Cho hàm số

C

- Hướng dẫn giải

Đáp án C

Gọi $M\left(x_0;\frac{2x_0-3}{x_0-2}\right)$ là tiếp tuyến của d với  (C) 

Ta có $y\text{'}=-\frac{1}{{\left(x-2\right)}^2}\Rightarrow y\text{'}\left(x_0\right)=-\frac{1}{{\left(x_0-2\right)}^2}$ 

Suy ra $d:y=-\frac{1}{{\left(x_0-2\right)}^2}\left(x-x_0\right)+\frac{2x_0-3}{x_0-2}\iff d:y=-\frac{1}{{\left(x_0-2\right)}^2}x+\frac{2x_0^2-6x_0+6}{{\left(x_0-2\right)}^2}$ 

Ta có $\left\{\begin{array}{l}d\cap \left(x=2\right)=A\left(2;\frac{2x_0-2}{x_0-2}\right)\\ d\cap \left(y=2\right)=B\left(2x_0-2;2\right)\end{array}\right.\Rightarrow AB=\sqrt{4{\left(x_0-2\right)}^2+\frac{4}{{\left(x_0-2\right)}^2}}$ 

Có $A{B}^2=4{\left(x_0-2\right)}^2+\frac{4}{{\left(x_0-2\right)}^2}\ge 24\sqrt{{\left(x_0-2\right)}^2\frac{4}{{\left(x_0-2\right)}^2}}=8\Rightarrow AB\ge 2\sqrt{2}\Rightarrow \min AB=2\sqrt{2}$