Gọi M, m lần lượt là giá trị lớn nhất...

C

- Hướng dẫn giải

Đáp án C.

Ta có: $f^2\left(x\right)=2+\sin x+\cos x+2\sqrt{\left(1+\sin x\right)\left(1+cosx\right)}$ 

$=2+\sin x+\cos x+2\sqrt{1+\sin x+\cos x+\sin x\cos x}$ 

Đặt $t=\sin x+\cos x=\sqrt{2}\sin \left(x+\frac{\mathrm{\pi }}{4}\right)\Rightarrow t\in \left[-\sqrt{2};\sqrt{2}\right]$. 

Suy ra $\sin x\cos x=\frac{t^2-1}{2}\Rightarrow f^2\left(x\right)=2+t+2\sqrt{1+t+\frac{t^2-1}{2}}=2+t+\sqrt{2}\sqrt{t^2+2t+1}$

$\Rightarrow f\left(t\right)=t+2+\sqrt{2}\left|t+1\right|=\left\{\begin{array}{l}t+2+\sqrt{2}\left(t+1\right) khi t\ge -1\\ t+2-\sqrt{2}\left(t+1\right) khi t<-1\end{array}\right.=\left\{\begin{array}{l}\left(1+\sqrt{2}\right)t+2+\sqrt{2} khi t\ge -1\\ \left(1-\sqrt{2}\right)t+2-\sqrt{2} khi t<-1\end{array}\right.$ 

Từ đó suy ra $1\le f^2\left(x\right)\le 4+2\sqrt{2}\iff f\left(x\right)\le \sqrt{4+2\sqrt{2}}\Rightarrow M-m=\sqrt{4+2\sqrt{2}}-1$.