Đáp án:
B
Giải thích các bước giải:
\(\begin{array}{l}
\frac{{4\sin 2x + \cos 2x + 17}}{{3\cos 2x + \sin 2x + m + 1}} \ge 2\,\,\forall x \in R\\
DKXD:\,\,3\cos 2x + \sin 2x + m + 1 \ne 0\,\,\forall x \in R\\
\Rightarrow {3^2} + {1^2} \le {\left( {m + 1} \right)^2} \Leftrightarrow {\left( {m + 1} \right)^2} \ge 10\\
\Leftrightarrow \left[ \begin{array}{l}
m + 1 \ge \sqrt {10} \\
m + 1 \le - \sqrt {10}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
m \ge - 1 + \sqrt {10} \\
m \le - 1 - \sqrt {10}
\end{array} \right.\\
BPT \Leftrightarrow 4\sin 2x + \cos 2x + 17 \ge 6\cos 2x + 2\sin 2x + 2m + 2\,\,\forall x\\
\Leftrightarrow 2\sin 2x - 5\cos 2x \ge 2m - 15\,\,\forall x\\
\Leftrightarrow 2m - 15 \le \min \left( {2\sin 2x - 5\cos 2x} \right)\\
Ta\,\,co:\,\, - \sqrt {{2^2} + {5^2}} \le 2\sin 2x - 5\cos 2x \le \sqrt {{2^2} + {5^2}} \\
\Leftrightarrow - \sqrt {29} \le 2\sin 2x - 5\cos 2x \le \sqrt {29} \\
\Rightarrow 2m - 15 \le - \sqrt {29} \Leftrightarrow 2m \le 15 - \sqrt {29} \Leftrightarrow m \le \frac{{15 - \sqrt {29} }}{2}\\
Vay\,\, - 1 + \sqrt {10} \le m \le \frac{{15 - \sqrt {29} }}{2}\,\,hoac\,\,m \le - 1 - \sqrt {10} \\
Chon\,\,B.
\end{array}\)
