Question

If the real valued function $f\left(x\right)=\frac{a^x-1}{x^n(a^x+1)}$ is even, then $n$ is equal to

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

Answer 1

Answer: (D)

$1$

$f\left(x\right)=\frac{a^x-1}{x^n(a^x+1)}$ is even
$\therefore f(-x)=f\left(x\right)$
i.e., $\frac{a^{-x}-1}{(-x{)}^n(a^{-x}+1)}=\frac{a^x-1}{x^n(a^x+1)}$
$\Rightarrow \frac{-1}{-1^n}=1\Rightarrow (-1{)}^n=(-1)$
Hence, $n=1$ can satisfy this equation.