The solution set of the inequality |3x| ≥ |6 - 3x| is:

The correct answer is Option (2) → \([1, \infty)\)

Both sides are non-negative, so squaring keeps the inequality:

    \(9x^2 \ge (6 - 3x)^2\)
    \(9x^2 \ge 36 - 36x + 9x^2\)
    \(36x \ge 36\)
    \(x \ge 1\)

The solution set is \([1, \infty)\); x = 1 is included because the two sides are equal there (3 = 3).

Shortcut: Dividing by 3, \(|x| \ge |x - 2|\): the distance of x from 0 is at least its distance from 2. That holds exactly for points on or to the right of the midpoint 1, so \(x \ge 1\).