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

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(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

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