Match List-I with List-II
| List-I [Inequation] | List-II [Solution set] | ||
| (A) | \( x+2\ge 5 \) | (I) | \( (-\infty,0] \) |
| (B) | \( |x|\ge 2 \) | (II) | \( (-\infty,-2]\cup[2,\infty) \) |
| (C) | \( x+1\le -0.5 \) | (III) | \( (-\infty,-1.5] \) |
| (D) | \( 2x+4\le 4 \) | (IV) | \( [3,\infty) \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) - (IV), (B) - (II), (C) - (III), (D) - (I)
(A) \( x+2\ge 5 \)
This matches (IV) [3, ∞).
\( \displaystyle x\ge 5-2=3 \)
(B) \( |x|\ge 2 \)
This matches (II) (−∞, −2] ∪ [2, ∞), by the rule for a modulus inequality.
\( \displaystyle |x|\ge a\ (a\gt 0)\ \Leftrightarrow\ x\le -a\ \text{or}\ x\ge a \)
\( \displaystyle x\le -2\ \text{or}\ x\ge 2 \)
(C) \( x+1\le -0.5 \)
This matches (III) (−∞, −1.5].
\( \displaystyle x\le -0.5-1=-1.5 \)
(D) \( 2x+4\le 4 \)
This matches (I) (−∞, 0].
\( \displaystyle 2x\le 0 \)
\( \displaystyle x\le 0 \)
So the matching is (A)-(IV), (B)-(II), (C)-(III), (D)-(I).