Question Number 180827 by sciencestudent last updated on 17/Nov/22
$${which}\:{is}\:{not}\:{range}\:{of}\:{f}\left({x}\right)=\frac{\mathrm{2}{x}+\mathrm{1}}{{x}−\mathrm{2}}? \\ $$$$\left.\mathrm{1}\left.\right)\left.\:\left.\mathrm{2}\:\:\:\:\:\:\:\:\mathrm{2}\right)\:\mathrm{3}\:\:\:\:\:\:\:\:\:\mathrm{3}\right)\mathrm{10}\:\:\:\:\:\:\:\mathrm{4}\right)\:\frac{\mathrm{2}}{\mathrm{3}} \\ $$
Commented by kaivan.ahmadi last updated on 17/Nov/22
$${we}\:{know}\:{that}\:{R}_{{f}} ={D}_{{f}^{−\mathrm{1}} } \\ $$$${y}=\frac{\mathrm{2}{x}+\mathrm{1}}{{x}−\mathrm{2}}\Rightarrow\mathrm{2}{x}+\mathrm{1}={yx}−\mathrm{2}{y}\Rightarrow\left({y}−\mathrm{2}\right){x}=\mathrm{1}+\mathrm{2}{y} \\ $$$$\Rightarrow{x}=\frac{\mathrm{1}+\mathrm{2}{y}}{{y}−\mathrm{2}}\Rightarrow{f}^{−\mathrm{1}} \left({x}\right)=\frac{\mathrm{1}+\mathrm{2}{x}}{{x}−\mathrm{2}} \\ $$$${D}_{{f}^{−\mathrm{1}} } =\mathbb{R}−\left\{\mathrm{2}\right\}={R}_{{f}} \\ $$$${so}\:\mathrm{2}\notin{R}_{{f}} \\ $$$$ \\ $$$$ \\ $$