= Solution
The stationary <Schrödinger equation> is
$$
-\frac{\hbar^2}{2m}\psi''(x)+V(x)\psi(x)=E\psi(x).
$$
A nonzero bound-state <wave function> has $\int_{\mathbb R}|\psi|^2dx<\infty$, so it can be rescaled to have integral one. Odd <parity> means $\psi(-x)=-\psi(x)$.
For $-V_0<E<0$, set $k=\sqrt{2m(V_0+E)}/\hbar>0$ and $\kappa=\sqrt{-2mE}/\hbar>0$. Oddness removes the interior cosine solution, and normalizability removes the growing exponentials outside. A continuous odd candidate can therefore be written
$$
\psi(x)=A\begin{cases}
-\sin(ka)e^{\kappa(x+a)},&x\leq-a,\\
\sin(kx),&|x|<a,\\
\sin(ka)e^{-\kappa(x-a)},&x\geq a.
\end{cases}
$$
<Continuity> of the derivative at $a$ gives $k\cos(ka)=-\kappa\sin(ka)$; oddness gives the other boundary condition. A nonzero solution cannot have $\sin(ka)=0$, since the derivative condition would then force its interior coefficient to vanish. Thus, with $\xi=ka$ and $\eta=\kappa a$,
$$
\boxed{\eta=-\xi\cot\xi,\qquad
\xi^2+\eta^2=\frac{2mV_0a^2}{\hbar^2}.}
$$
Physical <bound state>[bound states] require $\xi,\eta>0$. On $0<\xi<\pi/2$, the cotangent condition gives negative $\eta$, so no physical intersection occurs. On $(\pi/2,\pi)$, the curve $-\xi\cot\xi$ rises from zero to positive infinity, while the positive quadrant of the circle decreases. Their first intersection exists exactly when the circle radius $R=\sqrt{2mV_0a^2/\hbar^2}$ exceeds $\pi/2$.
\Image[/past-exam-of-the-options-ii-course-of-the-university-of-oxford/2010-odd-well-bound-states.png]
{title=Odd-parity finite-well energies from a circle and a cotangent curve}
{description=Only positive eta gives a bound state. A circle of radius below pi over two misses the positive cotangent branch, while a larger circle intersects it.}
In particular,
$$
\boxed{2mV_0<\left(\frac{\pi\hbar}{2a}\right)^2
\ \Longrightarrow\ \text{no odd-parity bound state}.}
$$
There is also no normalizable odd state at equality: the putative intersection has $\eta=0$, hence $E=0$, and its exterior tail does not decay. Higher positive branches begin beyond this first <interval> and require still larger $R$.
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