Relations and functions: what you bring from Class XI, and three new facts

A school in Pune runs three clubs, and every student may join as many clubs as they like, or none.

Write down every pair (student, club) for which the student belongs to the club, and you have a relation.

Some students appear in several pairs, some in none.

A function is the special, tidy case in which every student appears in exactly one pair.

This chapter studies relations and functions more closely: which relations behave like "equality", and which functions match the elements of two sets perfectly.

This section recalls what you met in Class XI, then adds three facts that recent board papers have asked about directly: how many relations there are between two finite sets, when a relation fails to be a function, and the domain of a sum of two functions.

In this unit you need to know:

• relations from a set $A$ to a set $B$ and relations in a set $A$, and how to count them;

• the types of relations: reflexive, symmetric, transitive and equivalence relations, together with the empty and universal relations;

• how to test a relation for each property, with a proof for general elements when it holds and one numerical counterexample when it fails;

• equivalence classes, and the smallest equivalence relation that contains a given relation;

• one-one (injective), onto (surjective) and bijective functions, and how to prove or disprove each for rules on $\mathbf{N}$, $\mathbf{Z}$ and $\mathbf{R}$;

• why, on a finite set, a function from the set to itself is one-one exactly when it is onto.

Relations and Functions shares the 8 marks of Unit I with Inverse Trigonometric Functions.

Recent board papers have given this chapter about 4 to 6 marks, and the 2026-27 sample paper gives it 5.

An ordered pair $(a, b)$ has a first entry and a second entry, and order matters: $(2, 5) \neq (5, 2)$.

The Cartesian product of sets $A$ and $B$ is

the set of all ordered pairs whose first entry comes from $A$ and second entry from $B$.

If $A$ has $m$ elements and $B$ has $n$ elements, then $A \times B$ has $mn$ elements, because each of the $m$ first entries can be paired with each of the $n$ second entries.

For example, if $A = \{1, 2\}$ and $B = \{x, y, z\}$, then $A \times B = \{(1, x), (1, y), (1, z), (2, x), (2, y), (2, z)\}$, which has $2 \times 3 = 6$ elements.

Notice that $B \times A$ is a different set: its pairs start with a letter.

Mathematics strips the idea of "being connected" down to its bare bones.

A relation from $A$ to $B$ is simply a collection of ordered pairs from $A \times B$: any subset at all.

You do not need a sensible rule.

The pairs are the relation.

The figure below shows a relation $R$ from a set $A$ of four students to a set $B$ of three clubs, "student $a$ is a member of club $b$".

Each arrow is one ordered pair.

In roster form, $R = \{(\text{Asha}, \text{Chess}), (\text{Asha}, \text{Music}), (\text{Bilal}, \text{Dance}), (\text{Chitra}, \text{Chess})\}$.

Three features of this picture matter for everything that follows:

• Asha has two arrows. A relation allows that.

• Dev has no arrow. A relation allows that too.

• Two students point to Chess. Again, no problem for a relation.

The domain of a relation is the set of first entries that actually occur, here $\{\text{Asha}, \text{Bilal}, \text{Chitra}\}$.

The range is the set of second entries that occur, here $\{\text{Chess}, \text{Dance}, \text{Music}\}$.

The whole of $B$ is the codomain.

When $A = B$, a relation from $A$ to $A$ is called a relation in $A$ (or on $A$).

It is a subset of $A \times A$.

Almost every relation in the next three sections is of this kind: "is a sibling of" on the students of a class, "divides" on a set of numbers, "is parallel to" on a set of lines.

Recent board papers have asked how many relations there are from one finite set to another.

The argument is short and worth understanding.

Suppose $A$ has $m$ elements and $B$ has $n$ elements.

Then $A \times B$ has $mn$ ordered pairs.

To build a relation, you go through these pairs one by one and make a yes-or-no decision for each: put it in, or leave it out.

Two choices for each of $mn$ pairs gives

This count includes the empty relation (every answer "no") and the whole of $A \times B$ (every answer "yes").

A relation in a set with $n$ elements is a subset of a set with $n^2$ pairs, so there are $2^{n^2}$ of them.

The same yes-or-no idea answers variations.

If a particular pair must be included, its decision is already made, and the other $mn - 1$ pairs give $2^{mn - 1}$ relations.

If you want non-empty relations, subtract the one empty relation: $2^{mn} - 1$.

A function $f$ from $A$ to $B$, written $f : A \to B$, is a relation from $A$ to $B$ in which every element of $A$ has exactly one image in $B$.

In arrow diagrams, every element of $A$ sends out exactly one arrow.

