Class 11 Maths | NCERT Exercise 2.1 | Solved Answers and Questions | #eduvictors

Class 11 Maths | NCERT Exercise 2.1 | Solved Answers and Questions

RELATIONS AND FUNCTIONS

Q1. If (x/3 + 1, y - 2/3) = (5/3, 1/3), find the values of x and y.

Answer: Since ordered pairs are equal, 

∴ x/3 + 1 = 5/3

⇒ x/3 = 5/3 - 1 = 2/3

⇒ x = 2

y - 2/3 = 1/3

y = 1/3 + 2/3

y = 3/3 = 1

Thus, x = 2 and y = 1

Q2. If the set A has 3 elements and the set B = {3, 4, 5}, then find the number of elements in (A×B)  

Answer: n(A) = 3 and n(B) = 3

Thus n(A×B) = 3×3 = 9

Q3. If G = {7, 8} and H = {5, 4, 2}, find G × H and H × G

Answer: 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.

Answer: 

(i) False

P × Q = {(m,n),(m,m),(n,n),(n,m)}

(ii) True

(iii) True (The Cartesian product of any set with the empty set is also the empty set.)

Q5. If A = {–1, 1}, find A × A × A.

Answer: A × A × A = {(a, b, c) : a, b, c ∈ A}. Here (a, b, c) is called an ordered triplet.

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

Answer: We know that the Cartesian product of two non-empty sets P and Q is defined as P × Q = {(p, q): p ∈ P, q ∈ Q}

∴ A is the set of all first elements and B is the set of all second elements.

Thus, 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.

Answer: 

(i) (B ∩ C) = φ

Thus LHS = A × (B ∩ C) = 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)}

RSH = (A × B) ∩ (A × C) = φ

∴ LHS= RHS

(ii) A × C = {(1,5), (1,6), (2,5), (2,6)}

B × 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

Answer: 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.

Answer: A×B has ordered pairs. Thus A = {x,y,z} and B={1,2}

Q10. The Cartesian product A × A has 9 elements, found (–1, 0) and (0,1). Find the set A and the remaining elements of A × A.

Answer: 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)

👉SEE ALSO:

Ch2: Relations and Functions (1 Mark Q & A) Part-1
Ch2: Cartesian Product of Two Sets (Important Points)
Ch2: Relations - Domain, Range and Co-Domain (Solved Problems)