Chapter 2 of 14

Updated 6 August 2026

NCERT Solutions

Relations and Functions — NCERT Solutions

Q: What are the important topics in Relations and Functions for CBSE Class 11 Mathematics?

Key topics in Relations and Functions include A pair of elements written, For two non, A relation from a non, The domain is the set. Study these first, then practise questions on each for Class 11 exams.

Q: Are these NCERT Solutions for Relations and Functions free?

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Q: How should I revise Relations and Functions for Class 11 exams?

Learn the core ideas first, then work through the 136 practice questions on Relations and Functions. Revise definitions regularly and use flashcards for quick recall before the exam.

CBSE · Class 11 · Mathematics

NCERT Solutions for Relations and Functions, CBSE Class 11 Mathematics: 36 textbook questions solved step by step.

136 questions56 flashcards9 formulas & key relations5 concepts

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36 Questions Solved · 4 Sections

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Exercise 2.1

1If

\left(\frac{x}{3} + 1, y - \frac{2}{3}\right) = \left(\frac{5}{3}, \frac{1}{3}\right)

, find the values of

x

and

y

.Show solution

Given: Two ordered pairs are equal: $\left(\dfrac{x}{3} + 1,\ y - \dfrac{2}{3}\right) = \left(\dfrac{5}{3},\ \dfrac{1}{3}\right)$ .

Concept: Two ordered pairs $(a, b)$ and $(c, d)$ are equal if and only if $a = c$ and $b = d$ .

Equating first elements:
$\frac{x}{3} + 1 = \frac{5}{3}$
$\frac{x}{3} = \frac{5}{3} - 1 = \frac{5-3}{3} = \frac{2}{3}$
$x = 2$

Equating second elements:
$y - \frac{2}{3} = \frac{1}{3}$
$y = \frac{1}{3} + \frac{2}{3} = \frac{3}{3} = 1$

Answer: $x = 2$ and $y = 1$ .

2If the set A has 3 elements and the set

\mathrm{B} = \{3, 4, 5\}

, then find the number of elements in

(\mathrm{A} \times \mathrm{B})

.Show solution

Given: $n(\mathrm{A}) = 3$ and $\mathrm{B} = \{3, 4, 5\}$ , so $n(\mathrm{B}) = 3$ .

Concept: If $n(\mathrm{A}) = p$ and $n(\mathrm{B}) = q$ , then $n(\mathrm{A} \times \mathrm{B}) = p \times q$ .

Calculation:
$n(\mathrm{A} \times \mathrm{B}) = n(\mathrm{A}) \times n(\mathrm{B}) = 3 \times 3 = 9$

Answer: The number of elements in $\mathrm{A} \times \mathrm{B}$ is $\mathbf{9}$ .

3If

\mathrm{G} = \{7, 8\}

and

\mathrm{H} = \{5, 4, 2\}

, find

\mathrm{G} \times \mathrm{H}

and

\mathrm{H} \times \mathrm{G}

.Show solution

Given: $\mathrm{G} = \{7, 8\}$ and $\mathrm{H} = \{5, 4, 2\}$ .

Concept: $\mathrm{A} \times \mathrm{B} = \{(a, b) : a \in \mathrm{A},\ b \in \mathrm{B}\}$ .

Finding $\mathrm{G} \times \mathrm{H}$ :
$\mathrm{G} \times \mathrm{H} = \{(7,5),\ (7,4),\ (7,2),\ (8,5),\ (8,4),\ (8,2)\}$

Finding $\mathrm{H} \times \mathrm{G}$ :
$\mathrm{H} \times \mathrm{G} = \{(5,7),\ (5,8),\ (4,7),\ (4,8),\ (2,7),\ (2,8)\}$

4State whether each of the following statements are true or false. If the statement is false, rewrite the given statement correctly.
(i) If

\mathrm{P} = \{m, n\}

and

\mathrm{Q} = \{n, m\}

, then

\mathrm{P} \times \mathrm{Q} = \{(m, n), (n, m)\}

.
(ii) If A and B are non-empty sets, then

\mathrm{A} \times \mathrm{B}

is a non-empty set of ordered pairs

(x, y)

such that

x \in \mathrm{A}

and

y \in \mathrm{B}

.
(iii) If

\mathrm{A} = \{1, 2\}

\mathrm{B} = \{3, 4\}

, then

\mathrm{A} \times (\mathrm{B} \cap \phi) = \phi

.Show solution

(i) False.

$\mathrm{P} = \{m, n\}$ and $\mathrm{Q} = \{n, m\} = \{m, n\}$ .

The correct Cartesian product is:
$\mathrm{P} \times \mathrm{Q} = \{(m,m),\ (m,n),\ (n,m),\ (n,n)\}$

The given statement lists only 2 pairs instead of 4, so it is False.

Correct statement: If $\mathrm{P} = \{m, n\}$ and $\mathrm{Q} = \{n, m\}$ , then $\mathrm{P} \times \mathrm{Q} = \{(m,n),\ (m,m),\ (n,n),\ (n,m)\}$ .

(ii) True.

