4. 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 $k$ 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
(B)
4. 【Analysis and Solution】Obviously, $k \geqslant 0$, and $k=0$ leads to $x=2 n$, so the original equation has only one root, hence $k>0$. 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 different 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 $k$ $\leqslant \frac{1}{\sqrt{2 n+1}}$. Therefore, the answer is (B).