China Mathematical Competition

First, we will find the domain of $f(x)$. From $x^2-2x-3>0$, we obtain $x<-1$ or $x>3$. So the domain of definition for $f(x)$ is $(-\infty ,-1)\cup (3,+\infty )$.

But $u=x^2-2x-3=(x-1{)}^2-4$ is monotone decreasing on $(-\infty ,-1)$, and monotone increasing on $(3,+\infty )$.

So $f(x)={\log }_{\frac{1}{2}}(x^2-2x-3)$ is monotone increasing on $(-\infty ,-1)$, and monotone decreasing on $(3,+\infty )$.

Answer: A.