JEE MAIN - Mathematics (2003 - No. 3)
A function $$f$$ from the set of natural numbers to integers defined by
$$$f\left( n \right) = \left\{ {\matrix{
{{{n - 1} \over 2},\,when\,n\,is\,odd} \cr
{ - {n \over 2},\,when\,n\,is\,even} \cr
} } \right.$$$
is
neither one -one nor onto
one-one but not onto
onto but not one-one
one-one and onto both
Explanation
We have $$f:N \to I$$
If $$x$$ and $$y$$ are two even natural numbers,
then $$f\left( x \right) = f\left( y \right) \Rightarrow {{ - x} \over 2} = {{ - y} \over 2} \Rightarrow x = y$$
Again if $$x$$ and $$y$$ are two odd natural numbers then
$$f\left( x \right) = f\left( y \right) \Rightarrow {{x - 1} \over 2} = {{y - 1} \over 2} \Rightarrow x = y$$
$$\therefore$$ $$f$$ is onto.
Also each negative integer is an image of even natural number and each positive integer is an image of odd natural number.
$$\therefore$$ $$f$$ is onto.
Hence $$f$$ is one one and onto both.
If $$x$$ and $$y$$ are two even natural numbers,
then $$f\left( x \right) = f\left( y \right) \Rightarrow {{ - x} \over 2} = {{ - y} \over 2} \Rightarrow x = y$$
Again if $$x$$ and $$y$$ are two odd natural numbers then
$$f\left( x \right) = f\left( y \right) \Rightarrow {{x - 1} \over 2} = {{y - 1} \over 2} \Rightarrow x = y$$
$$\therefore$$ $$f$$ is onto.
Also each negative integer is an image of even natural number and each positive integer is an image of odd natural number.
$$\therefore$$ $$f$$ is onto.
Hence $$f$$ is one one and onto both.
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