= Solution
Differentiating the <Gaussian variational bound for an attractive Gaussian well> gives
$$
E'(a)=\frac12-\frac{V_0}{2\sqrt a(1+a)^{3/2}}.
$$
Thus an interior <stationary point> obeys
$$
\boxed{\sqrt{a_*}(1+a_*)^{3/2}=V_0,\qquad a_*(1+a_*)^3=V_0^2.}
$$
The left side of the first equation increases strictly from zero to infinity, so there is exactly one such point. More explicitly,
$$
E''(a)=\frac{V_0(1+4a)}{4a^{3/2}(1+a)^{5/2}}>0,
$$
so $E$ is a <strictly convex function> on $a>0$ and $a_*$ is its global minimum. At that minimum the stationarity equation gives
$$
E(a_*)=\frac{a_*}{2}-a_*(1+a_*)=-\frac{a_*}{2}-a_*^2<0.
$$
There is a continuous limiting zero at $a=0$, although $y_T=1$ at exactly $a=0$ is not a <square-integrable function> and hence is not an admissible trial. For a positive zero, division by $\sqrt a$ and squaring gives $a(1+a)=4V_0^2$, so
$$
\boxed{a_z=\frac{\sqrt{1+16V_0^2}-1}{2}}
$$
is the unique positive zero. The other quadratic root is negative and inadmissible. Near zero,
$$
E(a)=-V_0\sqrt a+\frac a2+O(a^{3/2}),
$$
so the graph leaves the origin with slope $-\infty$, falls to its unique negative minimum, rises through $a_z$, and grows as $a/2-V_0+O(a^{-1})$ at infinity.
\Image[/past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2025/paper-1-gaussian-variational-energy.png]
{title=The <Gaussian variational bound for an attractive Gaussian well> at $V_0=1$, showing its negative minimum and positive zero}
By the <Rayleigh-Ritz variational principle>,
$$
\boxed{E_0\leq E(a_*)=-\frac{a_*}{2}-a_*^2<0.}
$$
Thus the attractive well supports a negative-energy <bound state> for every $V_0>0$.
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