Question

Let \( A = \{1,2,3\} \) and \( B = \{2,4,6\} \). A function \( f : A \to B \) is defined by \( f(x)=2x \). The function is:

• A. One-one only
• B. Onto only
• C. Both one-one and onto
• D. Neither one-one nor onto

Hint:

To verify one-one: Check if \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \).
To verify onto: Check if the range of the function equals its codomain. If \( |A| = |B| \) and the function is one-one, it must also be onto.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
In the study of functions, we categorize mappings based on how elements of the domain relate to elements of the codomain.
A function \(f: A \to B\) is defined as a rule that assigns each element in set \(A\) (the domain) to exactly one element in set \(B\) (the codomain).
There are two primary properties we examine here: Injectivity (One-one) and Surjectivity (Onto).

1. One-one (Injective): A function is one-one if every distinct element in the domain maps to a distinct element in the codomain.
Formally, \(f(x_1) = f(x_2) \implies x_1 = x_2\). If this holds for all \(x_1, x_2 \in A\), the function is injective.

2. Onto (Surjective): A function is onto if the range of the function is exactly equal to its codomain.
In other words, for every element \(y \in B\), there exists at least one element \(x \in A\) such that \(f(x) = y\).
If a function satisfies both these properties, it is called Bijective.
Step 2: Detailed Explanation:
Let us analyze the given sets and the function rule.
Domain \(A = \{1, 2, 3\}\)
Codomain \(B = \{2, 4, 6\}\)
Function rule: \(f(x) = 2x\)

First, we calculate the image for each element in the domain:
For \(x = 1\): \(f(1) = 2(1) = 2\).
For \(x = 2\): \(f(2) = 2(2) = 4\).
For \(x = 3\): \(f(3) = 2(3) = 6\).

Now, let's verify the properties:
Testing for One-one:
The set of images is \(\{2, 4, 6\}\). We observe that:
\(1 \to 2\)
\(2 \to 4\)
\(3 \to 6\)
Each element in the domain has a unique and distinct image in the codomain. No two different elements in \(A\) share the same result in \(B\).
Thus, the function is One-one.

Testing for Onto:
The codomain is given as \(B = \{2, 4, 6\}\).
From our calculations, the range (the set of actual outputs) is \(\{2, 4, 6\}\).
Since the \(\text{Range} = \text{Codomain}\), every element in set \(B\) is "covered" or mapped to by an element in set \(A\).
Specifically:
\(2\) is the image of \(1\).
\(4\) is the image of \(2\).
\(6\) is the image of \(3\).
Therefore, the function is Onto.
Step 3: Final Answer:
Since the function \(f(x) = 2x\) is both one-one (injective) and onto (surjective), it is a bijective function.
In the context of the given options, the most complete description is that it is both one-one and onto.