2. RELATIONS AND FUNCTIONS

Class 11 Mathematics Chapter 2. RELATIONS AND FUNCTIONS

Chapter 2. Relations and Functions

Class 11 Maths Chapter 2. Relations and Functions Exercise 2.1 Solutions :

1. If , find the values of and .

Solution : We have,

and

Therefore, the value of and are 2 and 1 .

2. If the set A has 3 elements and the set B = {3, 4, 5}, then find the number of elements in (A×B).
Solution : [Note : If n(A) = p and n(B) = q, then n(A × B) = pq.]

Give,

Here , and

3. If G = 7, 8 and H = 5, 4, 2 find G × H and H × G.

Solution : [Note : Cartesian product A × B of two sets A and B is given by A × B = {(a, b): a ∈ A, b ∈ B} ]

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

4. State whether each of the following statements are true or false. If the statement is false, rewrite the given statement correctly.
(i) If and then
(ii) If A and B are non-empty sets, then A × B is a non-empty set of ordered pairs (, ) such that ∈ A and ∈ B.
(iii) If A = 1, 2 B = 3, 4 then .

Solution : (i) False

Because, P = {m ,n} and Q = {n ,m} then P × Q ={(m ,m),(m, n),(n ,m),(n ,n)}

(ii) False

If A andB are non-empty sets, then A×B is a non-empty set of ordered pairs (x,y) such that x∈A and y∈B . Because, A×B is always non-empty; however, if either A or B is empty, then A×B would be empty.

(ii) True

If A={1,2} , B={3,4} , and represents the empty set, then is equivalent to , which is the empty set . This is because the intersection of any set with the empty set is the empty set.

5. If A = 1, 1 find A × A × A.

Solution : Given ,

={(– 1 ,–1 , –1), (–1 , 1 , –1), (1 , –1 , –1) ,(1 , 1 , –1) , (–1, –1 , 1), (–1 , 1 ,1), (1 , –1 , 1), (1 , 1 , 1)}
6. If . Find A and B.

Solution : [Cartesian product A × B of two sets A and B is given by A × B = {(a, b): a ∈ A, b ∈ B}]

We have,

and
7. Let and Verify that
(i) A × (B ∩ C) = (A × B) ∩ (A × C). (ii) A × C is a subset of B × D.

Solution : (i) A × (B ∩ C) = (A × B) ∩ (A × C)

Since B ∩ C = {1,2,3,4} ∩ {5,6} =

A × B = {1,2} × {1,2,3,4}

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

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

LHS : A × (B ∩ C)

A × (B ∩ C) = A × =

RHS : (A × B) ∩ (A × C)

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

LHS = RHS verified .

(ii) A × C is a subset of B × D :

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)}

Since every element in A × C is also in B × D, A × C is a subset of B × D.

Therefore, the statements is verified.

8. Let A = 1, 2 and B = 3, 4 Write A × B. How many subsets will A × B have? List them.

Solution : Here, and

A×B = {1 , 2} × {3 , 4}

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

A×B has 4 elements, so it will have subsets.

A × B = , {(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)}

So, A×B has 16 subsets .

9. Let A and B be two sets such that and . If are in A × B, find A and B, where and are distinct elements.

Solution : Given that , it means that set A has three distinct elements.

and , it means that set B has two distinct elements.

Now, the ordered pairs and are in .

i.e., and .

So, and , where and are distinct elements.

10. 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 (−1,0)and (0,1) are elements of A×A, so −1,0,1 must be the elements of A.

Therefore, A={−1,0,1}.

So, A×A={(−1,−1),(−1,0),(−1,1),(0,−1),(0,0),(0,1),(1,−1),(1,0),(1,1)}.

The remaining elements of A×B are (−1,−1) ,(−1,1) ,(0,−1),(0,0) ,(1,−1) , (1,1) .

Class 11 Maths Chapter 2. Relations and Functions Exercise 2.2 :

1. Let Define a relation from A to A by , where Write down its domain, co-domain and range.

Solution : Given,

, where

If then

Therefore, the domain is the set of all values in A .

