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Question Number 160848 by naka3546 last updated on 07/Dec/21

Find the value of a such that −2 < ((2x+a)/(x^2 +1)) < 2

$${Find}\:\:{the}\:\:{value}\:\:{of}\:\:{a}\:\:{such}\:\:{that} \\ $$ $$\:\:\:\:\:−\mathrm{2}\:<\:\frac{\mathrm{2}{x}+{a}}{{x}^{\mathrm{2}} +\mathrm{1}}\:<\:\mathrm{2} \\ $$

Answered by kowalsky78 last updated on 08/Dec/21

I think the correct answer is a=−(3/2).

$${I}\:{think}\:{the}\:{correct}\:{answer}\:{is}\:{a}=−\frac{\mathrm{3}}{\mathrm{2}}. \\ $$

Commented bynaka3546 last updated on 08/Dec/21

prove it, please.

$${prove}\:{it},\:{please}. \\ $$

Answered by mr W last updated on 08/Dec/21

−2<((2x+a)/(x^2 +1))<2 −2x^2 −2<2x+a<2x^2 +2 (1): −2x^2 −2<2x+a 2x^2 +2x+a+2>0 2(x+(1/2))^2 +a+(3/2)>0 a+(3/2)>0 ⇒a>−(3/2) (2): 2x+a<2x^2 +2 2x^2 −2x+2−a>0 2(x−(1/2))^2 +(3/2)−a>0 (3/2)−a>0 ⇒a<(3/2) (1)+(2): −(3/2)<a<(3/2)

$$−\mathrm{2}<\frac{\mathrm{2}{x}+{a}}{{x}^{\mathrm{2}} +\mathrm{1}}<\mathrm{2} \\ $$ $$−\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}<\mathrm{2}{x}+{a}<\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2} \\ $$ $$\left(\mathrm{1}\right): \\ $$ $$−\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}<\mathrm{2}{x}+{a} \\ $$ $$\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{x}+{a}+\mathrm{2}>\mathrm{0} \\ $$ $$\mathrm{2}\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +{a}+\frac{\mathrm{3}}{\mathrm{2}}>\mathrm{0} \\ $$ $${a}+\frac{\mathrm{3}}{\mathrm{2}}>\mathrm{0} \\ $$ $$\Rightarrow{a}>−\frac{\mathrm{3}}{\mathrm{2}} \\ $$ $$\left(\mathrm{2}\right): \\ $$ $$\mathrm{2}{x}+{a}<\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2} \\ $$ $$\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}−{a}>\mathrm{0} \\ $$ $$\mathrm{2}\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{2}}−{a}>\mathrm{0} \\ $$ $$\frac{\mathrm{3}}{\mathrm{2}}−{a}>\mathrm{0} \\ $$ $$\Rightarrow{a}<\frac{\mathrm{3}}{\mathrm{2}} \\ $$ $$\left(\mathrm{1}\right)+\left(\mathrm{2}\right): \\ $$ $$−\frac{\mathrm{3}}{\mathrm{2}}<{a}<\frac{\mathrm{3}}{\mathrm{2}} \\ $$

Commented bynaka3546 last updated on 08/Dec/21

Thank you, sir.

$${Thank}\:\:{you},\:\:{sir}. \\ $$

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