Câu hỏi 44

TXĐ: \(D = \mathbb{R},\,\,x =  - 1 \in D\).

Ta có:

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

Để hàm số liên tục tại \(x =  - 1\) thì \(\mathop {\lim }\limits_{x \to  - {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to  - {1^ - }} f\left( x \right) = f\left( { - 1} \right)\)\( \Leftrightarrow a - 1 = 1 - b \Leftrightarrow a + b = 2\).

Tính đạo hàm tại điểm \(x =  - 1\):

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

Thay \(a = 2 - b\) ta có:

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

Để hàm số có đạo hàm tại \(x =  - 1\) thì \(f'\left( { - {1^ + }} \right) = f'\left( { - {1^ - }} \right)\)\( \Leftrightarrow b - 2 = 2 \Leftrightarrow b = 4\)\( \Rightarrow a = -2\).

Vậy \(a = -2,\,\,b = 4\).

Chọn B.