By definition, if A and B are non-empty sets, $\mathrm{A} \times \mathrm{B}$ is indeed a non-empty set of ordered pairs $(x, y)$ with $x \in \mathrm{A}$ and $y \in \mathrm{B}$ .

(iii) True.

$\mathrm{B} \cap \phi = \phi$ (intersection of any set with the empty set is empty).

Therefore, $\mathrm{A} \times (\mathrm{B} \cap \phi) = \mathrm{A} \times \phi = \phi$ .

5If

\mathrm{A} = \{-1, 1\}

, find

\mathrm{A} \times \mathrm{A} \times \mathrm{A}

.Show solution

Given: $\mathrm{A} = \{-1, 1\}$ .

Concept: $\mathrm{A} \times \mathrm{A} \times \mathrm{A} = \{(a, b, c) : a, b, c \in \mathrm{A}\}$ .

Solution:
$\mathrm{A} \times \mathrm{A} \times \mathrm{A} = \{(-1,-1,-1),\ (-1,-1,1),\ (-1,1,-1),\ (-1,1,1),$
$(1,-1,-1),\ (1,-1,1),\ (1,1,-1),\ (1,1,1)\}$

There are $2^3 = 8$ elements in total.

6If

\mathrm{A} \times \mathrm{B} = \{(a, x), (a, y), (b, x), (b, y)\}

. Find A and B.Show solution

Given: $\mathrm{A} \times \mathrm{B} = \{(a, x),\ (a, y),\ (b, x),\ (b, y)\}$ .

Concept: In a Cartesian product, A is the set of all first elements and B is the set of all second elements of the ordered pairs.

Finding A: Set of all first elements $= \{a, b\}$
$\therefore \mathrm{A} = \{a, b\}$

Finding B: Set of all second elements $= \{x, y\}$
$\therefore \mathrm{B} = \{x, y\}$

7Let

\mathrm{A} = \{1, 2\}

\mathrm{B} = \{1, 2, 3, 4\}

\mathrm{C} = \{5, 6\}

and

\mathrm{D} = \{5, 6, 7, 8\}

. Verify that
(i)

\mathrm{A} \times (\mathrm{B} \cap \mathrm{C}) = (\mathrm{A} \times \mathrm{B}) \cap (\mathrm{A} \times \mathrm{C})

.
(ii)

\mathrm{A} \times \mathrm{C}

is a subset of

\mathrm{B} \times \mathrm{D}

.Show solution

Given: $\mathrm{A} = \{1,2\}$ , $\mathrm{B} = \{1,2,3,4\}$ , $\mathrm{C} = \{5,6\}$ , $\mathrm{D} = \{5,6,7,8\}$ .

(i) Verify $\mathrm{A} \times (\mathrm{B} \cap \mathrm{C}) = (\mathrm{A} \times \mathrm{B}) \cap (\mathrm{A} \times \mathrm{C})$ :

LHS:
$\mathrm{B} \cap \mathrm{C} = \{1,2,3,4\} \cap \{5,6\} = \phi$
$\mathrm{A} \times (\mathrm{B} \cap \mathrm{C}) = \mathrm{A} \times \phi = \phi$

RHS:
$\mathrm{A} \times \mathrm{B} = \{(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4)\}$
$\mathrm{A} \times \mathrm{C} = \{(1,5),(1,6),(2,5),(2,6)\}$
$(\mathrm{A} \times \mathrm{B}) \cap (\mathrm{A} \times \mathrm{C}) = \phi \quad (\text{no common elements})$

Since LHS $= \phi =$ RHS, the statement is verified.

(ii) Verify $\mathrm{A} \times \mathrm{C} \subseteq \mathrm{B} \times \mathrm{D}$ :

$\mathrm{A} \times \mathrm{C} = \{(1,5),(1,6),(2,5),(2,6)\}$
$\mathrm{B} \times \mathrm{D} = \{(1,5),(1,6),(1,7),(1,8),(2,5),(2,6),(2,7),(2,8),(3,5),\ldots,(4,8)\}$

Checking each element of $\mathrm{A} \times \mathrm{C}$ :

$(1,5) \in \mathrm{B} \times \mathrm{D}$ ✓ (since $1 \in \mathrm{B}$ , $5 \in \mathrm{D}$ )
$(1,6) \in \mathrm{B} \times \mathrm{D}$ ✓
$(2,5) \in \mathrm{B} \times \mathrm{D}$ ✓
$(2,6) \in \mathrm{B} \times \mathrm{D}$ ✓

Since every element of $\mathrm{A} \times \mathrm{C}$ belongs to $\mathrm{B} \times \mathrm{D}$ , we have $\mathrm{A} \times \mathrm{C} \subseteq \mathrm{B} \times \mathrm{D}$ . Verified.

8Let

\mathrm{A} = \{1, 2\}

and

\mathrm{B} = \{3, 4\}

. Write

\mathrm{A} \times \mathrm{B}

. How many subsets will

\mathrm{A} \times \mathrm{B}

have? List them.Show solution

Given: $\mathrm{A} = \{1,2\}$ , $\mathrm{B} = \{3,4\}$ .

