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

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