Past exam of the mathematics course of the University of Cambridge / 2017 / ib / Paper 1 / 11G / ii / Solution

True. If $|f^{\prime }(x)|\le K$ everywhere, the mean value theorem applied between any two distinct points givesThus $f$ is Lipschitz continuous and hence uniformly continuous. No continuity of $f^{\prime }$ is needed: differentiability of $f$ gives continuity on each finite closed interval and is exactly the hypothesis needed for the mean value theorem there.