Chapter 2 Relations and Functions of Class 11 Maths is categorized under the term â€“ I CBSE Syllabus for 2021-22. This chapter contains 3 exercises along with a miscellaneous exercise. Exercise 2.2 of NCERT Solutions for Class 11 Maths Chapter 2- Relations And Functions is based on Relations:

- A relation R from a non-empty set A to a non-empty set B is a subset of the cartesian product A Ã— B. The subset is derived by describing the relationship between the first element and the second element of the ordered pairs in A Ã— B. The second element is called the image of the first element.
- The set of all first elements of the ordered pairs in a relation R from a set A to a set B is called the domain of the relation R.
- The set of all second elements in a relation R from a set A to a set B is called the range of the relation R. The whole set B is called the codomain of the relation R. Note that range âŠ‚ codomain.
- A relation may be represented algebraically either by the Roster method or by the Set-builder method.
- An arrow diagram is a visual representation of a relation.

Every student aims at scoring excellent marks in the Maths examinations. Practising the NCERT textual questions by referring to the NCERT Solutions of Class 11 Maths, can undoubtedly help the students in fulfilling their aim of scoring high in the first term examination.

### Download PDF of NCERT Solutions for Class 11 Maths Chapter 2- Relations And Functions Exercise 2.2

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

Exercise 2.1 Solutions 10 Questions

Exercise 2.3 Solutions 5 Questions

Miscellaneous Exercise On Chapter 2 Solutions 12 Questions

### Access Solutions for Class 11 Maths Chapter 2.2 exercise

**1. Let A = {1, 2, 3, â€¦ , 14}. Define a relation R from A to A by R = {( x,Â y): 3xÂ â€“Â yÂ = 0, whereÂ x,Â yÂ âˆˆÂ A}. Write down its domain, codomain and range.**

**Solution: **

The relation R from A to A is given as:

R = {(*x*,Â *y*): 3*x*Â â€“Â *y*Â = 0, whereÂ *x*,Â *y*Â âˆˆÂ A}

= {(*x*,Â *y*): 3*x*Â =Â *y*, whereÂ *x*,Â *y*Â âˆˆÂ A}

So,

R = {(1, 3), (2, 6), (3, 9), (4, 12)}

Now,

The domain of R is the set of all first elementsÂ of the ordered pairs in the relation.

Hence, Domain of R = {1, 2, 3, 4}

The whole set A is the codomain of the relation R.

Hence, CodomainÂ of RÂ =Â A = {1, 2, 3, â€¦, 14}

The range of R is the set of all second elementsÂ of the ordered pairs in the relation.

Hence, Range of R = {3, 6, 9, 12}

**2. Define a relation R on the setÂ NÂ of natural numbers by R = {( x,Â y):Â yÂ =Â xÂ + 5,Â xÂ is a natural number less than 4;Â x,Â yÂ âˆˆÂ N}. Depict this relationship using roster form. Write down the domain and the range.**

**Solution:**

**The relation R is given by:**

R = {(*x*,Â *y*):Â *y*Â =Â *x*Â + 5,Â *x*Â is a natural number less than 4,Â *x*,Â *y*Â âˆˆÂ **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 elementsÂ of the ordered pairs in the relation.

Hence, Domain of R = {1, 2, 3}

The range of R is the set of all second elementsÂ of the ordered pairs in the relation.

Hence, Range of R = {6, 7, 8}

**3. A = {1, 2, 3, 5} and B = {4, 6, 9}. Define a relation R from A to B by R = {( x,Â y): the difference betweenÂ xÂ andÂ yÂ is odd;Â xÂ âˆˆÂ A,Â yÂ âˆˆÂ 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*,Â *y*): the difference betweenÂ *x*Â andÂ *y*Â is odd;Â *x*Â âˆˆÂ A,Â *yÂ *âˆˆÂ B}

Thus,

R = {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)}

**4. The figure shows a relationship between the sets P and Q. write this relation**

**(i) in set-builder form (ii) in roster form.**

**What is its domain and range?**

**Solution:**

From the given figure, itâ€™s seen that

P = {5, 6, 7}, Q = {3, 4, 5}

The relation between P and Q:

Set-builder form

(i) R = {(*x, y*):Â *y = x*Â â€“ 2;Â *x*Â âˆˆÂ P} or R = {(*x, y*):Â *y = x*Â â€“ 2 forÂ *x*Â = 5, 6, 7}

Roster form

(ii) R = {(5, 3), (6, 4), (7, 5)}

Domain of R = {5, 6, 7}

Range of R = {3, 4, 5}

**5. Let A = {1, 2, 3, 4, 6}. Let R be the relation on A defined by**

**{( a,Â b):Â a,Â bÂ âˆˆÂ A,Â bÂ is exactly divisible byÂ 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*,Â *b*):Â *a*,Â *b*Â âˆˆÂ A,Â *b*Â is exactly divisible byÂ *a*}

Hence,

(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}

**6. Determine the domain and range of the relation R defined by R = {( x,Â xÂ + 5):Â xÂ âˆˆÂ {0, 1, 2, 3, 4, 5}}.**

**Solution: **

Given,

Relation R = {(*x*,Â *x*Â + 5):Â *x*Â âˆˆÂ {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}

**7. Write the relation R = {( x,Â x^{3}):Â xÂ is a prime number less than 10} in roster form.**

**Solution: **

Given,

Relation R = {(*x*,Â *x*^{3}):Â *xÂ *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)}

**8. Let A = { x,Â y, z} and B = {1, 2}. Find the number of relations from A to B.**

**Solution: **

Given, A = {*x*,Â *y*, z} and B = {1, 2}.

Now,

A Ã— B = {(*x*, 1), (*x*, 2), (*y*, 1), (*y*, 2), (*z*, 1), (*z*, 2)}

AsÂ *n*(A Ã— B) = 6, the number of subsets of A Ã— B will be 2^{6}.

Thus, the number of relations from A to B is 2^{6}.

**9. Let R be the relation onÂ ZÂ defined by R = {( a,Â b):Â a,Â bÂ âˆˆÂ Z,Â aÂ â€“Â bÂ is an integer}. Find the domain and range of R.**

**Solution: **

Given,

Relation R = {(*a*,Â *b*):Â *a*,Â *b*Â âˆˆÂ Z,Â *aÂ *â€“Â *b*Â is an integer}

We know that the difference between any two integers is always an integer.

Therefore,

Domain of R =Â Z and Range of R =Â Z