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प्रश्न
Solve 1 ≤ |x – 2| ≤ 3.
बेरीज
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उत्तर
We have 1 ≤ |x – 2| ≤ 3
⇒ |x − 2| ≥ 1 and |x – 2| ≤ 3
⇒ (x – 2 ≤ –1 or x – 2 ≥ 1) and (–3 ≤ x – 2 ≤ 3)
⇒ (x ≤ 1 or x ≥ 3) and (–1 ≤ x ≤ 5)
⇒ x ∈ `(-oo, 1]` ∪ `[3, oo)` and x ∈ [–1, 5]
Combining the solutions of two inequalities, We have x ∈ [–1, 1] ∪ [3, 5].
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