Cho hàm số y=x/2+sin^2x, x thuộc [0;π] hỏi hàm số đồng biến trên khoảng nào

Đáp án: C. $\left({0;\dfrac{7\pi}{12}}\right)$ và $\left({\dfrac{7\pi}{12}};\dfrac{11\pi}{12}\right)$
Lời giải:

$\begin{array}{l} y = f\left( x \right) = \dfrac{x}{2} + {\sin ^2}x\\ f'\left( x \right) = \dfrac{1}{2} + 2\sin x\cos x = \dfrac{1}{2} + \sin 2x>0 \end{array}$

Xét $\dfrac12+\sin2x=0\Leftrightarrow\sin2x=-\dfrac12$

$\Leftrightarrow\left[\begin{array}{I}2x=-\dfrac{\pi}6+k2\pi\\2x=\pi+\dfrac{\pi}6+k2\pi\end{array}\right.\Leftrightarrow\left[\begin{array}{I}x=-\dfrac{\pi}{12}+k\pi\\x=\dfrac{7\pi}{12}+k\pi\end{array}\right.$

$x\in[0;\pi]\Leftrightarrow\left[\begin{array}{I}x=\dfrac{11\pi}{12}(0\le-\dfrac{\pi}{12}+k\pi\le\pi\Leftrightarrow k=1)\\x=\dfrac{7\pi}{12}(0\le\dfrac{7\pi}{12}+k\pi\le\pi\Leftrightarrow k=0)\end{array}\right.$

$ \dfrac{1}{2} + \sin 2x>0\Leftrightarrow \left[ \begin{array}{l} 0 < x < \dfrac{{7\pi }}{{12}}\\ \dfrac{{11\pi }}{{12}} < x < \pi \end{array} \right.$