Question

Let

\[f : R - \left\{ n \right\} \to R\]

\[f\left( x \right) = \frac{x - m}{x - n}, \text{where} \ m \neq n .\] Then,

 

विकल्प

•  f is one-one onto

•  f is one-one into

•  f is many one onto

• f is many one into

Answer 1

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) .