Cho hàm số y=1/3x^3-mx^2-2x-2m-1/3

* Hướng dẫn giải

Đáp án B

Xét hàm số: $y=\frac{1}{3}{x}^3+m{x}^2-2x-2m-\frac{1}{3}$

Có: $y\text{'}=x^2+2mx-2$

$y\text{'}=x^2+2mx-2=0\iff \left[\begin{array}{l}x=-m-\sqrt{m^2+2}\\ x=-m+\sqrt{m^2+2}\end{array}\right.$

Do $m\in \left(0;\frac{5}{6}\right)$ nên $\left\{\begin{array}{l}-m-\sqrt{m^2+2}<0\\ 0<-m+\sqrt{m^2+2}<2\end{array}\right.$

Và $\left\{\begin{array}{l}y\left(0\right)=-2m-\frac{1}{3}<0\\ y\left(2\right)=2m-\frac{5}{3}<0\end{array}\right.$

Suy ra $y<0,\forall x\in (0;2)$

Vậy S=4

$\begin{array}{l}\iff \underset{0}{\overset{2}{\int }}\left|\frac{1}{3}{x}^3+m{x}^2-2x-2m-\frac{1}{3}\left|dx=4\right.\right.\\ \iff -\underset{0}{\overset{2}{\int }}\left(\frac{1}{3}{x}^3+m{x}^2-2x-2m-\frac{1}{3}\right)dx=4\\ \iff \frac{4m+10}{3}=4\iff m=\frac{1}{2}.\end{array}$