= Solution
Write $\rho=a\sqrt{2mV_0}/\hbar>0$. On the first tangent branch, $0<\beta<\pi/2$, the matching relations are equivalent to
$$
\rho^2=\beta^2(1+\tan^2\beta),\qquad \rho=\frac\beta{\cos\beta}.
$$
The function $g(\beta)=\beta/\cos\beta$ is continuous and strictly increasing there:
$$
g'(\beta)=\sec\beta\left(1+\beta\tan\beta\right)>0,\qquad \lim_{\beta\to0^+}g(\beta)=0,\qquad \lim_{\beta\to(\pi/2)^-}g(\beta)=\infty.
$$
The <intermediate value theorem> therefore gives a unique first-branch root for \b[every $\rho>0$]. At this root $\beta<\rho$ and $\alpha=\beta\tan\beta>0$, so the <energy eigenvalue>
$$
E=-\frac{\hbar^2\alpha^2}{2ma^2}
$$
obeys $-V_0<E<0$. This proves the <even bound state in every attractive square well> result.
\Image[/past-exam-of-the-quantum-theory-course-of-the-university-of-oxford/2015-finite-well.png]
{title=Even-state matching in a <finite square well>: the rising tangent branch meets the decaying circle for shallow and deeper wells}
To verify all the requested properties, take the resulting even <wavefunction> with exterior tails written in terms of their boundary values:
$$
\psi(x)=B\begin{cases}\cos(\kappa x),&|x|<a,\\\cos\beta\,e^{-k(|x|-a)},&|x|\geq a.\end{cases}
$$
The matching equation makes both the values and first <derivatives> agree at the endpoints. The exterior tails are <square-integrable> because $k>0$, and the interior interval is finite. More explicitly,
$$
\|\psi\|^2=|B|^2\left[a+\frac{\sin(2\beta)}{2\kappa}+\frac{\cos^2\beta}{k}\right],
$$
so choosing $B$ as the positive inverse square root of the bracket gives <wavefunction normalization>. Thus \b[an even, continuously differentiable, normalizable stationary solution always exists], however small the positive depth.
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