Chapter 2 Relations and Functions

NCERT Solutions Class 11 Maths Chapter 2 – Relations and Functions

NCERT Solutions for Class 11 Maths Chapter 2 (Relations and Functions) presents one of the most influential concepts in mathematics. While Chapter 1 familiarized us with collections (Sets), this chapter focuses on how to link elements from one set to another. This principle serves as the foundation for Calculus, Trigonometry, and nearly all advanced mathematical modeling.

NCERT Solutions for Class 11 Maths Chapter 2 by ALLEN are prepared by expert faculty to help students understand relations and functions clearly. These solutions provide step-by-step explanations and exam-focused methods to help students prepare for CBSE and JEE.

A clear understanding of domain and range is important for solving questions in Class 11 Maths and competitive exams such as JEE Main and JEE Advanced. These solutions help students build their understanding from Cartesian products and relations to real-valued functions and their graphical representations.

1.0Class 11 Maths Chapter 2 Relations and Functions: Key Concepts

This chapter explains how elements of two sets are related and how functions connect elements of one set with elements of another set. The key concepts include:

Cartesian Product of Sets: If A and B are two non-empty sets, their Cartesian product A × B is the set of all ordered pairs (a, b), where a ∈ A and b ∈ B.
Number of Elements: If A and B are finite sets, then n(A × B) = n(A) × n(B).
Relations: A relation R from a set A to a set B is a subset of the Cartesian product A × B.
Domain: The domain of a relation is the set of all first elements of the ordered pairs in the relation.
Range: The range of a relation is the set of all second elements that occur in the ordered pairs.
Codomain: The codomain is the complete set B to which the elements of A are related.
Functions: A function is a special type of relation in which every element of set A has exactly one image in set B.
Real Functions and Their Graphs: The chapter introduces different types of real functions and their graphs:
Identity Function: f(x) = x
Constant Function: f(x) = k, where k is a constant.
Polynomial Function: f(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + ... + a₁x + a₀
Rational Function: f(x) = g(x)/h(x), where h(x) ≠ 0
Modulus Function: f(x) = |x|
Signum Function: f(x) = |x|/x, where x ≠ 0
Greatest Integer Function: f(x) = [x], which gives the greatest integer less than or equal to x.
Algebra of Real Functions: Students learn how to add, subtract, multiply, and divide two real functions.

2.0Detailed NCERT Textbook Solutions : Class 11 Maths Chapter 2

EXERCISE - 2.1

If $(\frac{x}{3} + 1, y - \frac{2}{3}) = (\frac{5}{3}, \frac{1}{3})$ , find the values of x and y. Sol. It is given that $(\frac{x}{3} + 1, y - \frac{2}{3}) = (\frac{5}{3}, \frac{1}{3})$ . Since the ordered pairs are equal, the corresponding elements will also be equal. Therefore $\frac{x}{3} + 1 = \frac{5}{3}$ and $y - \frac{2}{3} = \frac{1}{3}$

$\Rightarrow \Rightarrow \Rightarrow ∴ \frac{x}{3} = \frac{5}{3} - 1 \frac{x}{3} = \frac{2}{3} = \frac{2}{3} x = 2 x = 2 and y = 1 and \Rightarrow$

If the set A has 3 elements and the set $B = {3, 4, 5}$ , then find the number of elements in (A × B)? Sol. It is given that set A has 3 elements and the elements of set B are 3, 4, and 5. ⇒ Number of elements in set $B = 3$ Number of elements in (A × B) = (Number of elements in A) × (Number of elements in B) $= 3 \times 3 = 9$

Thus, the number of elements in $(A \times B)$ is 9 .

If $G = {7, 8}$ and $H = {5, 4, 2}$ , find $G \times H$ and $H \times G$ . Sol. $G = {7, 8}$ and $H = {5, 4, 2}$ $∴ G \times H = {(7, 5), (7, 4), (7, 2), (8, 5), (8, 4), (8, 2)}$ $H \times G = {(5, 7), (5, 8), (4, 7), (4, 8), (2, 7), (2, 8)}$
State whether each of the following statement are true or false. If the statement is false, rewrite the given statement correctly. (i) If $P = {m, n}$ and $Q = {n, m}$ , then $P \times Q = {(m, n), (n, m)}$ . (ii) If A and B are non-empty sets, then A × B is a non-empty set of ordered pairs (x, y) such that $x \in A$ and $y \in B$ . (iii) If $A = {1, 2}, B = {3, 4}$ , then $A \times (B \cap ϕ) = ϕ$ .

