Question

The domain of $f\left(x\right)=co{t}^{-1}\left[\frac{x}{\sqrt{x^2-\left[x^2\right]}}\right]$, $x\in R$ is

• A. $R$
• B. $R-\left[0\right]$
• C. $R-\left[\pm \sqrt{n},n\ge 0,\ne Z\right]$
• D. None of these

Answer 1

The correct option is C

$R-\left[\pm \sqrt{n},n\ge 0,\ne Z\right]$

Explanation for the correct option :

Find the domain of $f\left(x\right)$,

Consider the given Equation as,

$f\left(x\right)=co{t}^{-1}\left[\frac{x}{\sqrt{x^2-\left[x^2\right]}}\right]$

In the above Equation under root

$$\begin{gathered} \Rightarrow \sqrt{x^2-\left[x^2\right]}>0 \\ \Rightarrow x^2-\left[x^2\right]>0 \end{gathered}$$

Since $x\in I$ then $\left[x\right]=x$

If $x^2\in R-I$, then $x^2\ne I$

Then, $x^2>\left[x^2\right]$

$\begin{array}{rcl}& \Rightarrow & x^2=R-\left\{0,\pm 1,\pm 2,\pm 4......\right\}\\ & \Rightarrow & x^2=R-\left\{0,1,2,3,4......\right\}\\ & \Rightarrow & x=R-\left\{0,1,\sqrt{2},\sqrt{3},\sqrt{4}.........n\right\}\\ & \Rightarrow & x=R-\left\{\pm \sqrt{n},n\ge 0,n\in z\right\}\end{array}$

Hence, the correct answer is Option (C).