Algebra

Inequalities

Medium

applied

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

$(-\infty, 0)$

$(-\infty, 0) \cup (0, 1)$

$(-\infty, 1)$

$(1, \infty)$

✅ Correct Option: 2

The inequality to solve is $\frac{1}{x^2} - \frac{1}{x} > 0$ .

Since $x$ appears in denominators, $x \neq 0$ .

Finding a common denominator of $x^2$ :

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

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

The inequality becomes:

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

For the fraction to be positive, the numerator and denominator must have the same sign.

The denominator $x^2$ is always positive for any $x \neq 0$ .

The numerator $1-x$ is positive when:

$1 - x > 0$

$1 > x$

$x < 1$

Since $x^2$ is always positive, the condition required is:

$x < 1$ and $x \neq 0$

This gives the interval $(-\infty, 0) \cup (0, 1)$ .

Therefore, $x$ lies in the interval $(-\infty, 0) \cup (0, 1)$ .

2025: 21 May Shift 2

If a, b, c are positive real numbers, then the least value of $(a+b+c)(ab+bc+ca)$ is:

2025: 14 May Shift 1

Which of the following inequalities holds true?

(A) $\sqrt{5} + \sqrt{3} > \sqrt{6} + \sqrt{2}$ .

(B) If $a > b$ and $c < 0$ , then $\frac{a}{c} < \frac{b}{c}$ .

(D) If $a$ and $b$ are positive integers and $\frac{a - b}{6.25} = \frac{4}{2.5}$ , then $b > a$ .

Choose the correct answer from the options given below:

2026: 19 May Shift 1

The number that satisfies the inequality $\frac{29}{4} < \frac{x+20}{4} < 8$ is

2026: 31 May Shift 2

The solution set $\{x: x \in R, 3x-2 < x+2 \le 2x+6\}$ (Where $R$ is set of real numbers) is :

2026: 30 May Shift 2

Match List-I with List-II

List-I	List-II
(A) The solution set of the inequation $3x-6\le 0$ is	(I) $(3,\infty)$
(B) The solution set of the inequation $-5x+20<0$ is	(II) $[3,\infty)$
(C) The solution set of the inequation $3x-9\ge 0$ is	(III) $(-\infty,2]$
(D) The solution set of the inequation $8x+11>35$ is	(IV) $(4,\infty)$

Choose the correct answer from the options given below:

2025: 15 May Shift 1

The inequality $\frac{5x - 2}{3} - \frac{7x - 3}{5} > \frac{x}{4}$ holds when

If 1x2−1x>0\frac{1}{x^2} - \frac{1}{x} > 0x21​−x1​>0, then xxx lies in the interval

Solution

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Keyboard Shortcuts

CUET Mathematics Past Year Questions (Topic-Wise)

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