(THPT Yên Lạc – Vĩnh Phúc – 2022) Cho \(f(x) = 2023.\ln \left( {{e^{\frac{x}{{2023}}}} + {e^{\frac{1}{2}}}} \right)\). 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. \({e^{1011}}\).

Lời giải:

Chọn A

Ta có \(f(x) = 2023.\ln \left( {{e^{\frac{x}{{2023}}}} + {e^{\frac{1}{2}}}} \right) \Rightarrow f’\left( x \right) = \frac{{{e^{\frac{x}{{2023}}}}}}{{{e^{\frac{x}{{2023}}}} + {e^{\frac{1}{2}}}}}\).

Khi đó \(H = f’\left( 1 \right) + f’\left( 2 \right) + … + f’\left( {2022} \right) = \frac{{{e^{\frac{1}{{2023}}}}}}{{{e^{\frac{1}{{2023}}}} + {e^{\frac{1}{2}}}}} + \frac{{{e^{\frac{2}{{2023}}}}}}{{{e^{\frac{2}{{2023}}}} + {e^{\frac{1}{2}}}}} + … + \frac{{{e^{\frac{{2022}}{{2023}}}}}}{{{e^{\frac{{2022}}{{2023}}}} + {e^{\frac{1}{2}}}}}\)

Để ý \(\frac{{{e^x}}}{{{e^x} + \sqrt e }} + \frac{{{e^{1 – x}}}}{{{e^{1 – x}} + \sqrt e }} = \frac{{{e^x}}}{{{e^x} + \sqrt e }} + \frac{e}{{e + {e^x}\sqrt e }} = \frac{{{e^x}}}{{{e^x} + \sqrt e }} + \frac{{\sqrt e }}{{\sqrt e + {e^x}}} = \frac{{{e^x} + \sqrt e }}{{{e^x} + \sqrt e }} = 1\)

Từ đó

\(H = \left( {\frac{{{e^{\frac{1}{{2023}}}}}}{{{e^{\frac{1}{{2023}}}} + {e^{\frac{1}{2}}}}} + \frac{{{e^{\frac{{2022}}{{2023}}}}}}{{{e^{\frac{{2022}}{{2023}}}} + {e^{\frac{1}{2}}}}}} \right) + … + \left( {\frac{{{e^{\frac{2}{{2023}}}}}}{{{e^{\frac{2}{{2023}}}} + {e^{\frac{1}{2}}}}} + … + \frac{{{e^{\frac{{2020}}{{2023}}}}}}{{{e^{\frac{{2020}}{{2023}}}} + {e^{\frac{1}{2}}}}}} \right)\)

\(H = 1 + 1 + … + 1 = \frac{{2022}}{2} = 1011\).