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

The co-domain is the set A , as both and are taken from A .

Co-domain of R = {1,2,3,...,14}

Since , the range is the set of all multiples of 3 that are in the given set A .

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

2. Define a relation on the set of natural numbers by is a natural number less than Depict this relationship using roster form. Write down the domain and the range.

Solution : Given, is a natural number less than .

If , then

So, the domain of

The range of R is the set of all values in the ordered pairs .

In this case, is determined by , so the range is the set {6,7,8} .

The range of

3. and Define a relation from to by the difference between and is odd; Write R in roster form.

Solution : Given, and

the difference between and is odd;

4. 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 : In the given figure ,

and

(i) in set-builder form :

and

(ii) roster form :

So, domain and Range

5. Let Let be the relation on A defined by is exactly divisible by
(i) Write in roster form
(ii) Find the domain of
(iii) Find the range of .

Solution : Given,

is exactly divisible by

(i) R in roster form :

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) The domain of R = {1,2,3,4,6}

(iii) The range of R = {1,2,3,4,6}

6. Determine the domain and range of the relation defined by

Solution : Given , the relation

The domain of a relation is the set of all possible first components (in this case ,).

In this relation, takes values from the set {0,1,2,3,4,5} .

Domain of R = {0,1,2,3,4,5}

The range of a relation is the set of all possible second components (in this case, ).

For each in the domain, is the corresponding second component.

Range of R = {0+5,1+5,2+5,3+5,4+5,5+5} = {5,6,7,8,9,10}

7. Write the relation is a prime number less than in roster form.

Solution : Give, the relation is a prime number less than

For ,then

For , then

Therefore, the relation R in roster form :

8. Let and . Find the number of relations from A to B.

Solution : [Note : If and , then and the total number of relations is ]

Given, and

Here, and

Therefore, the number of relations from A to B .

9. Let R be the relation on Z defined by is an integer}. Find the domain and range of R.

Solution : Given, the relation is an integer}.

The domain of a relation is the set of all possible first components (in this case, ).

Since can be any integer, the domain is the set of all integers (Z).

Domain of R = Z .

The range of a relation is the set of all possible second components (in this case, ).

Since is an integer, can be any integer as well.

Therefore, the range is also the set of all integers (Z).

Range of R = Z .

Class 11 Maths Chapter 2. Relations and Functions Exercise 2.3 :

1. 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)} :

This relation is a function because each input (first component) is associated with only one output (second component).

The function is constant, where y = 1 for all x .

Domain = {2,5,8,11,14,17}

Range = {1}

(ii) {(2,1),(4,2),(6,3),(8,4),(10,5),(12,6),(14,7)} :

This relation is a function because each input is associated with a unique output.

Domain = {2,4,6,8,10,12,14}

Range = {1,2,3,4,5,6,7}

(iii) {(1,3),(1,5),(2,5)} :

This relation is not a function because the input x = 1 is associated with two different outputs ( y = 3 and y = 5).

2. Find the domain and range of the following real functions:
(i) (ii) .

Solution : (i) Given the function

Since is defined as , the domain of is also all real numbers.

So, the domain of is .

The function is the negation of the absolute value function.

The absolute value of any real number is always non-negative, and negating it will give a non-positive result.

So, the range of is the set of all non-positive real numbers, i.e., .

(ii) Given the function

The domain is the set of all real numbers x such that

The solution is

So, the domain is [−3,3] .

The function involves taking the square root of , and since , the square root will be real. The range includes all non-negative real numbers .So, the value of will lie between 0 and 3 .

Range of

3. A function is defined by . Write down the values of : (i) (ii) (iii)

Solution : The given function

(i)

(ii)

(iii)

4. The function ’ which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by .Find (i) (ii) (iii) ) (iv) The value of , when .

Solution : We have,

(i) Here,

(ii) Here,

(iii) Here ,

(iv) The value of , when .

