Cho hàm số y = |2x^3 - 3x^2 + 6(m^2 + 1)x + 2021|. Gọi S là tập hợp các giá trị

* Hướng dẫn giải

Xét hàm số $f\left(x\right)=2x^3-3x^2+6\left(m^2+1\right)x+2021$ ta có $f\text{'}\left(x\right)=6x^2-6x+6\left(m^2+1\right).$

Ta có $f\text{'}\left(x\right)=6\left(x^2-x+1+m^2\right)>0\forall x\in \left(-1;0\right),\forall m$ do đó hàm số f(x) đồng biến trên [-1; 0].

$\Rightarrow \underset{\left[-1;0\right]}{\min }f\left(x\right)=f\left(-1\right)=-6m^2+2010$

     $\underset{\left[-1;0\right]}{\max }f\left(x\right)=f\left(0\right)=2021$

$\Rightarrow \underset{\left[-1;0\right]}{\max }\left|f\left(x\right)\right|=\max \left\{\left|-m^2+2010\right|;2021\right\}.$

TH1:

$\left\{\begin{array}{l}\underset{\left[-1;0\right]}{\max }\left|f\left(x\right)\right|=\left|-m^2+2010\right|\\ \left|-6m^2+2010\right|\ge 2021\end{array}\right.\iff \left\{\begin{array}{l}\underset{\left[-1;0\right]}{\max }\left|f\left(x\right)\right|=\left|-m^2+2010\right|\\ \left[\begin{array}{l}-6m^2+2010\ge 2021\\ -6m^2+2010\le -2021\end{array}\right.\end{array}\right.$

$\iff \left\{\begin{array}{l}\underset{\left[-1;0\right]}{\max }\left|f\left(x\right)\right|=\left|-6m^2+2010\right|\\ \left[\begin{array}{l}6m^2\le -1\left(vonghiem\right)\\ -6m^2+2010\le -2010\end{array}\right.\end{array}\right.\iff \left\{\begin{array}{l}\underset{\left[-1;0\right]}{\max }\left|f\left(x\right)\right|=\left|-6m^2+2010\right|\\ m^2\ge \frac{4031}{6}\end{array}\right.$

$\Rightarrow -6m^2+2010\le -2010$

$\Rightarrow \underset{\left[-1;0\right]}{\max }\left|f\left(x\right)\right|\ge 2021\forall m^2\ge \frac{4031}{6}$

$\Rightarrow \min \left(\underset{\left[-1;0\right]}{\max }\left|f\left(x\right)\right|\right)=2021\iff m^2=\frac{4031}{6}$

TH2:

$\left\{\begin{array}{l}\underset{\left[-1;0\right]}{\max }\left|f\left(x\right)\right|=2021\\ 2021\ge \left|-6m^2+2010\right|\end{array}\right.\iff \left\{\begin{array}{l}\underset{\left[-1;0\right]}{\max }\left|f\left(x\right)\right|=2021\\ -2021\le 6m^2+2010\le 2021\end{array}\right.$

$\iff \left\{\begin{array}{l}\underset{\left[-1;0\right]}{\max }\left|f\left(x\right)\right|=2021\\ -\frac{1}{6}\le m^2\le \frac{4031}{6}\end{array}\right.\iff \left\{\begin{array}{l}\underset{\left[-1;0\right]}{\max }\left|f\left(x\right)\right|=2021\\ 0\le m^2\le \frac{4031}{6}\end{array}\right.$

$\Rightarrow \min \left(\underset{\left[-1;0\right]}{\max }\left|f\left(x\right)\right|\right)=2021\iff 0\le m^2\le \frac{4031}{6}$

Vậy $S=\left[-\sqrt{\frac{4031}{6}};\sqrt{\frac{4031}{6}}\right].$

Do S là tập đối xứng nên tổng các phần tử của S bằng 0.

Chọn B.