Question

Giá trị cực đại của hàm số \(y=x+sin2x\) trên \(\left(0;\pi\right)\)là

\(A.\frac{\pi}{6}+\frac{\sqrt{3}}{2}\)

\(B.\frac{2\pi}{3}+\frac{\sqrt{3}}{2}\)

\(C.\frac{2\pi}{3}-\frac{\sqrt{3}}{2}\)

\(D.\frac{\pi}{3}+\frac{\sqrt{3}}{2}\)

Answer 1

\(y=x+sin\left(2x\right)\)

\(y'=1+2cos\left(2x\right)\)

\(y'=0\Leftrightarrow1+cos\left(2x\right)=0\Leftrightarrow\orbr{\begin{cases}x=\frac{\pi}{3}\\x=\frac{2\pi}{3}\end{cases}}\)vì \(x\in\left(0,\pi\right)\).

\(y\left(\frac{\pi}{3}\right)=\frac{\pi}{3}+\frac{\sqrt{3}}{2},y\left(\frac{2\pi}{3}\right)=\frac{2\pi}{3}-\frac{\sqrt{3}}{2}\)

\(y\left(\frac{\pi}{3}\right)>y\left(\frac{2\pi}{3}\right)\)ta chọn D.