3. If the circle $x^{2}+y^{2}=k^{2}$, at least covers one maximum point and one minimum point of the function $f(x)=\sqrt{3} \sin \frac{\pi x}{k}$, then the range of $k$ is ( ).
A. $|k| \geqslant 3$
B. $|k| \geqslant 2$
C. $|k| \geqslant 1$
D. $1 \leqslant|k| \leqslant 2$
3. B

Since $f(x)=\sqrt{3} \sin \frac{x x}{k}$ is an odd function, its graph is symmetric about the origin, so the circle $x^{2}+y^{2}=k^{2}$ only needs to cover one of the extreme points of $f(x)$. Let $\frac{\pi x}{k}=\frac{\pi}{2}$, solving this gives the nearest maximum point to the origin $P\left(\frac{k}{2}, \sqrt{3}\right)$. From the problem, $k^{2} \geqslant\left(\frac{k}{2}\right)^{2}+(\sqrt{3})^{2}$, we get $|k| \geqslant 2$.