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Let f:N->N be defined by f(x)=x^2+x...

Let `f:N->N` be defined by `f(x)=x^2+x+1,x in N`. Then is `f` is

A

one-one and onto

B

many-one and onto

C

one-one but not onto

D

None of the above

Text Solution

Verified by Experts

The correct Answer is:
C
Let `x, y in N` such that `f(x)=f(y)`
`implies" "x^(2)+x+1=y^(2)+y+1`
`implies(x-y)(x+y+1)=0`
`implies" "x=y`
or `" "x=(-y-1)!inN`
So, f is one-one.
Also, f is not onto.
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