1. Given the set
$$
\begin{aligned}
M & =\left\{f(x) \mid f^{2}(x)-f^{2}(y)\right. \\
& =f(x+y) \cdot f(x-y), x, y \in \mathbf{R}\} .
\end{aligned}
$$

The following statements are given:
(1) If $f_{1}(x)=\left\{\begin{array}{cc}1, & x \geqslant 0, \\ -1, & x<0,\end{array}\right.$ then $f_{1}(x) \in$
$M$;
(2) If $f_{2}(x)=2 x$, then $f_{2}(x) \in M$;
(3) If $f_{3}(x)=\sin x$, then $f_{3}(x) \in M$;
(4) If $f_{4}(x)=\cos x$, then $f_{4}(x) \in M$;
(5) If $f(x) \in M$, then the graph of $\dot{y}=f(x)$ is symmetric about the origin;
(6) If $f(x) \in M$, then for any distinct real numbers $x_{1} 、 x_{2}$, we always have $\frac{f\left(x_{1}\right)-f\left(x_{2}\right)}{x_{1}-x_{2}}<0$;
(7) If $f(x) \in M$, then for any real numbers $x_{1}$ 、 $x_{2}$, we always have $f\left(\frac{x_{1}+x_{2}}{2}\right) \leqslant \frac{f\left(x_{1}\right)+f\left(x_{2}\right)}{2}$.
Among the correct statements, there are ( ) in total.
(A) 1
(B) 2
(C) 3
(D) 4
- 1.C.

It is easy to verify that (1) and (4) are incorrect, while (2) and (3) are correct.
Let $x=y=0$, we get $f(0)=0$;
Then let $x=0$, we have $f^{2}(0)-f^{2}(y)=f(y) f(-y)$, which simplifies to $f(y)(f(y)+f(-y))=0$.
Therefore, $f(y)=0$ or $f(y)+f(-y)=0$, so we always have
$$
f(-y)=-f(y)
$$

Thus, $f(x)$ is an odd function, and its graph is symmetric about the origin. Hence, (5) is correct.

Since $f(x)=\sin x \in M$, but $f(x)=\sin x$ is not monotonic on $\mathbf{R}$, (6) is incorrect.

Using the graph of $f(x)=\sin x$, we can see that the inequality does not hold. Hence, (7) is incorrect.