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Question Number 161817 by HongKing last updated on 22/Dec/21

Answered by aleks041103 last updated on 23/Dec/21

f(x)=(e^(−1) ○g○e)(x) e(x)=x^(2021) g(x)=(1/(1−x)) e^(−1) (x)=x^(1/2021) ⇒(f^n )(x)=(e^(−1) ○g^n ○e)(x) Ω_n =e^(−1) ((g^n )(e(x))) g^(−1) (x)=1−(1/x) g^2 (x)=(1/(1−(1/(1−x))))=((x−1)/x)=1−(1/x)=g^(−1) (x) ⇒g^3 (x)=x ⇒g^(2022) (x)=g^(3.674) (x)=(g^3 )^(674) (x)=id^(674) (x)=x ⇒Ω_(2022) =(e^(−1) ○id○e)(x)=(e^(−1) ○e)(x)= =id(x)=x=Ω_(2022) ⇒Ans. x=2021

$${f}\left({x}\right)=\left({e}^{−\mathrm{1}} \circ{g}\circ{e}\right)\left({x}\right) \\ $$$${e}\left({x}\right)={x}^{\mathrm{2021}} \\ $$$${g}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{1}−{x}} \\ $$$${e}^{−\mathrm{1}} \left({x}\right)={x}^{\mathrm{1}/\mathrm{2021}} \\ $$$$ \\ $$$$\Rightarrow\left({f}^{{n}} \right)\left({x}\right)=\left({e}^{−\mathrm{1}} \circ{g}^{{n}} \circ{e}\right)\left({x}\right) \\ $$$$\Omega_{{n}} ={e}^{−\mathrm{1}} \left(\left({g}^{{n}} \right)\left({e}\left({x}\right)\right)\right) \\ $$$${g}^{−\mathrm{1}} \left({x}\right)=\mathrm{1}−\frac{\mathrm{1}}{{x}} \\ $$$${g}^{\mathrm{2}} \left({x}\right)=\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}−{x}}}=\frac{{x}−\mathrm{1}}{{x}}=\mathrm{1}−\frac{\mathrm{1}}{{x}}={g}^{−\mathrm{1}} \left({x}\right) \\ $$$$\Rightarrow{g}^{\mathrm{3}} \left({x}\right)={x} \\ $$$$\Rightarrow{g}^{\mathrm{2022}} \left({x}\right)={g}^{\mathrm{3}.\mathrm{674}} \left({x}\right)=\left({g}^{\mathrm{3}} \right)^{\mathrm{674}} \left({x}\right)={id}^{\mathrm{674}} \left({x}\right)={x} \\ $$$$\Rightarrow\Omega_{\mathrm{2022}} =\left({e}^{−\mathrm{1}} \circ{id}\circ{e}\right)\left({x}\right)=\left({e}^{−\mathrm{1}} \circ{e}\right)\left({x}\right)= \\ $$$$={id}\left({x}\right)={x}=\Omega_{\mathrm{2022}} \\ $$$$\Rightarrow{Ans}.\:{x}=\mathrm{2021} \\ $$

Commented by HongKing last updated on 25/Dec/21

cool my dear Sir thank you so much

$$\mathrm{cool}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Sir}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\ $$

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