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, ∞)
Rate this question :
How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
view all courses
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
RELATED QUESTIONS :
Solve each of the
RD Sharma - MathematicsA solution is to
RD Sharma - MathematicsA company manufac
RD Sharma - Mathematics<span lang="EN-US
RS Aggarwal - Mathematics<span lang="EN-US
RS Aggarwal - Mathematics<span lang="EN-US
RS Aggarwal - MathematicsSolve each of the
RS Aggarwal - MathematicsSolve x + 5 >
RS Aggarwal - MathematicsFind the solution
RS Aggarwal - MathematicsIf |x – 1| > 5
Mathematics - Exemplar