Question Number 160848 by naka3546 last updated on 07/Dec/21
$${Find}\:\:{the}\:\:{value}\:\:{of}\:\:{a}\:\:{such}\:\:{that} \\ $$$$\:\:\:\:\:−\mathrm{2}\:<\:\frac{\mathrm{2}{x}+{a}}{{x}^{\mathrm{2}} +\mathrm{1}}\:<\:\mathrm{2} \\ $$
Answered by kowalsky78 last updated on 08/Dec/21
$${I}\:{think}\:{the}\:{correct}\:{answer}\:{is}\:{a}=−\frac{\mathrm{3}}{\mathrm{2}}. \\ $$
Commented by naka3546 last updated on 08/Dec/21
$${prove}\:{it},\:{please}. \\ $$
Answered by mr W last updated on 08/Dec/21
$$−\mathrm{2}<\frac{\mathrm{2}{x}+{a}}{{x}^{\mathrm{2}} +\mathrm{1}}<\mathrm{2} \\ $$$$−\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}<\mathrm{2}{x}+{a}<\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2} \\ $$$$\left(\mathrm{1}\right): \\ $$$$−\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}<\mathrm{2}{x}+{a} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{x}+{a}+\mathrm{2}>\mathrm{0} \\ $$$$\mathrm{2}\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +{a}+\frac{\mathrm{3}}{\mathrm{2}}>\mathrm{0} \\ $$$${a}+\frac{\mathrm{3}}{\mathrm{2}}>\mathrm{0} \\ $$$$\Rightarrow{a}>−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\left(\mathrm{2}\right): \\ $$$$\mathrm{2}{x}+{a}<\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}−{a}>\mathrm{0} \\ $$$$\mathrm{2}\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{2}}−{a}>\mathrm{0} \\ $$$$\frac{\mathrm{3}}{\mathrm{2}}−{a}>\mathrm{0} \\ $$$$\Rightarrow{a}<\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\left(\mathrm{1}\right)+\left(\mathrm{2}\right): \\ $$$$−\frac{\mathrm{3}}{\mathrm{2}}<{a}<\frac{\mathrm{3}}{\mathrm{2}} \\ $$
Commented by naka3546 last updated on 08/Dec/21
$${Thank}\:\:{you},\:\:{sir}. \\ $$