Với giá trị nào của $m$ thì hàm số sau đây xác định với mọi
$X\in R :$
$y=log\sqrt{cos2x+mcosx+4}              (1) $
Đặt
$\begin{array}{l}
t = \cos X, - 1 \le t \le 1\\
 \Rightarrow c{\rm{os}}2X + m.\cos X + 4 = 2{\cos ^2}X - 1 + m.\cos X + 4 = 2{t^2} + mt + 3
\end{array}$
Hàm số $(1)$ xác định
$\begin{array}{l}
\forall X \in R\\
 \Leftrightarrow 2{t^2} + mt + 3 > 0\\
\forall t \in \left[ { - 1,1} \right]
\end{array}$
Đặt
$f(t) > 0\forall t \in \left[ { - 1,1} \right]$
$ \Leftrightarrow \left[ \begin{array}{l}
\Delta  < 0(2)\\
\left\{ \begin{array}{l}
\Delta  \ge 0\\
\left[ \begin{array}{l}
 - 1 < 1 < {t_1} < {t_2}\\
{t_1} \le {t_2} <  - 1 < 1
\end{array} \right.
\end{array} \right.(3)
\end{array} \right.$
Ta có:
$\Delta  = {m^2} - 24;f(1) = m + 5;f( - 1) =  - m + 5$
(HV-18)
$\begin{array}{l}
(2) \Leftrightarrow  - 2\sqrt 6  < m < 2\sqrt 6 \\
(3) \Leftrightarrow \left\{ \begin{array}{l}
\left\{ {m \le  - 2\sqrt 6 } \right. \vee m \ge 2\sqrt 6  \vee m \ge 2\sqrt 6 \\
\left\{ \begin{array}{l}
f(1) > 0\\
\frac{s}{2} - 1 > 0
\end{array} \right. \vee \left\{ \begin{array}{l}
f( - 1) > 0\\
\frac{s}{2} + 1 < 0
\end{array} \right.
\end{array} \right.\\
\left\{ \begin{array}{l}
f(1) > 0\\
\frac{s}{2} - 1 > 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
m + 5 > 0\\
 - \frac{m}{4} - 1 > 0
\end{array} \right. \Leftrightarrow  - 5 < m <  - 4\\
\left\{ \begin{array}{l}
f( - 1) > 0\\
\frac{s}{2} + 1 < 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
 - m + 5 > 0\\
 - \frac{m}{4} + 1 < 0
\end{array} \right. \Leftrightarrow 4 < m < 5\\
suy\,\,ra\,\,\,(3) \Leftrightarrow \left[ \begin{array}{l}
 - 5 < m \le  - 2\sqrt 6 \\
2\sqrt 6  \le m \le 5
\end{array} \right.(5)
\end{array}$
Hợp các tập nghiệm ở $(4)$và $(5)$ ta có
$ - 5 < m < 5$.  Vậy ${\rm{D =  ( - 5, 5)}}$
Vote nhiều nhiều bạn nhé –  Robinhood 25-06-12 03:00 PM

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