Sol.

(i) False. If $P = {m, n}$ and $Q = {n, m}$ , then $P \times Q = {(m, m), (m, n), (n, m), (n, n)}$ (ii) True (iii) True

5. If $A = {- 1, 1}$ , find $A \times A \times A$ . Sol. It is known that for any non-empty set A, $A \times A \times A$ is defined as $A \times A \times A = {(a, b, c) : a, b, c \in A}$ It is given that $A = {- 1, 1}$ $∴ A \times A \times A = {(- 1, - 1, - 1), (- 1, - 1, 1)$ , $(- 1, 1, - 1), (- 1, 1, 1), (1, - 1, - 1), (1, - 1, 1)$ , (1, 1, -1), (1, 1, 1)}

6. If $A \times B = {(a, x), (a, y), (b, x), (b, y)}$ . Find A and B. Sol. It is given that $A \times B = {(a, x), (a, y), (b, x)$ , (b, y)} We know that the Cartesian product of two non-empty sets P and Q is defined as $P \times Q = {(p, q) : p \in P, q \in Q}$ $∴ A$ is the set of all first elements and B is the set of all second elements. Thus, $A = {a, b}$ and $B = {x, y}$

7. Let $A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6}$ and $D = {5, 6, 7, 8}$ . Verify that (i) $A \times (B \cap C) = (A \times B) \cap (A \times C)$ (ii) $A \times C$ is a subset of $B \times D$

Sol.

(i) To verify: $A \times (B \cap C) = (A \times B) \cap (A \times C)$ We have $B \cap C = {1, 2, 3, 4} \cap {5, 6} = ϕ$ $∴$ L.H.S. $= A \times (B \cap C) = A \times ϕ = ϕ$ $A \times B = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1)$ , (2, 2), (2, 3), (2, 4)} $A \times C = {(1, 5), (1, 6), (2, 5), (2, 6)}$ $∴$ R.H.S. $= (A \times B) \cap (A \times C) = ϕ$ $∴$ L.H.S. = R.H.S Hence, $A \times (B \cap C) = (A \times B) \cap (A \times C)$ (ii) To verify: $A \times C$ is a subset of $B \times D$ $A \times C = {(1, 5), (1, 6), (2, 5), (2, 6)}$ $A \times D = {(1, 5), (1, 6), (1, 7), (1, 8), (2, 5)$ , (2, 6), (2, 7), (2, 8), (3, 5), (3, 6), (3, 7), (3, 8), (4, 5), (4, 6), (4, 7), (4, 8)}

We can observe that all the elements of set A × C are the elements of set B × D. Therefore, A × C is a subset of B × D.

8. Let $A = {1, 2}$ and $B = {3, 4}$ . Write $A \times B$ . How many subsets will A × B have? List them. Sol. $A = {1, 2}$ and $B = {3, 4}$ $∴ A \times B = {(1, 3), (1, 4), (2, 3), (2, 4)}$ $\Rightarrow n (A \times B) = 4$ We know that if C is a set with $n (C) = m$ , Then, total no. of subsets are $= 2^{m}$ . Therefore, the set $A \times B$ has $2^{4} = 16$ subsets. These are $ϕ, {(1, 3)}, {(1, 4)}, {(2, 3)}, {(2, 4)}, {(1, 3)$ , $(1, 4)}, {(1, 3), (2, 3)}, {(1, 3), (2, 4)}, {(1, 4)$ , $(2, 3)}, {(1, 4), (2, 4)}, {(2, 3), (2, 4)}, {(1, 3)$ , $(1, 4), (2, 3)}, {(1, 3), (1, 4), (2, 4)}, {(1, 3)$ , $(2, 3), (2, 4)}, {(1, 4), (2, 3), (2, 4)}, {(1, 3)$ , (1, 4), (2, 3), $(2, 4)}$

