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Câu hỏi: Cho hàm số $y=f(x)=2020\ln \left( {{e}^{\dfrac{x}{2020}}}+\sqrt{e} \right)$. Tính giá trị biểu thức $A={f}'(1)+{f}'(2)+...+{f}'(2019)$
A. 2018
B. 1009
C. $\dfrac{2017}{2}$
D. $\dfrac{2019}{2}$
A. 2018
B. 1009
C. $\dfrac{2017}{2}$
D. $\dfrac{2019}{2}$
Ta có ${y}'={f}'(x)=2020\dfrac{{{\left( {{e}^{\dfrac{x}{2020}}}+\sqrt{e} \right)}^{\prime }}}{\left( {{e}^{\dfrac{x}{2020}}}+\sqrt{e} \right)}=\dfrac{{{e}^{\dfrac{x}{2020}}}}{{{e}^{\dfrac{x}{2020}}}+\sqrt{e}}$
$\Rightarrow {f}'(x)+{f}'(2020-x)=\dfrac{{{e}^{\dfrac{x}{2020}}}}{{{e}^{\dfrac{x}{2020}}}+\sqrt{e}}+\dfrac{{{e}^{\dfrac{2020-x}{2020}}}}{{{e}^{\dfrac{2020-x}{2020}}}+\sqrt{e}}=\dfrac{{{e}^{\dfrac{x}{2020}}}}{{{e}^{\dfrac{x}{2020}}}+\sqrt{e}}+\dfrac{{{e}^{1-\dfrac{x}{2020}}}}{{{e}^{1-\dfrac{x}{2020}}}+\sqrt{e}}$
$=\dfrac{{{e}^{\dfrac{x}{2020}}}}{{{e}^{\dfrac{x}{2020}}}+\sqrt{e}}+\dfrac{e}{e+\sqrt{e}.{{e}^{\dfrac{x}{2020}}}}=\dfrac{{{e}^{\dfrac{x}{2020}}}}{{{e}^{\dfrac{x}{2020}}}+\sqrt{e}}+\dfrac{\sqrt{e}}{\sqrt{e}+{{e}^{\dfrac{x}{2020}}}}=1$
Bởi vậy $2\text{A}=\left[ {f}'(1)+{f}'(2019) \right]+\left[ {f}'(2)+{f}'(2018) \right]+...+\left[ {f}'(2019)+{f}'(1) \right]=2019$.
Nên $A=\dfrac{2019}{2}$.
$\Rightarrow {f}'(x)+{f}'(2020-x)=\dfrac{{{e}^{\dfrac{x}{2020}}}}{{{e}^{\dfrac{x}{2020}}}+\sqrt{e}}+\dfrac{{{e}^{\dfrac{2020-x}{2020}}}}{{{e}^{\dfrac{2020-x}{2020}}}+\sqrt{e}}=\dfrac{{{e}^{\dfrac{x}{2020}}}}{{{e}^{\dfrac{x}{2020}}}+\sqrt{e}}+\dfrac{{{e}^{1-\dfrac{x}{2020}}}}{{{e}^{1-\dfrac{x}{2020}}}+\sqrt{e}}$
$=\dfrac{{{e}^{\dfrac{x}{2020}}}}{{{e}^{\dfrac{x}{2020}}}+\sqrt{e}}+\dfrac{e}{e+\sqrt{e}.{{e}^{\dfrac{x}{2020}}}}=\dfrac{{{e}^{\dfrac{x}{2020}}}}{{{e}^{\dfrac{x}{2020}}}+\sqrt{e}}+\dfrac{\sqrt{e}}{\sqrt{e}+{{e}^{\dfrac{x}{2020}}}}=1$
Bởi vậy $2\text{A}=\left[ {f}'(1)+{f}'(2019) \right]+\left[ {f}'(2)+{f}'(2018) \right]+...+\left[ {f}'(2019)+{f}'(1) \right]=2019$.
Nên $A=\dfrac{2019}{2}$.
Đáp án D.