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