= Solution
Above $T_B$, essentially all $N$ particles occupy the excited levels, so
$$
\boxed{\langle E\rangle\simeq N\epsilon,\qquad C\simeq0
\quad(T>T_B).}
$$
Below $T_B$, the ground-state condensate carries no energy and $\mu\simeq0$, hence
$$
\boxed{\langle E\rangle\simeq\frac{M\epsilon}{e^{\beta\epsilon}-1}
\quad(T<T_B).}
$$
Differentiating gives the <heat capacity>
$$
\boxed{C=\frac{\partial\langle E\rangle}{\partial T}
\simeq\frac{M\epsilon^2e^{\beta\epsilon}}
{kT^2(e^{\beta\epsilon}-1)^2}
\quad(T<T_B).}
$$
The displayed formula in the examination paper omits the factor $e^{\beta\epsilon}$. Its stated expression follows after the additional high-temperature approximation $\beta\epsilon\ll1$, for which $e^{\beta\epsilon}\simeq1$.
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