Question

Given below are two statements:

Assertion (A):
A function \(f:N\to N\), defined as \[ f(x)= \begin{cases} x+1, & \text{if } x \text{ is odd}\\ x-1, & \text{if } x \text{ is even}\\ \end{cases} \] is not surjective.
Reason (R):
A function \(f:X\to Y\) is said to be surjective if every element of \(Y\) is the image of some element of \(X\) under \(f\), i.e., for every \(y\in Y\), there exists \(x\in X\), such that \(f(x)=y\).

• A. Both (A) and (R) are correct and (R) is the correct explanation of (A)
• B. Both (A) and (R) are correct but (R) is not the correct explanation of (A)
• C. (A) is correct but (R) is not correct
• D. (A) is not correct but (R) is correct

Hint:

To check surjectivity, verify whether every element of the codomain is obtained as an output of the function.

Answer 1

The Correct Option is D

Concept:
A function is surjective or onto if every element of the codomain has at least one pre-image in the domain.

Step 1: Understand the function. \[ f(x)= \begin{cases} x+1, & x \text{ is odd}\\ x-1, & x \text{ is even} \end{cases} \]

Step 2: Find some values. \[ f(1)=1+1=2 \] \[ f(2)=2-1=1 \] \[ f(3)=3+1=4 \] \[ f(4)=4-1=3 \] \[ f(5)=5+1=6 \] \[ f(6)=6-1=5 \]

Step 3: Observe the mapping.
The function maps: \[ 1\leftrightarrow 2,\quad 3\leftrightarrow 4,\quad 5\leftrightarrow 6 \] Thus, every natural number appears as an image of some natural number. So, \(f\) is surjective. Hence Assertion (A), which says that \(f\) is not surjective, is false.

Step 4: Check Reason (R).
The reason gives the correct definition of a surjective function. Therefore, Assertion is false but Reason is true. \[ \therefore \text{Correct Answer is (D)} \]