Past exam of the mathematics course of the University of Cambridge / 2026 / ia / Paper 1 / 10E / c / Solution

Since $f\prime \prime \prime <0$, $f\prime \prime$ decreases from $0$, so $f\prime$ strictly decreases from $1$ and $f(x)<x$. If $f\prime$ never vanished, its decreasing positive limit would make $f$ stay positive, but $f\prime \prime <0$ bounded away from zero after any fixed point forces $f\prime$ negative eventually. Thus $f$ rises to one maximum and then decreases strictly to $-\infty$, crossing zero exactly once; it is positive before that crossing.