A) onto but not one-one.
B) one-one but, not onto.
C) both one-one and onto.
D) neither one-one nor onto.
Correct Answer: A
Solution :
\[\left. \begin{matrix} f(g(1))=1 \\ f(g(2))=1 \\ \end{matrix} \right\}many\,one\] |
\[f(g(2k))=k\] |
\[f(g(2k+1))=k+1\] |
\[\therefore \] Onto |
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