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Câu hỏi: Cho $f\left( x \right)=2023.\ln \left( {{e}^{\dfrac{x}{2023}}}+{{e}^{\dfrac{1}{2}}} \right)$. Tính giá trị biểu thức $H={f}'\left( 1 \right)+{f}'\left( 2 \right)+...+{f}'\left( 2022 \right)$
A. $1011$.
B. $2022$.
C. ${{e}^{2022}}$.
D. ${{e}^{1011}}$.
A. $1011$.
B. $2022$.
C. ${{e}^{2022}}$.
D. ${{e}^{1011}}$.
Ta có: $f\left( x \right)=2023.\ln \left( {{e}^{\dfrac{x}{2023}}}+{{e}^{\dfrac{1}{2}}} \right)$. Suy ra ${f}'\left( x \right)=2023.\dfrac{{{\left( {{e}^{\dfrac{x}{2023}}}+{{e}^{\dfrac{1}{2}}} \right)}^{\prime }}}{\left( {{e}^{\dfrac{x}{2023}}}+{{e}^{\dfrac{1}{2}}} \right)}=\dfrac{{{e}^{\dfrac{x}{2023}}}}{{{e}^{\dfrac{x}{2023}}}+{{e}^{\dfrac{1}{2}}}}$.
Do đó ${f}'\left( x \right)+{f}'\left( 2023-x \right)=\dfrac{{{e}^{\dfrac{x}{2023}}}}{{{e}^{\dfrac{x}{2023}}}+{{e}^{\dfrac{1}{2}}}}+\dfrac{{{e}^{\dfrac{2023-x}{2023}}}}{{{e}^{\dfrac{2023-x}{2023}}}+{{e}^{\dfrac{1}{2}}}}=\dfrac{{{e}^{\dfrac{x}{2023}}}}{{{e}^{\dfrac{x}{2023}}}+{{e}^{\dfrac{1}{2}}}}+\dfrac{{{e}^{1-\dfrac{x}{2023}}}}{{{e}^{1-\dfrac{x}{2023}}}+{{e}^{\dfrac{1}{2}}}}$
$=\dfrac{{{e}^{\dfrac{x}{2023}}}}{{{e}^{\dfrac{x}{2023}}}+{{e}^{\dfrac{1}{2}}}}+\dfrac{e}{e+{{e}^{\dfrac{1}{2}}}.{{e}^{\dfrac{x}{2023}}}}=\dfrac{{{e}^{\dfrac{x}{2023}}}}{{{e}^{\dfrac{x}{2023}}}+{{e}^{\dfrac{1}{2}}}}+\dfrac{{{e}^{\dfrac{1}{2}}}}{{{e}^{\dfrac{1}{2}}}+{{e}^{\dfrac{x}{2023}}}}=1$.
Vì vậy: $2H=\left[ {f}'\left( 1 \right)+{f}'\left( 2022 \right) \right]+\left[ {f}'\left( 2 \right)+{f}'\left( 2021 \right) \right]+...+\left[ {f}'\left( 2022 \right)+{f}'\left( 1 \right) \right]=2022$.
Vậy $H=\dfrac{2022}{2}=1011$.
Do đó ${f}'\left( x \right)+{f}'\left( 2023-x \right)=\dfrac{{{e}^{\dfrac{x}{2023}}}}{{{e}^{\dfrac{x}{2023}}}+{{e}^{\dfrac{1}{2}}}}+\dfrac{{{e}^{\dfrac{2023-x}{2023}}}}{{{e}^{\dfrac{2023-x}{2023}}}+{{e}^{\dfrac{1}{2}}}}=\dfrac{{{e}^{\dfrac{x}{2023}}}}{{{e}^{\dfrac{x}{2023}}}+{{e}^{\dfrac{1}{2}}}}+\dfrac{{{e}^{1-\dfrac{x}{2023}}}}{{{e}^{1-\dfrac{x}{2023}}}+{{e}^{\dfrac{1}{2}}}}$
$=\dfrac{{{e}^{\dfrac{x}{2023}}}}{{{e}^{\dfrac{x}{2023}}}+{{e}^{\dfrac{1}{2}}}}+\dfrac{e}{e+{{e}^{\dfrac{1}{2}}}.{{e}^{\dfrac{x}{2023}}}}=\dfrac{{{e}^{\dfrac{x}{2023}}}}{{{e}^{\dfrac{x}{2023}}}+{{e}^{\dfrac{1}{2}}}}+\dfrac{{{e}^{\dfrac{1}{2}}}}{{{e}^{\dfrac{1}{2}}}+{{e}^{\dfrac{x}{2023}}}}=1$.
Vì vậy: $2H=\left[ {f}'\left( 1 \right)+{f}'\left( 2022 \right) \right]+\left[ {f}'\left( 2 \right)+{f}'\left( 2021 \right) \right]+...+\left[ {f}'\left( 2022 \right)+{f}'\left( 1 \right) \right]=2022$.
Vậy $H=\dfrac{2022}{2}=1011$.
Đáp án A.