Question

Assertion : Let \( Z \) be the set of integers. A function \( f : Z \to Z \) defined as \( f(x) = 3x - 5 \), \( \forall x \in Z \), is a bijective.
Reason (R): A function is bijective if it is both surjective and injective.

• A. Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of the Assertion (A).
• B. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
• C. Assertion (A) is true, but Reason (R) is false.
• D. Assertion (A) is false, but Reason (R) is true.

Hint:

A function is bijective if it is both injective (one-to-one) and surjective (onto). For linear functions of the form \( f(x) = ax + b \) where \( a \neq 0 \), the function is always bijective.

Answer 1

The Correct Option is A

Assertion:
We are given the function \( f : Z \to Z \) defined by \( f(x) = 3x - 5 \), where \( Z \) is the set of integers. To determine if this function is bijective, we must verify if it is both injective (one-to-one) and surjective (onto).

Step 1: Checking for injectivity (one-to-one).
A function is injective if distinct inputs yield distinct outputs. Mathematically, if \( f(x_1) = f(x_2) \), then it must follow that \( x_1 = x_2 \).

Given:
\[ f(x_1) = 3x_1 - 5, \quad f(x_2) = 3x_2 - 5 \]
Assume \( f(x_1) = f(x_2) \). Then:
\[ 3x_1 - 5 = 3x_2 - 5 \]
Simplifying:
\[ 3x_1 = 3x_2 \quad \Rightarrow \quad x_1 = x_2 \]
As \( x_1 = x_2 \), the function is injective.

Step 2: Checking for surjectivity (onto).
A function is surjective if for every element \( y \) in the codomain \( Z \), there exists an element \( x \) in the domain \( Z \) such that \( f(x) = y \).

For any \( y \in Z \), we seek an \( x \in Z \) such that:
\[ f(x) = 3x - 5 = y \quad \Rightarrow \quad 3x = y + 5 \quad \Rightarrow \quad x = \frac{y + 5}{3} \]
Since \( y \) is an integer and \( 5 \) is an integer, their sum \( y + 5 \) is an integer. For \( x \) to be an integer, \( y + 5 \) must be divisible by 3. This condition holds for all integers \( y \). Therefore, the function is surjective.

Consequently, the function is both injective and surjective, rendering it bijective. Thus, the Assertion is true.

Reason (R):
Reason (R) states that a function is bijective if it is both surjective and injective. Since we have demonstrated that the function is both injective and surjective, Reason (R) is also true.

Conclusion:
Both the Assertion and Reason (R) are true, and Reason (R) provides a correct explanation for the truth of the Assertion. Therefore, the correct answer is .