4. 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
4. (B).

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 $k \leqslant \frac{1}{\sqrt{2 n+1}}$.