= Solution
{c}
Put $A=\sum_i x_i$, $B=\sum_i x_i^2$ and $M_2=B/n$. The <normal distribution> model gives
$$
L(\mu)=(2\pi)^{-n/2}\mu^{-n}\exp\!\left\{-\frac{B-2\mu A+n\mu^2}{2\mu^2}\right\},\qquad\mu>0,
$$
and, up to a constant independent of $\mu$, its <log-likelihood function> is
$$
\ell(\mu)=-n\log\mu-\frac{B}{2\mu^2}+\frac A\mu.
$$
The <score function> and its zero equation are
$$
\ell'(\mu)=\frac{B-A\mu-n\mu^2}{\mu^3},\qquad
n\mu^2+A\mu-B=0.
$$
For $B>0$, this <quadratic equation> has exactly one positive root, giving the <maximum-likelihood estimator>
$$
\boxed{\widehat\mu=\frac{-A+\sqrt{A^2+4nB}}{2n}
=\frac{-\overline X+\sqrt{\overline X^{\,2}+4M_2}}2.}
$$
The numerator $B-A\mu-n\mu^2$ of the <score function> is positive between its negative and positive roots and negative above its positive root. Thus the <likelihood function> increases up to $\widehat\mu$ and decreases afterwards: this is its unique global maximum on $(0,\infty)$. Also, $\ell(\mu)\to-\infty$ at both ends. The all-zero observation vector has <probability> zero; on that exceptional vector $L(\mu)\propto\mu^{-n}$ and there is no attained maximum.
The second <derivative> and the relevant <moments> are
$$
\ell''(\mu)=\frac n{\mu^2}-\frac{3B}{\mu^4}+\frac{2A}{\mu^3},\qquad
\mathbb E_\mu A=n\mu,\qquad
\mathbb E_\mu B=n(\mu^2+\mu^2)=2n\mu^2.
$$
Therefore the expected <Fisher information> is
$$
I_n(\mu)=-\mathbb E_\mu\ell''(\mu)=\frac{3n}{\mu^2}.
$$
The <support of a probability distribution> is the whole <real line>, independent of the positive <statistical parameter>. The <law of large numbers> gives $\overline X\to\mu$ and $M_2\to2\mu^2$ in <probability>, so the displayed <maximum-likelihood estimator> is a <consistent estimator>. The regular <asymptotic normality of a maximum likelihood estimator> now yields
$$
\boxed{\sqrt n(\widehat\mu-\mu)\xrightarrow dN(0,\mu^2/3),\qquad
\widehat\mu\mathrel{\dot\sim}N\!\left(\mu,\frac{\mu^2}{3n}\right).}
$$
Thus the requested leading <asymptotic variance> on the unscaled estimator's scale is $\mu^2/(3n)$. The <normal mean proportional to standard deviation model> provides a reusable version of this calculation.
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