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Câu hỏi: Cho hàm số f xác định, đơn điệu giảm, có đạo hàm liên tục trên $\mathbb{R}$ và thỏa mãn
$3{{\left[ f\left( x \right) \right]}^{2}}=\int\limits_{0}^{x}{\left[ 8{{\left( f\left( t \right) \right)}^{3}}+{{\left( {f}'\left( t \right) \right)}^{3}} \right]\text{d}t}+x,\forall x\in \mathbb{R}$. Tích phân $\int\limits_{0}^{12}{\left( 12+f\left( x \right) \right)}\text{d}x$ nhận giá trị trong khoảng nào trong các khoảng sau?
A. $\left( 10;11 \right).$
B. $\left( 11;12 \right).$
C. $\left( 12;13 \right).$
D. $\left( 13;14 \right).$
$3{{\left[ f\left( x \right) \right]}^{2}}=\int\limits_{0}^{x}{\left[ 8{{\left( f\left( t \right) \right)}^{3}}+{{\left( {f}'\left( t \right) \right)}^{3}} \right]\text{d}t}+x,\forall x\in \mathbb{R}$. Tích phân $\int\limits_{0}^{12}{\left( 12+f\left( x \right) \right)}\text{d}x$ nhận giá trị trong khoảng nào trong các khoảng sau?
A. $\left( 10;11 \right).$
B. $\left( 11;12 \right).$
C. $\left( 12;13 \right).$
D. $\left( 13;14 \right).$
Lấy đạo hàm 2 vế của phương trình giả thiết ta có:
$\begin{aligned}
& 6f\left( x \right).{f}'\left( x \right)=8{{\left( f\left( x \right) \right)}^{3}}+{{\left( {f}'\left( x \right) \right)}^{3}}+1 \\
& \Leftrightarrow \left[ {f}'\left( x \right)+2f\left( x \right)+1 \right]\left[ {{\left( {f}'\left( x \right) \right)}^{2}}-\left( 2f\left( x \right)+1 \right).{f}'\left( x \right)+\left( 4.{{\left( f\left( x \right) \right)}^{2}}-2f\left( x \right)+1 \right) \right]=0 \\
& \Leftrightarrow \left[ {f}'\left( x \right)+2f\left( x \right)+1 \right]\left[ {{\left( {f}'\left( x \right)-\left( f\left( x \right) \right) \right)}^{2}}+{{\left( f\left( x \right)-1 \right)}^{2}}+2{{\left( f\left( x \right) \right)}^{2}}-{f}'\left( x \right) \right]=0 \\
& \Leftrightarrow {f}'\left( x \right)+2f\left( x \right)+1=0,\left( do\begin{matrix}
{} & {f}'\left( x \right)<0 \\
\end{matrix} \right) \\
& \Leftrightarrow {{e}^{2x}}.{f}'\left( x \right)+2{{e}^{2x}}.f\left( x \right)=-{{e}^{2x}} \\
& \Leftrightarrow {{\left[ {{e}^{2x}}.f\left( x \right) \right]}^{\prime }}=-{{e}^{2x}}\Rightarrow {{e}^{2x}}.f\left( x \right)=-\int{{{e}^{2x}}\text{d}x}=-\dfrac{1}{2}{{e}^{2x}}+C \\
\end{aligned}$
Thay $x=0\Rightarrow f\left( 0 \right)=0\Rightarrow C=\dfrac{1}{2}\Rightarrow f\left( x \right)=\dfrac{1}{2{{e}^{2x}}}-\dfrac{1}{2}$.
Suy ra $\int\limits_{0}^{12}{\left( 12+f\left( x \right) \right)}\text{d}x=\int\limits_{0}^{12}{\left( 12+\dfrac{1}{2{{e}^{2x}}}-\dfrac{1}{2} \right)}\text{d}x=11.716\in \left( 11;12 \right)$.
$\begin{aligned}
& 6f\left( x \right).{f}'\left( x \right)=8{{\left( f\left( x \right) \right)}^{3}}+{{\left( {f}'\left( x \right) \right)}^{3}}+1 \\
& \Leftrightarrow \left[ {f}'\left( x \right)+2f\left( x \right)+1 \right]\left[ {{\left( {f}'\left( x \right) \right)}^{2}}-\left( 2f\left( x \right)+1 \right).{f}'\left( x \right)+\left( 4.{{\left( f\left( x \right) \right)}^{2}}-2f\left( x \right)+1 \right) \right]=0 \\
& \Leftrightarrow \left[ {f}'\left( x \right)+2f\left( x \right)+1 \right]\left[ {{\left( {f}'\left( x \right)-\left( f\left( x \right) \right) \right)}^{2}}+{{\left( f\left( x \right)-1 \right)}^{2}}+2{{\left( f\left( x \right) \right)}^{2}}-{f}'\left( x \right) \right]=0 \\
& \Leftrightarrow {f}'\left( x \right)+2f\left( x \right)+1=0,\left( do\begin{matrix}
{} & {f}'\left( x \right)<0 \\
\end{matrix} \right) \\
& \Leftrightarrow {{e}^{2x}}.{f}'\left( x \right)+2{{e}^{2x}}.f\left( x \right)=-{{e}^{2x}} \\
& \Leftrightarrow {{\left[ {{e}^{2x}}.f\left( x \right) \right]}^{\prime }}=-{{e}^{2x}}\Rightarrow {{e}^{2x}}.f\left( x \right)=-\int{{{e}^{2x}}\text{d}x}=-\dfrac{1}{2}{{e}^{2x}}+C \\
\end{aligned}$
Thay $x=0\Rightarrow f\left( 0 \right)=0\Rightarrow C=\dfrac{1}{2}\Rightarrow f\left( x \right)=\dfrac{1}{2{{e}^{2x}}}-\dfrac{1}{2}$.
Suy ra $\int\limits_{0}^{12}{\left( 12+f\left( x \right) \right)}\text{d}x=\int\limits_{0}^{12}{\left( 12+\dfrac{1}{2{{e}^{2x}}}-\dfrac{1}{2} \right)}\text{d}x=11.716\in \left( 11;12 \right)$.
Đáp án B.