11.5. Find the set of values of the function $y=\sqrt{x}-\sqrt{2-x}+2 \sin x$.
Answer: $[-\sqrt{2} ; \sqrt{2}+2 \sin 2]$. Solution. The domain of the function $y=f(x)$ is the interval [0,2]. We will prove that on the interval $(0 ; 2)$, the derivative $f'(x)$ is positive, and thus the function $f(x)$, continuous on $[0,2]$, is monotonically increasing. We have

$$
f'(x)=\frac{1}{2 \sqrt{x}}+\frac{1}{2 \sqrt{2-x}}+2 \cos x
$$

Obviously, $f'(x)>0$ for $x \in\left(0 ; \frac{\pi}{2}\right)$, since all terms are positive. Let's check that $f'(x)>0$ also holds on the half-interval $\left[\frac{\pi}{2} ; 2\right)$. For this, in the following estimates, we will use the obvious inequalities $\sqrt{2}3$, as well as the monotonic decrease of the function $\cos x$ in the second quadrant. For all $x \in\left[\frac{\pi}{2} ; 2\right)$, we will have
$\frac{1}{2 \sqrt{x}}>\frac{1}{2 \sqrt{2}}>\frac{1}{2 \cdot 1.5}=\frac{1}{3}, \frac{1}{2 \sqrt{2-x}} \geq \frac{1}{2 \sqrt{2-\frac{\pi}{2}}}>\frac{1}{\sqrt{2}}>\frac{1}{1.5}=\frac{2}{3}$ and $\cos x>\cos 2>\cos \frac{2 \pi}{3}=-\frac{1}{2}$. Thus, $f'(x)=\frac{1}{2 \sqrt{x}}+\frac{1}{2 \sqrt{2-x}}+2 \cos x>\frac{1}{3}+\frac{2}{3}+2 \cdot\left(-\frac{1}{2}\right)=0$ for all $x \in\left[\frac{\pi}{2} ; 2\right)$. Therefore, the continuous function $f(x)$ is monotonically increasing on $[0 ; 2]$, from which the answer follows.