Gọi hệ số bậc cao nhất của f(x) là \(k>0\)
\(f\left(x\right)-x\) là hàm bậc 3 có 2 nghiệm nên phải có 1 nghiệm kép
\(\Rightarrow f\left(x\right)-x=k\left(x-a\right)^2\left(x-b\right)\)
\(\Rightarrow f\left(x\right)=k\left(x-a\right)^2\left(x-b\right)+x\)
\(f\left(0\right)=0\Rightarrow ab=0\Rightarrow\left[{}\begin{matrix}a=0\\b=0\end{matrix}\right.\)
\(f'\left(x\right)=2k\left(x-a\right)\left(x-b\right)+k\left(x-a\right)^2+1\)
\(f'\left(1\right)=1\Rightarrow2k\left(1-a\right)\left(1-b\right)+k\left(1-a\right)^2=0\)
\(\Rightarrow k\left(1-a\right)\left(3-a-2b\right)=0\Rightarrow\left[{}\begin{matrix}a=1\\3-a-2b=0\end{matrix}\right.\)
- Nếu \(a=1\Rightarrow b=0\) (do ab=0) \(\Rightarrow f\left(x\right)=k.x\left(x-1\right)^2+x\Rightarrow f\left(x\right)+x=k.x\left(x-1\right)^2+2x\)
\(=x\left[kx^2-2kx+k+2\right]=0\) có nghiệm kép
\(\Rightarrow\Delta'=k^2-k\left(k+2\right)=0\Rightarrow k=0\) (loại)
- Nếu \(3-a-2b=0\Rightarrow\left[{}\begin{matrix}a=0;b=\dfrac{3}{2}\\b=0;a=3\end{matrix}\right.\)
Với \(a=0;b=\dfrac{3}{2}\Rightarrow f\left(x\right)=kx^2\left(x-\dfrac{3}{2}\right)+x\)
\(\Rightarrow f\left(x\right)+x=kx^2\left(x-\dfrac{3}{2}\right)+2x=x\left[kx^2-\dfrac{3}{2}kx+2\right]\)
\(\Rightarrow\left(\dfrac{3}{2}k\right)^2-8k=0\Rightarrow k=\dfrac{32}{9}\)
\(\Rightarrow f\left(x\right)=\dfrac{32}{9}x^2\left(x-\dfrac{3}{2}\right)+x\)
Với \(a=3;b=0\Rightarrow f\left(x\right)+x=kx\left(x-3\right)^2+2x\)
\(=x\left(kx^2-6kx+9k+2\right)\)
\(\Rightarrow\Delta'=9k^2-k\left(9k+2\right)=0\Rightarrow k=0\) (loại)
Vậy \(f\left(x\right)=\dfrac{32}{9}x^2\left(x-\dfrac{3}{2}\right)+x\)
