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Câu hỏi: Cho hàm số $f\left( x \right)$ thỏa mãn ${f}'\left( x \right)={{2018}^{x}}.ln2018-\cos x$ và $f\left( 0 \right)=2.$ Khẳng định nào đúng?
A. $f\left( x \right)={{2018}^{x}}+\sin x+1.$
B. $f\left( x \right)=\dfrac{{{2018}^{x}}}{\ln 2018}+\sin x+1.$
C. $f\left( x \right)=\dfrac{{{2018}^{x}}}{\ln 2018}-\sin x+1.$
D. $f\left( x \right)={{2018}^{x}}-\sin x+1.$
A. $f\left( x \right)={{2018}^{x}}+\sin x+1.$
B. $f\left( x \right)=\dfrac{{{2018}^{x}}}{\ln 2018}+\sin x+1.$
C. $f\left( x \right)=\dfrac{{{2018}^{x}}}{\ln 2018}-\sin x+1.$
D. $f\left( x \right)={{2018}^{x}}-\sin x+1.$
Ta có: $\left\{ \begin{aligned}
& f\left( x \right)=\int{\left( {{2018}^{x}}\ln 2018-\cos x \right)dx} \\
& f\left( 0 \right)=2 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& f\left( x \right)={{2018}^{x}}-\sin x+C \\
& 2={{2018}^{0}}-\sin 0+C \\
\end{aligned} \right.$
$\Leftrightarrow f\left( x \right)={{2018}^{x}}-\sin x+1.$
& f\left( x \right)=\int{\left( {{2018}^{x}}\ln 2018-\cos x \right)dx} \\
& f\left( 0 \right)=2 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& f\left( x \right)={{2018}^{x}}-\sin x+C \\
& 2={{2018}^{0}}-\sin 0+C \\
\end{aligned} \right.$
$\Leftrightarrow f\left( x \right)={{2018}^{x}}-\sin x+1.$
Đáp án D.