Question 37: Let $\mathrm{a}$ be a real number, and the function $\mathrm{f}(\mathrm{x})=\mathrm{a} \sqrt{1-\mathrm{x}^{2}}+\sqrt{1-\mathrm{x}}+\sqrt{1+\mathrm{x}}$ has a maximum value of $\mathrm{g}(\mathrm{a})$.
(1) Let $t=\sqrt{1-x}+\sqrt{1+x}$, find the range of $t$, and express $f(x)$ as a function $m(t)$ of $t$;
(2) Find $g(a)$;
(3) Find all real numbers $a$ that satisfy $g(a)=g\left(\frac{1}{a}\right)$.
Question 37, Solution: (1) Clearly $t \geq 0$, and $t^{2}=2+2 \sqrt{1-x^{2}} \in[2,4]$, so $t \in[\sqrt{2}, 2]$, and $f(x)=a \cdot \frac{t^{2}-2}{2}+t=\frac{a t^{2}}{2}+t-a$
(2) Let $\mathrm{h}(\mathrm{t})=\frac{\mathrm{at}^{2}}{2}+\mathrm{t}-\mathrm{a}$. According to the conclusion in (1), we have:
$$
g(a)=\max _{t \in[\sqrt{2}, 2]} h(t)
$$

When $a=0$, it is clear that $g(a)=h(2)=2$.
When $a>0$, $h(t)$ is decreasing on $\left(-\infty,-\frac{1}{a}\right)$ and increasing on $\left[-\frac{1}{a},+\infty\right)$; when $a>0$, $g(a)=h(2)=a+2$;
When $-\frac{1}{2} \leq a<0$, $g(a)=h(2)=a+2$;
When $-\frac{\sqrt{2}}{2}<a<-\frac{1}{2}$, $g(a)=h\left(-\frac{1}{a}\right)=-\frac{1}{2 a}-a$;
When $a \leq-\frac{\sqrt{2}}{2}$, $g(a)=h(\sqrt{2})=\sqrt{2}$.
In summary,
$$
g(a)=\left\{\begin{array}{c}
a+2, \text { when } a \geq-\frac{1}{2} \\
-\frac{1}{2 a}-a, \text { when }-\frac{\sqrt{2}}{2}<a<-\frac{1}{2} \\
\sqrt{2}, \text { when } a \leq-\frac{\sqrt{2}}{2}
\end{array}\right.
$$
(3) According to the conclusion in (2), $g(a)$ is a monotonically increasing function on $\left[-\frac{\sqrt{2}}{2},+\infty\right)$ and a constant function $\sqrt{2}$ on $\left(-\infty,-\frac{\sqrt{2}}{2}\right)$. Therefore,
$$
g(a)=g\left(\frac{1}{a}\right) \Leftrightarrow a=\frac{1}{a} \text { or }\left\{\begin{array}{l}
a \leq-\frac{\sqrt{2}}{2} \\
\frac{1}{a} \leq-\frac{\sqrt{2}}{2}
\end{array} \Leftrightarrow a=1 \text { or } a \in\left[-\sqrt{2},-\frac{\sqrt{2}}{2}\right]\right.
$$

Therefore, the solution set for $a$ is $\{1\} \cup\left[-\sqrt{2},-\frac{\sqrt{2}}{2}\right]$.