Finding $\mathrm{A} \times \mathrm{B}$ :
$\mathrm{A} \times \mathrm{B} = \{(1,3),\ (1,4),\ (2,3),\ (2,4)\}$

Number of subsets: $n(\mathrm{A} \times \mathrm{B}) = 4$ , so the number of subsets $= 2^4 = \mathbf{16}$ .

List of all subsets:

$\phi$
$\{(1,3)\}$
$\{(1,4)\}$
$\{(2,3)\}$
$\{(2,4)\}$
$\{(1,3),(1,4)\}$
$\{(1,3),(2,3)\}$
$\{(1,3),(2,4)\}$
$\{(1,4),(2,3)\}$
$\{(1,4),(2,4)\}$
$\{(2,3),(2,4)\}$
$\{(1,3),(1,4),(2,3)\}$
$\{(1,3),(1,4),(2,4)\}$
$\{(1,3),(2,3),(2,4)\}$
$\{(1,4),(2,3),(2,4)\}$
$\{(1,3),(1,4),(2,3),(2,4)\}$

9Let A and B be two sets such that

n(\mathrm{A}) = 3

and

n(\mathrm{B}) = 2

. If

(x, 1), (y, 2), (z, 1)

are in

\mathrm{A} \times \mathrm{B}

, find A and B, where

x, y

and

z

are distinct elements.Show solution

Given: $n(\mathrm{A}) = 3$ , $n(\mathrm{B}) = 2$ , and $(x,1),\ (y,2),\ (z,1) \in \mathrm{A} \times \mathrm{B}$ , where $x, y, z$ are distinct.

Finding A: The first elements of the ordered pairs in $\mathrm{A} \times \mathrm{B}$ belong to A.
First elements are $x, y, z$ (all distinct), and $n(\mathrm{A}) = 3$ .
$\therefore \mathrm{A} = \{x, y, z\}$

Finding B: The second elements of the ordered pairs in $\mathrm{A} \times \mathrm{B}$ belong to B.
Second elements are $1, 2, 1$ , i.e., the distinct values are $1$ and $2$ , and $n(\mathrm{B}) = 2$ .
$\therefore \mathrm{B} = \{1, 2\}$

10The Cartesian product

\mathrm{A} \times \mathrm{A}

has 9 elements among which are found

(-1, 0)

and

(0, 1)

. Find the set

\mathrm{A}

and the remaining elements of

\mathrm{A} \times \mathrm{A}

.Show solution

Given: $n(\mathrm{A} \times \mathrm{A}) = 9$ and $(-1, 0),\ (0, 1) \in \mathrm{A} \times \mathrm{A}$ .

Finding $n(\mathrm{A})$ :
$n(\mathrm{A} \times \mathrm{A}) = [n(\mathrm{A})]^2 = 9 \implies n(\mathrm{A}) = 3$

Finding A: Since $(-1, 0) \in \mathrm{A} \times \mathrm{A}$ , both $-1$ and $0$ must be in A. Since $(0, 1) \in \mathrm{A} \times \mathrm{A}$ , $1$ must also be in A. We already have 3 distinct elements $\{-1, 0, 1\}$ and $n(\mathrm{A}) = 3$ .
$\therefore \mathrm{A} = \{-1, 0, 1\}$

Finding all elements of $\mathrm{A} \times \mathrm{A}$ :
$\mathrm{A} \times \mathrm{A} = \{(-1,-1),\ (-1,0),\ (-1,1),\ (0,-1),\ (0,0),\ (0,1),\ (1,-1),\ (1,0),\ (1,1)\}$

Remaining elements (other than $(-1,0)$ and $(0,1)$ ) are:
$\{(-1,-1),\ (-1,1),\ (0,-1),\ (0,0),\ (1,-1),\ (1,0),\ (1,1)\}$

Exercise 2.2

1Let

\mathrm{A} = \{1, 2, 3, \dots, 14\}

. Define a relation

\mathrm{R}

from

\mathrm{A}

\mathrm{A}

\mathrm{R} = \{(x, y) : 3x - y = 0 \text{, where } x, y \in \mathrm{A}\}

. Write down its domain, codomain and range.Show solution

Given: $\mathrm{A} = \{1, 2, 3, \ldots, 14\}$ and $\mathrm{R} = \{(x, y) : 3x - y = 0,\ x, y \in \mathrm{A}\}$ .

Condition: $3x - y = 0 \Rightarrow y = 3x$ .

We need $x, y \in \mathrm{A}$ , i.e., both $x$ and $3x$ must lie in $\{1, 2, \ldots, 14\}$ .

