Câu hỏi: Cho hàm số
A.
B.
C.
D.
A.
B.
C.
D.
Ta có
Ta có
Theo giả thiết ta có phương trình
& g\left( n \right)=2 \\
& g\left( m \right)=5. \\
\end{aligned} \right.$$\underset{x\to -\infty }{\mathop{\lim }} g\left( x \right)=\underset{x\to -\infty }{\mathop{\lim }} \left( {{\text{e}}^{3x}}+a{{\text{e}}^{2x}}+b{{\text{e}}^{x}} \right)=0
& \left( -f\left( x \right)+5{f}'\left( x \right)+2{{\text{e}}^{3x}} \right){{g}^{2}}\left( x \right)={{g}^{3}}\left( x \right) \\
& \Leftrightarrow -f\left( x \right)+5{f}'\left( x \right)+2{{e}^{3x}}=g\left( x \right) \\
& \Leftrightarrow -\left( {{\text{e}}^{3x}}+a{{\text{e}}^{2x}}+b{{\text{e}}^{x}} \right)+5\left( \text{3}{{\text{e}}^{3x}}+2a{{\text{e}}^{2x}}+b{{\text{e}}^{x}} \right)+2{{\text{e}}^{3x}}=4{{\text{e}}^{3x}}+3a{{\text{e}}^{2x}}+2b{{\text{e}}^{x}} \\
& \Leftrightarrow 12{{\text{e}}^{3x}}+6a{{\text{e}}^{2x}}+2b{{\text{e}}^{x}}=0 \\
& \Leftrightarrow {g}'\left( x \right)=0\Leftrightarrow \left[ \begin{aligned}
& x=m \\
& x=n. \\
\end{aligned} \right. \\
\end{aligned}
& S=\left| \int\limits_{m}^{n}{\left[ \left( -f\left( x \right)+5{f}'\left( x \right)+2{{\text{e}}^{3x}} \right){{g}^{2}}\left( x \right)-{{g}^{3}}\left( x \right) \right]\text{d}x} \right| \\
& =\left| \int\limits_{m}^{n}{{{g}^{2}}\left( x \right)\left( -f\left( x \right)+5{f}'\left( x \right)+2{{\text{e}}^{3x}}-g\left( x \right) \right)}\text{d}x \right| \\
& =\left| \int\limits_{m}^{n}{{{g}^{2}}\left( x \right)\text{{g}'}\left( x \right)\text{d}x} \right|=\left| \int\limits_{m}^{n}{{{g}^{2}}\left( x \right)\text{dg}\left( x \right)} \right| \\
& =\left| \dfrac{1}{3}{{g}^{3}}\left( x \right)\left| \begin{aligned}
& n \\
& m \\
\end{aligned} \right. \right|=\dfrac{1}{3}\left| {{g}^{3}}\left( n \right)-{{g}^{3}}\left( m \right) \right|=\dfrac{117}{3}. \\
\end{aligned}$
Ta có
Theo giả thiết ta có phương trình
& g\left( n \right)=2 \\
& g\left( m \right)=5. \\
\end{aligned} \right.$$\underset{x\to -\infty }{\mathop{\lim }} g\left( x \right)=\underset{x\to -\infty }{\mathop{\lim }} \left( {{\text{e}}^{3x}}+a{{\text{e}}^{2x}}+b{{\text{e}}^{x}} \right)=0
& \left( -f\left( x \right)+5{f}'\left( x \right)+2{{\text{e}}^{3x}} \right){{g}^{2}}\left( x \right)={{g}^{3}}\left( x \right) \\
& \Leftrightarrow -f\left( x \right)+5{f}'\left( x \right)+2{{e}^{3x}}=g\left( x \right) \\
& \Leftrightarrow -\left( {{\text{e}}^{3x}}+a{{\text{e}}^{2x}}+b{{\text{e}}^{x}} \right)+5\left( \text{3}{{\text{e}}^{3x}}+2a{{\text{e}}^{2x}}+b{{\text{e}}^{x}} \right)+2{{\text{e}}^{3x}}=4{{\text{e}}^{3x}}+3a{{\text{e}}^{2x}}+2b{{\text{e}}^{x}} \\
& \Leftrightarrow 12{{\text{e}}^{3x}}+6a{{\text{e}}^{2x}}+2b{{\text{e}}^{x}}=0 \\
& \Leftrightarrow {g}'\left( x \right)=0\Leftrightarrow \left[ \begin{aligned}
& x=m \\
& x=n. \\
\end{aligned} \right. \\
\end{aligned}
& S=\left| \int\limits_{m}^{n}{\left[ \left( -f\left( x \right)+5{f}'\left( x \right)+2{{\text{e}}^{3x}} \right){{g}^{2}}\left( x \right)-{{g}^{3}}\left( x \right) \right]\text{d}x} \right| \\
& =\left| \int\limits_{m}^{n}{{{g}^{2}}\left( x \right)\left( -f\left( x \right)+5{f}'\left( x \right)+2{{\text{e}}^{3x}}-g\left( x \right) \right)}\text{d}x \right| \\
& =\left| \int\limits_{m}^{n}{{{g}^{2}}\left( x \right)\text{{g}'}\left( x \right)\text{d}x} \right|=\left| \int\limits_{m}^{n}{{{g}^{2}}\left( x \right)\text{dg}\left( x \right)} \right| \\
& =\left| \dfrac{1}{3}{{g}^{3}}\left( x \right)\left| \begin{aligned}
& n \\
& m \\
\end{aligned} \right. \right|=\dfrac{1}{3}\left| {{g}^{3}}\left( n \right)-{{g}^{3}}\left( m \right) \right|=\dfrac{117}{3}. \\
\end{aligned}$
Đáp án D.