A) One-one onto
B) Many one onto
C) One-one but not onto
D) None of these
Correct Answer: A
Solution :
Let \[x,\,y\in N\] such that \[f(x)=f(y)\] Then \[f(x)=f(y)\Rightarrow {{x}^{2}}+x+1={{y}^{2}}+y+1\] Þ \[(x-y)(x+y+1)=0\Rightarrow x=y\] or \[x=(-y-1)\notin N\] \ f is one-one. Again, since for each \[y\in N\], there exist \[x\in N\] \ f is onto.
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