The domain of the function f(x) = frac{cot^{- | vedclass

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(C) For the function $f(x) = \frac{\cot^{-1} x}{\sqrt{x^2 - [x^2]}}$ to be defined,the denominator must be non-zero and the expression inside the squa

Vedclass · Accepted Answer

$R - \{\pm \sqrt{n} : n \in I^+ \cup \{0\}\}$

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(C) For the function $f(x) = \frac{\cot^{-1} x}{\sqrt{x^2 - [x^2]}}$ to be defined,the denominator must be non-zero and the expression inside the square root must be positive.$1$. The expression inside the square root is $x^2 - [x^2]$,which is equal to the fractional part of $x^2$,denoted as $\{x^2\}$.$2$. The condition for the square root to be defined and non-zero is $\{x^2\} > 0$.$3$. We know that for any real number $y$,$0 \le \{y\}  0$ implies $\{x^2\} 
eq 0$.$4$. The fractional part $\{x^2\} = 0$ if and only if $x^2$ is an integer,i.e.,$x^2 = n$ for some non-negative integer $n \in \{0, 1, 2, \dots\}$.$5$. This means $x = \pm \sqrt{n}$ where $n \in \{0, 1, 2, \dots\}$.$6$. Therefore,the domain is all real numbers except where $x^2$ is an integer,which is $R - \{\pm \sqrt{n} : n \in I^+ \cup \{0\}\}$.

The domain of the function $f(x) = \frac{\cot^{-1} x}{\sqrt{x^2 - [x^2]}}$,where $[x]$ denotes the greatest integer not greater than $x$,is :

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