$7 \cdot 87$ Let $f(x)$ be a periodic function with period 2 defined on the set of real numbers, and it is an even function. It is known that when $x \in [2,3]$, $f(x) = x$. Then, when $x \in [-2,0]$, the analytical expression of $f(x)$ is
(A) $f(x) = x + 4$.
(B) $f(x) = 2 - x$.
(C) $f(x) = 3 - |x + 1|$.
(D) $f(x) = 2 + |x + 1|$.
(China High School Mathematics League, 1990)
[Solution 1] (1) Given $f(x)=x, 2 \leqslant x \leqslant 3$, and the period is 2, we have
$$
f(x)=\left\{\begin{array}{l}
x+4, -2 \leqslant x \leqslant -1 ; \\
x+2, 0 \leqslant x < 1 .
\end{array}\right.
$$
(2) Since $f(x)$ is an even function, we get
$$
f(x)=-x+2, -1 \leqslant x \leqslant 0 .
$$

Combining (1) and (2), we have
$$
f(x)=\left\{\begin{array}{l}
x+4, -2 \leqslant x \leqslant -1 \\
-x+2, -1 \leqslant x \leqslant 0 .
\end{array}\right.
$$

That is
$$
f(x)=3-|x+1| \text{. }
$$

Therefore, the answer is $(C)$.
[Solution 2] First, draw the graph of $f(x)=x$, where $x \in [2,3]$, i.e., the line segment $AB$.
Since $f(x)$ is a function with a period of 2, we can draw the graph of $y=f(x)$ for $x \in [0,1] \cup [-2,-1]$, i.e., the line segments $CD, EF$.

Since $f(x)$ is an even function, its graph is symmetric about the $y$-axis, so the graph of $f(x)$ on $[-1,0]$ is the line segment $FC$.

Thus, the graph of $f(x)$ on $[-2,0]$ is the broken line $EFC$.
It is easy to see that its analytical expression is
$$
f(x)=3-|x+1| .
$$

Therefore, the answer is $(C)$.