= Solution
Write $s=S_0$, assume the usual nonzero initial price vector, and define
$$
A=s^TV^{-1}s,\quad B=s^TV^{-1}\mu,\quad C=\mu^TV^{-1}\mu,\quad
D=AC-B^2.
$$
The <covariance matrix> $V$ is positive definite. If $s,\mu$ are linearly independent, then $D>0$ by the strict <Cauchy-Schwarz inequality>. The <Lagrangian> for <mean-variance optimization> can be written
$$
\mathcal L(\pi,\lambda,\eta)=\frac12\pi^TV\pi
-\lambda(s^T\pi-x)-\eta(\mu^T\pi-m).
$$
Its first-order equation gives $\pi=\lambda V^{-1}s+\eta V^{-1}\mu$. The two constraints then give
$$
\boxed{\pi=\frac{Cx-Bm}{D}\pi_A+\frac{Am-Bx}{D}\pi_B,\qquad
\pi_A=V^{-1}s,\quad\pi_B=V^{-1}\mu.}
$$
Strict convexity makes this the unique constrained minimizer. Its <variance> is
$$
\boxed{v_{\min}(m)=\frac{Cx^2-2Bxm+Am^2}{D}
=\frac{x^2}{A}+\frac A D\left(m-\frac{Bx}{A}\right)^2.}
$$
The <mean-variance efficient frontier> is the upper branch $m\ge Bx/A$: points on the lower branch have a larger-mean competitor of the same <variance>. The picture shows the generic nondegenerate case at a fixed initial budget. With only two stocks and independent budget and mean constraints, every feasible portfolio lies on the curve; additional independent portfolio directions allow larger variances at the same mean.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2010/ii/paper-4-frontier.png]
{title=Mean–variance frontier with its efficient upper branch and minimum-variance portfolio}
If $\mu=\kappa s$, then $D=0$, and only $m=\kappa x$ is feasible. The <variance> minimizer is $(x/A)V^{-1}s$, still in the stated span; no two-dimensional frontier exists. This degeneracy is not excluded merely by nonsingularity of $V$.
For exponential utility and Gaussian prices, the <moment-generating function> gives
$$
\mathbb EU(\pi^TS_1)
=-\exp\left(-\gamma\mu^T\pi+\frac{\gamma^2}2\pi^TV\pi\right).
$$
Maximizing this is equivalent to maximizing $\mu^T\pi-\gamma\pi^TV\pi/2$ subject to $s^T\pi=y$. Its strictly concave first-order equation gives
$$
\boxed{\pi=\frac1\gamma V^{-1}\mu-\frac\eta\gamma V^{-1}s,\qquad
\eta=\frac{B-\gamma y}{A}.}
$$
Thus the same two funds suffice.
For the <Gaussian two-fund theorem> with general utility, the Gaussian assumption and the exponential growth bounds justify the differentiations and expectations. Gaussian integration by parts applied to $S_1=\mu+V^{1/2}Z$ gives
$$
\mathbb E[S_1U'(\pi^TS_1)]
=\mu\,\mathbb EU'(\pi^TS_1)+V\pi\,\mathbb EU''(\pi^TS_1).
$$
At any finite maximizer the left side is a multiplier $\eta s$. If $U$ is concave and non-affine, and $\pi\ne0$, then $\pi^TS_1$ is a nondegenerate Gaussian with positive density everywhere. Since continuous $U''\le0$ is strictly negative on some interval, its expectation is strictly negative. Rearranging proves
$$
\boxed{\pi=
\frac{\eta V^{-1}s-\mathbb EU'(\pi^TS_1)V^{-1}\mu}
{\mathbb EU''(\pi^TS_1)}
\in\operatorname{span}\{\pi_A,\pi_B\}.}
$$
The zero portfolio already belongs to that span.
There is a genuine qualification in the final printed assertion: increasing and concave does not exclude affine utility. For $U(w)=w$, $s=\mu=(1,1,1)^T$, and $V=I$, every budget-feasible portfolio is optimal. With $y=1$, $(1,0,0)^T$ is optimal but is outside the common line spanned by $\pi_A,\pi_B$. If $\mu$ is not proportional to $s$, affine utility has no maximizer at all, because a zero-cost direction with positive mean can be scaled without bound. Thus the universal claim about the maximizing portfolio needs non-affine utility and existence of a finite maximizer. Under the literal weaker assumptions there is still a two-fund optimum whenever an optimum exists: the affine constant-mean case permits choosing the minimum-variance portfolio.
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