The first exercise of Relations and Functions revolves around the basic concepts of the chapter. Exercise 2.1 of NCERT Solutions for Class 11 Maths Chapter 2 â€“ Relations And Functions is based on the following topics:

- Introduction to Relations And Functions
- Cartesian Product of Sets: Given two non-empty sets P and Q. The cartesian product P Ã— Q is the set of all ordered pairs of elements from P and Q, i.e., P Ã— Q = { (p,q) : p âˆˆ P, q âˆˆ Q }. If either P or Q is the null set, then P Ã— Q will also be an empty set.

The best way to learn Maths is to do Maths. Practising the textbook problems usingÂ NCERT solutions of class 11 maths will help the students in understanding the concepts discussed in this exercise.

### Download PDF of NCERT Solutions for Class 11 Maths Chapter 2- Relations And Functions Exercise 2.1

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### Access other exercise solutions of Class 11 Maths Chapter 2- Relations And Functions

Exercise 2.2 Solutions 9 Questions

Exercise 2.3 Solutions 5 Questions

Miscellaneous Exercise On Chapter 2 Solutions 12 Questions

### Access Solutions for Class 11 Maths Chapter 2.1 exercise

**1. If , find the values ofÂ xÂ andÂ y.**

**Solution: **

Given,

As the ordered pairs are equal, the corresponding elements should also be equal.

Thus,

x/3 + 1 = 5/3 and y â€“ 2/3 = 1/3

Solving, we get

x + 3 = 5 and 3y â€“ 2 = 1 [Taking L.C.M and adding]

x = 2 and 3y = 3

Therefore,

x = 2 and y = 1

**2. If the set A has 3 elements and the set B = {3, 4, 5}, then find the number of elements in (A Ã— B)?**

**Solution:**

Given, set A has 3 elements and the elements of set B are {3, 4, and 5}.

So, the number of elements in set B = 3

Then, the number of elements in (A Ã— B) = (Number of elements in A) Ã— (Number of elements in B)

= 3 Ã— 3 = 9

Therefore, the number of elements in (A Ã— B) will be 9.

**3. If G = {7, 8} and H = {5, 4, 2}, find G Ã— H and H Ã— G.**

**Solution: **

Given, G = {7, 8} and H = {5, 4, 2}

We know that,

The Cartesian product of two non-empty sets P and Q is given as

P Ã— Q = {(*p*,Â *q*):Â *p *âˆˆÂ P,Â *q*Â âˆˆÂ Q}

So,

G Ã— H = {(7, 5), (7, 4), (7, 2), (8, 5), (8, 4), (8, 2)}

H Ã— G = {(5, 7), (5, 8), (4, 7), (4, 8), (2, 7), (2, 8)}

**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,Â n} and Q = {n,Â m}, then P Ã— Q = {(m,Â n), (n,Â m)}.**

**(ii) If A and B are non-empty sets, then A Ã— B is a non-empty set of ordered pairs ( x,Â y) such thatÂ xÂ âˆˆÂ A andÂ yÂ âˆˆÂ B.**

**(iii) If A = {1, 2}, B = {3, 4}, then A Ã— (BÂ âˆ©Â Î¦) =Â Î¦.**

**Solution: **

(i) The statement is False. The correct statement is:

If P = {*m*,Â *n*} and Q = {*n*,Â *m*}, then

P Ã— Q = {(*m*,Â *m*), (*m*,Â *n*), (*n,*Â *m*), (*n*,Â *n*)}

(ii) True

(iii) True

**5. If A = {â€“1, 1}, find A Ã— A Ã— A.**

**Solution: **

The A Ã— A Ã— A for a non-empty set A is given by

A Ã— A Ã— A = {(*a*,Â *b*,Â *c*):Â *a*,Â *b*,Â *cÂ *âˆˆÂ A}

Here, It is given A = {â€“1, 1}

So,

A Ã— A Ã— A = {(â€“1, â€“1, â€“1), (â€“1, â€“1, 1), (â€“1, 1, â€“1), (â€“1, 1, 1), (1, â€“1, â€“1), (1, â€“1, 1), (1, 1, â€“1), (1, 1, 1)}

**6. If A Ã— B = {( a,Â x), (a,Â y), (b,Â x), (b,Â y)}. Find A and B.**

**Solution:**

Â

Given,

A Ã— B = {(*a*,Â *x*), (*a,*Â *y*), (*b*,Â *x*), (*b*,Â *y*)}

We know that the Cartesian product of two non-empty sets P and Q is given by:

