The solution set of the inequation $\frac{3}{|x|+2}≥1$, is

[-1, 1] (-1, 1) (-∞, 1] [1, ∞)

[-1, 1]

We have, $\frac{3}{|x|+2}≥1$ $⇒3≥|x|+2$  [Multiplying both sides by |x| + 2 as it is positive] $⇒1≥|x|⇒|x|≤1⇒x∈[-1, 1]$