Question

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

The correct option is A continuous for every real $x$.

The given function can be continuous between two integers. But when we talk at integers, we need to check the continuity by putting limits on both sides of integers for this function.

$L.H.L.=\underset{x\to n^{-}}{lim}$$\left[x\right]\cos \left(\frac{2x-1}{2}\right)\pi$
$=(n-1)\cos \left(\frac{2n-1}{2}\right)\pi =0$

$R.H.L.=\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 $f\left(n\right)=0$

$\because f\left(n^{-}\right)=f\left(n^{+}\right)=f\left(0\right)$
Hence, the function is continuous for every real $x.$