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

Answered by aleks041103 last updated on 23/Dec/21

$$f\left(x\right)=(e^{-1}\circ g\circ e)\left(x\right)$$$$e\left(x\right)=x^{2021}$$$$g\left(x\right)=\frac{1}{1-x}$$$$e^{-1}\left(x\right)=x^{1/2021}$$$$$$$$\Rightarrow \left(f^n\right)\left(x\right)=(e^{-1}\circ g^n\circ e)\left(x\right)$$$${\mathrm{\Omega }}_n=e^{-1}\left(\left(g^n\right)\left(e\left(x\right)\right)\right)$$$$g^{-1}\left(x\right)=1-\frac{1}{x}$$$$g^2\left(x\right)=\frac{1}{1-\frac{1}{1-x}}=\frac{x-1}{x}=1-\frac{1}{x}=g^{-1}\left(x\right)$$$$\Rightarrow g^3\left(x\right)=x$$$$\Rightarrow g^{2022}\left(x\right)=g^{3.674}\left(x\right)={\left(g^3\right)}^{674}\left(x\right)={id}^{674}\left(x\right)=x$$$$\Rightarrow {\mathrm{\Omega }}_{2022}=(e^{-1}\circ id\circ e)\left(x\right)=(e^{-1}\circ e)\left(x\right)=$$$$=id\left(x\right)=x={\mathrm{\Omega }}_{2022}$$$$\Rightarrow Ans.x=2021$$

Commented by HongKing last updated on 25/Dec/21

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

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