Match List-I with List-II
| List-I | List-II | ||
| Inequation | Solution set | ||
| (A) | \( \displaystyle\frac{x}{x-2}\gt 0 \) | (I) | \( x\in(0,2) \) |
| (B) | \( \displaystyle\frac{x}{x-2}\le 0 \) | (II) | \( x\in[0,2) \) |
| (C) | \( \displaystyle\frac{x-2}{x}\ge 0 \) | (III) | \( x\in(-\infty,0)\cup(2,\infty) \) |
| (D) | \( \displaystyle\frac{x-2}{x}\lt 0 \) | (IV) | \( x\in(-\infty,0)\cup[2,\infty) \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) - (III), (B) - (II), (C) - (IV), (D) - (I)
Both fractions change sign only at x = 0 and x = 2. Their signs on the three intervals are the same (a fraction and its reciprocal have the same sign):
\( \displaystyle x\lt 0:\ \frac{(-)}{(-)}\gt 0 \)
\( \displaystyle 0\lt x\lt 2:\ \frac{(+)}{(-)}\ \text{or}\ \frac{(-)}{(+)}\lt 0 \)
\( \displaystyle x\gt 2:\ \frac{(+)}{(+)}\gt 0 \)
The difference is only at the end points: a fraction is 0 where its numerator is 0 and is undefined where its denominator is 0.
(A) \( \frac{x}{x-2}\gt 0 \)
This matches (III) \( x\in(-\infty,0)\cup(2,\infty) \). Strict inequality, so neither x = 0 nor x = 2 is included.
(B) \( \frac{x}{x-2}\le 0 \)
This matches (II) \( x\in[0,2) \). x = 0 makes the fraction 0 and is included; x = 2 makes it undefined and is excluded.
(C) \( \frac{x-2}{x}\ge 0 \)
This matches (IV) \( x\in(-\infty,0)\cup[2,\infty) \). x = 2 makes the fraction 0 and is included; x = 0 is excluded.
(D) \( \frac{x-2}{x}\lt 0 \)
This matches (I) \( x\in(0,2) \). Strict inequality, so both end points are excluded.
So the matching is (A)-(III), (B)-(II), (C)-(IV), (D)-(I).
Tip: In rational inequalities with ≤ or ≥, include the zeros of the numerator but never the zeros of the denominator.