Tìm a, b để hàm số \(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}{\frac{{{x^2} + 1}}{{x + 1}}\,\,khi\,x \ge 0}\\{ax + b\,\,khi\,x < 0}\end{array}} \right.\) có đạo hàm tại điểm x = 0.

</>

Trước tiên hàm số phải liên tục tại x = 0.

Ta có:

\[\begin{array}{*{20}{l}}{\mathop {\lim }\limits_{{\rm{x}} \to {0^ + }} {\rm{f}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to {0^ + }} \frac{{{{\rm{x}}^2} + 1}}{{{\rm{x}} + 1}} = 1 = {\rm{f}}\left( 0 \right)}\\{\mathop {\lim }\limits_{{\rm{x}} \to {0^ - }} {\rm{f}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to {0^ - }} \left( {{\rm{ax + b}}} \right) = b}\end{array}\]

Để hàm số liên tục tại x = 0 thì \[\mathop {\lim }\limits_{{\rm{x}} \to {0^ + }} {\rm{f}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to {0^ - }} {\rm{f}}\left( {\rm{x}} \right){\rm{ = f}}\left( {\rm{0}} \right) \Leftrightarrow {\rm{b}} = 1\]

Khi đó ta có \[{\rm{f'}}\left( 0 \right) = \mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{{\rm{f}}\left( {\rm{x}} \right) - {\rm{f}}\left( {\rm{0}} \right)}}{{\rm{x}}}\]

Ta có

\[\mathop {\lim }\limits_{{\rm{x}} \to {0^ + }} \frac{{{\rm{f}}\left( {\rm{x}} \right) - {\rm{f}}\left( 0 \right)}}{{\rm{x}}} = \mathop {\lim }\limits_{{\rm{x}} \to {0^ + }} \frac{{\frac{{{{\rm{x}}^2} + 1}}{{{\rm{x}} + 1}} - 1}}{{\rm{x}}} = \mathop {\lim }\limits_{{\rm{x}} \to {0^ + }} \frac{{{{\rm{x}}^2} - {\rm{x}}}}{{{\rm{x}}\left( {{\rm{x}} + 1} \right)}} = \mathop {\lim }\limits_{{\rm{x}} \to {0^ + }} \frac{{{\rm{x}} - 1}}{{{\rm{x}} + 1}} = - 1\]

\[\mathop {\lim }\limits_{{\rm{x}} \to {0^ - }} \frac{{{\rm{f}}\left( {\rm{x}} \right) - {\rm{f}}\left( 0 \right)}}{{\rm{x}}} = \mathop {\lim }\limits_{{\rm{x}} \to {0^ + }} \frac{{\left( {{\rm{ax}} + 1} \right) - 1}}{{\rm{x}}} = \mathop {\lim }\limits_{{\rm{x}} \to {0^ + }} {\rm{a = a}}\]

Để hàm số có đạo hàm tại x = 0 thì  \[\mathop {\lim }\limits_{{\rm{x}} \to {0^ + }} \frac{{{\rm{f}}\left( {\rm{x}} \right) - {\rm{f}}\left( {\rm{0}} \right)}}{{\rm{x}}} = \mathop {\lim }\limits_{{\rm{x}} \to {0^ - }} \frac{{{\rm{f}}\left( {\rm{x}} \right) - {\rm{f}}\left( {\rm{0}} \right)}}{{\rm{x}}} \Leftrightarrow {\rm{a}} = - 1\]

Vậy a = −1, b = 1.

Đáp án cần chọn là: D