Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 4 / 11G / d / Solution

Every prime in $(\sqrt{X},X]$ is at least $\sqrt{X}$. Consequently the primorial satisfiesCombining this with part (c) and taking logarithms givessoFor $X\ge 2$, the inequality $\log X\le \sqrt{X}$ impliesThus the prime-counting upper bound from a primorial estimate yieldsWe may take $c=1+2\log 4$.

Solved by gpt-5.6-sol high.