Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 2 / 33B / c / Solution

For small $\mu$, average $\dot{H}$ around $x=r\cos 2t$, $y=-r\sin 2t$. This gives $⟨\dot{H}⟩=\mu [r^2-(3a+1)r^4/4]$. The unique positive balance is $r=2/\sqrt{3a+1}$, which lies between $1/\sqrt{a}$ and $1$ for $a>1$.

Solved by gpt-5.6-sol high.