= Solution
At nonzero <band hybridization>, the crossing becomes an <avoided crossing>. The separation at $k_c$ is $2\alpha$. If $\alpha\ll E_{\rm fl}$, appreciable mixing occurs only where $|Ak^2-E_{\rm fl}|\lesssim\alpha$; away from that region each branch retains nearly its original orbital character. For $\alpha\sim E_{\rm fl}$, the avoided crossing is broad and both branches are substantially reshaped, as shown in the figure.
Use $E_-$ for the lower branch and $E_+$ for the upper one. For $\alpha>0$,
$$
E_-(k)<E_{\rm fl}<E_+(k),\qquad
\min_k E_+(k)=E_+(0)=\frac{E_{\rm fl}+\sqrt{E_{\rm fl}^2+4\alpha^2}}2.
$$
The lower branch increases towards $E_{\rm fl}$ from below. Thus an indirect gap exists in the exactly flat-band model. In the unbounded-parabola limit its lower bound is
$$
\boxed{\Delta_{\rm ind}=\frac{\sqrt{E_{\rm fl}^2+4\alpha^2}-E_{\rm fl}}2
\simeq\frac{\alpha^2}{E_{\rm fl}}\quad(\alpha\ll E_{\rm fl}).}
$$
A finite <Brillouin zone> gives a finite maximum of $E_-$ below $E_{\rm fl}$ and hence a slightly larger indirect gap. This <hybridization gap> differs from the local direct gap $2\alpha$.
The material can therefore be a <band insulator> when the lower band is full and the upper band empty. With two <spin>-degenerate orbitals per <unit cell>, the required filling is
$$
\boxed{2\text{ electrons per unit cell, i.e. half of the combined four-state capacity}.}
$$
With spin degeneracy removed, replace two by one for each independent spin sector. Other fillings are metallic in this noninteracting model.
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