5. If the function $f(x)=x^{3}-6 b x+3 b$ has a local minimum in $(0,1)$, then the range of the real number $b$ is ( ).
(A) $(0,1)$
(B) $(-\infty, 1)$
(C) $(0,+\infty)$
(D) $\left(0, \frac{1}{2}\right)$
5.D.

From the problem, we know $f^{\prime}(0)0$, which means
$$
-6 b0 \Rightarrow 0<b<\frac{1}{2} \text {. }
$$