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 :
\[f:N\to I\] \[f(1)=0,\,f(2)=-1,\,f(3)=1,\,f(4)=-2,\,f(5)=2\] and \[f(6)=-3\] so on. In this type of function every element of set A has unique image in set B and there is no element left in set B. Hence f is one-one and onto function.You need to login to perform this action.
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