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.
So, the answer is (b).
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