Question

Range of the function $y=\frac{x}{1+x^2}$ is

• A. $-\frac{1}{2}\le y\le \frac{1}{2}.$
• B. $-\frac{1}{4}\le y\le \frac{1}{2}.$
• C. $-\frac{1}{2}\le y\le \frac{1}{4}.$
• D. $-\frac{1}{1}\le y\le \frac{1}{1}.$

Answer 1

The correct option is A $-\frac{1}{2}\le y\le \frac{1}{2}.$

We have $y=0$ at $x=0$ and for $x\ne 0$
$y=\frac{x}{1+x^2}$ or $y{x}^2-x+y=0$
$\therefore x=\{1\pm \sqrt{1-4y^2}\}/2y$, Since x is real, the range of the function y is determined from the relation $1-4y^2\ge 0,$ or $4(y^2-\frac{1}{4})\le 0$ whence $-\frac{1}{2}\le y\le \frac{1}{2}.$