Practicing the AP Board Solutions Class 11 Maths and Chapter 2 Inter 1st Year Maths Relations and Functions Solutions Exercise 2b Pdf Download will help students to clear their doubts quickly.
Intermediate 1st Year Maths Relations and Functions Solutions Exercise 2b
Relations and Functions Exercise 2b Solutions
Relations and Functions Class 11 Exercise 2b Solutions – Relations and Functions 2b Exercise Solutions
I.
Question 1.
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 ∈ A, y ∈ B}. Write R in roster form.
Solution:
Given, A = (1, 2, 3, 5) and B = (4, 6, 9)
The relation from A to B is given as
R = {(x, y): the difference between x and y is odd x ∈ A, y ∈ B}
Thus, R = {(1, 4) (1, 6) (2, 9) (3, 4), (3, 6), (5, 4), (5, 6)}
Question 2.
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
R = {(x, x + 5): x ∈ (0, 1, 2, 3, 4, 5)}
Thus, R = {(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)}
So, Domain of R = (0, 1, 2, 3, 4, 5) and Range of R = (5, 6, 7, 8, 9, 10)
Question 3.
Write the relation R = {(x, x3): x is a prime number less than 10} in roster form.
Solution:
Given Relation
R = {((x, x3) : x is a prime number less than 10)}
The prime numbers less than 10 are 2, 3, 5, and 7.
Therefore, R = {(2, 8) (3, 27) (5, 125), (7, 343)}
Question 4.
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)
Now, A × B = {(x, 1) (x, 2), (y, 1) (y, 2), (z, 1), (z, 2)}
As n(A × B) = 6, the number of subsets of A × B will be 26.
Thus, the number of relations from A to B is 26.
Question 5.
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
R = {(a, b) : a, b ∈ Z, a – b is an integer}
We know that the difference between any two integers is always an integer.
Therefore, Domain of R = z and Range of R = z.
II.
Question 1.
Let A = {1, 2, 3,…, 14}. Define a relation R from A to A by R = {(x, y): 3x-y = 0, where x, y ∈ A}. Write down its domain, codomain, and range.
Solution:
The relation R from A to A is given as:
R = {(x, y) : 3x – y = 0, where x, y ∈ A)} = {(x, y) : 3x = y, where x, y ∈ A)}
So, R = (1, 3), (2, 6), (3, 9), (4, 12)
Now, the domain of R is the set of all first elements of the ordered pairs in the relation.
Hence, the domain of R = (1, 2, 3, 4)
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:
The relation R is given by:
R = {(x, y) : y = x + 5, x is a natural number less than 4, x, y ∈ N}
The natural numbers less than 4 are 1, 2, and 3.
So, R = {(1, 6), (2, 7), (3, 8)}
Now, the domain of R is the set of all first elements of the ordered pairs in the relation.
Hence, the domain of R = (1, 2, 3)
The range of R is the set of all second elements of the ordered pairs in the relation.
Hence, Range of R = (6, 7, 8).
Question 3.
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?
Solution:
From the given figure, it is seen that
P = (5, 6, 7), Q = (3, 4, 5)
The relation between P and Q:
(i) Set-builder form
R = {(x, y) : y = x – 2; x ∈ P)}
or
R = {(x, y) : y = x – 2 for 5, 6, 7)}
(ii) Roster form
R = {(5, 3), (6, 4), (7, 5)}
Domain of R = (5, 6, 7)
Range of R = (3, 4, 5)
Question 4.
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) and relation R = {(a, b): a, b ∈ A is exactly divisible by a}
Hence,
(i) R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 2) (2, 4), (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)