Cho hàm số f(x)=x^3-2ax^2+a^2x+b (a,b

* Hướng dẫn giải

Đáp án D

Ta có: $f\text{'}\left(x\right)=3x^2-4ax+a^2;f\text{'}\left(x\right)=0\iff 3x^2-4ax+a^2=0\left(*\right)$

Để hàm số có 2 điểm cực trị thì $\Delta \text{'}=a^2>0\iff a\ne 0$

Khi đó (*) $\iff \left[\begin{array}{l}x=\frac{2a+a}{3}=a\Rightarrow y=b\\ x=\frac{2a-a}{3}=\frac{a}{3}\Rightarrow y=\frac{4a^3}{27}+b\end{array}\right.\Rightarrow \left\{\begin{array}{l}A\left(a;b\right)\\ B\left(\frac{a}{3};\frac{4a^3}{27}+b\right)\end{array}\right.$

Đặt $c=\frac{a}{3}\ne 0\Rightarrow \left\{\begin{array}{l}A\left(3c;b\right)\\ B\left(c;4c^3+b\right)\end{array}\right.\Rightarrow \left\{\begin{array}{l}A{B}^2=4c^2+16c^6\\ O{A}^2=9c^2+b^2\\ O{B}^2=c^2+16c^6+8c^{3}b+b^2\end{array}\right.$

Tam giác OAB vuông cân $\iff \left\{\begin{array}{l}O{A}^2=O{B}^2\\ A{B}^2=2O{A}^2\end{array}\right.\iff \left\{\begin{array}{l}{c}^2=2c^6+c^{3}b\\ 8c^6=7c^2+b^2\end{array}\right.$

$\iff \left\{\begin{array}{l}2c^4+bc=1\\ 8c^6=7c^2+b^2\end{array}\right.\iff \left[\begin{array}{l}b=1;c=-1\\ b=-1;c=1\end{array}\right.\Rightarrow \left[\begin{array}{l}b=1;a=-3\\ b=-1;a=3\end{array}\right.\Rightarrow T=a^2+b^2=10$