\(f: R → R\) is defined as \(f(x)=x^4\).
Let \(x, y ∈ R\) such that \(f(x) = f(y)\).
\(⇒ x^4=y^4\)
\(⇒ x=±y\)
∴ \(f(x_1)=f(x_2)\) does not imply that \(x_1=x_2\)
For instance,
\(f(1)=f(-1)=1\)
∴ f is not one-one.
Consider an element 2 in co-domain R. It is clear that there does not exist any x in domain R such that \(f(x) = 2\).
∴ f is not onto.
Hence, function f is neither one-one nor onto.
The correct answer is D.
A school is organizing a debate competition with participants as speakers and judges. $ S = \{S_1, S_2, S_3, S_4\} $ where $ S = \{S_1, S_2, S_3, S_4\} $ represents the set of speakers. The judges are represented by the set: $ J = \{J_1, J_2, J_3\} $ where $ J = \{J_1, J_2, J_3\} $ represents the set of judges. Each speaker can be assigned only one judge. Let $ R $ be a relation from set $ S $ to $ J $ defined as: $ R = \{(x, y) : \text{speaker } x \text{ is judged by judge } y, x \in S, y \in J\} $.
The correct IUPAC name of \([ \text{Pt}(\text{NH}_3)_2\text{Cl}_2 ]^{2+} \) is:
A function is said to be one to one function when f: A → B is One to One if for each element of A there is a distinct element of B.
A function which maps two or more elements of A to the same element of set B is said to be many to one function. Two or more elements of A have the same image in B.
If there exists a function for which every element of set B there is (are) pre-image(s) in set A, it is Onto Function.
A function, f is One – One and Onto or Bijective if the function f is both One to One and Onto function.
Read More: Types of Functions