= Solution
Use adjacent segments $[-\ell,0]$ and $[0,\ell]$, with $\ell=2\pi/k_1$. Integrate the threshold <probability density> from part (d). The left segment contains two complete density periods, so its average $\cos^2$ is $1/2$:
$$
P_-\propto\int_{-\ell}^{0}4|A|^2\cos^2(k_1x)\,dx
=2|A|^2\ell=\frac{4\pi|A|^2}{k_1}.
$$
The right-hand density is constant:
$$
P_+\propto\int_0^\ell4|A|^2\,dx
=4|A|^2\ell=\frac{8\pi|A|^2}{k_1}.
$$
The common stationary-state normalization cancels, yielding
$$
\boxed{\frac{P_-}{P_+}=\frac12,\qquad P_-:P_+=1:2.}
$$
The larger local probability weight on the right is compatible with zero transmitted <probability current> at threshold: density and flux measure different quantities.
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