Cho hàm số \(y = f\left( x \right) = \left\{ \begin{array}{l}\frac{{\sqrt {{x^2} + x + 3} - 3}}{{4 - 2x}},\,\,x < 2\\mx + 5\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x \ge 2\end{array} \right.\) và với \(m\)là tham số. Tìm \(m\) để hàm số liên tục tại \(x = 2\).    

 \(\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ - }} \frac{{\sqrt {{x^2} + x + 3}  - 3}}{{4 - 2x}}\)

\( = \mathop {\lim }\limits_{x \to {2^ - }} \frac{{{{\left( {\sqrt {{x^2} + x + 3} } \right)}^2} - 9}}{{\left( {4 - 2x} \right)\left( {\sqrt {{x^2} + x + 3}  + 3} \right)}} = \mathop {\lim }\limits_{x \to {2^ - }} \frac{{{x^2} + x - 6}}{{\left( {4 - 2x} \right)\left( {\sqrt {{x^2} + x + 3}  + 3} \right)}}\)

\( = \mathop {\lim }\limits_{x \to {2^ - }} \frac{{\left( {x - 2} \right)\left( {x + 3} \right)}}{{2\left( {2 - x} \right)\left( {\sqrt {{x^2} + x + 3}  + 3} \right)}} = \mathop {\lim }\limits_{x \to {2^ - }} \frac{{x + 3}}{{ - 2\left( {\sqrt {{x^2} + x + 3}  + 3} \right)}} =  - \frac{5}{{12}}\)

\(\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ + }} \left( {mx + 5} \right) = 2m + 5\)

Hàm số liên tục tại \(x = 2\) thì \(\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = f\left( 2 \right)\)

\( \Leftrightarrow  - \frac{5}{{12}} = 2m + 5 \Leftrightarrow m =  - \frac{{65}}{{24}}\) . Chọn A.