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Q.
Let $f : R \to R$ be a function defined by $f\left(x\right) = \frac{x-m}{x-n}$, where $m \ne n$ , then
Relations and Functions - Part 2
Solution:
Let $f : R \to R$ be a function defined by
$f\left(x\right) = \frac{x-m}{x-n}$
For any $\left(x, y\right) \in R$
Let $f \left(x\right) = f \left(y\right)$
$\Rightarrow \frac{x-m}{x-n} = \frac{y-m}{y-n} \Rightarrow x = y$
$\therefore $ f is one - one
Let $\alpha\in R$ such that $f \left(x\right) = \alpha$
$\Rightarrow \alpha = \frac{x-m}{x-n} \quad\Rightarrow \quad\left(x - n\right) \alpha = x - m$
$\Rightarrow x \alpha - n \alpha = x - m \quad\Rightarrow x\alpha - x = n\alpha -m$
$\Rightarrow x\left(\alpha -1\right) = n\alpha -m$
$\Rightarrow x = \frac{n\alpha-m}{\alpha-1}$. for $\alpha = 1, \,x\notin R$
So, f is not onto.