Write the solution set of the inequation

We have:

$\left|\frac{1}{x}-2\right|>4$

Here, two cases arise.

CASE 1: When $\frac{1}{x}-2>0$, then $\left|\frac{1}{x}-2\right|=\frac{1}{x}-2$

$\therefore \frac{1}{x}-2>4$

$\Rightarrow \frac{1}{x}-2-4>0$

$\Rightarrow \frac{1}{x}>6$

$\Rightarrow x \in\left(0, \frac{1}{6}\right)$             ...(i)

CASE $2:$ When $\frac{1}{x}-2<0$, then $\left|\frac{1}{x}-2\right|=-\left(\frac{1}{x}-2\right)$

$\therefore-\frac{1}{x}+2>4$

$\Rightarrow-\frac{1}{x}>2$

$\Rightarrow \frac{1}{x}<-2$

$\Rightarrow x \in\left(-\infty, \frac{-1}{2}\right) \quad \ldots($ ii $)$

Hence, the solution set of the given inequation is the union of (i) and (ii).

$\therefore x \in\left(-\infty, \frac{-1}{2}\right) \cup\left(0, \frac{1}{6}\right)$