Solve: $\frac{-1}{|x|-2}≥1$

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Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Relations and Functions

Question:

Solve: $\frac{-1}{|x|-2}≥1$

Options:

$(-2,-1]$

$(2,-1]$

$(2,1]∪[-1,2)$

$(-2,-1]∪[1,2)$

Correct Answer:

$(-2,-1]∪[1,2)$

Explanation:

Given $\frac{-1}{|x|-2}≥1$

$⇒\frac{-1}{|x|-2}-1≥0$

$⇒\frac{-1(|x|-2)}{|x|-2}≥0$

$⇒\frac{1-|x|}{|x|-2}≥0$

$⇒\frac{|x|-1}{|x|-2}≤0$

$⇒\frac{y-1}{y-2}≤0$, where $y = |x|$

$⇒1≤y<2$

$⇒1≤|x|<2$

$⇒x∈(-2,-1]∪[1,2)$