Question

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

Then, $f\left(100\right)$ is equal to

• A. $0$ (Zero)
• B. $1$
• C. $100$
• D. $10$

Answer 1

The correct option is A $0$ (Zero)

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

Taking $x$ common from $C_2$, $(x+1)$ common from $C_3$ and $(x-1)$ common from $R_3$, we get

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

Apply $C_2\to C_2-C_1,C_3\to C_3-C_1$

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

$\Rightarrow f\left(x\right)=(x-1)x(x+1)\times 0=0\therefore f\left(100\right)=0$