5. For any $x \in \mathbf{R}, [x]$ represents the greatest integer not exceeding the real number $x$. Then the set of $x$ that satisfies $\left[\left|x^{2}-1\right|\right]=10$ is $(\quad)$.
(A) $(-2 \sqrt{3},-\sqrt{11})$
(B) $[\sqrt{11}, 2 \sqrt{3}]$
(C) $(-2 \sqrt{3},-\sqrt{11}] \cup[\sqrt{11}, 2 \sqrt{3})$
(D) $[-2 \sqrt{3},-\sqrt{11}) \cup(\sqrt{11}, 2 \sqrt{3}]$
5. C.

It is known that when $[x] \leqslant x1$, we have
$$
\begin{array}{l}
10 \leqslant x^{2}-1<11 \Rightarrow 11 \leqslant x^{2}<12 \\
\Rightarrow \sqrt{11} \leqslant|x|<2 \sqrt{3} \\
\Rightarrow \sqrt{11} \leqslant x<2 \sqrt{3} \text { or }-2 \sqrt{3}<x \leqslant-\sqrt{11} .
\end{array}
$$