The greatest and the least value of the function, $\mathrm{f}\left(\mathrm{x}\right)=\sqrt{1-2\mathrm{x}+{\mathrm{x}}^2}-\sqrt{1+2\mathrm{x}+{\mathrm{x}}^2}, \mathrm{x}\in \left(-\infty ,\infty \right)$

The given function is , $\mathrm{f}\left(\mathrm{x}\right)=\sqrt{1-2\mathrm{x}+{\mathrm{x}}^2}-\sqrt{1+2\mathrm{x}+{\mathrm{x}}^2}$ 

As we know ${\left(\mathrm{x}+1\right)}^2={\mathrm{x}}^2+2\mathrm{x}+1$ and ${\left(\mathrm{x}-1\right)}^2={\mathrm{x}}^2-2\mathrm{x}+1$

$\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=|1-\mathrm{x}|-|1+\mathrm{x}|$

Splitting the function at $\mathrm{x}=1 , \mathrm{x}=-1$,

$\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{l}\left(1-\mathrm{x}\right)+\left(1+\mathrm{x}\right), \mathrm{x}<-1\\ \left(1-\mathrm{x}\right)-\left(1+\mathrm{x}\right), -1\le \mathrm{x}<1\\ -\left(1-\mathrm{x}\right)-\left(1+\mathrm{x}\right),\mathrm{x}\ge 1\end{array}\right.$

So our function becomes,

$\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{l}2, \mathrm{x}<-1\\ -2x, -1\le \mathrm{x}<1\\ -2,\mathrm{x}\ge 1\end{array}\right.$

So, The maximum value of $f\left(x\right)$ is $2$ and the minimum value of $f\left(x\right)$ is $-2$.

Hence option-$1$ is the correct answer.