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Câu hỏi: Cho hàm số $y=f\left( x \right)$ 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 $f\left( 0 \right)=f'\left( 0 \right)=2.$ Tính giá trị của $T={{f}^{2}}\left( 2 \right)$
A. $\dfrac{268}{15}.$
B. $\dfrac{160}{15}.$
C. $\dfrac{268}{30}.$
D. $\dfrac{4}{15}.$
A. $\dfrac{268}{15}.$
B. $\dfrac{160}{15}.$
C. $\dfrac{268}{30}.$
D. $\dfrac{4}{15}.$
Ta có: $VT=\left[ f\left( x \right).f'\left( x \right) \right]'=f'\left( x \right).f'\left( x \right)+f\left( x \right).f''\left( x \right)={{\left[ f'\left( x \right) \right]}^{2}}+f\left( x \right).f''\left( x \right)$
$\Rightarrow \left[ f'\left( x \right).f\left( x \right) \right]'={{x}^{3}}-2x \left( * \right)$
Nguyên hàm hai vế của (*) ta được: $f'\left( x \right).f\left( x \right)=\dfrac{{{x}^{4}}}{4}-{{x}^{2}}+C \left( 1 \right)$
Lại có: $f'\left( 0 \right)=f\left( 0 \right)=2\Rightarrow C=2.2=4\Rightarrow \left( 1 \right)\Leftrightarrow f\left( x \right).f'\left( x \right)=\dfrac{{{x}^{4}}}{4}-{{x}^{2}}+4$
$\begin{aligned}
& \Rightarrow \int{f\left( x \right)f'\left( x \right)dx}=\int{\left( \dfrac{{{x}^{4}}}{4}-{{x}^{2}}+4 \right)dx}\Leftrightarrow \int{f\left( x \right)df\left( x \right)}=\dfrac{{{x}^{5}}}{20}-\dfrac{{{x}^{3}}}{3}+4x+A \\
& \Leftrightarrow \dfrac{{{f}^{2}}\left( x \right)}{2}=\dfrac{{{x}^{5}}}{20}-\dfrac{{{x}^{3}}}{3}+4x+A\Leftrightarrow {{f}^{2}}\left( x \right)=\dfrac{{{x}^{5}}}{10}-\dfrac{2{{x}^{3}}}{3}+8x+2A \\
\end{aligned}$
Có $f\left( 0 \right)=2\Rightarrow 4=2A\Leftrightarrow A=2\Rightarrow {{f}^{2}}\left( x \right)=\dfrac{{{x}^{5}}}{10}-\dfrac{2{{x}^{3}}}{3}+8x+4$
$\Rightarrow {{f}^{2}}\left( x \right)=\dfrac{{{2}^{5}}}{10}-\dfrac{{{2.2}^{3}}}{3}+8.2+4=\dfrac{268}{15}$
$\Rightarrow \left[ f'\left( x \right).f\left( x \right) \right]'={{x}^{3}}-2x \left( * \right)$
Nguyên hàm hai vế của (*) ta được: $f'\left( x \right).f\left( x \right)=\dfrac{{{x}^{4}}}{4}-{{x}^{2}}+C \left( 1 \right)$
Lại có: $f'\left( 0 \right)=f\left( 0 \right)=2\Rightarrow C=2.2=4\Rightarrow \left( 1 \right)\Leftrightarrow f\left( x \right).f'\left( x \right)=\dfrac{{{x}^{4}}}{4}-{{x}^{2}}+4$
$\begin{aligned}
& \Rightarrow \int{f\left( x \right)f'\left( x \right)dx}=\int{\left( \dfrac{{{x}^{4}}}{4}-{{x}^{2}}+4 \right)dx}\Leftrightarrow \int{f\left( x \right)df\left( x \right)}=\dfrac{{{x}^{5}}}{20}-\dfrac{{{x}^{3}}}{3}+4x+A \\
& \Leftrightarrow \dfrac{{{f}^{2}}\left( x \right)}{2}=\dfrac{{{x}^{5}}}{20}-\dfrac{{{x}^{3}}}{3}+4x+A\Leftrightarrow {{f}^{2}}\left( x \right)=\dfrac{{{x}^{5}}}{10}-\dfrac{2{{x}^{3}}}{3}+8x+2A \\
\end{aligned}$
Có $f\left( 0 \right)=2\Rightarrow 4=2A\Leftrightarrow A=2\Rightarrow {{f}^{2}}\left( x \right)=\dfrac{{{x}^{5}}}{10}-\dfrac{2{{x}^{3}}}{3}+8x+4$
$\Rightarrow {{f}^{2}}\left( x \right)=\dfrac{{{2}^{5}}}{10}-\dfrac{{{2.2}^{3}}}{3}+8.2+4=\dfrac{268}{15}$
Đáp án A.