= Solution
{c}
Regard $V(x)$ as an <electron> <potential energy>, and take $G=2\pi/a$. Each cosine supplies half its coefficient to each of its two <Fourier coefficients>. The nonzero coefficients are
$$
V_{\pm G}=-\pi V_0\delta,\qquad
V_{\pm2G}=\frac{V_0}{2},\qquad
V_{\pm3G}=+\pi V_0\delta.
$$
The <Fourier coefficient> connecting the two <plane waves> at $k=\pm\pi/a$ is $V_G$, so the two-wave <band gap> is
$$
\boxed{\Delta=2\pi|V_0|\,|\delta|,\qquad
\epsilon=\frac{\pi|V_0|}{E_{\rm ZB}}|\delta|.}
$$
At $\delta=0$ the actual <periodic potential> has period $a/2$ and no Fourier harmonic at $G$. The crossing at the artificially halved <Brillouin zone> boundary then has zero <band gap>. Higher-order weak-potential corrections can change the proportionality coefficient at nonzero $\delta$; the displayed coefficient belongs to the given two-wave approximation.
There is also a small-displacement qualification in the source: the claimed positions of the <potential energy> minima are correct to first order in $\delta$, not exactly. Put $t=2\pi x/a$. Then
$$
\frac1{V_0}\frac{dV}{dt}=-2\sin2t+2\pi\delta(\sin t-3\sin3t).
$$
At $\delta=0.01$ and $x/a=1/4-\delta=0.24$, this equals approximately $-2.802\times10^{-3}$, rather than zero. For $V_0>0$, expanding about $x/a=1/4$ and $3/4$ gives minima at $1/4-\delta+O(\delta^3)$ and $3/4+\delta+O(\delta^3)$. This does not alter the leading <Fourier coefficient> or the linear opening of the <band gap>.
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