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

The Correct Option is B

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