= Solution
{c}
For the stated inward heating, take $Q\geq0$ and put $\tau=\pi^2kt/(16\ell^2)$. Since $|\cos(\mu_nx)|\leq1$, the transient from <5/b/ii> satisfies
$$
|T(x,t)-T_s(x)|\leq\frac{16Q\ell}{k\pi^2}
\sum_{n=0}^\infty\frac{e^{-(2n+1)^2\tau}}{(2n+1)^2}.
$$
Equality holds at $x=0$, where every cosine is one and all coefficients have the same sign. Thus the maximum is attained at the heated end, and its exact value and late-time form are
$$
\begin{aligned}
\max_{0\leq x\leq2\ell}|T(x,t)-T_s(x)|
&=\frac{16Q\ell}{k\pi^2}e^{-\tau}
\left(1+\frac19e^{-8\tau}+O(e^{-24\tau})\right)\\
&\sim\boxed{\frac{16}{\pi^2}\frac{Q\ell}{k}
\exp\!\left(-\frac{\pi^2}{16}\frac{kt}{\ell^2}\right)}.
\end{aligned}
$$
This is dominance of the slowest-decaying <eigenfunction> when $kt\gg\ell^2$. More explicitly, the omitted relative correction is nonnegative and bounded by $(\pi^2/8-1)e^{-8\tau}$, using $\sum_{n\geq0}(2n+1)^{-2}=\pi^2/8$. For a signed imposed <heat flux>, the maximum-deviation formula uses $|Q|$.
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