1.
Sai, vì hàm \(f\left(x\right)\) ko phải hàm tuần hoàn.
\(f\left(x+2\pi\right)=\left|x+2\pi\right|.sin\left(x+2\pi\right)=\left|x+2\pi\right|.sinx\ne f\left(x\right)\)
2.
Khi \(\dfrac{\pi}{3}\le x\le\dfrac{5\pi}{6}\Rightarrow-\dfrac{\sqrt{3}}{2}\le cosx\le\dfrac{1}{2}\)
Đặt \(cosx=t\Rightarrow y=4t^2-4t+3\) với \(t\in\left[-\dfrac{\sqrt{3}}{2};\dfrac{1}{2}\right]\)
Ta có: \(-\dfrac{b}{2a}=-\dfrac{-4}{2.4}=\dfrac{1}{2}\)
\(y\left(-\dfrac{\sqrt{3}}{2}\right)=6+2\sqrt{3}\); \(y\left(\dfrac{1}{2}\right)=2\)
Vậy \(y_{min}=2\) khi \(cosx=\dfrac{1}{2}\Rightarrow x=\dfrac{\pi}{3}\)
\(y_{max}=6+2\sqrt{3}\) khi \(cosx=-\dfrac{\sqrt{3}}{2}\Rightarrow x=\dfrac{5\pi}{6}\)
