Question

Cho \(\frac{1}{3}< a< 1\)Tìm GTLN

\(Q=\left(1-a\right)\left(3a-1\right)\)

Answer 1

\(Q=-3a^2+4a-1=-3\left(a^2-2.a.\frac{2}{3}+\frac{4}{9}\right)+\frac{1}{3}=-3\left(a-\frac{2}{3}\right)^2+\frac{1}{3}\le\frac{1}{3}\)

\(\Rightarrow Q_{max}=\frac{1}{3}\) khi \(a=\frac{2}{3}\)