A constant function $ f:A \to B $ will be onto if
A. $ n\left( A \right) = n\left( B \right) $
B. $ n\left( A \right) = 1 $
C. $ n\left( B \right) = 1 $
D. $ n\left( A \right) > n\left( B \right) $

Hint: A constant function is a function whose output value is the same for all the input values and the range of the constant function does not change no matter what the domain is. For a function to be onto, the co-domain of the function must be equal to its range.

Complete step-by-step answer:
We are given that $ f:A \to B $ maps from A to B is a constant function.
We have to find on which conditions the above given constant function will be an onto function.
Domain is the set of input values (A), co-domain is the set of values that are present in the function definition (B) which can possibly be the output values and range is the set of values that actually will be output values.
Range is always the subset of Co-domain.
Let’s say the function $ f:A \to B $ is said to be onto, then for every element in set B there is at least one matching element in set A. An onto function is also known as a surjective function. The co-domain and range of an onto function are equal.
So for the given function $ f:A \to B $ to be a constant function and onto, the co-domain must have only 1 value. The co-domain is the set B.
Therefore, $ n\left( B \right) $ must be equal to 1.
The correct answer is Option C, $ n\left( B \right) = 1 $

So, the correct answer is “Option C”.

Note: The sets which contain one element are called singleton sets. Subset means being a part of another larger set with relevant elements. A one-to-one function is a function where each range value is unique. If a function is both one-to-one and onto, then the function is said to be a bijection so do not confuse a bijection with a surjection.