📌 Snapshot
- Master ordered pairs, relations and functions, domain and range, and standard graphs with CUET practice.
- How the Cartesian product A × B is built from two sets, and why counting its pairs gives the formula pq.
- Relations as subsets of A × B, the 2 to the power pq count, and the one test that separates a function from a relation.
- Domain and range of the modulus, signum, greatest integer and rational functions, and the algebra of functions.
📖 Detailed Notes
Written and reviewed by Devendra Dubey, Mathematics faculty at UniDrill (M.Sc. Mathematics, Osmania University · 10 years teaching Mathematics). Based on NCERT Mathematics Class XI, Chapter 2.
Why does a roll number point to exactly one student?
In your class, every roll number belongs to exactly one student. Ask "who has roll number 14?" and you always get a single answer. Now ask "which mobile numbers does Asha have?" and the answer may be two numbers, or none. The first rule is a function: every input has exactly one output. The second is only a relation: a link between two collections with no promise about how many outputs each input gets. This chapter builds both ideas from the sets you met in the last chapter, and then looks at the functions CUET asks about most.
Ordered pairs and the Cartesian product
An ordered pair $(p, q)$ is two objects in a fixed order. Order matters here, unlike in a set: $\{2, 3\} = \{3, 2\}$ but $(2, 3) \ne (3, 2)$. You already know why. On a graph, $(2, 3)$ and $(3, 2)$ are different points. Two ordered pairs are equal only when they match place by place:
$$(a, b) = (c, d) \iff a = c \text{ and } b = d$$
The Cartesian product of two sets collects every possible pair, with the first member from the first set and the second from the second set:
$$A \times B = \{(a, b) : a \in A,\ b \in B\}$$
If A or B is empty, then $A \times B = \varnothing$. Otherwise, if $n(A) = p$ and $n(B) = q$, you have p choices for the first place and q for the second, so
$$n(A \times B) = pq, \qquad n(A \times A \times A) = \big(n(A)\big)^3$$
If either set is infinite, so is the product. The plane is $\mathbb{R} \times \mathbb{R}$ and space is $\mathbb{R} \times \mathbb{R} \times \mathbb{R}$. A useful law is $A \times (B \cup C) = (A \times B) \cup (A \times C)$.
Take $A = \{1, 2\}$ and $B = \{a, b, c\}$. The two products have the same size and no common pair.
| Product | Pairs | Count |
|---|---|---|
| $A \times B$ | $(1,a), (1,b), (1,c), (2,a), (2,b), (2,c)$ | 6 |
| $B \times A$ | $(a,1), (b,1), (c,1), (a,2), (b,2), (c,2)$ | 6 |
So $A \times B = B \times A$ only when $A = B$ (or when one of them is empty).
Worked example: equal ordered pairs. If $(3x + 1,\ y - 2) = (x + 5,\ 2y - 6)$, find x and y.
| Step | What we do | Result |
|---|---|---|
| 1 | Equal pairs match in the first place | $3x + 1 = x + 5$, so $x = 2$ |
| 2 | They also match in the second place | $y - 2 = 2y - 6$, so $y = 4$ |
| 3 | Check both pairs | $(7, 2) = (7, 2)$ |
Common wrong step. Equating a first entry with a second entry, such as $3x + 1 = y - 2$. Ordered pairs match position by position: first with first, second with second.
Pause and predict: if $n(A) = 3$ and $n(B) = 4$, what are $n(A \times B)$ and $n(B \times A)$? Are the two sets equal?
Check your answer: both counts are 12, but the sets are not equal. Every pair in $A \times B$ has its first member in A, while every pair in $B \times A$ has its first member in B. Equal size never means equal sets.
Relations: any subset of A × B
A relation R from A to B is any subset of $A \times B$. If $(a, b) \in R$, we say a is related to b, and b is the image of a. Three sets describe a relation:
| Term | Meaning |
|---|---|
| Domain | the set of all first members of the pairs in R |
| Range | the set of all second members of the pairs in R |
| Codomain | the whole target set B |
The range is always a subset of the codomain, but it need not fill it. A relation can be shown by listing pairs (roster form), by a rule (set-builder form) or by an arrow diagram, which draws an arrow from each element to its image.
