A) \[-1,-2\]
B) \[-1,\,2\]
C) \[1,-2\]
D) \[1,\,2\]
Correct Answer: B
Solution :
\[\left| \,\begin{matrix} 1 & 4 & 20 \\ 1 & -2 & 5 \\ 1 & 2x & 5{{x}^{2}} \\ \end{matrix}\,\, \right|\,=0\] \[\Rightarrow \] \[\left| \,\begin{matrix} 0 & 6 & 15 \\ 0 & -2-2x & 5(1-{{x}^{2}}) \\ 1 & 2x & 5{{x}^{2}} \\ \end{matrix}\, \right|\,=0\] \[\left( \begin{align} & {{R}_{1}}\to {{R}_{1}}-{{R}_{2}} \\ & {{R}_{2}}\to {{R}_{2}}-{{R}_{3}} \\ \end{align} \right)\] \[\Rightarrow \]\[3\,.\,2\,.\,5.\,\left| \,\begin{matrix} 0 & 1 & 1 \\ 0 & -(1+x) & 1-{{x}^{2}} \\ 1 & x & {{x}^{2}} \\ \end{matrix}\, \right|=0\] \[\Rightarrow \] \[(1+x)\,\left| \,\begin{matrix} 0 & 1 & 1 \\ 0 & -1 & 1-x \\ 1 & x & {{x}^{2}} \\ \end{matrix}\, \right|\,=0\] \[\Rightarrow \] \[x+1=0\] or \[x-2=0\] \[\Rightarrow \] \[x=-1,\,2\]. Trick: Obviously by inspection, \[x=-1,\,2\] satisfy the equation. At \[x=-1,\,\]\[\left| \,\begin{matrix} 1 & 4 & 20 \\ 1 & -2 & 5 \\ 1 & -2 & 5 \\ \end{matrix}\, \right|\,=0\] as \[{{R}_{2}}\equiv {{R}_{3}}\] At \[x=2\], \[\left| \,\begin{matrix} 1 & 4 & 20 \\ 1 & -2 & 5 \\ 1 & 4 & 20 \\ \end{matrix}\, \right|=0\] as \[{{R}_{1}}\equiv {{R}_{3}}\].
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