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Câu hỏi: Tìm m để hàm $y=\sqrt{\cos 3x-9\cos x-m}$ có tập xác định.
A. $m\ge -8$
B. $m=3$
C. $m<-8$
D. $m\le -8$
A. $m\ge -8$
B. $m=3$
C. $m<-8$
D. $m\le -8$
Ta có: $y=\sqrt{\cos 3x-9\cos x-m}=\sqrt{4{{\cos }^{3}}x-12\cos x-m}=\sqrt{4{{t}^{3}}-12t-m},t=\cos x$.
Theo bài ra $4{{t}^{3}}-12t-m\ge 0,\forall t\in \left[ -1;1 \right]\Rightarrow \min \left( 4{{t}^{3}}-12t-m \right)\ge 0,\forall t\in \left[ -1;1 \right]$.
${f}'\left( t \right)=12{{t}^{2}}-12=0\Leftrightarrow t\in \left\{ -1;1 \right\}\Rightarrow \min f\left( t \right)=f\left( 1 \right)=-8-m\Rightarrow -8-m\ge 0\Leftrightarrow m\le -8$.
Theo bài ra $4{{t}^{3}}-12t-m\ge 0,\forall t\in \left[ -1;1 \right]\Rightarrow \min \left( 4{{t}^{3}}-12t-m \right)\ge 0,\forall t\in \left[ -1;1 \right]$.
${f}'\left( t \right)=12{{t}^{2}}-12=0\Leftrightarrow t\in \left\{ -1;1 \right\}\Rightarrow \min f\left( t \right)=f\left( 1 \right)=-8-m\Rightarrow -8-m\ge 0\Leftrightarrow m\le -8$.
Đáp án D.