A) x
B) \[{{x}^{3}}\]
C) \[14+{{x}^{2}}\]
D) \[{{x}^{5}}\]
Correct Answer: C
Solution :
\[\left| \,\begin{matrix} 4+{{x}^{2}} & -6 & -2 \\ -6 & 9+{{x}^{2}} & 3 \\ -2 & 3 & 1+{{x}^{2}} \\ \end{matrix} \right|={{x}^{4}}(14+{{x}^{2}})\]\[=x.{{x}^{3}}(14+{{x}^{2}})\] Hence, the determinant is divisible by x,\[{{x}^{3}}\] and \[(14+{{x}^{2}})\], but not divisible by \[{{x}^{5}}\].
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