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Câu hỏi: Cho hàm số $f\left( x \right)$ liên tục trên $\mathbb{R}$, đồ thị hàm số $y=f\left( x \right)$ đi qua điểm $A\left( 1;0 \right)$ và nhận điểm $I\left( 2;2 \right)$ làm tâm đối xứng. Giá trị của $\int\limits_{1}^{3}{x\left( x-2 \right)\left[ f\left( x \right)+{f}'\left( x \right) \right]dx}$ bằng
A. $-\dfrac{8}{3}$.
B. $-\dfrac{16}{3}$.
C. $\dfrac{16}{3}$.
D. $\dfrac{8}{3}$.
A. $-\dfrac{8}{3}$.
B. $-\dfrac{16}{3}$.
C. $\dfrac{16}{3}$.
D. $\dfrac{8}{3}$.
Cách 1 : Đặt $t=4-x\Rightarrow dt=-dx$.
Đồ thị hàm số $y=f\left( x \right)$ có $I\left( 2;2 \right)$ là tâm đối xứng nên $\dfrac{f\left( x \right)+f\left( 4-x \right)}{2}=2$.
Như vậy $f\left( x \right)+f\left( 4-x \right)=4\Rightarrow {f}'\left( x \right)-{f}'\left( 4-x \right)=0\Leftrightarrow {f}'\left( x \right)={f}'\left( 4-x \right),\forall x\in \mathbb{R}$.
Ta có $I=\int\limits_{1}^{3}{x\left( x-2 \right)\left[ f\left( x \right)+{f}'\left( x \right) \right]dx}$ $=\int\limits_{1}^{3}{\left( 4-t \right)\left( 2-t \right)\left[ f\left( 4-t \right)+{f}'\left( 4-t \right) \right]dt}$
$=\int\limits_{1}^{3}{\left( {{t}^{2}}-6t+8 \right)\left[ 4-f\left( t \right)+{f}'\left( t \right) \right]dt}$
$=\int\limits_{1}^{3}{4\left( {{t}^{2}}-6t+8 \right)dt}-\int\limits_{1}^{3}{\left[ {{t}^{2}}-2t-4\left( t-2 \right) \right].\left[ f\left( t \right)+{f}'\left( t \right) \right]dt}+\int\limits_{1}^{3}{2\left( {{t}^{2}}-6t+8 \right){f}'\left( t \right)dt}$
$=2.\int\limits_{1}^{3}{\left( {{t}^{2}}-6t+8 \right)dt}+\int\limits_{1}^{3}{2\left( t-2 \right)\left[ f\left( t \right)+{f}'\left( t \right) \right]dt}+\int\limits_{1}^{3}{\left( {{t}^{2}}-6t+8 \right){f}'\left( t \right)dt}$
$=\int\limits_{1}^{3}{\left( {{t}^{2}}-6t+8 \right)dt+\int\limits_{1}^{3}{\left( 2t-4 \right)f\left( t \right)dt}+\int\limits_{1}^{3}{\left( {{t}^{2}}-4t+4 \right){f}'\left( t \right)dt}}$
$=\left. \left( \dfrac{{{t}^{3}}}{3}-3{{t}^{2}}+8t \right) \right|_{1}^{3}+\left. \left( {{t}^{2}}-4t+4 \right)f\left( t \right) \right|_{1}^{3}=\dfrac{16}{3}$.
Cách 2 : Chọn hàm $f\left( x \right)=a{{\left( x-2 \right)}^{3}}+2$. Ta có $A\left( 1;0 \right)\in $ đồ thị $\left( C \right):y=f\left( x \right)$.
Khi đó $a{{\left( -1 \right)}^{3}}+2=0\Leftrightarrow a=2\Rightarrow f\left( x \right)=2{{\left( x-2 \right)}^{3}}+2$ và có ${f}'\left( x \right)=6{{\left( x-2 \right)}^{2}}$.
Do đó $\int\limits_{1}^{3}{x\left( x-2 \right)\left[ f\left( x \right)+{f}'\left( x \right) \right]dx}=\int\limits_{1}^{3}{x\left( x-2 \right)\left[ 2{{\left( x-2 \right)}^{3}}+6{{\left( x-2 \right)}^{2}} \right]dx}=\dfrac{16}{3}$.
Đồ thị hàm số $y=f\left( x \right)$ có $I\left( 2;2 \right)$ là tâm đối xứng nên $\dfrac{f\left( x \right)+f\left( 4-x \right)}{2}=2$.
Như vậy $f\left( x \right)+f\left( 4-x \right)=4\Rightarrow {f}'\left( x \right)-{f}'\left( 4-x \right)=0\Leftrightarrow {f}'\left( x \right)={f}'\left( 4-x \right),\forall x\in \mathbb{R}$.
Ta có $I=\int\limits_{1}^{3}{x\left( x-2 \right)\left[ f\left( x \right)+{f}'\left( x \right) \right]dx}$ $=\int\limits_{1}^{3}{\left( 4-t \right)\left( 2-t \right)\left[ f\left( 4-t \right)+{f}'\left( 4-t \right) \right]dt}$
$=\int\limits_{1}^{3}{\left( {{t}^{2}}-6t+8 \right)\left[ 4-f\left( t \right)+{f}'\left( t \right) \right]dt}$
$=\int\limits_{1}^{3}{4\left( {{t}^{2}}-6t+8 \right)dt}-\int\limits_{1}^{3}{\left[ {{t}^{2}}-2t-4\left( t-2 \right) \right].\left[ f\left( t \right)+{f}'\left( t \right) \right]dt}+\int\limits_{1}^{3}{2\left( {{t}^{2}}-6t+8 \right){f}'\left( t \right)dt}$
$=2.\int\limits_{1}^{3}{\left( {{t}^{2}}-6t+8 \right)dt}+\int\limits_{1}^{3}{2\left( t-2 \right)\left[ f\left( t \right)+{f}'\left( t \right) \right]dt}+\int\limits_{1}^{3}{\left( {{t}^{2}}-6t+8 \right){f}'\left( t \right)dt}$
$=\int\limits_{1}^{3}{\left( {{t}^{2}}-6t+8 \right)dt+\int\limits_{1}^{3}{\left( 2t-4 \right)f\left( t \right)dt}+\int\limits_{1}^{3}{\left( {{t}^{2}}-4t+4 \right){f}'\left( t \right)dt}}$
$=\left. \left( \dfrac{{{t}^{3}}}{3}-3{{t}^{2}}+8t \right) \right|_{1}^{3}+\left. \left( {{t}^{2}}-4t+4 \right)f\left( t \right) \right|_{1}^{3}=\dfrac{16}{3}$.
Cách 2 : Chọn hàm $f\left( x \right)=a{{\left( x-2 \right)}^{3}}+2$. Ta có $A\left( 1;0 \right)\in $ đồ thị $\left( C \right):y=f\left( x \right)$.
Khi đó $a{{\left( -1 \right)}^{3}}+2=0\Leftrightarrow a=2\Rightarrow f\left( x \right)=2{{\left( x-2 \right)}^{3}}+2$ và có ${f}'\left( x \right)=6{{\left( x-2 \right)}^{2}}$.
Do đó $\int\limits_{1}^{3}{x\left( x-2 \right)\left[ f\left( x \right)+{f}'\left( x \right) \right]dx}=\int\limits_{1}^{3}{x\left( x-2 \right)\left[ 2{{\left( x-2 \right)}^{3}}+6{{\left( x-2 \right)}^{2}} \right]dx}=\dfrac{16}{3}$.
Đáp án B.