The function $\text{⨍(x)=}\left|2\mathrm{sgn}2\mathrm{x}\right|\text{+2 }$ has

Concept: The Signum function outputs the sign for the specified $x$ values. When x is more than zero, the output has a value of +1, when x is zero, it has a value of zero.
$\underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{x\to 0^{-}}{\lim }|2sgn(2x\left)\right|+2$
For $x\to 0^{-}$, value of $sgn\left(2x\right)=-1$
$\underset{x\to 0^{-}}{\lim }|2sgn(2x\left)\right|+2=\left|2\right(-1\left)\right|+2=2+2=4$
$\underset{x\to 0^{+}}{\lim }f\left(x\right)=\underset{x\to 0^{+}}{\lim }|2sgn(2x\left)\right|+2$
For $x\to 0^{+}$, value of $sgn\left(2x\right)=1$

$\underset{x\to 0^{+}}{\lim }|2sgn(2x\left)\right|+2=\left|2\right(1\left)\right|+2=2+2=4$
Thus, $\underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{x\to 0^{+}}{\lim }f\left(x\right)=4$
At x=0 we have
$f\left(0\right)=|2sgn(2\left(0\right)\left)\right|+2=0+2=2$
This implies f(x) is discontinuous at x=0
$\underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{x\to 0^{+}}{\lim }f\left(x\right)\ne f\left(0\right)$
As x=0 is the only point of discontinuity so we can say that f(x) has a removable discontinuity.
Hence, the correct answer is removable discontinuity.