Question

Let $f\left(x\right)=4{\cos }^2\sqrt{x^2-\frac{{\mathrm{\pi }}^2}{9}}$, then

• A. $D_f=[\frac{\mathrm{\pi }}{3},\infty ),R_f=\left[-1,1\right]$
• B. $D_f=[\frac{\mathrm{\pi }}{3},\infty ),R_f=\left[-2,2\right]$
• C. $D_f=(-\infty ,-\frac{\mathrm{\pi }}{3}]\cup [\frac{\mathrm{\pi }}{4},\infty )$
• D. $D_f=\left(-\infty ,\frac{\mathrm{\pi }}{3}\right),R_f=(0,4]$

Answer 1

The correct option is C

$D_f=(-\infty ,-\frac{\mathrm{\pi }}{3}]\cup [\frac{\mathrm{\pi }}{4},\infty )$

Explanation of correct answer :

Finding domain of the given function :

Given, $f\left(x\right)=4{\cos }^2\sqrt{x^2-\frac{{\mathrm{\pi }}^2}{9}}$

As we know, the term under square root must always be non- negative .

So that, $x^2-\frac{{\mathrm{\pi }}^2}{9}\ge 0$

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

So, $x\le -\frac{\mathrm{\pi }}{3}\text{or}x\ge \frac{\mathrm{\pi }}{3}$

On comparining the values of $x$ from the options, we get

Thus, the domain of function is $D_f=(-\infty ,-\frac{\mathrm{\pi }}{3}]\cup [\frac{\mathrm{\pi }}{4},\infty )$.

Hence, the correct answer is Option (C).