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Câu hỏi: Cho hàm số f(x) thỏa mãn ${{\left( f'\left( x \right) \right)}^{2}}+f\left( x \right).f''\left( x \right)={{x}^{3}}-2x,\forall x\in \mathbb{R}$ và $f\left( 0 \right)=f'\left( 0 \right)=1$. Tính giá trị của $T={{f}^{2}}\left( 2 \right)$.
A. $\dfrac{43}{30}$.
B. $\dfrac{16}{15}$.
C. $\dfrac{43}{15}$.
D. $\dfrac{26}{15}$.
A. $\dfrac{43}{30}$.
B. $\dfrac{16}{15}$.
C. $\dfrac{43}{15}$.
D. $\dfrac{26}{15}$.
$\begin{aligned}
& {{\left( f'(x) \right)}^{2}}+f(x).f''(x)={{x}^{3}}-2x \\
& \Leftrightarrow \left[ f(x).f'(x) \right]'={{x}^{3}}-2x \\
& \Leftrightarrow f(x).f'(x)=\dfrac{{{x}^{4}}}{4}-{{x}^{2}}+C \\
\end{aligned}$
Ta có $f(0)=f'(0)=1$ nên $C=1$
$\Rightarrow f(x).f'(x)=\dfrac{{{x}^{4}}}{4}-{{x}^{2}}+1\Leftrightarrow \left[ \dfrac{1}{2}{{f}^{2}}(x) \right]'=\dfrac{{{x}^{4}}}{4}-{{x}^{2}}+1\Leftrightarrow \dfrac{1}{2}{{f}^{2}}(x)=\dfrac{{{x}^{5}}}{20}-\dfrac{{{x}^{2}}}{3}+x+{{C}_{1}}$
Ta có $f(0)=1$ nên ${{C}_{1}}=\dfrac{1}{2}$
$\Rightarrow \dfrac{1}{2}{{f}^{2}}(x)=\dfrac{{{x}^{5}}}{20}-\dfrac{{{x}^{2}}}{3}+x+\dfrac{1}{2}\Leftrightarrow {{f}^{2}}(x)=\dfrac{{{x}^{5}}}{10}-\dfrac{2{{x}^{3}}}{3}+2x+1\Rightarrow {{f}^{2}}(2)=\dfrac{43}{15}$
Vậy ${{f}^{2}}(2)=\dfrac{43}{15}.$
& {{\left( f'(x) \right)}^{2}}+f(x).f''(x)={{x}^{3}}-2x \\
& \Leftrightarrow \left[ f(x).f'(x) \right]'={{x}^{3}}-2x \\
& \Leftrightarrow f(x).f'(x)=\dfrac{{{x}^{4}}}{4}-{{x}^{2}}+C \\
\end{aligned}$
Ta có $f(0)=f'(0)=1$ nên $C=1$
$\Rightarrow f(x).f'(x)=\dfrac{{{x}^{4}}}{4}-{{x}^{2}}+1\Leftrightarrow \left[ \dfrac{1}{2}{{f}^{2}}(x) \right]'=\dfrac{{{x}^{4}}}{4}-{{x}^{2}}+1\Leftrightarrow \dfrac{1}{2}{{f}^{2}}(x)=\dfrac{{{x}^{5}}}{20}-\dfrac{{{x}^{2}}}{3}+x+{{C}_{1}}$
Ta có $f(0)=1$ nên ${{C}_{1}}=\dfrac{1}{2}$
$\Rightarrow \dfrac{1}{2}{{f}^{2}}(x)=\dfrac{{{x}^{5}}}{20}-\dfrac{{{x}^{2}}}{3}+x+\dfrac{1}{2}\Leftrightarrow {{f}^{2}}(x)=\dfrac{{{x}^{5}}}{10}-\dfrac{2{{x}^{3}}}{3}+2x+1\Rightarrow {{f}^{2}}(2)=\dfrac{43}{15}$
Vậy ${{f}^{2}}(2)=\dfrac{43}{15}.$
Đáp án C.