Question

The function$f\left(x\right)=\left[x\right]\cos \left[\right(2x-1)/2]\pi ,[.]$ 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

The correct option is A

All $x$

Explanation for the correct option

$f\left(x\right)=\left[x\right]\cos \left[\frac{(2x-1)}{2}\right]\pi$

For, a function to be continuous

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

Let the $0\le x<1$

$f\left(x\right)=[x]\cos \frac{(2x-1)}{2}\pi ,0\le x<1$

Solving L.H.L

$$\begin{gathered} \Rightarrow \underset{x\to 0^{-}}{\lim }f\left(x\right) \\ \Rightarrow \underset{x\to 0^{-}}{\lim }\left[x\right]\cos \left(\frac{2x-1}{2}\right)\mathrm{\pi } \\ \Rightarrow 0\times \cos \left(\frac{-1}{2}\right)\mathrm{\pi } \\ \Rightarrow 0 \end{gathered}$$

Solving R.H.L

$$\begin{gathered} \Rightarrow \underset{x\to 1^{+}}{\lim }f\left(x\right) \\ \Rightarrow \underset{x\to 1^{+}}{\lim }\left[x\right]\cos \left(\frac{2x-1}{2}\right)\mathrm{\pi } \\ \Rightarrow \left[1\right]\cos \left(\frac{2-1}{2}\right)\mathrm{\pi } \\ \Rightarrow \cos \frac{\mathrm{\pi }}{2}=0 \end{gathered}$$

Now, $f\left(x\right)=f\left(0\right)$

$\Rightarrow f\left(x\right)=\left[x\right]\cos \left[\frac{(2x-1)}{2}\right]\pi$

$\Rightarrow f\left(0\right)=0\cos \left(\frac{-1}{2}\right)\mathrm{\pi }$

$\Rightarrow f\left(0\right)=0$

$\therefore$L.H.L$=$R.H.L$=f\left(x\right)$

So, $f\left(x\right)$ is continuous for all $x$.

Hence, the correct option is A.