Đáp án đúng: B
Giải chi tiết:
Ta có \(f\left( x \right) = \left| {1 + x} \right| - \left| {1 - x} \right| = \left\{ \begin{array}{l}2\,\,\,\,\,\,\,\,\,\,khi\,\,\,\,x \ge 1\\2x\,\,\,\,\,\,\,\,khi\,\,\,\, - \,1 \le x < 1\\ - \,2\,\,\,\,\,\,\,khi\,\,\,\,\,x < - \,1\end{array} \right. \Rightarrow F\left( x \right) = \left\{ \begin{array}{l}2x + {C_1}\,\,\,\,\,\,\,\,khi\,\,\,\,x \ge 1\\{x^2} + {C_2}\,\,\,\,\,\,\,\,khi\,\,\,\, - \,1 \le x < 1\\ - \,2x + {C_3}\,\,\,\,khi\,\,\,\,\,x < - \,1\end{array} \right.\)
Theo đề bài ta có \(\left\{ \begin{array}{l}F\left( 1 \right) = 3\\F\left( { - 1} \right) = 2\\F\left( { - 2} \right) = 4\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}2 + {C_1} = 3\\1 + {C_2} = 2\\4 + {C_3} = 4\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{C_1} = 1\\{C_2} = 1\\{C_3} = 0\end{array} \right.\)
\(\begin{array}{l} \Rightarrow F\left( x \right) = \left\{ \begin{array}{l}2x + 1\,\,\,\,\,\,\,\,khi\,\,\,\,x \ge 1\\{x^2} + 1\,\,\,\,\,\,\,\,khi\,\,\,\, - \,1 \le x < 1\\ - \,2x\,\,\,\,khi\,\,\,\,\,x < - \,1\end{array} \right. \Rightarrow \left\{ \begin{array}{l}F\left( 2 \right) = 2.2 + 1 = 5\\F\left( 0 \right) = 1\\F\left( { - 3} \right) = - 2.\left( { - 3} \right) = 6.\end{array} \right.\\ \Rightarrow T = 5 + 1 + 6 = 12.\end{array}\)
Chọn B.
