The function $$f\left( x \right) = \left[ x \right]\cos \left( {\frac{{2x - 1}}{2}} \right)\pi ,\,\left[ . \right]$$      denotes the greatest integer function, is discontinuous at-

Solution :
When $$x$$ is not an integer, both the functions $$\left[ x \right]$$ and $$\cos \left( {\frac{{2x - 1}}{2}} \right)\pi $$    are continuous.
$$\therefore \,f\left( x \right)$$   continuous on all non integral points.
For $$x = n \in I$$
$$\eqalign{ & \mathop {\lim }\limits_{x \to {n^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {n^ - }} \left[ x \right]\cos \left( {\frac{{2x - 1}}{2}} \right)\pi \cr & = \left( {n - 1} \right)\cos \left( {\frac{{2n - 1}}{2}} \right)\pi = 0 \cr & \mathop {\lim }\limits_{x \to {n^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {n^ + }} \left[ x \right]\cos \left( {\frac{{2x - 1}}{2}} \right)\pi \cr & = n\cos \left( {\frac{{2n - 1}}{2}} \right)\pi = 0 \cr & {\text{Also }}f\left( n \right) = n\,\cos \frac{{\left( {2n - 1} \right)\pi }}{2} = 0 \cr} $$
$$\therefore \,f$$  is continuous at all integral pts. as well.
Thus, $$f$$ is continuous everywhere.