4. The equation $\left|x^{4}-4\right|-a-1=0$ has 4 solutions
(A) $a=0$
(B) $-1<a \leqslant 0$
(C) $-1<a<3$
(D) $a \geq 3$
4. (C).

From $\left|x^{4}-4\right|=a+1$, we have $a+1 \geqslant 0$, which means $a \geqslant-1$. At this point, $x^{4}=4 \pm(a+1)$. For the equation to have 4 real roots, we also need $a+1 \neq 0$ and $4-(a+1)>0$, which means $-1<a<3$. Therefore, the answer is (C).