Question

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

• A. Graph of g(x) is symmetrical about origin.
• B. Graph of g(x) is symmetrical about y - axis.
• C. ${\frac{d^{4}g\left(x\right)}{d{x}^4}|}_{x-0}=0$
• D. $f\left(x\right)=g\left(x\right)\times log\left(\frac{a-x}{a+x}\right)$ is an odd function.

Answer 1

Answer: (A), (C)

The correct options are
A Graph of g(x) is symmetrical about origin.
C ${\frac{d^{4}g\left(x\right)}{d{x}^4}|}_{x-0}=0$

$g\left(x\right)=\left|\begin{array}{ccc}{a}^{-x}& a^x& x^2\\ a^{-3x}& a^{3x}& x^4\\ a^{-5x}& a^{5x}& 1\end{array}\right|g(-x)=\left|\begin{array}{ccc}{a}^x& a^{-x}& x^2\\ a^{3x}& a^{-3x}& x^4\\ a^{5x}& a^{-5x}& 1\end{array}\right|$
g(x) + g(-x) = 0
Hence odd function
Hence options A, C are correct.