Let f: R to R be a function defined by f(x) = | vedclass

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(B) For any $x, y \in R \setminus \{n\}$,we have $f(x) = f(y)$.$\frac{x - m}{x - n} = \frac{y - m}{y - n}$$(x - m)(y - n) = (y - m)(x - n)$$xy - nx -

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$f$ is one-one into

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(B) For any $x, y \in R \setminus \{n\}$,we have $f(x) = f(y)$.$\frac{x - m}{x - n} = \frac{y - m}{y - n}$$(x - m)(y - n) = (y - m)(x - n)$$xy - nx - my + mn = xy - ny - mx + mn$$-nx - my = -ny - mx$$mx - nx = my - ny$$x(m - n) = y(m - n)$Since $m 
e n$,we have $x = y$.Therefore,$f$ is one-one.Now,let $f(x) = y$ where $y \in R$.$y = \frac{x - m}{x - n} \Rightarrow y(x - n) = x - m$$yx - yn = x - m \Rightarrow x(y - 1) = yn - m$$x = \frac{yn - m}{y - 1}$.For $y = 1$,there is no $x \in R$ such that $f(x) = 1$. Thus,the range of $f$ is $R \setminus \{1\} 
eq R$.Therefore,$f$ is not onto. Hence,$f$ is one-one into.

Let $f: R \to R$ be a function defined by $f(x) = \frac{x - m}{x - n}$,where $m \ne n$. Then

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