A) 1
B) 0
C) 3
D) 2
Correct Answer: D
Solution :
\[f(x)=\left| \begin{matrix} 1+{{a}^{2}}x & (1+{{b}^{2}})x & (1+{{c}^{2}})x \\ (1+{{a}^{2}})x & 1+{{b}^{2}}x & (1+{{c}^{2}})x \\ (1+{{a}^{2}})x & (1+{{b}^{2}})x & 1+{{c}^{2}}x \\ \end{matrix} \right|\] Applying\[{{C}_{1}}\to {{C}_{1}}+{{C}_{2}}+{{C}_{3}}\] \[=\left| \begin{matrix} 1+{{a}^{2}}x+x+{{b}^{2}}x+x+{{c}^{2}}x & (1+{{b}^{2}})x & (1+{{c}^{2}})x \\ x+{{a}^{2}}x+1+{{b}^{2}}x+x+{{c}^{2}}x & 1+{{b}^{2}}x & (1+{{c}^{2}})x \\ x+{{a}^{2}}x+x+{{b}^{2}}x+1+{{c}^{2}}x & (1+{{b}^{2}})x & 1+{{c}^{2}}x \\ \end{matrix} \right|\] \[=\left| \begin{matrix} 0 & (1+{{b}^{2}})x & (1+{{c}^{2}})x \\ 1 & 1+{{b}^{2}}x & (1+{{c}^{2}})x \\ 1 & (1+{{b}^{2}})x & 1+{{c}^{2}}x \\ \end{matrix} \right|\] \[[\because {{a}^{2}}+{{b}^{2}}+{{c}^{2}}=-2\] (given)] Applying \[{{R}_{1}}\to {{R}_{1}}-{{R}_{3}},{{R}_{2}}\to {{R}_{2}}-{{R}_{3}}\] \[=\left| \begin{matrix} 0 & 0 & x-1 \\ 0 & 1-x & x-1 \\ 1 & (1+{{b}^{2}})x & 1+{{c}^{2}}x \\ \end{matrix} \right|\] \[=1[0-(x-1)(1-x)]={{(x-1)}^{2}}\] Hence, power of\[f(x)\]is 2.
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