How many relations are there? Each pair of $A \times B$ is either in R or out of it. With $n(A) = p$ and $n(B) = q$ there are pq pairs, each with two choices, so the number of relations is the number of subsets of $A \times B$:
$$2^{pq}$$
For $p = q = 2$ that is $2^4 = 16$. The exponent is the product pq, never the sum.
Worked example: reading a relation. Let $A = \{1, 2, 3, 4, 5, 6\}$ and $R = \{(x, y) : y = x + 1\}$ from A to A. Write R in roster form and give its domain, range and codomain.
| Step | What we do | Result |
|---|---|---|
| 1 | Test each x in A: does $x + 1$ lie in A? | For $x = 1, \dots, 5$ yes. For $x = 6$, $x + 1 = 7 \notin A$ |
| 2 | List the pairs | $R = \{(1,2), (2,3), (3,4), (4,5), (5,6)\}$ |
| 3 | First members | domain $= \{1, 2, 3, 4, 5\}$ |
| 4 | Second members | range $= \{2, 3, 4, 5, 6\}$ |
| 5 | The target set | codomain $= A = \{1, 2, 3, 4, 5, 6\}$ |
Common wrong step. Calling the domain $\{1, \dots, 6\}$ because A is the set the relation goes "from". The domain holds only the first members that really appear, and 6 has no partner here.
Pause and predict: how many relations are there from $\{1, 2\}$ to $\{a, b, c\}$?
Check your answer: $n(A \times B) = 6$, so there are $2^6 = 64$ relations. The tempting $2^{2+3} = 32$ uses the sum and is wrong.
Functions: one input, exactly one output
A function f from A to B, written $f : A \to B$, is a relation in which every element of A has exactly one image in B. We write $f(a) = b$, call b the image of a and call a a preimage of b. For a function the domain is all of A. The range is the set of images actually used, and it can be smaller than the codomain B.
There are two ways to fail, and they are not the same:
| Failure | What goes wrong | Example from $A = \{1, 2, 3\}$ |
|---|---|---|
| Missing input | some element of A has no image | $\{(1, a), (2, b)\}$, where 3 is left out |
| Split output | some element of A has two images | $\{(1, a), (1, b), (2, c), (3, c)\}$ |
Two different inputs sharing one output is perfectly allowed. One input with two outputs is not.
Worked example: function or not? Let $A = \{1, 2, 3\}$ and $B = \{a, b, c\}$. Decide which of these relations from A to B are functions: (i) $\{(1,a), (2,b), (3,a)\}$, (ii) $\{(1,a), (1,b), (2,c), (3,c)\}$, (iii) $\{(1,a), (2,b)\}$, (iv) $\{(1,a), (2,a), (3,a)\}$.
| Step | What we do | Result |
|---|---|---|
| 1 | For each, check that 1, 2 and 3 each appear as a first member exactly once | (i) yes, (ii) 1 appears twice, (iii) 3 is missing, (iv) yes |
| 2 | Conclude | (i) and (iv) are functions. (ii) and (iii) are not |
Common wrong step. Rejecting (i) because "a is used twice". That is allowed. In (iv) all inputs share one image, which is the constant function. The rule is "one image per input", not "one input per image".
A real-valued function has its range inside $\mathbb{R}$. A real function has both domain and range inside $\mathbb{R}$. Unless a question says otherwise, the domain of a formula is its natural domain, the largest set of real numbers for which the formula makes sense.
When are two functions equal? Same rule is not enough. They need the same domain and codomain too. The function $f(x) = x$ on $\mathbb{R}$ and $g(x) = \dfrac{x^2}{x}$ are different, because g is undefined at $x = 0$. Likewise $f(x) = x^2$ on $\mathbb{R}$ and $g(x) = x^2$ on $[0, \infty)$ are different functions, even though the formulas agree.
Pause and predict: is $f(x) = x$ the same function as $g(x) = \dfrac{x^2}{x}$?
Check your answer: no. Dividing by x is impossible at $x = 0$, so the domain of g is $\mathbb{R} - \{0\}$ while the domain of f is all of $\mathbb{R}$. For every $x \ne 0$ their values agree.
