{c}

Suppose distinct <fixed points> $x<y$ existed. Then $h(t)=f(t)-t$ satisfies $h(x)=h(y)=0$. <Rolle's theorem> supplies $c\in(x,y)$ with $h'(c)=0$, hence $f'(c)=1$, contrary to the hypothesis. Thus \b[there is at most one <fixed point>]. The argument does not require $f'$ to be <continuous>.

Solution

Past exam of the mathematics course of the University of Oxford / 2023 / Prelims / Mathematics II / 6 / b / i / Solution

 Solution (source code)