The function f(x) = frac{sec^{-1}x}{sqrt{x - | vedclass

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

Question

(B) The function $f(x)$ is defined if the following conditions are met:$1$. The domain of $\sec^{-1}x$ is $(-\infty, -1] \cup [1, \infty)$,which can b

Vedclass · Accepted Answer

$R - ((-1, 1) \cup \{n \mid n \in Z\})$

Vedclass · Answer

(B) The function $f(x)$ is defined if the following conditions are met:$1$. The domain of $\sec^{-1}x$ is $(-\infty, -1] \cup [1, \infty)$,which can be written as $R - (-1, 1)$.$2$. The denominator $\sqrt{x - [x]}$ must be defined and non-zero. The expression inside the square root must be strictly positive: $x - [x] > 0$.$3$. Since $x - [x] = \{x\}$ (the fractional part of $x$),the condition $x - [x] > 0$ implies that $x$ cannot be an integer. Thus,$x \in R - Z$.$4$. Combining these two conditions,the function is defined for $x \in (R - (-1, 1)) \cap (R - Z)$.$5$. This intersection is equivalent to $R - ((-1, 1) \cup \{n \mid n \in Z\})$.

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

Similar Questions

The domain and range for the function $f(x) = e^{|x| \sin x}$ are:

The domain of the function $f(x) = \sqrt{\log_{0.5} x!}$ is

If the domain of the function $\log _5(18 x-x^2-77)$ is $(\alpha, \beta)$ and the domain of the function $\log _{(x-1)}\left(\frac{2 x^2+3 x-2}{x^2-3 x-4}\right)$ is $(\gamma, \delta)$,then $\alpha^2+\beta^2+\gamma^2$ is equal to :

The mid-point of the domain of the function $f(x) = \sqrt{4 - \sqrt{2x + 5}}$ for real $x$ is

Is it true that $x = e^{\log x}$ for all real $x$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module