Question

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

• A. All x
• B. All integer points
• C. No x
• D. x which is not an integer

Answer 1

Answer: (C)

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