If $f$ be a real valued function defined for all real numbers $x$ such that for some fixed $a > 0$, it satisfies $f(x + a) = \dfrac{1}{2} + \sqrt{f(x) - (f(x))^2} \, \forall x$, then $f(x)$ is periodic with period

Given $f(x + a) = \dfrac{1}{2} + \sqrt{f(x) - (f(x))^2}$

Rearranging the terms, we get:

$f(x + a) - \dfrac{1}{2} = \sqrt{f(x) - (f(x))^2}$

Squaring both sides:

$\left(f(x + a) - \dfrac{1}{2}\right)^2 = f(x) - (f(x))^2$

$\left(f(x + a) - \dfrac{1}{2}\right)^2 = \dfrac{1}{4} - \left(f(x) - \dfrac{1}{2}\right)^2$

$\left(f(x + a) - \dfrac{1}{2}\right)^2 + \left(f(x) - \dfrac{1}{2}\right)^2 = \dfrac{1}{4}$

Replacing $x$ by $x + a$, we obtain:

$\left(f(x + 2a) - \dfrac{1}{2}\right)^2 + \left(f(x + a) - \dfrac{1}{2}\right)^2 = \dfrac{1}{4}$

Subtracting the first equation from the second equation:

$\left(f(x + 2a) - \dfrac{1}{2}\right)^2 - \left(f(x) - \dfrac{1}{2}\right)^2 = 0$

$\left(f(x + 2a) - \dfrac{1}{2}\right)^2 = \left(f(x) - \dfrac{1}{2}\right)^2$

From the given equation, $f(x + a) = \dfrac{1}{2} + \sqrt{f(x) - (f(x))^2} \ge \dfrac{1}{2}$ for all $x$. Thus, $f(x) \ge \dfrac{1}{2}$ for all $x$, which implies $f(x) - \dfrac{1}{2} \ge 0$.

Taking the positive square root on both sides:

$f(x + 2a) - \dfrac{1}{2} = f(x) - \dfrac{1}{2}$

$f(x + 2a) = f(x)$

Therefore, $f(x)$ is a periodic function with period $2a$.

Answer: $2a$