China Mathematical Competition (Hainan)

For any $m\in \mathbb{R}$ we have $M\cap N\ne \varnothing$, which means point $(0,b)$ is on or in the ellipsoid $\frac{x^2}{3}+\frac{2y^2}{3}=1$. Therefore $$\frac{2b^2}{3}\le 1,\text{ or }-\frac{\sqrt{6}}{2}\le b\le \frac{\sqrt{6}}{2}.$$ Answer: A.