If Z and R denote set of integers and set of real numbers respectively, then match List I with List II.
| List-I | List-II | ||
| (A) | \(5x-3\lt 3x + 1, x \in Z\) | (I) | \(x \in (-\infty,-3]\) |
| (B) | \(3x+17\le 2(1-x),x \in R\) | (II) | \(x \in (-\infty,-1)\) |
| (C) | \(13x + 17 \lt 2(1-x),x \in R\) | (III) | \(\{\ldots\ldots -4,-3,\ldots\ldots,0,1\}\) |
| (D) | \(\displaystyle\frac{2x + 3}{5}-2\gt\frac{3(x-2)}{5},x \in Z\) | (IV) | \(\{\ldots\ldots -4, -3, -2\}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(III), (B)-(I), (C)-(II), (D)-(IV)
Each inequality is solved for x; where \(x \in Z\), only the integers are kept. An integer solution set is written as a list; a real solution set as an interval.
(A) \(5x-3\lt 3x + 1, x \in Z\)
\(2x \lt 4,\quad x \lt 2\)
Integers less than 2: \(\{\ldots, -4, -3, \ldots, 0, 1\}\)
This matches (III).
(B) \(3x+17\le 2(1-x),x \in R\)
\(3x + 17 \le 2 - 2x\)
\(5x \le -15,\quad x \le -3\)
This matches (I) \((-\infty, -3]\).
(C) \(13x + 17 \lt 2(1-x),x \in R\)
\(13x + 17 \lt 2 - 2x\)
\(15x \lt -15,\quad x \lt -1\)
This matches (II) \((-\infty, -1)\).
(D) \(\displaystyle\frac{2x + 3}{5}-2\gt\frac{3(x-2)}{5},x \in Z\)
Multiplying by 5:
\(2x + 3 - 10 \gt 3x - 6\)
\(-x \gt 1,\quad x \lt -1\)
Integers less than −1: \(\{\ldots, -4, -3, -2\}\)
This matches (IV).
So the matching is (A)-(III), (B)-(I), (C)-(II), (D)-(IV).