(PUC-MG) - Se A= e B= , é igual a:a) b) c) d) e) Resolução, por favor!.... Pergunta de ideia deWiseNath

May 2019 1 94 Report

(PUC-MG) - Se A= $\frac{2}{e^{ x} + e ^{ - x} }$ e B= $\frac{ {e}^{x} - \: {e}^{- x} }{ {e}^{x} \: + \: {e}^{- x} }$ , ${a}^{2} + {b}^{2}$ é igual a:

a) $1$
b) $2$
c) $4$
d) $e$
e) ${e}^{2}$

Resolução, por favor!

Lista de comentários

DanJR

Resposta:

$\boxed{\mathtt{a}}$

Explicação passo-a-passo:

$\\ \displaystyle \mathsf{a^2 + b^2 = \left ( \frac{2}{e^x + e^{- x}} \right )^2 + \left ( \frac{e^x - e^{- x}}{e^x + e^{- x}} \right )^2} \\\\\\ \mathsf{\qquad \ \quad = \frac{2^2}{(e^x + e^{- x})^2} + \frac{(e^x)^2 - 2 \cdot e^x \cdot e^{- x} + (- e^{- x})^2}{(e^x + e^{- x})^2}} \\\\\\ \mathsf{\qquad \ \quad = \frac{4}{(e^x + e^{- x})^2} + \frac{e^{2x} - 2 \cdot e^0 + e^{- 2x}}{(e^x + e^{- x})^2}} \\\\\\ \mathsf{\qquad \ \quad = \frac{4 + e^{2x} - 2 + e^{- 2x}}{(e^x + e^{- x})^2}} \\\\\\ \mathsf{\qquad \ \quad = \frac{e^{2x} + 2 + e^{- 2x}}{(e^x + e^{- x})^2}}$

$\displaystyle \\ \mathsf{\qquad \ \quad = \frac{(e^{x})^2 + 2 \cdot 1 + (e^{-x})^2}{(e^x + e^{- x})^2}} \\\\\\ \mathsf{\qquad \ \quad = \frac{(e^{x})^2 + 2 \cdot \left ( e^x \cdot e^{- x} \right ) + (e^{-x})^2}{(e^x + e^{- x})^2}} \\\\\\ \mathsf{\qquad \ \quad = \frac{\left ( e^x + e^{- x} \right )^2}{(e^x + e^{- x})^2}} \\\\\\ \mathsf{\qquad \ \quad = \left ( \frac{e^x + e^{- x}}{e^x + e^{- x}} \right )^2} \\\\\\ \mathsf{\qquad \ \quad = \left ( 1 \right )^2} \\\\\\ \boxed{\boxed{\mathsf{a^2 + b^2 = 1}}}$