A) 3
B) 2
C) 1
D) 0
Correct Answer: B
Solution :
Applying \[{{C}_{1}}\to {{C}_{1}}+{{C}_{2}}+{{C}_{3}}\] \[f(x)=\left| \,\begin{matrix} 1 & (1+{{b}^{2}})x & (1+{{c}^{2}})x \\ 1 & 1+{{b}^{2}}x & (1+{{c}^{2}})x \\ 1 & (1+{{b}^{2}})x & 1+{{c}^{2}}x \\ \end{matrix}\, \right|\], \[(\because {{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2=0\]) [Applying \[{{R}_{2}}\to {{R}_{2}}-{{R}_{1}}\],\[{{R}_{3}}\to {{R}_{3}}-{{R}_{1}}\]] \[=\left| \,\begin{matrix} 1 & (1+{{b}^{2}})x & (1+{{c}^{2}})x \\ 0 & 1-x & 0 \\ 0 & 0 & 1-x \\ \end{matrix}\, \right|={{(1-x)}^{2}}.\] Hence degree of f(x) = 2.
You need to login to perform this action.
You will be redirected in
3 sec
