Q.Is the given relation a function? Give reasons for your answer.

A relation is a function if every element in the domain maps to exactly one element in the codomain. The test: no input appears twice with different outputs.

(i) $h = \{(4, 6), (3, 9), (-11, 6), (3, 11)\}$

The input $3$ maps to both $9$ and $11$. This violates the function definition.

Not a function.

(ii) $f = \{(x, x) \mid x \text{ is a real number}\}$

Each real number $x$ maps to itself — one output per input.

Is a function.

(iii) $g = \left\{\left(n,\ \frac{1}{n}\right) \mid n \text{ is a positive integer}\right\}$

Each positive integer $n$ maps to the unique value $\frac{1}{n}$.

Is a function.

(iv) $s = \{(n,\ n^2) \mid n \text{ is a positive integer}\}$ …

The heart of the function concept is single-valuedness: each input must produce exactly one output. Think of a function as a reliable machine—feed it the same input, and it must always return the same result. If an input could yield two different outputs, the relation breaks this contract and fails to be a function.

The quickest diagnostic is to scan the domain (the set of first coordinates). If any domain element appears in two different ordered pairs with different second coordinates, the relation is not a function. If every domain element appears with a unique output—or the same output repeated—it passes.

(i) $h = \{(4, 6), (3, 9), (-11, 6), (3, 11)\}$

1. List the domain elements: $4, 3, -11, 3$. Notice that $3$ appears twice.

2. Check the outputs for $3$: The pair $(3, 9)$ says "$3$ maps to $9$," while $(3, 11)$ says "$3$ maps to $11$."

3. Verdict: The input $3$ is assigned two different outputs. This violates the definition of a function.

Conclusion for (i): $h$ is not a function.

(ii) $f = \{(x, x) \mid x \text{ is a real number}\}$

1. Understand the rule: Every real number $x$ is paired with itself. For instance, $(2, 2)$, $(-\sqrt{3}, -\sqrt{3})$, $(\pi, \pi)$ all belong to $f$.

2. Check uniqueness: Each $x$ in the domain appears in exactly one pair, $(x, x)$, so there is no ambiguity about the output.

3. Verdict: Every input has a single, well-defined output (itself).

Conclusion for (ii): $f$ is a function (the identity function on $\mathbb{R}$).

(iii) $g = \left\{\left(n,\ \frac{1}{n}\right) \mid n \text{ is a positive integer}\right\}$

1. Understand the rule: Each positive integer $n$ is paired with its reciprocal $\frac{1}{n}$. For example, $(1, 1)$, $(2, \frac{1}{2})$, $(3, \frac{1}{3})$, and so on.

2. Check uniqueness: For any given $n$, there is exactly one value $\frac{1}{n}$. No positive integer appears twice in the domain with different outputs.

3. Verdict: Each input determines a unique output.

Conclusion for (iii): $g$ is a function.

(iv) $s = \{(n, n^2) \mid n \text{ is a positive integer}\}$

1. Understand the rule: Each positive integer $n$ is paired with its square. For instance, $(1, 1)$, $(2, 4)$, $(3, 9)$, $(4, 16)$, etc. …