= Solution
{c}
Let the sample area be $A$, set the band touching <energy> to zero, and write $m^*=m_e^*=m_h^*$. Assume one <electronic valley> in each <energy band> with two <spin> states. The near-edge <parabolic dispersions> are
$$
E_c(k)=\frac{\hbar^2k^2}{2m^*},\qquad E_v(k)=-\frac{\hbar^2k^2}{2m^*}.
$$
Under <periodic boundary conditions>, each <spin> occupies reciprocal-space area $(2\pi)^2/A$. Counting the <quantum states> inside a circle gives
$$
\mathcal N_c(E)=2\frac{A}{(2\pi)^2}\pi k^2
=\frac{Am^*}{\pi\hbar^2}E\qquad(E>0).
$$
Its <derivative> is the constant <two-dimensional free-electron density of states>. Defining $D=Am^*/(\pi\hbar^2)$, the signed-<energy> <density of states> is
$$
\boxed{g_c(E)=D\Theta(E),\qquad g_v(E)=D\Theta(-E).}
$$
The total electronic <density of states> is therefore $D$ on either side of zero, not $2D$ at every signed <energy>. If instead one counts positive-<energy> excitations, the <electron> and <hole> species each have <density of states> $D$, giving $2D$ when summed. A common additional <valley degeneracy> multiplies all these quantities by that factor. The <effective mass> approximation assumes thermally sampled <energies> remain within the near-edge regions of both <energy bands>.
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