$x$	$y = 3x$	In A?
1	3	Yes
2	6	Yes
3	9	Yes
4	12	Yes
5	15	No (15 > 14)

So the relation in roster form is:
$\mathrm{R} = \{(1,3),\ (2,6),\ (3,9),\ (4,12)\}$

Domain = set of first elements $= \{1, 2, 3, 4\}$

Codomain = A $= \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14\}$

Range = set of second elements $= \{3, 6, 9, 12\}$

2Define a relation

\mathbf{R}

on the set

\mathbf{N}

of natural numbers by

\mathbf{R} = \{(x, y) : y = x + 5, x \text{ is a natural number less than } 4; x, y \in \mathbf{N}\}

. Depict this relationship using roster form. Write down the domain and the range.Show solution

Given: $\mathrm{R} = \{(x, y) : y = x + 5,\ x \in \mathbf{N},\ x < 4\}$ .

Natural numbers less than 4: $x = 1, 2, 3$ .

$x$	$y = x + 5$
1	6
2	7
3	8

Roster form:
$\mathrm{R} = \{(1, 6),\ (2, 7),\ (3, 8)\}$

Domain $= \{1, 2, 3\}$

Range $= \{6, 7, 8\}$

\mathrm{A} = \{1, 2, 3, 5\}

and

\mathrm{B} = \{4, 6, 9\}

. Define a relation

\mathrm{R}

from

\mathrm{A}

\mathrm{B}

\mathrm{R} = \{(x, y) : \text{the difference between } x \text{ and } y \text{ is odd; } x \in \mathrm{A}, y \in \mathrm{B}\}

. Write

\mathrm{R}

in roster form.Show solution

Given: $\mathrm{A} = \{1, 2, 3, 5\}$ , $\mathrm{B} = \{4, 6, 9\}$ .

Condition: $|x - y|$ is odd (i.e., $x - y$ is odd), which happens when one of $x, y$ is even and the other is odd.

Checking all pairs $(x, y)$ with $x \in \mathrm{A}$ , $y \in \mathrm{B}$ :

$x=1$ (odd): $y=4$ (even) → $1-4=-3$ odd ✓; $y=6$ (even) → odd ✓; $y=9$ (odd) → even ✗
$x=2$ (even): $y=4$ (even) → even ✗; $y=6$ (even) → even ✗; $y=9$ (odd) → odd ✓
$x=3$ (odd): $y=4$ → odd ✓; $y=6$ → odd ✓; $y=9$ → even ✗
$x=5$ (odd): $y=4$ → odd ✓; $y=6$ → odd ✓; $y=9$ → even ✗

$\mathrm{R} = \{(1,4),\ (1,6),\ (2,9),\ (3,4),\ (3,6),\ (5,4),\ (5,6)\}$

4The Fig 2.7 shows a relationship between the sets P and Q. Write this relation (i) in set-builder form (ii) roster form. What is its domain and range?Show solution

Note: The figure (Fig 2.7) is not visible in the OCR text. Based on the standard NCERT textbook, Fig 2.7 shows the relation where elements of P = {5, 6, 7} are related to elements of Q = {3, 4, 5} by the rule $x - y = 2$ (i.e., $y = x - 2$ ), giving pairs (5,3), (6,4), (7,5).

(i) Set-builder form:
$\mathrm{R} = \{(x, y) : y = x - 2,\ x \in \mathrm{P},\ y \in \mathrm{Q}\}$
or equivalently $\mathrm{R} = \{(x,y): x - y = 2,\ x \in \mathrm{P},\ y \in \mathrm{Q}\}$ .

(ii) Roster form:
$\mathrm{R} = \{(5, 3),\ (6, 4),\ (7, 5)\}$

Domain $= \{5, 6, 7\}$

Range $= \{3, 4, 5\}$

5Let

\mathrm{A} = \{1, 2, 3, 4, 6\}

. Let

\mathrm{R}

be the relation on

\mathrm{A}

defined by

\{(a,b): a, b \in \mathrm{A}, b \text{ is exactly divisible by } a\}

.
(i) Write R in roster form
(ii) Find the domain of R
(iii) Find the range of R.Show solution

Given: $\mathrm{A} = \{1, 2, 3, 4, 6\}$ , $\mathrm{R} = \{(a,b) : a,b \in \mathrm{A},\ a \mid b\}$ .

(i) Roster form: We list all pairs $(a, b)$ where $b$ is exactly divisible by $a$ :

$a=1$ : $b$ can be $1, 2, 3, 4, 6$ → $(1,1),(1,2),(1,3),(1,4),(1,6)$
$a=2$ : $b$ can be $2, 4, 6$ → $(2,2),(2,4),(2,6)$
$a=3$ : $b$ can be $3, 6$ → $(3,3),(3,6)$
$a=4$ : $b$ can be $4$ → $(4,4)$
$a=6$ : $b$ can be $6$ → $(6,6)$

$\mathrm{R} = \{(1,1),(1,2),(1,3),(1,4),(1,6),(2,2),(2,4),(2,6),(3,3),(3,6),(4,4),(6,6)\}$

(ii) Domain = set of all first elements $= \{1, 2, 3, 4, 6\}$

(iii) Range = set of all second elements $= \{1, 2, 3, 4, 6\}$

6Determine the domain and range of the relation

\mathbf{R}

defined by

\mathrm{R} = \{(x, x + 5) : x \in \{0, 1, 2, 3, 4, 5\}\}

.Show solution

Given: $\mathrm{R} = \{(x, x+5) : x \in \{0,1,2,3,4,5\}\}$ .

