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### Answers (1)

Answer:

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

Given:

$A=R-\left \{ 3 \right \},B=R-\left \{ 1 \right \},f:A\rightarrow B$    defined by $f\left ( x \right )=\frac{x-2}{x-3}$.

Hint:

One-one means every domain has a distinct range. One means every image has some preimage in the domain function.

Solution:

We have,$A=R-\left \{ 3 \right \},B=R-\left \{ 1 \right \}$.

The function $f:A\rightarrow B$ defined by $f\left ( x \right )=\frac{x-2}{x-3}$.

Let  $x,y\epsilon A$ such that $f\left ( x \right )=f\left ( y \right )$ then

$\frac{x-2}{x-3}=\frac{y-2}{y-3}$

$xy-3x-2y+6=xy-2x-3y+6$

$-x=-y$

$x=y$

$f$is one-one.

Let $y\epsilon B$, then $y\neq 1$.

The function $f$is onto if there exists $x\epsilon A$ such that $f\left ( x \right )=y$.

$\Rightarrow \frac{x-2}{x-3}=y$

$x-2=xy-3y$

$x-xy=2-3y$

$x=\frac{2-3y}{1-y}\epsilon A\left ( y\neq 1 \right )$

Thus, for any $y\epsilon B$, there exists $\frac{2-3y}{1-y}\epsilon A$ such that

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

$f$ is onto.

So, $f$ is one-one and onto function.

$x=\frac{2-3y}{1-y}$

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

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