Past exam of the mathematics course of the University of Cambridge / 2026 / ia / Paper 1 / 10E (Analysis I)

If $f$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a)=f(b)$, then some $c\in (a,b)$ has $f\prime (c)=0$. A maximum or minimum is attained; unless $f$ is constant, one extremum lies inside, where the derivative vanishes.

The mean value theorem gives $f(b)-f(a)=(b-a)f\prime (c)>0$.

Since $f\prime \prime \prime <0$, $f\prime \prime$ decreases from $0$, so $f\prime$ strictly decreases from $1$ and $f(x)<x$. If $f\prime$ never vanished, its decreasing positive limit would make $f$ stay positive, but $f\prime \prime <0$ bounded away from zero after any fixed point forces $f\prime$ negative eventually. Thus $f$ rises to one maximum and then decreases strictly to $-\infty$, crossing zero exactly once; it is positive before that crossing.

$\sin x={\sum }_{n\ge 0}(-1{)}^n{x}^{2n+1}/(2n+1)!$ and $\cos x={\sum }_{n\ge 0}(-1{)}^n{x}^{2n}/(2n)!$. Here $\sin \prime \prime \prime =-\cos$ is negative while $\cos >0$ near zero; applying the preceding shape argument until the first zero of $\cos$ and then the addition identities establishes a first positive zero $x_0$ with positivity before it.

From the addition formula and $\sin x_0=0$, $\cos x_0=-1$, so $2x_0$ is a period. Repeatedly subtract $2x_0$ from a positive period $2x_1$; the remainder in $[0,2x_0)$ is also a period, and positivity plus the addition identities force it to be $0$. Thus $x_1=n{x}_0$; the converse follows from periodicity and oddness.