Question

The mapping $ f : N \to N $ given by $ f(n) = n^3 + 3, n \in N $ where $ N $ is the set of natural number, is

• A. One to one and onto
• B. One to one but not onto
• C. Onto but not one to one
• D. Neither one to one nor onto

Answer 1

The Correct Option is B

The correct option is(B): One to one but not onto.

We have, \(f : N \to N\) given by \(f (n)=n^{3}+3\)
Let \(f (n_{1})=f (n_{2})\)
\(\Rightarrow n^{3}_{1}+3=n^{3}_{2}+3\)
\(\Rightarrow n^{3}_{1}=n^{3}_{2}\)
\(\Rightarrow n_{1}=n_{2}\)
So, \(f (n)\) is one to one mapping 
Let \(y=f (n) = n^{3}+3\)
\(\Rightarrow n=(y-3)^{1/3}\)
Now, \(\forall y \in\,N, n \notin\,N\)
so, \(f (n)\) is not onto