Official solution
Given, \[f(x) = [x - 1]\cos \left( {{{2x - 1} \over 2}} \right)\pi \] where [ . ] is greatest integer function and f : R \[\to\] R
\[\because\] It is a greatest integer function then we need to check its continuity at x \[\in\] I except these it is continuous.
Let, x = n where n \[\in\] I
Then
LHL = \[\mathop {\lim }\limits_{x \to {n^ - }} [x - 1]\cos \left( {{{2x - 1} \over 2}} \right)\pi \]
\[ = (n - 2)\cos \left( {{{2x - 1} \over 2}} \right)\pi = 0\]
RHL = \[\mathop {\lim }\limits_{x \to {n^ + }} [x - 1]\cos \left( {{{2x - 1} \over 2}} \right)\pi \]
\[ = (n - 1)\cos \left( {{{2x - 1} \over 2}} \right)\pi = 0\]
and f(n) = 0
Here, \[\mathop {\lim }\limits_{x \to {n^ - }} f(x) = \mathop {\lim }\limits_{x \to {n^ + }} f(x) = f(n)\]
\[\therefore\] It is continuous at every integers.
Therefore, the given function is continuous for all real x.