5. Find the range of each of the following functions.
(i) .
(ii) is a real number.
(iii) is a real number.

Solution : (i) .

The given function is for belonging to the set of real numbers R where .

Since , the function is decreasing. Therefore, the range is all real numbers such that .

Range of

(ii) is a real number.

Since is always non-negative for real numbers , the minimum value of is 2. Therefore, the range is all real numbers greater than or equal to 2 .

Range of or

(iii) is a real number.

The range of this linear function is the set of all real numbers.

Range of .

Class 11 Maths Chapter 2. Relations and Functions Miscellaneous Exercise :

1. The relation f is defined by . The relation g is defined by . Show that f is a function and g is not a function.

Solution : Given,

For any value of in the given domain (), the function is well-defined. In the interval [0, 3], takes the form , and in the interval [3, 10], takes the form .

Therefore, for each in the domain, there is a unique -value, making a function.

and

Now, the interval .

For , , and for , . Both 4 and 9 are associated with and , which violates the definition of a function. Therefore, is not a function.

2. If , find .

Solution : Given,

Now,

3. Find the domain of the function .

Solution : Given, the function ,

Since,

, the function is defined for all real numbers except at and.

Hence the domain of is .

4. Find the domain and the range of the real function defined by .

Solution : For the real function

The expression inside the square root must be non-negative, so the domain is restricted to values of for which

Therefore, the domain of is all real numbers such that .

Domain of

The square root of a real number is always non-negative,

i.e.,

So, the range is all non-negative real numbers.

Range of or and

5. Find the domain and the range of the real function defined by .

Solution : The function is the absolute value function.

The absolute value function is defined for all real numbers, so the domain of is R , which means can be any real number.

The range of is the set of non-negative real numbers.

The absolute value of any real number is non-negative, so takes values greater than or equal to zero.

Therefore, the range is .

6. Let be a function from into . Determine the range of .

Solution : Given, the function is defined for all real numbers .

The second component of is

Since, for all real .

Therefore, , and the fraction will lies between 0 and 1 .

So, the range of f is the set of all real numbers such that .

Therefore, the range is [0,1] .

7. Let be defined, respectively by , . Find and .

Solution : Given, be defined by

and be defined by

Now,

Again,

and

8. Let be a function from to defined by , for some integers . Determine .

Solution : Given,

The function f is defined as , and we are given the values of for certain values:

So, the values of and are and .

9. Let be a relation from to defined by and Are the following true?
(i) , for all .

(ii) , implies .
(iii) implies . Justify your answer in each case.

Solution : (i) No, for all . The relation R is defined as and .

This means that would only be in R if , is a perfect square, but not for all natural numbers.

(ii) No, does not imply .

For example, if , it does not mean that because the relation R specifically involves the square of the first element being equal to the second element.

(iii) No, and does not imply .

For example, if and , it does not mean that . The relation is only defined for elements where the first element is the square of the second element.

10. Let and Are the following true?
(i) is a relation from A to B (ii) is a function from to . Justify your answer in each case.

Solution : Given , and

(i) is a relation from A to B :

Yes, because every ordered pair in has its first element in set A and its second element in set B.

(ii) is a function from to :

No, is not a function from A to B , because there are elements in A (specifically, 2) that map to more than one element in B .

11. Let be the subset of defined by Is a function from to ? Justify your answer.

Solution : The set is defined as

For to be a function from Z to Z , each element in Z should be related to a unique element in Z under .

Suppose and , then and .

Again, and , then and .

So, maps different elements in Z (0 and 1) to the same element (1) in Z . Therefore, is not a function from Z to Z .

12. Let and let be defined by f(n)= the highest prime factor of . Find the range of .

Solution : Given, A = {9, 10, 11, 12, 13}

(since 3 is the highest prime factor of 9)

(since 5 is the highest prime factor of 10)

(since 11 is the highest prime factor of 11)

(since 3 is the highest prime factor of 12)

(since 13 is a prime number, and it is the only prime factor of 13)

Range

So, the range of the function is {3, 5, 11, 13}.