9. Let A and B be two sets such that $n (A) = 3$ and $n (B) = 2$ . If (x, 1), (y, 2), (z, 1) are in $A \times B$ , find A and B, where x, y and z are distinct elements. Sol. It is given that $n (A) = 3$ and $n (B) = 2$ ; and $(x, 1), (y, 2), (z, 1)$ are in A × B. We know that A = Set of first elements of the ordered pair elements of A × B B = Set of second elements of the ordered pair elements of A × B. ∴ x, y and z are the elements of A; and 1 and 2 are the elements of B. Since $n (A) = 3$ and $n (B) = 2$ , it is clear that A $= {x, y, z}$ and $B = {1, 2}$ .

10. The Cartesian product A × A has 9 elements among which are found (-1, 0) and (0, 1). Find the set A and the remaining elements of $A \times A$ . Sol. We know that if $n (A) = p$ and $n (B) = q$ , then $n (A \times B) = pq$ . $∴ n (A \times A) = n (A) \times n (A)$ It is given that $n (A \times A) = 9$ $∴ n (A) \times n (A) = 9$ $\Rightarrow n (A) = 3$ The ordered pairs $(- 1, 0)$ and $(0, 1)$ are two of the nine elements of $A \times A$ . We know that $A \times A = {(a, a) : a \in A}$ . Therefore, -1, 0, and 1 are elements of A. Since $n (A) = 3$ , it is clear that $A = {- 1, 0, 1}$ . The remaining elements of set $A \times A$ are (-1, -1), (-1, 1), (0, -1), (0, 0), (1, -1), $(1, 0)$ and $(1, 1)$ .

EXERCISE - 2.2

Let $A = {1, 2, 3 \dots ..14}$ . Define a relation R from A to A by $R = {(x, y) : 3 x - y = 0$ , where $x, y \in A}$ . Write down its domain, codomain and range. Sol. The relation R from A to A is given as $R = {(x, y) : 3 x - y = 0$ , where $x, y \in A}$ i.e., $R = {(x, y) : 3 x = y$ , where $x, y \in A}$ $∴ R = {(1, 3), (2, 6), (3, 9), (4, 12)}$ The domain of R is the set of all first elements of the ordered pairs in the relation. ∴ Domain of $R = {1, 2, 3, 4}$ The whole set A is the codomain of the relation R. ∴ Codomain of $R = A = {1, 2, 3 \dots ..14}$ The range of R is the set of all second elements of the ordered pairs in the relation. ∴ Range of $R = {3, 6, 9, 12}$
Define a relation R on the set $N$ of natural numbers by $R = {(x, y) : y = x + 5, x$ is a natural number less than $4; x, y \in N}$ . Depict this relationship using roster form. Write down the domain and the range. Sol. $R = {(x, y) : y = x + 5, x$ is a natural number less than $4, x, y \in N$ } The natural numbers less than 4 are 1,2, and 3. $∴ R = {(1, 6), (2, 7), (3, 8)}$ The domain of R is the set of all first elements of the ordered pairs in the relation. $∴$ Domain of $R = {1, 2, 3}$

The range of R is the set of all second elements of the ordered pairs in the relation.

∴ Range of $R = {6, 7, 8}$

3. $A = {1, 2, 3, 5}$ and $B = {4, 6, 9}$ . Define a relation R from A to B by $R = {(x, y)$ : the difference between x and y is odd; $x \in A$ , $y \in B}$ . Write R in roster form. Sol. $A = {1, 2, 3, 5}$ and $B = {4, 6, 9}$ $R = {(x, y)$ : the difference between x and y is odd; $x \in A, y \in B}$ $∴ R = {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4)$ , $(5, 6)}$ 4. The given figure shows a relationship between the sets P and Q . write this relation (i) in setbuilder form (ii) in roster form. What is its domain and range?

