Consider the set,$ \ A$ of continuous functions $f:[0,1]\rightarrow \mathbb R$ that satisfy : $\int\_0^{1}f(x)(1-f(x))dx=\frac{1}{4}$,then what is the cardinality of $ \ A$?

I wanted to know how to initiate this problem,I'm thinking of only "**hit and trial,$f(x)=\frac{1}{2}$**" method but it seems very much tedious.

I just wanted to know is there any specific method to solve $\int\_0^{1}f(x)dx=\frac{1}{4}+\int\_0^{1}(f(x))^2dx$...

Please give your suggestions...

thank you

**My attempt**

Let $g(x)=f(x)(1-f(x))$,then clearly $g(x)$ satisfies Rolle's theorem on $[0,1]$,so there exists $c\in [0,1]$ such that $g'(c)=0$

Now,$g'(c)=f'(c)-2f(c)f'(c)=0\implies f'(c)[1-2f(c)]=0\implies f'(c)=0 $ or $f(c)=\frac{1}{2}\ $

Now,if we choose $f(x)=\frac{1}{2}$,then it will satisfy $\int\_0^{1}f(x)(1-f(x))dx=\frac{1}{4}$.

So,**one possibility** is $f(x)=\frac{1}{2}$,but it does not rule out the other possibilities.

*My Confusion:* (1) *How does Rolle's theorem is directing to this possiblity or it is just a mere hoax?*

(2) *If Rolle's theorem is correctly used then why it is not exploring the other possibilities?*