Question Number 143997 by liberty last updated on 20/Jun/21
$$\:{The}\:{maximum}\:{value}\:{of}\: \\ $$$${y}\:=\:\sqrt{\left({x}−\mathrm{3}\right)^{\mathrm{2}} +\left({x}^{\mathrm{2}} −\mathrm{2}\right)^{\mathrm{2}} }−\sqrt{{x}^{\mathrm{2}} +\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${is}\:\left({A}\right)\:\sqrt{\mathrm{10}}\:\:\:\:\:\:\:\left({C}\right)\:\mathrm{4}\: \\ $$$$\:\:\:\:\:\:\left({B}\right)\:\mathrm{2}\sqrt{\mathrm{5}}\:\:\:\:\:\left({D}\right)\:\mathrm{10}\: \\ $$
Answered by mitica last updated on 20/Jun/21
$$\left({A}\right)\:\sqrt{\mathrm{10}} \\ $$
Answered by mitica last updated on 20/Jun/21
$${A}\left({x},{x}^{\mathrm{2}} \right);{B}\left(\mathrm{3},\mathrm{2}\right);{C}\left(\mathrm{0},\mathrm{1}\right) \\ $$$${y}={AB}−{AC}\leqslant{BC}\Rightarrow{y}\leqslant\sqrt{\left(\mathrm{3}−\mathrm{0}\right)^{\mathrm{2}} +\left(\mathrm{2}−\mathrm{1}\right)^{\mathrm{2}} }=\sqrt{\mathrm{10}} \\ $$$${y}=\sqrt{\mathrm{10}}\Leftrightarrow\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}−\mathrm{0}}=\frac{\mathrm{2}−\mathrm{1}}{\mathrm{3}−\mathrm{0}}\Leftrightarrow\mathrm{3}{x}^{\mathrm{2}} −{x}−\mathrm{3}=\mathrm{0}\Leftrightarrow{x}=\frac{\mathrm{1}\pm\sqrt{\mathrm{37}}}{\mathrm{6}} \\ $$