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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