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Câu hỏi: Cho hàm số $f\left( x \right)>0,\forall x\in \left[ 0;+\infty \right)$ và có đạo hàm cấp hai liên tục trên nửa khoảng $\left[ 0;+\infty \right)$ thỏa mãn $f''\left( x \right).f\left( x \right)-2{{\left[ f'\left( x \right) \right]}^{2}}+2x{{f}^{3}}\left( x \right)=0,f'\left( 0 \right)=0,f\left( 0 \right)=1.$ Tính $f\left( 1 \right)$
A. $\dfrac{7}{5}$
B. $\dfrac{5}{4}$
C. $\dfrac{3}{4}$
D. $\dfrac{5}{7}$
A. $\dfrac{7}{5}$
B. $\dfrac{5}{4}$
C. $\dfrac{3}{4}$
D. $\dfrac{5}{7}$
Ta có
$f''\left( x \right).f\left( x \right)-2{{\left[ f'\left( x \right) \right]}^{2}}+2x{{f}^{3}}\left( x \right)=0$
$\Leftrightarrow \dfrac{f''\left( x \right).f\left( x \right)-2{{\left[ f'\left( x \right) \right]}^{2}}}{{{f}^{3}}\left( x \right)}=-2x$
$\Leftrightarrow \dfrac{f''\left( x \right).{{f}^{2}}\left( x \right)-2{{\left[ f'\left( x \right) \right]}^{2}}.f\left( x \right)}{{{f}^{4}}\left( x \right)}=-2x$
$\Leftrightarrow \left[ \dfrac{f'\left( x \right)}{{{f}^{2}}\left( x \right)} \right]'=-2x$
$\Rightarrow \dfrac{f'\left( x \right)}{{{f}^{2}}\left( x \right)}=-{{x}^{2}}+C$
Giả thiết $f'\left( 0 \right)=0,f\left( 0 \right)=1$ nên $C=0\Rightarrow \dfrac{f'\left( x \right)}{{{f}^{2}}\left( x \right)}=-{{x}^{2}}\Rightarrow \dfrac{1}{f\left( x \right)}=\dfrac{{{x}^{3}}}{3}+{{C}_{1}}$
Vì $f\left( 0 \right)=1\Leftrightarrow 1=0+{{C}_{1}}\Leftrightarrow {{C}_{1}}=1\Rightarrow \dfrac{1}{f\left( x \right)}=\dfrac{{{x}^{3}}}{3}+1$
Vậy $f\left( 1 \right)=\dfrac{3}{4}.$
$f''\left( x \right).f\left( x \right)-2{{\left[ f'\left( x \right) \right]}^{2}}+2x{{f}^{3}}\left( x \right)=0$
$\Leftrightarrow \dfrac{f''\left( x \right).f\left( x \right)-2{{\left[ f'\left( x \right) \right]}^{2}}}{{{f}^{3}}\left( x \right)}=-2x$
$\Leftrightarrow \dfrac{f''\left( x \right).{{f}^{2}}\left( x \right)-2{{\left[ f'\left( x \right) \right]}^{2}}.f\left( x \right)}{{{f}^{4}}\left( x \right)}=-2x$
$\Leftrightarrow \left[ \dfrac{f'\left( x \right)}{{{f}^{2}}\left( x \right)} \right]'=-2x$
$\Rightarrow \dfrac{f'\left( x \right)}{{{f}^{2}}\left( x \right)}=-{{x}^{2}}+C$
Giả thiết $f'\left( 0 \right)=0,f\left( 0 \right)=1$ nên $C=0\Rightarrow \dfrac{f'\left( x \right)}{{{f}^{2}}\left( x \right)}=-{{x}^{2}}\Rightarrow \dfrac{1}{f\left( x \right)}=\dfrac{{{x}^{3}}}{3}+{{C}_{1}}$
Vì $f\left( 0 \right)=1\Leftrightarrow 1=0+{{C}_{1}}\Leftrightarrow {{C}_{1}}=1\Rightarrow \dfrac{1}{f\left( x \right)}=\dfrac{{{x}^{3}}}{3}+1$
Vậy $f\left( 1 \right)=\dfrac{3}{4}.$
Đáp án C.