 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 givesFor $0<C<1$, 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 $\xi =x/a$:As $\beta \to 0$, $\theta \to \xi (1-\xi )/2$, giving the check

The graph holds the normalization $q_0{a}^2/k$ 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.