Elements of $B$ may receive many arrows, one, or none.

So a relation fails to be a function in exactly two ways:

1. some element of $A$ has no image (no arrow leaves it); or

2. some element of $A$ has two or more images (two arrows leave it).

The club relation above fails on both counts: Dev has no image and Asha has two.

The figure below shows the difference on two small sets.

In the left diagram, 1 and 2 both go to $u$.

That is allowed: a function may send two inputs to the same output.

Nothing goes to $v$, which is also allowed.

Each element of $P$ has one arrow, so this is a function.

In the right diagram, 1 has two images and 3 has none, so it is not a function.

Recent board papers have used exactly this test inside case studies: if one element has two images, the relation is not a function, so the question "is it one-one?"

does not even arise.

For a function you keep the same words with sharper meanings.

The domain is all of $A$ (every element has an image).

The codomain is $B$, the set you declared the outputs live in.

The range is the set of images actually used, $f(A) = \{f(x) : x \in A\}$, which is a subset of the codomain.

Whether the range fills the codomain is the whole question of "onto" later in this chapter.

Counting functions uses the same choice-by-choice idea.

Each of the $m$ elements of $A$ must choose exactly one of the $n$ elements of $B$, so there are $n \times n \times \cdots \times n = n^m$ functions from $A$ to $B$.

Compare this with $2^{mn}$ relations: functions are a small part of all relations.

For a real function (domain and codomain subsets of $\mathbf{R}$), the graph is the set of points $(x, f(x))$.

A curve in the $xy$-coordinate system is the graph of a function of $x$ exactly when every vertical line meets it at most once; a vertical line meeting it twice means one $x$ with two $y$ values.

The circle $x^2 + y^2 = 9$ fails this test, so it is a relation in $\mathbf{R}$ but not a function.

When a real function is given only by a formula, its domain is taken to be every real $x$ for which the formula gives a real number.

Two Class XI rules decide most cases:

• a square root $\sqrt{g(x)}$ needs $g(x) \geq 0$;

• a fraction $\dfrac{p(x)}{q(x)}$ needs $q(x) \neq 0$.

For real functions $f$ and $g$, Class XI defined

To compute $f(x) + g(x)$ you need both $f(x)$ and $g(x)$ to exist.

So

and the same holds for $f - g$ and $fg$.

For $\dfrac{f}{g}$ you must also throw out the points where $g(x) = 0$.

Recent board papers have set Assertion-Reason items on exactly this point, with a wrong domain (the union, or one function's domain) offered as the assertion.

1. Let $A = \{a, b\}$ and $B = \{1, 2, 3, 4, 5\}$. The number of relations from $A$ to $B$ is

(A) $10$

(B) $32$

(C) $128$

(D) $1024$

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2. Which of the following relations from $P = \{1, 2, 3\}$ to $Q = \{4, 5\}$ is a function from $P$ to $Q$?

(A) $\{(1, 4), (2, 5)\}$

(B) $\{(1, 4), (1, 5), (2, 4), (3, 5)\}$

(C) $\{(1, 5), (2, 5), (3, 5)\}$

(D) $\{(4, 1), (5, 2), (5, 3)\}$

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3. Assertion (A): For $f(x) = \sqrt{x + 2}$ and $g(x) = \dfrac{1}{x - 1}$, the domain of $f + g$ is $[-2, 1) \cup (1, \infty)$.

Reason (R): The domain of $f + g$ is $D_f \cap D_g$, the set of points where both $f$ and $g$ are defined.

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

(B) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).

(C) Assertion (A) is true but Reason (R) is false.

(D) Assertion (A) is false but Reason (R) is true.

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4. Let $A = \{1, 2\}$ and $B = \{3, 4, 5\}$. Find the number of relations from $A$ to $B$, and the number of these relations that are functions from $A$ to $B$.

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5. (a) Let $A = \{2, 3, 4, 5\}$, $B = \{3, 6, 7, 10, 15\}$ and $R = \{(a, b) : a \in A, \ b \in B, \ a \text{ divides } b\}$. Write $R$ in roster form. Find its domain and range, and check whether $R$ is a function from $A$ to $B$.

OR

(b) Let $f(x) = \sqrt{x - 2}$ and $g(x) = \sqrt{8 - x}$. Find the domains of $f + g$, $fg$ and $\dfrac{f}{g}$, and find $(f + g)(6)$.

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6. Find the number of non-empty relations in the set $A = \{p, q, r\}$, and the number of relations in $A$ that contain both $(p, q)$ and $(q, p)$.

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