Question

The range of the function $f\left(x\right)=tan\sqrt{\left(\frac{{\pi }^2}{9}\right)-x^2}$ is

• A. $[0,3]$
• B. $[0,\surd 3]$
• C. $(-\infty ,\infty )$
• D. None of these

Answer 1

The correct option is B

$[0,\surd 3]$

Step- 1: Find the domain of the given function:

Given, $f\left(x\right)=tan\sqrt{\left(\frac{{\pi }^2}{9}\right)-x^2}$

For terms under the square root, it must be $\ge 0$

$$\begin{gathered} \Rightarrow \frac{{\mathrm{\pi }}^2}{9}-x^2\ge 0 \\ \Rightarrow x^2-\frac{{\mathrm{\pi }}^2}{9}\le 0 \\ \Rightarrow x^2\le \frac{{\mathrm{\pi }}^2}{9} \\ \Rightarrow -\frac{\mathrm{\pi }}{3}\le x\le \frac{\mathrm{\pi }}{3} \end{gathered}$$

Domain of $x\in \left[\frac{-\mathrm{\pi }}{3},\frac{\mathrm{\pi }}{3}\right]$

Step- 2: Find the range of the given function:

$$\begin{gathered} \Rightarrow \frac{-\mathrm{\pi }}{3}\le x\le \frac{\mathrm{\pi }}{3} \\ \Rightarrow 0\le x^2\le \frac{{\mathrm{\pi }}^2}{9} \end{gathered}$$

Multiplying both sides by $-1$

$\Rightarrow \frac{-{\mathrm{\pi }}^2}{9}\le -x^2\le 0$

Adding both sides by $\frac{{\mathrm{\pi }}^2}{9}$

$$\begin{gathered} \Rightarrow 0\le \frac{{\mathrm{\pi }}^2}{9}-x^2\le \frac{{\pi }^2}{9} \\ \Rightarrow 0\le \sqrt{\frac{{\mathrm{\pi }}^2}{9}-x^2}\le \frac{\mathrm{\pi }}{3} \end{gathered}$$

Taking $\tan$ on both the sides

$\Rightarrow 0\le \tan \left(\sqrt{\frac{{\mathrm{\pi }}^2}{9}-x^2}\right)\le \sqrt{3}$

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

Hence, the correct answer is $B$.