Question-204179

Question Number 204179 by mr W last updated on 08/Feb/24

Answered by AST last updated on 08/Feb/24

f(x)=f(y)⇒f(f(x))=f(f(y))⇒x^2 −x=y^2 −y ⇒(x−y)(x+y)=x−y⇒x=1−y or x=y ⇒f(x)=f(1−x)⇒f(0)=f(1) f(f(f(x)))=[f(x)]^2 −f(x)+1=f(f(f(x))) =f(x^2 −x+1) ⇒f(1)=[f(1)]^2 −f(1)+1⇒f(1)=1=f(0)

$${f}\left({x}\right)={f}\left({y}\right)\Rightarrow{f}\left({f}\left({x}\right)\right)={f}\left({f}\left({y}\right)\right)\Rightarrow{x}^{\mathrm{2}} −{x}={y}^{\mathrm{2}} −{y} \\ $$$$\Rightarrow\left({x}−{y}\right)\left({x}+{y}\right)={x}−{y}\Rightarrow{x}=\mathrm{1}−{y}\:{or}\:{x}={y} \\ $$$$\Rightarrow{f}\left({x}\right)={f}\left(\mathrm{1}−{x}\right)\Rightarrow{f}\left(\mathrm{0}\right)={f}\left(\mathrm{1}\right) \\ $$$${f}\left({f}\left({f}\left({x}\right)\right)\right)=\left[{f}\left({x}\right)\right]^{\mathrm{2}} −{f}\left({x}\right)+\mathrm{1}={f}\left({f}\left({f}\left({x}\right)\right)\right) \\ $$$$={f}\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right) \\ $$$$\Rightarrow{f}\left(\mathrm{1}\right)=\left[{f}\left(\mathrm{1}\right)\right]^{\mathrm{2}} −{f}\left(\mathrm{1}\right)+\mathrm{1}\Rightarrow{f}\left(\mathrm{1}\right)=\mathrm{1}={f}\left(\mathrm{0}\right) \\ $$

Commented by mr W last updated on 08/Feb/24

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Answered by witcher3 last updated on 08/Feb/24

f(f(0))=1 f(f(1))=1 f(0)=a;f(1)=b ⇒f(a)=1⇒f(f(a))=f(1) a^2 −a+1=b f(1)=b⇒f(f(1))=f(b)=1 b^2 −b+1=b⇒b=1 a(a−1)=0⇒a∈{0,1} f(0)=0⇒f(f(0))=f(0)⇒1=0 impossible⇒f(0)=1

$$\mathrm{f}\left(\mathrm{f}\left(\mathrm{0}\right)\right)=\mathrm{1} \\ $$$$\mathrm{f}\left(\mathrm{f}\left(\mathrm{1}\right)\right)=\mathrm{1} \\ $$$$\mathrm{f}\left(\mathrm{0}\right)=\mathrm{a};\mathrm{f}\left(\mathrm{1}\right)=\mathrm{b} \\ $$$$\Rightarrow\mathrm{f}\left(\mathrm{a}\right)=\mathrm{1}\Rightarrow\mathrm{f}\left(\mathrm{f}\left(\mathrm{a}\right)\right)=\mathrm{f}\left(\mathrm{1}\right) \\ $$$$\mathrm{a}^{\mathrm{2}} −\mathrm{a}+\mathrm{1}=\mathrm{b} \\ $$$$\mathrm{f}\left(\mathrm{1}\right)=\mathrm{b}\Rightarrow\mathrm{f}\left(\mathrm{f}\left(\mathrm{1}\right)\right)=\mathrm{f}\left(\mathrm{b}\right)=\mathrm{1} \\ $$$$\mathrm{b}^{\mathrm{2}} −\mathrm{b}+\mathrm{1}=\mathrm{b}\Rightarrow\mathrm{b}=\mathrm{1} \\ $$$$\mathrm{a}\left(\mathrm{a}−\mathrm{1}\right)=\mathrm{0}\Rightarrow\mathrm{a}\in\left\{\mathrm{0},\mathrm{1}\right\} \\ $$$$\mathrm{f}\left(\mathrm{0}\right)=\mathrm{0}\Rightarrow\mathrm{f}\left(\mathrm{f}\left(\mathrm{0}\right)\right)=\mathrm{f}\left(\mathrm{0}\right)\Rightarrow\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{impossible}\Rightarrow\mathrm{f}\left(\mathrm{0}\right)=\mathrm{1} \\ $$$$ \\ $$

Commented by mr W last updated on 08/Feb/24

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