The real-valued function f(x) = frac{operator | vedclass

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

Question

(B) For the function $f(x) = \frac{\operatorname{cosec}^{-1} x}{\sqrt{x - [x]}}$ to be defined,two conditions must be satisfied:$1$. The domain of $\o

Vedclass · Accepted Answer

all non-integers except the interval $[-1, 1]$

Vedclass · Answer

(B) For the function $f(x) = \frac{\operatorname{cosec}^{-1} x}{\sqrt{x - [x]}}$ to be defined,two conditions must be satisfied:$1$. The domain of $\operatorname{cosec}^{-1} x$ is $(-\infty, -1] \cup [1, \infty)$.$2$. The denominator $\sqrt{x - [x]}$ must be non-zero and real. Since $x - [x] = \{x\}$ (the fractional part of $x$),we require $\{x\} > 0$,which means $x$ cannot be an integer.Combining these conditions,$x$ must be in $(-\infty, -1] \cup [1, \infty)$ and $x 
otin \mathbb{Z}$.This corresponds to all non-integers in the set $(-\infty, -1] \cup [1, \infty)$,which is equivalent to all non-integers except the interval $(-1, 1)$.

The real-valued function $f(x) = \frac{\operatorname{cosec}^{-1} x}{\sqrt{x - [x]}}$,where $[x]$ denotes the greatest integer less than or equal to $x$,is defined for all $x$ belonging to:

Similar Questions

For the function $f(x) = (1 + \frac{1}{x})^x$,the domain of $f(x)$ is:

Find the domain and the range of the real function $f$ defined by $f(x) = \sqrt{x-1}$.

If the equation $\frac{1}{x} + \frac{1}{x - 1} + \frac{1}{x - 2} = 3x^3$ has $k$ real roots,then $k$ is equal to -

Let $f:(1,3) \rightarrow R$ be a function defined by $f(x)=\frac{x[x]}{1+x^{2}},$ where $[x]$ denotes the greatest integer $\leq x.$ Then the range of $f$ is

The domain of the real-valued function $f(x) = \frac{3}{4-x^2} + \log_{10}(x^3-x)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module