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