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Câu hỏi: Cho hàm số $f\left( x \right)$ liên tục trên $\mathbb{R}$ thỏa $f\left( x \right)=4f\left( -2x+3 \right)$. Gọi $F\left( x \right)$ là nguyên hàm của $f\left( x \right)$ trên $\mathbb{R}$ và thỏa mãn $F\left( 2 \right)-F\left( 4 \right)=24$. Khi đó $\int\limits_{-1}^{5}{f}\left( x \right)\text{d}x$ bằng
A. $10$.
B. $12$.
C. $-10$.
D. $-12$.
A. $10$.
B. $12$.
C. $-10$.
D. $-12$.
Ta có: $f\left( x \right)=4f\left( -2x+3 \right)\Rightarrow \int{f\left( x \right)\text{d}x=}4\int{f\left( -2x+3 \right)\text{d}x}$ $\Rightarrow F\left( x \right)=-2F\left( -2x+3 \right)+C$
Từ đó có: $\left\{ \begin{aligned}
& F\left( 2 \right)=-2F\left( 1 \right)+C \\
& F\left( 4 \right)=-2F\left( 5 \right)+C \\
\end{aligned} \right.$$\Rightarrow F\left( 2 \right)-F\left( 4 \right)=2\left( F\left( 5 \right)-F\left( 1 \right) \right) $ $ \Rightarrow F\left( 5 \right)-F\left( 1 \right)=12$
Vậy $\int\limits_{1}^{5}{f\left( x \right)\text{d}x}=\left. F\left( x \right) \right|_{1}^{5}=F\left( 5 \right)-F\left( 1 \right)=12$.
Từ đó có: $\left\{ \begin{aligned}
& F\left( 2 \right)=-2F\left( 1 \right)+C \\
& F\left( 4 \right)=-2F\left( 5 \right)+C \\
\end{aligned} \right.$$\Rightarrow F\left( 2 \right)-F\left( 4 \right)=2\left( F\left( 5 \right)-F\left( 1 \right) \right) $ $ \Rightarrow F\left( 5 \right)-F\left( 1 \right)=12$
Vậy $\int\limits_{1}^{5}{f\left( x \right)\text{d}x}=\left. F\left( x \right) \right|_{1}^{5}=F\left( 5 \right)-F\left( 1 \right)=12$.
Đáp án B.