If f(x) = left|begin{array}{ccc} 1 & x & x+1 | vedclass

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(A) Given, $f(x) = \left|\begin{array}{ccc} 1 & x & x+1 \ 2x & x(x-1) & x(x+1) \ 3x(x-1) & x(x-1)(x-2) & (x-1)x(x+1) \end{array}ight|$.Taking $x$

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(A) Given, $f(x) = \left|\begin{array}{ccc} 1 & x & x+1 \ 2x & x(x-1) & x(x+1) \ 3x(x-1) & x(x-1)(x-2) & (x-1)x(x+1) \end{array}ight|$.Taking $x$ common from $R_2$ and $x(x-1)$ common from $R_3$:$f(x) = x \cdot x(x-1) \left|\begin{array}{ccc} 1 & x & x+1 \ 2 & x-1 & x+1 \ 3 & x-2 & x+1 \end{array}ight|$.Taking $(x+1)$ common from $C_3$:$f(x) = x^2(x-1)(x+1) \left|\begin{array}{ccc} 1 & x & 1 \ 2 & x-1 & 1 \ 3 & x-2 & 1 \end{array}ight|$.Since $C_1$ and $C_3$ are not identical, let us perform row operations. Applying $R_2 ightarrow R_2 - R_1$ and $R_3 ightarrow R_3 - R_1$:$f(x) = x^2(x^2-1) \left|\begin{array}{ccc} 1 & x & 1 \ 1 & -1 & 0 \ 2 & -2 & 0 \end{array}ight|$.Expanding along $C_3$:$f(x) = x^2(x^2-1) \cdot 1 \cdot \left|\begin{array}{cc} 1 & -1 \ 2 & -2 \end{array}ight| = x^2(x^2-1) \cdot (-2 - (-2)) = x^2(x^2-1) \cdot 0 = 0$.Thus, $f(x) = 0$ for all $x$.Therefore, $f(2012) = 0$.

If $f(x) = \left|\begin{array}{ccc} 1 & x & x+1 \\ 2x & x(x-1) & x(x+1) \\ 3x(x-1) & x(x-1)(x-2) & (x-1)x(x+1) \end{array}\right|$, then $f(2012)$ is equal to:

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