Let $f\in C[0,1]$ such that $\int_{0}^{1} f(x)(1-f(x))={1\over4}$. Does it imply $f(x)={1\over2}$?

Question

My question is does there exist any such real valued continuous function $f$ on $[0,1]$ satisfying the property $\int_{0}^{1} f(x)(1-f(x))={1\over4}$ other than the function $x\mapsto{1\over2}$?
By intuition, it seems to be no. Observe the function $g(x)=x(1-x)$ on $[0,1]$ has maximum value ${1\over4}$ at ${1\over2}$.
So, $g(f(x))=f(x)(1-f(x))\le{1\over4}\implies \int_{0}^{1} g(f(x))=f(x)(1-f(x))\le{1\over4}$
Now, if $f(x)>{1\over2}$ for some $x\in[0,1]$, there is some interval in $[0,1]$ containing $x$ where $f(x)>{1\over2}$. I want to show in such cases the required integral should be STRICTLY less than ${1\over4}$. But I cannot show it.
Is there any way out to solve the problem? I even don't know whether my guess is correct or not. Thanks for assistance in advance.

Answer 1

\begin{align*} \int_{0}^{1}f(x)(1-f(x))dx=\int_{0}^{1}-(f(x)-1/2)^{2}+1/4dx, \end{align*} if it were the integral to be $1/4$, then \begin{align*} \int_{0}^{1}-(f(x)-1/2)^{2}dx=0. \end{align*} So $f(x)=1/2$ by the continuity.