Class 11 Maths Chapter 2 Relations and Functions

Exercise 2.1

Question 1.
If $\left( \frac { x }{ 3 } +1,y-\frac { 2 }{ 3 } \right) =\left( \frac { 5 }{ 3 } ,\frac { 1 }{ 3 } \right)$ , find the values of x and y.
Solution.
Since the ordered pairs are equal. So, the corresponding elements are equal
∴ $\frac { x }{ 3 } +1=\frac { 5 }{ 3 }$ and $y-\frac { 2 }{ 3 } =\frac { 1 }{ 3 }$
⇒ $\frac { x }{ 3 } =\frac { 5 }{ 3 } -1$ and $y=\frac { 1 }{ 3 } +\frac { 2 }{ 3 }$ ⇒ x = 2 and y = 1.

Question 2.
If the set A has 3 elements and the set B {3, 4, 5}, then find the number of elements in (A x B).
Solution.
According to question, n(A) = 3 and n(B) = 3.
∴ n(A x B) = n(A) x n(B) = 3 x 3 = 9
∴ There are total 9 elements in (A x B).

Question 3.
If G = {7, 8} and H = {5, 4, 2}, find G x H and H x G.
Solution.
We have G = {7, 8} and H = {5, 4, 2} Then, by the definition of the cartesian product, we have
G x H = {(7, 5), (7, 4), (7, 2), (8, 5), (8, 4), (8, 2)}
H x G = {(5, 7), (5, 8), (4, 7), (4, 8), (2, 7), (2, 8)}.

Question 4.
State whether each of the following statements 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 x Q = {(m, n), (n, m)}.
(ii) If A and B are non-empty sets, then Ax B is a non-empty set of ordered pairs (x, y) such
that x ∈ A and y ∈ B.
(iii) If A = {1, 2}, B = {3, 4}, then A x (B ∩φ) = φ
Solution.
(i) False, if P = {m, n} and Q = {n, m}
Then P x Q = {(m, n), (m, m), (n, n), (n, m)}.
(ii) True, by the definition of cartesian product.
(iii) True, We have A = {1, 2} and B = {3, 4}
Now, B ∩ φ = φ ∴ A x (B ∩ φ) = A x φ = φ.

Question 5.
If A = {-1, 1},find A x A x A.
Solution.
A = {-1, 1}
Then, A x A = {-1, 1} x {-1, 1} = {(-1, -1), (-1,1),(1,-1), (1,1)}
A x A x A = ((-1,-1),(-1,1),(1,-1),(1,1)} x {-1,1}
= {(-1, -1, -1), (-1, -1, 1), (-1, 1, -1), (-1, 1,1), (1, -1, -1), (1, -1,1), (1,1,-1), (1,1,1)}

Question 6.
If A x B = {(a, x), (a, y), (b, x), (b, y)}. Find A and B.
Solution.
Given, A x B = {(a, x), (a, y), (b, x), (b, y)}
If {p, q) ∈ A x B, then p ∈ A and q ∈ B
∴ A = {a, b} and B = {x, y}.

Question 7.
Let A = {1, 2}, B = (1, 2, 3, 4), C = {5, 6} and D = {5, 6, 7, 8}. Verify that
(i) A x (B ∩ C = (A x B) ∩ (AxC)
(ii) A x C is a subset of B x D.
Solution.
Given, A = {1, 2}, B ={1, 2, 3, 4}, C = {5, 6}, D = (5, 6, 7, 8}
NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions Ex 2.1 1
tiwari academy class 11 maths Chapter 2 Relations and Functions Ex 2.1 2

Question 8.
Let A = {1, 2} and B = {3, 4}. Write 4 x B. How many subsets will 4 x B have? List them.
Solution.
Given, A = {1, 2} and B = {3, 4}
Then, A x B = {(1, 3), (1,4), (2, 3), (2, 4)}
i. e., A x B has 4 elements. So, it has 24 i.e. 16 subsets.
The subsets of A x B are as follows :
φ, {(1, 3)1, ((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, 4), (2, 3), (2,4)},{(1, 3), (2, 3), (2, 4)}, {(1, 3), (1, 4), (2, 3), (2,4)}.

Question 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 x B, find A and B, where x, y and z are distinct elements.
Solution.
Given, n(A) = 3 and n(B) = 2
Now (x, 1) ∈ A x B ⇒ x ∈ A and 1 ∈ B,
(y, 2) ∈ A x B ⇒ y ∈ A and 2 ∈ B
(z, 1) ∈ A x B ⇒z ∈ A and 1 ∈ B
∴ x, y, z ∈ A and 1, 2 ∈ B
Hence, A = {x, y, z} and B = {1, 2}.

