Question

A função f: R → R, definida por f(x) = ax2 - √2, com a >
0, é tal que f[f(√2)] = -√2. O valor de a é
A) 2-√2/2 B) 1/2 C) 2 - √2 D) √2/2 E) 2+ √2/2

Answer 1

Resposta:

D

Explicação passo a passo:

$f[f(\sqrt{2})] = - \sqrt{2}. (I)\\\\f[(\sqrt{2})] = a.(\sqrt{2})^2-\sqrt{2} = 2a - \sqrt{2}.(II)\\\f[f(\sqrt{2})] = a.(2a-\sqrt{2})^2-\sqrt{2} = a.(4a^2-2.2a.\sqrt{2}+\sqrt{2})-\sqrt{2}\\= a.(4a^2-4a\sqrt{2}+2)-\sqrt{2}.(III)\\(I) = (III)\\\\a.(4a^2-4a\sqrt{2}+2)-\sqrt{2}=-\sqrt{2}\\a.(4a^2-4a\sqrt{2}+2)=0\\a=0; impossivel\\raizes-de-->4a^2-4a\sqrt{2}+2=0\\x=\frac{-(-4\sqrt{2})+-\sqrt{(-4\sqrt{2})^2-4.4.2}}{2.4} \\=\frac{4\sqrt{2}+-\sqrt{(32-32)}}{2.4}=\frac{4\sqrt{2}}{2.4}=\frac{\sqrt{2}}{2}$