2. Relations and Functions Mathematics class 11 exercise Miscellaneous Exercise
2. Relations and Functions Mathematics class 11 exercise Miscellaneous Exercise ncert book solution in english-medium
NCERT Books Subjects for class 11th Hindi Medium
Exercise 2.1
Exercise 2.1
Comparing both side as order pairs are equal, so corresponding elements also will be equal,
Q2. If the set A has 3 elements and the set B = {3, 4, 5}, then find the number of elements in (A×B).
Solution:
Given : n(A) = 3 and set B = {3, 4, 5}
∴ n(B) = 3
Number of elements in (A × B)
= (Number of elements in A) × (Number of elements in B)
= n(A) × n(B)
= 3 × 3
= 9
Q3. If G = {7, 8} and H = {5, 4, 2}, find G × H and H × G.
Solution:
Given: G = {7, 8} and H = {5, 4, 2}
We know that the Cartesian product P × Q of two non-empty sets P and Q is defined as P × Q = {(p, q): p∈ P, q ∈ Q}
∴ G × H = {(7, 5), (7, 4), (7, 2), (8, 5), (8, 4), (8, 2)}
H × G = {(5, 7), (5, 8), (4, 7), (4, 8), (2, 7), (2, 8)}
Q4. 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 × 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 ∈ A and y ∈ B.
(iii) If A = {1, 2}, B = {3, 4}, then A × (B ∩ φ) = φ.
Solution:
(i) If P = {m, n} and Q = { n, m}, then P × Q = {(m, n),(n, m)}.
False
If P = {m, n} and Q = {n, m},
P × Q = {(m, m), (m, n), (n, m), (n, n)}
Solution:
(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 ∈ A and y ∈ B.
(ii) True
Solution:
(iii) If A = {1, 2}, B = {3, 4}, then A × (B ∩ φ) = φ.
(iii) True
Q5. If A = {–1, 1}, find A × A × A.
Solution:
It is known that for any non-empty set A, A × A × A is defined as;
A × A × A = {(a, b, c): a, b, c ∈ A}
Given that A = {–1, 1}
∴ A × A × 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)}
Q6. If A × B = {(a, x),(a , y), (b, x), (b, y)}. Find A and B.
Solution:
The cartesian product of two non-empty sets P and Q is defined as P × Q = {(p, q): p ∈ P, q ∈ Q}
Given that: A × B = {(a, x),(a , y), (b, x), (b, y)}
∴ a, b ∈ A abd x, y ∈ B
So, A = {a, b} and B = {x, y}
Q7. Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}.
Verify that:
(i) A × (B ∩ C) = (A × B) ∩ (A × C).
(ii) A × C is a subset of B × D.
Solution:
(i) To verify: A × (B ∩ C) = (A × B) ∩ (A × C)
We have B ∩ C = {1, 2, 3, 4} ∩ {5, 6} = Φ
∴ L.H.S. = A × (B ∩ C) = A × Φ = Φ
A × B = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4)}
A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}
∴ R.H.S. = (A × B) ∩ (A × C) = Φ
∴ L.H.S. = R.H.S
Hence, A × (B ∩ C) = (A × B) ∩ (A × C)
(ii) To verify: A × C is a subset of B × D
Solution:
A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}
A × 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)}
All the elements of set A × C are the elements of set B × D.
Therefore, A × C is a subset of B × D.
Q8. Let A = {1, 2} and B = {3, 4}. Write A × B. How many subsets will A × B have? List them.
Solution:
A = {1, 2} and B = {3, 4}
∴ A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}
⇒ n(A × B) = 4
We know that if C is a set with n(C) = m,
then n[P(C)] = 2m.
Therefore, the set A × B has 24 = 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)}
Q9. 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 × B, find A and B, where x, y and z are distinct elements.
Solution:
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,
∴ A = {x, y, z} and B = {1, 2}.
Q10. 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 × A.
Solution:
We know that if n(A) = p and n(B) = q,
then n(A × B) = pq.
∴ n(A × A) = n(A) × n(A)
It is given that n(A × A) = 9
∴ n(A) × n(A) = 9 ⇒ n(A) = 3
The ordered pairs (–1, 0) and (0, 1) are two of the nine elements of A×A.
We know that A × A = {(a, a): a ∈ 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 × A are (–1, –1), (–1, 1), (0, –1), (0, 0), (1, –1), (1, 0), and (1, 1).
