$33 \cdot 31$ Let $[x]$ denote the greatest integer not exceeding $x$. If $M=\sqrt{[\sqrt{x}]}$ and $N=[\sqrt{\sqrt{x}}]$ (where $x \geqslant 1$). Then it must be true that
(A) $M>N$.
(B) $M=N$.
(C) $M<N$.
(D) None of the above. (Beijing High School Mathematics Competition, 1985)
[Solution] Let $x=1$, then $M=1, N=1$, so $M=N$. Therefore, options $(A),(C)$ are eliminated.
Let $x=4$, then $M=\sqrt{2}, N=[\sqrt{2}]=1, M>N$. Hence, option $(B)$ is eliminated. For any $x \geqslant 1$, there exists a natural number $a$ such that $a^{4} \leqslant x<(a+1)^{4}$. Therefore,
$$
a^{2} \leqslant \sqrt{x}<(a+1)^{2},
$$

From the above, on one hand, we get $a \leqslant \sqrt{\sqrt{x}}<a+1$, so $N=[\sqrt{\sqrt{x}}]=a$; on the other hand, $a^{2} \leqslant[\sqrt{x}]<(a+1)^{2}$, so $a \leqslant \sqrt{[\sqrt{x}]}<a+1$. It is clear that $[M]=[\sqrt{[x]}]=a$, i.e., $[M]=N$, so $M \geqslant N$. Therefore, the correct choice is $(D)$.