Q.(i) Let f : {1, 3, 4} → {1, 2, 5} and g : {1, 2, 5} → {1, 3} be given by f = {(1, 2), (3, 5), (4, 1)} and g = {(1, 3), (2, 3), (5, 1)}. Write down gof.

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 $x$: …