Question

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

• A. $\left[-\frac{1}{2},\frac{1}{2}\right]$
• B. $\left[0, \frac{1}{2}\right]$
• C. $R^{+}$
• D. $R^{+ }-Z^{+}$

Answer 1

Answer: (B)

$\left[0, \frac{1}{2}\right]$

Answer 2

We have,
$f(x)=\sqrt{\frac{x-|x|}{x-[x]}}$
$\therefore x-|x| \geq 0$ and $x-[x] > 0$
$\Rightarrow x > |x|$ and $x > [x]$
$\therefore x \in R^{+}-\\{\text {all integers }\\}$
Again,
$g(x)=\frac{2 x}{4+x^{2}}$
Let, $y = \frac{2x}{4 + x^2}$
$\Rightarrow 4y + x^2y = 2x$
$\Rightarrow yx^2 - 2x + 4y = 0$
$\Rightarrow x = \frac{2 \pm \sqrt{4 - 16y^2}}{2y}$
$\therefore 4 - 16 y^2 \ge 0$ and $y \ne 0$
$\Rightarrow 1 - 4y^2 \ge 0$ and $y \ne 0$
$\Rightarrow y \in \left[ - \frac{1}{2}, \frac{1}{2} \right] - \\{0\\}$
$\therefore$ Range of $g(x) = \left[ -\frac{1}{2}, \frac{1}{2}\right]$
$\therefore D=R^{+}- \\{\text {all integers }\\}$ and $C=\left[-\frac{1}{2}, \frac{1}{2}\right]$
$\therefore D \cap C=\left(0, \frac{1}{2}\right]$