= Solution
At the maximum found in part(a), $e^{-\mu x_m}=C$ and $(1-e^{-\beta})x_m/a=C\mu x_m=-C\ln C$. Substitution into the <exponential slab heating> solution gives
$$
\boxed{T_{\max}-T_0=\frac{q_0}{k\mu^2}\left(1-C+C\ln C\right),
\qquad C=\frac{1-e^{-\mu a}}{\mu a}.}
$$
For $0<C<1$, the function $1-C+C\ln C$ is positive: it vanishes at $C=1$ and has derivative $\ln C<0$. Thus the interior <temperature> exceeds the common surface <temperature>, consistently with positive internal heating.
A convenient dimensionless form uses $\xi=x/a$:
$$
\theta(\xi)=\frac{k(T-T_0)}{q_0a^2}
=\frac{1-e^{-\beta\xi}-(1-e^{-\beta})\xi}{\beta^2}.
$$
As $\beta\to0$, $\theta\to\xi(1-\xi)/2$, giving the check
$$
\boxed{T_{\max}-T_0\longrightarrow\frac{q_0a^2}{8k}.}
$$
\Image[/past-exam-of-the-mphil-in-nuclear-energy-course-of-the-university-of-cambridge/2020/paper-ne2-slab-temperature.png]
{title=Original normalized slab <temperature> profiles calculated from the <exponential slab heating> solution; markers indicate the maxima}
The graph holds the normalization $q_0a^2/k$ fixed. Its decreasing peaks with increasing $\beta$ reflect a decreasing source away from the illuminated face under that comparison; they do not assert the same trend at fixed total absorbed power.
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