= Solution
\b[Occupation and the two-dimensional <Fermi energy>.] For an independent single-particle state, the <Fermi-Dirac distribution> is
$$
\boxed{n(\varepsilon)=\frac1{e^{\beta(\varepsilon-\mu)}+1},\qquad\beta=\frac1{k_BT}.}
$$
The <chemical potential> enforces the chosen mean <particle number> through $N=\sum_j n(\varepsilon_j)$, including <spin> states in the sum. It is common to all states in particle-exchange equilibrium and must generally change with <temperature> or field when the total mean number is held fixed. At zero <temperature>, the distribution becomes a <Heaviside step function> at the <Fermi energy>.
Let the sample <area> be $\mathcal S$ and let $g_s$ be its internal <spin> <spin degeneracy>. In the continuum limit, filling the disk $k<k_F$ gives
$$
N=g_s\frac{\mathcal S}{(2\pi)^2}\pi k_F^2,
\qquad k_F^2=\frac{4\pi N}{g_s\mathcal S}.
$$
Therefore
$$
\boxed{\mu(T=0)=\varepsilon_F=\frac{\hbar^2k_F^2}{2m}
=\frac{2\pi\hbar^2}{g_s m}\frac N{\mathcal S}.}
$$
For spin-half <fermions> with both <spin> states available, $g_s=2$ and $\varepsilon_F=\pi\hbar^2N/(m\mathcal S)$; for a spin-polarized gas $g_s=1$. The question does not specify the <area> or <spin degeneracy>, so these conventions must be stated rather than extracting an <energy> from $N$ alone. Although the original text calls $\mathbf k$ <momentum>, the supplied $\hbar^2k^2/(2m)$ dispersion makes it a <wave vector>; physical <momentum> is $\mathbf p=\hbar\mathbf k$. The result agrees with the constant <two-dimensional free-electron density of states>. Discrete level spacings in a finite sample would require a specified boundary spectrum.
\b[One level occupied on average.] Write $\delta=\mu_BB$ and $z=e^{\beta(\varepsilon-\mu)}>0$. The two mean <spin> occupations are
$$
N_+=\frac1{1+ze^{\beta\delta}},\qquad
N_-=\frac1{1+ze^{-\beta\delta}}.
$$
Imposing $N_++N_-=1$ gives $2+z(e^{\beta\delta}+e^{-\beta\delta})=1+z(e^{\beta\delta}+e^{-\beta\delta})+z^2$, hence $z^2=1$. Thus, for every positive <temperature>,
$$
\boxed{\mu=\varepsilon,\qquad
N_+-N_-=-\tanh\left(\frac{\beta\mu_BB}{2}\right).}
$$
The total <magnetic moment> is $\mathcal M=-\mu_B(N_+-N_-)$. Consequently
$$
\boxed{\chi=\left(\frac{\partial\mathcal M}{\partial B}\right)_{B=0}
=\frac{\mu_B^2}{2k_BT}.}
$$
This is the <grand-canonical spin susceptibility of a half-filled single level>. Empty, singly occupied and doubly occupied orbital states are allowed in the <grand canonical ensemble>. At zero field, the singly occupied configurations together have <probability> $1/2$, explaining the factor of two relative to one permanent spin-half moment. If exactly one <Electron> were required at all times, a canonical two-state <partition function> would instead give $\mathcal M=\mu_B\tanh(\beta\mu_BB)$ and $\chi=\mu_B^2/(k_BT)$. That different answer does not implement the stated average-number <constraint>.
\b[Density-of-states formula for the susceptibility.] Here $g$ is the zero-field <single-spin density of states>. Regard its argument $E$ as the full spin-shifted <energy>. The shifted functions are $g_\pm(E)=g(E\mp\mu_BB)$, so
$$
N_\pm(B)=\int_{-\infty}^{\infty}g(E\mp\mu_BB)n(E;\mu(B),T)\,dE.
$$
The total number is even in $B$; thus $\mu(B)=\mu(0)+O(B^2)$ at fixed total number on a smooth branch. More directly, terms from $\mu'(0)$ cancel in the <spin> difference because the two densities coincide at zero field. To first order,
$$
N_+-N_-=-2\mu_BB\int_{-\infty}^{\infty}g'(E)n(E;\mu(0),T)\,dE+O(B^3).