Listing all ordered pairs:

$x$	$x+5$
0	5
1	6
2	7
3	8
4	9
5	10

$\mathrm{R} = \{(0,5),(1,6),(2,7),(3,8),(4,9),(5,10)\}$

Domain $= \{0, 1, 2, 3, 4, 5\}$

Range $= \{5, 6, 7, 8, 9, 10\}$

7Write the relation

\mathrm{R} = \{(x, x^3) : x \text{ is a prime number less than } 10\}

in roster form.Show solution

Given: $\mathrm{R} = \{(x, x^3) : x \text{ is a prime number less than } 10\}$ .

Prime numbers less than 10: $2, 3, 5, 7$ .

$x$	$x^3$
2	8
3	27
5	125
7	343

$\mathrm{R} = \{(2, 8),\ (3, 27),\ (5, 125),\ (7, 343)\}$

8Let

\mathrm{A} = \{x, y, z\}

and

\mathrm{B} = \{1, 2\}

. Find the number of relations from

\mathrm{A}

\mathrm{B}

.Show solution

Given: $\mathrm{A} = \{x, y, z\}$ , $\mathrm{B} = \{1, 2\}$ .

$n(\mathrm{A}) = 3$ , $n(\mathrm{B}) = 2$ .

$n(\mathrm{A} \times \mathrm{B}) = 3 \times 2 = 6$

Concept: The number of relations from A to B = number of subsets of $\mathrm{A} \times \mathrm{B} = 2^{n(\mathrm{A} \times \mathrm{B})}$ .

$\text{Number of relations} = 2^6 = \mathbf{64}$

9Let

\mathbf{R}

be the relation on

\mathbf{Z}

defined by

\mathrm{R} = \{(a, b) : a, b \in \mathbf{Z}, a - b \text{ is an integer}\}

. Find the domain and range of

\mathbf{R}

Given: $\mathrm{R} = \{(a, b) : a, b \in \mathbf{Z},\ a - b \in \mathbf{Z}\}$ .

Observation: For any two integers $a$ and $b$ , their difference $a - b$ is always an integer. Therefore, every pair $(a, b)$ with $a, b \in \mathbf{Z}$ satisfies the condition.

This means $\mathrm{R} = \mathbf{Z} \times \mathbf{Z}$ .

Domain = set of all first elements $= \mathbf{Z}$

Range = set of all second elements $= \mathbf{Z}$

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Exercise 2.3

1Which of the following relations are functions? Give reasons. If it is a function, determine its domain and range.
(i)

\{(2,1),(5,1),(8,1),(11,1),(14,1),(17,1)\}

(ii)

\{(2,1),(4,2),(6,3),(8,4),(10,5),(12,6),(14,7)\}

(iii)

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

Concept: A relation is a function if every element in the domain has exactly one image (no first element is repeated with different second elements).

(i) $\{(2,1),(5,1),(8,1),(11,1),(14,1),(17,1)\}$

Each first element $2, 5, 8, 11, 14, 17$ appears exactly once. So this is a function.

Domain $= \{2, 5, 8, 11, 14, 17\}$
Range $= \{1\}$

(ii) $\{(2,1),(4,2),(6,3),(8,4),(10,5),(12,6),(14,7)\}$

Each first element $2, 4, 6, 8, 10, 12, 14$ appears exactly once. So this is a function.

Domain $= \{2, 4, 6, 8, 10, 12, 14\}$
Range $= \{1, 2, 3, 4, 5, 6, 7\}$

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

The element $1$ appears as the first element in two pairs: $(1,3)$ and $(1,5)$ , with two different images $3$ and $5$ . So this is NOT a function.

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2Find the domain and range of the following real functions:
(i)

f(x) = -|x|

(ii)

f(x) = \sqrt{9 - x^2}

(i) $f(x) = -|x|$

Domain: $|x|$ is defined for all real numbers.
$\text{Domain} = \mathbf{R}$

Range: For any $x \in \mathbf{R}$ , $|x| \geq 0$ , so $-|x| \leq 0$ .
The function takes all values $\leq 0$ .
$\text{Range} = (-\infty, 0] = \{y \in \mathbf{R} : y \leq 0\}$

(ii) $f(x) = \sqrt{9 - x^2}$

Domain: The expression under the square root must be non-negative:
$9 - x^2 \geq 0 \implies x^2 \leq 9 \implies -3 \leq x \leq 3$
$\text{Domain} = [-3, 3]$

Range: When $x \in [-3, 3]$ , $x^2 \in [0, 9]$ , so $9 - x^2 \in [0, 9]$ , and $\sqrt{9-x^2} \in [0, 3]$ .
$\text{Range} = [0, 3]$

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3A function

f

is defined by

f(x) = 2x - 5

. Write down the values of
(i)

f(0)

(ii)

f(7)

(iii)

f(-3)

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4The function

t

which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by

t(\mathrm{C}) = \dfrac{9\mathrm{C}}{5} + 32

. Find (i)

t(0)

(ii)

t(28)

(iii)

t(-10)

(iv) The value of

C

, when

t(C) = 212

Given: $t(\mathrm{C}) = \dfrac{9\mathrm{C}}{5} + 32$ .

