Question
If \[g\left( x \right) = \left| {\begin{array}{*{20}{c}}
{{a^{ - x}}}&{{e^{x{{\log }_e}a}}}&{{x^2}}\\
{{a^{ - 3x}}}&{{e^{3x{{\log }_e}a}}}&{{x^4}}\\
{{a^{ - 5x}}}&{{e^{5x{{\log }_e}a}}}&1
\end{array}} \right|,\] then
A.
$$g\left( x \right) + g\left( { - x} \right) = 0$$
B.
$$g\left( x \right) - g\left( { - x} \right) = 0$$
C.
$$g\left( x \right) \times g\left( { - x} \right) = 0$$
D.
None of these
Answer :
$$g\left( x \right) + g\left( { - x} \right) = 0$$
Solution :
\[\begin{array}{l}
g\left( x \right) = \left| {\begin{array}{*{20}{c}}
{{a^{ - x}}}&{{e^{x{{\log }_e}a}}}&{{x^2}}\\
{{a^{ - 3x}}}&{{e^{3x{{\log }_e}a}}}&{{x^4}}\\
{{a^{ - 5x}}}&{{e^{5x{{\log }_e}a}}}&1
\end{array}} \right|\\
= \left| {\begin{array}{*{20}{c}}
{{a^{ - x}}}&{{e^x}}&{{x^2}}\\
{{a^{ - 3x}}}&{{e^{3x}}}&{{x^4}}\\
{{a^{ - 5x}}}&{{e^{5x}}}&1
\end{array}} \right|\left( {{e^{\log {a^x}}} = {a^x}} \right)\\
\Rightarrow g\left( { - x} \right) = \left| {\begin{array}{*{20}{c}}
{{a^x}}&{{a^{ - x}}}&{{x^2}}\\
{{a^{3x}}}&{{a^{ - 3x}}}&{{x^4}}\\
{{a^{5x}}}&{{a^{ - 5x}}}&1
\end{array}} \right|\\
= - \left| {\begin{array}{*{20}{c}}
{{a^{ - x}}}&{{a^x}}&{{x^4}}\\
{{a^{ - 3x}}}&{{a^{3x}}}&{{x^4}}\\
{{a^{ - 5x}}}&{{a^{5x}}}&1
\end{array}} \right|\left( \begin{array}{l}
{\rm{Interchanging }}\,\,I\\
{\rm{and }}\,\,II\,\,{\rm{ columns}}
\end{array} \right)
\end{array}\]
$$\eqalign{
& = - g\left( x \right) \cr
& \Rightarrow g\left( x \right) + g\left( { - x} \right) = 0 \cr} $$