The critical points of $f(x)=\frac{|2-x|}{x^2}$ is/are:

none of these

We have,

$f(x)=\frac{|2-x|}{x^2}=\left\{\begin{array}{l} \frac{x-2}{x^2}, x \geq 2 \\ \frac{2-x}{x^2}, \text { if } x<2 \end{array}\right. $

$\Rightarrow f^{\prime}(x)=\left\{\begin{array}{l}-\frac{1}{x^2}+\frac{4}{x^3}, \text { if } x>2 \\ \frac{1}{x^2}-\frac{4}{x^3}, \text { if } x<2\end{array}\right.$

$\Rightarrow f^{\prime}(x)= \begin{cases}\frac{4-x}{x^3}, & x>2 \\ \frac{x-4}{x^3}, & x<2\end{cases}$

Clearly, $f^{\prime}(x)=0$ at $x=4$ and $f^{\prime}(x)$ changes its sign from positive to negative in the neighbourhood of $x=4$. Therefore, $x=4$ is a point of local maximum.

Also, $f(x) \rightarrow 0$ as $x \rightarrow 2$ and $f(x)>0$ for all $x \neq 2$.

Therefore, $x=2$ is a point of local minimum.

Clearly, $f(x)$ is not differentiable at $x=0$.

Hence, $x=0,2,4$ are critical points of $f(x)$.