26-40 Given the sets $M=\left\{(x, y) \mid y \geqslant x^{2}\right\}, N=\left\{(x, y) \mid x^{2}+\right.$ $\left.(y-a)^{2} \leqslant 1\right\}$. Then, the necessary and sufficient condition for $M \cap N=N$ to hold is
(A) $a \geqslant 1 \frac{1}{4}$.
(B) $a=1 \frac{1}{4}$.
(C) $a \geqslant 1$.
(D) $0<a<1$.
(China High School Mathematics League, 1983)
[Solution] The necessary and sufficient condition for $M \cap N=N$ is $N \subset M$. Since set $M$ is the interior (including the parabola $y=x^{2}$) of the parabola shown in the right figure, and $N$ is the circle with center $(0, a)$ and radius 1, when $a$ is a sufficiently large positive number, it is obvious that $N \subset M$, thus negating $(B)$ and $(D)$.
When $a=1$, it is easy to find that the parabola $y=x^{2}$ and the circle $x^{2}+(y-1)^{2}=1$ have three intersection points $(0,0)$ and $(\pm 1,1)$. Therefore, $N \not \subset M$, negating $(C)$.
Thus, the correct choice is $(A)$.
Note: The rigorous proof of the correctness of $(A)$ is as follows:
From $x^{2}+(y-a)^{2} \leqslant 1$, we get
$$
x^{2} \leqslant y-y^{2}+(2 a-1) y+\left(1-a^{2}\right).
$$

Therefore, when
$$
-y^{2}+(2 a-1) y+\left(1-a^{2}\right) \leqslant 0
$$

it must be that $y \geqslant x^{2}$, i.e., $N \subseteq M$. The condition for (1) to hold is
$$
y_{\max }=\frac{-4\left(1-a^{2}\right)-(2 a-1)^{2}}{-4} \leqslant 0,
$$

i.e.,
$$
4\left(1-a^{2}\right)+(2 a-1)^{2} \leqslant 0 \text{ or } a \geqslant 1 \frac{1}{4}.
$$

Below, we prove that for $a<1 \frac{1}{4}$, we get $1-2 a<0, a^{2}-1 \geqslant 0$, so equation (2) has two distinct non-negative roots.
Thus, $y=x^{2}$ and $x^{2}+(y-a)^{2}=1$ have three or four intersection points. Hence, $N \not \subset M$.