Question

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

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

Answer 1

The correct option is C $[0,\frac{3}{\sqrt{2}}]$

Given function is $y=3\sin \sqrt{\frac{{\pi }^2}{16}-x^2}$
Here, domain is $[0,\frac{\pi }{4}]$

When, $x=0$

$y=3\sin \left(\sqrt{\frac{{\pi }^2}{16}-0}\right)$

$=3\sin \left(\frac{\pi }{4}\right)=3\times \frac{1}{\sqrt{2}}=\frac{3}{\sqrt{2}}$

When $x=\frac{\pi }{4},y=3\sin \left(\sqrt{\frac{{\pi }^2}{16}-\frac{{\pi }^2}{16}}\right)$

$=3\sin \left(0\right)=0$

Hence, range of given function is $[0,\frac{3}{\sqrt{2}}]$.