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Câu hỏi: Cho hàm số $f(x)$ thỏa mãn $\sqrt{x \cdot f^{\prime}(x)}=-f(x), \forall x \geq 1$ và $f(\mathrm{e})=-\dfrac{1}{2}$. Giá trị $f\left(\mathrm{e}^{2020}\right)$ bằng
A. -2021 .
B. $-\dfrac{1}{2020}$.
C. -2020 .
D. $-\dfrac{1}{2021}$.
A. -2021 .
B. $-\dfrac{1}{2020}$.
C. -2020 .
D. $-\dfrac{1}{2021}$.
- Ta có: $\sqrt{x . f^{\prime}(x)}=-f(x), \forall x \geq 1 \Leftrightarrow\left\{\begin{array}{l}f(x) \leq 0 \\ x . f^{\prime}(x)=f^2(x)\end{array}, \forall x \geq 1\right.$
$\Rightarrow \dfrac{f^{\prime}(x)}{f^2(x)}=\dfrac{1}{x} \Rightarrow \int \dfrac{f^{\prime}(x)}{f^2(x)} \mathrm{d} x=\int \dfrac{1}{x} \mathrm{~d} x \Rightarrow-\dfrac{1}{f(x)}=\ln x+C, \forall x \geq 1$.
Lại do: $f(\mathrm{e})=-\dfrac{1}{2} \Rightarrow C=1 \Rightarrow f(x)=-\dfrac{1}{1+\ln x}$ (thỏa mãn điều kiện $f(x) \leq 0, \forall x \geq 1$ ).
Vậy $f\left(\mathrm{e}^{2020}\right)=-\dfrac{1}{2021}$.
$\Rightarrow \dfrac{f^{\prime}(x)}{f^2(x)}=\dfrac{1}{x} \Rightarrow \int \dfrac{f^{\prime}(x)}{f^2(x)} \mathrm{d} x=\int \dfrac{1}{x} \mathrm{~d} x \Rightarrow-\dfrac{1}{f(x)}=\ln x+C, \forall x \geq 1$.
Lại do: $f(\mathrm{e})=-\dfrac{1}{2} \Rightarrow C=1 \Rightarrow f(x)=-\dfrac{1}{1+\ln x}$ (thỏa mãn điều kiện $f(x) \leq 0, \forall x \geq 1$ ).
Vậy $f\left(\mathrm{e}^{2020}\right)=-\dfrac{1}{2021}$.
Đáp án D.