11. (14th "Hope Cup" Senior High School Competition) The maximum value of the function $y=\sin x+\cos x+\frac{1}{\sqrt{1+|\sin 2 x|}}$ is $\qquad$ , and the minimum value is $\qquad$ .
11. It is known that $y=\sin x+\cos x+\frac{1}{\sqrt{1+|\sin 2 x|}}$ can reach its maximum value on $\left[0, \frac{\pi}{2}\right]$, at this time
$$
\begin{array}{l}
1 \leqslant \sin x+\cos x=\sqrt{2} \sin \left(x+\frac{\pi}{4}\right) \leqslant \sqrt{2}, \\
1 \leqslant \sqrt{1+|\sin 2 x|}=\sin x+\cos x \leqslant \sqrt{2} .
\end{array}
$$

If $y=x+\frac{1}{x}$, then $y^{\prime}=1-\frac{1}{x^{2}}$,
when $x>1$, $y^{\prime}=1-\frac{1}{x^{2}}>0$, i.e., $y=x+\frac{1}{x}$ is increasing on $(1,+\infty)$.
Also, $1+\frac{1}{1}=2<\sqrt{2}+\frac{1}{\sqrt{2}}$,
so the maximum value of $y=\sin x+\cos x+\frac{1}{\sqrt{1+|\sin 2 x|}}$ is $\sqrt{2}+\frac{1}{\sqrt{2}}=\frac{3 \sqrt{2}}{2}$, at this time $x=\frac{\pi}{4}$.
Next, find the minimum value of $y=\sin x+\cos x+\frac{1}{\sqrt{1+|\sin 2 x|}}$:
Since the minimum value of $\sin x+\cos x$ is $-\sqrt{2}$.
The minimum value of $\frac{1}{\sqrt{1+|\sin 2 x|}}$ is $\frac{1}{\sqrt{2}}$. Also, when $x=\frac{5}{4} \pi$, $\sin x+\cos x$ can achieve its minimum value $-\sqrt{2}$, and $\frac{1}{\sqrt{1+|\sin 2 x|}}$ can achieve its minimum value $\frac{1}{\sqrt{2}}$. Therefore, the minimum value of $y=\sin x+\cos x+\frac{1}{\sqrt{1+|\sin 2 x|}}$ is $-\sqrt{2}+\frac{1}{\sqrt{2}}=-\frac{\sqrt{2}}{2}$.