= Solution
Set $a=\alpha_s(\mu^2)/\pi$, $b=\beta_0/4$, $C=1.99-0.11n_f$ and $L=\ln\mu^2$, treating $n_f$ as fixed. The notation $\alpha_s(\mu^2)$ means the coupling evaluated at energy scale $\mu$, using the squared-scale label. The <one-loop running of the strong coupling> gives $da/dL=-ba^2$. Write $\ell=\ln(q^2/\mu^2)$, so $d\ell/dL=-1$ and
$$
K=1+a+a^2(C-b\ell).
$$
Differentiating both explicit and implicit scale dependence gives
$$
\frac{dK}{dL}=-ba^2-2ba^3(C-b\ell)+ba^2=-2ba^3(C-b\ell).
$$
For nonzero coupling and $\beta_0\ne0$, the <principle of minimal sensitivity> applied to this truncated expression with one-loop running therefore selects
$$
\boxed{\mu_*^2=q^2\exp\left[-\frac{4(1.99-0.11n_f)}{\beta_0}\right]=q^2\exp\left[-\frac{12(1.99-0.11n_f)}{33-2n_f}\right]\quad\text{for QCD}.}
$$
This is the <one-loop stationary scale for the hadronic annihilation correction>. For five active flavours it is approximately $\mu_*^2=0.472q^2$, or $\mu_*=0.687\sqrt{q^2}$. It is meaningful within a fixed-flavour perturbative region; crossing a threshold requires the appropriate matched coupling.
The derivative cancels identically through order $a^2$. The prescription retains the residual order-$a^3$ term generated by differentiating the truncated observable with the one-loop running. If one instead discards every order-$a^3$ contribution, no unique scale is determined: every scale is stationary to the retained accuracy. Two-loop running and uncomputed higher observable coefficients can shift the optimum. For $\beta_0=0$, or the trivial $a=0$ limit, this leading approximation likewise selects no unique scale.
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