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in Linear Inequations by (23.6k points)
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If \(\frac{-1}{|x|-2}\) ≥ 1, where x ∈ R, x ≠ ± 2 then the solution set of x is

(a) (– ∞, 2) ∪ (1, ∞) 

(b) (– 2, – 1] 

(c) [1, ∞) 

(d) (–2, –1] ∪ [1, 2)

1 Answer

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Best answer

(d) (– 2, – 1] ∪ [1, 2)

Examining \(\frac{y-1}{y-2}\) on the real number line shown below:

We see that \(\frac{y-1}{y-2}\) < 0 when 1 < y < 2

⇒ 1 < |x| < 2 ⇒ |x| > 1 and |x| < 2 

⇒ x < –1 or x > 1 and – 2 < \(x\) < 2 

\(x\) ∈ (– 2, – 1] ∪ [1, 2).

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