1. Given the function
$$
f(x)=\left|8 x^{3}-12 x-a\right|+a
$$

has a maximum value of 0 on the interval $[0,1]$. Then the maximum value of the real number $a$ is $\qquad$
$-1 .-2 \sqrt{2}$.
Let $g(x)=8 x^{3}-12 x$. Then $g^{\prime}(x)=24 x^{2}-12$.
Thus, $g(x)_{\min }=g\left(\frac{\sqrt{2}}{2}\right)=-4 \sqrt{2}$, $g(x)_{\max }=\max \{g(0), g(1)\}=0$.
Therefore, $\max _{1 \leqslant x \leqslant 2}\{f(x)\}$
$$
\begin{array}{l}
=\max \{|-4 \sqrt{2}-a|+a,|0-a|+a\}=0 \\
\Rightarrow\left\{\begin{array}{l}
a \leqslant 0, \\
|0-a| \leqslant-a \\
|-4 \sqrt{2}-a| \leqslant-a
\end{array}\right. \\
\Rightarrow a \leqslant-2 \sqrt{2} \Rightarrow a_{\max }=-2 \sqrt{2}
\end{array}
$$