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Since x2 is always greater than 0 so mod of x2 is always positive
So f(x) becomes x4
which is decreasing when x<0 and increasing when x>0
f(x)=|x-2|/x^2 for x>2=2-x/x^2 for x<2f'(x)=(4/x^3)-(1/x^2) for x>2=-((4/x^3)-(1/x^2)) for x<2for strictly increasing f'(x)>0(4/x^3)-(1/x^2) >0 for x>2x>4 for x>2 imply 2<x<4-((4/x^3)-(1/x^2)) >0 for x<21/x^2<-4/x^3 for x<0 imply x<-4for strictly decreasing f'(x)<0(4/x^3)-(1/x^2) <0 for x>2x<4 for x>2 imply x>4-((4/x^3)-(1/x^2)) <0 for x<21/x^2<-4/x^3 for x<2 imply -4<x<2Sher MohammadB.Tech, IIT Delhi.
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