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Câu hỏi: Tất cả các giá trị thực của m để bất phương trình $x\sqrt{x}+\sqrt{x+12}\le m{{\log }_{5-\sqrt{4-x}}}3$ có nghiệm:
A. $m>2\sqrt{3}$.
B. $m>12{{\log }_{3}}5$.
C. $m\ge 2\sqrt{3}$.
D. $2<m<12{{\log }_{3}}5$.
A. $m>2\sqrt{3}$.
B. $m>12{{\log }_{3}}5$.
C. $m\ge 2\sqrt{3}$.
D. $2<m<12{{\log }_{3}}5$.
Điều kiện xác định: $\left\{ \begin{matrix}
x\ge 0 \\
x+12\ge 0 \\
4-x\ge 0 \\
5-\sqrt{4-x}>0 \\
5-\sqrt{4-x}\ne 1 \\
\end{matrix} \right.\Leftrightarrow 0\le x\le 4$.
Ta có $0\le x\le 4\Leftrightarrow 0\le \sqrt{4-x}\le 2\Leftrightarrow 3\le 5-\sqrt{4-x}\le 5\Rightarrow 0<{{\log }_{5-\sqrt{4-x}}}3\le 1$
Khi đó $x\sqrt{x}+\sqrt{x+12}\le m{{\log }_{5-\sqrt{4-x}}}3\Leftrightarrow m\ge \dfrac{x\sqrt{x}+\sqrt{x+12}}{{{\log }_{5-\sqrt{4-x}}}3}=\left( x\sqrt{x}+\sqrt{x+12} \right){{\log }_{3}}\left( 5-\sqrt{4-x} \right)$
Xét hàm số $g\left( x \right)=\left( x\sqrt{x}+\sqrt{x+12} \right){{\log }_{3}}\left( 5-\sqrt{4-x} \right)$
$\Rightarrow {g}'\left( x \right)=\left( \sqrt{x}+\dfrac{1}{2}\sqrt{x}+\dfrac{1}{2\sqrt{x+12}} \right){{\log }_{3}}\left( 5-\sqrt{4-x} \right)+\left( x\sqrt{x}+\sqrt{x+12} \right)\dfrac{1}{2\sqrt{4-x}\left( 5-\sqrt{4-x} \right)\ln 3}$
Ta có: $\left\{ \begin{matrix}
{{\log }_{3}}\left( 5-\sqrt{4-x} \right)>0 \\
\dfrac{1}{2\sqrt{4-x}\left( 5-\sqrt{4-x} \right)\ln 3}>0 \\
\end{matrix} \right.{{\forall }^{{}}}x\in \left[ 0;4 \right]\Rightarrow {g}'\left( x \right)>{{0}^{{}}}{{\forall }^{{}}}x\in \left[ 0;4 \right]$
$\Rightarrow $ $g\left( x \right)$ đồng biến trên $\left[ 0;4 \right]$.
Để phương trình $x\sqrt{x}+\sqrt{x+12}\le m{{\log }_{5-\sqrt{4-x}}}3$ khi và chỉ khi $m\ge \underset{\left[ 0;4 \right]}{\mathop{\min }} g\left( x \right)=g\left( 0 \right)=2\sqrt{3}$.
x\ge 0 \\
x+12\ge 0 \\
4-x\ge 0 \\
5-\sqrt{4-x}>0 \\
5-\sqrt{4-x}\ne 1 \\
\end{matrix} \right.\Leftrightarrow 0\le x\le 4$.
Ta có $0\le x\le 4\Leftrightarrow 0\le \sqrt{4-x}\le 2\Leftrightarrow 3\le 5-\sqrt{4-x}\le 5\Rightarrow 0<{{\log }_{5-\sqrt{4-x}}}3\le 1$
Khi đó $x\sqrt{x}+\sqrt{x+12}\le m{{\log }_{5-\sqrt{4-x}}}3\Leftrightarrow m\ge \dfrac{x\sqrt{x}+\sqrt{x+12}}{{{\log }_{5-\sqrt{4-x}}}3}=\left( x\sqrt{x}+\sqrt{x+12} \right){{\log }_{3}}\left( 5-\sqrt{4-x} \right)$
Xét hàm số $g\left( x \right)=\left( x\sqrt{x}+\sqrt{x+12} \right){{\log }_{3}}\left( 5-\sqrt{4-x} \right)$
$\Rightarrow {g}'\left( x \right)=\left( \sqrt{x}+\dfrac{1}{2}\sqrt{x}+\dfrac{1}{2\sqrt{x+12}} \right){{\log }_{3}}\left( 5-\sqrt{4-x} \right)+\left( x\sqrt{x}+\sqrt{x+12} \right)\dfrac{1}{2\sqrt{4-x}\left( 5-\sqrt{4-x} \right)\ln 3}$
Ta có: $\left\{ \begin{matrix}
{{\log }_{3}}\left( 5-\sqrt{4-x} \right)>0 \\
\dfrac{1}{2\sqrt{4-x}\left( 5-\sqrt{4-x} \right)\ln 3}>0 \\
\end{matrix} \right.{{\forall }^{{}}}x\in \left[ 0;4 \right]\Rightarrow {g}'\left( x \right)>{{0}^{{}}}{{\forall }^{{}}}x\in \left[ 0;4 \right]$
$\Rightarrow $ $g\left( x \right)$ đồng biến trên $\left[ 0;4 \right]$.
Để phương trình $x\sqrt{x}+\sqrt{x+12}\le m{{\log }_{5-\sqrt{4-x}}}3$ khi và chỉ khi $m\ge \underset{\left[ 0;4 \right]}{\mathop{\min }} g\left( x \right)=g\left( 0 \right)=2\sqrt{3}$.
Đáp án C.