If $\frac{1}{x^2}-\frac{1}{x}>0$, then $x$ lies in the interval
Answer & explanation
Correct answer: option 2
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 $(-\infty,0)\cup(0,1)$.