Calculemos a inversa de $f$. Seja $y = f^{-1} (x)$ , temos: $x = \Large{\frac{100}{1+2^{-y}}}$ $\implies$ $2^{-y} = \Large{\frac{100-x}{x}}$ $\implies$ $y = f^{-1} (x)$ $=$ $- \log_{2} \left(\Large{\frac{100-x}{x}} \right)$. A composta $g\space \circ \space f^{-1}(90) = g(f^{-1}(90))$ . Assim: $f^{-1}(90) = - \log_{2} \left(\Large{\frac{100-90}{90}} \right)$ $=$ $- \log_{2} \left(\Large{\frac{1}{9}} \right)$ $=$ $2\log_{2} 3$ $g\space \circ \space f^{-1}(90)$ $=$ $g(2\log_{2} 3)$ $=$ $2^{\large{\frac{2\log_{2}3}{2}}}$ $=$ $2^{\large{\log_{2} 3}}$ $=$ $\Large{3}$ Alternativa $\mathbb{(B)}$