Let \(f : R \to (-2, 2)\) be defined by
\( \displaystyle f(x) = \frac{2x}{1 + |x|} \)
where R is the set of real numbers. Then f is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → one-one and onto
Writing f without the modulus sign:
\( \displaystyle f(x) = \frac{2x}{1 + x} \text{ for } x \ge 0, \qquad f(x) = \frac{2x}{1 - x} \text{ for } x \lt 0 \)
So \(f(x) \ge 0\) when \(x \ge 0\) and \(f(x) \lt 0\) when \(x \lt 0\).
One-one: inputs of opposite signs give outputs of opposite signs. For \(p, q \ge 0\),
\( \displaystyle \frac{2p}{1 + p} = \frac{2q}{1 + q} \)
\(2p + 2pq = 2q + 2pq\)
\(p = q\)
The same steps with \(1 - p\) and \(1 - q\) work for negative inputs.
Onto: solving \(f(x) = y\) for y in \((-2, 2)\):
\( \displaystyle 0 \le y \lt 2: \quad x = \frac{y}{2 - y} \ge 0 \)
\( \displaystyle -2 \lt y \lt 0: \quad x = \frac{y}{2 + y} \lt 0 \)
Every y in \((-2, 2)\) has a pre-image, so f is one-one and onto.