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 $g^{\prime}(x)=24 x^{2}-12$.
$$
\begin{array}{l}
\text { Thus, } g(x)_{\min }=g\left(\frac{\sqrt{2}}{2}\right)=-4 \sqrt{2}, \\
g(x)_{\max }=\max \{g(0), g(1)\}=0 . \\
\text { Therefore, } \max _{1 \leqslant x \leqslant 2}\{f(x)\} \\
=\max \{|-4 \sqrt{2}-a|+a,|0-a|+a\}=0 \\
\Rightarrow\left\{\begin{array}{l}
a \leqslant 0, \\
|0-a| \leqslant-a, \\
|-4 \sqrt{2}-a| \leqslant-a
\end{array}\right. \\
\Rightarrow a \leqslant-2 \sqrt{2} \Rightarrow a_{\max }=-2 \sqrt{2} .
\end{array}
$$