$2 \cdot 90$ Given the equation $|x-2 n|=k \sqrt{x} \ (n \in N)$ has two distinct real roots in the interval $(2 n-1,2 n+1]$. Then the range of $k$ is
(A) $k>0$.
(B) $0<k \leqslant \frac{1}{\sqrt{2 n+1}}$.
(C) $\frac{1}{2 n+1}<k \leqslant \frac{1}{\sqrt{2 n+1}}$.
(D) None of the above.
(China High School Mathematics League, 1995)
［Solution］From the given, it is obvious that $k \geqslant 0$.
If $k=0$, then $x=2 n$, and the original equation has only one root, so $k>0$.
Also, from $(x-2 n)^{2}=k^{2} x$, we know that the parabola $y=(x-2 n)^{2}$ intersects the line $y=k^{2} x$ at two distinct points in the interval $(2 n-1,2 n+1]$,
Therefore, when $x=2 n-1$, we have $(x-2 n)^{2}>k^{2} x$, and when $x=2 n+1$, we have $(x-2 n)^{2} \geqslant k^{2} x$.
Thus, $k^{2}(2 n+1) \leqslant 1$, which means $k \leqslant \frac{1}{\sqrt{2 n+1}}$. Hence, the answer is $(B)$.