Tìm \(a,b\) để hàm số $f(x) = \left\{ \begin{array}{l}a{x^2} + bx + 1\,\,\,\,\,\,\,\,\,\,\,\,\,khi\,\,\,x \ge 0\\a{\mathop{\rm s}\nolimits} {\rm{in }}x + b\cos x\,\,\,\,\,khi\,\,\,x < 0\end{array} \right.$ có đạo hàm tại điểm \({x_0} = 0\)

Ta có: \(f(0) = 1\)

\(\begin{array}{l}\mathop {\lim }\limits_{x \to {0^ + }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} (a{x^2} + bx + 1) = 1\\\mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{x \to {0^ - }} (a{\mathop{\rm s}\nolimits} {\rm{in }}x + b\cos x) = b\end{array}\)

Để hàm số liên tục thì \(b = 1\)

\(\begin{array}{l}\mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{f(x) - f(0)}}{x} = \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{a{x^2} + x + 1 - 1}}{x} = 1\\\mathop {\lim }\limits_{x \to {0^ - }} \dfrac{{f(x) - f(0)}}{x} = \mathop {\lim }\limits_{x \to {0^ - }} \dfrac{{a\sin x + b\cos x - 1}}{x} = \mathop {\lim }\limits_{x \to {0^ - }} \dfrac{{2a\sin \dfrac{x}{2}\cos \dfrac{x}{2} - 2{{\sin }^2}\dfrac{x}{2}}}{x}\\ = \mathop {\lim }\limits_{x \to {0^ - }} \dfrac{{\sin \dfrac{x}{2}}}{{\dfrac{x}{2}}}.\mathop {\lim }\limits_{x \to {0^ - }} \left( {a\cos \dfrac{x}{2}} \right) - \mathop {\lim }\limits_{x \to {0^ - }} \dfrac{{\sin \dfrac{x}{2}}}{{\dfrac{x}{2}}}.\mathop {\lim }\limits_{x \to {0^ - }} \sin \dfrac{x}{2} = a\end{array}\)

Để tồn tại \(f'\left( 0 \right)\) thì \(\mathop {\lim }\limits_{x \to {0^ - }} \dfrac{{f(x) - f(0)}}{x} = \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{f(x) - f(0)}}{x} \Leftrightarrow a = 1\)