Odisha State Board CHSE Odisha Class 11 Math Solutions Chapter 3 Relations And Functions Ex 3(b) Textbook Exercise Questions and Answers.

## CHSE Odisha Class 11 Math Solutions Chapter 3 Relations And Functions Exercise 3(b)

Question 1.

Let A = {a, b, c }, |B| = {1, 2}

(a) Determine all the relations from A to B and determine the domain, range, and inverse of each relation.

(b) Determine all the relations from B to A.

(c) Is there any relationship that is both a relation from A to B and B to A? How many?

(d) Of all the relations from A to B, identify which relations are many ones, one-many, and one-one and represent this diagrammatically.

Solution:

(a) A = {a, b, c}, B = {1, 2}

∴ A × B = {(a, 1), (a, 2), (b, 1), (b, 2), (c, 1), (c, 2)}

∴ |A × B| = 6

∴ |P(A × B)| = 2^{6} = 64

∴ There are 64 relations from A to B as any subset of A × B. The domain of these relations is any subset of A. The inverse of these relations is any subset of B × A.

(b) There are 64 relations from B to A as any sub-set of B x A is a relation from B to A.

(c) Φ is the only relation that is from A to B and from B to A.

(d) Some many-one relations are {(a, 1), (b, 1), (c, 1), (b, 2) (c, 2)}, {(a, 2), (b, 2), (c, 2)}.

Question 2.

Are the following sets related?

(i) Φ from A to B.

(ii) A × B from A to B.

(iii) A × Φ from A to Φ.

(iv) Φ × B from Φ to B.

(v) Φ × Φ from Φ to Φ.

(vi) Φ × C from A to B.

(vii) Φ × Φ from A to B.

Determine the domain range and inverse of each of the relations mentioned above

Solution :

(i) Φ from A to B is a relation.

(ii) A × B from A to B is a relation.

(iii) A × Φ from A to Φ is a relation.

(iv) Φ × B from Φ to B is a relation.

(v) Φ × Φ from Φ to Φ is a relation.

(vi) Φ × C from A to B is a relation.

(vii) Φ × Φ from A to B is a relation.

∴ Domain of Φ i.e. D_{Φ} = Φ

Range of Φ i.e., R_{Φ} = Φ

Similarly, D_{A × B} = A, R_{A × B} = β

D_{A}_{ × Φ} = Φ, R_{A × Φ} = Φ

D_{ Φ × B} = Φ = Φ, R_{ Φ × B} = Φ

D_{ Φ × Φ} = Φ, R_{ Φ × C} = Φ

D_{ Φ × C} = Φ, R_{ Φ × C} = Φ

D_{ Φ × Φ} = Φ, R_{ Φ × Φ} = Φ

The inverse of the above relations is Φ, B × A, Φ × A, B × Φ, Φ × Φ, C × Φ, and Φ × Φ respectively.

Question 3.

Express the following relations on A to B in each case in tabular form :

(i) A = {n ∈ N : n ≤ 10}, B = N

f = {(x, y) ∈ A × B : y = x^{2}}

Solution:

A = {n ∈ N : n ≤ 10}

= {1, 2, 3,…..10}, B = N

∴ B = {1, 2, 3}

∴ f ={(x, y) ∈ A × B : y = x^{2}}

= {(1, 1), (2, 4), (3, 9)…..(10, 100)}

(ii) A = B = R

∴ f = {(x, y) : x^{2} + y^{2} = 1 and |x – y| = 1}

Solution:

A = B = R

∴ f = {(x, y) : x^{2} + y^{2} = 1 and |x – y| = 1}

={(0, 1) (1, 0), (-1, 0), (0, -1)}

(iii) (1, 2, 3, 4), B = {1, 2, 3, 4, 5}

f = {x, y) : 2 divides 3x+y}

Solution:

A = {1, 2, 3, 4}, B = {1, 2, 3, 4, 5}

∴ f = {(x, y) : 2 divides 3x+y}

={(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (3, 1), (3, 3), (3,5), (4, 2), (4, 4)}

Question 4.

A and B are non-empty sets such that |A| = m, |B| = n. How many relations can be defined from A to B ? (Remember that the number of relations is the number of subsets of (A × B).

Solution:

|A| = m, |B| = n

⇒ |A × B| = mn

A relation is a subset of A to B

∴ Number of relations from A to B

= Number of subsets of A × B

= 2^{mn} (∴ |A × B| = mn)

Question 5.

Give an example of a relation f such that

(i) dom f – rng f (ii) dom f ⊂ rng f

(iii) dom f ⊃ rng f

(iv) f ∪ f^{-1} = Φ

(v) f = f^{-1}

(vi) f ∪ f^{-1} ≠ Φ

Solution:

Let A = { 1, 2, 3} = B

(i) Let f = {(x, y) ∈ A × B : x = y}

∴ Dom f = {1, 2, 3} = Range f

(ii) Let f = {(1, 1), (1, 2), (2, 3)}

on A = (1, 2, 3}

∴ Dom f = {1, 2} ⊂ { 1, 2, 3} = Range f

(iii) Do yourself

(iv) Let f = Φ

∴ f^{-1} = Φ = f ∪ f^{-1} = Φ

(v) Let f = {(x, y) ∈ A × B; x^{2} + y^{2} = 1}, where A = B = {1, – 1, 0}

= {(1, 0), (0, 1), (-1, 0), (0, -1)}

f^{-1} = {(0, 1) (1, 0), (0, -1), (-1, 0)}

=f

(vi) Let f = {(1, 3), (3, 1)} on A = { 1, 2, 3}

∴ f^{-1} = {3, 1), (1, 3)},

so that f ∩ f^{-1} = Φ.

Question 6.

Let R = {(a, a^{3}) I a is a prime number less than 10}

Fine (i) R, (ii) dom R, (iii) rng R (iv) R^{-1} (v) dom R^{-1} (vi) rng R^{-1}

Solution:

R = {(a, a^{3})} a is a prime number less than 10}

(i) R = {(2, 8), (3, 27), (5, 125), (7, 343)}

(ii) dom R = {2, 3, 5, 7}

(iii) rng R = {8, 27, 125, 343}

(iv) R^{-1} = {(8, 2), (27, 3), (125, 5), (343, 7)}

(v) Dom R^{-1} = {8, 27, 125, 343} = rng R

(vi) rng R^{-1} = {2, 3, 5, 7} = dom R

Question 7.

Let A = {1, 2, 3, 4, 5, 6} and Let R be a relation on A defined by R = {(a, b)} a divides b

Find (i) R, (ii) dom R, (iii) rng R (iv) R^{-1}, (v) Dom R^{-1} (vi) rng R^{-1}

Solution:

A = {1, 2, 3, 4, 6}

R on A is defined by

R = {(a, b) | a divides b}

(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) dom R = {1,2, 3, 4, 6} = A

(iii) rng R = {1, 2, 3, 4, 6} = A

(iv) R^{-1} = {(1, 1), (2, 1), (3, 1), (4, 1), (6, 1), (2, 2) (4, 2), (6, 2), (3, 3), (6, 3), (4, 4), (6, 6)}

(v) dom R^{-1} = {1, 2, 3, 4, 6} = A

(vi) rng R^{-1} = {l, 2, 3, 4, 6} = A