The range of the function y=3 sin left(sqrt{f | vedclass

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(C) Given function is $y=3 \sin \left(\sqrt{\frac{\pi^{2}}{16}-x^{2}}\right)$.For the function to be defined, we must have $\frac{\pi^{2}}{16}-x^{2} \

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$[0, 3/\sqrt{2}]$

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(C) Given function is $y=3 \sin \left(\sqrt{\frac{\pi^{2}}{16}-x^{2}}ight)$.For the function to be defined, we must have $\frac{\pi^{2}}{16}-x^{2} \geq 0$, which implies $x^{2} \leq \frac{\pi^{2}}{16}$, so $x \in \left[-\frac{\pi}{4}, \frac{\pi}{4}ight]$.Let $u = \sqrt{\frac{\pi^{2}}{16}-x^{2}}$. As $x$ varies from $0$ to $\frac{\pi}{4}$, $u$ varies from $\frac{\pi}{4}$ to $0$.Thus, the range of $u$ is $\left[0, \frac{\pi}{4}ight]$.Now, $y = 3 \sin(u)$. Since $u \in \left[0, \frac{\pi}{4}ight]$, $\sin(u)$ varies from $\sin(0) = 0$ to $\sin\left(\frac{\pi}{4}ight) = \frac{1}{\sqrt{2}}$.Therefore, $y$ varies from $3 	imes 0 = 0$ to $3 	imes \frac{1}{\sqrt{2}} = \frac{3}{\sqrt{2}}$.Hence, the range of the function is $\left[0, \frac{3}{\sqrt{2}}ight]$.

The range of the function $y=3 \sin \left(\sqrt{\frac{\pi^{2}}{16}-x^{2}}\right)$ is

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