# Ex :EF

• existe t-il une fonction $f :{\mathbb R}\longrightarrow{\mathbb R}$ telle que : $f(f(x))+xf(x)=1$ pour tout $x\in {\mathbb R}$ ?

• $No$ Because $P(0)$ $\Rightarrow f(f(0))=1$
$P(f(0))\Rightarrow f(0)+f(1)=1$
$P(1) \Rightarrow f(f(1))+f(1)=1$
So $f(f(1)=f(0)$ So $f(f(f(1)))=1$
$P(f(1)) \Rightarrow f(1)*f(f(1)=0$
If $f(1)=0$ Then $f(0)=1$ Then $f(1)=f(f(0))=1$ Absurde!
If $f(f(1))=0$ then $f(1)=1$ Then $f(f(1)=f(1)=1$ Absurde!
So there is no function that satisfy the given conditions

Looks like your connection to Expii Forum was lost, please wait while we try to reconnect.