How many continuous function $f(x)$ exist such that $\int_{0}^{1}f(x)\big(1-f(x)\big)\mathrm dx = \frac{1}{4}$? [closed]

How many continuous function $f(x)$ exist such that $$\int_{0}^{1}f(x)\big(1-f(x)\big)\mathrm dx = \frac{1}{4}\text?$$

I want to find the number of such functions but I don't know how to proceed. It would be of great help if someone can tell me which theorem to apply or provide with any other hint. Any help is highly appreciated.

$t(1-t) \leq \frac 1 4$ for all $t$. [ This follows from $(t-\frac 1 2)^{2} \geq 0$].

Hence $\frac 1 4 = \int f(1-f) \leq \int \frac 1 4=\frac 1 4$ which implies that equality holds throughout. In particular we must have $f(x)(1-f(x)=\frac 1 4$ and this give $f(x)=\frac 1 2$ for all $x$.

Kavi Rama Murthy's solution was first and I think it is beautiful. I'd like to propose a bit different approach, which requires less insight.

Obviously the constant function $f(x) = 1/2$ works. Let's find whether this is the only solution. Write $$\Delta(x) = f(x)-\frac 12.$$

The equality given says that $$\frac 14 = \int\limits_0^1 \left(\frac 12+\Delta(x) \right)\left(\frac 12-\Delta(x)\right)\, dx = \frac 14 - \int\limits_0^1 \ \Delta(x)^2\, dx.$$ Hence, as $\Delta^2$ is non-negative and continuous, it is $0$ everywhere.