Class-11-Maths-chap-2-exer-2.2-ques-4-ncert-sol

Sol. According to the given figure, $P = {5, 6, 7}$ , Q $= {3, 4, 5}$ (i) $R = {(x, y) : y = x - 2; x \in P, y \in Q}$ (ii) $R = {(5, 3), (6, 4), (7, 5)}$ Domain of $R = {5, 6, 7}$ Range of $R = {3, 4, 5}$

5. Let $A = {1, 2, 3, 4, 6}$ . Let R be the relation on A defined by ${(a, b) : a, b \in A, b$ is exactly divisible by a}. (i) Write R in roster form (ii) Find the domain of R (iii) Find the range of $R$ . Sol. $A = {1, 2, 3, 4, 6}, R = {(a, b) : a, b \in A, b$ is exactly divisible by a} (i) $R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 2)$ , (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (6, 6)} (ii) Domain of $R = {1, 2, 3, 4, 6}$ (iii) Range of $R = {1, 2, 3, 4, 6}$

6. Determine the domain and range of the relation R defined by

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

Sol. $R = {(x, x + 5) : x \in {0, 1, 2, 3, 4, 5}}$ $∴ R = {(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)}$ $∴$ Domain of $R = {0, 1, 2, 3, 4, 5}$ Range of $R = {5, 6, 7, 8, 9, 10}$

7. Write the relation $R = {(x, x^{3}) : x$ is a prime number less than 10 $}$ in roster form. Sol. $R = {(x, x^{3}) : x$ is a prime number less than 10 $}$ The prime numbers less than 10 are 2, 3, 5, and 7. $∴ R = {(2, 8), (3, 27), (5, 125), (7, 343)}$

8. Let $A = {x, y, z}$ and $B = {1, 2}$ . Find the number of relations from A to B. Sol. It is given that $A = {x, y, z}$ and $B = {1, 2}$ . $∴ A \times B = {(x, 1), (x, 2), (y, 1), (y, 2), (z, 1), (z, 2)}$ Since $n (A \times B) = 6$ , the number of subsets of $A \times B$ is $2^{6}$ . Therefore, the number of relations from A to B is $2^{6}$ .

9. Let R be the relation on $Z$ defined by $R = {(a, b) : a, b \in Z, a - b$ is an integer $}$ . Find the domain and range of R. Sol. $R = {(a, b) : a, b \in Z, a - b$ is an integer $}$ It is known that the difference between any two integers is always an integer. ∴ Domain of $R = Z$ ; Range of $R = Z$

EXERCISE - 2.3

Which 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) ${(1, 3), (1, 5), (2, 5)}$

Sol.

(i) ${(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}$ Since $2, 5, 8, 11, 14$ , and 17 are the elements of the domain of the given relation having their unique images, this relation is a function. Here, domain $= {2, 5, 8, 11, 14, 17}$ and range $= {1}$

(ii) ${(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6)$ , $(14, 7)}$ Since 2, 4, 6, 8, 10, 12, and 14 are the elements of the domain of the given relation having their unique images, this relation is a function. Here, domain $= {2, 4, 6, 8, 10, 12, 14}$ and range $= {1, 2, 3, 4, 5, 6, 7}$ (iii) ${(1, 3), (1, 5), (2, 5)}$ Since the same first element i.e., 1 corresponds to two different images i.e., 3 and 5, this relation is not a function.

2. Find the domain and range of the following real function: (i) $f (x) = - ∣ x ∣$ (ii) $f (x) = 9 - x^{2}$ Sol. (i) $f (x) = - ∣ x ∣, x \in R$ We know that $∣ x ∣ = {x, - x, if x \geq 0 if x < 0$ $∴ f (x) = - ∣ x ∣ = {- x, x, if x \geq 0 if x < 0$ Since $f (x)$ is defined for $x \in R$ , the domain of f is $R$ . It can be observed that the range of $f (x) = - ∣ x ∣$ is all real numbers except positive real numbers. $∴$ The range of f is $(- \infty, 0]$ . (ii) Domain : $∵ f (x) = 9 - x^{2}$ Since $f (x) = x$ is defined $\forall x \geq 0$

$∴ \Rightarrow 9 - x^{2} \geq 0 (x - 3) (x + 3) \leq 0 So, domain = [- 3, 3] \Rightarrow \Rightarrow x^{2} - 9 \leq 0 x \in [- 3, 3] .$

Range :

$\Rightarrow \Rightarrow \Rightarrow ∵ \Rightarrow \Rightarrow ∴ Let y = f (x) = 9 - x^{2} [∵ y \geq 0] y^{2} = 9 - x^{2} x^{2} = 9 - y^{2} x = 9 - y^{2} 9 - y^{2} \geq 0 y^{2} - 9 \leq 0 y \in [- 3, 3] but y \geq 0 y \in [0, 3]$

Hence, the range of f(x) is [0, 3].

