Tìm giá trị nhỏ nhât của hàm số 

D

- Hướng dẫn giải

Đáp án D

$\begin{array}{l}y=2\cos \frac{x}{2}+\mathrm{s}i\text{nx}+1\\ y\text{'}=-\sin \frac{x}{2}+\cos x=-2{\sin }^2\frac{x}{2}-\sin \frac{x}{2}+1\\ y\text{'}=0\iff \left[\begin{array}{l}\sin \frac{x}{2}=-1\\ \sin \frac{x}{2}=\frac{1}{2}\end{array}\right.\\ \iff \left[\begin{array}{l}\frac{x}{2}=-\frac{\pi }{2}+k2\pi \\ \frac{x}{2}=\frac{\pi }{6}+k2\pi \\ \frac{x}{2}=\frac{5\pi }{6}+k2\pi \end{array}\right.\iff \left[\begin{array}{l}x=-\pi +k4\pi \\ x=\frac{\pi }{3}+k4\pi \\ x=\frac{5\pi }{3}+k4\pi \end{array}\right.\\ y(-\pi )=1\\ y\left(0\right)=3\\ y\left(\frac{\pi }{3}\right)=\frac{2+3\sqrt{3}}{2}\\ y\left(\frac{5\pi }{3}\right)=\frac{2-3\sqrt{3}}{2}\\ y\left(\pi \right)=1\\ \Rightarrow \min y=\frac{2-3\sqrt{3}}{2}\end{array}$