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Câu hỏi: Cho hàm số $f\left( x \right)$ thỏa mãn $f\left( x \right)>0,\forall x>\dfrac{1}{2}$ và có đạo hàm ${f}'\left( x \right)$ liên tục trên khoảng $\left( \dfrac{1}{2};+\infty \right)$ thỏa mãn ${f}'\left( x \right)+8x{{f}^{2}}\left( x \right)=0,\forall x>\dfrac{1}{2}$ và $f\left( 1 \right)=\dfrac{1}{3}$. Tính $f\left( 1 \right)+f\left( 2 \right)+...+f\left( 1011 \right)$.
A. $\dfrac{1}{2}.\dfrac{2022}{2023}$.
B. $\dfrac{2021}{2043}$.
C. $\dfrac{2022}{4045}$.
D. $\dfrac{1}{2}.\dfrac{2021}{2022}$.
A. $\dfrac{1}{2}.\dfrac{2022}{2023}$.
B. $\dfrac{2021}{2043}$.
C. $\dfrac{2022}{4045}$.
D. $\dfrac{1}{2}.\dfrac{2021}{2022}$.
Ta có: ${f}'\left( x \right)+8x{{f}^{2}}\left( x \right)=0\Leftrightarrow \dfrac{-{f}'\left( x \right)}{{{f}^{2}}\left( x \right)}=8x\Leftrightarrow \int{\dfrac{-{f}'\left( x \right)}{{{f}^{2}}\left( x \right)}\text{d}x}=\int{8x}\text{d}x\Leftrightarrow \dfrac{1}{f\left( x \right)}=4{{x}^{2}}+C$.
Mà $f\left( 1 \right)=\dfrac{1}{3}\Rightarrow C=-1\Rightarrow f\left( x \right)=\dfrac{1}{4{{x}^{2}}-1}=\dfrac{1}{2}\left( \dfrac{1}{2x-1}-\dfrac{1}{2x+1} \right)$.
Ta có:
$\left. \begin{aligned}
& f\left( 1 \right)=\dfrac{1}{2}\left( 1-\dfrac{1}{3} \right) \\
& f\left( 2 \right)=\dfrac{1}{2}\left( \dfrac{1}{3}-\dfrac{1}{5} \right) \\
& .... \\
& f\left( 1011 \right)=\dfrac{1}{2}\left( \dfrac{1}{2021}-\dfrac{1}{2023} \right) \\
\end{aligned} \right\}\Rightarrow T=f\left( 1 \right)+f\left( 2 \right)+...+f\left( 1011 \right)=\dfrac{1}{2}\left( 1-\dfrac{1}{2023} \right)=\dfrac{1}{2}.\dfrac{2022}{2023}$.
Mà $f\left( 1 \right)=\dfrac{1}{3}\Rightarrow C=-1\Rightarrow f\left( x \right)=\dfrac{1}{4{{x}^{2}}-1}=\dfrac{1}{2}\left( \dfrac{1}{2x-1}-\dfrac{1}{2x+1} \right)$.
Ta có:
$\left. \begin{aligned}
& f\left( 1 \right)=\dfrac{1}{2}\left( 1-\dfrac{1}{3} \right) \\
& f\left( 2 \right)=\dfrac{1}{2}\left( \dfrac{1}{3}-\dfrac{1}{5} \right) \\
& .... \\
& f\left( 1011 \right)=\dfrac{1}{2}\left( \dfrac{1}{2021}-\dfrac{1}{2023} \right) \\
\end{aligned} \right\}\Rightarrow T=f\left( 1 \right)+f\left( 2 \right)+...+f\left( 1011 \right)=\dfrac{1}{2}\left( 1-\dfrac{1}{2023} \right)=\dfrac{1}{2}.\dfrac{2022}{2023}$.
Đáp án A.