Question Number 213953 by efronzo1 last updated on 22/Nov/24
Answered by mehdee7396 last updated on 22/Nov/24
$${let}\:\:\:{f}\left({x}\right)=\frac{{ax}+{b}}{{cx}+{d}}\Rightarrow{f}\left({f}\left({x}\right)\right)=\frac{{a}\frac{{ax}+{b}}{{cx}+{d}}+{b}}{{c}\frac{{ax}+{b}}{{cx}+{d}}+{d}} \\ $$$$=\frac{\frac{{a}^{\mathrm{2}} {x}+{ab}}{{cx}+{d}}+{b}}{\frac{{acx}+{bc}}{{cx}+{d}}+{d}}=\frac{\left({a}^{\mathrm{2}} +{bc}\right){x}+{ab}+{bd}}{\left({ac}+{cd}\right)+{bc}+{d}^{\mathrm{2}} } \\ $$$$\Rightarrow{a}^{\mathrm{2}} +{bc}=\mathrm{1}\:\:\&\:\:{ac}+{cd}=\mathrm{1}\:\&\:\:{ab}+{bd}=\mathrm{1}\:\:\&\:\:{bc}+{d}^{\mathrm{2}} =\mathrm{2} \\ $$$${bc}==\mathrm{1}−{a}^{\mathrm{2}} \Rightarrow\mathrm{1}−{a}^{\mathrm{2}} +{d}^{\mathrm{2}} =\mathrm{2}\Rightarrow{d}^{\mathrm{2}} −{a}^{\mathrm{2}} =\mathrm{1} \\ $$$${abc}+{bcd}={b}\:\:\&\:\:{abc}+{bcd}={c}\Rightarrow{b}={c} \\ $$$$\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{1}\:\:\&\:\:{b}^{\mathrm{2}} +{d}^{\mathrm{2}} =\mathrm{2} \\ $$$$\:{for}\:\:{a}=\mathrm{0}\:\&\:{b}={d}={c}=\mathrm{1} \\ $$$${if}\:\:\:{f}\left({x}\right)=\frac{\mathrm{1}}{{x}+\mathrm{1}}\Rightarrow{f}\left({f}\left({x}\right)\right)=\frac{{x}+\mathrm{1}}{{x}+\mathrm{2}} \\ $$$$\Rightarrow{f}\left(\mathrm{3}\right)=\frac{\mathrm{1}}{\mathrm{4}}\: \\ $$$$ \\ $$