Exercise 2.2
Exercise 2.2
Q1. 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}
Given, R = {(x, y): 3x - y = 0, where x, y ∈ A}
3x - y = 0
∴ 3x = y
Putting value x= 1, 2, 3, 4 from A we have y = 3, 6, 9, 12
∴ 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 R.
∴ Domain of R = {1, 2, 3, 4}
The whole set A is the codomain of the relation R.
∴ Co-domain of R = {1, 2, 3… 14}
The range of R is the set of all second elements of the ordered pairs in the relation R.
∴Range of R = {3, 6, 9, 12}
Q2. 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:
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.
∴ 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 R.
∴ Domain of R = {1, 2, 3}
The range of R is the set of all second elements of the ordered pairs in the relation R.
∴ Range of R = {6, 7, 8}
Q3. 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:
A = {1, 2, 3, 5} and B = {4, 6, 9}
R = {(x, y): the difference between x and y is odd; x ∈ A, y ∈ B}
∴ R = {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)}
Q4. The 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?
Solution: From the given figure, we have
A = {5, 6, 7}, B = {3, 4, 5}
(i)R = {(x, y): y = x – 2; x ∈ A and y ∈ B} or R = {(x, y): y = x – 2 for x = 5, 6, 7}
(ii) R = {(5, 3), (6, 4), (7, 5)} Domain of R = {5, 6, 7} Range of R = {3, 4, 5}
Q5. 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:
A = {1, 2, 3, 4, 6}, R = {(a, b): a, b ∈ 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}
Q6. Determine the domain and range of the relation R defined by R = {(x, x + 5) : x ∈ {0, 1, 2, 3, 4, 5}}.
Solution:
R = {(x, x + 5): x ∈ {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}
Q7. Determine the domain and range of the relation R defined by R = {(x, x + 5) : x ∈ {0, 1, 2, 3, 4, 5}}.
Solution:
R = {(x, x3): 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)}
Q8. 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}.
∴ A × B = {(x, 1), (x, 2), (y, 1), (y, 2), (z, 1), (z, 2)}
Since n(A × B) = 6,
The number of subsets of A × B is 26.
∴ the number of relations from A to B is 26.
Q9. 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:
R = {(a, b): a, b ∈ 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
Exercise 2.3
Q1. 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) {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}
Let be R = {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}
There is no same component in the domain of given relation, so this relation is a function.
Here, domain = {2, 5, 8, 11, 14, 17} and range = {1}
Solution:
(ii) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}
Let be R = {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}
There are also unique domain components or order pair in the given relation, so this relation is a function.
Here, domain = {2, 4, 6, 8, 10, 12, 14} and range = {1, 2, 3, 4, 5, 6, 7}
Solution:
(iii) {(1, 3), (1, 5), (2, 5)}
Since the domain has the same first element i.e., 1 corresponds to two different images i.e.,
3 and 5, this relation is not a function.
Q2. Find the domain and range of the following real functions:
(i) f(x) = – |x|
Solution2:
Q3: A function f is defined by f(x) = 2x – 5
(i) f (0) (ii) f(7) (iii) f(-3)
Solution 3:
The given function is f(x) = 2x – 5
Therefore,
(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
Solution:
Solution:
(iii) f(x) = x is real number.
It is clear that the range of f is the set of all real number.
Therefore, range of f = R
Miscellaneous Exercise
MISCELLANEOUS EXERCISE
Select Class for NCERT Books Solutions
NCERT Solutions
NCERT Solutions for class 6th
NCERT Solutions for class 7th
NCERT Solutions for class 8th
NCERT Solutions for class 9th
NCERT Solutions for class 10th
NCERT Solutions for class 11th
NCERT Solutions for class 12th
sponder's Ads
Mathematics Chapter List
1. Sets
2. Relations and Functions
3. Trigonometric Functions
4. Principle Of Mathematical Induction
5. Complex Numbers and Quadratic Equations
6. Linear Inequalities
7. Permutations and Combinations
8. Binomial Theorem
9. Sequences and Series
10. Straight Lines
11. Conic Sections
12. Introduction to Three Dimensional Geometry
13. Limits and Derivatives
14. Mathematical Reasoning
15. Statistics
16. Probability
sponser's ads