Which of the following are correct? (Where \( R \) is set of real numbers)
(A) If \( x\le 8 \), then \( -x\ge -8 \)
(B) If \( x\in R \), \( |x|\le 3 \), then \( -3\le x\le 3 \)
(C) If \( x\in R \), \( |x|\lt -4 \), then \( x\in\phi \)
(D) \( x=|x| \) iff \( x\ge 0 \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A), (B), (C) and (D)
Multiplying both sides of an inequality by a negative number reverses the sign. For a > 0, \( |x|\le a \) is equivalent to \( -a\le x\le a \). Also |x| = x when x ≥ 0 and |x| = −x when x < 0, so |x| ≥ 0 for every real x.
So multiplying x ≤ 8 by −1 gives −x ≥ −8, |x| ≤ 3 means −3 ≤ x ≤ 3, |x| is never negative, and |x| = x exactly when x ≥ 0.
Checking the statements
(A) If \( x\le 8 \), then \( -x\ge -8 \) — Correct. Multiplying by −1 reverses the inequality.
(B) If \( x\in R \), \( |x|\le 3 \), then \( -3\le x\le 3 \) — Correct. \( |x|\le 3 \) is the same as \( -3\le x\le 3 \).
(C) If \( x\in R \), \( |x|\lt -4 \), then \( x\in\phi \) — Correct. |x| ≥ 0 can never be less than −4, so the solution set is empty.
(D) \( x=|x| \) iff \( x\ge 0 \) — Correct. For x < 0, |x| = −x ≠ x.
Tip: An absolute value is never negative: \( |x|\lt a \) with a ≤ 0 has no solution, while \( |x|\gt a \) with a < 0 holds for every real x.