Lời giải của Tự Học 365

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 $\left\{ \begin{array}{l} F\left( 1 \right) = 3\\ F\left( { - 1} \right) = 2\\ F\left( { - 2} \right) = 4 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} 2 + {C_1} = 3\\ 1 + {C_2} = 2\\ 4 + {C_3} = 4 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} {C_1} = 1\\ {C_2} = 1\\ {C_3} = 0 \end{array} \right.$

$\begin{array}{l} \Rightarrow F\left( x \right) = \left\{ \begin{array}{l} 2x + 1\,\,\,\,\,\,\,\,khi\,\,\,\,x \ge 1\\ {x^2} + 1\,\,\,\,\,\,\,\,khi\,\,\,\, - \,1 \le x < 1\\ - \,2x\,\,\,\,khi\,\,\,\,\,x < - \,1 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} F\left( 2 \right) = 2.2 + 1 = 5\\ F\left( 0 \right) = 1\\ F\left( { - 3} \right) = - 2.\left( { - 3} \right) = 6. \end{array} \right.\\ \Rightarrow T = 5 + 1 + 6 = 12. \end{array}$

Đáp án cần chọn là: b