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Câu hỏi: Cho hàm số $f\left( x \right)$ có đạo hàm cấp hai liên tục trên $\mathbb{R}$ và thoả mãn $f\left( 0 \right)=0,{f}'\left( 0 \right)=1, {{f}'}'\left( x \right)=f\left( x \right)+\left( 3x+4 \right){{e}^{2x}}$ với mọi $x\in \mathbb{R}$. Giá trị của $f\left( 1 \right)$ bằng
A. ${{e}^{2}}$.
B. $2{{e}^{4}}$.
C. $2{{e}^{2}}$.
D. ${{e}^{4}}$.
A. ${{e}^{2}}$.
B. $2{{e}^{4}}$.
C. $2{{e}^{2}}$.
D. ${{e}^{4}}$.
Ta có:
$\begin{aligned}
& {{f}'}'\left( x \right)=f\left( x \right)+\left( 3x+4 \right){{e}^{2x}}\Leftrightarrow {{f}'}'\left( x \right)-f\left( x \right)=\left( 3x+4 \right){{e}^{2x}} \\
& \Leftrightarrow \dfrac{{{f}'}'\left( x \right)-f\left( x \right)}{{{e}^{x}}}=\left( 3x+4 \right){{e}^{x}}\Leftrightarrow \dfrac{{{f}'}'\left( x \right){{e}^{x}}-f\left( x \right){{e}^{x}}}{{{e}^{2x}}}=\left( 3x+4 \right){{e}^{x}} \\
& \Leftrightarrow \dfrac{{{f}'}'\left( x \right){{e}^{x}}-{f}'\left( x \right){{e}^{x}}}{{{e}^{2x}}}+\dfrac{{f}'\left( x \right){{e}^{x}}-f\left( x \right){{e}^{x}}}{{{e}^{2x}}}=\left( 3x+4 \right){{e}^{x}} \\
& \Leftrightarrow {{\left( \dfrac{{f}'\left( x \right)}{{{e}^{x}}} \right)}^{\prime }}+{{\left( \dfrac{f\left( x \right)}{{{e}^{x}}} \right)}^{\prime }}=\left( 3x+4 \right){{e}^{x}} \\
\end{aligned}$
Lấy nguyên hàm hai vế ta được:
$\dfrac{{f}'\left( x \right)}{{{e}^{x}}}+\dfrac{f\left( x \right)}{{{e}^{x}}}=\int{\left( 3x+4 \right){{e}^{x}}\text{d}x}=\left( 3x+1 \right){{e}^{x}}+C$
Khi: $x=0\Rightarrow C=0$
Suy ra: $\dfrac{{f}'\left( x \right)}{{{e}^{x}}}+\dfrac{f\left( x \right)}{{{e}^{x}}}=\left( 3x+1 \right){{e}^{x}}\Leftrightarrow {f}'\left( x \right)+f\left( x \right)=\left( 3x+1 \right){{e}^{2x}}$
$\begin{aligned}
& \Leftrightarrow {f}'\left( x \right){{e}^{x}}+f\left( x \right){{e}^{x}}=\left( 3x+1 \right){{e}^{3x}}\Leftrightarrow {{\left[ f\left( x \right){{e}^{x}} \right]}^{\prime }}=\left( 3x+1 \right){{e}^{3x}} \\
& \Rightarrow f\left( x \right){{e}^{x}}=\int{\left( 3x+1 \right){{e}^{3x}}\text{d}x}=x{{e}^{3x}}+{{C}_{1}} \\
\end{aligned}$
Lại có: $x=0\Rightarrow {{C}_{1}}=0$
Vậy: $f\left( x \right)=x{{e}^{2x}}\Rightarrow f\left( 1 \right)={{e}^{2}}$.
$\begin{aligned}
& {{f}'}'\left( x \right)=f\left( x \right)+\left( 3x+4 \right){{e}^{2x}}\Leftrightarrow {{f}'}'\left( x \right)-f\left( x \right)=\left( 3x+4 \right){{e}^{2x}} \\
& \Leftrightarrow \dfrac{{{f}'}'\left( x \right)-f\left( x \right)}{{{e}^{x}}}=\left( 3x+4 \right){{e}^{x}}\Leftrightarrow \dfrac{{{f}'}'\left( x \right){{e}^{x}}-f\left( x \right){{e}^{x}}}{{{e}^{2x}}}=\left( 3x+4 \right){{e}^{x}} \\
& \Leftrightarrow \dfrac{{{f}'}'\left( x \right){{e}^{x}}-{f}'\left( x \right){{e}^{x}}}{{{e}^{2x}}}+\dfrac{{f}'\left( x \right){{e}^{x}}-f\left( x \right){{e}^{x}}}{{{e}^{2x}}}=\left( 3x+4 \right){{e}^{x}} \\
& \Leftrightarrow {{\left( \dfrac{{f}'\left( x \right)}{{{e}^{x}}} \right)}^{\prime }}+{{\left( \dfrac{f\left( x \right)}{{{e}^{x}}} \right)}^{\prime }}=\left( 3x+4 \right){{e}^{x}} \\
\end{aligned}$
Lấy nguyên hàm hai vế ta được:
$\dfrac{{f}'\left( x \right)}{{{e}^{x}}}+\dfrac{f\left( x \right)}{{{e}^{x}}}=\int{\left( 3x+4 \right){{e}^{x}}\text{d}x}=\left( 3x+1 \right){{e}^{x}}+C$
Khi: $x=0\Rightarrow C=0$
Suy ra: $\dfrac{{f}'\left( x \right)}{{{e}^{x}}}+\dfrac{f\left( x \right)}{{{e}^{x}}}=\left( 3x+1 \right){{e}^{x}}\Leftrightarrow {f}'\left( x \right)+f\left( x \right)=\left( 3x+1 \right){{e}^{2x}}$
$\begin{aligned}
& \Leftrightarrow {f}'\left( x \right){{e}^{x}}+f\left( x \right){{e}^{x}}=\left( 3x+1 \right){{e}^{3x}}\Leftrightarrow {{\left[ f\left( x \right){{e}^{x}} \right]}^{\prime }}=\left( 3x+1 \right){{e}^{3x}} \\
& \Rightarrow f\left( x \right){{e}^{x}}=\int{\left( 3x+1 \right){{e}^{3x}}\text{d}x}=x{{e}^{3x}}+{{C}_{1}} \\
\end{aligned}$
Lại có: $x=0\Rightarrow {{C}_{1}}=0$
Vậy: $f\left( x \right)=x{{e}^{2x}}\Rightarrow f\left( 1 \right)={{e}^{2}}$.
Đáp án A.
