2. Given the equation $|x-2 n|=k \sqrt{x}(n \in \mathbf{N})$ has two distinct real roots in the interval $(2 n-1,2 n+1]$, then the range of values for $k$ is $\qquad$
2. Solution: $00$. 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 $x=2 n+1$, $(x-2 n)^{2} \geqslant k^{2} x$, thus $k^{2}(2 n+1) \leqslant 1$, which means $k \leqslant \frac{1}{\sqrt{2 n+1}}$.