The number of critical points of $f\left(x\right)=\frac{|x-1|}{x^2}$ is

The given function is $f\left(x\right)=\frac{|x-1|}{x^2}$

The function is not defined at $x=0$ and the function is not differentiable at $x=1$

$f\text{'}\left(x\right)=\frac{d}{dx}\left(\frac{x-1}{x^2}\right)=\frac{x^2-2x\left(x-1\right)}{x^4}=-\frac{x-2}{x^3}$

It implies that $f\text{'}\left(2\right)=0$

Therefore, the set of critical points is $\{0,1,2\}$