若关于 $x$ 的不等式 $\left(1+k^{2}\right) x \leqslant k^{4}+4$ 的解集是 $M$, 则对任意实常数 $k$, 总有 ( )
$\because\left(1+k^{2}\right) x \leqslant k^{4}+4, \quad \therefore x \leqslant \frac{k^{4}+4}{1+k^{2}}$.

$\because \frac{k^{4}+4}{1+k^{2}}=\frac{\left(1+k^{2}\right)^{2}-2\left(1+k^{2}\right)+5}{1+k^{2}}=\left(1+k^{2}\right)+\frac{5}{1+k^{2}}-2 \geqslant 2 \sqrt{5}-2$.

$\therefore x \leqslant 2 \sqrt{5}-2, M=\{x \mid x \leqslant 2 \sqrt{5}-2\}, \quad \therefore 2 \in M, 0 \in M$.