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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

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

Vedclass · Answer

(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 :

Similar Questions

Let $A = \{x \in R, x \neq 0, -4 \leq x \leq 4\}$ and $f: A \rightarrow R$ be defined by $f(x) = \frac{|x|}{x}$ for $x \in A$. Then, the range of $f$ is

The domain of the function $f: R \rightarrow R$ defined by $f(x) = \sqrt{x^{2}-7x+12}$ is

The domain of $f(x) = [\sin x] \cos \left( \frac{\pi}{[x - 1]} \right)$ is (where $[.]$ denotes the Greatest Integer Function $G.I.F.$).

Let $D = \{x \in R : f(x) = \sqrt{\frac{x-|x|}{x-[x]}} \text{ is defined} \}$ and $C$ be the range of the real function $g(x) = \frac{2x}{4+x^2}$. Then $D \cap C =$

The domain of the function $f(x) = \log(\log(\log(...\log(x)...)))$ where the logarithm is applied $n$ times (base $10$) is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module