The standard real functions
These seven appear in almost every CUET paper. Know the rule, the domain, the range and the shape of the graph.
| Function | Rule | Domain | Range | Graph in words |
|---|---|---|---|---|
| Identity | $f(x) = x$ | $\mathbb{R}$ | $\mathbb{R}$ | straight line through the origin at 45 degrees |
| Constant | $f(x) = c$ | $\mathbb{R}$ | $\{c\}$ | horizontal line at height c |
| Polynomial | $a_0 + a_1 x + \dots + a_n x^n$, n a whole number | $\mathbb{R}$ | depends on the polynomial | $x^2$ is a parabola, $x^3$ an S-curve through the origin |
| Rational | $\dfrac{f(x)}{g(x)}$, $g(x) \ne 0$ | all x with $g(x) \ne 0$ | depends | $\dfrac{1}{x}$ has two separate branches |
| Modulus | $\lvert x \rvert = x$ if $x \ge 0$, and $-x$ if $x < 0$ | $\mathbb{R}$ | $[0, \infty)$ | V shape with its point at the origin |
| Signum | $1$ if $x > 0$, $0$ if $x = 0$, $-1$ if $x < 0$ | $\mathbb{R}$ | $\{-1, 0, 1\}$ | two flat pieces and a point at the origin |
| Greatest integer | $[x]$ = greatest integer $\le x$ | $\mathbb{R}$ | $\mathbb{Z}$ | staircase with a jump at every integer |
A polynomial needs whole-number powers only. So $x^3 - x^2 + 2$ is a polynomial, while $x^{2/3} + 2x$, $\dfrac{1}{x}$ and $\sqrt{x}$ are not. The linear function $f(x) = mx + c$ is the simplest non-constant polynomial.
Why the range of $1/x$ is $\mathbb{R} - \{0\}$. From $y = \dfrac{1}{x}$ we get $x = \dfrac{1}{y}$, which exists for every y except 0. So every non-zero y is an output, and 0 never is. The domain is also $\mathbb{R} - \{0\}$, since you cannot divide by zero.
The greatest integer function. $[x] = n$ whenever $n \le x < n + 1$. It rounds down, towards negative infinity, not towards zero:
| x | 3.7 | 5 | 0.999 | $-0.5$ | $-2.3$ |
|---|---|---|---|---|---|
| $[x]$ | 3 | 5 | 0 | $-1$ | $-3$ |
For $-2.3$ the integers around it are $-3$ and $-2$, and $-3 \le -2.3 < -2$, so $[-2.3] = -3$.
Pause and predict: what is $[-0.5] + [0.5]$?
Check your answer: $-1$. Since $-1 \le -0.5 < 0$ we get $[-0.5] = -1$, and $[0.5] = 0$. The answer is not 0, because the function rounds down and not towards zero.
Beyond the textbook. For the exam: $[x]$ is the greatest integer not exceeding x, so $[-2.3] = -3$. In reality: this is the floor function from computing. Most programming languages have two different tools, one that truncates (drops the decimal part, so $-2.3$ becomes $-2$) and one that floors (so $-2.3$ becomes $-3$). Only the floor matches $[x]$. Why it matters: negative inputs are exactly where exam options are placed to catch you.
Finding the domain and range
To find a natural domain, ask what could go wrong with the formula. Only two things can: dividing by zero, and taking an even root of a negative number.
| Expression | What must hold | Domain |
|---|---|---|
| $\dfrac{1}{x - 3}$ | $x \ne 3$ | $\mathbb{R} - \{3\}$ |
| $\sqrt{x - 2}$ | $x - 2 \ge 0$ | $[2, \infty)$ |
| $\sqrt{x^2 - 4}$ | $x^2 \ge 4$ | $(-\infty, -2] \cup [2, \infty)$ |
| $\dfrac{x^2 + 1}{x^2 - 3x + 2}$ | $(x - 1)(x - 2) \ne 0$ | $\mathbb{R} - \{1, 2\}$ |
Worked example: a domain with two conditions. Find the domain of $f(x) = \dfrac{\sqrt{x^2 - 4}}{x - 3}$.
| Step | What we do | Result |
|---|---|---|
| 1 | The root needs $x^2 - 4 \ge 0$. Test points: $x = 0$ gives $-4$ (fails), $x = 3$ gives $5$ (works), $x = -3$ gives $5$ (works) | $x \le -2$ or $x \ge 2$ |
| 2 | The denominator needs $x - 3 \ne 0$ | $x \ne 3$ |
| 3 | Keep what satisfies both | $(-\infty, -2] \cup [2, 3) \cup (3, \infty)$ |
Common wrong step. Writing $x^2 \ge 4$ as $-2 \le x \le 2$. That is the solution of $x^2 \le 4$. The point $x = 0$ gives $\sqrt{-4}$, which is not real.
