Tìm a, b để hàm số \(f\left( x \right)\left\{ {\begin{array}{*{20}{c}}{a{x^2} + bx + 1\,\,khi\,\,x \ge 0}\\{a\sin x + b\cos x\,\,khi\,x < 0}\end{array}} \right.\) có đạo hàm tại điểm x₀= 0

Để hàm số có đạo hàm tại x = 1 thì trước hết hàm số phải liên tục tại x = 1.

Ta có: f(0) = 1

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

\[\mathop {\lim }\limits_{{\rm{x}} \to {0^ - }} {\rm{f}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to {0^ - }} \left( {asinx + bcosx} \right) = b\]

Để hàm số liên tục tại x = 1 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: \[{\rm{f'}}\left( 0 \right) = \mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{{\rm{f}}\left( {\rm{x}} \right) - {\rm{f}}\left( 0 \right)}}{{{\rm{x}} - 0}}\]

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

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

\[ = \mathop {\lim }\limits_{{\rm{x}} \to {0^ - }} \frac{{{\rm{2asin}}\frac{{\rm{x}}}{{\rm{2}}}{\rm{cos}}\frac{{\rm{x}}}{{\rm{2}}} - {\rm{2si}}{{\rm{n}}^{\rm{2}}}\frac{{\rm{x}}}{{\rm{2}}}}}{{\rm{x}}}\]

\[ = \mathop {\lim }\limits_{{\rm{x}} \to {0^ - }} \frac{{\sin \frac{{\rm{x}}}{2}}}{{\frac{{\rm{x}}}{2}}}\mathop {\lim }\limits_{{\rm{x}} \to {0^ - }} \left( {{\rm{acos}}\frac{{\rm{x}}}{{\rm{2}}} - {\rm{2sin}}\frac{{\rm{x}}}{{\rm{2}}}} \right) = {\rm{a}}\]

Để tồn tại \[{\rm{f'}}\left( 0 \right) \Leftrightarrow \mathop {\lim }\limits_{{\rm{x}} \to {0^ + }} \frac{{{\rm{f}}\left( {\rm{x}} \right) - {\rm{f}}\left( {\rm{0}} \right)}}{{{\rm{x}} - {\rm{0}}}} = \mathop {\lim }\limits_{{\rm{x}} \to {0^ - }} \frac{{{\rm{f}}\left( {\rm{x}} \right) - {\rm{f}}\left( {\rm{0}} \right)}}{{{\rm{x}} - {\rm{0}}}} \Leftrightarrow {\rm{a}} = 1.\]

Đáp án cần chọn là: A