If |3x|≥|6-3x|, x∈ R, then x lies in

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

$|3x|≥|6-3x|$   ($|3x| → (I), |6-3x| → (II)$)

$⇒|3x|-|6-3x|≥0$

Case 1: $x<0$

$-3x+6-3x≥0$

$-6x≥-6$

$x≥+1$ → not possible

Case 2: $0≤x<2

$3x+6-3x≥0$

$6≥0$ Not possible

Case 3: $x≥2$

$3x-6+3x≥0$

$6x-6≥0$

$x≥1$

$x∈[1, ∞)$