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Question

Let f : NN defined by f(n)=⎪ ⎪⎪ ⎪n+12if nis oddn2ifnis even
then f is.

A
Many-one and onto
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B
One-one and not onto
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C
Onto but not one-one
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D
Neither one-one nor onto
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Solution

The correct option is A Many-one and onto
By substituting values of n, the elements in the range of function are,
{1,2,3,4,5,6,...............}, i.e. set of all natural number
So the range and co-domain are same, ergo function is onto.
Also f(1)=f(2), so the function is many one.

Hence the given function is many-one and onto

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