Lời giải:Ta có:
\(\begin{array}{l}\,\,\,\,\,\,\,f'\left( x \right) - 2018f\left( x \right) = 2018{x^{2017}}{e^{2018x}}\\ \Leftrightarrow {e^{ - 2018x}}f'\left( x \right) - 2018{e^{ - 2018x}}.f\left( x \right) = 2018{x^{2017}}\end{array}\)
\( \Rightarrow {\left( {{e^{ - 2018x}}f\left( x \right)} \right)^\prime } = 2018{x^{2017}} \Rightarrow {e^{ - 2018x}}f\left( x \right)\) là một nguyên hàm của \(2018{x^{2017}}\)
Ta có:
\(\int {2018{x^{2017}}} dx = {x^{2018}} + C\)\( \Rightarrow {e^{ - 2018x}}f\left( x \right) = {x^{2018}} + {C_0}\)
Mà \(f\left( 0 \right) = 2018\)\( \Rightarrow 2018 = {C_0}\, \Rightarrow {e^{ - 2018x}}f\left( x \right) = {x^{2018}} + 2018 \Leftrightarrow f\left( x \right) = {x^{2018}}{e^{2018x}} + 2018{e^{2018x}}\)
\( \Rightarrow f\left( 1 \right)\)\( = {e^{2018}} + 2018{e^{2018}} = 2019{e^{2018}}\).
Chọn: A
