If \(\displaystyle\frac{1}{x^2}-\frac{1}{x} \gt 0\), then \(x\) lies in the interval
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \((-\infty, 0) \cup (0, 1)\)
The expression is defined only for \(x \ne 0\). Taking the common denominator:
\( \displaystyle \frac{1}{x^2} - \frac{1}{x} = \frac{1 - x}{x^2} \gt 0 \)
For \(x \ne 0\) the denominator \(x^2\) is always positive, so the sign depends only on the numerator:
\(1 - x \gt 0,\quad x \lt 1\)
Together with \(x \ne 0\), the solution is \((-\infty, 0) \cup (0, 1)\).
Why the other options are incorrect
Option (1) \((-\infty, 0)\) — Incorrect. It leaves out (0, 1); for example, \(x = \frac{1}{2}\) gives \(4 - 2 = 2 \gt 0\).
Option (3) \((-\infty, 1)\) — Incorrect. It contains \(x = 0\), where the expression is not defined.
Option (4) \((1, \infty)\) — Incorrect. For \(x \gt 1\), \(1 - x \lt 0\), so the expression is negative.