A) \[g(x)+g(-x)=0\]
B) \[g(x)-g(-x)=0\]
C) \[g(x)\times g(-x)=0\]
D) None of these
Correct Answer: A
Solution :
[a] \[g(x)=\left| \begin{matrix} {{a}^{-x}} & {{e}^{{{\log }_{e}}{{a}^{x}}}} & {{x}^{2}} \\ {{a}^{-3x}} & {{e}^{{{\log }_{e}}{{a}^{3x}}}} & {{x}^{4}} \\ {{a}^{-5x}} & {{e}^{{{\log }_{e}}{{a}^{5x}}}} & 1 \\ \end{matrix} \right|\] \[=\left| \begin{matrix} {{a}^{-x}} & {{e}^{x}} & {{x}^{2}} \\ {{a}^{-3x}} & {{e}^{3x}} & {{x}^{4}} \\ {{a}^{-5x}} & {{e}^{5x}} & 1 \\ \end{matrix} \right|\left( {{e}^{\log \,\,{{a}^{x}}}}={{a}^{x}} \right)\] \[\Rightarrow g(-x)=\left| \begin{matrix} {{a}^{x}} & {{a}^{-x}} & {{x}^{2}} \\ {{a}^{3x}} & {{a}^{-3x}} & {{x}^{4}} \\ {{a}^{5x}} & {{a}^{-5x}} & 1 \\ \end{matrix} \right|\] \[=-\left| \begin{matrix} {{a}^{-}}^{x} & {{a}^{x}} & {{x}^{2}} \\ {{a}^{-3x}} & {{a}^{3x}} & {{x}^{4}} \\ {{a}^{-5x}} & {{a}^{5x}} & 1 \\ \end{matrix} \right|\left( \begin{align} & \text{Interchaning}\,\, \\ & \text{and}\,\,\text{II}\,\,\text{columns} \\ \end{align} \right)\] \[=-g(x)\] \[\Rightarrow g(x)+g(-x)=0\]
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