= Solution
A <Hydrostatic equilibrium>[hydrostatic] <pressure> interval contains $dp/(gM)$ moles of background gas per area. Its absorbing column is $x(p)\,dp/(gM)$, so for a nadir ray
$$
d\tau=\frac{k_0x(p)}{p_0gM}\,p\,dp,
\qquad \mathfrak T(p)=e^{-\tau(p)}.
$$
The assumed Planck-temperature power and <dry adiabat> give $B(p)/B_0=[T(p)/T_0]^{3.5}=p/p_0$. To first order in <optical depth>, $-d\mathfrak T\simeq d\tau$, and the <radiance> solution becomes
$$
\boxed{L\simeq B_0+\frac{k_0B_0}{p_0gM}\int_0^{p_0}
x(p)\left(\frac p{p_0}-1\right)p\,dp.}
$$
For a uniform mixing ratio this also evaluates to $L\simeq B_0[1-k_0xp_0/(6gM)]$. Its total <optical depth> is $\tau_0=k_0xp_0/(2gM)$, so the deficit is $B_0\tau_0/3$.
A pressure-resolved perturbation defines the <infrared absorber sensitivity kernel> by $\delta L=\int K_p(p)\delta x(p)\,dp$, with
$$
\boxed{K_p(p)=\frac{k_0B_0}{p_0gM}p\left(\frac p{p_0}-1\right).}
$$
It is zero at the top, negative in the interior, and zero again at the surface. The magnitude is largest where $dK_p/dp=0$, namely
$$
\boxed{p_{\rm strongest}=p_0/2,\qquad K_p(p_0/2)=-\frac{k_0B_0}{4gM}.}
$$
\b[Sign and coordinate qualification.] Literally the maximum of the signed kernel is zero at the endpoints; $p_0/2$ is its minimum and the maximum of its absolute value. The derivative with respect to a single vertically uniform $x$ is the integrated constant $-k_0B_0p_0/(6gM)$ and has no <pressure> location. For equal-width logarithmic <pressure> perturbations the kernel is instead $K_{\log p}=pK_p$ and its magnitude peaks at $2p_0/3$. The stated pressure-interval convention is therefore essential.
\Image[/past-exam-of-the-physics-of-atmospheres-and-oceans-course-of-the-university-of-oxford/2017-infrared-absorber-sensitivity.png]
{title=Signed infrared absorber sensitivity per pressure interval and per logarithmic pressure interval}
{description=Original normalized kernels vanish at the top and surface. The pressure-interval kernel reaches its most negative value at half the surface pressure; logarithmic-pressure weighting moves the largest-magnitude sensitivity to two thirds of the surface pressure.}
The low top-of-atmosphere sensitivity reflects the small pressure-broadened <absorption coefficient> and small molecular column per $dp$ there; the low surface sensitivity reflects small thermal contrast. The printed phrase that <temperature> decreases with <pressure> reverses the actual stipulated relation: with positive exponent, <temperature> decreases as <pressure> decreases upward. The derivation uses the explicit adiabatic formula.
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