Question

If $f(x)$ = $ \begin{vmatrix}x+1&x&1\\ x\left(x+1\right)&x\left(x-1\right)&2x\\ x\left(x+1\right)\left(x-1\right)&x\left(x-1\right)\left(x-2\right)&3x\left(x-1\right)\end{vmatrix} $ ,
then $ f(1000) $ is equal to

• A. 1
• B. 1000
• C. -1000
• D. 0

Answer 1

Answer: (D)

Given,
$f(x)= \begin{vmatrix}x+1 & x & 1 \\x(x+1) & x(x-1) & 2 x \\x(x+1)(x-1) & x(x-1)(x-2) & 3 x(x-1)\end{vmatrix}$
Taking $x$ common from $R_{2}$ and $x(x-1)$ common from $R_{3}$,
$=x \times x(x-1) \begin{vmatrix}x+1 & x & 1 \\x+1 & x-1 & 2 \\x+1 & x-2 & 3
\end{vmatrix} $
$=x^{2}(x-1)(x+1) \begin{vmatrix}1 & x & 1 \\1 & x-1 & 2 \\1 & x-2 & 3\end{vmatrix}$
Applying $R_{2} \rightarrow R_{2}-R_{1}$ and $R_{3} \rightarrow R_{3}-R_{1}$,
$=x^{2}(x-1)(x+1)\begin{vmatrix}1 & x & 1 \\ 0 & -1 & 1 \\ 0 & -2 & 2\end{vmatrix}$
$ \Rightarrow =x^{2}\left(x^{2}-1\right)[1(-2+2)]=0 $
$ \therefore f(x) =0 $
$ \therefore f(1000) =0 $