Question

Show that the function f : R { x ∊ R : -1 < x < 1} defined by f(x) = x/1 + |x|, x ∊ R is one-one and onto function.

Answer 1

We have, f(x) = \(\begin{cases}\frac{x}{1+x},if\ x\ge0\\\frac{x}{1-x},if\ x<0\end{cases}\)

Now, we consider the following cases

Case 1: when x \(\ge\) 0 , we have f(x) = \(\frac{x}{1+x}\)

Injectivity: let x, y ∊ R⁺ ∪ {0} such that f(x) = f(y), then

⇒ x - xy = y + xy ⇒ x - y = 2xy, here LHS > 0 but RHS < 0, which is inadmissible.

Hence , f(x) ≠ f(y) when x ≠ y.

Hence f is one-one and onto function.