= Solution
The commutator follows directly on a test <function>:
$$
[f,p]=i\hbar f'.
$$
Expansion gives
$$
V_+=\frac{f^2-\hbar f'}{2m},\qquad V_-=\frac{f^2+\hbar f'}{2m}.
$$
To make $V_-=2m$, take
$$
f(x)=2m\tanh(2mx/\hbar).
$$
Then
$$
V_+(x)=2m-4m\operatorname{sech}^2(2mx/\hbar),
$$
which tends to $2m$ at both ends.
Applying $p+if$ to $e^{ikx}$ gives a partner scattering state with asymptotic amplitudes $\hbar k-2mi$ at $-\infty$ and $\hbar k+2mi$ at $+\infty$, and no reflected wave. Their <moduli> agree. Subtracting the common constant $2m$ therefore shows that the displayed $-4m\operatorname{sech}^2$ potential is reflectionless:
$$
\boxed{\mathcal R=0,\qquad \mathcal T=1.}
$$
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