Solution of the inequality |2x| ≥ |8-2x| is

The correct answer is Option (3) → \([2, \infty)\)

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

    \( (2x)^2 \ge (8-2x)^2 \)

    \( 4x^2 \ge 64 - 32x + 4x^2 \)

    \( 32x \ge 64 \)

    \( x \ge 2 \)

So the solution set is \([2, \infty)\).

Shortcut: Dividing by 2 gives \(|x| \ge |x - 4|\): the distance of x from 0 is at least its distance from 4. This holds exactly on and to the right of the midpoint 2.