4. Define the operation $\otimes$ on $\mathbf{R}$ as $x \otimes y=x(1-y)$. If the inequality $(x-a) \otimes(x+a)<1$ holds for all real numbers $x$, then
A. $-1<a<1$
B. $0<a<2$
C. $-\frac{1}{2}<a<\frac{3}{2}$
D. $-\frac{3}{2}<a<\frac{1}{2}$
4. $\mathrm{C}$ According to the definition of the new operation $\otimes$, $(x-a) \otimes(x+a)=(x-a)(1-x-a)$, so $(x-a) \otimes(x+a)0$ holds for any real number $x$, thus the problem is converted into a quadratic inequality that always holds. By its necessary and sufficient condition, we get $1-4\left(-a^{2}+a+1\right)<$ 0, hence solving yields $-\frac{1}{2}<a<\frac{3}{2}$, so the answer is $\mathrm{C}$.