Question 47, Find the maximum value of the function $f(x)=\frac{\sqrt{2} \sin x+\cos x}{\sin x+\sqrt{1-\sin x}}(0 \leq x \leq \pi)$.
Question 47, Answer: Since $(1-\sin x)^{2} \geq 0 \Rightarrow \cos x \leq \sqrt{2} \cdot \sqrt{1-\sin x}$, therefore $f(x)=$ $\frac{\sqrt{2} \sin x+\cos x}{\sin x+\sqrt{1-\sin x}} \leq \frac{\sqrt{2} \sin x+\sqrt{2} \cdot \sqrt{1-\sin x}}{\sin x+\sqrt{1-\sin x}}=\sqrt{2}$, so the maximum value of $f(x)=\frac{\sqrt{2} \sin x+\cos x}{\sin x+\sqrt{1-\sin x}}(0 \leq x \leq \pi)$ is $\sqrt{2}$, which is achieved when $x=\frac{\pi}{2}$.