Question-213953

Question Number 213953 by efronzo1 last updated on 22/Nov/24

Answered by mehdee7396 last updated on 22/Nov/24

let f(x)=((ax+b)/(cx+d))⇒f(f(x))=((a((ax+b)/(cx+d))+b)/(c((ax+b)/(cx+d))+d)) =((((a^2 x+ab)/(cx+d))+b)/(((acx+bc)/(cx+d))+d))=(((a^2 +bc)x+ab+bd)/((ac+cd)+bc+d^2 )) ⇒a^2 +bc=1 & ac+cd=1 & ab+bd=1 & bc+d^2 =2 bc==1−a^2 ⇒1−a^2 +d^2 =2⇒d^2 −a^2 =1 abc+bcd=b & abc+bcd=c⇒b=c ⇒a^2 +b^2 =1 & b^2 +d^2 =2 for a=0 & b=d=c=1 if f(x)=(1/(x+1))⇒f(f(x))=((x+1)/(x+2)) ⇒f(3)=(1/4)

$${let}\:\:\:{f}\left({x}\right)=\frac{{ax}+{b}}{{cx}+{d}}\Rightarrow{f}\left({f}\left({x}\right)\right)=\frac{{a}\frac{{ax}+{b}}{{cx}+{d}}+{b}}{{c}\frac{{ax}+{b}}{{cx}+{d}}+{d}} \\ $$$$=\frac{\frac{{a}^{\mathrm{2}} {x}+{ab}}{{cx}+{d}}+{b}}{\frac{{acx}+{bc}}{{cx}+{d}}+{d}}=\frac{\left({a}^{\mathrm{2}} +{bc}\right){x}+{ab}+{bd}}{\left({ac}+{cd}\right)+{bc}+{d}^{\mathrm{2}} } \\ $$$$\Rightarrow{a}^{\mathrm{2}} +{bc}=\mathrm{1}\:\:\&\:\:{ac}+{cd}=\mathrm{1}\:\&\:\:{ab}+{bd}=\mathrm{1}\:\:\&\:\:{bc}+{d}^{\mathrm{2}} =\mathrm{2} \\ $$$${bc}==\mathrm{1}−{a}^{\mathrm{2}} \Rightarrow\mathrm{1}−{a}^{\mathrm{2}} +{d}^{\mathrm{2}} =\mathrm{2}\Rightarrow{d}^{\mathrm{2}} −{a}^{\mathrm{2}} =\mathrm{1} \\ $$$${abc}+{bcd}={b}\:\:\&\:\:{abc}+{bcd}={c}\Rightarrow{b}={c} \\ $$$$\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{1}\:\:\&\:\:{b}^{\mathrm{2}} +{d}^{\mathrm{2}} =\mathrm{2} \\ $$$$\:{for}\:\:{a}=\mathrm{0}\:\&\:{b}={d}={c}=\mathrm{1} \\ $$$${if}\:\:\:{f}\left({x}\right)=\frac{\mathrm{1}}{{x}+\mathrm{1}}\Rightarrow{f}\left({f}\left({x}\right)\right)=\frac{{x}+\mathrm{1}}{{x}+\mathrm{2}} \\ $$$$\Rightarrow{f}\left(\mathrm{3}\right)=\frac{\mathrm{1}}{\mathrm{4}}\: \\ $$$$ \\ $$

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Question-213953

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