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Câu hỏi: Cho hàm số $y=f\left( x \right)$ có đạo hàm trên $\left[ 0;3 \right]$, thỏa mãn $\left\{ \begin{aligned}
& f\left( 3-x \right).f\left( x \right)=1 \\
& f\left( x \right)\ne -1 \\
\end{aligned} \right. $ với mọi $ x\in \left[ 0;3 \right] $ và $ f\left( 0 \right)=\dfrac{1}{2} $. Tính tích phân $ I=\int\limits_{0}^{3}{\dfrac{x{f}'\left( x \right)}{{{\left[ 1+f\left( 3-x \right) \right]}^{2}}.{{f}^{2}}\left( x \right)}dx}$.
A. $I=\dfrac{1}{2}$.
B. $I=1$.
C. $I=\dfrac{3}{2}$.
D. $I=\dfrac{5}{2}$.
& f\left( 3-x \right).f\left( x \right)=1 \\
& f\left( x \right)\ne -1 \\
\end{aligned} \right. $ với mọi $ x\in \left[ 0;3 \right] $ và $ f\left( 0 \right)=\dfrac{1}{2} $. Tính tích phân $ I=\int\limits_{0}^{3}{\dfrac{x{f}'\left( x \right)}{{{\left[ 1+f\left( 3-x \right) \right]}^{2}}.{{f}^{2}}\left( x \right)}dx}$.
A. $I=\dfrac{1}{2}$.
B. $I=1$.
C. $I=\dfrac{3}{2}$.
D. $I=\dfrac{5}{2}$.
Từ giả thiết $\left\{ \begin{aligned}
& f\left( 3-x \right).f\left( x \right)=1 \\
& f\left( x \right)=\dfrac{1}{2} \\
\end{aligned} \right.\Rightarrow f\left( 3 \right)=2$.
Ta có: ${{\left[ 1+f\left( 3-x \right) \right]}^{2}}.{{f}^{2}}\left( x \right)={{\left[ f\left( x \right)+f\left( 3-x \right).f\left( x \right) \right]}^{2}}={{\left[ f\left( x \right)+1 \right]}^{2}}$
+ Tính $I=\int\limits_{0}^{3}{\dfrac{x{f}'\left( x \right)}{{{\left[ 1+f\left( x \right) \right]}^{2}}}dx}=-\int\limits_{0}^{3}{xd\left( \dfrac{1}{1+f\left( x \right)} \right)}=-\dfrac{x}{1+f\left( x \right)}\left| \begin{aligned}
& 3 \\
& 0 \\
\end{aligned} \right.+\int\limits_{0}^{3}{\dfrac{1}{1+f\left( x \right)}dx=-1+J}$
+ Tính $J=\int\limits_{0}^{3}{\dfrac{1}{1+f\left( x \right)}dx}\overset{t=3-x}{\mathop{=}} -\int\limits_{3}^{0}{\dfrac{1}{1+f\left( 3-t \right)}dt}=\int\limits_{0}^{3}{\dfrac{1}{1+f\left( 3-t \right)}dt}=\int\limits_{0}^{3}{\dfrac{1}{1+f\left( 3-x \right)}dx}$
$\Rightarrow 2J=\int\limits_{0}^{3}{\dfrac{1}{1+f\left( x \right)}dx+\int\limits_{0}^{3}{\dfrac{1}{1+f\left( 3-x \right)}dx}=\int\limits_{0}^{3}{dx\left( f\left( 3-x \right).f\left( x \right)=1 \right)}=3\Rightarrow J=\dfrac{3}{2}}$
Vậy $I=\int\limits_{0}^{3}{\dfrac{x{f}'\left( x \right)}{{{\left[ 1+f\left( 3-x \right) \right]}^{2}}.{{f}^{2}}\left( x \right)}dx}=\dfrac{1}{2}$
& f\left( 3-x \right).f\left( x \right)=1 \\
& f\left( x \right)=\dfrac{1}{2} \\
\end{aligned} \right.\Rightarrow f\left( 3 \right)=2$.
Ta có: ${{\left[ 1+f\left( 3-x \right) \right]}^{2}}.{{f}^{2}}\left( x \right)={{\left[ f\left( x \right)+f\left( 3-x \right).f\left( x \right) \right]}^{2}}={{\left[ f\left( x \right)+1 \right]}^{2}}$
+ Tính $I=\int\limits_{0}^{3}{\dfrac{x{f}'\left( x \right)}{{{\left[ 1+f\left( x \right) \right]}^{2}}}dx}=-\int\limits_{0}^{3}{xd\left( \dfrac{1}{1+f\left( x \right)} \right)}=-\dfrac{x}{1+f\left( x \right)}\left| \begin{aligned}
& 3 \\
& 0 \\
\end{aligned} \right.+\int\limits_{0}^{3}{\dfrac{1}{1+f\left( x \right)}dx=-1+J}$
+ Tính $J=\int\limits_{0}^{3}{\dfrac{1}{1+f\left( x \right)}dx}\overset{t=3-x}{\mathop{=}} -\int\limits_{3}^{0}{\dfrac{1}{1+f\left( 3-t \right)}dt}=\int\limits_{0}^{3}{\dfrac{1}{1+f\left( 3-t \right)}dt}=\int\limits_{0}^{3}{\dfrac{1}{1+f\left( 3-x \right)}dx}$
$\Rightarrow 2J=\int\limits_{0}^{3}{\dfrac{1}{1+f\left( x \right)}dx+\int\limits_{0}^{3}{\dfrac{1}{1+f\left( 3-x \right)}dx}=\int\limits_{0}^{3}{dx\left( f\left( 3-x \right).f\left( x \right)=1 \right)}=3\Rightarrow J=\dfrac{3}{2}}$
Vậy $I=\int\limits_{0}^{3}{\dfrac{x{f}'\left( x \right)}{{{\left[ 1+f\left( 3-x \right) \right]}^{2}}.{{f}^{2}}\left( x \right)}dx}=\dfrac{1}{2}$
Đáp án A.