Question 10.
The Cartesian product 4×4 has 9 elements among which are found (-1, 0) and (0, 1). Find the set 4 and the remaining elements of 4 x 4.
Solution.
Since, we have n(A x A) = 9
⇒ n(A) x n(A) = 9 [ ∵ n (A x B) = n(A) x n(B)]
⇒ (n(A))2 = 9 ⇒ n(A) = 3
Also, given (-1, 0) ∈ A x A ⇒ -1, 0 ∈ A ,
and (0,1) ∈ A x A ⇒ 0, 1 ∈ A
∴ -1, 0,1 ∈ A
Hence, A = {-1, 0, 1} (∵ n(A) = 3)
and the remaining elements of A x A are (-1, -1), (-1,1), (0, -1), (0,0), (1, -1), (1,0), (1,1).

Exercise 2.2

Question 1.
Let 4 = {1,2,3, ,14}. Define a relation R from A to A by R = {(x,y): 3x – y = 0, where x, y ∈ 4}.
Write down its domain, codomain and range.
Solution.
We have A = (1, 2, 3,……..,14)
Given relation R = {(x, y) : 3x – y = 0, where x, y ∈ A}
= {(x, y): y = 3x, where x, y ∈ A)
= {(x, 3x), where x, 3x ∈ A}
= {(1, 3), (2, 6), (3, 9), (4,12)}
[∵ 1 ≤ 3x ≤ 14, ∴ $\frac { 1 }{ 3 } \le x\le \frac { 14 }{ 3 }$ ⇒ x = 1, 2, 3, 4 ]
Domain of R = {1, 2, 3, 4}
Codomain of R = {1, 2,……, 14}
Range of R = {3, 6, 9, 12}.

Question 2.
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 ∈ N}. Depict this relationship using roster form. Write down the domain and the range.
Solution.
Given relation R = {(x, y): y = x + 5, x < 4 and x, y ∈ N)
= {(x, y): y = x + 5, x ∈ (1, 2, 3) & y ∈ N}
= {(x, x + 5): x = 1, 2, 3}
Thus, R = {(1, 6), (2, 7), (3, 8)}.
Domain of R = {1, 2, 3}, Range of R = {6, 7, 8}.

Question 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 ∈ 4, y ∈B}. Write R in roster form.
Solution.
We have, A = {1, 2, 3, 5} and B = {4, 6, 9} R = {(x, y): difference between x and y is odd;
x ∈ A, y ∈B}
= {(x, y): y – x = odd; x ∈ A, y ∈ B}
Hence R = {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)}.

Question 4.
The figure 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?
NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions Ex 2.2 1
Solution.
(i) Its set builder form is
R = {(x, y): x – y = 2; x ∈ P, y ∈ Q}
i.e., R = {(x, y): y = x – 2 for x = 5, 6, 7)}

(ii) Roster form is R = {(5, 3), (6, 4), (7, 5)}
Domain of R = {5, 6, 7} = P,
Range of R = {3, 4, 5} = Q.

Question 5.
Let A = {1,2,3,4,6}. Let R be the relation on A defined by {(a, b): a, b ∈ 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?.
Solution.
Given A = {1, 2, 3, 4, 6}
Given relation is R = {(a, b):a,b ∈ A, b is exactly divisible by a}
(i) Roster form of 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} = A.
(iii) Range of R = {1, 2, 3, 4, 6} = A.

Question 6.
Determine the domain and range of the relation R defined by R = {(x, x + 5): x ∈ {0, 1, 2, 3, 4, 5}}.
Solution.
Given relation is R = {(x, x + 5): x ∈ {0, 1, 2, 3, 4, 5)}
= {(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5,10)}
∴Domain of R = {0,1, 2, 3, 4, 5} and
Range of R = {5, 6, 7, 8, 9, 10}.

Question 7.
Write the relation R = {(x, x3): x is a prime number less than 10} in roster form.
Solution.
Given relation is R = {(x, x3): x is a prime number less than 10)
= {(x, x3): x ∈ {2, 3, 5, 7}}
= {(2, 23), (3, 33), (5, 53), (7, 73)}
= {(2, 8), (3, 27), (5, 125), (7, 343)}.

