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Question Number 133989 by Mamifere last updated on 26/Feb/21
xx2+1<arctan(x)<x
Answered by Ñï= last updated on 26/Feb/21
Letf(x)=tan−1x wehavef(x)−f(0)=f(ξ)′(x−0) tan−1x=f(ξ)′x f(ξ)′=11+ξ2 1)::Ifx>0Then0<ξ<x ∴1<1+ξ2<1+x2 ⇒11+x2<11+ξ2<1 ⇒x1+x2<x1+ξ2<x ∴x1+x2<tan−1x<x 2)::Ifx<0Thenx<ξ<0 ∴1<1+ξ2<1+x2 ⇒11+x2<11+ξ2<1 ⇒x1+x2>x1+ξ2>x ∴x1+x2>tan−1x>x
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