Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2024 / Paper 2 / 3

For the natural logarithm, $f^{\prime }(x)=1/x$. The chord slope is $(\log 2-\log 1)/(2-1)=\log 2$. Hence the mean value theorem point satisfies $1/\xi =\log 2$, givingIt lies strictly between one and two. For example, $\log 2={\int }_1^{2}du/u$ lies strictly between $1/2$ and one.

The chord through the endpoints is the straight line $y=(x-1)\log 2$. At $\xi =1/\log 2$, the tangent line isThese two straight lines have the same slope, as required by the mean value theorem. Since $f^{\prime \prime }(x)=-1/x^2<0$, the natural logarithm is a concave function: its graph lies above the chord and below this tangent line.