P Ã— Q = {(*p*,Â *q*):Â *p*Â âˆˆÂ P,Â *q*Â âˆˆÂ Q}

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

Therefore, A = {*a*,Â *b*} and B = {*x*,Â *y*}

**7. Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify that**

**(i) A Ã— (BÂ âˆ©Â C) = (A Ã— B)Â âˆ©Â (A Ã— C)**

**(ii) A Ã— C is a subset of B Ã— D**

**Solution: **

Given,

A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}

(i) To verify: A Ã— (BÂ âˆ©Â C) = (A Ã— B)Â âˆ©Â (A Ã— C)

Now, BÂ âˆ©Â C = {1, 2, 3, 4}Â âˆ©Â {5, 6} =Â Î¦

Thus,

L.H.S. = A Ã— (BÂ âˆ©Â C) = A Ã—Â Î¦Â =Â Î¦

Next,

A Ã— B = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4)}

A Ã— C = {(1, 5), (1, 6), (2, 5), (2, 6)}

Thus,

R.H.S. = (A Ã— B)Â âˆ©Â (A Ã— C) =Â Î¦

Therefore, L.H.S. = R.H.S

â€“ Hence verified

(ii) To verify: A Ã— C is a subset of B Ã— D

First,

A Ã— C = {(1, 5), (1, 6), (2, 5), (2, 6)}

And,

B Ã— 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)}

Now, itâ€™s clearly seen that all the elements of set A Ã— C are the elements of set B Ã— D.

Thus,Â A Ã— C is a subset of B Ã— D.

â€“ Hence verified

**8. Let A = {1, 2} and B = {3, 4}. Write A Ã— B. How many subsets will A Ã— B have? List them.**

**Solution: **

Given,

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

So,

A Ã— B = {(1, 3), (1, 4), (2, 3), (2, 4)}

Number of elements in A Ã— B isÂ *n*(A Ã— B) = 4

We know that,

If C is a set withÂ *n*(C) =Â *m*, thenÂ *n*[P(C)] = 2* ^{m}*.

Thus, the set A Ã— B has 2^{4}Â = 16 subsets.

And, these subsets are as below:

Î¦, {(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)}

**9. Let A and B be two sets such thatÂ n(A) = 3 andÂ nÂ (B) = 2. If (x, 1), (y, 2), (z, 1) are in A Ã— B, find A and B, whereÂ x,Â yÂ andÂ zÂ are distinct elements.**

**Solution: **

Given,

*n*(A) = 3 andÂ *n*(B) = 2; and (*x*, 1), (*y*, 2), (*z*, 1) are in A Ã— B.

We know that,

A = Set of first elements of the ordered pair elements of A Ã— B

B = Set of second elements of the ordered pair elements of A Ã— B.

So, clearly *x*,Â *y*, andÂ *z*Â are the elements of A; and

1 and 2 are the elements of B.

AsÂ *n*(A) = 3 andÂ *n*(B) = 2, it is clear that set A = {*x*,Â *y*,Â *z*} and set B = {1, 2}.

**10. The Cartesian product A Ã— A has 9 elements among which are found (â€“1, 0) and (0, 1). Find the set A and the remaining elements of A Ã— A.**

**Solution:**

We know that,

IfÂ *n*(A) =Â *pÂ *andÂ *n*(B) =Â *q,Â *thenÂ *n*(A Ã— B) =Â *pq*.

Also,Â *n*(A Ã— A) =Â *n*(A) Ã—Â *n*(A)

Given,

*n*(A Ã— A) = 9

So, *n*(A) Ã—Â *n*(A) = 9

Thus, *n*(A) = 3

Also given that, the ordered pairs (â€“1, 0) and (0, 1) are two of the nine elements of A Ã— A.

And, we know in A Ã— A = {(*a, a*):Â *a*Â âˆˆÂ A}.

Thus, â€“1, 0, and 1 has to be the elements of A.

AsÂ *n*(A) = 3, clearlyÂ A = {â€“1, 0, 1}.

Hence, the remaining elements of set A Ã— A are as follows:

(â€“1, â€“1), (â€“1, 1), (0, â€“1), (0, 0), (1, â€“1), (1, 0), and (1, 1)