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Câu hỏi: Cho hàm số $f\left( x \right)$ liên tục trên đoạn $\left[ 0;2 \right]$ thoả mãn $f\left( 0 \right)=1$ và $f\left( x \right).f\left( 2-x \right)={{e}^{2{{x}^{2}}-4x}},\forall x\in \left[ 0;2 \right]$. Tích phân $\int\limits_{0}^{2}{\dfrac{\left( {{x}^{3}}-3{{x}^{2}} \right){f}'\left( x \right)}{f\left( x \right)}\text{d}x}$ có giá trị bằng
A. $-\dfrac{14}{3}$.
B. $-\dfrac{32}{5}$.
C. $-\dfrac{16}{3}$.
D. $-\dfrac{16}{5}$.
A. $-\dfrac{14}{3}$.
B. $-\dfrac{32}{5}$.
C. $-\dfrac{16}{3}$.
D. $-\dfrac{16}{5}$.
Thay $x=0$ vào đẳng thức, ta có $f\left( 0 \right).f\left( 2 \right)=1\Rightarrow f\left( 2 \right)=1.$
Sử dụng tích chất: $\int\limits_{a}^{b}{f\left( x \right)dx=\int\limits_{a}^{b}{f\left( a+b-x \right)dx}}$, ta có: $I=\int\limits_{0}^{2}{\dfrac{\left( {{x}^{3}}-3{{x}^{2}} \right){f}'\left( x \right)}{f\left( x \right)}}\text{d}x$
và $I=\int\limits_{0}^{2}{\dfrac{\left( {{\left( 2-x \right)}^{3}}-3{{\left( 2-x \right)}^{2}} \right){f}'\left( 2-x \right)}{f\left( 2-x \right)}}\text{d}x=\int\limits_{0}^{2}{\dfrac{\left( -{{x}^{3}}+3{{x}^{2}}-4 \right){f}'\left( 2-x \right)}{f\left( 2-x \right)}}\text{d}x$.
Cộng vế theo vế, ta được: $2I=\int\limits_{0}^{2}{\left( {{x}^{3}}-3{{x}^{2}} \right)\left( \dfrac{{f}'\left( x \right)}{f\left( x \right)}-\dfrac{{f}'\left( 2-x \right)}{f\left( 2-x \right)} \right)}dx-4\int\limits_{0}^{2}{\dfrac{{f}'\left( 2-x \right)}{f\left( 2-x \right)}dx}.$
Trong đó $\int\limits_{0}^{2}{\dfrac{{f}'\left( 2-x \right)}{f\left( 2-x \right)}dx=\ln }\left| f\left( 2-x \right) \right|\left| _{0}^{2} \right.=\ln \dfrac{f\left( 0 \right)}{f\left( 2 \right)}=\ln 1=0$.
Do đó, $I=\dfrac{1}{2}\int\limits_{0}^{2}{\left( {{x}^{3}}-3{{x}^{2}} \right)\left( \dfrac{{f}'\left( x \right)}{f\left( x \right)}-\dfrac{{f}'\left( 2-x \right)}{f\left( 2-x \right)} \right)}dx$.
Đạo hàm hai vế của đẳng thức, ta có: ${f}'\left( x \right)f\left( 2-x \right)-{f}'\left( 2-x \right)f\left( x \right)=\left( 4x-4 \right){{e}^{2{{x}^{2}}-4}}$.
$\Leftrightarrow \dfrac{{f}'\left( x \right)f\left( 2-x \right)-{f}'\left( 2-x \right)f\left( x \right)}{f\left( x \right).f\left( 2-x \right)}=\dfrac{\left( 4x-4 \right){{e}^{2{{x}^{2}}-4}}}{f\left( x \right).f\left( 2-x \right)}=4x-4\Leftrightarrow \dfrac{{f}'\left( x \right)}{{f}'\left( x \right)}-\dfrac{{f}'\left( 2-x \right)}{f\left( 2-x \right)}=4x-4$
Do đó, $I=\dfrac{1}{2}\int\limits_{0}^{2}{\left( {{x}^{3}}-3{{x}^{2}} \right)\left( \dfrac{{f}'\left( x \right)}{f\left( x \right)}-\dfrac{{f}'\left( 2-x \right)}{f\left( 2-x \right)} \right)}dx=\dfrac{1}{2}\int\limits_{0}^{2}{\left( {{x}^{3}}-3{{x}^{2}} \right)\left( 4x-4 \right)dx=-\dfrac{16}{5}}$.
