Let A = IR – 3 and B = IR – 1 . Consider the function f: A B defi… - Vidyadip

Question

Let $A = IR – \{3\}$ and $B = IR – \{1\}.$ Consider the function $f: A \rightarrow B$ defined by $\text{f(x)}=\Bigg(\frac{\text{x - 2}}{\text{x - 3}}\Bigg)$. Show that fis one-one and onto and hence find $f^{–1}.$

✓

Answer

Let $x_1, x_2 \in A$ and$ f(x_1) = f(x_2)\Rightarrow \frac{\text{x}_{1}- 2}{\text{x}_{1}-3}=\frac{\text{x}_{2}-2}{\text{x}_{2}-3}$
$\therefore x_1 x_2 – 2x_2 – 3x_1 = x_1 x_2 – 2x_1 – 3x_2$
$\Rightarrow x_1= x_2$
Hence f is $1 – 1$
Let $y \in B, \therefore y = f(x) \Rightarrow \text{y}=\frac{\text{x - 2}}{\text{x - 3}}\Rightarrow xy – 3y = x – 2$
Or $\text{x}=\frac{\text{3y - 2}}{\text{y - 1}}$
since y $\neq$ 1 and $\frac{\text{3y - 2}}{\text{y - 1}}\neq3\therefore\text{x}\in\text{A}$
Hence f is $ONTO$
and $f ^{–1}(y) = \frac{\text{3y - 2}}{\text{y - 1}}$.

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