If $$f: R \to R$$ is a function defined by $$f(x) = x - 1\cos\frac{2x-1}{2}\pi$$, where $$[\cdot]$$ denotes the greatest integer function, then $$f$$ is:

We have $$f(x) = [x] \cos\left(\frac{2x - 1}{2}\pi\right)$$, where $$[x]$$ denotes the greatest integer function.

We know that $$[x]$$ is discontinuous at every integer. So we check the behaviour of $$f$$ at any integer $$n$$.

At $$x = n$$ (an integer): $$f(n) = [n] \cdot \cos\left(\frac{2n - 1}{2}\pi\right) = n \cdot \cos\left(n\pi - \frac{\pi}{2}\right)$$.

Now $$\cos\left(n\pi - \frac{\pi}{2}\right) = \cos(n\pi)\cos\left(\frac{\pi}{2}\right) + \sin(n\pi)\sin\left(\frac{\pi}{2}\right) = 0 + 0 = 0$$.

So $$f(n) = 0$$ for every integer $$n$$.

As $$x \to n$$, we have $$[x] \to n$$ or $$n - 1$$ (from the left), which is bounded, and $$\cos\left(\frac{2x - 1}{2}\pi\right) \to \cos\left(\frac{2n - 1}{2}\pi\right) = 0$$.

So $$\lim_{x \to n} f(x) = (\text{bounded}) \times 0 = 0 = f(n)$$.

Since the only possible points of discontinuity are at integers, and the function is continuous at every integer, $$f$$ is continuous for every real $$x$$.

Hence, the correct answer is Option D.