Phương pháp:

- Giả sử $A\left(a;a^3+a^2-4\right),B\left(b;b^3+b^2-4\right).$

- Vì OA=2OB nên $\left[\begin{array}{l}\overrightarrow{OA}=2\overrightarrow{OB}\\ \overrightarrow{OA}=-2\overrightarrow{OB}\end{array}\right.$, giải hệ phương trình bằng phương pháp thế.

Cách giải:

Giả sử $A\left(a;a^3+a^2-4\right),B\left(b;b^3+b^2-4\right).$

- Vì OA= 2OB nên $\left[\begin{array}{l}\overrightarrow{OA}=2\overrightarrow{OB}\\ \overrightarrow{OA}=-2\overrightarrow{OB}\end{array}\right.\iff \left[\begin{array}{l}\left(a;a^3+a^2-4\right)=2\left(b;b^3+b^2-4\right)\\ \left(a;a^3+a^2-4\right)=-2\left(b+b^3+b^2-4\right)\end{array}\right.$

$\iff \left[\begin{array}{l}\left\{\begin{array}{l}a=2b\\ a^3+a^2-4=2b^3+2b^2-8\end{array}\right.\\ \left\{\begin{array}{l}a=-2b\\ a^3+a^2-4=-2b^3-2b^2+8\end{array}\right.\end{array}\right.\iff \left[\begin{array}{l}\left\{\begin{array}{l}a=2b\\ 8b^3+4b^2-4=2b^3+2b^2-8\end{array}\right.\\ \left\{\begin{array}{l}a=-2b\\ -8b^3+4b^2-4=-2b^3-2b^3+8\end{array}\right.\end{array}\right.$

$\iff \left[\begin{array}{l}\left\{\begin{array}{l}a=2b\\ 6b^3+2b^2+4=0\end{array}\right.\\ \left\{\begin{array}{l}a=-2b\\ 6b^3-6b^2+12=0\end{array}\right.\end{array}\right.\iff \left[\begin{array}{l}\left\{\begin{array}{l}a-2b\\ b=-1\end{array}\right.\\ \left\{\begin{array}{l}a=-2b\\ b=-1\end{array}\right.\end{array}\right.\iff \left[\begin{array}{l}\left\{\begin{array}{l}a=-2\\ b=-1\end{array}\right.\\ \left\{\begin{array}{l}a=2\\ b=-1\end{array}\right.\end{array}\right.$

Vậy có 2 cặp điểm A,B thỏa mãn yêu cầu bài toán.

Chọn A.