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Câu hỏi: Cho hàm số $f\left( x \right)$ xác định và có đạo hàm $f'\left( x \right)$ liên tục trên đoạn $\left[ 1;3 \right]$ và $f\left( x \right)\ne 0$ với mọi $x\in \left[ 1;3 \right]$, đồng thời $f'\left( x \right)+{{\left( 1+f\left( x \right) \right)}^{2}}={{\left[ {{\left( f\left( x \right) \right)}^{2}}\left( x-1 \right) \right]}^{2}}$ và $f\left( 1 \right)=-1.$ Biết rằng $\int\limits_{1}^{3}{f\left( x \right)dx}=a\ln 3+b,a,b\in \mathbb{Z}.$ Tính tổng $S=a+{{b}^{2}}.$
A. $S=-1.$
B. $S=2.$
C. $S=0.$
D. $S=-4.$
A. $S=-1.$
B. $S=2.$
C. $S=0.$
D. $S=-4.$
Ta có: $f'(x){{(1+f(x))}^{2}}={{\!\![\!\!{{(f(x))}^{2}}(x-1)\!\!]\!\!}^{2}}\Leftrightarrow \dfrac{f'(x){{(1+f(x))}^{2}}}{{{f}^{4}}(x)}={{(x-1)}^{2.}}$
Lấy nguyên hàm 2 vế ta được $\int{\dfrac{f'(x){{(1+f(x))}^{2}}}{{{f}^{4}}(x)}dx=\int{{{(x-1)}^{2}}dx}}$
$\begin{aligned}
& \Leftrightarrow \int{\dfrac{(1+2f(x)+{{f}^{2}}(x))f'(x)}{{{f}^{4}}(x)}dx}=\int{{{(x-1)}^{2}}dx} \\
& \Leftrightarrow \int{\left( \dfrac{1}{{{f}^{4}}(x)}+2\dfrac{1}{{{f}^{3}}(x)}+\dfrac{1}{{{f}^{2}}(x)} \right)}d(f(x))=\dfrac{{{(x-1)}^{3}}}{3}+C \\
& \Leftrightarrow -\dfrac{1}{3{{f}^{3}}(x)}-\dfrac{1}{{{f}^{2}}(x)}-\dfrac{1}{f(x)}=\dfrac{{{(x-1)}^{3}}}{3}+C \\
& \Leftrightarrow -\dfrac{1+3f(x)+3{{f}^{2}}(x)}{3{{f}^{3}}(x)}=\dfrac{{{(x-1)}^{3}}}{3}+C \\
\end{aligned}$
Mà $f(1)=-1\Rightarrow -\dfrac{1-3+3}{-3}=C\Rightarrow C=\dfrac{1}{3}$.
$\begin{aligned}
& \Rightarrow -\dfrac{1+3f(x)+3{{f}^{2}}(x)}{3{{f}^{3}}(x)}=\dfrac{{{(x-1)}^{3}}}{3}+\dfrac{1}{3} \\
& \Leftrightarrow \dfrac{1+3f(x)+3{{f}^{2}}(x)}{3{{f}^{3}}(x)}+\dfrac{1}{3}=-\dfrac{{{(x-1)}^{3}}}{3} \\
& \Leftrightarrow \dfrac{{{(1+f(x))}^{3}}}{{{f}^{3}}(x)}=-{{(x-1)}^{3}} \\
& \Leftrightarrow {{\left( 1+\dfrac{1}{f(x)} \right)}^{3}}={{(1-x)}^{3}} \\
& \Leftrightarrow f(x)=\dfrac{-1}{x}. \\
\end{aligned}$
Vậy $\int\limits_{1}^{3}{f(x)dx}=\int\limits_{1}^{3}{\dfrac{-1}{x}dx}=-\ln |x|\left| \begin{aligned}
& 3 \\
& 1 \\
\end{aligned} \right.=-\ln 3 $. Suy ra $ a=-1;b=0 $ hay $ a+b=-1$.
Lấy nguyên hàm 2 vế ta được $\int{\dfrac{f'(x){{(1+f(x))}^{2}}}{{{f}^{4}}(x)}dx=\int{{{(x-1)}^{2}}dx}}$
$\begin{aligned}
& \Leftrightarrow \int{\dfrac{(1+2f(x)+{{f}^{2}}(x))f'(x)}{{{f}^{4}}(x)}dx}=\int{{{(x-1)}^{2}}dx} \\
& \Leftrightarrow \int{\left( \dfrac{1}{{{f}^{4}}(x)}+2\dfrac{1}{{{f}^{3}}(x)}+\dfrac{1}{{{f}^{2}}(x)} \right)}d(f(x))=\dfrac{{{(x-1)}^{3}}}{3}+C \\
& \Leftrightarrow -\dfrac{1}{3{{f}^{3}}(x)}-\dfrac{1}{{{f}^{2}}(x)}-\dfrac{1}{f(x)}=\dfrac{{{(x-1)}^{3}}}{3}+C \\
& \Leftrightarrow -\dfrac{1+3f(x)+3{{f}^{2}}(x)}{3{{f}^{3}}(x)}=\dfrac{{{(x-1)}^{3}}}{3}+C \\
\end{aligned}$
Mà $f(1)=-1\Rightarrow -\dfrac{1-3+3}{-3}=C\Rightarrow C=\dfrac{1}{3}$.
$\begin{aligned}
& \Rightarrow -\dfrac{1+3f(x)+3{{f}^{2}}(x)}{3{{f}^{3}}(x)}=\dfrac{{{(x-1)}^{3}}}{3}+\dfrac{1}{3} \\
& \Leftrightarrow \dfrac{1+3f(x)+3{{f}^{2}}(x)}{3{{f}^{3}}(x)}+\dfrac{1}{3}=-\dfrac{{{(x-1)}^{3}}}{3} \\
& \Leftrightarrow \dfrac{{{(1+f(x))}^{3}}}{{{f}^{3}}(x)}=-{{(x-1)}^{3}} \\
& \Leftrightarrow {{\left( 1+\dfrac{1}{f(x)} \right)}^{3}}={{(1-x)}^{3}} \\
& \Leftrightarrow f(x)=\dfrac{-1}{x}. \\
\end{aligned}$
Vậy $\int\limits_{1}^{3}{f(x)dx}=\int\limits_{1}^{3}{\dfrac{-1}{x}dx}=-\ln |x|\left| \begin{aligned}
& 3 \\
& 1 \\
\end{aligned} \right.=-\ln 3 $. Suy ra $ a=-1;b=0 $ hay $ a+b=-1$.
Đáp án A.