Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Relations and Functions

Question:

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

Options:

$\left(-\frac{1}{2},\frac{1}{2}\right)$

$\left(-\frac{1}{2},\frac{1}{2}\right]$

$\left[-\frac{1}{2},\frac{1}{2}\right)$

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

Correct Answer:

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

Explanation:

y = 0 at x = 0 and for x ≠ 0

$y=\frac{x}{1+x^2}$ or $yx^2 − x + y = 0$

$∴x=\frac{1±\sqrt{1-4y^2}}{2y}$

Since x is real, the range of the function y is determined from the relation $1-4y^2 ≥ 0, -\frac{1}{2}≤y≤\frac{1}{2}$.