Example 3 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 values for $k$ 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
(1995, National High School Mathematics Competition)
Analysis: Let $y_{1}=\mid x - 2 n \mid, y_{2}=k \sqrt{x}$. Then the original problem is transformed into: the two curves intersect at two different points in $x \in (2 n-1, 2 n+1]$ for the range of $k$. As shown in the figure, $k$ must and only needs to satisfy $0 < k \sqrt{2 n+1} \leqslant |(2 n+1) - n|$. 
Therefore, we have $0 < k \leqslant \frac{1}{\sqrt{2 n+1}}$.