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Câu hỏi: Cho hàm số $f\left( x \right)$ xác định và có đạo hàm cấp hai trên $\left( 0;+\infty \right)$ thỏa mãn $f\left( 0 \right)=0$, $\underset{x\to 0}{\mathop{\lim }} \dfrac{f\left( x \right)}{x}=1$ và $f''\left( x \right)+{{\left[ f'\left( x \right) \right]}^{2}}+{{x}^{2}}=1+2xf'\left( x \right)$. Tính $f\left( 2 \right)$.
A. $1+\ln 3$.
B. $2+\ln 3$.
C. $2-\ln 3$.
D. $1-\ln 3$.
A. $1+\ln 3$.
B. $2+\ln 3$.
C. $2-\ln 3$.
D. $1-\ln 3$.
Do $\underset{x\to 0}{\mathop{\lim }} \dfrac{f\left( x \right)}{x}=1\Leftrightarrow \underset{x\to 0}{\mathop{\lim }} \dfrac{f\left( x \right)-f\left( 0 \right)}{x-0}=1\Leftrightarrow f'\left( 0 \right)=1$.
Ta có: $f''\left( x \right)+{{\left[ f'\left( x \right) \right]}^{2}}+{{x}^{2}}=1+2xf'\left( x \right)\Leftrightarrow {{\left[ f'\left( x \right)-x \right]}^{2}}=-\left( f''\left( x \right)-1 \right)$, (1)
Đặt $g\left( x \right)=f'\left( x \right)-x\Rightarrow g'\left( x \right)=f''\left( x \right)-1$, nên (1) trở thành ${{g}^{2}}\left( x \right)=-g'\left( x \right)\Rightarrow \dfrac{g'\left( x \right)}{{{g}^{2}}\left( x \right)}=-1.$
Lấy nguyên hàm hai vế, ta được $-\dfrac{1}{g\left( x \right)}=-x+C\Rightarrow g\left( x \right)=\dfrac{1}{x-C}\Rightarrow f'\left( x \right)=x+\dfrac{1}{x-C}$
Cho $x=0\Rightarrow f'\left( 0 \right)=\dfrac{1}{-C}\Rightarrow C=-1$. Do đó $f'\left( x \right)=x+\dfrac{1}{x+1}\Rightarrow f\left( x \right)=\dfrac{{{x}^{2}}}{2}+\ln \left| x+1 \right|+{{C}_{1}}$
Mặt khác $f\left( 0 \right)=0\Rightarrow {{C}_{1}}=0$. Suy ra $f\left( x \right)=\dfrac{{{x}^{2}}}{2}+\ln \left| x+1 \right|$. Vậy $f\left( 2 \right)=2+\ln 3$.
Ta có: $f''\left( x \right)+{{\left[ f'\left( x \right) \right]}^{2}}+{{x}^{2}}=1+2xf'\left( x \right)\Leftrightarrow {{\left[ f'\left( x \right)-x \right]}^{2}}=-\left( f''\left( x \right)-1 \right)$, (1)
Đặt $g\left( x \right)=f'\left( x \right)-x\Rightarrow g'\left( x \right)=f''\left( x \right)-1$, nên (1) trở thành ${{g}^{2}}\left( x \right)=-g'\left( x \right)\Rightarrow \dfrac{g'\left( x \right)}{{{g}^{2}}\left( x \right)}=-1.$
Lấy nguyên hàm hai vế, ta được $-\dfrac{1}{g\left( x \right)}=-x+C\Rightarrow g\left( x \right)=\dfrac{1}{x-C}\Rightarrow f'\left( x \right)=x+\dfrac{1}{x-C}$
Cho $x=0\Rightarrow f'\left( 0 \right)=\dfrac{1}{-C}\Rightarrow C=-1$. Do đó $f'\left( x \right)=x+\dfrac{1}{x+1}\Rightarrow f\left( x \right)=\dfrac{{{x}^{2}}}{2}+\ln \left| x+1 \right|+{{C}_{1}}$
Mặt khác $f\left( 0 \right)=0\Rightarrow {{C}_{1}}=0$. Suy ra $f\left( x \right)=\dfrac{{{x}^{2}}}{2}+\ln \left| x+1 \right|$. Vậy $f\left( 2 \right)=2+\ln 3$.
Đáp án B.