MCQ
The function$f(x) = [x]\cos \left[ {\frac{{2x - 1}}{2}} \right]\pi ,\,$ where$[.]$ denotes the greatest integer function, is discontinuous at
- AAll $x$
- BNo $x$
- ✓All integer points
- D$x$ which is not an integer
Since $g(x) = [x]$ is always discontinuous at all integral values of points.
Hence $f(x)$ is discontinuous for all integral points.
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$[A]$ $x=-1$ $[B]$ $x=0$ $[C]$ $x=2$ $[D] x=1$