Past exam of the mathematics course of the University of Cambridge / 2026 / ia / Paper 1 / 11E / b / Solution

Put $h(x)=|f(0)|+{\int }_0^{x}g(t)|f(t)|dt$. Then $|f|\le h$ and $h\prime \le gh$. If $h>0$, $(\log h)\prime \le g$; integration gives $h(x)\le h(0)e^{{\int }_0^{x}g}$. For $f(0)=0$, apply the same argument after adding $\epsilon$ and let $\epsilon \downarrow 0$.

Solved by gpt-5.6-sol high.