Question

Let the function $f:R-\left\{-b\right\}\to R-\left\{1\right\}$ be defined by

$f\left(x\right)=\frac{x+a}{x+b},a\ne b.\text{T}\mathrm{hen},$

(a) f is one-one but not onto
(b) f is onto but not one-one
(c) f is both one-one and onto
(d) 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
$$\begin{gathered} f\left(x\right)=f\left(y\right) \\ \Rightarrow \frac{x+a}{x+b}=\frac{y+a}{y+b} \\ \Rightarrow \left(x+a\right)\left(y+b\right)=\left(x+b\right)\left(y+a\right) \\ \Rightarrow xy+bx+ay+ab=xy+ax+by+ab \\ \Rightarrow bx+ay=ax+by \\ \Rightarrow \left(a-b\right)x=\left(a-b\right)y \\ \Rightarrow x=y \end{gathered}$$
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

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