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If f: NvecZf(n)={(n-1)/2,w h e nni sod d...

If `f: NvecZf(n)={(n-1)/2,w h e nni sod d-n/2,i d e n t ifyt h ew h e nni se v e n`

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The correct Answer is:
bijective
When n is even, let
`f(2m_(1))=f(2m_(2))`
or `-(2m_(1))/(2)=-(2m_(2))/(2)`
or `m_(1)=m_(2)`
When n is odd, let
`f(2m_(1)+1)=f(2m_(2)+1)`

or `(2m_(1)+1-1)/(2)=(2m_(2)+1-1)/(2) " or "m_(1)=m_(2)`
Therefore, `f(x)` is one-one.
Also, when n is even, `-(n)/(2)=-(2m)/(2)= -m.`
When n is odd, `(n-1)/(2)=(2m+1-1)/(2)=m.`
Hence, the range of the function is Z.
Therefore, function is onto.
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