Functions satisfying f(x)f(-x)=1

Detailed Solution

Setting $x=0$ gives $f(0)^2=1$, so $f(0)=1$ or $f(0)=-1$. Also, $f(x)f(-x)=1\neq0$ for every $x$, so $f$ is never zero; since $f$ is continuous and never zero on the connected set $\mathbb R$, the Intermediate Value Theorem forces $f$ to keep a single, constant sign throughout — positive everywhere if $f(0)=1$, or negative everywhere if $f(0)=-1$ (for example, the constant function $f\equiv-1$ satisfies the equation with $f(0)=-1$ and is negative everywhere). This rules out (A), which wrongly assumes only the positive case is possible. For (C) and (D): the function $f(x)=e^x$ satisfies $f(x)f(-x)=e^x\cdot e^{-x}=1$, yet $e^x$ is neither odd ($e^x\neq-e^{-x}$) nor even ($e^x\neq e^{-x}$ in general), so both (C) and (D) fail for this example. Thus the correct option is $\boxed{\mathrm{B}}$.