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Câu hỏi: Cho hàm số $f\left( x \right)$ nhận giá trị dương, có đạo hàm liên tục trên $\left[ 0;2 \right]$. Biết $f\left( 0 \right)=1$ và $f\left( x \right)f\left( 2-x \right)={{e}^{2{{\text{x}}^{2}}-4\text{x}}}$ với mọi $x\in \left[ 0;2 \right]$. Tích phân $I=\int\limits_{0}^{2}{\dfrac{\left( {{x}^{3}}-3{{\text{x}}^{2}} \right){f}'\left( x \right)}{f\left( x \right)}}d\text{x}$ bằng
A. $I=-\dfrac{14}{3}.$
B. $I=-\dfrac{32}{5}.$
C. $I=-\dfrac{16}{3}.$
D. $I=-\dfrac{16}{5}.$
A. $I=-\dfrac{14}{3}.$
B. $I=-\dfrac{32}{5}.$
C. $I=-\dfrac{16}{3}.$
D. $I=-\dfrac{16}{5}.$
Ta có $f\left( x \right)f\left( 2-x \right)={{e}^{2{{\text{x}}^{2}}-4\text{x}}}$ suy ra $f\left( 0 \right)f\left( 2 \right)={{e}^{0}}\Rightarrow f\left( 2 \right)=1$.
Đặt $\left\{ \begin{aligned}
& u={{x}^{3}}-3{{x}^{2}} \\
& dv=\dfrac{{f}'\left( x \right)}{f\left( x \right)}dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=\left( 3{{x}^{2}}-6x \right)dx \\
& v=\ln f\left( x \right) \\
\end{aligned} \right.$
Khi đó $I=\left( {{x}^{3}}-3{{x}^{2}} \right)\ln f\left( x \right)\mathop{|}_{0}^{2}=\int\limits_{0}^{2}{\left( 3{{x}^{2}}-6x \right)\ln f\left( x \right)dx=-\int\limits_{0}^{2}{\left( 3{{x}^{2}}-6x \right)\ln f\left( x \right)dx=-J.}}$
Tính J : Đặt $x=2-t\Rightarrow J=\int\limits_{0}^{2}{\left( 3{{t}^{2}}-6t \right)\ln f\left( 2-t \right)dt=\int\limits_{0}^{2}{\left( 3{{x}^{2}}-6x \right)\ln f\left( 2-x \right)dx}}$.
Vậy $I+J=\int\limits_{0}^{2}{\left( 3{{x}^{2}}-6x \right)\ln f\left( x \right)dx+\int\limits_{0}^{2}{\left( 3{{x}^{2}}-6x \right)\ln f\left( 2-x \right)dx}}$
$=\int\limits_{0}^{2}{\left( 3{{x}^{2}}-6x \right)\ln \left[ f\left( x \right)f\left( 2-x \right) \right]dx=\int\limits_{0}^{2}{\left( 3{{x}^{2}}-6x \right)\ln \left( {{e}^{2{{x}^{2}}-4x}} \right)dx}}$
$=\int\limits_{0}^{2}{\left( 3{{x}^{2}}-6x \right)\left( 2{{x}^{2}}-4x \right)dx=\dfrac{32}{5}}$ mà $I=-J\Rightarrow -2J=\dfrac{32}{5}\Leftrightarrow I=-\dfrac{16}{5}$.
Đặt $\left\{ \begin{aligned}
& u={{x}^{3}}-3{{x}^{2}} \\
& dv=\dfrac{{f}'\left( x \right)}{f\left( x \right)}dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=\left( 3{{x}^{2}}-6x \right)dx \\
& v=\ln f\left( x \right) \\
\end{aligned} \right.$
Khi đó $I=\left( {{x}^{3}}-3{{x}^{2}} \right)\ln f\left( x \right)\mathop{|}_{0}^{2}=\int\limits_{0}^{2}{\left( 3{{x}^{2}}-6x \right)\ln f\left( x \right)dx=-\int\limits_{0}^{2}{\left( 3{{x}^{2}}-6x \right)\ln f\left( x \right)dx=-J.}}$
Tính J : Đặt $x=2-t\Rightarrow J=\int\limits_{0}^{2}{\left( 3{{t}^{2}}-6t \right)\ln f\left( 2-t \right)dt=\int\limits_{0}^{2}{\left( 3{{x}^{2}}-6x \right)\ln f\left( 2-x \right)dx}}$.
Vậy $I+J=\int\limits_{0}^{2}{\left( 3{{x}^{2}}-6x \right)\ln f\left( x \right)dx+\int\limits_{0}^{2}{\left( 3{{x}^{2}}-6x \right)\ln f\left( 2-x \right)dx}}$
$=\int\limits_{0}^{2}{\left( 3{{x}^{2}}-6x \right)\ln \left[ f\left( x \right)f\left( 2-x \right) \right]dx=\int\limits_{0}^{2}{\left( 3{{x}^{2}}-6x \right)\ln \left( {{e}^{2{{x}^{2}}-4x}} \right)dx}}$
$=\int\limits_{0}^{2}{\left( 3{{x}^{2}}-6x \right)\left( 2{{x}^{2}}-4x \right)dx=\dfrac{32}{5}}$ mà $I=-J\Rightarrow -2J=\dfrac{32}{5}\Leftrightarrow I=-\dfrac{16}{5}$.
Đáp án D.