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Question Number 133989 by Mamifere last updated on 26/Feb/21

(x/(x2+1))<arctan(x)<x

$$\frac{{x}}{{x}\mathrm{2}+\mathrm{1}}<{arctan}\left({x}\right)<{x} \\ $$ $$ \\ $$

Answered by Ñï= last updated on 26/Feb/21

Let f(x)=tan^(−1) x we have f(x)−f(0)=f(ξ)′(x−0) tan^(−1) x=f(ξ)′x f(ξ)′=(1/(1+ξ^2 )) 1)::If x>0 Then 0<ξ<x ∴1<1+ξ^2 <1+x^2 ⇒(1/(1+x^2 ))<(1/(1+ξ^2 ))<1 ⇒(x/(1+x^2 ))<(x/(1+ξ^2 ))<x ∴(x/(1+x^2 ))<tan^(−1) x<x 2)::If x<0 Then x<ξ<0 ∴1<1+ξ^2 <1+x^2 ⇒(1/(1+x^2 ))<(1/(1+ξ^2 ))<1 ⇒(x/(1+x^2 ))>(x/(1+ξ^2 ))>x ∴(x/(1+x^2 ))>tan^(−1) x>x

$${Let}\:{f}\left({x}\right)={tan}^{−\mathrm{1}} {x} \\ $$ $${we}\:{have}\:{f}\left({x}\right)−{f}\left(\mathrm{0}\right)={f}\left(\xi\right)'\left({x}−\mathrm{0}\right) \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{tan}^{−\mathrm{1}} {x}={f}\left(\xi\right)'{x} \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{f}\left(\xi\right)'=\frac{\mathrm{1}}{\mathrm{1}+\xi^{\mathrm{2}} } \\ $$ $$\left.\mathrm{1}\right)::{If}\:\:{x}>\mathrm{0}\:\:{Then}\:\mathrm{0}<\xi<{x} \\ $$ $$\therefore\mathrm{1}<\mathrm{1}+\xi^{\mathrm{2}} <\mathrm{1}+{x}^{\mathrm{2}} \\ $$ $$\Rightarrow\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }<\frac{\mathrm{1}}{\mathrm{1}+\xi^{\mathrm{2}} }<\mathrm{1} \\ $$ $$\Rightarrow\frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} }<\frac{{x}}{\mathrm{1}+\xi^{\mathrm{2}} }<{x} \\ $$ $$\therefore\frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} }<{tan}^{−\mathrm{1}} {x}<{x} \\ $$ $$\left.\mathrm{2}\right)::{If}\:{x}<\mathrm{0}\:{Then}\:{x}<\xi<\mathrm{0} \\ $$ $$\therefore\mathrm{1}<\mathrm{1}+\xi^{\mathrm{2}} <\mathrm{1}+{x}^{\mathrm{2}} \\ $$ $$\Rightarrow\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }<\frac{\mathrm{1}}{\mathrm{1}+\xi^{\mathrm{2}} }<\mathrm{1} \\ $$ $$\Rightarrow\frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} }>\frac{{x}}{\mathrm{1}+\xi^{\mathrm{2}} }>{x} \\ $$ $$\therefore\frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} }>{tan}^{−\mathrm{1}} {x}>{x} \\ $$

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