Q103 / 200 · WBJEE 2014 · Mathematics
The range of the function $y=3 \sin \left(\sqrt{\frac{\pi^{2}}{16}-x^{2}}\right)$ is
- A$[0, \sqrt{3 / 2}]$
- B[0,1]
- Correct answerC$[0,3 / \sqrt{2}]$
- D$[0, \infty)$
Solution
Given function is $y=3 \sin \left(\sqrt{\frac{\pi^{2}}{16}-x^{2}}\right)$
Here, domain is $\left[0, \frac{\pi}{4}\right]$
$$
\begin{array}{l}
\begin{aligned}
x=& 0 \\
y &=3 \sin \left(\sqrt{\frac{\pi^{2}}{16}-0}\right) \\
&=3 \sin \left(\frac{\pi}{4}\right)=3 \times \frac{1}{\sqrt{2}}=\frac{3}{\sqrt{2}} \\
\text { When } x=\frac{\pi}{4}, y &=3 \sin \left(\sqrt{\frac{\pi^{2}}{16}-\frac{\pi^{2}}{16}}\right) \\
&=3 \sin (0)=0
\end{aligned}
\end{array}
$$
Hence, range of given function is $\left[0, \frac{3}{\sqrt{2}}\right]$
Hence, the correct answer is option: 'C'.