Question

The largest interval in which the function $f\left(x\right)=3\sin \left[\sqrt{\left(\frac{{\mathrm{\pi }}^2}{9}\right)-x^2}\right]$ assumes values is

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

Answer 1

The correct option is C

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

The Explanation for the correct option.

Solve for the range of the given function $f\left(x\right)=3\sin \left[\sqrt{\left(\frac{{\mathrm{\pi }}^2}{9}\right)-x^2}\right]$

$f\left(x\right)=3\sin \left[\sqrt{\left(\frac{{\mathrm{\pi }}^2}{9}\right)-x^2}\right]$

Here, $\sqrt{\left(\frac{{\mathrm{\pi }}^2}{9}\right)-x^2}\ge 0$

$$\begin{gathered} \Rightarrow x^2\le \frac{{\mathrm{\pi }}^2}{9} \\ \Rightarrow \left|x\right|=\frac{\mathrm{\pi }}{3} \end{gathered}$$

The value of $f\left(\frac{-\mathrm{\pi }}{3}\right)=0$

$f\left(0\right)=3\sin \left(\frac{\mathrm{\pi }}{3}\right)=\frac{3\sqrt{3}}{2}$

The range of $f\left(x\right)=3\sin \left[\sqrt{\left(\frac{{\mathrm{\pi }}^2}{9}\right)-x^2}\right]$ is $\left[0,\frac{3\sqrt{3}}{2}\right]$.

Hence, option (C) is the correct answer.