Question

If F is a function defined by f ∶ R− {2} → R − {1} such that f(x) = x− 1 x − 2 . Show that f is bijective.​

Answer 1

Appropriate Question :-

If f is a function defined by f ∶ R− {2} → R − {1} such that f(x) = (x− 1)/( x − 2). Show that f is bijective.

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

Given that

$\red{\rm :\longmapsto\:f : R - \{2 \} \: \rm \longmapsto\:R - \{1 \} \: defined \: by \: }$

$\red{\rm :\longmapsto\:f(x) = \dfrac{x - 1}{x - 2}}$

To show that f(x) is bijective, we have to prove that f(x) is one - one as well as onto.

One - one

Let assume that

$\red{\rm :\longmapsto\:x, \: y \: \in \: R - \{2 \} \: such \: that \: f(x) = f(y)}$

$\rm :\longmapsto\:\dfrac{x - 1}{x - 2} = \dfrac{y - 1}{y - 2}$

$\rm :\longmapsto\:(x - 1)(y - 2) = (x - 2)(y - 1)$

$\rm :\longmapsto\:xy - 2x - y + 2 = xy - x - 2y + 2$

$\rm :\longmapsto\: - 2x - y = - x - 2y$

$\rm :\longmapsto\: - 2x + x = - 2y + y$

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

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

$\bf\implies \:f(x) \: is \: one \: - \: one.$

Onto :-

Let if possible there exist an element

$\red{\rm :\longmapsto\:y \: \in \: R - \{1 \} \: such \: that \: f(x) = y}$

$\rm :\longmapsto\:\dfrac{x - 1}{x - 2} = y$

$\rm :\longmapsto\:x - 1 = y(x - 2)$

$\rm :\longmapsto\:x - 1 = yx - 2y$

$\rm :\longmapsto\:x - yx = 1 - 2y$

$\rm :\longmapsto\:x(1 - y) = 1 - 2y$

$\rm :\longmapsto\:x = \dfrac{1 - 2y}{1 - y} \: \in \: R - \{2 \}$

$\red{\bigg \{ \because \:when \: x = 2, \: we \: get \: 2 - 2y = 1 - 2y \: \bigg \}} \\ \red{\bigg \{ \:which \: gives \: 1 = 2 \: which \: is \: impossible \bigg \}}$

$\bf\implies \:f(x) \: is \: onto .$

So,

$\rm :\longmapsto\:f(x) \: is \: one - one \: and \: onto$

$\bf\implies \:f(x) \: is \: bijective .$

Basic Concept Used :-

One - one :-

In order to show that f(x) is one - one, we have to choose two elements x and y belongs to domain such that f(x) = f(y), if on simplifying we get x = y, then f(x) is one - one otherwise its not one - one.

Onto :-

In order to show that f(x) is onto, we have to choose an element y belongs to co - domain such that f(x) = y. Then represent x as a function of g(y). If for every y, x exist, then f(x) is onto.