Question

If $g\left(x\right)=\left|\begin{array}{ccc}{a}^{-x}& e^{x{\log }_{e}a}& x^2\\ a^{-3x}& e^{3x{\log }_{e}a}& x^4\\ a^{-5x}& e^{5x{\log }_{e}a}& 1\end{array}\right|$, then

• A. $g\left(x\right)+g(-x)=0$
• B. $g\left(x\right)-g(-x)=0$
• C. $g\left(x\right)\times g(-x)=0$
• D. None of these

Answer 1

The correct option is A $g\left(x\right)+g(-x)=0$

$e^{x\log e^x}=e^{{\log }_e{a}^x}=a^x$

Similarly for other elements in $C_2,$

$y\left(n\right)=\left|\begin{array}{ccc}{a}^{-x}& a^x& x^2\\ a^{-3x}& a^{3x}& x^4\\ a^{-bx}& a^{5x}& 1\end{array}\right|$

$g(-x)=\left|\begin{array}{ccc}{a}^x& a^{-x}& x^2\\ a^{3x}& a^{-3x}& x^4\\ a^{5x}& a^{-5x}& 1\end{array}\right|$

Interchanging $C_1$ and $C_2$, sign of $g(-x)$ would change

$\Rightarrow g(-x)=-\left|\begin{array}{ccc}{a}^{-x}& a^x& x^2\\ a^{-3x}& a^{3x}& x^4\\ a^{-5x}& a^{5x}& 1\end{array}\right|$

$\Rightarrow g(-x)=-g\left(x\right)$

$\Rightarrow g(-x)+g\left(x\right)=0$

$\Rightarrow \left(A\right)$