= Solution
Write $t_s=|\beta_s|$ and $t_p=|\beta_p|$, so
$$
E_s=\alpha_s-2t_s\cos(ka),\qquad
E_p=\alpha_p+2t_p\cos(ka).
$$
The labelled key points of the two <dispersion relations> are
$$
\begin{array}{c|c|c}
k&E_s&E_p\\\hline
0&\alpha_s-2t_s&\alpha_p+2t_p\\
\pi/(2a)&\alpha_s&\alpha_p\\
\pi/a&\alpha_s+2t_s&\alpha_p-2t_p
\end{array}
$$
The $s$ band rises from centre to boundary and the axial $p$ band falls. The separated-band assumption requires $\alpha_p-\alpha_s>2(t_s+t_p)$. The upper-left panel shows a representative pair, with the other panels supplying the requested comparisons in part (b).
\Image[/past-exam-of-the-chemistry-a-course-of-the-university-of-cambridge/2022/paper-2-sp-bands.png]
{title=Original one-dimensional s and axial-p band sketches, comparing separated bands, mixing, and the equal-on-site-energy special case}
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