Question

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

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

Answer 1

The correct option is C

$\left[0,\frac{3}{\sqrt{2}}\right]$

Step 1: Find the domain of the given function

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

So, the term under the square root must be $\ge 0$

$\Rightarrow \frac{{\mathrm{\pi }}^2}{16}-x^2\ge 0$

$\Rightarrow x^2\le \frac{{\mathrm{\pi }}^2}{16}$

$\Rightarrow \frac{-\mathrm{\pi }}{4}\le x\le \frac{\mathrm{\pi }}{4}$

So, $x\in \left[\frac{-\mathrm{\pi }}{4},\frac{\mathrm{\pi }}{4}\right]$

Step 2: Find the range of the given function

$-\frac{\pi }{4}\le x\le \frac{\pi }{4}$

$\Rightarrow 0\le x^2\le \frac{{\mathrm{\pi }}^2}{16}$

multiply by $-1$ both the sides,

$\Rightarrow \frac{-{\mathrm{\pi }}^2}{16}\le -x^2\le 0$

Adding $\frac{{\mathrm{\pi }}^2}{16}$ both the sides

$$\begin{gathered} \Rightarrow 0\le \frac{{\pi }^2}{16}-x^2\le \frac{{\pi }^2}{16} \\ \Rightarrow 0\le \sqrt{\frac{{\pi }^2}{16}-x^2}\le \frac{\mathrm{\pi }}{4} \end{gathered}$$

Taking $\sin$ both the sides

$$\begin{gathered} \Rightarrow \sin \left(0\right)\le \sin \left(\sqrt{\frac{{\mathrm{\pi }}^2}{16}-x^2}\right)\le \sin \left(\frac{\mathrm{\pi }}{4}\right) \\ \Rightarrow 0\le 3\sin \left(\sqrt{\frac{{\mathrm{\pi }}^2}{16}-x^2}\right)\le \frac{3}{\sqrt{2}} \end{gathered}$$

So, the range is $\left[0,\frac{3}{\sqrt{2}}\right]$

Hence, the correct answer is $C$, $\left[0,\frac{3}{\sqrt{2}}\right]$.