Câu hỏi 10

Ta có:

\(\begin{array}{l}f'\left( x \right) - 2018f\left( x \right) = 2018{x^{2017}}{e^{2018x}} \Leftrightarrow f'\left( x \right){e^{ - 2018x}} - 2018{e^{ - 2018x}}f\left( x \right) = 2018{x^{2017}}\\ \Rightarrow \left[ {f\left( x \right){e^{ - 2018x}}} \right]' = \left( {{x^{2018}}} \right)' \Rightarrow \int {\left[ {f\left( x \right){e^{ - 2018x}}} \right]'dx}  = \int {\left( {{x^{2018}}} \right)'dx}  \Rightarrow f\left( x \right){e^{ - 2018x}} = {x^{2018}} + C\end{array}\) 

Do \(f\left( 0 \right) = 2018 \Rightarrow f\left( 0 \right).{e^0} = C \Leftrightarrow C = 2018\)

\(\begin{array}{l} \Rightarrow f\left( x \right) = {x^{2018}}{e^{2018x}} + 2018{e^{2018x}}\\ \Rightarrow f\left( 1 \right) = {e^{2018}} + 2018{e^{2018}} = 2019{e^{2018}}.\end{array}\)

Chọn D.