Câu hỏi 48

TXĐ: \(D = \mathbb{R}\).

Ta có:

\(\begin{array}{l}\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} \left( {a{x^3} + 2{x^2} + 3bx + 2} \right) = a + 3b + 4\\\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ - }} \left( {\sqrt {5 - 4x}  - 2x} \right) =  - 1 = f\left( 1 \right)\end{array}\)

Để hàm số có đại hàm tại \(x = 1\) thì hàm số phải liên tục tại \(x = 1\)\( \Rightarrow a + 3b + 4 =  - 1 \Leftrightarrow a + 3b =  - 5\,\,\,\left( 1 \right)\).

Ta có:

\(\begin{array}{l}f'\left( {{1^ - }} \right) = \mathop {\lim }\limits_{x \to {1^ - }} \dfrac{{f\left( x \right) - f\left( 1 \right)}}{{x - 1}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to {1^ - }} \dfrac{{\sqrt {5 - 4x}  - 2x + 1}}{{x - 1}} =  - 4\\f'\left( {{1^ + }} \right) = \mathop {\lim }\limits_{x \to {1^ + }} \dfrac{{f\left( x \right) - f\left( 1 \right)}}{{x - 1}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to {1^ + }} \dfrac{{a{x^3} + 2{x^2} + 3bx + 2 + 1}}{{x - 1}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to {1^ + }} \dfrac{{\left( { - 5 - 3b} \right){x^3} + 2{x^2} + 3bx + 3}}{{x - 1}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to {1^ + }} \dfrac{{\left( { - 5 - 3b} \right){x^3} + 2{x^2} - 2 + 3bx - 3b + 3b + 5}}{{x - 1}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to {1^ + }} \dfrac{{\left( { - 5 - 3b} \right)\left( {{x^3} - 1} \right) + 2\left( {x - 1} \right)\left( {x + 1} \right) + 3b\left( {x - 1} \right)}}{{x - 1}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to {1^ + }} \left[ {\left( { - 5 - 3b} \right)\left( {{x^2} + x + 1} \right) + 2\left( {x + 1} \right) + 3b} \right]\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 3\left( { - 5 - 3b} \right) + 4 + 3b =  - 6b - 11\end{array}\)

Để hàm số có đạo hàm tại \(x = 1\) thì \(f'\left( {{1^ + }} \right) = f'\left( {{1^ - }} \right)\)\( \Leftrightarrow  - 6b - 11 =  - 4 \Leftrightarrow b = \dfrac{{ - 7}}{6}\).

Thay vào (1) ta có: \(a + 3.\dfrac{{ - 7}}{6} =  - 5 \Leftrightarrow a =  - \dfrac{3}{2}\)

Vậy \(ab =  - \dfrac{3}{2}.\dfrac{{ - 7}}{6} = \dfrac{7}{4}\).

Chọn C.