Q: For |x – 1| + |x – 2| + |x – 3| ≤ 6, x belongs to
(A) 0 ≤ x ≤ 4
(B) x ≤ 0 or x ≤ 4
(C) x ≤ – 2 or x ≥ 4
(D) none of these
Sol. We have |x – 1+x – 2 + x – 3| ≤ |x – 1| + |x – 2| + |x – 3| ≤ 6
⇒ |3x – 6| ≤ 6
⇒ – 6 ≤ 3x – 6 ≤ 6
⇒ 0 ≤ x ≤ 4
Hence (A) is the correct answer.