Question

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

• A. $\left[0,\sqrt{3/2}\right]$
• B. $[0,1]$
• C. $\left[0,3/\sqrt{2}\right]$
• D. $[0,\infty )$

Answer 1

Answer: (C)

Given function is $y=3\sin \left(\sqrt{\frac{{\pi }^2}{16}-x^2}\right)$
Here, domain is $\left[0,\frac{\pi }{4}\right]$
When, $x=0$
$y=3\sin \left(\sqrt{\frac{{\pi }^2}{16}-0}\right)$
When $x=\frac{\pi }{4},y=3\sin \left(\sqrt{\frac{{\pi }^2}{16}-\frac{{\pi }^2}{16}}\right)$
$=3\sin \left(\frac{\pi }{4}\right)=3\times \frac{1}{\sqrt{2}}=\frac{3}{\sqrt{2}}$
$=3\sin (0)=0$
Hence, range of given function is $\left[0,\frac{3}{\sqrt{2}}\right]$.