Question

$D=\left\{x\in \mathbb{R}:f\left(x\right)=\sqrt{\frac{x-|x|}{x-\left[x\right]}}\text{is defined}\right\}$ 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)

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