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
- One to one and onto
- One to one but not onto
- Onto but not one to one
- Neither one to one nor onto
The Correct Option is B
Solution and Explanation
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
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Concepts Used:
Functions
A function is a relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output. Let A & B be any two non-empty sets, mapping from A to B will be a function only when every element in set A has one end only one image in set B.
Kinds of Functions
The different types of functions are -
One to One Function: When elements of set A have a separate component of set B, we can determine that it is a one-to-one function. Besides, you can also call it injective.
Many to One Function: As the name suggests, here more than two elements in set A are mapped with one element in set B.
Moreover, if it happens that all the elements in set B have pre-images in set A, it is called an onto function or surjective function.
Also, if a function is both one-to-one and onto function, it is known as a bijective. This means, that all the elements of A are mapped with separate elements in B, and A holds a pre-image of elements of B.
Read More: Relations and Functions