3. A function f is defined by $f (x) = 2 x - 5$ . Write down the values of (i) f(0), (ii) f(7), (iii) $f (- 3)$

Sol. The given function is $f (x) = 2 x - 5$ . Therefore,

(i) $f (0) = 2 \times 0 - 5 = 0 - 5 = - 5$ (ii) $f (7) = 2 \times 7 - 5 = 14 - 5 = 9$ (iii) $f (- 3) = 2 \times (- 3) - 5 = - 6 - 5 = - 11$

4. The function 't' which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by $t (C) = \frac{9 C}{5} + 32$ . Find (i) $t (0)$ (ii) $t (28)$ (iii) $t (- 10)$ (iv) The value of C, when $t (C) = 212$ Sol. The given function is $t (C) = \frac{9 C}{5} + 32$ Therefore, (i) $t (0) = \frac{9 \times 0}{5} + 32 = 0 + 32 = 32$ (ii) $t (28) = \frac{9 \times 28}{5} + 32 = \frac{252 + 160}{5} = \frac{412}{5}$ (iii) $t (- 10) = \frac{9 \times ( - 10 )}{5} + 32$ $= 9 \times (- 2) + 32 = - 18 + 32 = 14$

(iv) It is given that $t (C) = 212$ $∴ 212 = \frac{9 C}{5} + 32$ $\Rightarrow \frac{9 C}{5} = 212 - 32$ $\Rightarrow \frac{9 C}{5} = 180$ $\Rightarrow 9 C = 180 \times 5$ $\Rightarrow C = \frac{180 \times 5}{9} = 100$ Thus, the value of t , when $t (C) = 212$ , is 100.

5. Find the range of each of the following functions. (i) $f (x) = 2 - 3 x, x \in R, x > 0$ . (ii) $f (x) = x^{2} + 2, x$ , is a real number. (iii) $f (x) = x, x$ is a real number

Sol.

(i) $f (x) = 2 - 3 x, x \in R, x > 0$ The values of $f (x)$ for various values of real numbers $x > 0$ can be written in the tabular form as

x	f(x)
0.01	1.97
0.1	1.7
0.9	-0.7
1	-1
2	-4
2.5	-5.5
4	-10
5	-13
........	......

Thus, it can be clearly observed that the range of f is the set of all real numbers less than 2. i.e., range of $f = (- \infty, 2)$

Alter:

$∵ x > 0$ $\Rightarrow 3 x > 0$ $\Rightarrow - 3 x < 0$ $\Rightarrow 2 - 3 x < 2$ $\Rightarrow f (x) < 2$ ∴ Range of $f = (- \infty, 2)$ (ii) $f (x) = x^{2} + 2, x$ , is a real number. The values of f(x) for various values of real numbers x can be written in the tabular form as

x	f(x)
0	2
±0.3	2.09
±0.8	2.64
$\pm 1$	3
$\pm 2$	6
$\pm 3$	11
........	......

Thus, it can be clearly observed that the range of f is the set of all real numbers greater than 2. i.e., range of $f = [2, \infty)$ Alter: Let x be any real number. Accordingly,

$x^{2} \geq 0$

$\Rightarrow x^{2} + 2 \geq 0 + 2$ $\Rightarrow x^{2} + 2 \geq 2$ $\Rightarrow f (x) \geq 2$ ∴ Range of $f = [2, \infty)$ (iii) $f (x) = x, x$ is a real number. It is clear that the range of f is the set of all real numbers. $∴$ Range of $f = R$

MISCELLANEOUS EXERCISE

The relation f is defined by

$f (x) = {x^{2}, 3 x, 0 \leq x \leq 3 3 \leq x \leq 10$

The relation g is defined by

$g (x) = {x^{2}, 3 x, 0 \leq x \leq 2 2 \leq x \leq 10$

Show that f is a function and g is not a function. Sol. The relation f is defined as