Question 8.
Let A = {x, y, z} and B = {1, 2}. Find the number of relations from A to B.
Solution.
Given A = {x, y, z} and B = {1, 2}
∴ n(A) = 3 & n(B) = 2
Since n(A x B) = n(A) x n(B)
∴ n(A x B) = 3 x 2 = 6
Number of relations from A to B is equal to the number of subsets of A x B.
Since A x B contains 6 elements.
⇒ Number of subsets of A x B = 26 = 64
So, there are 64 relations from A to B.

Question 9.
Let R be the relation on Z defined by R = {{a, b): a, b ∈ Z, a – b is an integer}. Find the domain and range of R.
Solution.
Given relation is R = {(a, b): a, b ∈ Z, a – b is an integer}
If a, b ∈ Z, then a- b ∈ Z ⇒ Every ordered pair of integers is contained in R.
R = {(a, b) :a,b ∈ Z}
So, Range of R = Domain of R = Z.

Exercise 2.3

Question 1.
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)}.
Solution.
(i) We have a relation R = {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)} Since 2, 5, 8, 11, 14, 17 are the elements of domain of R having their unique images.
∴ The given relation is a function.
Hence domain = {2, 5, 8, 11, 14, 17) and Range = {1}.

(ii) We have a relation
R = {(2,1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}
Since 2, 4, 6, 8, 10, 12, 14 are the elements of domain of R having their unique images.
∴ The given relation is a function.
Hence domain = {2, 4, 6, 8, 10, 12, 14} and Range = {1, 2, 3, 4, 5, 6, 7}.

(iii) We have a relation R = {(1, 3), (1, 5), (2, 5)}
Since the distinct ordered pairs (1, 3) and (1, 5) have same first element i.e., 1 does not have a unique image under R.
∴ It is not a function.

Question 2.
Find the domain and range of the following real functions:
(i) f(x) = $-\left| x \right|$
(ii) f(x) = $\sqrt { 9-{ x }^{ 2 } }$
Solution.
NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions Ex 2.3 1

Question 3.
A function f is defined by f (x) = 2x – 5. Write down the values of
(i) f (0)
(ii) f (7)
(iii) f (-3)
Solution.
We are given f (x) = 2x – 5
(i) f (0) = 2(0) – 5 = 0- 5 = -5
(ii) f (7) = 2(7) – 5 = 14- 5 = 9
(iii) f (-3) = 2(-3) – 5 = -6 – 5 = -11.

Question 4.
The function T which maps temperature in degree Celsius into temperature in degree by
$t(C)=\frac { 9C }{ 5 } +32$
Find
(i) t (0)
(ii) t (28)
(iii) t (-10)
(iv) The value of C, when t (C = 212
Solution.
study rankers class 11 maths Chapter 2 Relations and Functions Ex 2.3 3
Question 5.
Find the range of each of the following functions.
(i) f(x) = 2 – 3x, x ∈ R, x>0.
(ii) f(x)=x2+ 2, x is a real number.
(iii) f (x) = x, x is a real number.
Solution.
(i) Given f (x) = 2 – 3x, x ∈ R, x > 0
∵ x > 0 ⇒ -3x < 0 ⇒ 2 – 3x < 2 + 0 ⇒ f (x) < 2
∴ The range of f (x) is (-2).

(ii) Given f (x) = x2 + 2, x is a real number
We know x2≥ 0 ⇒ x2 + 2 ≥ 0 + 2
⇒ x2 + 2 > 2 ∴ f (x) ≥ 2
∴ The range of f (x) is [2, ∞).

(iii) Given f (x) = x, x is a real number.
Let y =f (x) = x ⇒ y = x
∴ Range of f (x) = Domain of f (x)
∴ Range of f (x) is R.

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Verbs What is a verb? Verbs are the action words in a sentence that describe what the subject is doing. Along with nouns, verbs are the main part of a sentence or phrase, telling a story about what is taking place. Verbs are words that express action or state of being. There are three types of verbs: action verbs, linking verbs, and helping verbs . Action verbs are words that express action (give, eat, walk, etc.) or possession (have, own, etc.). Action verbs can be either transitive or intransitive. List of Verb in hindi Present Hindi Meaning Past Past Participle Buy खरीदना Bought Bought Build बांधना Built Built burn जलना Burnt Burnt Bend झुकना Bent Bent Bring लाना brought brought Become होना Became Become Come आना Came Come Catch पकड़ना Caught Caught Do करना Did Done Dream ख्वाब देखना Dreamt Dreamt Arise उठना / जागना Arose Arisen Be होना Was, were Been Bear सहन करना Bore Bore Beat मारना Beat Beat Bite काटना Bit Bitten Break तोडना Broke Broken Choose चुनना Chose Chosen Draw चि

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