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Câu hỏi: Cho hàm số $f\left( x \right)$ có đạo hàm liên tục và nhận giá trị dương trên $\left( 0;+\infty \right)$ thỏa mãn điều kiện $\dfrac{1}{{{f}^{2}}\left( x \right)}=\dfrac{1}{{{x}^{2}}}+\dfrac{2\text{x}{f}'\left( x \right)}{{{f}^{3}}\left( x \right)}$ với mọi $x\in \left( 1;+\infty \right)$ đồng thời $f\left( 2 \right)=1$. Giá trị của $f\left( 4 \right)$ là
A. $\dfrac{2\sqrt{3}}{3}$
B. $\dfrac{\sqrt{2}}{3}$
C. $\dfrac{4}{3}$
D. $\dfrac{16}{9}$
A. $\dfrac{2\sqrt{3}}{3}$
B. $\dfrac{\sqrt{2}}{3}$
C. $\dfrac{4}{3}$
D. $\dfrac{16}{9}$
Ta có $\dfrac{1}{{{f}^{2}}\left( x \right)}=\dfrac{1}{{{x}^{2}}}+\dfrac{2\text{x}{f}'\left( x \right)}{{{f}^{3}}\left( x \right)}\Leftrightarrow \dfrac{{{f}^{2}}\left( x \right)-2\text{x}f\left( x \right).{f}'\left( x \right)}{{{f}^{4}}\left( x \right)}=\dfrac{1}{{{x}^{2}}}\Leftrightarrow {{\left( \dfrac{x}{{{f}^{2}}\left( x \right)} \right)}^{\prime }}=\dfrac{1}{{{x}^{2}}}$.
Suy ra $\int{{{\left( \dfrac{x}{{{f}^{2}}\left( x \right)} \right)}^{\prime }}d\text{x}}=\int{\dfrac{1}{{{x}^{2}}}d\text{x}}\Leftrightarrow \dfrac{x}{{{f}^{2}}\left( x \right)}=-\dfrac{1}{x}+C$.
Lại có $f\left( 2 \right)=1$ nên $C=\dfrac{5}{2}$.
Do đó: $\dfrac{x}{{{f}^{2}}\left( x \right)}=\dfrac{5}{2}-\dfrac{1}{x}=\dfrac{5\text{x}-2}{2\text{x}}\Rightarrow {{f}^{2}}\left( x \right)=\dfrac{2{{\text{x}}^{2}}}{5\text{x}-2}$.
Suy ra ${{f}^{2}}\left( 4 \right)=\dfrac{16}{9}\Rightarrow f\left( 4 \right)=\dfrac{4}{3}$.
Suy ra $\int{{{\left( \dfrac{x}{{{f}^{2}}\left( x \right)} \right)}^{\prime }}d\text{x}}=\int{\dfrac{1}{{{x}^{2}}}d\text{x}}\Leftrightarrow \dfrac{x}{{{f}^{2}}\left( x \right)}=-\dfrac{1}{x}+C$.
Lại có $f\left( 2 \right)=1$ nên $C=\dfrac{5}{2}$.
Do đó: $\dfrac{x}{{{f}^{2}}\left( x \right)}=\dfrac{5}{2}-\dfrac{1}{x}=\dfrac{5\text{x}-2}{2\text{x}}\Rightarrow {{f}^{2}}\left( x \right)=\dfrac{2{{\text{x}}^{2}}}{5\text{x}-2}$.
Suy ra ${{f}^{2}}\left( 4 \right)=\dfrac{16}{9}\Rightarrow f\left( 4 \right)=\dfrac{4}{3}$.
Đáp án C.