$f (x) = {x^{2}, 3 x, 0 \leq x \leq 3 3 \leq x \leq 10$

It is obszerved that for

$0 \leq x \leq 3, 3 \leq x \leq 10, f (x) = x^{2} f (x) = 3 x$

Also, at $x = 3, f (x) = 3^{2} = 9$ or $f (x) = 3 \times 3 = 9$ i.e., at $x = 3, f (x) = 9$ Therefore, for $0 \leq x \leq 10$ , the images of f(x) are unique. Thus, the given relation is a function. The relation g is defined as

$g (x) = {x^{2}, 3 x, 0 \leq x \leq 2 2 \leq x \leq 10$

It can be observed that for $x = 2$ , $g (x) = 2^{2} = 4$ and $g (x) = 3 \times 2 = 6$ Hence, element 2 of the domain of the relation g corresponds to two different images i.e., 4 and 6. Hence, this relation is not a function.

2. If $f (x) = x^{2}$ , find $\frac{f ( 1.1 ) - f ( 1 )}{( 1.1 - 1 )}$ .

Sol. $f (x) = x^{2}$

$∴ \frac{f ( 1.1 ) - f ( 1 )}{( 1.1 - 1 )} = \frac{( 1.1 ) ^{2} - ( 1 ) ^{2}}{( 1.1 - 1 )} = \frac{1.21 - 1}{0.1} = \frac{0.21}{0.1} = 2.1$

Find the domain of the function

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

Sol. The given function is

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

$∵ x^{2} - 8 x + 12 \neq = 0$

$\Rightarrow (x - 6) (x - 2) \neq = 0$

$\Rightarrow x \neq = 2, 6$

$∴ Domain = R - {2, 6}$

Find the domain and the range of the real function f defined by $f (x) = (x - 1)$ Sol. The given real function is $f (x) = (x - 1)$ It can be seen that $(x - 1)$ is defined for $x \geq 1$ . Therefore, the domain of f is the set of all real numbers greater than or equal to 1 i.e., the domain of $f = [1, \infty)$ . As $x \geq 1$

$\Rightarrow (x - 1) \geq 0 \Rightarrow (x - 1) \geq 0$

Therefore, the range of f is the set of all real numbers greater than or equal to 0 i.e., the range of $f = [0, \infty)$ .

5. Find the domain and the range of the real function f defined by $f (x) = ∣ x - 1∣$ . Sol. The given real function is $f (x) = ∣ x - 1∣$ . It is clear that $∣ x - 1∣$ is defined for all real numbers. ∴ Domain of $f = R$ Also, for $x \in R, ∣ x - 1∣$ assumes all real numbers. Hence, the range of f is the set of all nonnegative real numbers.

6. Let $f = {(x, \frac{x ^{2}}{1 + x ^{2}}) : x \in R}$ be a function from R into R. Determine the range of f. Sol. $f = {(x, \frac{x ^{2}}{1 + x ^{2}}) : x \in R}$

$= {(0, 0), (\pm 0.5, \frac{1}{5}), (\pm 1, \frac{1}{2}), (\pm 1.5, \frac{9}{13}), (\pm 2, \frac{4}{5}), (\pm 3, \frac{9}{10}), (\pm 4, \frac{16}{17}), \dots\dots}$

The range of f is the set of all second elements. It can be observed that all these elements are greater than or equal to 0 but less than 1 . [Denominator is greater numerator] Thus, range of $f = [0, 1)$

Let $f, g : R \to R$ be defined, respectively by $f (x) = x + 1, g (x) = 2 x - 3$ . Find $f + g$ , $f - g$ and $\frac{f}{g}$ . Sol. f, g: $R \to R$ is defined as

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

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

$∴ ∴ (\frac{f}{g}) (x) = \frac{f ( x )}{g ( x )}, g (x) \neq = 0, x \in R (\frac{f}{g}) (x) = \frac{x + 1}{2 x - 3}, 2 x - 3 \neq = 0 or 2 x \neq = 3 (\frac{f}{g}) (x) = \frac{x + 1}{2 x - 3}, x \neq = \frac{3}{2}$

