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प्रश्न
Let
\[f : R - \left\{ n \right\} \to R\]
पर्याय
f is one-one onto
f is one-one into
f is many one onto
f is many one into
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उत्तर
Injectivity:
Let x and y be two elements in the domain R-{n}, such that
\[f\left( x \right) = f\left( y \right)\]
\[ \Rightarrow \frac{x - m}{x - n} = \frac{y - m}{y - n}\]
\[ \Rightarrow \left( x - m \right)\left( y - n \right) = \left( x - n \right)\left( y - m \right)\]
\[ \Rightarrow xy - nx - my + mn = xy - mx - ny + mn\]
\[ \Rightarrow \left( m - n \right)x = \left( m - n \right)y\]
\[ \Rightarrow x = y\]
So, f is one-one.
Surjectivity:
Let y be an element in the co domain R, such that
\[f\left( x \right) = y\]
\[ \Rightarrow \frac{x - m}{x - n} = y\]
\[ \Rightarrow x - m = xy - ny\]
\[ \Rightarrow ny - m = xy - x\]
\[ \Rightarrow ny - m = x\left( y - 1 \right)\]
\[ \Rightarrow x = \frac{ny - m}{y - 1}, \text{which is not defined for } y=1\]
\[So, 1 \in R\left( co domain \right)\text{has no pre image in }R-\left\{ n \right\}\]
\[\Rightarrow\] is not onto
Thus, the answer is (b) .
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