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Câu hỏi: Cho hàm số $y=f\left( x \right)$ có đạo hàm liên tục trên $\mathbb{R},f\left( 0 \right)=0,{f}'\left( 0 \right)\ne 0$ và thỏa mãn hệ thức $2f\left( x \right).{f}'\left( x \right)+18{{x}^{2}}-\left( 2{{x}^{2}}+3x \right).{f}'\left( x \right)=\left( 4x+3 \right)f\left( x \right),\forall x\in \mathbb{R}$.
Biết $\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( x \right)\cos f\left( x \right)dx=-\dfrac{a\pi +b}{6}}$ biết $a,b\in \mathbb{Z}$. Tính giá trị $S=2022a-2023b$ ?
A. $S=2021$.
B. $S=2023$.
C. $S=2022$.
D. $S=2020$.
Biết $\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( x \right)\cos f\left( x \right)dx=-\dfrac{a\pi +b}{6}}$ biết $a,b\in \mathbb{Z}$. Tính giá trị $S=2022a-2023b$ ?
A. $S=2021$.
B. $S=2023$.
C. $S=2022$.
D. $S=2020$.
Ta có: $2f\left( x \right).{f}'\left( x \right)+18{{x}^{2}}-\left( 2{{x}^{2}}+3x \right).{f}'\left( x \right)=\left( 4x+3 \right)f\left( x \right)$
$\begin{aligned}
& \Leftrightarrow 2f\left( x \right).{f}'\left( x \right)+18{{x}^{2}}=\left( 4x+3 \right)f\left( x \right)+\left( 2{{x}^{2}}+3x \right).{f}'\left( x \right) \\
& \Leftrightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{\left( 2f\left( x \right).{f}'\left( x \right)+18{{x}^{2}} \right)dx=}\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \left( 4x+3 \right)f\left( x \right)+\left( 2{{x}^{2}}+3x \right).{f}'\left( x \right) \right)dx} \\
& \Leftrightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{2f\left( x \right).{f}'\left( x \right)dx+\int\limits_{0}^{x}{18{{x}^{2}}dx=}}\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \left( 4x+3 \right)f\left( x \right)+\left( 2{{x}^{2}}+3x \right).{f}'\left( x \right) \right)dx} \\
\end{aligned}$
$\begin{aligned}
& \Leftrightarrow \left. \left( {{\left[ f\left( x \right) \right]}^{2}} \right) \right|_{0}^{\dfrac{\pi }{2}}+\left. \left( 6{{x}^{3}} \right) \right|_{0}^{\dfrac{\pi }{2}}=\left. \left( \left( 2{{x}^{2}}+3x \right).f\left( x \right) \right) \right|_{0}^{\dfrac{\pi }{2}} \\
& \Leftrightarrow {{f}^{2}}\left( x \right)+6{{x}^{3}}=\left( 2{{x}^{2}}+3x \right).f\left( x \right) \\
& \Leftrightarrow \left( f\left( x \right)-3x \right)\left( f\left( x \right)-2{{x}^{2}} \right)=0 \\
& \Leftrightarrow \left[ \begin{aligned}
& f\left( x \right)-3x=0 \\
& f\left( x \right)-2{{x}^{2}}=0 \\
\end{aligned} \right. \\
\end{aligned}$
Mặt khác $f\left( 0 \right)=0,{f}'\left( 0 \right)\ne 0\Rightarrow f\left( x \right)=3x$
Khi đó
$\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( x \right)\cos f\left( x \right)dx=}\int\limits_{0}^{\dfrac{\pi }{2}}{3x\cos 3xdx}=x\sin 3x+\left. \left( \dfrac{1}{3}\cos 3x \right) \right|_{0}^{\dfrac{\pi }{2}}=-\dfrac{\pi }{2}-\dfrac{1}{3}=-\dfrac{3\pi +2}{6}$
Suy ra $a=3;b=2$
$S=2022.3-2023.2=2020$
$\begin{aligned}
& \Leftrightarrow 2f\left( x \right).{f}'\left( x \right)+18{{x}^{2}}=\left( 4x+3 \right)f\left( x \right)+\left( 2{{x}^{2}}+3x \right).{f}'\left( x \right) \\
& \Leftrightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{\left( 2f\left( x \right).{f}'\left( x \right)+18{{x}^{2}} \right)dx=}\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \left( 4x+3 \right)f\left( x \right)+\left( 2{{x}^{2}}+3x \right).{f}'\left( x \right) \right)dx} \\
& \Leftrightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{2f\left( x \right).{f}'\left( x \right)dx+\int\limits_{0}^{x}{18{{x}^{2}}dx=}}\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \left( 4x+3 \right)f\left( x \right)+\left( 2{{x}^{2}}+3x \right).{f}'\left( x \right) \right)dx} \\
\end{aligned}$
$\begin{aligned}
& \Leftrightarrow \left. \left( {{\left[ f\left( x \right) \right]}^{2}} \right) \right|_{0}^{\dfrac{\pi }{2}}+\left. \left( 6{{x}^{3}} \right) \right|_{0}^{\dfrac{\pi }{2}}=\left. \left( \left( 2{{x}^{2}}+3x \right).f\left( x \right) \right) \right|_{0}^{\dfrac{\pi }{2}} \\
& \Leftrightarrow {{f}^{2}}\left( x \right)+6{{x}^{3}}=\left( 2{{x}^{2}}+3x \right).f\left( x \right) \\
& \Leftrightarrow \left( f\left( x \right)-3x \right)\left( f\left( x \right)-2{{x}^{2}} \right)=0 \\
& \Leftrightarrow \left[ \begin{aligned}
& f\left( x \right)-3x=0 \\
& f\left( x \right)-2{{x}^{2}}=0 \\
\end{aligned} \right. \\
\end{aligned}$
Mặt khác $f\left( 0 \right)=0,{f}'\left( 0 \right)\ne 0\Rightarrow f\left( x \right)=3x$
Khi đó
$\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( x \right)\cos f\left( x \right)dx=}\int\limits_{0}^{\dfrac{\pi }{2}}{3x\cos 3xdx}=x\sin 3x+\left. \left( \dfrac{1}{3}\cos 3x \right) \right|_{0}^{\dfrac{\pi }{2}}=-\dfrac{\pi }{2}-\dfrac{1}{3}=-\dfrac{3\pi +2}{6}$
Suy ra $a=3;b=2$
$S=2022.3-2023.2=2020$
Đáp án D.