Let $f = {(1, 1), (2, 3), (0, - 1), (- 1, - 3)}$ be a function from Z to Z defined by $f (x) = ax + b$ , for some integers $a, b$ . Determine $a, b$ . Sol. $f = {(1, 1), (2, 3), (0, - 1), (- 1, - 3)}$ and $f (x) = a x + b$

$(1, 1) \in f \Rightarrow f (1) = 1$

$\Rightarrow a \times 1 + b = 1 \Rightarrow a + b = 1$

$(0, - 1) \in f \Rightarrow f (0) = - 1$

$\Rightarrow a \times 0 + b = - 1 \Rightarrow b = - 1$ On substituting $b = - 1$ in $a + b = 1$ We obtain $a + (- 1) = 1$ $\Rightarrow a = 1 + 1 = 2$ .

Thus, the respective values of a and b are 2 and -1.

Let R be a relation from N to N defined by $R = {(a, b) : a, b \in N$ and $a = b^{2}}$ . Are the following true? (i) $(a, a) \in R$ , for all $a \in N$ (ii) $(a, b) \in R$ , implies $(b, a) \in R$ (iii) $(a, b) \in R, (b, c) \in R$ implies $(a, c) \in R$ . Justify your answer in each case.

Sol. $R = {(a, b) : a, b \in N$ and $a = b^{2}}$

(i) It can be seen that $2 \in N$ ; however, $2 \neq = 2^{2} = 4$ . Therefore, the statement "(a, a) $\in R$ , for all $a \in N^{''}$ is not true. (ii) It can be seen that $(9, 3) \in N$ because $9, 3 \in N$ and $9 = 3^{2}$ . Now, $3 \neq = 9^{2} = 81$ ; therefore, $(3, 9) \in / N$ Therefore, the statement "(a, b) $\in R$ , implies $(b, a) \in R^{''}$ is not true. (iii) It can be seen that $(9, 3) \in R, (16, 4) \in R$ because $9, 3, 16, 4 \in N$ and $9 = 32$ and $16 = 42$ . Now, $9 \neq = 42 = 16$ ; therefore, $(9, 4) \in / N$ Therefore, the statement "(a, b) $\in R, (b, c) \in R$ implies $(a, c) \in R$ " is not true.

10. Let $A = {1, 2, 3, 4}, B = {1, 5, 9, 11, 15, 16}$ 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. Sol. $A = {1, 2, 3, 4}$ and $B = {1, 5, 9, 11, 15, 16}$ $∴ A \times B = {(1, 1), (1, 5), (1, 9), (1, 11), (1, 15)$ , (1, 16), (2, 1), (2, 5), (2, 9), (2, 11), (2, 15), (2, 16), (3, 1), (3, 5), (3, 9), (3, 11), (3, 15), (3, 16), (4,1), (4,5), (4,9), (4,11), $(4, 15)$ , $(4$ , 16)} It is given that $f = {(1, 5), (2, 9), (3, 1), (4, 5)$ , $(2, 11)$ } (i) A relation from a non-empty set A to a nonempty set B is a subset of the Cartesian product A × B. It is observed that f is a subset of $A \times B$ . Thus, f is a relation from A to B . (ii) Since the same first element i.e., 2 corresponds to two different images i.e., 9 and 11, relation f is not a function.

Let f be the subset of $Z \times Z$ defined by $f = {(ab, a + b) : a, b \in Z}$ . Is f a function from $Z$ to $Z$ : justify your answer. Sol. The relation f is defined as $f = {(ab, a + b) : a, b \in Z}$ We know that a relation f from a set A to a set B is said to be a function if every element of set A has unique images in set B. Since 2, 6, -2, -6 $\in Z, (2 \times 6, 2 + 6)$ , $(- 2 \times - 6, - 2 + (- 6)) \in f$ i.e., $(12, 8), (12, - 8) \in f$ It can be seen that the same first element i.e., 12 corresponds to two different images i.e., 8 and -8. Thus, relation f is not a function.
Let $A = {9, 10, 11, 12, 13}$ and let $f : A \to N$ be defined by $f (n) =$ the highest prime factor of n. Find the range of f. Sol. $A = {9, 10, 11, 12, 13} f : A \to N$ is defined as $f (n) =$ The highest prime factor of n. Prime factor of $9 = 3$ Prime factors of $10 = 2, 5$ Prime factor of $11 = 11$ Prime factors of $12 = 2, 3$ Prime factor of $13 = 13$ $∴ f (9) =$ The highest prime factor of $9 = 3$ $f (10) =$ The highest prime factor of $10 = 5$ $f (11) =$ The highest prime factor of $11 = 11$ $f (12) =$ The highest prime factor of $12 = 3$ $f (13) =$ The highest prime factor of $13 = 13$ The range of f is the set of all f(n), where n $\in$ A.

