$\therefore$ The function $y=\sqrt{x-m}$ and its inverse function have common points, then the range of values for $m$ is
(A) $m \geqslant \frac{1}{4}$
(B) $m \leqslant \frac{1}{4}$
(C) $m \geqslant 0$
(D) $m \leqslant 0$
4. (B).
From the problem, the original function is $y^{2}=x \cdots m$, and its inverse function is $x^{2}=y-m$. From $(x-y)(x+y-1)=$ 0, we get $x-y$ or $x+y-1=0$.
When $x=y$, if $y^{2}-y+m=0$ has real solutions, then $\Delta_{1}=(-1)^{2} \cdot 4 m \geqslant 0$, which gives $m \leqslant \frac{1}{4}$. At this time, the equation has non-negative real solutions not less than $m$, so there are common points.
When $x+y-1=0$, $y^{2}+y+1+m=0$ has real solutions, then $\Delta_{2}=1^{2}-4(m+1) \geqslant 0$, which gives $m \leqslant-\frac{3}{4}$.
According to the problem, when $m \leqslant \frac{1}{4}$, the function $y=\sqrt{x-m}$ and its inverse function have common points.