1. Given $a$ is a positive real number, and $f(x)=\frac{1}{a}-\frac{1}{a^{x}+1}$ is an odd function, then the range of $f(x)$ is
Solve for $\left(-\frac{1}{2}, \frac{1}{2}\right)$.
Analysis From the fact that $f(x)$ is an odd function, we know $\frac{1}{a}-\frac{1}{a^{x}+1}=-\frac{1}{a}+\frac{1}{a^{-x}+1}$, solving this yields $a=2$, thus $f(x)=\frac{1}{2}-\frac{1}{2^{x}+1}$. From this, we get the range of $f(x)$ as $\left(-\frac{1}{2}, \frac{1}{2}\right)$.