Question
If $${a^2} + {b^2} + {c^2} = - 2\,{\text{and}}$$ \[f\left( x \right) = \left| \begin{array}{l}
\,\,\,1 + {a^2}x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {1 + {b^2}} \right)x\,\,\,\,\,\,\,\left( {1 + {c^2}} \right)x\\
\left( {1 + {a^2}} \right)x\,\,\,\,\,\,\,\,\,\,\,\,1 + {b^2}x\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {1 + {c^2}} \right)x\\
\left( {1 + {a^2}} \right)x\,\,\,\,\,\,\,\,\,\left( {1 + {b^2}} \right)x\,\,\,\,\,\,\,\,\,\,\,1 + {c^2}x
\end{array} \right|,\] then $$f\left( x \right)$$ is a polynomial of degree
A.
1
B.
0
C.
3
D.
2
Answer :
2
Solution :
Applying, $${C_1} \to {C_1} + {C_2} + {C_3}\,\,{\text{we get}}$$
\[f\left( x \right) = \left| \begin{array}{l}
1 + \left( {{a^2} + {b^2} + {c^2} + 2} \right)x\,\,\,\,\,\,\,\,\,\,\left( {1 + {b^2}} \right)x\,\,\,\,\,\,\,\,\left( {1 + {c^2}} \right)x\\
1 + \left( {{a^2} + {b^2} + {c^2} + 2} \right)x\,\,\,\,\,\,\,\,\,\,\,1 + {b^2}x\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {1 + {c^2}} \right)x\\
1 + \left( {{a^2} + {b^2} + {c^2} + 2} \right)x\,\,\,\,\,\,\,\,\,\,\left( {1 + {b^2}} \right)x\,\,\,\,\,\,\,\,\,\,\,\,1 + {c^2}x
\end{array} \right|\]
\[\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left| \begin{array}{l}
1\,\,\,\,\,\,\,\left( {1 + {b^2}} \right)x\,\,\,\,\,\,\,\,\,\left( {1 + {c^2}} \right)x\\
1\,\,\,\,\,\,\,\,\,1 + {b^2}x\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {1 + {c^2}} \right)x\\
1\,\,\,\,\,\,\,\left( {1 + {b^2}} \right)x\,\,\,\,\,\,\,\,\,\,\,\,1 + {c^2}x
\end{array} \right|\]
$$\eqalign{
& \left[ {{\text{As given that }}{a^2} + {b^2} + {c^2} = - 2} \right] \cr
& \therefore \,\,{a^2} + {b^2} + {c^2} + 2 = 0 \cr
& {\text{Applying }}{R_1} \to {R_1} - {R_2},\,\,\,{R_2} \to {R_2} - {R_3} \cr} $$
\[\therefore \,\,f\left( x \right) = \left| \begin{array}{l}
0\,\,\,\,\,\,\,\,\,x - 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\\
0\,\,\,\,\,\,\,\,\,1 - x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x - 1\\
1\,\,\,\,\,\,\left( {1 + {b^2}} \right)x\,\,\,\,\,\,\,\,\,1 + {c^2}x
\end{array} \right|\]
$$f\left( x \right) = {\left( {x - 1} \right)^2}\,\,\,\,\,\,\,{\text{Hence degree}} = 2.$$