Question

Show that the function $\sf f \rightarrow R : R$ and $\sf x \in R$ defined by $\sf f(x) = \dfrac{x}{1 + |x|}$ is a bijective function. ​

Answer 1

Basic Concept Used

In order to prove that f(x) is bijective, it is sufficient to show that f(x) is one - one as well as onto.

One - One function:-

A function f(x) defined from R to R,

Then to show function is one-one, we choose x,y belongs to R such that f(x) = f(y), if on simplifying x = y, then f(x) is one - one.

Onto Function :-

Let if possible there exist an element y belongs to R such that y = f(x), then express x = g(y), if x belongs to R, then f(x) is onto.

Let's solve the problem now!!!

$\green{\large\underline{\sf{Solution-}}}$

Let us first define the function f(x).

$\blue{\begin{gathered}\begin{gathered}\bf\: f(x) = \begin{cases} &\sf{\dfrac{x}{1 - x} \: \: when \: x < 0} \\ &\sf{\dfrac{x}{1 + x} \: \: \: when \: x \geqslant 0} \end{cases}\end{gathered}\end{gathered}}$

$\bf :\longmapsto\: \: Range \: of \: f(x) \in \: ( - 1, \: 1)$

So,

Let us consider the First case

$\red{\rm :\longmapsto\: \bf \: When \: x \: < \: 0}$

$\red{\rm :\longmapsto\: \bf \:\: f(x) \: = \: \dfrac{x}{1 - x} }$

One - One

Let us consider two elements x < 0 and y < 0 such that

$\rm :\longmapsto\:f(x) = f(y)$

$\rm :\longmapsto\:\dfrac{x}{1 - x} = \dfrac{y}{1 - y}$

$\rm :\longmapsto\:x - xy = y - xy$

$\bf\implies \:x = y$

$\bf :\longmapsto\:Hence, \: f(x) \: is \: one - one$

Onto :-

Let if possible there exist an element y belongs (- 1, 0),

When x < 0,

$\rm :\longmapsto\:f(x) = \dfrac{x}{1 - x} < 0$

Also,

$\rm :\longmapsto\:f(x) = \dfrac{x}{1 - x} = \dfrac{x - 1 + 1}{1 - x}$

$\rm \: = \: \: \dfrac{ - (1 - x) + 1}{1 - x}$

$\rm \: = \: \: \dfrac{ - (1 - x)}{1 - x} + \dfrac{1}{1 - x}$

$\rm \: = \: \: - 1 + \dfrac{1}{1 - x}$

$\bf\implies \:f(x) > - 1$

$\bf :\longmapsto\:Hence, \: - 1 < f(x) < 0 - - (1)$

$\bf :\longmapsto\:Hence, \: f(x) \: is \: onto$

Let us Consider the second case :-

$\red{\rm :\longmapsto\: \bf \: When \: x \: \geqslant \: 0}$

$\red{\rm :\longmapsto\: \bf \:\: f(x) \: = \: \dfrac{x}{1 + x} }$

One - One

$\sf \: Let \: us \: consider \: two \: elements \: x \: and \: y \: \geqslant \: 0 \: such \: that$

$\rm :\longmapsto\:f(x) = f(y)$

$\rm :\longmapsto\:\dfrac{x}{1 + x} = \dfrac{y}{1 + y}$

$\rm :\longmapsto\:x + xy = y + xy$

$\bf\implies \:x = y$

$\bf :\longmapsto\:Hence, \: f(x) \: is \: one - one$

Onto :-

Let if possible there exist an element y belongs [0, 1),

$\rm :\longmapsto\:When \: x \geqslant 0$

$\rm :\longmapsto\:f(x) = \dfrac{x}{1 + x} \geqslant 0$

Also,

$\rm :\longmapsto\:f(x) = \dfrac{x}{1 + x} = \dfrac{x + 1 - 1}{1 + x}$

$\rm \: = \: \: \dfrac{x + 1}{x + 1} - \dfrac{1}{x + 1}$

$\rm \: = \: \: 1 - \dfrac{1}{x + 1}$

$\bf\implies \:f(x) < 1$

$\bf :\longmapsto\:Hence, \: 0 \leqslant f(x) < 1 - - (2)$

$\bf :\longmapsto\:Hence, \: f(x) \: is \: onto$

So,

From equation (1) and equation (2), we get

$\rm :\longmapsto\: - 1 < f(x) < 1$

$\bf :\longmapsto\:Hence, \: f(x) \: is \: one - one \: and \: onto$

$\bf :\longmapsto\:Hence, \: f(x) \: is \: bijective$

$\rm :\implies\:f : R\to \: ( - 1,1)\:given\: by\:f(x)=\dfrac{x}{1+|x|}\:is\:bijective$