Dissertativa 3 - 2ª Fase - Dia 1 - IME 2026

Como x está no intervalo [0, 1[, f(x) representa a soma de uma PG infinita convergente.

$\mathrm{f}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=1}^{\infty } {\mathrm{x}}^{\frac{\mathrm{n}}{2}}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\frac{\sqrt{\mathrm{x}}}{1-\sqrt{\mathrm{x}}}$

Logo:

$\left(a, \mathrm{f}\left(a\right)\right)=\left(\mathrm{\alpha }, \frac{\sqrt{a}}{1-\sqrt{a}}\right) \mathrm{e} \left(\mathrm{x}, \mathrm{f}\left(\mathrm{x}\right)\right)=\left(\mathrm{x}, \frac{\sqrt{\mathrm{x}}}{1-\sqrt{\mathrm{x}}}\right)$

A inclinação será: 

$$\begin{gathered} \frac{\mathrm{f}\left(a\right)-\mathrm{f}\left(\mathrm{x}\right)}{a-\mathrm{x}}=\frac{{\displaystyle \frac{\sqrt{a}}{1-\sqrt{a}}-\frac{\sqrt{\mathrm{x}}}{1-\sqrt{\mathrm{x}}}}}{a-\mathrm{x}}=\frac{\sqrt{a}-\cancel{\sqrt{ax}}-\sqrt{x}+\cancel{\sqrt{ax}}}{\left(1-\sqrt{a}\right)\left(1-\sqrt{\mathrm{x}}\right)\left(a-\mathrm{x}\right)} \\ \frac{\mathrm{f}\left(a\right)-\mathrm{f}\left(\mathrm{x}\right)}{a-\mathrm{x}}=\frac{1}{\left(1-\sqrt{a}\right)\left(1-\sqrt{\mathrm{x}}\right)\left(\sqrt{a}+\sqrt{\mathrm{x}}\right)}=\frac{1}{\left(1-\sqrt{a}\right)\left[-\mathrm{x}+\left(1-\sqrt{a}\right)\sqrt{\mathrm{x}}+\sqrt{a}\right]} \end{gathered}$$

Pretende-se maximizar $-\mathrm{x}+\left(1-\sqrt{a}\right)\sqrt{\mathrm{x}}+\sqrt{a}$, uma função do 2º grau em $\sqrt{\mathrm{x}}$.

$\mathrm{\Delta }={\left(1-\sqrt{a}\right)}^2+4\sqrt{a}=a+2\sqrt{a}+1={\left(\sqrt{a}+1\right)}^2$

Seu valor máximo é $\frac{-\mathrm{\Delta }}{-4}=\frac{\mathrm{\Delta }}{4}$.

Logo, $\mathrm{coef} {\mathrm{ang}}_{\min }=\frac{4}{\left(1-\sqrt{a}\right){\left(\sqrt{a}+1\right)}^2}$.

Como o coeficiente angular de $\mathrm{ky}-4\mathrm{x}=2026$ é $\frac{\mathrm{k}}{4}$:

$\frac{4}{\mathrm{k}}=\frac{4}{\left(1-\sqrt{a}\right){\left(\sqrt{a}+1\right)}^2}\Rightarrow \boxed{\mathrm{k}=\left(1-a\right)\left(1+\sqrt{a}\right)}$