Bộ 24 đề thi cuối kì 1 Toán 11 Cánh diều (2023 - 2024) có đáp án - Đề 12

Chọn B

Từ $f(x) + 3f\left( {\frac{1}{x}} \right) = 8x,\,\,x \ne 0$, thay $x$ bởi $\frac{1}{x}$ ta được $f\left( {\frac{1}{x}} \right) + 3f\left( x \right) = \frac{8}{x}$. Suy ra

$$\left\{ \begin{array}{l}f\left( {\frac{1}{x}} \right) + 3f\left( x \right) = \frac{8}{x}\\f\left( x \right) + 3f\left( {\frac{1}{x}} \right) = 8x\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}3f\left( {\frac{1}{x}} \right) + 9f\left( x \right) = \frac{{24}}{x}\\f\left( x \right) + 3f\left( {\frac{1}{x}} \right) = 8x\end{array} \right. \Rightarrow 8f\left( x \right) = 8\left( {\frac{3}{x} - x} \right)$$

$\Rightarrow f\left( x \right) = \frac{3}{x} - x = \frac{{3 - {x^2}}}{x}$. Do đó

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