Cho hàm số \(f\left( x \right) = \left\{ \begin{ar...

C

- Hướng dẫn giải

Trước hết ta xét tính liên tục của hàm số tại x = 1

Ta có

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

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

Tính \(f'(1)\) 

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