Q.Assertion (A) : Let $\mathbb{Z}$ be the set of integers. A function $f : \mathbb{Z} \to \mathbb{Z}$ defined as $f(x) = 3x - 5$, $\forall x \in \mathbb{Z}$ is a bijective. Reason (R) : A function is a 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.

Let’s unpack this carefully. The question gives us an Assertion (A) and a Reason (R), and we need to decide if each is true, and if (R) correctly explains (A).

The function is $f: \mathbb{Z} \to \mathbb{Z}$, $f(x) = 3x - 5$. The domain and codomain are both the set of all integers. The Reason (R) simply states the definition of a bijection — that’s a standard fact, so it’s true. The real work is in checking whether $f$ is actually bijective.

Why the bijection proof approach works

To test if $f$ is bijective, we check two properties separately:

1. One-one (injective): If $f(a) = f(b)$, does it force $a = b$?
2. Onto (surjective): For every integer $y$ in the codomain, does there exist an integer $x$ such that $f(x) = y$?

If either fails, the function is not bijective. Let’s go step by step.

Step-by-step reasoning

1. Check one-one (injectivity) Suppose $f(a) = f(b)$ for integers $a, b$. Then:

Adding 5 to both sides: $3a = 3b$. Dividing by 3 (which is allowed over integers): $a = b$.

So $f$ is one-one. No integer can map to the same output from two different inputs.

2. Check onto (surjectivity) We need: for any integer $y$, can we find an integer $x$ such that $3x - 5 = y$? Solve for $x$:

For $x$ to be an integer, $y+5$ must be divisible by 3. That means $y \equiv 1 \pmod{3}$ (since $y+5 \equiv 0 \pmod{3}$ implies $y \equiv -5 \equiv 1 \pmod{3}$).

So only integers of the form $y = 3k + 1$ (for integer $k$) are hit. For example, $y = 0$ gives $x = \frac{5}{3}$, not an integer — so 0 is not in the range. …

To determine if a function is bijective, we must check two properties: injectivity (one-to-one) and surjectivity (onto). A function is bijective if and only if it possesses both these properties.

1. Evaluate Reason (R):

   Reason (R) states: "A function is a bijective if it is both surjective and injective."

   This statement is the fundamental definition of a bijective function. A function that is both injective and surjective is indeed called a bijection.

   Therefore, Reason (R) is true.

2. Evaluate Assertion (A):

   Assertion (A) states: "Let $\mathbb{Z}$ be the set of integers. A function $f : \mathbb{Z} \to \mathbb{Z}$ defined as $f(x) = 3x - 5$, $\forall x \in \mathbb{Z}$ is a bijective."

   We need to check if $f(x) = 3x - 5$ is both injective and surjective.

   • Check for Injectivity (One-to-one): A function $f: A \to B$ is injective if for any $x_1, x_2 \in A$, $f(x_1) = f(x_2)$ implies $x_1 = x_2$. Let $x_1, x_2 \in \mathbb{Z}$ such that $f(x_1) = f(x_2)$.

    Add 5 to both sides:

    Divide by 3:

    Since $f(x_1) = f(x_2)$ implies $x_1 = x_2$, the function $f$ is injective.

*   **Check for Surjectivity (Onto):**
    A function $f: A \to B$ is surjective if for every element $y$ in the codomain $B$, there exists at least one element $x$ in the domain $A$ such that $f(x) = y$.
    Here, the codomain is $\mathbb{Z}$. We need to check if for every $y \in \mathbb{Z}$, there exists an $x \in \mathbb{Z}$ such that $f(x) = y$.
    Set $f(x) = y$:

    Solve for $x$: …