A) \[10(b-a)(c-a)\]
B) \[100(b-a)(c-b)(a-c)\]
C) \[100\,abc\]
D) 0
Correct Answer: D
Solution :
\[f(x)=\left| \begin{matrix} 1+a & 1+ax & 1+a{{x}^{2}} \\ 1+b & 1+bx & 1+b{{x}^{2}} \\ 1+c & 1+cx & 1+c{{x}^{2}} \\ \end{matrix} \right|\] Applying\[{{C}_{2}}\to {{C}_{2}}-{{C}_{1}}\]and\[{{C}_{3}}\to {{C}_{3}}-{{C}_{2}}\] \[\Rightarrow \]\[f(x)=\left| \begin{matrix} 1+a & a(x-1) & ax(x-1) \\ 1+b & b(x-1) & dx(x-1) \\ 1+c & c(x-1) & cx(x-1) \\ \end{matrix} \right|\] \[=(x-1)x(x-1)\left| \begin{matrix} 1+a & a & a \\ 1+b & b & b \\ 1+c & c & c \\ \end{matrix} \right|\] (\[\because \]Two columns are same.) =0
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