Past exam of the mathematics course of the University of Cambridge / 2023 / ia / Paper 1 / 10E / Solution

The intermediate value theorem says that a continuous $f:[a,b]\to \mathbb{R}$ takes every value between $f(a)$ and $f(b)$. For $f(a)<y<f(b)$, the closed sets where $f\le y$ and $f\ge y$ cannot separate the connected interval; equivalently, taking the supremum of $\{x:f(x)\le y\}$ and using continuity gives a point with value $y$.

Set $\varphi (a)=0$ and $\varphi (x)=\sin (1/(x-a))$ for $x>a$. Every interval $[a,b]$ contains a zero $c<b$; continuity on $[c,b]$ makes $\varphi$ take every value between $0=\varphi (a)$ and $\varphi (b)$, yet $\varphi$ is discontinuous at $a$.

A monotone function can be discontinuous only by a jump. If it had a jump at $c$, any number strictly between the left and right limits would lie between $f(a)$ and $f(b)$ but would not be attained, contrary to the hypothesis. Thus it is continuous.

For the last assertion, pass to subsequences with $x_n,y_n$ both within $1/n$ of $a$ and $g(x_n)$ near $l$, $g(y_n)$ near $L$. For $\lambda \in (l,L)$, the intermediate value theorem on the interval joining $x_n$ and $y_n$ gives $z_n$ with $g(z_n)=\lambda$; then $z_n\to a$. The endpoint cases use the original sequences.

Solved by gpt-5.6-sol high.