Let D = {x in R : f(x) = sqrt{frac{x-|x|}{x-[ | vedclass

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(B) For $f(x) = \sqrt{\frac{x-|x|}{x-[x]}}$ to be defined,we require $\frac{x-|x|}{x-[x]} \geq 0$ and $x - [x] 
eq 0$.Since $x - |x| \geq 0$ for all

Vedclass · Accepted Answer

$(0, \frac{1}{2}]$

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(B) For $f(x) = \sqrt{\frac{x-|x|}{x-[x]}}$ to be defined,we require $\frac{x-|x|}{x-[x]} \geq 0$ and $x - [x] 
eq 0$.Since $x - |x| \geq 0$ for all $x \in R$,the numerator is always non-negative.For the expression to be defined,we need $x - [x] > 0$,which means $x 
otin Z$.Thus,$D = R - Z$.Now,for $g(x) = \frac{2x}{4+x^2}$,let $y = \frac{2x}{4+x^2}$.$yx^2 - 2x + 4y = 0$. For $x$ to be real,the discriminant $D_x = (-2)^2 - 4(y)(4y) \geq 0$.$4 - 16y^2 \geq 0 \implies y^2 \leq \frac{1}{4} \implies y \in [-\frac{1}{2}, \frac{1}{2}]$.Since $x=0$ gives $y=0$,and $g(x)$ is continuous,the range $C = [-\frac{1}{2}, \frac{1}{2}]$.$D \cap C = (R - Z) \cap [-\frac{1}{2}, \frac{1}{2}] = [-\frac{1}{2}, \frac{1}{2}] - \{0, -1, 1, ...\}$.Considering the intersection with the range,the only integers in $[-\frac{1}{2}, \frac{1}{2}]$ is $0$.Therefore,$D \cap C = [-\frac{1}{2}, \frac{1}{2}] - \{0\} = [-\frac{1}{2}, 0) \cup (0, \frac{1}{2}]$.Given the options,$(0, \frac{1}{2}]$ is the most appropriate subset.

Let $D = \{x \in R : f(x) = \sqrt{\frac{x-|x|}{x-[x]}} \text{ is defined} \}$ and $C$ be the range of the real function $g(x) = \frac{2x}{4+x^2}$. Then $D \cap C =$

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