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Câu hỏi: Cho hàm số $y=f\left( x \right)$ liên tục trên $\mathbb{R}$ đồng thời thỏa mãn điều kiện $f\left( 0 \right)<0$ và $\left[ f\left( x \right)+6{{x}^{3}}-2 \right]f\left( x \right)+9{{x}^{6}}=4{{x}^{4}}+6{{x}^{3}}+12{{x}^{2}}+8, \forall x\in \mathbb{R}$. Gọi $M, m$ lần lượt là giá trị lớn nhất và giá trị nhỏ nhất của hàm số $y=f\left( x+\sqrt{1-{{x}^{2}}} \right)$ trên đoạn $\left[ -1; 1 \right].$ Khi đó, tổng $M+m$ bằng
A. $-7-6\sqrt{2}$.
B. $7-6\sqrt{2}$.
C. $-6-6\sqrt{2}$.
D. $6-6\sqrt{2}$.
A. $-7-6\sqrt{2}$.
B. $7-6\sqrt{2}$.
C. $-6-6\sqrt{2}$.
D. $6-6\sqrt{2}$.
Ta có: $\left[ f\left( x \right)+6{{x}^{3}}-2 \right]f\left( x \right)+9{{x}^{6}}=4{{x}^{4}}+6{{x}^{3}}+12{{x}^{2}}+8\Leftrightarrow {{\left[ f\left( x \right)+3{{x}^{3}}-1 \right]}^{2}}={{\left( 2{{x}^{2}}+3 \right)}^{2}}$
$\Leftrightarrow \left[ \begin{aligned}
& f\left( x \right)=-3{{x}^{3}}+2{{x}^{2}}+4 \left( l \right) \\
& f\left( x \right)=-3{{x}^{3}}-2{{x}^{2}}-2\left( t/m \right) \\
\end{aligned} \right. $(do $ f\left( 0 \right)<0$).
$f\left( x \right)=-3{{x}^{3}}-2{{x}^{2}}-2\Rightarrow {f}'\left( x \right)=-9{{x}^{2}}-4x=0\Leftrightarrow \left[ \begin{aligned}
& x=0 \\
& x=-\dfrac{4}{9} \\
\end{aligned} \right.$.
$y=f\left( x+\sqrt{1-{{x}^{2}}} \right)\Rightarrow {y}'=\left( 1-\dfrac{x}{\sqrt{1-{{x}^{2}}}} \right).{f}'\left( x+\sqrt{1-{{x}^{2}}} \right)$.
${y}'=0\Leftrightarrow \left[ \begin{aligned}
& 1-\dfrac{x}{\sqrt{1-{{x}^{2}}}}=0 \\
& x+\sqrt{1-{{x}^{2}}}=0 \\
& x+\sqrt{1-{{x}^{2}}}=\dfrac{-4}{9} \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& x=\pm \dfrac{\sqrt{2}}{2} \\
& x=\dfrac{-4-\sqrt{146}}{18} \\
\end{aligned} \right.\in \left( -1;1 \right)$.
$y\left( -1 \right)=f\left( -1 \right)=-1; y\left( 1 \right)=f\left( 1 \right)=-7; y\left( \dfrac{\sqrt{2}}{2} \right)=f\left( \sqrt{2} \right)=-6-6\sqrt{2}; y\left( -\dfrac{\sqrt{2}}{2} \right)=f\left( 0 \right)=-2;$
$y\left( \dfrac{-4-\sqrt{146}}{18} \right)=f\left( -\dfrac{4}{9} \right)=-\dfrac{518}{243}$.
Vậy $M=-1, m=-6-6\sqrt{2}\Rightarrow M+m=-7-6\sqrt{2}$.
$\Leftrightarrow \left[ \begin{aligned}
& f\left( x \right)=-3{{x}^{3}}+2{{x}^{2}}+4 \left( l \right) \\
& f\left( x \right)=-3{{x}^{3}}-2{{x}^{2}}-2\left( t/m \right) \\
\end{aligned} \right. $(do $ f\left( 0 \right)<0$).
$f\left( x \right)=-3{{x}^{3}}-2{{x}^{2}}-2\Rightarrow {f}'\left( x \right)=-9{{x}^{2}}-4x=0\Leftrightarrow \left[ \begin{aligned}
& x=0 \\
& x=-\dfrac{4}{9} \\
\end{aligned} \right.$.
$y=f\left( x+\sqrt{1-{{x}^{2}}} \right)\Rightarrow {y}'=\left( 1-\dfrac{x}{\sqrt{1-{{x}^{2}}}} \right).{f}'\left( x+\sqrt{1-{{x}^{2}}} \right)$.
${y}'=0\Leftrightarrow \left[ \begin{aligned}
& 1-\dfrac{x}{\sqrt{1-{{x}^{2}}}}=0 \\
& x+\sqrt{1-{{x}^{2}}}=0 \\
& x+\sqrt{1-{{x}^{2}}}=\dfrac{-4}{9} \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& x=\pm \dfrac{\sqrt{2}}{2} \\
& x=\dfrac{-4-\sqrt{146}}{18} \\
\end{aligned} \right.\in \left( -1;1 \right)$.
$y\left( -1 \right)=f\left( -1 \right)=-1; y\left( 1 \right)=f\left( 1 \right)=-7; y\left( \dfrac{\sqrt{2}}{2} \right)=f\left( \sqrt{2} \right)=-6-6\sqrt{2}; y\left( -\dfrac{\sqrt{2}}{2} \right)=f\left( 0 \right)=-2;$
$y\left( \dfrac{-4-\sqrt{146}}{18} \right)=f\left( -\dfrac{4}{9} \right)=-\dfrac{518}{243}$.
Vậy $M=-1, m=-6-6\sqrt{2}\Rightarrow M+m=-7-6\sqrt{2}$.
Đáp án A.