= Solution
The stationary <Schrödinger equation> is $-\hbar^2\psi''/(2m)+V\psi=E\psi$. A nonzero bound-state function is <square-integrable> on the <real line> and can be scaled to $\int_{\mathbb R}|\psi|^2dx=1$. Odd <parity operator>[parity] means $\psi(-x)=-\psi(x)$.
Set $k=\sqrt{2m(V_0+E)}/\hbar$ and $\kappa=\sqrt{-2mE}/\hbar$, both positive. <parity operator>[Oddness] and decay give
$$
\psi(x)=\begin{cases}-B e^{\kappa(x+a)},&x\le-a,\\A\sin(kx),&|x|<a,\\B e^{-\kappa(x-a)},&x\ge a.\end{cases}
$$
<Continuity> at $a$ gives $B=A\sin(ka)$, and <derivative> <continuity> gives $-\kappa B=Ak\cos(ka)$. A nonzero solution cannot have $\sin(ka)=0$, since that would force both the value and <derivative> there to vanish. Dividing therefore yields $\kappa=-k\cot(ka)$. With $\eta=\kappa a$ and $\xi=ka$,
$$
\boxed{\eta=-\xi\cot\xi,\qquad\eta^2+\xi^2=2mV_0a^2/\hbar^2.}
$$
The matching at $-a$ follows automatically by odd <parity operator>[parity].
In the required sketch, the second equation is a first-quadrant circle arc of radius $R=\sqrt{2mV_0a^2}/\hbar$. The curve $-\xi\cot\xi$ starts with limiting value $-1$ at zero, rises to zero at $\pi/2$, and then rises to $+\infty$ as $\xi\uparrow\pi$. Thus a positive-$\eta$ intersection in this range must have $\xi>\pi/2$. The accompanying plot shows the circle below threshold and one above it.
\Image[/past-exam-of-the-probability-statistics-and-options-ii-course-of-the-university-of-oxford/2010-as2-odd-well.png]
{title=Odd bound-state condition and the threshold circle}
Consequently $R<\pi/2$ rules out all odd states: the first positive branch already requires that radius, and later branches require still larger $\xi$. This is precisely $\boxed{2mV_0<(\pi\hbar/(2a))^2}$. At equality the intersection has $\eta=0$ and would not decay at infinity, so it too is not a <normalizable> <bound state>. An odd state first appears only for a strict increase beyond threshold.
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