$+1 \mid=k$ has four distinct branches. Find the range of $k$.

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Three, first $k \geqslant 0$. Secondly, consider two cases:
(1) When $x^{2}-2 \sqrt{3} x+1=k$,
$$
\begin{array}{l}
\Delta=12-4(1-k)>0, \\
\therefore k>-2 .
\end{array}
$$
$$
\begin{array}{l}
\text { (2) When } x^{2}-2 \sqrt{3} x+1=-k \text {, } \\
\quad \Delta=12-4(1+k)>0, \\
\therefore k<2 .
\end{array}
$$

Combining the results, we get $0 \leqslant k<2$.