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Câu hỏi: Cho hàm số $f\left( x \right)$ có đạo hàm trên $\mathbb{R}$, thỏa mãn ${f}'\left( x \right)-2018f\left( x \right)=2018{{\text{x}}^{2017}}{{e}^{2018\text{x}}}$ và $f\left( 0 \right)=2018$. Tính giá trị $f\left( 1 \right)$.
A. $f\left( 1 \right)=2018{{\text{e}}^{-2018}}$
B. $f\left( 1 \right)=2017{{\text{e}}^{2018}}$
C. $f\left( 1 \right)=2018{{\text{e}}^{2018}}$
D. $f\left( 1 \right)=2019{{\text{e}}^{2018}}$
A. $f\left( 1 \right)=2018{{\text{e}}^{-2018}}$
B. $f\left( 1 \right)=2017{{\text{e}}^{2018}}$
C. $f\left( 1 \right)=2018{{\text{e}}^{2018}}$
D. $f\left( 1 \right)=2019{{\text{e}}^{2018}}$
Nhân cả hai vế với ${{e}^{-2018\text{x}}}$, ta được:
${f}'\left( x \right).{{e}^{-2018\text{x}}}-2018f\left( x \right).{{e}^{-2018\text{x}}}=2018{{\text{x}}^{2017}}\Leftrightarrow {{\left[ f\left( x \right).{{e}^{-2018\text{x}}} \right]}^{\prime }}=2018{{\text{x}}^{2017}}$
Lấy nguyên hàm hai vế, ta được:
$\int{{{\left[ f\left( x \right).{{e}^{-2018\text{x}}} \right]}^{\prime }}d\text{x}}=\int{2018{{\text{x}}^{2017}}d\text{x}}\Rightarrow f\left( x \right).{{e}^{-2018\text{x}}}={{x}^{2018}}+C$.
Do $f\left( 0 \right)=2018$, nên ta có $f\left( 0 \right).{{e}^{-2018.0}}={{0}^{2018}}+C\Leftrightarrow C=2018$.
Suy ra: $f\left( x \right)=\left( {{x}^{2018}}+2018 \right){{e}^{2018\text{x}}}$.
Vậy $f\left( 1 \right)=2019{{e}^{2018}}$.
${f}'\left( x \right).{{e}^{-2018\text{x}}}-2018f\left( x \right).{{e}^{-2018\text{x}}}=2018{{\text{x}}^{2017}}\Leftrightarrow {{\left[ f\left( x \right).{{e}^{-2018\text{x}}} \right]}^{\prime }}=2018{{\text{x}}^{2017}}$
Lấy nguyên hàm hai vế, ta được:
$\int{{{\left[ f\left( x \right).{{e}^{-2018\text{x}}} \right]}^{\prime }}d\text{x}}=\int{2018{{\text{x}}^{2017}}d\text{x}}\Rightarrow f\left( x \right).{{e}^{-2018\text{x}}}={{x}^{2018}}+C$.
Do $f\left( 0 \right)=2018$, nên ta có $f\left( 0 \right).{{e}^{-2018.0}}={{0}^{2018}}+C\Leftrightarrow C=2018$.
Suy ra: $f\left( x \right)=\left( {{x}^{2018}}+2018 \right){{e}^{2018\text{x}}}$.
Vậy $f\left( 1 \right)=2019{{e}^{2018}}$.
Đáp án D.