$5 \cdot 33$ On the coordinate plane, there are two regions $M$ and $N. M$ is determined by the three inequalities $y \geqslant 0, y \leqslant x$ and $y \leqslant 2-x$. $N$ is a region that varies with $t$, determined by the inequality $t \leqslant x \leqslant t+1$, where the range of $t$ is $0 \leqslant t \leqslant 1$. Let the common area of $M$ and $N$ be the function $f(t)$, then $f(t)$ is
(A) $-t^{2}+t+\frac{1}{2}$.
(B) $-2 t^{2}+2 t$.
(C) $1-\frac{1}{2} t^{2}$.
(D) $\frac{1}{2}(t-2)^{2}$.
(China High School Mathematics League, 1981)
[Solution] According to the problem, the common part of $M$ and $N$ is the shaded area in the right figure, and its area is
$$
\begin{aligned}
f(t) & =1-\frac{1}{2} t^{2}-\frac{1}{2}(1-t)^{2} \\
& =-t^{2}+t+\frac{1}{2} \quad(0 \leqslant t \leqslant 1) .
\end{aligned}
$$

Therefore, the answer is $(A)$.