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Many-One onto Function

Show that the...

Question

Show that the function $f$ in A = $R = \frac{2}{3}$ defined by $f (x) = \frac{3 x + 2}{5 x + 3},$ $\neq \frac{- 3}{5}$ is one-one and onto.Hence find $f^{- 1} (x)$

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Solution

$f (x) = \frac{3 x + 2}{5 x + 3}, x \neq \frac{- 3}{5}$

One One:

$\frac{3 x_{1} + 2}{5 x_{1} + 3} = \frac{3 x_{2} + 2}{5 x_{2} + 3}$

$\Rightarrow 15 x_{1} x_{2} + 9 x_{1} + 10 x_{2} + 6 = 15 x_{1} x_{2} + 9 x_{2} + 10 x_{1} + 6$

$\Rightarrow x_{2} = x_{1}$

f is one-one

Onto:

Let $y = \frac{3 x + 2}{5 x + 3}$

$\Rightarrow 5 x y + 3 y = 3 x + 2$

$\Rightarrow (5 y - 3) x = 2 - 3 y$

$\Rightarrow x = \frac{2 - 3 y}{5 y - 3}$

Now, $f (\frac{2 - 3 y}{5 y - 3}) = \frac{3 (\frac{2 - 3 y}{5 y - 3}) + 2}{5 (\frac{2 - 3 y}{5 y - 3}) + 3}$

$= \frac{6 - 9 y + 10 y - 6}{10 - 15 y + 15 y - 9}$

$= y$

f is onto

and $f^{- 1} (x) = \frac{2 - 3 x}{5 x - 3}$

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