Cho hàm số \(f\left( x \right) = \left\{ \begin{array}{l}\frac{{\sqrt {a{x^2} + 1} - bx - 2}}{{4{x^3} - 3x + 1}}\,\,\,\,{\rm{khi}}\,\,x \ne \frac{1}{2}\\\frac{c}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{khi}}\,\,x = \frac{1}{2}\end{array} \right.\,,\,\,\left( {a,\,b,\,c \in \mathbb{R}} \right)\). Biết hàm số liên tục tại \(x = \frac{1}{2}\)Tính \(S = abc\).

Chọn B

Vì hàm số \(f\left( x \right)\) liên tục tại \(x = \frac{1}{2}\) nên \(\mathop {\lim }\limits_{x \to \frac{1}{2}} f\left( x \right) = \mathop {\lim }\limits_{x \to \frac{1}{2}} \frac{{\sqrt {a{x^2} + 1} - bx - 2}}{{4{x^3} - 3x + 1}} = \mathop {\lim }\limits_{x \to \frac{1}{2}} \frac{{\sqrt {a{x^2} + 1} - bx - 2}}{{{{\left( {2x - 1} \right)}^2}\left( {x + 1} \right)}} = \frac{c}{2}\). Do đó tử thức nhận \(x = \frac{1}{2}\)làm nghiệm kép.

Vậy yêu cầu bài toán tương đương với \(a,\,\,b\) phải thoả mãn các điều kiện: \(\sqrt {\frac{{a + 4}}{4}} - \frac{b}{2} - 2 = 0\) (1) và \(\sqrt {a{x^2} + 1} - bx - 2 = 0\) (2) có nghiệm kép \(x = \frac{1}{2}\)

(2) \( \Leftrightarrow \sqrt {a{x^2} + 1} - bx - 2 = 0 \Leftrightarrow \sqrt {a{x^2} + 1} = bx + 2 \Leftrightarrow \left\{ \begin{array}{l}a{x^2} + 1 = {\left( {bx + 2} \right)^2}\\\frac{b}{2} + 2 \ge 0\end{array} \right.\)có nghiệm kép\( \Leftrightarrow \left\{ \begin{array}{l}a{x^2} + 1 = {\left( {bx + 2} \right)^2}\\\frac{b}{2} + 2 \ge 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a{x^2} + 1 = {b^2}{x^2} + 4bx + 4\\\frac{b}{2} \ge - 2\end{array} \right.\)có nghiệm kép.

 Pt \(\left( {a - {b^2}} \right){x^2} - 4bx - 3 = 0\) có nghiệm kép \( \Leftrightarrow \left\{ \begin{array}{l}a \ne {b^2}\\4{b^2} + 3\left( {a - {b^2}} \right) = 0\end{array} \right.\) \( \Leftrightarrow {b^2} + 3a = 0\) (3). Kết hợp (1) và (3) ta có hpt \(\left\{ \begin{array}{l}\frac{{a + 4}}{4} = {\left( {\frac{b}{2} + 2} \right)^2}\\{b^2} + 3a = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = {b^2} + 8b + 12\\{b^2} + 3a = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = {b^2} + 8b + 12\\{\left( {b + 3} \right)^2} = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = - 3\\b = - 3\end{array} \right.\).

Khi đó \(\mathop {\lim }\limits_{x \to \frac{1}{2}} f\left( x \right) = \mathop {\lim }\limits_{x \to \frac{1}{2}} \frac{{\sqrt { - 3{x^2} + 1} + 3x - 2}}{{4{x^3} - 3x + 1}} = \mathop {\lim }\limits_{x \to \frac{1}{2}} \frac{{\sqrt { - 3{x^2} + 1} + 3x - 2}}{{{{\left( {2x - 1} \right)}^2}\left( {x + 1} \right)}}\)\(\mathop {\lim }\limits_{x \to \frac{1}{2}} \frac{{\sqrt { - 3{x^2} + 1} + 3x - 2}}{{{{\left( {2x - 1} \right)}^2}\left( {x + 1} \right)}} = \mathop {\lim }\limits_{x \to \frac{1}{2}} \frac{{\left( {\sqrt { - 3{x^2} + 1} + 3x - 2} \right)\left( {\sqrt { - 3{x^2} + 1} - 3x + 2} \right)}}{{{{\left( {2x - 1} \right)}^2}\left( {x + 1} \right)\left( {\sqrt { - 3{x^2} + 1} - 3x + 2} \right)}}\)\( = \mathop {\lim }\limits_{x \to \frac{1}{2}} \frac{{ - 3{x^2} + 1 - 9{x^2} + 12x - 4}}{{{{\left( {2x - 1} \right)}^2}\left( {x + 1} \right)\left( {\sqrt { - 3{x^2} + 1} - 3x + 2} \right)}} = \mathop {\lim }\limits_{x \to \frac{1}{2}} \frac{{ - 12{x^2} + 12x - 3}}{{{{\left( {2x - 1} \right)}^2}\left( {x + 1} \right)\left( {\sqrt { - 3{x^2} + 1} - 3x + 2} \right)}}\)\( = \mathop {\lim }\limits_{x \to \frac{1}{2}} \frac{{ - 3{{\left( {2x - 1} \right)}^2}}}{{{{\left( {2x - 1} \right)}^2}\left( {x + 1} \right)\left( {\sqrt { - 3{x^2} + 1} - 3x + 2} \right)}} = \mathop {\lim }\limits_{x \to \frac{1}{2}} \frac{{ - 3}}{{\left( {x + 1} \right)\left( {\sqrt { - 3{x^2} + 1} - 3x + 2} \right)}} = - 2\).

Vậy \(c = - 4\), \(abc = - 36\).