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. $3$

Answer 1

The correct option is D $3$

$f\left(x\right)=\frac{a^x-1}{x^n(a^x+1)}$
$\Rightarrow f(-x)=\frac{a^{-x}-1}{(-x{)}^n(a^{-x}+1)}$
$=\frac{1-a^x}{(-1{)}^n{x}^n(a^x+1)}$
Since, f(x) is even function.
Therefore, $f\left(x\right)=f(-x)$
$\Rightarrow \frac{a^x-1}{x^n(a^x+1)}=\frac{1-a^x}{(-1{)}^n{x}^n(a^x+1)}$
$\Rightarrow (-1{)}^n=(-1)$
It satisfies, if $n$ is odd
Hence, $n=3$