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Câu hỏi: Cho hàm số $y=f\left( x \right)$ có đạo hàm, nhận giá trị dương trên $\left( 0;+\infty \right)$ và thỏa mãn $2{f}'\left( {{x}^{2}} \right)=9\text{x}\sqrt{f\left( {{x}^{2}} \right)}$ với mọi $x\in \left( 0;+\infty \right)$. Biết $f\left( \dfrac{2}{3} \right)=\dfrac{2}{3}$, tính giá trị $f\left( \dfrac{1}{3} \right)$.
A. $\dfrac{1}{4}$
B. $\dfrac{1}{3}$
C. $\dfrac{1}{12}$
D. $\dfrac{1}{6}$
A. $\dfrac{1}{4}$
B. $\dfrac{1}{3}$
C. $\dfrac{1}{12}$
D. $\dfrac{1}{6}$
Ta có $2{f}'\left( {{x}^{2}} \right)=9\text{x}\sqrt{f\left( {{x}^{2}} \right)}$
$\Leftrightarrow \dfrac{2\text{x}{f}'\left( x \right)}{2\sqrt{f\left( {{x}^{2}} \right)}}=\dfrac{9}{2}{{x}^{2}}\Leftrightarrow \dfrac{{{\left[ f\left( {{x}^{2}} \right) \right]}^{\prime }}}{2\sqrt{f\left( {{x}^{2}} \right)}}=\dfrac{9}{2}{{x}^{2}}\Leftrightarrow {{\left[ \sqrt{f\left( {{x}^{2}} \right)} \right]}^{\prime }}=\dfrac{9}{2}{{x}^{2}}$
Do đó $\sqrt{f\left( {{x}^{2}} \right)}=\int{\dfrac{9}{2}{{x}^{2}}d\text{x}}=\dfrac{3}{2}{{x}^{3}}+C$
Mà $f\left( \dfrac{2}{3} \right)=\dfrac{2}{3}\Rightarrow \sqrt{\dfrac{2}{3}}=\dfrac{3}{2}.\dfrac{2}{3}\sqrt{\dfrac{2}{3}}+C\Leftrightarrow C=0$.
Suy ra $f\left( {{x}^{2}} \right)=\dfrac{9}{4}{{x}^{6}}\Leftrightarrow f\left( x \right)=\dfrac{9}{4}{{x}^{3}}\Rightarrow f\left( \dfrac{1}{3} \right)=\dfrac{9}{4}.{{\left( \dfrac{1}{3} \right)}^{3}}=\dfrac{1}{12}$.
$\Leftrightarrow \dfrac{2\text{x}{f}'\left( x \right)}{2\sqrt{f\left( {{x}^{2}} \right)}}=\dfrac{9}{2}{{x}^{2}}\Leftrightarrow \dfrac{{{\left[ f\left( {{x}^{2}} \right) \right]}^{\prime }}}{2\sqrt{f\left( {{x}^{2}} \right)}}=\dfrac{9}{2}{{x}^{2}}\Leftrightarrow {{\left[ \sqrt{f\left( {{x}^{2}} \right)} \right]}^{\prime }}=\dfrac{9}{2}{{x}^{2}}$
Do đó $\sqrt{f\left( {{x}^{2}} \right)}=\int{\dfrac{9}{2}{{x}^{2}}d\text{x}}=\dfrac{3}{2}{{x}^{3}}+C$
Mà $f\left( \dfrac{2}{3} \right)=\dfrac{2}{3}\Rightarrow \sqrt{\dfrac{2}{3}}=\dfrac{3}{2}.\dfrac{2}{3}\sqrt{\dfrac{2}{3}}+C\Leftrightarrow C=0$.
Suy ra $f\left( {{x}^{2}} \right)=\dfrac{9}{4}{{x}^{6}}\Leftrightarrow f\left( x \right)=\dfrac{9}{4}{{x}^{3}}\Rightarrow f\left( \dfrac{1}{3} \right)=\dfrac{9}{4}.{{\left( \dfrac{1}{3} \right)}^{3}}=\dfrac{1}{12}$.
Đáp án C.