Question

$f\left(x\right)=\left|\begin{array}{ccc}1& x& x+1\\ 2x& x(x-1)& x(x+1)\\ 3x(x-1)& x(x-1)(x-2)& (x-1)x(x+1)\end{array}\right|\Rightarrow f\left(2012\right)=$

• A. $0$
• B. $1$
• C. $-500$
• D. $500$

Answer 1

The correct option is A $0$

$f\left(x\right)=\left|\begin{array}{ccc}1& x& x+1\\ 2x& x(x-1)& x(x+1)\\ 3x(x-1)& x(x-1)(x-2)& (x-1)x(x+1)\end{array}\right|$

$=(x+1)\left|\begin{array}{ccc}1& x& 1\\ 2x& x(x-1)& x\\ 3x(x-1)& x(x-1)(x-2)& (x-1)x\end{array}\right|$

$=x^2(x+1)\left|\begin{array}{ccc}1& x& 1\\ 2& (x-1)& 1\\ 3(x-1)& (x-2)(x-1)& x-1\end{array}\right|$

$=x^2(x+1)(x-1)\left|\begin{array}{ccc}1& x& 1\\ 2& x-1& 1\\ 3& (x-2)& 1\end{array}\right|$

$=x^2(x^2-1)\left|\begin{array}{ccc}1& x& 1\\ 1& -1& 0\\ 2& -2& 0\end{array}\right|$

$=x^2(x^2-1)[0+x\times 0+(-2+2)]$

$=0$

$\Rightarrow f\left(x\right)=0$

$\therefore f\left(2012\right)=0$