1. Let the function $f(x)=\sqrt{1-x^{2}}\left(|x-2|+\frac{4}{2^{x}-1}\right)$.

Consider the proposition $p: f(x)$ is an odd function; proposition $q: f(x)$ is an even function. Then the correct conclusion is ( ).
(A) $p 、 q$
(B) $p \backsim \neg q$
(C) $\neg p 、 q$
(D) $\neg p \vee \neg q$
- 1. B.

From the domain of $f(x)$ being $[-1,0) \cup(0,1]$, we know
$$
\begin{array}{l}
f(x)=\sqrt{1-x^{2}}\left(2-x+\frac{4}{2^{x}-1}\right) \\
=\sqrt{1-x^{2}}\left(-x+2 \cdot \frac{2^{x}+1}{2^{x}-1}\right)
\end{array}
$$

is an odd function, not an even function.