= Solution
For the <symmetric double-delta potential>, take an odd state at $E=-\hbar^2\lambda^2/(2M)$, $\lambda>0$. On $0<x<a$ odd <parity> selects $B\sinh(\lambda x)$. For $x>a$, decay and <continuity> give $B\sinh(\lambda a)e^{-\lambda(x-a)}$; extend to negative $x$ by oddness. At $x=a$ the jump condition is
$$
-\lambda B\sinh(\lambda a)-\lambda B\cosh(\lambda a)=-\frac{B\sinh(\lambda a)}{\Delta}.
$$
For a nonzero state this means $\lambda\Delta[1+\coth(\lambda a)]=1$, and therefore
$$
\boxed{E=-\frac{\hbar^2\lambda^2}{2M},\qquad\tanh(\lambda a)=\frac{\lambda\Delta}{1-\lambda\Delta}}.
$$
Oddness makes the jump at $-a$ follow automatically: both its derivative jump and its required <wavefunction> value have the corresponding opposite signs.
There is a positive solution precisely when $\Delta<a$. To see this, rewrite the equation as
$$
\Delta=\frac{1-e^{-2a\lambda}}{2\lambda}.
$$
The right side decreases strictly from $a$ to zero as $\lambda$ increases from zero to infinity: its derivative has the sign of $(1+2a\lambda)e^{-2a\lambda}-1<0$. Hence for $0<\Delta<a$ there is exactly one positive solution, while $\Delta=a$ is a nonnormalizable zero-energy threshold and $\Delta>a$ gives no odd <bound state>. This verifies the stated sufficiently-small-$\Delta$ condition and identifies its exact boundary, consistent with the <odd bound state of a symmetric double-delta potential>.
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