In $\mathrm{Az} A B C$ triangle, $\angle A C B=120^{\circ}$ and $A C+C B=2$. Within what limits can the length of side $A B$ vary?
Let's write down the cosine theorem for side $AB$:

$$
\begin{gathered}
A B^{2}=A C^{2}+B C^{2}-2 A C \cdot B C \cos 120^{\circ}=A C^{2}+B C^{2}+A C \cdot B C= \\
=(A C+B C)^{2}-A C \cdot B C=4-A C \cdot B C
\end{gathered}
$$

Now, we just need to notice that $0 < A C \cdot B C \leq \left(\frac{A C+B C}{2}\right)^{2}=1$, from which it follows that $\sqrt{3} \leq A B < 2$.