Past exam of the mathematics course of the University of Cambridge / 2018 / ia / Paper 1 / 10F / b / Solution

The mean value theorem says that if $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then $f(b)-f(a)=f^{\prime }(c)(b-a)$ for some $c\in (a,b)$.

For $F(t)=\cos (e^{-t})$ and $t\ge 0$,The mean value theorem gives $|\cos (e^{-x})-\cos (e^{-y})|\le |x-y|$.

For the second inequality put $t=\sqrt{1+x}\ge 1$. It becomes $2\log t\le t-t^{-1}$. The difference $G(t)=t-t^{-1}-2\log t$ satisfies $G(1)=0$ andThus $\log (1+x)\le x/\sqrt{1+x}$.