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Câu hỏi: Cho hàm số $f\left( x \right)$ có đạo hàm dương, liên tục trên đoạn $\left[ 0;2 \right]$ thỏa mãn điều kiện $f\left( 0 \right)=3$ và $225\int\limits_{0}^{2}{{f}'\left( x \right){{f}^{2}}\left( x \right)dx}+8\le 60\int\limits_{0}^{2}{\sqrt{{f}'\left( x \right)}f\left( x \right)dx}$. Tích phân $\int\limits_{0}^{2}{{{f}^{3}}\left( x \right)dx}$ bằng
A. $\dfrac{274}{5}$.
B. $\dfrac{4068}{75}$.
C. $\dfrac{4058}{75}$.
D. $\dfrac{274}{75}$.
A. $\dfrac{274}{5}$.
B. $\dfrac{4068}{75}$.
C. $\dfrac{4058}{75}$.
D. $\dfrac{274}{75}$.
Ta có $225\int\limits_{0}^{2}{{f}'\left( x \right){{f}^{2}}\left( x \right)dx}+8\le 60\int\limits_{0}^{2}{\sqrt{{f}'\left( x \right)}f\left( x \right)dx}$
$\Leftrightarrow 225\int\limits_{0}^{2}{{f}'\left( x \right){{f}^{2}}\left( x \right)dx}-60\int\limits_{0}^{2}{\sqrt{{f}'\left( x \right)}f\left( x \right)dx}+\int\limits_{0}^{2}{4dx}\le 0$
$\Leftrightarrow \int\limits_{0}^{2}{{{\left[ 15\sqrt{{f}'\left( x \right)}.f\left( x \right)-2 \right]}^{2}}dx}\le 0\Rightarrow 15\sqrt{{f}'\left( x \right)}.f\left( x \right)-2=0\Leftrightarrow \sqrt{{f}'\left( x \right)}.f\left( x \right)=\dfrac{2}{15}$
$\Leftrightarrow {f}'\left( x \right).{{f}^{2}}\left( x \right)=\dfrac{4}{225}\Leftrightarrow \int{{f}'\left( x \right).{{f}^{2}}\left( x \right)dx}=\dfrac{4x}{225}+C\Leftrightarrow \dfrac{{{f}^{3}}\left( x \right)}{3}=\dfrac{4x}{225}+C$
Lại có $f\left( 0 \right)=3\Rightarrow C=9\Rightarrow {{f}^{3}}\left( x \right)=\dfrac{12x}{225}+27\Rightarrow \int\limits_{0}^{2}{{{f}^{3}}\left( x \right)dx}=\dfrac{4058}{75}$.
$\Leftrightarrow 225\int\limits_{0}^{2}{{f}'\left( x \right){{f}^{2}}\left( x \right)dx}-60\int\limits_{0}^{2}{\sqrt{{f}'\left( x \right)}f\left( x \right)dx}+\int\limits_{0}^{2}{4dx}\le 0$
$\Leftrightarrow \int\limits_{0}^{2}{{{\left[ 15\sqrt{{f}'\left( x \right)}.f\left( x \right)-2 \right]}^{2}}dx}\le 0\Rightarrow 15\sqrt{{f}'\left( x \right)}.f\left( x \right)-2=0\Leftrightarrow \sqrt{{f}'\left( x \right)}.f\left( x \right)=\dfrac{2}{15}$
$\Leftrightarrow {f}'\left( x \right).{{f}^{2}}\left( x \right)=\dfrac{4}{225}\Leftrightarrow \int{{f}'\left( x \right).{{f}^{2}}\left( x \right)dx}=\dfrac{4x}{225}+C\Leftrightarrow \dfrac{{{f}^{3}}\left( x \right)}{3}=\dfrac{4x}{225}+C$
Lại có $f\left( 0 \right)=3\Rightarrow C=9\Rightarrow {{f}^{3}}\left( x \right)=\dfrac{12x}{225}+27\Rightarrow \int\limits_{0}^{2}{{{f}^{3}}\left( x \right)dx}=\dfrac{4058}{75}$.
Đáp án C.