Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 314 / 3 / d / Solution

For fixed $0<{\mathcal{M}}_0<1$, eliminate $\alpha$ byThe remaining equation isIts left side diverges at both ends and has one minimum, atFor $0<{\mathcal{M}}_0<1$ this minimum lies strictly below the right side, so there are exactly two positive values of $x$, and hence exactly two values of $\alpha$. The branches merge at the marginal point $(\alpha ,{\mathcal{M}}_0)=(1/2,1)$.

On the very hot branch, $\alpha \to \infty$ and ${\mathcal{M}}_0=O({\alpha }^{-2})$. Dropping $1/\alpha$ and ${\mathcal{M}}_0^2$ in the relation from part b gives