The solution set of the inequation $\frac{3}{|x|+2}≥1$, is
Answer & explanation
Correct answer: option 1
We have,
$\frac{3}{|x|+2}≥1$
$⇒3≥|x|+2$ [Multiplying both sides by |x| + 2 as it is positive]
$⇒1≥|x|⇒|x|≤1⇒x∈[-1, 1]$
The solution set of the inequation $\frac{3}{|x|+2}≥1$, is
Correct answer: option 1
We have,
$\frac{3}{|x|+2}≥1$
$⇒3≥|x|+2$ [Multiplying both sides by |x| + 2 as it is positive]
$⇒1≥|x|⇒|x|≤1⇒x∈[-1, 1]$