Functions Question 645
Question: Let $ f:N\to N $ defined by $ f(x)=x^{2}+x+1 $ , $ x\in N $ , then f is
[AMU 2000]
Options:
A) One-to-one onto
B) Many one onto
C) One-to-one but not onto
D) None of these
Show Answer
Answer:
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 \mathbb{N} $ \ f is one-one. Again, since for each $ y\in \mathbb{N} $ , there exist $ x\in \mathbb{N} $ \ f is onto.
BETA
