Answer :
|x – 1| ≤ 3
⇒ -3 ≤ x – 1 ≤ 3
⇒ -2 ≤ x ≤ 4
|x – 1 | ≤ 1
⇒ -1 ≤ x – 1 ≤ 1
⇒ 0 ≤ x ≤ 2
We have to take intersection of x ∈ [-2, 4] and x ∈ [0, 2]
So, final answer is x ∈ [0, 2]
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