Đáp án đúng: C
Giải chi tiết:Ta có :
\(\begin{array}{l}f'\left( x \right) = 2018{\left( {{e^x} + {x^3}\cos x} \right)^{2017}}.{\left( {{e^x} + {x^3}\cos x} \right)^\prime }\\\,\,\,\,\,\,\,\,\,\,\,\,\, = 2018{\left( {{e^x} + {x^3}\cos x} \right)^{2017}}.\left( {{e^x} + 3{x^2}\cos x - {x^3}\sin x} \right)\end{array}\)
\(\begin{array}{l}f''\left( x \right) = {\left( {f'\left( x \right)} \right)^\prime } = {\left[ {2018{{\left( {{e^x} + {x^3}\cos x} \right)}^{2017}}.\left( {{e^x} + 3{x^2}\cos x - {x^3}\sin x} \right)} \right]^\prime }\\\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2018.2017.{\left( {{e^x} + {x^3}\cos x} \right)^{2016}}.{\left( {{e^x} + {x^3}\cos x} \right)^\prime }\left( {{e^x} + 3{x^2}\cos x - {x^3}\sin x} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 2018.{\left( {{e^x} + {x^3}\cos x} \right)^{2017}}.{\left( {{e^x} + 3{x^2}\cos x - {x^3}\sin x} \right)^\prime }\\\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2018.2017.{\left( {{e^x} + {x^3}\cos x} \right)^{2016}}.{\left( {{e^x} + 3{x^2}\cos x - {x^3}\sin x} \right)^2}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 2018.{\left( {{e^x} + {x^3}\cos x} \right)^{2017}}.\left( {{e^x} + 6x\cos x - 3{x^2}\sin x - 3{x^2}\sin x - {x^3}\cos x} \right)\end{array}\)
Khi đó \(f''\left( 0 \right) = 2018.2017.1.1 + 2018.1.1 = {2018^2}.\)
Chọn C.
