Câu hỏi 41

Hàm số liên tục trên các khoảng \(\left( { - \infty ;0} \right);\,\,\left( {0;2} \right) ;\,\left( {2; + \infty } \right)\) .

Ta có: \(f\left( 0 \right) = b;\,\,\,f\left( 2 \right) = a.\)

 \(\begin{array}{l}\mathop {\lim }\limits_{x \to 0} f\left( x \right) = \mathop {\lim }\limits_{x \to 0} \frac{{{x^3} - 3{x^2} + 2x}}{{x(x - 2)}} = \mathop {\lim }\limits_{x \to 0} \frac{{x\left( {x - 1} \right)\left( {x - 2} \right)}}{{x\left( {x - 2} \right)}} = \mathop {\lim }\limits_{x \to 0} \left( {x - 1} \right) =  - 1\\\mathop {\lim }\limits_{x \to 2} f\left( x \right) = \mathop {\lim }\limits_{x \to 2} \frac{{{x^3} - 3{x^2} + 2x}}{{x(x - 2)}} = \mathop {\lim }\limits_{x \to 2} \frac{{x\left( {x - 1} \right)\left( {x - 2} \right)}}{{x\left( {x - 2} \right)}} = \mathop {\lim }\limits_{x \to 2} \left( {x - 1} \right) = 1\end{array}\)

\( \Rightarrow \) Hàm số liên tục trên \(\mathbb{R} \Leftrightarrow \left\{ \begin{array}{l}\mathop {\lim }\limits_{x \to 0} f\left( x \right) = f\left( 0 \right)\\\mathop {\lim }\limits_{x \to 2} f\left( x \right) = f\left( 2 \right)\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = 1\\b =  - 1\end{array} \right..\)

Khi đó:  \({a^3} + {b^3} = {1^3} + {\left( { - 1} \right)^3} = 0.\)

Chọn D.