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Câu hỏi: Cho hàm số $f\left( x \right)$ nhận giá trị dương và có đạo hàm liên tục trên đoạn $\left[ 0;1 \right]$ sao cho $f\left( 1 \right)=1$ và $f\left( x \right).f\left( 1-x \right)={{e}^{{{x}^{2}}-x}},$ $\forall x\in \left[ 0;1 \right]$. Tính $I=\int\limits_{0}^{1}{\dfrac{\left( 2{{x}^{3}}-3{{x}^{2}} \right){f}'\left( x \right)}{f\left( x \right)}dx}$.
A. $I=-\dfrac{1}{10}$.
B. $I=\dfrac{2}{5}$.
C. $I=-\dfrac{1}{60}$.
D. $I=\dfrac{1}{10}$.
A. $I=-\dfrac{1}{10}$.
B. $I=\dfrac{2}{5}$.
C. $I=-\dfrac{1}{60}$.
D. $I=\dfrac{1}{10}$.
Ta có $f\left( x \right).f\left( 1-x \right)={{e}^{{{x}^{2}}-x}}\Leftrightarrow \ln f\left( x \right)+\ln f\left( 1-x \right)={{x}^{2}}-x$
$\Leftrightarrow \left( {{x}^{2}}-x \right)\ln f\left( x \right)+\left( {{x}^{2}}-x \right)\ln f\left( 1-x \right)={{\left( {{x}^{2}}-x \right)}^{2}}$
$\Rightarrow \int\limits_{0}^{1}{\left( {{x}^{2}}-x \right)\ln f\left( x \right)\text{d}x}+\int\limits_{0}^{1}{\left( {{x}^{2}}-x \right)\ln f\left( 1-x \right)\text{d}x}=\int\limits_{0}^{1}{{{\left( {{x}^{2}}-x \right)}^{2}}\text{d}x}$
$\Leftrightarrow \int\limits_{0}^{1}{\left( {{x}^{2}}-x \right)\ln f\left( x \right)\text{d}x}+\int\limits_{0}^{1}{x\left( x-1 \right)\ln f\left( 1-x \right)\text{d}x}=\int\limits_{0}^{1}{{{\left( {{x}^{2}}-x \right)}^{2}}\text{d}x}$
Đặt $t=1-x\Leftrightarrow x=1-t\Rightarrow \left\{ \begin{matrix}
\text{d}x=-\text{d}t \\
x=0\Rightarrow t=1 \\
x=1\Rightarrow t=0 \\
\end{matrix} \right.$
$\Rightarrow I=\int\limits_{0}^{1}{x\left( x-1 \right)\ln f\left( 1-x \right)\text{d}x}=\int\limits_{1}^{0}{\left( 1-t \right)t\ln f\left( t \right)\text{d}t}=\int\limits_{0}^{1}{x\left( x-1 \right)\ln f\left( x \right)\text{d}x}=\int\limits_{0}^{1}{\left( {{x}^{2}}-x \right)\ln f\left( x \right)\text{d}x}$
$\Rightarrow 2\int\limits_{0}^{1}{\left( {{x}^{2}}-x \right)\ln f\left( x \right)\text{d}x}=\int\limits_{0}^{1}{{{\left( {{x}^{2}}-x \right)}^{2}}\text{d}x}=\dfrac{1}{30}\Leftrightarrow \int\limits_{0}^{1}{\left( {{x}^{2}}-x \right)\ln f\left( x \right)\text{d}x}=\dfrac{1}{60}$
$I=\int\limits_{0}^{1}{\dfrac{\left( 2{{x}^{3}}-3{{x}^{2}} \right){f}'\left( x \right)}{f\left( x \right)}\text{d}x}=\int\limits_{0}^{1}{\left( 2{{x}^{3}}-3{{x}^{2}} \right)\text{d}\ln }f\left( x \right)$
$=\left. \left( 2{{x}^{3}}-3{{x}^{2}} \right)\ln f\left( x \right) \right|_{0}^{1}-6\int\limits_{0}^{1}{\left( {{x}^{2}}-x \right)\ln f\left( x \right)\text{d}x}=-6\int\limits_{0}^{1}{\left( {{x}^{2}}-x \right)\ln f\left( x \right)\text{d}x}=\dfrac{-1}{10}$.
$\Leftrightarrow \left( {{x}^{2}}-x \right)\ln f\left( x \right)+\left( {{x}^{2}}-x \right)\ln f\left( 1-x \right)={{\left( {{x}^{2}}-x \right)}^{2}}$
$\Rightarrow \int\limits_{0}^{1}{\left( {{x}^{2}}-x \right)\ln f\left( x \right)\text{d}x}+\int\limits_{0}^{1}{\left( {{x}^{2}}-x \right)\ln f\left( 1-x \right)\text{d}x}=\int\limits_{0}^{1}{{{\left( {{x}^{2}}-x \right)}^{2}}\text{d}x}$
$\Leftrightarrow \int\limits_{0}^{1}{\left( {{x}^{2}}-x \right)\ln f\left( x \right)\text{d}x}+\int\limits_{0}^{1}{x\left( x-1 \right)\ln f\left( 1-x \right)\text{d}x}=\int\limits_{0}^{1}{{{\left( {{x}^{2}}-x \right)}^{2}}\text{d}x}$
Đặt $t=1-x\Leftrightarrow x=1-t\Rightarrow \left\{ \begin{matrix}
\text{d}x=-\text{d}t \\
x=0\Rightarrow t=1 \\
x=1\Rightarrow t=0 \\
\end{matrix} \right.$
$\Rightarrow I=\int\limits_{0}^{1}{x\left( x-1 \right)\ln f\left( 1-x \right)\text{d}x}=\int\limits_{1}^{0}{\left( 1-t \right)t\ln f\left( t \right)\text{d}t}=\int\limits_{0}^{1}{x\left( x-1 \right)\ln f\left( x \right)\text{d}x}=\int\limits_{0}^{1}{\left( {{x}^{2}}-x \right)\ln f\left( x \right)\text{d}x}$
$\Rightarrow 2\int\limits_{0}^{1}{\left( {{x}^{2}}-x \right)\ln f\left( x \right)\text{d}x}=\int\limits_{0}^{1}{{{\left( {{x}^{2}}-x \right)}^{2}}\text{d}x}=\dfrac{1}{30}\Leftrightarrow \int\limits_{0}^{1}{\left( {{x}^{2}}-x \right)\ln f\left( x \right)\text{d}x}=\dfrac{1}{60}$
$I=\int\limits_{0}^{1}{\dfrac{\left( 2{{x}^{3}}-3{{x}^{2}} \right){f}'\left( x \right)}{f\left( x \right)}\text{d}x}=\int\limits_{0}^{1}{\left( 2{{x}^{3}}-3{{x}^{2}} \right)\text{d}\ln }f\left( x \right)$
$=\left. \left( 2{{x}^{3}}-3{{x}^{2}} \right)\ln f\left( x \right) \right|_{0}^{1}-6\int\limits_{0}^{1}{\left( {{x}^{2}}-x \right)\ln f\left( x \right)\text{d}x}=-6\int\limits_{0}^{1}{\left( {{x}^{2}}-x \right)\ln f\left( x \right)\text{d}x}=\dfrac{-1}{10}$.
Đáp án A.