Questão 73 - Prova Medicina - FMABC 2025

Sendo $\mathrm{f}\left(\mathrm{x}\right)=2^{\mathrm{mx}-\mathrm{p}}$ e $\mathrm{g}\left(\mathrm{x}\right)=\mathrm{m}-{\log }_2\mathrm{x}$ e, pelo gráfico, se x = 1 $\Rightarrow$y = 2 e f(1) = g(10) = 2

$$\begin{gathered} \mathrm{f}\left(1\right) = 2^{\mathrm{m}-\mathrm{p}} = \mathrm{g}\left(1\right) = \mathrm{m} - \cancel{{\log }_2}1 = 2 \Rightarrow 2^{\mathrm{m}-\mathrm{p}} = \mathrm{m} = 2 \Rightarrow \left\{\begin{array}{l}\begin{array}{c}\mathrm{m}=2\\ \mathrm{e}\end{array}\\ 2^{\mathrm{m}-\mathrm{p}}=2\end{array}\Rightarrow 2^{2-\mathrm{p}}=2^1\Rightarrow \mathrm{p}=1\right. \\ \mathrm{Se} \mathrm{f}\left(2\right) = 2^{2\mathrm{m}-\mathrm{p}} = \mathrm{k} \Rightarrow \mathrm{f}\left(2\right) = 2^{2\left(2\right) - \left(1\right)} = \mathrm{k} \Rightarrow 2^3 = \mathrm{k} \Rightarrow \mathrm{k} = 8 \\ \mathrm{g}\left(\mathrm{k}\right) = \mathrm{m} - {\log }_2\mathrm{k} \Rightarrow \mathrm{g}\left(8\right) = \mathrm{m} - {\log }_{2}8 \Rightarrow \mathrm{g}\left(8\right) = 2 - 3 \Rightarrow \boxed{\mathrm{g}\left(8\right)=-1} \end{gathered}$$