(i) $t(0)$ :
$t(0) = \frac{9 \times 0}{5} + 32 = 0 + 32 = 32$

(ii) $t(28)$ :
$t(28) = \frac{9 \times 28}{5} + 32 = \frac{252}{5} + 32 = 50.4 + 32 = 82.4$

(iii) $t(-10)$ :
$t(-10) = \frac{9 \times (-10)}{5} + 32 = \frac{-90}{5} + 32 = -18 + 32 = 14$

(iv) Value of $C$ when $t(C) = 212$ :
$\frac{9C}{5} + 32 = 212$
$\frac{9C}{5} = 212 - 32 = 180$
$9C = 900$
$C = 100$

Answer: $t(0) = 32$ , $t(28) = 82.4$ , $t(-10) = 14$ , and $C = 100$ when $t(C) = 212$ .

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5Find the range of each of the following functions.
(i)

f(x) = 2 - 3x, x \in \mathbf{R}, x > 0.

(ii)

f(x) = x^{2} + 2, x

is a real number.
(iii)

f(x) = x, x

is a real number.

(i) $f(x) = 2 - 3x,\ x \in \mathbf{R},\ x > 0$ :

Since $x > 0$ , we have $3x > 0$ , so $-3x < 0$ , thus $2 - 3x < 2$ .
As $x \to 0^+$ , $f(x) \to 2$ (but $x = 0$ is excluded, so $f(x)$ approaches but never equals 2).
As $x \to +\infty$ , $f(x) \to -\infty$ .

$\text{Range} = (-\infty, 2) = \{y \in \mathbf{R} : y < 2\}$

(ii) $f(x) = x^2 + 2,\ x \in \mathbf{R}$ :

For all real $x$ , $x^2 \geq 0$ , so $x^2 + 2 \geq 2$ .
The minimum value is $2$ (attained at $x = 0$ ), and $f(x)$ can be arbitrarily large.

$\text{Range} = [2, +\infty) = \{y \in \mathbf{R} : y \geq 2\}$

(iii) $f(x) = x,\ x \in \mathbf{R}$ :

This is the identity function. For every real number $y$ , there exists $x = y$ such that $f(x) = y$ .

$\text{Range} = \mathbf{R}$

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Miscellaneous Exercise on Chapter 2

1The relation

f

is defined by

f(x) = \begin{cases} x^2, & 0 \leq x \leq 3 \\ 3x, & 3 \leq x \leq 10 \end{cases}

. The relation

g

is defined by

g(x) = \begin{cases} x^2, & 0 \leq x \leq 2 \\ 3x, & 2 \leq x \leq 10 \end{cases}

. Show that

f

is a function and

g

is not a function.

For $f$ to be a function: Every element in the domain must have a unique image.

The domain of $f$ is $[0, 10]$ . The two pieces overlap only at $x = 3$ .

At $x = 3$ : From the first piece, $f(3) = 3^2 = 9$ . From the second piece, $f(3) = 3 \times 3 = 9$ .

Both pieces give the same value at $x = 3$ . For all other $x$ , only one piece applies. Therefore, every element in $[0, 10]$ has a unique image.

$\therefore f \text{ is a function.}$

For $g$ : The two pieces overlap at $x = 2$ .

At $x = 2$ : From the first piece, $g(2) = 2^2 = 4$ . From the second piece, $g(2) = 3 \times 2 = 6$ .

The two pieces give different values ( $4 \neq 6$ ) at $x = 2$ , so the element $2$ in the domain has two different images.

$\therefore g \text{ is not a function.}$

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2If

f(x) = x^2

, find

\dfrac{f(1.1) - f(1)}{(1.1 - 1)}

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3Find the domain of the function

f(x) = \dfrac{x^2 + 2x + 1}{x^2 - 8x + 12}

Given: $f(x) = \dfrac{x^2 + 2x + 1}{x^2 - 8x + 12}$ .

Concept: The function is defined for all real $x$ except where the denominator is zero.

Setting denominator $= 0$ :
$x^2 - 8x + 12 = 0$
$(x - 6)(x - 2) = 0$
$x = 6 \quad \text{or} \quad x = 2$

The function is undefined at $x = 2$ and $x = 6$ .

$\text{Domain} = \mathbf{R} \setminus \{2, 6\} = \{x \in \mathbf{R} : x \neq 2 \text{ and } x \neq 6\}$

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4Find the domain and the range of the real function

f

defined by

f(x) = \sqrt{x - 1}

Given: $f(x) = \sqrt{x-1}$ .

