A function $f: R→ \{x∈ R: −1 < x < 1\}$ is defined as $f(x) =\frac{x}{1+|x|}$, then $f$ is:

both one-one and onto

The correct answer is Option (4) → both one-one and onto

Given: $f: \mathbb{R} \to \{x \in \mathbb{R} : -1 < x < 1\}$

$f(x) = \frac{x}{1 + |x|}$

For $x \ge 0$: $f(x) = \frac{x}{1 + x}$, strictly increasing for $x \ge 0$.

For $x < 0$: $f(x) = \frac{x}{1 - x}$, strictly increasing for $x < 0$.

Left and right limits at $x = 0$ match, so $f(x)$ is strictly increasing on $\mathbb{R}$.

Hence, $f$ is one-one.

Range check:

As $x \to +\infty$, $f(x) \to 1^{-}$; as $x \to -\infty$, $f(x) \to -1^{+}$.

For any $y \in (-1, 1)$, solving $\frac{x}{1+|x|} = y$ gives a real $x$, so range is $(-1, 1)$.

Thus, $f$ is onto $\{x \in \mathbb{R} : -1 < x < 1\}$.

Answer: Both one-one and onto