Question

Let A = ℝ − {3}, B = ℝ − {1}. Let f : A → B be defined by \[f\left( x \right) = \frac{x - 2}{x - 3}, \forall x \in A\] Show that f is bijective. Also, find
(i) x, if f⁻¹(x) = 4
(ii) f⁻¹(7)

Answer 1

We have,
A = R  \[-\] {3} and B = R \[-\]  {1}
The function f : A  \[\to\] B defined by f(x) =  \[\frac{x - 2}{x - 3}\]

\[\text { Let }x, y \text { in A such that }f\left( x \right) = f\left( y \right) . \text { Then }, \]

\[\frac{x - 2}{x - 3} = \frac{y - 2}{y - 3}\]

\[ \Rightarrow xy - 3x - 2y + 6 = xy - 2x - 3y + 6\]

\[ \Rightarrow - x = - y\]

\[ \Rightarrow x = y\]

\[ \therefore\text { f is one - one } .\] 

\[\text { Let y } \in B . \text { Then, y } \neq 1 . \]

\[\text { The function f is onto if there exists } x \text { in A such that f }\left( x \right) = y . \]

\[\text {Now,}\]

\[f\left( x \right) = y\]

\[ \Rightarrow \frac{x - 2}{x - 3} = y\]

\[ \Rightarrow x - 2 = xy - 3y\]

\[ \Rightarrow x - xy = 2 - 3y\]

\[ \Rightarrow x\left( 1 - y \right) = 2 - 3y\]

\[ \Rightarrow x = \frac{2 - 3y}{1 - y} \in A \left[ y \neq 1 \right]\]

\[\text { Thus, for any } y \text { in B, there exists } \frac{2 - 3y}{1 - y} \text { in A such that }\]

\[f\left( \frac{2 - 3y}{1 - y} \right) = \frac{\left( \frac{2 - 3y}{1 - y} \right) - 2}{\left( \frac{2 - 3y}{1 - y} \right) - 3} = \frac{2 - 3y - 2 + 2y}{2 - 3y - 3 + 3y} = \frac{- y}{- 1} = y\]

\[ \therefore \text { f is onto } .\]

\[\text { So, f is one - one and onto fucntion } . \]

\[\text { Now }, \]

\[\text { As,} x = \left( \frac{2 - 3y}{1 - y} \right)\]

\[\text { So }, f^{- 1} \left( x \right) = \left( \frac{2 - 3x}{1 - x} \right) = \frac{3x - 2}{x - 1}\]

Therefore,f is bijective.

(i) \[f^{- 1} \left( x \right) = 4\]

\[ \Rightarrow \frac{3x - 2}{x - 1} = 4\]

\[ \Rightarrow 3x - 2 = 4x - 4\]

\[ \Rightarrow x = 2\]

\[f^{- 1} \left( x \right) = \frac{3x - 2}{x - 1}\]

\[ \Rightarrow f^{- 1} \left( 7 \right) = \frac{3\left( 7 \right) - 2}{7 - 1} = \frac{19}{6}\]