CUET · MATHS · PYQ PAPER 2025
Match List I with List II
| List-I (Inequality) | List-II (Solution Set) |
| (A) \(2 x-3<x+2 \leq 3 x+5, x \in R\) | (I) \(x \in(-1, \infty)\) |
| (B) \(|2 x+3|<7, x \in R\) | (II) \(x \in(-\infty, 120]\) |
| (C) \(\frac{1}{2}\left(\frac{3}{5} x+4\right) \geq \frac{1}{3}(x-6), x \in R\) | (III) \(x \in(-5,2)\) |
| (D) \(\frac{|x+1|}{x+1}>0, x \in R -\{1\}\) | (IV) \(x \in\left[-\frac{3}{2}, 5\right)\) |
- A (А) - (II), (В) - (І), (С) - (III), (D) - (IV)
- B (А) - (І), (B) - (II), (C) - (III), (D) - (IV)
- C (А) - (III), (B) - (IV), (C) - (II), (D) - (I)
- D (A) - (IV), (B) - (III), (С) - (II), (D) - (I)
Answer & Solution
Correct Answer
(D) (A) - (IV), (B) - (III), (С) - (II), (D) - (I)
Step-by-step Solution
Detailed explanation
(A)\(2x-3 < x+2 \leq 3x+5\) \(2x-3 < x+2 \implies x < 5\) \(x+2 \leq 3x+5 \implies -3 \leq 2x \implies x \geq -\frac{3}{2}\) \(x \in \left[-\frac{3}{2}, 5\right)\) matches (IV) (B) \(|2x+3| < 7\) \(-7 < 2x+3 < 7\) \(-10 < 2x < 4\) \(-5 < x < 2\) \(x \in (-5, 2)\) matches (III)…
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