To find a range, you have three tools. You can solve $y = f(x)$ for x and see which y give a real x. You can complete the square for a quadratic. Or you can read it from the graph.
Worked example: two ranges. Find the range of (a) $f(x) = x^2 - 4x + 7$ and (b) $g(x) = \dfrac{x + 1}{x - 2}$.
| Step | What we do | Result |
|---|---|---|
| 1 | (a) Complete the square | $f(x) = (x - 2)^2 + 3$ |
| 2 | A square is never negative and equals 0 at $x = 2$ | $f(x) \ge 3$, and 3 is reached, so the range is $[3, \infty)$ |
| 3 | (b) Solve $y = \dfrac{x + 1}{x - 2}$ for x | $y(x - 2) = x + 1$, so $x(y - 1) = 2y + 1$ |
| 4 | x exists exactly when the factor $y - 1$ is not zero | $x = \dfrac{2y + 1}{y - 1}$ needs $y \ne 1$ |
| 5 | The value $y = 1$ would need $x + 1 = x - 2$, which is impossible | range $= \mathbb{R} - \{1\}$ |
Common wrong step. Stating the range of (b) as $\mathbb{R} - \{2\}$. The number 2 is excluded from the domain because it makes the denominator zero. The range is the set of outputs, and here the missing output is 1.
The algebra of functions
Real functions can be combined, pointwise, on any set X where both are defined:
| Combination | Rule | Condition |
|---|---|---|
| Sum | $(f + g)(x) = f(x) + g(x)$ | none beyond the common domain |
| Difference | $(f - g)(x) = f(x) - g(x)$ | none beyond the common domain |
| Scalar multiple | $(\alpha f)(x) = \alpha f(x)$ | $\alpha \in \mathbb{R}$ |
| Product | $(fg)(x) = f(x)\,g(x)$ | none beyond the common domain |
| Quotient | $\left(\dfrac{f}{g}\right)(x) = \dfrac{f(x)}{g(x)}$ | remove every x with $g(x) = 0$ |
The domain of the result is the overlap of the two domains, and for the quotient you also delete the zeros of g.
Worked example: combining two functions. Let $f(x) = \sqrt{x}$ and $g(x) = x - 4$. Find $(f + g)(9)$, $(fg)(9)$, $\left(\dfrac{f}{g}\right)(9)$ and the domain of $\dfrac{f}{g}$.
| Step | What we do | Result |
|---|---|---|
| 1 | Evaluate each function at 9 | $f(9) = 3$, $g(9) = 5$ |
| 2 | Sum and product | $(f + g)(9) = 8$, $(fg)(9) = 15$ |
| 3 | Quotient | $\left(\dfrac{f}{g}\right)(9) = \dfrac{3}{5}$ |
| 4 | Domain of f is $[0, \infty)$, of g is $\mathbb{R}$. Then remove the zero of g, which is $x = 4$ | $[0, 4) \cup (4, \infty)$ |
Common wrong step. Giving $[0, \infty)$ as the domain of the quotient. At $x = 4$ the quotient has a zero denominator, even though $\sqrt{4}$ is perfectly fine.
Beyond the textbook. For the exam: a function is a rule with a formula and a graph. In reality: modern mathematics defines a function as nothing but its set of ordered pairs, and an ordered pair $(a, b)$ can itself be built from sets, as $\{\{a\}, \{a, b\}\}$, so that order is stored using only membership. Why it matters: this is why a relation is "a subset of $A \times B$" and why two functions are equal only if they have the same pairs, the same domain and the same codomain. A formula is only one way to describe the pairs.
Beyond the textbook. For the exam: the modulus is $\lvert x \rvert = x$ or $-x$ by sign. In reality: $\lvert x - a \rvert$ is the distance between x and a on the number line, which is why it appears in error tolerances and "within 0.5 of the target" conditions. The modulus and signum functions also split any real number into size and direction: $x = \lvert x \rvert \cdot \text{sgn}(x)$ for every x. Why it matters: "within 2 of 5" becomes the single statement $\lvert x - 5 \rvert \le 2$, and you can read it as a distance instead of splitting into cases.
