绝对值函数最大值求a

v6mm131

若 $a>0, h(x)=\left|2 \sqrt{2-x^2}+|x-a|-4 \sqrt{2}\right|$ 的最大值为 $2 \sqrt{2}$ ，则 $a=$

Aluminiumor

$h(x)$ 的定义域为 $[-\sqrt{2},\sqrt{2}]$
$$h(\sqrt{2})=\big||a-\sqrt{2}|-4\sqrt{2}\big|\leq2\sqrt{2}\Longrightarrow3\sqrt{2}\leq a\leq7\sqrt{2}$$
故
$$h(x)=|x-2\sqrt{2-x^2}+4\sqrt{2}-a|$$
令
$$f(x)=x-2\sqrt{2-x^2},f'(x)=1+\frac{2x}{\sqrt{2-x^2}}$$
故 $f(x)$ 在 $(-\sqrt{2},-\frac{\sqrt{10}}{5})\downarrow$ ，在 $(-\frac{\sqrt{10}}{5},\sqrt{2})\uparrow$
而
$$f(-\sqrt{2})=-\sqrt{2},f(-\frac{\sqrt{10}}{5})=-\sqrt{10},f(\sqrt{2})=\sqrt{2}$$
要使 $h(x)_{\max}=2\sqrt{2}$，则 $4\sqrt{2}-a=\sqrt{2}$ 或 $-2\sqrt{2}+\sqrt{10}$
故 $a=3\sqrt{2}$ 或 $6\sqrt{2}-\sqrt{10}$