# Q. 10 Let $A = R - \{3\}$ and $B = R - \{1\}$. Consider the function$f : A\rightarrow B$ defined by          $f(x) = \left (\frac{x-2}{x-3} \right )$ . Is f one-one and onto? Justify your answer.

$A = R - \{3\}$

$B = R - \{1\}$

$f : A\rightarrow B$

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

Let $a,b \in A$ such that  $f(a)=f(b)$

$\left (\frac{a-2}{a-3} \right ) = \left ( \frac{b-2}{b-3} \right )$

$(a-2)(b-3)=(b-2)(a-3)$

$ab-3a-2b+6=ab-2a-3b+6$

$-3a-2b=-2a-3b$

$3a+2b= 2a+3b$

$3a-2a= 3b-2b$

$a=b$

$\therefore$  f is one-one.

Let,  $b \in B = R - \{1\}$      then   $b\neq 1$

$a \in A$ such that   $f(a)=b$

$\left (\frac{a-2}{a-3} \right ) =b$

$(a-2)=(a-3)b$

$a-2 = ab -3b$

$a-ab = 2 -3b$

$a(1-b) = 2 -3b$

$a= \frac{2-3b}{1-b}\, \, \, \, \in A$

For any $b \in B$ there exists  $a= \frac{2-3b}{1-b}\, \, \, \, \in A$   such that

$f(\frac{2-3b}{1-b}) = \frac{\frac{2-3b}{1-b}-2}{\frac{2-3b}{1-b}-3}$

$=\frac{2-3b-2+2b}{2-3b-3+3b}$

$=\frac{-3b+2b}{2-3}$

$= b$

$\therefore$  f is onto

Hence, the function is one-one and onto.

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