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Question Number 7600 by Rohit last updated on 05/Sep/16
solve ∣((x^2 −3x−1)/(x^2 +x+1))∣<3 give solution
$${solve}\:\mid\frac{{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{1}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}\mid<\mathrm{3}\:\:{give}\:{solution} \\ $$
Answered by Yozzia last updated on 05/Sep/16
∣u(x)∣<a⇒−a<u(x)<a. ∴∣((x^2 −3x−1)/(x^2 +x+1))∣<3⇒−3<((x^2 −3x−1)/(x^2 +x+1))<3. Now, x^2 +x+1>0 ∀x∈R. To see this observe that x^2 +x+1=(x+(1/2))^2 +(3/4) and for ∀x∈R, min((x+(1/2))^2 +(3/4))=(3/4)>0. ⇒x^2 +x+1>0 ∀x∈R. ∴ −3(x^2 +x+1)<x^2 −3x−1<3(x^2 +x+1) −−−−−−−−−−−−−−−−−−−−−−−−−−−− For x^2 −3x−1>−3x^2 −3x−3 4x^2 +2>0⇒2x^2 +1>0 which is true ∀x∈R. −−−−−−−−−−−−−−−−−−−−−−−−− For x^2 −3x−1<3x^2 +3x+3 2x^2 +6x+4>0 x^2 +3x+2>0 (x+1)(x+2)>0⇒ x>−1 or x<−2 −−−−−−−−−−−−−−−−−−−−−−−−−− In all x>−1 or x<−2.
$$\mid{u}\left({x}\right)\mid<{a}\Rightarrow−{a}<{u}\left({x}\right)<{a}. \\ $$$$\therefore\mid\frac{{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{1}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}\mid<\mathrm{3}\Rightarrow−\mathrm{3}<\frac{{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{1}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}<\mathrm{3}. \\ $$$${Now},\:{x}^{\mathrm{2}} +{x}+\mathrm{1}>\mathrm{0}\:\forall{x}\in\mathbb{R}. \\ $$$${To}\:{see}\:{this}\:{observe}\:{that}\: \\ $$$${x}^{\mathrm{2}} +{x}+\mathrm{1}=\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}\:{and}\:{for}\:\forall{x}\in\mathbb{R}, \\ $$$${min}\left(\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}\right)=\frac{\mathrm{3}}{\mathrm{4}}>\mathrm{0}.\:\Rightarrow{x}^{\mathrm{2}} +{x}+\mathrm{1}>\mathrm{0}\:\forall{x}\in\mathbb{R}. \\ $$$$\therefore\:−\mathrm{3}\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)<{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{1}<\mathrm{3}\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right) \\ $$$$−−−−−−−−−−−−−−−−−−−−−−−−−−−− \\ $$$${For}\:{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{1}>−\mathrm{3}{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{3} \\ $$$$\mathrm{4}{x}^{\mathrm{2}} +\mathrm{2}>\mathrm{0}\Rightarrow\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1}>\mathrm{0}\:{which}\:{is}\:{true}\:\forall{x}\in\mathbb{R}. \\ $$$$−−−−−−−−−−−−−−−−−−−−−−−−− \\ $$$${For}\:{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{1}<\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{3} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{4}>\mathrm{0} \\ $$$${x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{2}>\mathrm{0} \\ $$$$\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)>\mathrm{0}\Rightarrow\:{x}>−\mathrm{1}\:{or}\:{x}<−\mathrm{2} \\ $$$$−−−−−−−−−−−−−−−−−−−−−−−−−− \\ $$$${In}\:{all}\:{x}>−\mathrm{1}\:{or}\:{x}<−\mathrm{2}. \\ $$

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