Exam traps
| The trap | Why it's tempting | The truth |
|---|---|---|
| $A \times B = B \times A$ | Both have pq pairs | Equal only when $A = B$ or one is empty. The pairs are ordered |
| Relations from A to B number $2^{p+q}$ | Adding feels natural | There are $2^{pq}$, one in/out choice for each of the pq pairs |
| A repeated image means "not a function" | Repetition looks like a defect | Only a repeated input with different outputs breaks it |
| Domain of a relation equals A | A is the "from" set | Domain holds only first members that appear. A function's domain is all of A |
| Range equals codomain | Both are target sets | Range is a subset of the codomain and may be smaller |
| Signum has range $[-1, 1]$ | A line from $-1$ to $1$ seems natural | Range is exactly $\{-1, 0, 1\}$ |
| $[-0.5] = 0$ and $[0.999] = 1$ | Rounding habit | $[x]$ rounds down: $[-0.5] = -1$ and $[0.999] = 0$ |
| $f(a + b) = f(a) + f(b)$ | Looks like distributing | True only for special f. For $f(x) = x^2 + 1$ it fails |
Formula sheet
| Quantity | Formula | When to use |
|---|---|---|
| Equal ordered pairs | $(a, b) = (c, d) \iff a = c,\ b = d$ | solving for unknowns in pairs |
| Size of a product | $n(A \times B) = pq$ | counting pairs |
| Size of a triple product | $n(A \times A \times A) = \big(n(A)\big)^3$ | counting triplets |
| Number of relations | $2^{pq}$ | relations from A to B |
| Modulus | $\lvert x \rvert = x$ for $x \ge 0$, $-x$ for $x < 0$ | range $[0, \infty)$ |
| Greatest integer | $[x] = n$ for $n \le x < n + 1$ | rounding down |
| Domain of $\sqrt{f(x)}$ | $f(x) \ge 0$ | root functions |
| Domain of $\dfrac{f}{g}$ | domain overlap with $g(x) \ne 0$ | quotients |
| Range of $ax^2 + bx + c$ | $\ge c - \dfrac{b^2}{4a}$ if $a > 0$, $\le$ that if $a < 0$ | quadratics, by completing the square |
60-second revision
| Topic | Lock this in |
|---|---|
| Ordered pair | Order matters. Equal means first with first, second with second |
| $A \times B$ | pq pairs. Not equal to $B \times A$ in general |
| Relation | Any subset of $A \times B$. There are $2^{pq}$ of them |
| Domain, range, codomain | First members, second members, whole target set |
| Function | Every element of A has exactly one image |
| Constant, identity | Range $\{c\}$, range $\mathbb{R}$ |
| Modulus, signum | Range $[0, \infty)$ and $\{-1, 0, 1\}$ |
| Greatest integer | Rounds down: $[-2.3] = -3$. Range $\mathbb{Z}$ |
| Domain rules | Denominator not 0, even root argument not negative |
| Quotient $f/g$ | Domain overlap, minus the zeros of g |
Where this chapter connects
- Previous: Sets supplies the language. A relation is a subset of a Cartesian product, and domain and range are sets.
- Next: Trigonometric Functions defines sine, cosine and the rest as real functions with their own domains, ranges and periods.
- Class XII link: Relations and Functions goes further and asks which functions are one to one, onto or invertible.
🎯 Practice MCQs
First 3 questions free · create a free account to unlock the rest — answers & explanations included, no payment needed
Q1. If $(2x - 1,\ y + 3) = (x + 4,\ 3y - 5)$, then the values of x and y are
▸ Show answer & explanation
Answer: A
Equate first entries: $2x - 1 = x + 4$ gives $x = 5$. Equate second entries: $y + 3 = 3y - 5$ gives $2y = 8$, so $y = 4$. Option B swaps the two values.
Q2. Let $A = \{1, 2\}$ and $B = \{2, 3\}$. Consider the statements. I: $n(A \times B) = n(B \times A)$. II: $A \times B = B \times A$. III: The pair $(2, 2)$ belongs to both $A \times B$ and $B \times A$. Which are correct?
