2 Given $M=\left\{(x, y) \mid x^{2}+2 y^{2}=3\right\}, N=\{(x, y) \mid y=$ $m x+b\}$. If for all $m \in \mathbf{R}$, there is $M \cap N \neq \varnothing$, then the range of $b$ is ( ).
(A) $\left[-\frac{\sqrt{6}}{2}, \frac{\sqrt{6}}{2}\right]$
(B) $\left(-\frac{\sqrt{6}}{2}, \frac{\sqrt{6}}{2}\right)$
(C) $\left(-\frac{2 \sqrt{3}}{3}, \frac{2 \sqrt{3}}{3}\right]$
(D) $\left[-\frac{2 \sqrt{3}}{3}, \frac{2 \sqrt{3}}{3}\right]$
For any $m \in \mathbf{R}$, $M \cap N \neq \varnothing$ is equivalent to the point $(0, b)$ being on or inside the ellipse $\frac{x^{2}}{3}+\frac{2 y^{2}}{3}=1$, so
$$
\frac{2 b^{2}}{3} \leqslant 1 \text {, i.e., }-\frac{\sqrt{6}}{2} \leqslant b \leqslant \frac{\sqrt{6}}{2} \text {. }
$$

Therefore, the correct choice is A.