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Câu hỏi: Cho hàm số y = f(x) thỏa mãn $f'\left( x \right).f\left( x \right)={{x}^{4}}+{{x}^{2}}$ với mọi số thực x, biết $f\left( 0 \right)=2$. Tính ${{f}^{2}}\left( 2 \right)$.
A. ${{f}^{2}}\left( 2 \right)=\dfrac{313}{15}$.
B. ${{f}^{2}}\left( 2 \right)=\dfrac{332}{15}$.
C. ${{f}^{2}}\left( 2 \right)=\dfrac{324}{15}$.
D. ${{f}^{2}}\left( 2 \right)=\dfrac{323}{15}$.
A. ${{f}^{2}}\left( 2 \right)=\dfrac{313}{15}$.
B. ${{f}^{2}}\left( 2 \right)=\dfrac{332}{15}$.
C. ${{f}^{2}}\left( 2 \right)=\dfrac{324}{15}$.
D. ${{f}^{2}}\left( 2 \right)=\dfrac{323}{15}$.
Ta có:
$f'\left( x \right).f\left( x \right)={{x}^{4}}+{{x}^{2}}\Rightarrow \int{f'\left( x \right).f\left( x \right)dx}=\int{\left( {{x}^{4}}+{{x}^{2}} \right)dx}$
$\Leftrightarrow \int{f\left( x \right).d\left( f\left( x \right) \right)}=\dfrac{{{x}^{5}}}{5}+\dfrac{{{x}^{3}}}{3}+C\Leftrightarrow \dfrac{{{f}^{2}}\left( x \right)}{2}=\dfrac{{{x}^{5}}}{5}+\dfrac{{{x}^{3}}}{3}+C$
mà $f\left( 0 \right)=2\Rightarrow {{f}^{2}}\left( 0 \right)=4\Rightarrow C=2\Rightarrow \dfrac{{{x}^{5}}}{5}+\dfrac{{{x}^{3}}}{3}+2$
Vậy ${{f}^{2}}\left( 2 \right)=2\left( \dfrac{{{x}^{5}}}{5}+\dfrac{{{x}^{3}}}{3}+2 \right)\left| \begin{aligned}
& \\
& x=2 \\
\end{aligned} \right.=\dfrac{332}{15}$
$f'\left( x \right).f\left( x \right)={{x}^{4}}+{{x}^{2}}\Rightarrow \int{f'\left( x \right).f\left( x \right)dx}=\int{\left( {{x}^{4}}+{{x}^{2}} \right)dx}$
$\Leftrightarrow \int{f\left( x \right).d\left( f\left( x \right) \right)}=\dfrac{{{x}^{5}}}{5}+\dfrac{{{x}^{3}}}{3}+C\Leftrightarrow \dfrac{{{f}^{2}}\left( x \right)}{2}=\dfrac{{{x}^{5}}}{5}+\dfrac{{{x}^{3}}}{3}+C$
mà $f\left( 0 \right)=2\Rightarrow {{f}^{2}}\left( 0 \right)=4\Rightarrow C=2\Rightarrow \dfrac{{{x}^{5}}}{5}+\dfrac{{{x}^{3}}}{3}+2$
Vậy ${{f}^{2}}\left( 2 \right)=2\left( \dfrac{{{x}^{5}}}{5}+\dfrac{{{x}^{3}}}{3}+2 \right)\left| \begin{aligned}
& \\
& x=2 \\
\end{aligned} \right.=\dfrac{332}{15}$
Đáp án B.