▸ Show answer & explanation
Answer: B
Here $A \times B = \{(1,2), (1,3), (2,2), (2,3)\}$ and $B \times A = \{(2,1), (2,2), (3,1), (3,2)\}$. Both have 4 pairs, so I is true. They are different sets, so II is false. The pair $(2, 2)$ is in both, so III is true.
Q3. If $n(A) = 3$ and $n(B) = 4$, the number of relations from A to B is
▸ Show answer & explanation
Answer: C
$n(A \times B) = 12$ and every subset of $A \times B$ is a relation, so there are $2^{12} = 4096$ relations. Option A is $2^{3+4}$, which wrongly uses the sum, and option D is just $n(A \times B)$.
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Q4. Which of the following relations from $A = \{1, 2, 3\}$ to $B = \{1, 2, 3, \dots, 8\}$ is a function?
▸ Show answer & explanation
Answer: B
In B each of 1, 2, 3 has exactly one image, and sharing the image 4 is allowed. In A the element 1 has two images. In C the element 3 has no image. In D the element 2 has two images.
Q5. The domain of $f(x) = \sqrt{x^2 - 9}$ is
▸ Show answer & explanation
Answer: B
We need $x^2 - 9 \ge 0$, which means $x \le -3$ or $x \ge 3$. Option A solves $x^2 \le 9$ instead. Option C misses the negative values: $x = -4$ gives $\sqrt{7}$, which is real.
Q6. The range of $f(x) = x^2 - 6x + 11$ is
▸ Show answer & explanation
Answer: A
Completing the square, $f(x) = (x - 3)^2 + 2$. A square is at least 0, so $f(x) \ge 2$, with the value 2 reached at $x = 3$. Option B is the value at $x = 0$, not the minimum.
Q7. The value of $[-2.5] + [3.9] + [-0.1]$, where $[x]$ is the greatest integer function, is
▸ Show answer & explanation
Answer: B
$[-2.5] = -3$, $[3.9] = 3$ and $[-0.1] = -1$, so the sum is $-3 + 3 - 1 = -1$. Rounding towards zero would give $-2 + 3 + 0 = 1$, which is the trap in option C.
Q8. Match List I with List II. List I: (i) $f(x) = \lvert x \rvert$ (ii) the signum function (iii) $f(x) = [x]$ (iv) $f(x) = \dfrac{1}{x}$. List II: (p) range $\mathbb{R} - \{0\}$ (q) range $[0, \infty)$ (r) range $\{-1, 0, 1\}$ (s) range $\mathbb{Z}$
▸ Show answer & explanation
Answer: A
The modulus is never negative, so its range is $[0, \infty)$. The signum function takes only the three values $-1$, $0$ and $1$. The greatest integer function takes integer values only. The function $1/x$ never equals 0.
Q9. Assertion (A): The relation $\{(1, 2), (2, 2), (3, 2)\}$ is a function from $\{1, 2, 3\}$ to $\{2\}$. Reason (R): In a function, two different elements of the domain cannot have the same image.
▸ Show answer & explanation
Answer: C
Each of 1, 2, 3 has exactly one image, so A is true. It is a constant function. R is false because different inputs may share an image. A function forbids only one input having two images.
Q10. If $f(x) = \sqrt{x}$ and $g(x) = x - 4$, the domain of $\dfrac{f}{g}$ is
▸ Show answer & explanation
Answer: C
The function f needs $x \ge 0$ and g is defined everywhere. For the quotient we also need $g(x) \ne 0$, so $x \ne 4$. Option A forgets to remove 4. Option B ignores the root.
Q11. If $f(x) = x^2 + 1$, then $f(a + b) - f(a) - f(b)$ equals
▸ Show answer & explanation
Answer: C
$f(a + b) = (a + b)^2 + 1 = a^2 + 2ab + b^2 + 1$. Subtracting $f(a) = a^2 + 1$ and $f(b) = b^2 + 1$ leaves $2ab + 1 - 2 = 2ab - 1$. The function is not additive.
Q12. Which of the following is a polynomial function?
▸ Show answer & explanation
Answer: B
A polynomial allows only whole-number powers of x. Option A has the power 2/3, option C contains $x^{-1}$ and option D contains $x^{1/2}$.
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