(THPT Kim Liên – Hà Nội – 2022) Cho hàm số \(y = f\left( x \right)\) thỏa mãn \(f\left( x \right) > 0,\forall x > \frac{1}{2}\) và có đạo hàm \(f’\left( x \right)\) liên tục trên khoảng \(\left( {\frac{1}{2}; + \infty } \right)\) thỏa mãn \(f’\left( x \right) + 8x{f^2}\left( x \right) = 0,\forall x > \frac{1}{2}\) và \(f\left( 1 \right) = \frac{1}{3}\). Tính \(f\left( 1 \right) + f\left( 2 \right) + \;…\; + f\left( {1011} \right)\).

A. \(\frac{1}{2}.\frac{{2022}}{{2023}}\).

B. \(\frac{{2021}}{{2043}}\).

C. \(\frac{{2022}}{{4045}}\).

D. \(\frac{1}{2}.\frac{{2021}}{{2022}}\).

Lời giải:

Chọn A

Ta có \(f’\left( x \right) + 8x{f^2}\left( x \right) = 0 \Leftrightarrow \frac{{f’\left( x \right)}}{{{f^2}\left( x \right)}} = – 8x\).

Lấy nguyên hàm hai vế ta được:

\(\int {\frac{{f’\left( x \right)}}{{{f^2}\left( x \right)}}} dx = \int { – 8x} dx \Leftrightarrow \int {\frac{1}{{{f^2}\left( x \right)}}} d\left( {f\left( x \right)} \right) = \int { – 8x} dx \Leftrightarrow – \frac{1}{{f\left( x \right)}} = – 4{x^2} + C\).

Mà \(f\left( 1 \right) = \frac{1}{3} \Rightarrow – 3 = – 4 + C \Leftrightarrow C = 1\).

Khi đó \(f\left( x \right) = \frac{1}{{4{x^2} – 1}} = \frac{1}{2}\left[ {\frac{1}{{2x – 1}} – \frac{1}{{2x + 1}}} \right]\).

\(f\left( 1 \right) = \frac{1}{2}\left[ {1 – \frac{1}{3}} \right],f\left( 2 \right) = \frac{1}{2}\left[ {\frac{1}{3} – \frac{1}{5}} \right],f\left( 3 \right) = \frac{1}{2}\left[ {\frac{1}{5} – \frac{1}{7}} \right]\;,\;…\;f\left( {1011} \right) = \frac{1}{2}\left[ {\frac{1}{{2001}} – \frac{1}{{2023}}} \right].\)

Vậy \(f\left( 1 \right) + f\left( 2 \right) + … + f\left( {1011} \right) = \frac{1}{2}\left( {1 – \frac{1}{{2023}}} \right) = \frac{1}{2}.\frac{{2022}}{{2023}}\).