Domain: The expression under the square root must be non-negative:
$x - 1 \geq 0 \implies x \geq 1$
$\text{Domain} = [1, +\infty) = \{x \in \mathbf{R} : x \geq 1\}$

Range: When $x \geq 1$ , $x - 1 \geq 0$ , so $\sqrt{x-1} \geq 0$ . As $x$ increases from $1$ to $\infty$ , $f(x)$ increases from $0$ to $\infty$ .
$\text{Range} = [0, +\infty) = \{y \in \mathbf{R} : y \geq 0\}$

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5Find the domain and the range of the real function

f

defined by

f(x) = |x - 1|

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6Let

f = \left\{\left(x, \dfrac{x^2}{1 + x^2}\right) : x \in \mathbf{R}\right\}

be a function from

\mathbf{R}

into

\mathbf{R}

. Determine the range of

f

Given: $f(x) = \dfrac{x^2}{1 + x^2}$ , $x \in \mathbf{R}$ .

Let $y = \dfrac{x^2}{1+x^2}$ .

Step 1: Note that $x^2 \geq 0$ and $1 + x^2 \geq 1 > 0$ , so $y \geq 0$ .

Step 2: Also, $y = \dfrac{x^2}{1+x^2} = 1 - \dfrac{1}{1+x^2}$ .

Since $1 + x^2 \geq 1$ , we have $\dfrac{1}{1+x^2} \leq 1$ , so $y = 1 - \dfrac{1}{1+x^2} < 1$ .

Step 3: As $x \to \infty$ , $y \to 1$ (but never equals 1). At $x = 0$ , $y = 0$ .

For any $y_0 \in [0,1)$ , we can solve: $y_0 = \dfrac{x^2}{1+x^2} \Rightarrow x^2 = \dfrac{y_0}{1-y_0} \geq 0$ , which has a real solution.

$\text{Range of } f = [0, 1) = \left\{y \in \mathbf{R} : 0 \leq y < 1\right\}$

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7Let

f, g: \mathbf{R} \to \mathbf{R}

be defined, respectively, by

f(x) = x + 1

g(x) = 2x - 3

. Find

f + g, f - g

and

\dfrac{f}{g}

Given: $f(x) = x + 1$ , $g(x) = 2x - 3$ , both from $\mathbf{R}$ to $\mathbf{R}$ .

$(f + g)(x)$ :
$(f+g)(x) = f(x) + g(x) = (x+1) + (2x-3) = 3x - 2$

$(f - g)(x)$ :
$(f-g)(x) = f(x) - g(x) = (x+1) - (2x-3) = x + 1 - 2x + 3 = -x + 4$

$\left(\dfrac{f}{g}\right)(x)$ :
$\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{x+1}{2x-3}, \quad x \neq \frac{3}{2}$

Summary:
$f + g = 3x - 2, \quad f - g = 4 - x, \quad \frac{f}{g} = \frac{x+1}{2x-3},\ x \neq \frac{3}{2}$

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8Let

f = \{(1,1), (2,3), (0,-1), (-1, -3)\}

be a function from

\mathbf{Z}

\mathbf{Z}

defined by

f(x) = ax + b

, for some integers

a, b

. Determine

a, b

Given: $f(x) = ax + b$ and the function contains $(1,1)$ and $(2,3)$ (among others).

Using $(1, 1)$ : $f(1) = 1$
$a(1) + b = 1 \implies a + b = 1 \quad \cdots (1)$

Using $(2, 3)$ : $f(2) = 3$
$a(2) + b = 3 \implies 2a + b = 3 \quad \cdots (2)$

Subtracting (1) from (2):
$a = 2$

Substituting in (1):
$2 + b = 1 \implies b = -1$

Verification: $f(0) = 2(0) + (-1) = -1$ ✓; $f(-1) = 2(-1) - 1 = -3$ ✓.

$\therefore a = 2,\ b = -1$

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9Let

\mathbf{R}

be a relation from

\mathbf{N}

\mathbf{N}

defined by

\mathbf{R} = \{(a, b) : a, b \in \mathbf{N} \text{ and } a = b^2\}

. Are the following true?
(i)

(a, a) \in \mathbf{R}

, for all

a \in \mathbf{N}

(ii)

(a, b) \in \mathbf{R}

, implies

(b, a) \in \mathbf{R}

(iii)

(a, b) \in \mathbf{R}

(b, c) \in \mathbf{R}

, implies

(a, c) \in \mathbf{R}

Given: $\mathrm{R} = \{(a,b) : a, b \in \mathbf{N},\ a = b^2\}$ .

(i) Is $(a, a) \in \mathbf{R}$ for all $a \in \mathbf{N}$ ?

For $(a,a) \in \mathrm{R}$ , we need $a = a^2$ , i.e., $a(a-1) = 0$ , so $a = 0$ or $a = 1$ . Since $0 \notin \mathbf{N}$ , this holds only for $a = 1$ , not for all $a \in \mathbf{N}$ (e.g., $a = 2$ : $2 \neq 4$ ).

False.

(ii) Does $(a,b) \in \mathrm{R}$ imply $(b,a) \in \mathrm{R}$ ?

If $(a,b) \in \mathrm{R}$ , then $a = b^2$ . For $(b,a) \in \mathrm{R}$ , we need $b = a^2 = (b^2)^2 = b^4$ , i.e., $b^4 - b = 0$ , i.e., $b(b^3-1)=0$ , so $b=1$ .

