The function

Question:

The function $f:[0, \infty) \rightarrow R$ given by $f(x)=\frac{x}{x+1}$ is

(a) one-one and onto
(b) one-one but not onto
(c) onto but not one-one
(d) onto but not one-one

Solution:

Injectivity:
Let x and y be two elements in the domain, such that

$f(x)=f(y)$

$\Rightarrow \frac{x}{x+1}=\frac{y}{y+1}$

$\Rightarrow x y+x=x y+y$

$\Rightarrow x=y$

So, f is one-one.

Surjectivity:
Let y be an element in the co domain R, such that

$y=f(x)$

$\Rightarrow y=\frac{x}{x+1}$

$\Rightarrow x y+y=x$

$\Rightarrow x(y-1)=-y$

$\Rightarrow x=\frac{-y}{y-1}$

Range of $f=R-\{1\} \neq$ co domain $(R)$

$\Rightarrow f$ is not onto.