If $\frac{1}{x^2}-\frac{1}{x}>0$, then $x$ lies in the interval

$(-∞,0) ∪ (0,1)$

The correct answer is Option (2) → $(-∞,0) ∪ (0,1)$ **

Given: $\frac{1}{x^2}-\frac{1}{x}>0$

$\frac{1}{x^2}-\frac{1}{x}=\frac{1 - x}{x^2}$

So inequality becomes:

$\frac{1 - x}{x^2} > 0$

$x^2 > 0$ for all $x \neq 0$ (always positive)

Thus sign depends on numerator:

$1 - x > 0 \Rightarrow x < 1$

And $x \neq 0$.

Hence, the solution interval is $(-\infty,0)\cup(0,1)$.