Câu hỏi 19

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

Để hàm số liên tục tại x = 1 thì \(\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)=f\left( 1 \right)\Leftrightarrow a+b=1\,\,\,\left( 1 \right)\)

Khi đó ta có: \(f'\left( 1 \right)=\underset{x\to 1}{\mathop{\lim }}\,\frac{f\left( x \right)-f\left( 1 \right)}{x-1}\)

\(\begin{array}{l}\mathop {\lim }\limits_{x \to {1^ + }} \frac{{f\left( x \right) - f\left( 1 \right)}}{{x - 1}} = \mathop {\lim }\limits_{x \to {1^ + }} \frac{{a{x^2} + bx - \left( {a + b} \right)}}{{x - 1}} = \mathop {\lim }\limits_{x \to {1^ + }} \frac{{a\left( {{x^2} - 1} \right) + b\left( {x - 1} \right)}}{{x - 1}} = \mathop {\lim }\limits_{x \to {1^ + }} \left[ {a\left( {x + 1} \right) + b} \right] = 2a + b\\\mathop {\lim }\limits_{x \to {1^ - }} \frac{{f\left( x \right) - f\left( 1 \right)}}{{x - 1}} = \mathop {\lim }\limits_{x \to {1^ - }} \frac{{2x - 1 - \left( {a + b} \right)}}{{x - 1}} = \mathop {\lim }\limits_{x \to {1^ - }} \frac{{2x - 2}}{{x - 1}} = 2\end{array}\)

Để hàm số có đạo hàm tại x = 1 thì \(f'\left( 1 \right)=\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,\frac{f\left( x \right)-f\left( 1 \right)}{x-1}=\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,\frac{f\left( x \right)-f\left( 1 \right)}{x-1}\Leftrightarrow 2a+b=2\,\,\,\left( 2 \right)\)

Từ (1) và (2) ta có hệ: \(\left\{ \begin{array}{l}a + b = 1\\2a + b = 2\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = 1\\b = 0\end{array} \right.\)

Chọn C.