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Câu hỏi: Cho hàm số $f\left( x \right)$ thỏa mãn ${{\left[ {f}'\left( x \right) \right]}^{2}}+f\left( x \right).{f}''\left( x \right)=4{{x}^{3}}+2x$ với mọi $x\in \mathbb{R}$ và $f\left( 0 \right)=0$. Giá trị của ${{f}^{2}}\left( 1 \right)$ bằng
A. $\dfrac{5}{2}$.
B. $\dfrac{9}{2}$.
C. $\dfrac{16}{15}$.
D. $\dfrac{8}{15}$.
A. $\dfrac{5}{2}$.
B. $\dfrac{9}{2}$.
C. $\dfrac{16}{15}$.
D. $\dfrac{8}{15}$.
Ta có: ${{\left[ {f}'\left( x \right) \right]}^{2}}+f\left( x \right).{f}''\left( x \right)={{\left[ f\left( x \right).{f}'\left( x \right) \right]}^{\prime }}$.
Từ giả thiết ta có: ${{\left[ f\left( x \right).{f}'\left( x \right) \right]}^{\prime }}=4{{x}^{3}}+2x$.
Suy ra: $f\left( x \right).{f}'\left( x \right)=\int\limits_{{}}^{{}}{\left( 4{{x}^{3}}+2x \right)dx}={{x}^{4}}+{{x}^{2}}+C$. Với $f\left( 0 \right)=0\Rightarrow C=0$
Nên ta có: $f\left( x \right).{f}'\left( x \right)={{x}^{4}}+{{x}^{2}}$
Suy ra: $\int\limits_{0}^{1}{f\left( x \right).{f}'\left( x \right)dx}=\int\limits_{0}^{1}{\left( {{x}^{4}}+{{x}^{2}} \right)dx}\Leftrightarrow \dfrac{{{f}^{2}}\left( x \right)}{2}\left| \begin{aligned}
& 1 \\
& 0 \\
\end{aligned} \right.=\dfrac{8}{15}\Rightarrow {{f}^{2}}\left( 1 \right)=\dfrac{16}{15}$.
Từ giả thiết ta có: ${{\left[ f\left( x \right).{f}'\left( x \right) \right]}^{\prime }}=4{{x}^{3}}+2x$.
Suy ra: $f\left( x \right).{f}'\left( x \right)=\int\limits_{{}}^{{}}{\left( 4{{x}^{3}}+2x \right)dx}={{x}^{4}}+{{x}^{2}}+C$. Với $f\left( 0 \right)=0\Rightarrow C=0$
Nên ta có: $f\left( x \right).{f}'\left( x \right)={{x}^{4}}+{{x}^{2}}$
Suy ra: $\int\limits_{0}^{1}{f\left( x \right).{f}'\left( x \right)dx}=\int\limits_{0}^{1}{\left( {{x}^{4}}+{{x}^{2}} \right)dx}\Leftrightarrow \dfrac{{{f}^{2}}\left( x \right)}{2}\left| \begin{aligned}
& 1 \\
& 0 \\
\end{aligned} \right.=\dfrac{8}{15}\Rightarrow {{f}^{2}}\left( 1 \right)=\dfrac{16}{15}$.
Đáp án C.