A) \[n\]
B) \[n-1\]
C) \[n+1\]
D) None of these
Correct Answer: C
Solution :
[c] Taking \[{{x}^{5}}\] common from last row, we get, \[{{x}^{5}}\left| \begin{matrix} {{x}^{n}} & {{x}^{n+2}} & {{x}^{2n}} \\ 1 & {{x}^{a}} & a \\ {{x}^{n}} & {{x}^{a+1}} & {{x}^{2n}} \\ \end{matrix} \right|=0\,\,\,\,x\in R,\] \[\Rightarrow a+1=n+2\Rightarrow a=n+1\] (as it will make first and third row identical)
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