Domain of f(x) = (x^2 - 1)^{-1/2} is | vedclass

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

Question

(A) The function is defined as $f(x) = \frac{1}{\sqrt{x^2 - 1}}$.For $f(x)$ to be defined,the expression inside the square root must be strictly great

Vedclass · Accepted Answer

$( - \infty, -1) \cup (1, \infty)$

Vedclass · Answer

(A) The function is defined as $f(x) = \frac{1}{\sqrt{x^2 - 1}}$.For $f(x)$ to be defined,the expression inside the square root must be strictly greater than zero because it is in the denominator.Therefore,$x^2 - 1 > 0$.This implies $x^2 > 1$.Solving the inequality $x^2 > 1$,we get $|x| > 1$.This means $x > 1$ or $x In interval notation,this is expressed as $x \in ( - \infty, -1) \cup (1, \infty)$.

Domain of $f(x) = (x^2 - 1)^{-1/2}$ is

Similar Questions

The domain of the function $f(x) = \log_{3 + x}(x^2 - 1)$ is

The domain of the function $y = \frac{1}{\sqrt{|x| - x}}$ is

If $f(x) = \cos([\pi^2]x) + \cos([- \pi^2]x)$,then which of the following is true?

Let $S = \{1, 2, 3, \dots, 100\}$. The number of non-empty subsets $A$ of $S$ such that the product of elements in $A$ is even is

The domain of the function $f(x) = \sqrt{2 - x} - \frac{1}{\sqrt{9 - x^2}}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module