= Solution
Monovalency gives one <electron> per atom, so the two-spin-state band is half filled. At zero <temperature>, occupying $-k_F<k<k_F$ gives $2(Na/2\pi)(2k_F)=N$. Hence
$$
\boxed{k_F=\frac{\pi}{2a},\qquad E_F=E(k_F)=\epsilon_0.}
$$
Relative to the band minimum, the <Fermi energy> is $2t$.
The <density of states> obeys $g(\epsilon_0+x)=g(\epsilon_0-x)$. At $\mu=\epsilon_0$, the <Fermi-Dirac distribution> satisfies
$$
f(\epsilon_0+x)+f(\epsilon_0-x)=1.
$$
Pairing the energies above and below the band centre therefore leaves exactly half of the two available states per atom occupied, at every positive <temperature>. Increasing $\mu$ strictly increases the mean particle number, so this solution at fixed half filling is unique. Consequently
$$
\boxed{\mu(T)=\epsilon_0\quad\text{for every }T\text{ in this isolated symmetric-band model}.}
$$
This <particle-hole symmetry of a half-filled bipartite band> argument uses the symmetric <tight binding> band and fixed half filling; it is not a general temperature-independence property of <chemical potential> in metals with additional bands or an asymmetric <density of states>.
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