Sử dụng tích chất: $\int\limits_{a}^{b}{f\left( x \right)dx=\int\limits_{a}^{b}{f\left( a+b-x \right)dx}}$, ta có: $I=\int\limits_{0}^{2}{\dfrac{\left( {{x}^{3}}-3{{x}^{2}} \right){f}'\left( x \right)}{f\left( x \right)}}\text{d}x$
và $I=\int\limits_{0}^{2}{\dfrac{\left( {{\left( 2-x \right)}^{3}}-3{{\left( 2-x \right)}^{2}} \right){f}'\left( 2-x \right)}{f\left( 2-x \right)}}\text{d}x=\int\limits_{0}^{2}{\dfrac{\left( -{{x}^{3}}+3{{x}^{2}}-4 \right){f}'\left( 2-x \right)}{f\left( 2-x \right)}}\text{d}x$.
Cộng vế theo vế, ta được: $2I=\int\limits_{0}^{2}{\left( {{x}^{3}}-3{{x}^{2}} \right)\left( \dfrac{{f}'\left( x \right)}{f\left( x \right)}-\dfrac{{f}'\left( 2-x \right)}{f\left( 2-x \right)} \right)}dx-4\int\limits_{0}^{2}{\dfrac{{f}'\left( 2-x \right)}{f\left( 2-x \right)}dx}.$
Trong đó $\int\limits_{0}^{2}{\dfrac{{f}'\left( 2-x \right)}{f\left( 2-x \right)}dx=\ln }\left| f\left( 2-x \right) \right|\left| _{0}^{2} \right.=\ln \dfrac{f\left( 0 \right)}{f\left( 2 \right)}=\ln 1=0$.
Do đó, $I=\dfrac{1}{2}\int\limits_{0}^{2}{\left( {{x}^{3}}-3{{x}^{2}} \right)\left( \dfrac{{f}'\left( x \right)}{f\left( x \right)}-\dfrac{{f}'\left( 2-x \right)}{f\left( 2-x \right)} \right)}dx$.
Đạo hàm hai vế của đẳng thức, ta có: ${f}'\left( x \right)f\left( 2-x \right)-{f}'\left( 2-x \right)f\left( x \right)=\left( 4x-4 \right){{e}^{2{{x}^{2}}-4}}$.
$\Leftrightarrow \dfrac{{f}'\left( x \right)f\left( 2-x \right)-{f}'\left( 2-x \right)f\left( x \right)}{f\left( x \right).f\left( 2-x \right)}=\dfrac{\left( 4x-4 \right){{e}^{2{{x}^{2}}-4}}}{f\left( x \right).f\left( 2-x \right)}=4x-4\Leftrightarrow \dfrac{{f}'\left( x \right)}{{f}'\left( x \right)}-\dfrac{{f}'\left( 2-x \right)}{f\left( 2-x \right)}=4x-4$
Do đó, $I=\dfrac{1}{2}\int\limits_{0}^{2}{\left( {{x}^{3}}-3{{x}^{2}} \right)\left( \dfrac{{f}'\left( x \right)}{f\left( x \right)}-\dfrac{{f}'\left( 2-x \right)}{f\left( 2-x \right)} \right)}dx=\dfrac{1}{2}\int\limits_{0}^{2}{\left( {{x}^{3}}-3{{x}^{2}} \right)\left( 4x-4 \right)dx=-\dfrac{16}{5}}$.
Đáp án D.