If f(x) = begin{vmatrix} 1 & x & x+1 2x & x( | vedclass

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(A) Given $f(x) = \begin{vmatrix} 1 & x & x+1 \ 2x & x(x-1) & x(x+1) \ 3x(x-1) & x(x-1)(x-2) & x(x+1)(x-1) \end{vmatrix}$.Taking $x$ common from $R_

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(A) Given $f(x) = \begin{vmatrix} 1 & x & x+1 \ 2x & x(x-1) & x(x+1) \ 3x(x-1) & x(x-1)(x-2) & x(x+1)(x-1) \end{vmatrix}$.Taking $x$ common from $R_2$ and $x(x-1)$ common from $R_3$, we get:$f(x) = x \cdot x(x-1) \begin{vmatrix} 1 & x & x+1 \ 2 & x-1 & x+1 \ 3 & x-2 & x+1 \end{vmatrix}$.Taking $(x+1)$ common from $C_3$, we get:$f(x) = x^2(x-1)(x+1) \begin{vmatrix} 1 & x & 1 \ 2 & x-1 & 1 \ 3 & x-2 & 1 \end{vmatrix}$.Applying $R_1 ightarrow R_1 - R_2$ and $R_2 ightarrow R_2 - R_3$:$f(x) = x^2(x-1)(x+1) \begin{vmatrix} -1 & 1 & 0 \ -1 & 1 & 0 \ 3 & x-2 & 1 \end{vmatrix}$.Since two rows ($R_1$ and $R_2$) are identical, the value of the determinant is $0$.Therefore, $f(x) = 0$ for all $x$, which implies $f(100) = 0$.

If $f(x) = \begin{vmatrix} 1 & x & x+1 \\ 2x & x(x-1) & (x+1)x \\ 3x(x-1) & x(x-1)(x-2) & (x+1)x(x-1) \end{vmatrix}$, then $f(100)$ is equal to:

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