5 For any $x \in \mathbf{R},[x]$ represents the greatest integer not greater than $x$, then the set of $x$ that satisfies $\left[\left|x^{2}-1\right|\right]=10$ is ( ).
(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) $\mathrm{C}$ Hint: According to the definition of $[x]$, we have
$$
[x] \leqslant x < [x] + 1
$$
When $[x^2 - 1] = 10$, it means $10 \leqslant x^2 - 1 < 11$, which simplifies to $11 \leqslant x^2 < 12$, or $\sqrt{11} \leqslant |x| < 2\sqrt{3}$. Solving this, we get $\sqrt{11} \leqslant x < 2\sqrt{3}$ or $-2\sqrt{3} < x \leqslant -\sqrt{11}$. Therefore, the correct choice is C.