Question

If $f\left(x\right)=\left|\begin{array}{ccc}1& 2(x-1)& 3(x-1)(x-2)\\ x-1& (x-1)(x-2)& (x-1)(x-2)(x-3)\\ x& (x-1)x& (x-1)(x-2)x\end{array}\right|$ then $f\left(41\right)$is

• A. $41$
• B. $-41$
• C. $0$
• D. $Noneofthese$

Answer 1

The correct option is A $41$

$f\left(x\right)=\left|\begin{array}{ccc}1& 2(x-1)& 3(x-1)(x-2)\\ x-1& (x-1)(x-2)& (x-1)(x-2)(x-3)\\ x& (x-1)x& x(x-1)(x-2)\end{array}\right|f\left(x\right)=(x-1)\left|\begin{array}{ccc}1& 2& 3(x-1)(x-2)\\ (x-1)& (x-2)& (x-1)(x-2)(x-3)\\ x& x& x(x-1)(x-2)\end{array}\right|=x{(x-1)}^2(x-2)\left|\begin{array}{ccc}1& 2& 3\\ (x-1)& (x-2)& (x-3)\\ 1& 1& 1\end{array}\right|=x{(x-1)}^2(x-2)\left[\right(1)-(4)+(3\left)\right]=x{(x-1)}^2(x-2)\left(0\right)=0\therefore f\left(x\right)=0\Rightarrow f\left(41\right)=0$