$ \frac{1}{2} \int_{0}^{2} f(x) d x
Let $f:[0,2] \rightarrow \mathbb{R}$ be a continuous function such that
$$
\frac{1}{2} \int_{0}^{2} f(x) d x<f(2)
$$
Then which of the following statements must be true?
(A) $f$ must be strictly inereasing.
(B) $f$ must attain a maximum value at $x=2$
(C) $f$ cannot have a minimum at $x=2$
(D)None of the above.
My work
I have taken an example and tried to solve.Like $x^{2}$. which is satisfying the condition. So I am getting option (A) as answer. But in the meantime I have also came across a counter example $|x-1|$. It also satisfies the condition but not strictly increasing.

I am looking for:
$1$.Is there any general approach to solve the problem?
$2$.Ans of this question
 A: I think this is the easy way to see that (C) is the answer:
Assume that $f$ does have a minimum at $x=2$ and let's denote $f(2)=k$, so that $k \leq f(x) \forall x \in [0, 2]$.
Then $2k=\int_0^2 k~dx \leq \int_0^2 f(x) ~ dx \Rightarrow k =f(2) \leq \frac 12 \int_0^2f(x)~dx$, contradicting our hypothesis.
A: $\bullet~ $ $\textbf{Lemma:}~$ Let $f$ be a real-valued function defined on a ${\textbf{metric space}}$ $S$ to $~\mathbb{R}^{k}.~$Assume that $f$ is continuous on a $~\textbf{compact subset }$ $X$ of $S$. Then the function $f$ always attends it's maximum and minimum on $X$, namely $$~\sup(f(X)) ~\textit{ and }~\inf(f(X))$$
$\bullet~$ $\textbf{In this case:}$
The function $f$ is defined on $[0, 2]$ $\subseteq_{\mathrm{C}}$ $\mathbb{R}.$ [where $\subseteq_{\mathrm{C}}$ means compact subset]
Now by the Fundamental Theorem of Calculus, there exists $F$ such that
$$ F'(x) = f(x) ~\text{ for } x \in [0, 2]$$
Therefore we have
$$ \int_{0}^{2} f(t)dt = \int_{0}^{2} F'(t)dt = F(2) - F(0) $$
Now by FTC, we have that $F$ is continuous on $[0, 2]$ and differentiable on $[0, 2]$. Therefore by LMVT, we have that
$$ \frac{1}{2} \int_{0}^{2} f(t)dt = \frac{F(2) - F(0)}{2} = F'(c) = f(c) \quad \text{for some } c \in (0, 2) $$
Now the given condition is reduced to
$$ f(c) < f(2) \quad \text{for some } c \in (0, 2) $$
Now as $f$ is continuous on $[0, 2]$, a compact subset of $\mathbb{R}$, by our $\textbf{Lemma},$ we have
$$ \inf\{f(x) : x \in [0, 2] \} \leqslant f(x) \leqslant \sup\{f(x) : x \in [0,2] \}$$
$\circ$ But we have $f(c) < f(2)$ for some fixed $c$ $\in$ $(0,2)$, not any arbitrary $c$. Therefore we can't comment on the monotonocity of $f$. $\implies$ $(a)$ is not true.
$\circ$ As $c$ is not arbitrary, then by our $\textbf{Lemma}$ we can say that
$$ f(c) < f(2) \leqslant \sup\{f(x) : x \in [0,2] \} \quad \text{for some } c \in (0, 2)$$
But can't surely say that $f(2) = \sup \{ f(x) : x \in [0, 2] \}$ $\implies$ $(b)$ is not true.
$\circ$ By our $\textbf{Lemma}$ we have that
$$ \inf \{ f(x) : x \in [0. 2] \} \leqslant f(c) < f(2) \quad \text{for some } c \in (0, 2)  $$
Which implies $(c)$ is true.
Hence we are done!
I hope this helps @Integral Calculus.
