Cho hàm số \(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}{\frac{{{x^2}}}{2}\,\,khi\,\,x \le 1}\\{ax + b\,\,khi\,\,x > 1}\end{array}} \right.\). Tìm tất cả các giá trị của các tham số a, b sao cho f(x) có đạo hàm tại điểm x = 1.

Hàm số có đạo hàm tại x = 1, do đó hàm số liên tục tại x = 1.

\[ \Rightarrow \mathop {\lim }\limits_{{\rm{x}} \to {1^ + }} {\rm{f}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to {1^ - }} {\rm{f}}\left( {\rm{x}} \right) = {\rm{f}}\left( 1 \right) \Leftrightarrow \mathop {\lim }\limits_{{\rm{x}} \to {1^ + }} \left( {{\rm{ax + b}}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to {1^ - }} \frac{{{{\rm{x}}^{\rm{2}}}}}{{\rm{2}}} = \frac{1}{2} \Leftrightarrow {\rm{a + b}} = \frac{1}{2}\] (1)

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

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

Hàm số có đạo hàm tại \[{\rm{x}} = 1 \Leftrightarrow \mathop {\lim }\limits_{{\rm{x}} \to {1^ + }} \frac{{{\rm{f}}\left( {\rm{x}} \right) - {\rm{f}}\left( {\rm{1}} \right)}}{{{\rm{x}} - {\rm{1}}}} = \mathop {\lim }\limits_{{\rm{x}} \to {1^ - }} \frac{{{\rm{f}}\left( {\rm{x}} \right) - {\rm{f}}\left( {\rm{1}} \right)}}{{{\rm{x}} - {\rm{1}}}} \Leftrightarrow {\rm{a}} = 1\] (2)

Từ (1) và (2), ta có \[{\rm{a}} = 1,\;{\rm{b}} = - \frac{1}{2}\]

Đáp án cần chọn là: A