Let $f: R \rightarrow R$ be defined by $f(x)=\frac{x}{1+x^{2}}, x \in R$. Then the range of $f$ is:

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

(C) $\mathrm{R}-\left[-\frac{1}{2}, \frac{1}{2}\right]$

JEE Main · 2019 · 11 Jan 2019 · Shift 1 · Sets Relations and Functions

Let

f : R \to R

be defined by

f (x) = \frac{x}{1 + x ^{2}}, x \in R

. Then the range of

f

is:

Answer: (A)

Set

y = \frac{x}{1 + x ^{2}}

and clear the denominator to get

y x^{2} - x + y = 0

, a quadratic in

x

. For a real

x

to exist the discriminant must be non-negative:

1 - 4 y^{2} \geq 0

, so

y^{2} \leq \frac{1}{4}

, that is

- \frac{1}{2} \leq y \leq \frac{1}{2}

. Both endpoints are actually reached, at

x = \pm 1

, so the range is the closed interval

[- \frac{1}{2}, \frac{1}{2}]

JEE Main · 2019 · 11 Jan 2019 · Shift 1 · Single correct option