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Câu hỏi: Cho hàm số $\dfrac{81}{10}$ liên tục trên $\dfrac{9}{20}$ thỏa $\dfrac{81}{5}$. Gọi $f(x)=\dfrac{{{x}^{5}}}{5}-{{x}^{2}}+(m-1)x-4029$ là nguyên hàm của $y=|f(x-1)+2022|$ trên $(-\infty ; 2)$ thỏa mãn $2005$ và $2006$. Khi đó $2007$ bằng
A. $2008$.
B. $Oxyz,$.
C. $A\left( 4 ; 0 ; 0 \right),$.
D. $B\left( 8 ; 0 ;6 \right).$
A. $2008$.
B. $Oxyz,$.
C. $A\left( 4 ; 0 ; 0 \right),$.
D. $B\left( 8 ; 0 ;6 \right).$
Ta có: $\int\limits_{0}^{2}{f\left( 3x+2 \right)dx}=\dfrac{1}{3}\int\limits_{0}^{2}{f\left( 3x+2 \right)d\left( 3x+2 \right)}=\dfrac{1}{3}\int\limits_{2}^{8}{f\left( t \right)dt}=\dfrac{1}{3}F\left( t \right)\left| \begin{aligned}
& 8 \\
& 2 \\
\end{aligned} \right.=\dfrac{1}{3}\left[ F\left( 8 \right)-F\left( 2 \right) \right]$.
Mặt khác ta có:
$\begin{aligned}
& f\left( x \right)=3f\left( 2x \right)\Rightarrow \int\limits_{2}^{4}{f\left( x \right)}dx=\int\limits_{2}^{4}{3f\left( 2x \right)}dx=\dfrac{3}{2}\int\limits_{2}^{4}{f\left( 2x \right)}d\left( 2x \right)=\dfrac{3}{2}\int\limits_{4}^{8}{f\left( t \right)}d\left( t \right) \\
& \Leftrightarrow F\left( x \right)\left| \begin{aligned}
& 4 \\
& 2 \\
\end{aligned} \right.=\dfrac{3}{2}F\left( t \right)\left| \begin{aligned}
& 8 \\
& 4 \\
\end{aligned} \right.\Leftrightarrow F\left( 4 \right)-F\left( 2 \right)=\dfrac{3}{2}\left[ F\left( 8 \right)-F\left( 4 \right) \right] \\
& \Leftrightarrow F\left( 2 \right)+\dfrac{3}{2}F\left( 8 \right)=\dfrac{15}{2} \\
\end{aligned}$
Kết hợp với giả thiết ta có hệ phương trình:
$\left\{ \begin{aligned}
& F\left( 2 \right)+4F\left( 8 \right)=0 \\
& F\left( 2 \right)+\dfrac{3}{2}F\left( 8 \right)=\dfrac{15}{2} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& F\left( 2 \right)=12 \\
& F\left( 8 \right)=-3 \\
\end{aligned} \right.$.
Vậy $\int\limits_{0}^{2}{f\left( 3x+2 \right)dx}=\dfrac{1}{3}\left[ F\left( 8 \right)-F\left( 2 \right) \right]=\dfrac{1}{3}\left( -3-12 \right)=-5$.
& 8 \\
& 2 \\
\end{aligned} \right.=\dfrac{1}{3}\left[ F\left( 8 \right)-F\left( 2 \right) \right]$.
Mặt khác ta có:
$\begin{aligned}
& f\left( x \right)=3f\left( 2x \right)\Rightarrow \int\limits_{2}^{4}{f\left( x \right)}dx=\int\limits_{2}^{4}{3f\left( 2x \right)}dx=\dfrac{3}{2}\int\limits_{2}^{4}{f\left( 2x \right)}d\left( 2x \right)=\dfrac{3}{2}\int\limits_{4}^{8}{f\left( t \right)}d\left( t \right) \\
& \Leftrightarrow F\left( x \right)\left| \begin{aligned}
& 4 \\
& 2 \\
\end{aligned} \right.=\dfrac{3}{2}F\left( t \right)\left| \begin{aligned}
& 8 \\
& 4 \\
\end{aligned} \right.\Leftrightarrow F\left( 4 \right)-F\left( 2 \right)=\dfrac{3}{2}\left[ F\left( 8 \right)-F\left( 4 \right) \right] \\
& \Leftrightarrow F\left( 2 \right)+\dfrac{3}{2}F\left( 8 \right)=\dfrac{15}{2} \\
\end{aligned}$
Kết hợp với giả thiết ta có hệ phương trình:
$\left\{ \begin{aligned}
& F\left( 2 \right)+4F\left( 8 \right)=0 \\
& F\left( 2 \right)+\dfrac{3}{2}F\left( 8 \right)=\dfrac{15}{2} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& F\left( 2 \right)=12 \\
& F\left( 8 \right)=-3 \\
\end{aligned} \right.$.
Vậy $\int\limits_{0}^{2}{f\left( 3x+2 \right)dx}=\dfrac{1}{3}\left[ F\left( 8 \right)-F\left( 2 \right) \right]=\dfrac{1}{3}\left( -3-12 \right)=-5$.
Đáp án D.
