Câu hỏi 28

Trước hết ta xét tính liên tục của hàm số tại \(x = 1\).

Ta có:

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

\( \Rightarrow \) Hàm số liên tục tại \(x = 1\).

Tính \(f'\left( 1 \right)\).

\(\begin{array}{l} \Rightarrow f'\left( 1 \right) = \mathop {\lim }\limits_{x \to 1} \dfrac{{f\left( x \right) - f\left( 1 \right)}}{{x - 1}} = \dfrac{{\dfrac{{\sqrt {3x + 1}  - 2x}}{{x - 1}} + \dfrac{5}{4}}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \dfrac{{4\sqrt {3x + 1}  - 8x + 5x - 5}}{{4{{\left( {x - 1} \right)}^2}}}\\ = \mathop {\lim }\limits_{x \to 1} \dfrac{{4\sqrt {3x + 1}  - 3x - 5}}{{4{{\left( {x - 1} \right)}^2}}} = \mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {4\sqrt {3x + 1}  - 3x - 5} \right)\left( {4\sqrt {3x + 1}  + 3x + 5} \right)}}{{4{{\left( {x - 1} \right)}^2}\left( {4\sqrt {3x + 1}  + 3x + 5} \right)}}\\ = \mathop {\lim }\limits_{x \to 1} \dfrac{{16\left( {3x + 1} \right) - \left( {9{x^2} + 30x + 25} \right)}}{{4{{\left( {x - 1} \right)}^2}\left( {4\sqrt {3x + 1}  + 3x + 5} \right)}} = \mathop {\lim }\limits_{x \to 1} \dfrac{{ - 9{x^2} + 18x - 9}}{{4{{\left( {x - 1} \right)}^2}\left( {4\sqrt {3x + 1}  + 3x + 5} \right)}}\\ = \mathop {\lim }\limits_{x \to 1} \dfrac{{ - 9{{\left( {x - 1} \right)}^2}}}{{4{{\left( {x - 1} \right)}^2}\left( {4\sqrt {3x + 1}  + 3x + 5} \right)}} = \mathop {\lim }\limits_{x \to 1} \dfrac{{ - 9}}{{4\left( {4\sqrt {3x + 1}  + 3x + 5} \right)}} = \dfrac{{ - 9}}{{64}}\end{array}\)

Chọn C.