Match List-I with List-II
| List-I | List-II | ||
| Inequation | Solution set | ||
| (A) | \( 2(2x+3)-10\lt 6(x-2) \) | (I) | \( x\in(-\infty,3.9) \) |
| (B) | \( \displaystyle\frac{2x-1}{12}-\displaystyle\frac{x-1}{3}\lt\displaystyle\frac{3x+1}{4} \) | (II) | \( x\in\left(-\infty,\displaystyle\frac{-1}{2}\right)\cup\left[\displaystyle\frac{1}{2},\infty\right) \) |
| (C) | \( \displaystyle\frac{x}{2x+1}\ge\displaystyle\frac{1}{4} \) | (III) | \( x\in(0,\infty) \) |
| (D) | \( \displaystyle\frac{2x-3}{4}+6\gt 2+\displaystyle\frac{4x}{3} \) | (IV) | \( x\in(4,\infty) \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) - (IV), (B) - (III), (C) - (II), (D) - (I)
(A) \( 2(2x+3)-10\lt 6(x-2) \)
This matches (IV) \( x\in(4,\infty) \).
\( \displaystyle 4x-4\lt 6x-12 \)
\( \displaystyle 8\lt 2x\ \Rightarrow\ x\gt 4 \)
(B) \( \frac{2x-1}{12}-\frac{x-1}{3}\lt\frac{3x+1}{4} \)
This matches (III) \( x\in(0,\infty) \). Multiplying by 12 (positive, so the sign stays):
\( \displaystyle (2x-1)-4(x-1)\lt 3(3x+1) \)
\( \displaystyle -2x+3\lt 9x+3 \)
\( \displaystyle 0\lt 11x\ \Rightarrow\ x\gt 0 \)
(C) \( \frac{x}{2x+1}\ge\frac{1}{4} \)
This matches (II) \( x\in\left(-\infty,\frac{-1}{2}\right)\cup\left[\frac{1}{2},\infty\right) \). The sign of 2x + 1 is unknown, so everything goes to one side instead of cross-multiplying:
\( \displaystyle \frac{x}{2x+1}-\frac{1}{4}=\frac{4x-(2x+1)}{4(2x+1)} \)
\( \displaystyle =\frac{2x-1}{4(2x+1)}\ge 0 \)
The fraction is \( \ge 0 \) when numerator and denominator have the same sign (the numerator may be 0, the denominator may not):
\( \displaystyle x\ge\frac{1}{2}\ \text{or}\ x\lt-\frac{1}{2} \)
(D) \( \frac{2x-3}{4}+6\gt 2+\frac{4x}{3} \)
This matches (I) \( x\in(-\infty,3.9) \). Multiplying by 12:
\( \displaystyle 3(2x-3)+72\gt 24+16x \)
\( \displaystyle 6x+63\gt 24+16x \)
\( \displaystyle 39\gt 10x\ \Rightarrow\ x\lt 3.9 \)
So the matching is (A)-(IV), (B)-(III), (C)-(II), (D)-(I).