If $\left[\mathrm{x}\right]$ denotes the greatest integer not exceeding the number $\mathrm{x},$ then $\text{⨍ (x)}$ defined by 

$\text{⨍ (x)=}\left\{\begin{array}{ll}\left[\mathrm{x}\right]\text{,}& \mathrm{i}\text{⨍ x<2}\\ \left[\mathrm{x}\right]\text{-1,}& \mathrm{i}\text{⨍ x≥2}\end{array}\right.$

$$\begin{gathered} At x=2 \\ LHL=\underset{x\to 2^{-}}{\lim }f\left(x\right)=\underset{x\to 2^{-}}{\lim }\left[x\right]=1 \\ RHL=\underset{x\to 2^{+}}{\lim }f\left(x\right)=\underset{x\to 2^{+}}{\lim }\left[x\right]-1=2-1=1 \\ \therefore f is continuous at x=2 \\ f\left(1\right)=1,\underset{x\to 1^{+}}{\lim }f\left(x\right)=\underset{x\to 1^{+}}{\lim }\left[x\right]=1 \\ \Rightarrow f is continuous in [1,3) \end{gathered}$$