Cho hàm số

A

- Hướng dẫn giải

Đáp án A.

Điều kiện $x\in ℝ$

 $\begin{array}{l}y=\cos x+\cos \left(x-\frac{\pi }{3}\right)\\ =\cos x+\cos x.\cos \frac{\pi }{3}+\sin x.\sin \frac{\pi }{3}\\ =\cos x+\frac{1}{2}\cos x+\frac{\sqrt{3}}{2}\sin x\end{array}$

$=\frac{3}{2}\cos x+\frac{\sqrt{3}}{2}\sin x$

Cách 1: $y=\sqrt{3}\left(\frac{\sqrt{3}}{2}\cos x+\frac{1}{2}\sin x\right)$$=\sqrt{3}\sin \left(x+\frac{\pi }{3}\right)$ Suy ra $-\sqrt{3}\le y\le \sqrt{3}$

Vậy  $m=-\sqrt{3};M=\sqrt{3}$và do đó $M^2+m^2=6$

Cách 2:

Áp dụng bất đẳng thức Bunyakovsky ta có:

 ${\left(\frac{3}{2}\cos x+\frac{\sqrt{3}}{2}\sin x\right)}^2$$\le \left[{\left(\frac{3}{2}\right)}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2\right]\left[{\left(\cos x\right)}^2+{\left(\sin x\right)}^2\right]$

 $\begin{array}{l}\iff {\left(\frac{3}{2}\cos x+\frac{\sqrt{3}}{2}\sin x\right)}^2\le 3\\ \iff -\sqrt{3}\le y\le \sqrt{3}\end{array}$

 $\Rightarrow M=\sqrt{3}$khi  $\left\{\begin{array}{l}\frac{2}{3}\cos x=\frac{2}{\sqrt{3}}\sin x\\ \frac{3}{2}\cos x+\frac{\sqrt{3}}{2}\sin x=\sqrt{3}\end{array}\right.$

Tương tự ta có $m=-\sqrt{3}$   khi  $\left\{\begin{array}{l}\frac{2}{3}\cos x=\frac{2}{\sqrt{3}}\sin x\\ \frac{3}{2}\cos x+\frac{\sqrt{3}}{2}\sin x=-\sqrt{3}\end{array}\right.$

$\Rightarrow M^2+m^2={\left(\sqrt{3}\right)}^2+{\left(-\sqrt{3}\right)}^2=6$

Vậy ta chọn A.