Question

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

• A. all $x$
• B. all integer points
• C. no $x$
• D. $x$ which is not integer

Answer 1

Answer: (C)

no $x$

$f\left(x\right)=\left[x\right]\cos \left(\frac{2x-1}{2}\right)\pi$
$\therefore f\left(x\right)\{\begin{array}{c}\begin{array}{c}-\cos \left(\frac{2x-1}{2}\right)\pi \\ 0\end{array}\begin{array}{c}-1\le x<0\\ 0\le x<1\end{array}\\ \begin{array}{c}\cos \left(\frac{2x-1}{2}\right)\pi \\ 2\cos \left(\frac{2x-1}{2}\right)\pi \end{array}\begin{array}{c}1\le x<2\\ 2\le x<3\end{array}\end{array}$
Which shows that $RHL=LHL$ at $x=nϵ$ integer
as if $x=1$
$\Rightarrow \underset{x\to 1+}{lim}\cos \left(\frac{2x-1}{2}\right)\pi =0$ and ${lim}_{x\to 1-}0=0$
Also $f\left(1\right)=0$
Therefore continuous at $x=1$
Similarly, when $x=2$
$\Rightarrow \underset{x\to 2+}{lim}f\left(x\right)=\underset{x\to 2-}{lim}f\left(x\right)=0$

similarly at other x values it will be continuous

Thus this function is discontinuous at no $x$.