A) one-one but not onto
B) onto but not one-one
C) one-one and onto both
D) neither one-one nor onto
Correct Answer: C
Solution :
\[f(n)=\left\{ \begin{align} & \frac{n-1}{2},when\text{ }n\text{ }is\text{ }odd \\ & -\frac{n}{2},\,when\text{ }n\text{ }is\text{ }even \\ \end{align} \right.\] and f: \[N\to l\], where N is the set of natural numbers and \[l\] is the set of integers. Let \[x,\,y\in N\] and both are even. Then, \[f(x)=f(y)\] \[\Rightarrow \] \[-\frac{x}{2}=-\frac{y}{2}\Rightarrow x=y\] Again let \[x,y\in N\] and both are odd. Then, \[f(x)=f(y)\] \[\Rightarrow \] \[\frac{x-1}{2}=\frac{y-1}{2}\Rightarrow x=y\] i.e., mapping is one-one. Since, each negative integer is an image of even natural number and positive integer is an image of odd natural number. So, mapping is onto. Hence, mapping is one-one and onto.
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