1. Given the following two propositions:

Proposition $P$ : There exist functions $f(x)$, $g(x)$, and an interval $I$, such that $f(x)$ is an increasing function on $I$, $g(x)$ is also an increasing function on $I$, but $f(g(x))$ is a decreasing function on $I$;

Proposition $Q$ : There exist an odd function $f(x)(x \in A)$ and an even function $g(x)(x \in B)$, such that the function
$$
f(x) g(x)(x \in A \cap B)
$$
is an even function.
Then, ( ).
(A) $P$ and $Q$ are both true
(B) $P$ and $Q$ are both false
(C) $P$ is true and $Q$ is false
(D) $Q$ is true and $P$ is false
$-1 . A$
Take $f(x)=1-x^{2}, g(x)=\left(\frac{1}{2}\right)^{\frac{1}{x}}$, interval $I=(-\infty, 0)$.

Then $f(x)$ is an increasing function on the interval $(-\infty, 0)$, and $g(x)$ is also an increasing function on the interval $(-\infty, 0)$.

But $f(g(x))=1-\left(\frac{1}{2}\right)^{\frac{2}{x}}$ is a decreasing function on $I$ (since $\left(\frac{1}{2}\right)^{\frac{1}{x}}>0), P$ is true.
Take $f(x)=-x(x \in \mathbf{R}), g(x)=0(x \in \mathbf{R})$.
Then $f(x) g(x)=0(x \in \mathbf{R})$ is an even function, $Q$ is true.