1. If there exists $x_{0} \in[0,1]$, such that $2^{x_{0}}\left(3 x_{0}+a\right)<1$,
then the range of real number $a$ is $\qquad$
$$
\text { I, 1. } a<1 \text {. }
$$

Notice that,
$$
2^{x_{0}}\left(3 x_{0}+a\right)<1 \Leftrightarrow a<\left(\frac{1}{2}\right)^{x_{0}}-3 x_{0} \text {. }
$$

Given that $f(x)=\left(\frac{1}{2}\right)^{x}-3 x$ is decreasing on $x \in[0,1]$, we know that $a<f(x)_{\text {max }}=1$.