Observe the following lists:
List - I

A)  $\text{⨍ (x)=x+}\left[\mathrm{x}\right]$ at $\mathrm{x}=\frac{1}{2}$ then $\mathrm{f}\left(\mathrm{x}\right)$ is

B)  $⨍ (\mathrm{x})=\frac{\left|\mathrm{x}\right|}{\mathrm{x}}, \mathrm{x}\ne 0, ⨍ (0)=1$

C)  $⨍ (\mathrm{x})=\frac{{\mathrm{Tan}}^{-1}(\mathrm{x}+2)}{\left|\mathrm{x}+2\right|}, \mathrm{x}\ne -2$ and

D)  $⨍ (\mathrm{x})=\mathrm{x} \cos \left(\frac{1}{{\mathrm{x}}^2}\right), \mathrm{x}\ne 0 ⨍ (0)=1$

List - II

I) Continuous

II) right continuous then$\text{⨍ (x)}$ at $\mathrm{x}=0$ is
III) Left continuous $⨍ (-2)=-1,$ then at $\mathrm{x}=-2, ⨍ (\mathrm{x})$ is
IV) Discontinuous then $⨍ (\mathrm{x})$ at $\mathrm{x}=0$ is
The correct match is :

$$\begin{gathered} \left(A\right)f\left(x\right)=x+\left[x\right] at x=\frac{1}{2} \\ f\left(\frac{1}{2}\right)=\frac{1}{2}+\left[\frac{1}{2}\right]=\frac{1}{2}+0=\frac{1}{2} \\ \underset{x\to \frac{1}{2}}{\lim }f\left(x\right)=\underset{x\to \frac{1}{2}}{\lim }x+\left[x\right]=\frac{1}{2} \\ f is continuous \end{gathered}$$

$$\begin{gathered} \left(B\right)f\left(x\right)=\frac{\left|x\right|}{x}, if x\ne 0,f\left(0\right)=1 \\ RHL=\underset{x\to 0^{+}}{\lim }f\left(x\right)=\underset{x\to 0^{+}}{\lim }\frac{\left|x\right|}{x}=\underset{x\to 0^{+}}{\lim }\frac{x}{x}=1 \\ LHL=\underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{x\to 0^{-}}{\lim }\frac{\left|x\right|}{x}=\underset{x\to 0^{-}}{\lim }\frac{-x}{x}=-1 \\ f\left(0\right)=1=\underset{x\to 0^{+}}{\lim }f\left(x\right) \\ \therefore f is right continuous at x=0 \end{gathered}$$

$$\begin{gathered} f\left(x\right)=\frac{{\mathrm{Tan}}^{-1}(\mathrm{x}+2)}{\left|\mathrm{x}+2\right|}, \mathrm{x}\ne -2 \\ \Rightarrow \underset{x\to -2-}{\lim }f\left(x\right)=\underset{x\to -2-}{\lim }\frac{{\mathrm{Tan}}^{-1}(\mathrm{x}+2)}{\left|\mathrm{x}+2\right|}=\underset{x\to -2-}{\lim }\frac{{\mathrm{Tan}}^{-1}(\mathrm{x}+2)}{-\left(x+2\right)}=-1 \\ \therefore f is left continuous if f\left(-2\right)=-1 \end{gathered}$$

$$\begin{gathered} f\left(x\right)=\mathrm{x} \cos \left(\frac{1}{{\mathrm{x}}^2}\right), \mathrm{x}\ne 0, f\left(0\right)=1 \\ \underset{x\to 0}{\lim }f\left(x\right)=\underset{x\to 0}{\lim }\mathrm{x} \cos \left(\frac{1}{{\mathrm{x}}^2}\right)=0\left(finite value\right)=0 \\ f\left(0\right)=1=1\Rightarrow \underset{x\to 0}{\lim }f\left(x\right)\ne f\left(0\right) \\ f is discontinuous \end{gathered}$$