Cho hàm số

- Hướng dẫn giải

Đáp án là C.

$y^{\text{'}}=3{\text{x}}^2-3$ ;

 $y^{\text{'}}\left(x_A\right)=3{\text{x}}_A^2-3$;

$y^{\text{'}}\left(x_B\right)=3{\text{x}}_B^2-3$

Tiếp tuyến tại A,B song song nên$y^{\text{'}}\left(x_A\right)=y^{\text{'}}\left(x_B\right)\iff 3{\text{x}}_A^2-3=3{\text{x}}_B^2-3$

$\iff \left[\begin{array}{l}{x}_A=x_B\left(\text{lo¹i do\hspace{0.17em}\hspace{0.17em}}{x}_A>x_B\right)\\ x_A=-x_B\left(\text{chän}\right)\end{array}\right.$

Ta có :

$$\begin{gathered} A{B}^2={\left(x_B-x_A\right)}^2+{\left(y_B-y_A\right)}^2 \\ ={\left(x_B-x_A\right)}^2+{\left[x_B^3-3{\text{x}}_B+1-\left(x_A^3-3{\text{x}}_A+1\right)\right]}^2 \end{gathered}$$

$=4{\text{x}}_B^6-24{\text{x}}_B^4+40{\text{x}}_B^2$

Giả thiết $AB=6\sqrt{37}$

$$\begin{gathered} \iff 4{\text{x}}_B^6-24{\text{x}}_B^4+40{\text{x}}_B^2=36.37 \\ \iff {\left(x_B^2\right)}^3-6{\left(x_B^2\right)}^2+10\left(x_B^2\right)-333=0 \end{gathered}$$

$\iff x_B^2=9\iff \left[\begin{array}{l}{x}_B=3\Rightarrow x_A=-3\left(\text{lo¹i}\right)\\ x_B=-3\Rightarrow x_A=3\left(\text{chän}\right)\end{array}\right.$

Vậy $S=2{\text{x}}_A-3{\text{x}}_B=2.3-3\left(-3\right)=15$.