The solution set of the inequality |3 x| ≥|6-3 x| is : Option 1: (-∞, 1] Option 2: [1, ∞) Option 3: (-∞, 1) (1, ∞) Option 4: (-∞,-1) (-1, ∞)

The answer is '[1, ∞)'. View full solution:

CUET Mathematics 2024 - The solution set of the inequality |3 x| ≥|6-3 x| is : | PYQs + Solutions

2026: 22nd May Shift 1

Match List-I with List-II

List-I	List-II
Inequalities	Solution set
(A) If $\left\vert 3 - 5x \right\vert < 7$ , then $x \in$	(I) $\left[-\dfrac{5}{2}, \infty\right)$
(B) If $3x - 7 > 5x - 1$ , then $x \in$	(II) $(-\infty, -3)$
(C) If $\dfrac{x}{3} > \dfrac{x}{2} + 1$ , then $x \in$	(III) $\left(-\dfrac{4}{5}, 2\right)$
(D) If $5x + 4 \leq 7x + 9$ , then $x \in$	(IV) $(-\infty, -6)$

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2026: 19th May Shift 1

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

2026: 30th May Shift 2

The solution set of the inequality $3x+4y<12$ is

2025: 22 May Shift 1

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

2025: 30 May Shift 2

The solution set of the inequality $\frac{2x+3}{x-1} < 0$ 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 $\frac{a}{c} < \frac{b}{c}$ .

(C) $\frac{1}{x^2} > \frac{1}{x} > 1$ , if $0 < x < 1$ .

(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: