If $\left[.\right]$ denotes the greatest integer function then the number of points where

$\text{⨍ }\left(\mathrm{x}\right)\text{=}\left[\mathrm{x}\right]\text{+}\left[\mathrm{x}+\frac{1}{3}\right]\text{+}\left[\mathrm{x}+\frac{2}{3}\right]$ is discontinuous for $\mathrm{x}\in \left(0,3\right)$ are

$\mathrm{f}\left(\mathrm{x}\right)=\left[\mathrm{x}\right]+\left[\mathrm{x}+\frac{1}{3}\right]+\left[\mathrm{x}+\frac{2}{3}\right]$

$\mathrm{f}\left(\mathrm{x}\right)=\left[3\mathrm{x}\right]$

$\boxed{\left[\mathrm{x}\right]+\left[\mathrm{x}+\frac{1}{\mathrm{n}}\right]+\left[\mathrm{x}+\frac{2}{\mathrm{n}}\right]+,,,\left[\mathrm{x}+\frac{\mathrm{n}-1}{\mathrm{n}}\right]=\left[\mathrm{nx}\right]}$

$0<\mathrm{x}<3\Rightarrow 0<3\mathrm{x}<9$

$\therefore \mathrm{f}\left(\mathrm{x}\right)=\left[3\mathrm{x}\right]$ is discontinuous at $3x\in \left\{1,2,3\dots .8\right\}$

$\left(\mathrm{i}.\mathrm{e}\right)\mathrm{x}=\frac{1}{3},\frac{2}{3},\dots \frac{.8}{3}$

No of points of discontinuity = 8

Option (3) is correct