Ensino Superior ⇒ (Bulgária) Funções Tópico resolvido

[SkyWalker17]

(Bulgária) Sejam f₁(x) = 1 + [tex3]\frac{1}{x}[/tex3] e f_n+1(x) = f₁(f_n(x)), para todo n [tex3]\in [/tex3] [tex3]\mathbb{N}[/tex3].

Quantas soluções reais e distintas tem a equação x = f₂₀₀₁(x)?

Gabarito: 2 raízes reais e distintas.

[rcompany]

[tex3]

\text{Seja }u_0=0,\,u_1=1 \text{ e }u_{n+2}=u_{n+1}+u_{n}\\
f_1(x)=1+\frac{1}{x}=\frac{1+x}{x}=\frac{u_1+u_2x}{u_0+u_1x}\\
\text{Seja }n\in\mathbb{N}^*:\\
f_n(x)=\frac{u_n+u_{n+1}x}{u_{n-1}+u_nx}\implies f_{n+1}(x)=\!1\!\!+\!\!\frac{u_{n-1}\!\!+\!\!u_nx}{u_n\!\!+\!\!u_{n+1}x}=\frac{u_n+u_{n-1}\!\!+\!\!(u_n\!\!+\!\!u_{n+1})x}{u_n\!\!+\!\!u_{n+1}x}=\frac{u_{n+1}\!\!+\!\!u_{n+2}x}{u_n\!\!+\!\!u_{n+1}x}\\
\text{e portanto }\forall n\in\mathbb{N}^*,\,f_n(x)=f_n(x)=\frac{u_n\!+\!u_{n+1}x}{u_{n-1}\!+\!u_nx}\\[24pt]
\begin{aligned}
\forall n \in \mathbb{N}^*, f_n(x)=x&\iff\frac{u_n+u_{n+1}x}{u_{n-1}+u_nx}=x\\[6pt]
&\iff u_nx^2+(u_{n-1}-u_{n+1})x-u_n=u_nx^2-u_nx-u_n=0\\[6pt]
&\iff\left\{\begin{array}{l}x=\dfrac{u_n+\sqrt{u_n^2+4u_n^2}}{2u_n}=\dfrac{1+\sqrt{5}}{2}\quad\text{já que }\forall n\in\mathbb{N}^*,\,u_n>0\\\text{ou}\\x=\dfrac{1-\sqrt{5}}{2}\end{array}\right.
\end{aligned}
[/tex3]

[tex3]
\boxed{\\\hspace{1cm}\\[3pt]\hspace{1cm}\forall n\in\mathbb{N}^*,\,f_n(x)=x\text{ tem 2 raízes reais distintas}\hspace{1cm}\\}
[/tex3]