Match List-I with List-II, where \(x\) is a real number.
| List-I (Inequality) | List-II (Solution set) | ||
| (A) | \(5 - 2x \gt 11\) | (I) | \([3, \infty)\) |
| (B) | \(|x + 1| \ge 2\) | (II) | \((-3, 3]\) |
| (C) | \(-3 \lt 2x + 3 \le 9\) | (III) | \((-\infty, -3)\) |
| (D) | \(4x - 1 \ge 2x + 5\) | (IV) | \((-\infty, -3] \cup [1, \infty)\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(III), (B)-(IV), (C)-(II), (D)-(I)
(A) \(5 - 2x \gt 11\)
This matches (III) \((-\infty, -3)\). Dividing by a negative number reverses the inequality:
\(-2x \gt 6\)
\(x \lt -3\)
(B) \(|x + 1| \ge 2\)
This matches (IV) \((-\infty, -3] \cup [1, \infty)\). \(|y| \ge a\) means \(y \le -a\) or \(y \ge a\):
\(x + 1 \le -2\) or \(x + 1 \ge 2\)
\(x \le -3\) or \(x \ge 1\)
(C) \(-3 \lt 2x + 3 \le 9\)
This matches (II) \((-3, 3]\).
\(-6 \lt 2x \le 6\)
\(-3 \lt x \le 3\)
(D) \(4x - 1 \ge 2x + 5\)
This matches (I) \([3, \infty)\).
\(2x \ge 6\)
\(x \ge 3\)