- Hướng dẫn giải

Ta có

$$\begin{gathered} {\sin }^5\left(x\right)\le {\sin }^4\left(x\right) \\ \Rightarrow y\le {\sin }^4\left(x\right)+\sqrt{3}\cos \left(x\right) \end{gathered}$$

Áp dụng bất đẳng thức Cauchy ta có:

$$\begin{gathered} \left(1-\cos \left(x\right)\right)\left(1+\cos \left(x\right)\right)\left(1+\cos \left(x\right)\right) \\ =\frac{1}{2}\left(2-2\cos \left(x\right)\right)\left(1+\cos \left(x\right)\right)\left(1+\cos \left(x\right)\right) \end{gathered}$$

$$\begin{gathered} \le \frac{1}{2}{\left(\frac{2-2\cos \left(x\right)+{(1+\cos \left(x\right))}^2}{3}\right)}^3 \\ =\frac{32}{27}<\sqrt{3} \end{gathered}$$

$$\begin{gathered} \Rightarrow \sqrt{3}-\left(1-\cos x\right){\left(1+\cos x\right)}^2>0 \\ \Rightarrow \left(1-\cos x\right)\left[\sqrt{3}-\left(1-\cos x\right){\left(1+\cos x\right)}^2\right]\ge 0 \\ \Rightarrow \sqrt{3}\left(1-\cos x\right)-{\sin }^4\left(x\right)\ge 0 \\ \iff {\sin }^4\left(x\right)+\sqrt{3}\cos x\le \sqrt{3} \end{gathered}$$

M = maxy = $\sqrt{3}$$\iff$cos(x) = 1

$\iff$$x=k2\mathrm{\pi }, \mathrm{k}\in \mathrm{\mathbb{Z}}$

Ta lại có

$y\ge -{\sin }^4\left(x\right)+\sqrt{3}\cos \left(x\right)$

Tương tự như trên, áp dụng bất đẳng thức Cauchy ta có:

$$\begin{gathered} \left(1+\cos x\right)\left(1-\cos x\right)\left(1-\cos x\right) \\ =\frac{1}{2}\left(2+2\cos x\right){\left(1-\cos x\right)}^2 \\ \ge \frac{32}{27}\le \sqrt{3} \\ \Rightarrow \sqrt{3}-\left(1+\cos x\right){\left(1-\cos x\right)}^2>0 \\ \Rightarrow \left(1+\cos x\right)\left[\sqrt{3}-\left(1+\cos x\right){\left(1-\cos x\right)}^2\right] \\ \iff {\sin }^4\left(x\right)+\sqrt{3}\cos \left(x\right)\ge -\sqrt{3} \\ m=miny=-\sqrt{3} \\ \iff \cos x=-1 \\ \iff x=k2\mathrm{\pi },\mathrm{k}\in \mathrm{\mathbb{Z}} \end{gathered}$$

Do đó $\frac{M}{m}=-1$. Vì vậy, mệnh đề D sai.

Đáp án cần chọn là D