Answer :




Case I : x2 – 2x + 1 ≥ 0 and x > 0


(x – 1)2 ≥ 0 and x > 0


So, by taking intersection x > 0


Case II : : x2 – 2x + 1 ≤ 0 and x < 0


(x – 1)2 ≤ 0 and x < 0


Square term is always positive so case II is irrelevant.


Then, the final answer of question is x (0, ∞)


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