Let D = x ∈ ℝ :   f x = x - | x | x - [ x ] is defined } and C be the range of the real…

The given function is

$f\left(x\right)=\sqrt{\frac{x-\left|x\right|}{x-\left[x\right]}}$

Now, $x-\left|x\right| \ge 0 \& x-\left[x\right]>0$

$\Rightarrow x\ge \left|x\right|$ and $x>\left[x\right]$

Therefore, $x\mathit{ }\mathit{\in }{ℝ}^{\mathit{+}}\mathit{-}$(all integers)

Again, $g\left(x\right)=\frac{2x}{4+x^2}$

$\Rightarrow y=\frac{2x}{4+x^2}$

Rewrite the above equation.

$y{\mathrm{x}}^2-2\mathrm{x}+4\mathrm{y}=0$

$\Rightarrow x=\frac{2\pm \sqrt{4-16{\mathrm{y}}^2}}{2\mathrm{y}}$

From the above equation

$4-16y^2\ge 0 \& y\ne 0$

$1-4y^2\ge 0 \& y\ne 0$

Therefore, $y\in \left[-\frac{1}{2},\frac{1}{2}\right]-\left\{0\right\}$

Range of $g\left(\mathrm{x}\right)$ is $\left[-\frac{1}{2},\frac{1}{2}\right]-\left\{0\right\}$

The $D=$${\mathrm{ℝ}}^{+}-$(all integers) and $C=\left[-\frac{1}{2},\frac{1}{2}\right]-\left\{0\right\}$

Thus $D\cap C=\left(0,\frac{1}{2}\right]$