A function \(f: R \to \{x \in R: -1 \lt x \lt 1\}\) is defined as \(f(x) = \displaystyle\frac{x}{1+|x|}\), then \(f\) is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → both one-one and onto
Since \(1 + |x| \gt 0\), f(x) has the same sign as x, and \(|f(x)| \lt 1\).
One-one: let \(f(x_1) = f(x_2)\). Then \(x_1\) and \(x_2\) have the same sign (or are both 0).
If both are \(\ge 0\):
\( \displaystyle \frac{x_1}{1+x_1} = \frac{x_2}{1+x_2} \)
\(x_1 + x_1x_2 = x_2 + x_1x_2\)
\(x_1 = x_2\)
If both are negative, \(\frac{x_1}{1-x_1} = \frac{x_2}{1-x_2}\) gives \(x_1 = x_2\) in the same way. So f is one-one.
Onto: take any y with \(-1 \lt y \lt 1\).
If \(0 \le y \lt 1\), let \(x = \frac{y}{1-y} \ge 0\):
\( \displaystyle f(x) = \frac{\frac{y}{1-y}}{1 + \frac{y}{1-y}} = \frac{\frac{y}{1-y}}{\frac{1}{1-y}} = y \)
If \(-1 \lt y \lt 0\), let \(x = \frac{y}{1+y} \lt 0\):
\( \displaystyle f(x) = \frac{\frac{y}{1+y}}{1 - \frac{y}{1+y}} = \frac{\frac{y}{1+y}}{\frac{1}{1+y}} = y \)
Every element of the codomain has a pre-image, so f is onto. Hence f is a bijection.
NCERT (2026-27 edition):
Relations and Functions
