Given, $f:[\frac{1}{2},2] \to \mathbb{R} $ a strictly increasing function.
Let, $g(x)=f(x)+f(\frac{1}{x})$ ,$\forall x \in [1,2]$
(1) there is a suitable $f$ such that $U(P,g)=L(P,g)$, where $U(P,g)$ is upper Riemann sum and $L(P,g)$ is lower Riemann sum.
(2) there is suitable $f$ such that $U(P,g)\neq L(P,g)$
Option (1) is clearly true by taking $f=x$.
But what about option (2)?
As, $f$ is strictly increasing, it can have atmost countable number of discontinies.
So, then, $g(x)$ also have atmost countable number of discontinies $\implies$ $g$ is Riemann integrable.
But, I don't understand,how option (2) is given as true in answer?
Edit : in comment, I have seen option (1) is true only if $f$ is constant. But, $f$ is strictly increasing, so, $f$ can't be constant. But option (1) is also given as true in answer.
Please help me by giving an example so that option (1) is also true.
Actually, if $f(x)=x$, then $f(x)=x+\frac1x$, which is strictly increasing on $[1,2]$. Therefore, for any partition $P$ of $[1,2]$, $L(P,g)<U(P,g)$. This shows that the first option is false and that the second option is true.