Let [x] denote the greatest integer not more | vedclass

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

Question

(B) For $f(x)=\frac{x-[x]}{\sqrt{|x|-x}}$ to be defined, the denominator must be non-zero and the expression inside the square root must be positive:

Vedclass · Accepted Answer

$A \cap B=\phi$

Vedclass · Answer

(B) For $f(x)=\frac{x-[x]}{\sqrt{|x|-x}}$ to be defined, the denominator must be non-zero and the expression inside the square root must be positive: $|x|-x > 0 \Rightarrow |x| > x$. This inequality holds for all $x For $g(x)=\frac{x-[x]}{\sqrt{|x|+x}}$ to be defined, the denominator must be non-zero and the expression inside the square root must be positive: $|x|+x > 0 \Rightarrow |x| > -x$. This inequality holds for all $x > 0$. Thus, $B = (0, \infty)$. Since $A = (-\infty, 0)$ and $B = (0, \infty)$, their intersection is empty: $A \cap B = \phi$.

Let $[x]$ denote the greatest integer not more than $x$. If $A$ and $B$ are the domains of the functions $f(x)=\frac{x-[x]}{\sqrt{|x|-x}}$ and $g(x)=\frac{x-[x]}{\sqrt{|x|+x}}$ respectively, then

Similar Questions

Find the domain of the function $f(x) = \frac{1}{\sqrt{(x + 1)(e^x - 1)(x - 4)(x + 5)(x - 6)}}$.

If a real valued function $f:[-1,2] \rightarrow B$ defined by $f(x) = \begin{cases} 1-x, & -1 \leq x \leq 1 \\ x-1, & 1 < x \leq 2 \end{cases}$ is a surjection,then $B=$

The range of the function $f(x) = -\sqrt{-x^2-6x-5}$ is

The domain of a function $f(y) = \frac{\cos^{-1}(y-5)}{\sqrt{25-y^2}}$ is

The domain of the function $f(x) = \sqrt{\frac{x-7}{9-x}}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module