1. Tardigrade
  2. Question
  3. Mathematics
  4. If g(x)=|a-x ex log e a x2 a-3 x e3 x log e a x4 a-5 x e5 x log e a 1|, then

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Q. If $g(x)=\begin{vmatrix}a^{-x} & e^{x \log _{e} a} & x^{2} \\ a^{-3 x} & e^{3 x \log _{e} a} & x^{4} \\ a^{-5 x} & e^{5 x \log _{e} a} & 1\end{vmatrix}$, then

Determinants

Solution:

$g(x)=\begin{vmatrix}a^{-x} & e^{\log _{e} a^{x}} & x^{2} \\ a^{-3 x} & e^{\log _{e} a^{3 x}} & x^{4} \\ a^{-5 x} & e^{\log _{e} a^{5 x}} & 1\end{vmatrix}=\begin{vmatrix}a^{-x} & a^{x} & x^{2} \\ a^{-3 x} & a^{3 x} & x^{4} \\ a^{-5 x} & a^{5 x} & 1\end{vmatrix}\left(e^{\log a^{x}=a^{x}}\right)$
$\Rightarrow g(-x)=\begin{vmatrix}a^{x} & a^{-x} & x^{2} \\ a^{3 x} & a^{-3 x} & x^{4} \\ a^{5 x} & a^{-5 x} & 1\end{vmatrix}$
$=\begin{vmatrix}a^{-x} & a^{x} & x^{2} \\ a^{-3 x} & a^{3 x} & x^{4} \\ a^{-5 x} & a^{5 x} & 1\end{vmatrix}$
(Interchanging I and II columns )
$=-g(x)$
$\Rightarrow g(x)+g(-x)=0$