Question

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

• A. all real $x$
• B. $x=\frac{n}{2},nϵI-\left\{1\right\}$
• C. $xϵN$
• D. $x\mathrm{ϵ/}I-\left\{0\right\}$

Answer 1

The correct option is B

$x=\frac{n}{2},nϵI-\left\{1\right\}$

Case 1 : when $n$ is an integer.

${lim}_{x\to n^{+}}f\left(x\right)=n\cos \left[\right(n-\frac{1}{2}\left)\pi \right]$

${lim}_{x\to n^{-}}f\left(x\right)=(n-1)\cos \left[\right(n-\frac{1}{2}\left)\pi \right]$

The limit exists as $\cos \left[\right(n-\frac{1}{2}\left)\pi \right]$ is always zero. $[\because \cos (2m+1)\frac{\pi }{2}=0,mϵI]$

$\therefore f\left(x\right)$ is continuous at all $xϵI$.

Case 2 : when $n$ is not an integer

${lim}_{x\to n^{+}}f\left(x\right)=\left[n\right]\cos \left[\right(n-\frac{1}{2}\left)\pi \right]$

${lim}_{x\to n^{-}}f\left(x\right)=\left[n\right]\cos \left[\right(n-\frac{1}{2}\left)\pi \right]$

Since, left-hand limit is equal to right-hand limit, the function is continous.

Hence, the function is continuous everywhere.