= Solution
With length measured in units of $a=1$, plot the real function
$$
f(k)=\cos k+10\frac{\sin k}{k},\qquad f(0)=11.
$$
Only the portions with $|f(k)|\leq1$ can equal $\cos\phi$, so the strip between the horizontal lines $+1$ and $-1$ identifies allowed <energy bands>. The <band edges of a repulsive delta-comb potential> have the following approximate first intervals:
$$
\boxed{\begin{aligned}
2.6277&\leq k\leq\pi,\\
5.3073&\leq k\leq2\pi,\\
8.0671&\leq k\leq3\pi,\\
10.9087&\leq k\leq4\pi.
\end{aligned}}
$$
Each interval corresponds to <energy> $E=\hbar^2k^2/(2m)$; the intervening intervals are forbidden. The exact upper endpoints follow from $f(j\pi)=(-1)^j$. The lower endpoints solve $f(k)=(-1)^{j-1}$ inside $((j-1)\pi,j\pi)$. The divergence suggested by $\sin k/k$ at the origin is removable, and $f(0)=11>1$, so there is no zero-energy state in the repulsive model.
\Image[/past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2019/paper-2-c11-repulsive-bands.png]
{title=Calculated delta-comb discriminant for barrier strength 10, with allowed bands marked by the strip between minus one and plus one}
The periodic barriers produce separated <energy bands> and <band gaps>, unlike the continuous free-particle spectrum. A finite $N$-atom loop has discrete states within these bands. Whether the material conducts still depends on <band filling>: one <Electron> per primitive cell half-fills a spin-degenerate band, while two <Electrons> per cell can fill the first band and place the <Fermi energy> in a gap. Band gaps alone do not establish that every filling is insulating. At high $k$, the perturbing term $10\sin k/k$ becomes relatively small and the spectrum approaches free-electron behaviour in relative terms.
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