Lời giải chi tiết:

+) \(a \le 0\):

\(\begin{array}{l}\,\,\,\,\,\mathop {\lim }\limits_{x \to  + \infty } \left[ {\sqrt {4{x^2} - 3x + 1}  - \left( {ax + b} \right)} \right]\\ = \mathop {\lim }\limits_{x \to  + \infty } x\left[ {\sqrt {4 - \dfrac{3}{x} + \dfrac{1}{{{x^2}}}}  - a - \dfrac{b}{x}} \right] =  + \infty \end{array}\)

(do \(\mathop {\lim }\limits_{x \to  + \infty } x =  + \infty ,\,\,\mathop {\lim }\limits_{x \to  + \infty } \left[ {\sqrt {4 - \dfrac{3}{x} + \dfrac{1}{{{x^2}}}}  - a - \dfrac{b}{x}} \right] = 4 - a > 0\))

\( \Rightarrow \) Loại

+) \(a > 0\):

\(\begin{array}{l}\,\,\,\,\,\mathop {\lim }\limits_{x \to  + \infty } \left[ {\sqrt {4{x^2} - 3x + 1}  - \left( {ax + b} \right)} \right]\\ = \mathop {\lim }\limits_{x \to  + \infty } \dfrac{{4{x^2} - 3x + 1 - {a^2}{x^2} - 2abx - {b^2}}}{{\sqrt {4{x^2} - 3x + 1}  + \left( {ax + b} \right)}}\\ = \mathop {\lim }\limits_{x \to  + \infty } \dfrac{{\left( {4 - {a^2}} \right){x^2} - \left( {3 + 2ab} \right)x + 1 - {b^2}}}{{\sqrt {4{x^2} - 3x + 1}  + \left( {ax + b} \right)}}\\ = \mathop {\lim }\limits_{x \to  + \infty } \dfrac{{\left( {4 - {a^2}} \right)x - \left( {3 + 2ab} \right) + \dfrac{{1 - {b^2}}}{x}}}{{\sqrt {4 - \dfrac{3}{x} + \dfrac{1}{{{x^2}}}}  + a + \dfrac{b}{x}}}\\\mathop {\lim }\limits_{x \to  + \infty } \left( {\sqrt {4 - \dfrac{3}{x} + \dfrac{1}{{{x^2}}}}  + a + \dfrac{b}{x}} \right) = 2 + a\end{array}\)

Để \(\mathop {\lim }\limits_{x \to  + \infty } \left[ {\sqrt {4{x^2} - 3x + 1}  - \left( {ax + b} \right)} \right] = 0\) thì \(\left\{ \begin{array}{l}4 - {a^2} = 0\\3 + 2{\rm{a}}b = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = 2\\b =  - \dfrac{3}{4}\end{array} \right.\) (do \(a > 0\))

\( \Rightarrow a - 4b = 2 - 4.\dfrac{{ - 3}}{4} = 5\).

Chọn: A