Question Number 201547 by Calculusboy last updated on 08/Dec/23
Answered by mr W last updated on 08/Dec/23
$${let}\:{t}=\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}+{x} \\ $$$$\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}−{x}\right){t}=\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}−{x}\:\right)\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}+{x}\right)=\mathrm{2} \\ $$$$\Rightarrow\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}−{x}=\frac{\mathrm{2}}{{t}} \\ $$$${f}\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}+{x}\right)=\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}−{x} \\ $$$$\equiv{f}\left({t}\right)=\frac{\mathrm{2}}{{t}} \\ $$$${f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)×{f}\left(\frac{\mathrm{1}}{\mathrm{47}}\right)=\mathrm{2}×\mathrm{2}×\mathrm{2}×\mathrm{47}=\mathrm{376}\:\checkmark \\ $$
Commented by Calculusboy last updated on 08/Dec/23
$$\boldsymbol{{thanks}}\:\boldsymbol{{sir}} \\ $$