Question

If the graph of $f\left(x\right)=\frac{a^x-1}{x^a(a^x+1)}$ is symmetrical about $y$ axis, then $n=$

• A. $2$
• B. $\frac{2}{3}$
• C. $\frac{1}{4}$
• D. $\frac{-1}{3}$

Answer 1

Answer: (D)

$\frac{-1}{3}$

Given

$f\left(x\right)=\frac{a^x-1}{x^n(a^x+1)}$ is symmetrical about the y-axis.

such functions where $f(-x)=f\left(x\right)$ are termed as even functions.

Hence $f\left(x\right)$ is an even function.

$f(-x)=\frac{a^{-x}-1}{{(-x)}^n(a^{-x}+1)}-\frac{1-a^x}{{(-1)}^n{x}^n(1+a^x)}$

Since it is an even function we have

$f\left(x\right)=f(-x)\Rightarrow \frac{a^x-1}{x^n(a^x+1)}=\frac{1-a^x}{{(-1)}^n{x}^n(1+a^x)}\Rightarrow (a^x-1)=\frac{-(a^x-1)}{{(-1)}^n}\Rightarrow {(-1)}^n=(-1)$

this is possible only when $n$ is odd.

$\therefore$ for $\left(b\right)\frac{2}{3}$

then ${(-1)}^{\frac{2}{3}}={\left(1\right)}^{\frac{1}{3}}=1\ne -1$ hence this too is not possible .

$\therefore \left(d\right)\frac{-1}{3}$