∴ Range of $f = {3, 5, 11, 13}$

3.0Class 11 Maths NCERT Solutions – Chapter-wise Links

Access chapter-wise NCERT Solutions for Class 11 Maths with clear answers to exercise questions, important formulas, and easy steps to solve problems.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Sets
Chapter 3	Trigonometric Functions
Chapter 4	Complex Numbers and Quadratic Equations
Chapter 5	Linear Inequalities
Chapter 6	Permutations and Combinations
Chapter 7	Binomial Theorem
Chapter 8	Sequence and Series
Chapter 9	Straight Lines
Chapter 10	Conic Sections
Chapter 11	Introduction to Three-dimensional Geometry
Chapter 12	Limits and Derivatives
Chapter 13	Statistics
Chapter 14	Probability

4.0Class 11 Maths Chapter 2 Relations and Functions: Exercise-wise Questions and Topics

Exercises	Number of Questions	Important Topics Covered
Exercise 2.1	10 Questions & Solutions	Relations, types and properties of relations, domain and range, and representation of relations
Exercise 2.2	9 Questions & Solutions	Functions and their types, domain and range, graphical representation, and inverse of a function
Exercise 2.3	5 Questions & Solutions	Composition of functions, types and properties of composition, and one-one and onto functions

5.0Key Features of NCERT Solutions for Class 11 Maths Chapter 2

Simple Explanation of Relations and Functions: The solutions explain relations and functions in simple language. Arrow diagrams and mappings help students understand how relations and functions are represented and how they differ from each other.
Clear Understanding of Domain and Range: Step-by-step solutions explain how to find the domain and range of a function. Students can understand which values can be used as inputs and which values are obtained as outputs.
Easy Explanation of Types of Functions: The solutions explain different types of functions with suitable examples. Students can learn how to identify and differentiate between functions based on their properties.
Clear Explanation of Graphs: The solutions explain how functions are represented on the Cartesian plane. Graphs help students understand the values and behaviour of different functions.
Prepared by ALLEN Subject Experts: These NCERT Solutions for Class 11 Maths Chapter 2 are prepared by ALLEN subject experts and follow the NCERT syllabus. They are useful for Class 11 Maths preparation and revision.
Step-by-Step Answers: Each question is solved step by step using simple language and correct mathematical notation. This helps students understand the method used to reach the answer.
Coverage of Important Chapter Topics: The solutions cover the important concepts from Class 11 Maths Chapter 2 – Relations and Functions, helping students understand the chapter, check their answers, and practise NCERT questions.

Frequently Asked Questions

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Frequently Asked Questions

What topics are covered in Class 11 Maths Chapter 2 – Relations and Functions?

What is the difference between a relation and a function?

How do you find the domain and range of a function?

What are the important types of functions in Class 11 Maths Chapter 2?

How do Class 11 Maths NCERT Solutions for Chapter 2 help in board, JEE and BITSAT preparation?

What is the formula for the number of elements in the Cartesian product of two finite sets?

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NCERT Solutions Class 11 Maths Chapter 2 – Relations and Functions

1.0Class 11 Maths Chapter 2 Relations and Functions: Key Concepts

2.0Detailed NCERT Textbook Solutions : Class 11 Maths Chapter 2

EXERCISE - 2.1

EXERCISE - 2.2

EXERCISE - 2.3

MISCELLANEOUS EXERCISE

3.0Class 11 Maths NCERT Solutions – Chapter-wise Links

4.0Class 11 Maths Chapter 2 Relations and Functions: Exercise-wise Questions and Topics

5.0Key Features of NCERT Solutions for Class 11 Maths Chapter 2

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