$$
It follows that
$$
\boxed{\chi=2\mu_B^2\int_{-\infty}^{\infty}g'(E)n(E)_{B=0}\,dE.}
$$
For a physical <energy band>, take $g$ to vanish below its lower edge. Provided the endpoint term vanishes, <integration by parts> also gives
$$
\boxed{\frac\chi{2\mu_B^2}=\int_{-\infty}^{\infty}g(E)[-n'(E)]\,dE.}
$$
This form displays the positive response and the narrow thermal window around $\mu$. A sharp band edge is included by the corresponding distributional <derivative> in the first expression. The susceptibility defined here is the <derivative> of total <magnetic moment> with respect to $B$, with units $\mathrm{J\,T^{-2}}$; it is not automatically a dimensionless SI <volume (thermodynamics)> susceptibility. A density-of-states convention per unit <volume (thermodynamics)> likewise gives a response per unit <volume (thermodynamics)>. These are the <field conventions for magnetic susceptibility>.
\b[<Chemical potential> at fixed number.] Define the integrated single-spin state count
$$
G(E)=\int_{-\infty}^E g(u)\,du,
\qquad K_T(E)=-n'(E).
$$
The <probability> kernel $K_T$ has unit <integral>, mean $\mu$, and <variance> $a(k_BT)^2$. Using <integration by parts>,
$$
\frac N2=\int g(E)n(E)\,dE=\int G(E)K_T(E)\,dE.
$$
Expand $G$ about $\mu$. The linear term integrates to zero, while $G''=g'$ gives
$$
\frac N2=G(\mu)+\frac a2(k_BT)^2g'(\mu)+O(T^4).
$$
This is the number form of the <Sommerfeld expansion>. Since $N/2=G(\varepsilon_F)$ at zero <temperature>, set $\mu=\varepsilon_F+\delta\mu$ with $\delta\mu=O(T^2)$:
$$
0=g(\varepsilon_F)\delta\mu+\frac a2(k_BT)^2g'(\varepsilon_F)+O(T^4).
$$
Therefore
$$
\boxed{\mu=\varepsilon_F-\frac a2\frac{g'(\varepsilon_F)}{g(\varepsilon_F)}(k_BT)^2+O(T^4).}
$$
The coefficient can be obtained directly. With $y=(E-\mu)/(k_BT)$,
$$
a=\int_{-\infty}^{\infty}\frac{y^2e^y}{(1+e^y)^2}dy
=4\sum_{\ell=1}^{\infty}\frac{(-1)^{\ell+1}}{\ell^2}
=\boxed{\frac{\pi^2}{3}}.
$$
For example, the sum follows by expanding $e^{-y}/(1+e^{-y})^2$ on $y>0$, using the evenness of the kernel and integrating each exponential. Thus $a/2=\pi^2/6$. A <density of states> increasing with <energy> gives a downward quadratic chemical-potential shift. For the constant two-dimensional <density of states> that shift vanishes to this order; a finite lower band edge still allows exponentially small corrections.
\b[Low-temperature susceptibility at fixed number.] Apply the same kernel expansion to $g$ itself:
$$
\frac\chi{2\mu_B^2}
=g(\mu)+\frac a2(k_BT)^2g''(\mu)+O(T^4).
$$
The shift of $\mu$ must be retained in the first term. Writing $g_F=g(\varepsilon_F)$ and similarly for its <derivatives>,
$$
g(\mu)=g_F+g_F'\delta\mu+O(T^4)
=g_F-\frac a2\frac{g_F'^2}{g_F}(k_BT)^2+O(T^4).
$$
Consequently
$$
\boxed{\frac\chi{2\mu_B^2}
=g_F-\frac a2\left(\frac{g_F'^2}{g_F}-g_F''\right)(k_BT)^2+O(T^4),
\qquad a=\frac{\pi^2}{3}.}
$$
This is the <fixed-number low-temperature Pauli susceptibility>. The $g_F''$ contribution is thermal broadening at fixed <chemical potential>; the $-g_F'^2/g_F$ contribution enforces fixed total number. Both are needed. If the <density of states> were instead defined with both <spin> directions included, the prefactor $2$ would be absorbed into that definition. Only the <spin> response has been calculated; no orbital magnetic response is included.
The expansion assumes $g_F>0$, sufficient smoothness near the <Fermi energy>, and a <temperature> small compared with the <energy> scale over which $g$ changes or encounters a band edge. It does not apply unmodified at a singular <density of states>. For the ideal two-dimensional parabolic band, $g_F'=g_F''=0$, so the leading Pauli response is <temperature> independent, consistently with the earlier state count.
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