqrt{\frac{3}{2}} \sin \theta$.
Thus, $f(x, y)=\frac{5}{2} \sin \theta\left(\sin \theta-\frac{6}{5} \sqrt{6}\right)$, where $-1 \leqslant \sin \theta \leqslant 1$.
It is easy to see that the range of $f(x, y)$ is
$$
\left[\frac{5}{2}-3 \sqrt{6}, \frac{5}{2}+3 \sqrt{6}\right] \text {. }
$$