Question

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

• A. continuous for every real $x$
• B. discontinuous only at $x=0$
• C. discontinuous only at non-zero integral values of $x$
• D. continuous only at $x=0$

Answer 1

Answer: (A)

continuous for every real $x$

For $n\in I$
$\underset{x\to n+}{lim}f\left(x\right)=\underset{x\to n+}{lim}\left[x\right]\cos \left(\frac{2x-1}{2}\right)\pi =n\cos \left(\frac{2n-1}{2}\right)\pi =0$

And $\underset{x\to n-}{lim}f\left(x\right)=\underset{x\to n-}{lim}\left[x\right]\cos \left(\frac{2x-1}{2}\right)\pi =(n-1)\cos \left(\frac{2n-1}{2}\right)\pi =0$

Thus $f$ is continuous for $x=n\in I$ ........ (1)

Since the function $g\left(x\right)=\left[x\right]$ and $h\left(x\right)=\cos \left(\frac{2x-1}{2}\right)\pi$ are continuous on $x\in R-I$, ......... (2)

From, (1) and (2), we get

$f$ is continuous everywhere