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Q.
Let A = R - {3}, B = R - {1}. Let f : A $\to$ B be defined by f(x) = $\frac{x -2 }{x-3}$ Then :
Relations and Functions - Part 2
Solution:
Let $f(x) = \frac{x-2}{x-3}$
To show : f (x) is one - one
Let f (x) = f (y)
$\Rightarrow \, \frac{x-2}{x-3} = \frac{y-2}{y-3} $ To show : x = y
$\Rightarrow \, xy - 3x - 2y + 6 = xy - 2x - 3y + 6 $
$\Rightarrow \, -3x - 2y = -2x - 3y $
$\Rightarrow \, y = x $ (proved)
Hence, f (x) is one - one
To show : f (x) is an onto function
Let y $\in$ Co-domain (B) (To show : There exist x $\in$ Domain such that f(x) = y) consider f (x) = y
$\Rightarrow \, \frac{x-2}{x-3} = y \, \Rightarrow \, x - 2 = xy - 3y $
$\Rightarrow \, x(1 - y) = 2 - 3y \, \Rightarrow \, x = \frac{2 - 3y}{1 - y} \in A $
Clearly, for all y $\in$ B we have
$x = \frac{2 - 3y}{1- y} \in A$
$\therefore $ f(x) is an onto function
Thus f(x) is one - one and onto function.
$\Rightarrow $ f(x) is Bijective function.