Câu hỏi 21

Ta có: \(f''\left( x \right) = 12{x^2} + 6x - 4\)

\(\begin{array}{l} \Rightarrow f'\left( x \right) = \int {\left( {12{x^2} + 6x - 4} \right)dx}  = 4{x^3} + 3{x^2} - 4x + C\\ \Rightarrow f\left( x \right) = \int {f'\left( x \right)dx}  = \int {\left( {4{x^3} + 3{x^2} - 4x + C} \right)dx} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {x^4} + {x^3} - 2{x^2} + Cx + C'.\end{array}\)

Lại có: \(\left\{ \begin{array}{l}f\left( 0 \right) = 1\\f\left( 1 \right) = 3\end{array} \right. \Rightarrow \left\{ \begin{array}{l}C' = 1\\1 + 1 - 2 + C + C' = 3\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}C' = 1\\C = 2\end{array} \right..\)

\(\begin{array}{l} \Rightarrow f\left( x \right) = {x^4} + {x^3} - 2{x^2} + 2x + 1.\\ \Rightarrow f\left( { - 1} \right) = {\left( { - 1} \right)^4} + {\left( { - 1} \right)^3} - 2{\left( { - 1} \right)^2} + 2\left( { - 1} \right) + 1 =  - 3.\end{array}\)

Chọn  B.