Match List-I with List-II
| List-I | List-II | ||
| (Inequality) | (Solution Set) | ||
| (A) | \(2x - 3 \lt x + 2 \le 3x + 5,\quad x \in \mathrm{R}\) | (I) | \(x \in (-1, \infty)\) |
| (B) | \(|2x + 3| \lt 7,\quad x \in \mathrm{R}\) | (II) | \(x \in (-\infty, 120]\) |
| (C) | \(\displaystyle\frac{1}{2}\left(\frac{3}{5}x + 4\right) \ge \frac{1}{3}(x - 6),\quad x \in \mathrm{R}\) | (III) | \(x \in (-5, 2)\) |
| (D) | \(\displaystyle\frac{|x + 1|}{x + 1} \gt 0,\quad x \in \mathrm{R} - \{1\}\) | (IV) | \(x \in \left[-\displaystyle\frac{3}{2}, 5\right)\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(IV), (B)-(III), (C)-(II), (D)-(I)
(A) \(2x - 3 \lt x + 2 \le 3x + 5\)
This matches (IV). A double inequality means both parts hold:
\(2x - 3 \lt x + 2\) gives \(x \lt 5\)
\(x + 2 \le 3x + 5\) gives \(-3 \le 2x\)
\( \displaystyle x \ge -\frac{3}{2} \)
\( \displaystyle x \in \left[-\frac{3}{2}, 5\right) \)
(B) \(|2x + 3| \lt 7\)
This matches (III). Since \(|t| \lt a\) means \(-a \lt t \lt a\):
\(-7 \lt 2x + 3 \lt 7\)
\(-10 \lt 2x \lt 4\)
\(-5 \lt x \lt 2\)
(C) \(\frac{1}{2}\left(\frac{3}{5}x + 4\right) \ge \frac{1}{3}(x - 6)\)
This matches (II). Multiplying both sides by 30 (positive, so the sign stays):
\( \displaystyle 15\left(\frac{3}{5}x + 4\right) \ge 10(x - 6) \)
\(9x + 60 \ge 10x - 60\)
\(x \le 120\)
(D) \(\frac{|x + 1|}{x + 1} \gt 0\)
This matches (I), by elimination. The expression is not defined at x = −1. Elsewhere:
\(x \gt -1\): \(|x + 1| = x + 1\), so the expression is 1
\(x \lt -1\): \(|x + 1| = -(x + 1)\), so the expression is −1
It is positive exactly when \(x \gt -1\), that is \(x \in (-1, \infty)\). With the domain as printed, \(\mathrm{R} - \{1\}\), the point x = 1 would also be left out, and the set would be \((-1, 1) \cup (1, \infty)\) rather than (I) exactly.
So the matching is (A)-(IV), (B)-(III), (C)-(II), (D)-(I).
Note: It seems that there is a mistake in the original NTA paper, which prints the domain in (D) as \(x \in \mathrm{R} - \{1\}\) when it should read \(x \in \mathrm{R} - \{-1\}\), the point where the expression is not defined. The question can still be solved because (A) matches only (IV) and (B) only (III), and Option (4) is the only option containing (A)-(IV) and (B)-(III).