Cho hai đường cong

C

- Hướng dẫn giải

Đáp án C

Xét đường cong

$\left(C_1\right):f\left(x\right)=3^x\left(3^x-m+2\right)+m^2-3m$

Và đường cong $\left(C_2\right):g\left(x\right)=3^x+1.$

Để $\left(C_1\right)$ tiếp xúc với $\left(C_2\right)$

$\iff \left\{\begin{array}{l}f\text{'}\left(x\right)=g\text{'}\left(x\right)\\ f\left(x\right)=g\left(x\right)\end{array}\right.$ 

$$\begin{gathered} \iff \left\{\begin{array}{l}2\ln 3{\left(3^x\right)}^2-\left(m-2\right)\ln {3.3}^x=3^x.\ln 3\\ {\left(3^x\right)}^2-\left(m-2\right){.3}^x+m^2-3m=3^x+1\end{array}\right. \\ \iff \left\{\begin{array}{l}{23}^x-m+2=1\\ {\left(3^x\right)}^2-\left(m-2\right){.3}^x+m^2-3m=3^x+1\end{array}\right. \end{gathered}$$

$$\begin{gathered} \iff \left\{\begin{array}{l}{3}^x=\frac{m-1}{2}\\ {\left(3^x\right)}^2-\left(m-2\right){.3}^x+m^2-3m=3^x+1\end{array}\right. \\ \Rightarrow {\left(\frac{m-1}{2}\right)}^2-\left(m-2\right).\frac{m-1}{2}+m^2-3m \\ =\frac{m-1}{2}+1\left(*\right) \end{gathered}$$

 vì $3^x>0\Rightarrow m>1,$

do đó $\left(*\right)\iff m=\frac{5+2\sqrt{10}}{3}$