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Câu hỏi: Cho $F\left( x \right)$ là một nguyên hàm của hàm số $f\left( x \right)=\left| 1+x \right|-\left| 1-x \right|$ trên $\mathbb{R}$ và thỏa mãn $F\left( 1 \right)=3$. Tính tổng $F\left( 0 \right)+F\left( 2 \right)$
A. $3$.
B. $2$.
C. $7$.
D. $5$.
A. $3$.
B. $2$.
C. $7$.
D. $5$.
Ta có: $f\left( x \right)=\left\{ \begin{aligned}
& -\left( 1+x \right)-\left( 1-x \right) , \text{khi }x<-1 \\
& \left( 1+x \right)-\left( 1-x \right) , \text{khi}-1\le x\le 1 \\
& \left( 1+x \right)+\left( 1-x \right) , \text{khi }x>1 \\
\end{aligned} \right.\Leftrightarrow f\left( x \right)=\left\{ \begin{aligned}
& -2 , \text{khi }x<-1 \\
& 2x , \text{khi}-1\le x\le 1 \\
& 2 , \text{khi }x>1 \\
\end{aligned} \right.$.
Ta có: $F\left( 0 \right)-F\left( 1 \right)+F\left( 2 \right)-F\left( 1 \right)=\int\limits_{1}^{0}{f\left( x \right)dx}+\int\limits_{1}^{2}{f\left( x \right)dx=}\int\limits_{1}^{0}{2xdx}+\int\limits_{1}^{2}{2dx=} 1$.
$\Rightarrow F\left( 0 \right)+F\left( 2 \right)=1+2F\left( 1 \right)=1+2\cdot 3=7$.
& -\left( 1+x \right)-\left( 1-x \right) , \text{khi }x<-1 \\
& \left( 1+x \right)-\left( 1-x \right) , \text{khi}-1\le x\le 1 \\
& \left( 1+x \right)+\left( 1-x \right) , \text{khi }x>1 \\
\end{aligned} \right.\Leftrightarrow f\left( x \right)=\left\{ \begin{aligned}
& -2 , \text{khi }x<-1 \\
& 2x , \text{khi}-1\le x\le 1 \\
& 2 , \text{khi }x>1 \\
\end{aligned} \right.$.
Ta có: $F\left( 0 \right)-F\left( 1 \right)+F\left( 2 \right)-F\left( 1 \right)=\int\limits_{1}^{0}{f\left( x \right)dx}+\int\limits_{1}^{2}{f\left( x \right)dx=}\int\limits_{1}^{0}{2xdx}+\int\limits_{1}^{2}{2dx=} 1$.
$\Rightarrow F\left( 0 \right)+F\left( 2 \right)=1+2F\left( 1 \right)=1+2\cdot 3=7$.
Đáp án C.