Question

Let the function

\[f : R - \left\{ - b \right\} \to R - \left\{ 1 \right\}\]

\[f\left( x \right) = \frac{x + a}{x + b}, a \neq b .\text{Then},\]

 

पर्याय

•  f is one-one but not onto

• f is onto but not one-one

• f is both one-one and onto

• None of these

Answer 1

c) f is both one-one and onto
Injectivity:
Let x and y be two elements in the domain R- {-b}, such that

  \[f\left( x \right) = f\left( y \right)\]
\[ \Rightarrow \frac{x + a}{x + b} = \frac{y + a}{y + b}\]
\[ \Rightarrow \left( x + a \right)\left( y + b \right) = \left( x + b \right)\left( y + a \right)\]
\[ \Rightarrow xy + bx + ay + ab = xy + ax + by + ab\]
\[ \Rightarrow bx + ay = ax + by\]
\[ \Rightarrow \left( a - b \right)x = \left( a - b \right)y\]

\[ \Rightarrow x = y\]

So, f is one-one.
Surjectivity:
Let y be an element in the co-domain of f, i.e. R-{1}, such that f (x)=y

  \[f\left( x \right) = y\]
\[ \Rightarrow \frac{x + a}{x + b} = y\]
\[ \Rightarrow x + a = yx + yb\]
\[ \Rightarrow x - yx = yb - a\]
\[ \Rightarrow x\left( 1 - y \right) = yb - a\]
\[ \Rightarrow x = \frac{yb - a}{1 - y} \in R - \left\{ - b \right\}\]
So,  f is onto.