Counterexample: $(4, 2) \in \mathrm{R}$ since $4 = 2^2$ , but $(2, 4) \notin \mathrm{R}$ since $2 \neq 4^2 = 16$ .

False.

(iii) Does $(a,b) \in \mathrm{R}$ and $(b,c) \in \mathrm{R}$ imply $(a,c) \in \mathrm{R}$ ?

If $(a,b) \in \mathrm{R}$ : $a = b^2$ . If $(b,c) \in \mathrm{R}$ : $b = c^2$ . Then $a = b^2 = (c^2)^2 = c^4$ . For $(a,c) \in \mathrm{R}$ , we need $a = c^2$ , but we have $a = c^4$ . These are equal only if $c = 1$ .

Counterexample: $(16, 4) \in \mathrm{R}$ (since $16 = 4^2$ ) and $(4, 2) \in \mathrm{R}$ (since $4 = 2^2$ ), but $(16, 2) \notin \mathrm{R}$ since $16 \neq 2^2 = 4$ .

False.

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10Let

A = \{1,2,3,4\}

B = \{1,5,9,11,15,16\}

and

f = \{(1,5), (2,9), (3,1), (4,5), (2,11)\}

. Are the following true?
(i)

f

is a relation from A to B
(ii)

f

is a function from A to B.
Justify your answer in each case.

Given: $A = \{1,2,3,4\}$ , $B = \{1,5,9,11,15,16\}$ , $f = \{(1,5),(2,9),(3,1),(4,5),(2,11)\}$ .

(i) Is $f$ a relation from A to B?

A relation from A to B is any subset of $A \times B$ . We check that all first elements belong to A and all second elements belong to B:

$(1,5)$ : $1 \in A$ , $5 \in B$ ✓
$(2,9)$ : $2 \in A$ , $9 \in B$ ✓
$(3,1)$ : $3 \in A$ , $1 \in B$ ✓
$(4,5)$ : $4 \in A$ , $5 \in B$ ✓
$(2,11)$ : $2 \in A$ , $11 \in B$ ✓

Since $f \subseteq A \times B$ , $f$ is a relation from A to B. True.

(ii) Is $f$ a function from A to B?

For $f$ to be a function, every element of A must have exactly one image in B. However, the element $2 \in A$ appears in two pairs: $(2,9)$ and $(2,11)$ , giving two different images $9$ and $11$ .

Also, the element $4 \in A$ has image $5$ , but we should check all elements: $1 \mapsto 5$ , $2 \mapsto 9$ and $2 \mapsto 11$ (two images), $3 \mapsto 1$ , $4 \mapsto 5$ .

Since element $2$ has two images, $f$ is NOT a function from A to B. False.

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11Let

f

be the subset of

\mathbf{Z} \times \mathbf{Z}

defined by

f = \{(ab, a + b) : a, b \in \mathbf{Z}\}

. Is

f

a function from

\mathbf{Z}

\mathbf{Z}

? Justify your answer.

Given: $f = \{(ab, a+b) : a, b \in \mathbf{Z}\}$ .

For $f$ to be a function from $\mathbf{Z}$ to $\mathbf{Z}$ , each element of $\mathbf{Z}$ (as a first element) must have a unique image.

Counterexample: Consider the integer $2$ as the first element (i.e., $ab = 2$ ).

Take $a = 1, b = 2$ : $ab = 2$ , $a + b = 3$ → pair $(2, 3)$ .
Take $a = 2, b = 1$ : $ab = 2$ , $a + b = 3$ → pair $(2, 3)$ . (Same)
Take $a = -1, b = -2$ : $ab = 2$ , $a + b = -3$ → pair $(2, -3)$ .

So the element $2$ has two different images: $3$ and $-3$ .

Since a single first element ( $ab = 2$ ) corresponds to two different second elements ( $3$ and $-3$ ), $f$ is not a function from $\mathbf{Z}$ to $\mathbf{Z}$ .

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12Let

\mathbf{A} = \{9,10,11,12,13\}

and let

f: \mathbf{A} \to \mathbf{N}

be defined by

f(n) =

the highest prime factor of

n

. Find the range of

f

Given: $\mathrm{A} = \{9, 10, 11, 12, 13\}$ and $f(n)$ = highest prime factor of $n$ .

Finding the highest prime factor for each element:

$n$	Prime factorisation	Highest prime factor
9	$3^2$	3
10	$2 \times 5$	5
11	$11$ (prime)	11
12	$2^2 \times 3$	3
13	$13$ (prime)	13

So:

$f(9) = 3$
$f(10) = 5$
$f(11) = 11$
$f(12) = 3$
$f(13) = 13$

$\text{Range of } f = \{3, 5, 11, 13\}$

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Frequently Asked Questions

What are the important topics in Relations and Functions for CBSE Class 11 Mathematics?

Key topics in Relations and Functions include A pair of elements written, For two non, A relation from a non, The domain is the set. Study these first, then practise questions on each for Class 11 exams.

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Exercise 2.1

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