2. Given $0 \leqslant x \leqslant 1$, the inequality $x^{2}+a x+3-a$
$>0$ always holds. Then the range of the real number $a$ is $\qquad$
2. $\left\{\begin{array}{l}0 \leqslant -\frac{a}{2} \leqslant 1, \\ f\left(-\frac{a}{2}\right)=3-\frac{a^{2}}{4}-a>0\end{array}\right.$ or $\left\{\begin{array}{l}-\frac{a}{2}>1, \\ f(1)=4>0 .\end{array}\right.$
From this, we get $0<a<3$ or $-2 \leqslant a \leqslant 0$ or $a<-2$. In summary, we have $a<3$.