Question

If the function f: N \(\rightarrow\) N, is defined by f(x) = x - 1 for all x > 2 and f(1) = f(2) = 1, then f is

• A. one-one and onto
• B. onto but not one-one
• C. many one but not onto
• D. neither one-one nor onto

Hint:

To quickly check if a function is one-one, look for any two distinct inputs that produce the same output. To check if it's onto, determine the set of all possible output values (the range) and see if it covers the entire specified codomain.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
One-one (Injective): A function f is one-one if every distinct element in the domain maps to a distinct element in the codomain. That is, if f(x\(_1\)) = f(x\(_2\)), then x\(_1\) = x\(_2\).
Onto (Surjective): A function f from a set A to a set B is onto if for every element 'y' in the codomain B, there is at least one element 'x' in the domain A such that f(x) = y. In other words, the range of the function is equal to its codomain.
Step 2: Detailed Explanation:
The function is defined as f: N \(\rightarrow\) N.
Domain = N = \{1, 2, 3, 4, ...\}
Codomain = N = \{1, 2, 3, 4, ...\}
Function definition: \[ f(x) = \begin{cases} 1 & \text{if } x = 1 \text{ or } x = 2
x-1 & \text{if } x>2 \end{cases} \] Checking for One-one property:
From the definition, we have f(1) = 1 and f(2) = 1.
Here, we have two different elements in the domain (1 and 2) that map to the same element in the codomain (1).
Since f(1) = f(2) but 1 \(\neq\) 2, the function is not one-one. It is a many-one function.
Checking for Onto property:
We need to check if the range of the function is equal to the codomain (N).
Let's find the range:
f(1) = 1
f(2) = 1
f(3) = 3 - 1 = 2
f(4) = 4 - 1 = 3
f(5) = 5 - 1 = 4
The set of output values (range) is \{1, 2, 3, 4, ...\}, which is the set of all natural numbers, N.
Since the Range = Codomain (N), the function is onto.
Step 3: Final Answer:
The function is onto but not one-one. So, option (B) is correct.