$L e t D = {x \in R; f (x) = \sqrt{\frac{x - | x |}{x - [x]}} i s d e f i n e d} a n d c b e t h e r a n g e o f t h e r e a l f u n c t i o n g (x) = \frac{2 x}{4 + x^{2}} . T h e n D \cap C =$

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$[\frac{- 1}{2}, \frac{1}{2}]$

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

$R^{+}$

$R^{+} - Z^{+}$

answer is B.

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Detailed Solution

$\begin{array}{l} f (x) = \sqrt{\frac{x - | x |}{x - [x]}} i s d e f i n e d i f f \\ \frac{x - | x |}{x - [x]} \geq 0, a n d x - [x] \neq 0 \\ \Rightarrow x - | x | \geq 0, [x] \neq x \\ \Rightarrow | x | \leq x, \forall x \in z \Rightarrow [x] = x \\ x \in R^{+}, x \in R - z \Rightarrow D = R^{+} - Z^{+} \\ L e t g (x) = \frac{2 x}{4 + x^{2}} = y \\ \Rightarrow 2 x = 4 y + x^{2} y \\ \Rightarrow x^{2} y - 2 x + 4 y = 0 \\ \forall x \in R Δ \geq 0 \Rightarrow {(- 2)}^{2} - 4 (y) (4 y) \geq 0 \\ \Rightarrow 1 - 4 y^{2} \geq 0 \\ \Rightarrow 4 y^{2} - 1 \leq 0 \\ \Rightarrow (2 y + 1) (2 y - 1) \leq 0 \\ C = y \in [\frac{- 1}{2}, \frac{1}{2}] \\ D \cap C = (0, \frac{1}{2}] \end{array}$