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Câu hỏi: Cho hàm số $f\left( x \right)$ có đạo hàm liên tục trên $\mathbb{R}$ thỏa:
$f\left( x \right)={{x}^{2}}-3x+2\int_{0}^{1}{f\left( x \right){f}'\left( x \right)\mathrm{d}x}, \forall x\in \mathbb{R}$. Tìm giá trị thực dương của $a$ để $\int_{0}^{a}{f\left( x \right)\mathrm{d}x}=\dfrac{4}{5}a$.
A. $\dfrac{9}{2}$.
B. $\dfrac{3}{2}$.
C. $\dfrac{1}{2}$.
D. $2$.
$f\left( x \right)={{x}^{2}}-3x+2\int_{0}^{1}{f\left( x \right){f}'\left( x \right)\mathrm{d}x}, \forall x\in \mathbb{R}$. Tìm giá trị thực dương của $a$ để $\int_{0}^{a}{f\left( x \right)\mathrm{d}x}=\dfrac{4}{5}a$.
A. $\dfrac{9}{2}$.
B. $\dfrac{3}{2}$.
C. $\dfrac{1}{2}$.
D. $2$.
Đặt $m=\int_{0}^{1}{f\left( x \right){f}'\left( x \right)\mathrm{d}x}$. Khi đó $f\left( x \right)={{x}^{2}}-3x+2m, \forall x\in \mathbb{R}$. Suy ra ${f}'\left( x \right)=2x-3$.
Vậy $m=\int_{0}^{1}{\left( {{x}^{2}}-3x+2m \right)\left( 2x-3 \right)\mathrm{d}x}$. Đặt $t={{x}^{2}}-3x+2m\Rightarrow \mathrm{d}t=\left( 2x-3 \right)\mathrm{d}x$.
Do đó $m=-\int_{2m-2}^{2m}{t\mathrm{d}t}\Leftrightarrow m=-\left. \dfrac{{{t}^{2}}}{2} \right|_{2m-2}^{2m}\Leftrightarrow m=-4m+2\Leftrightarrow m=\dfrac{2}{5}$.
Vậy $f\left( x \right)={{x}^{2}}-3x+\dfrac{4}{5}$.
Ta có $\int_{0}^{a}{f\left( x \right)\mathrm{d}x}=\dfrac{4}{5}a\Leftrightarrow \int_{0}^{a}{\left( {{x}^{2}}-3x+\dfrac{4}{5} \right)\mathrm{d}x}=\dfrac{4}{5}a\Leftrightarrow \dfrac{1}{3}{{a}^{3}}-\dfrac{3}{2}{{a}^{2}}+\dfrac{4}{5}a=\dfrac{4}{5}a\Leftrightarrow a=\dfrac{9}{2}$.
Vậy $m=\int_{0}^{1}{\left( {{x}^{2}}-3x+2m \right)\left( 2x-3 \right)\mathrm{d}x}$. Đặt $t={{x}^{2}}-3x+2m\Rightarrow \mathrm{d}t=\left( 2x-3 \right)\mathrm{d}x$.
Do đó $m=-\int_{2m-2}^{2m}{t\mathrm{d}t}\Leftrightarrow m=-\left. \dfrac{{{t}^{2}}}{2} \right|_{2m-2}^{2m}\Leftrightarrow m=-4m+2\Leftrightarrow m=\dfrac{2}{5}$.
Vậy $f\left( x \right)={{x}^{2}}-3x+\dfrac{4}{5}$.
Ta có $\int_{0}^{a}{f\left( x \right)\mathrm{d}x}=\dfrac{4}{5}a\Leftrightarrow \int_{0}^{a}{\left( {{x}^{2}}-3x+\dfrac{4}{5} \right)\mathrm{d}x}=\dfrac{4}{5}a\Leftrightarrow \dfrac{1}{3}{{a}^{3}}-\dfrac{3}{2}{{a}^{2}}+\dfrac{4}{5}a=\dfrac{4}{5}a\Leftrightarrow a=\dfrac{9}{2}$.
Đáp án A.