# CLASS 11 MATHS CHAPTER-2 RELATIONS AND FUNCTIONS

### Exercise 2.1

Question1. Iffind the values of x and y

Solution :

Here

x/3  + 1 = 5/3 and  y – 2/3 = 1/3

x/3 + 1 = 5/3 and   3y -2 = 1

x = 2        and        3y = 3

x = 2 and y =1and

Question2.If the set A has 3 elements and the set B = {3, 4, 5}, then find the number of elements in (A×B).

Solution :

Number of elements in set A = 3 and Number of elements in set B = 3

∴ Number of elements in A × B = 3 × 3 = 9

Question3. If G = {7, 8} and H = {5, 4, 2}, find G × H and H × G.

Solution :

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

∴ G × H = {(7, 5), (7, 4), (7, 2), (8, 5), (8, 4), (8, 2)}

And H × G = {(5, 7), (4, 7), (2, 7), (5, 8), (4, 8), (2, 8)}

Question4. 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

Solution :

(i) Here P = {m, n} and Q = { n, m}

Number of elements in set P = 2 and Number of elements in set Q = 2

∵ Number of elements in P × Q = 2 × 2 = 4

But P × Q = {(m,n),(),}and here number of elements in P × Q = 2

Therefore, statement is false.

(ii) True

(iii) True

Question5. If A = {–1, 1}, find A × A × A.

Solution :

It is known that for any non-empty set A, A × A × A is defined as

A × A × A = {(a, b, c): a, b, c ∈ A}

It is given that A = {–1, 1}

∴ 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)}

Question6. If A × B = {(a, x),(a , y), (b, x), (b, y)}. Find A and B

Solution :

It is given that A × B = {(a, x), (a, y), (b, x), (b, y)}

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}

Question7. 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.

Solution :

Given: A = {1, 2}, B = {1, 2, 3, 4}, C

= {5, 6} and D = {5, 6, 7, 8}

Question8. Let A = {1, 2} and B = {3, 4}. Write A × B. How many subsets will A × B have?

List them.

Solution :

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)}

Question9: 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.

Solution :We are provided with the fact that n(A)=3 and n(B)=2 ; and(x,1),(y,2),(z,1)
are in A×B

.

We also know that, A
is the set of all the first elements and B

is the set of all the second elements.

So, we can conclude, A
having elements x,y,z and B having elements 1,2

.

Thus, we get, n(A)=3,n(B)=2

.

So, A={x,y,z},B={1,2}

Question10: 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 are provided with, n(A×A)=9

.

We also know that, if n(A)=a,n(B)=b
, then n(A×B)=ab

As it is given that, n(A×A)=9

It can be written as,

n(A)×n(A)=9

⇒n(A)=3

And it is also given that (−1,0),(0,1)
are the two elements of A×A

.

Again, the fact is also known that, A×A={(a,a):a∈A}
. And also −1,0,1 are the elements of A

.

Also, n(A)=3
, implies A={−1,0,1}

.

So, (−1,−1),(−1,1),(0,−1),(0,0),(1,−1),(1,0),(1,1)
are the remaining elements of A×A .

### Exercise 2.2

Question1. Let A = {1, 2, 3,…,14}. Define a relation R from A to A by R = {(x, y) : 3x – y = 0, where x, y ∈ A}. Write down its domain, codomain and range.

Solution :
Given: A = {1, 2, 3, ……….., 14}

The ordered pairs which satisfy 3x – y=0  are (1, 3), (2, 6), (3, 9) and (4, 12).

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

Domain = {1, 2, 3, 4}

Range = {3, 6, 9, 12}

Co-domain = {1, 2, 3, ……….., 14}

Question2. Define a relation R on the set N of natural numbers by R = {(x, y) : y = x + 5, x is a natural number less than 4; x, y ∈N}. Depict this relationship using roster form. Write down the domain and the range.

Solution :

R = {(x, y): y = x + 5, x is a natural number less than 4, x, y∈ N}

The natural numbers less than 4 are 1, 2, and 3.

∴R = {(1, 6), (2, 7), (3, 8)}

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

∴ Domain of R = {1, 2, 3}

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

∴ Range of R = {6, 7, 8}

Question3. A = {1, 2, 3, 5} and B = {4, 6, 9}. Define a relation R from A to B by R = {(x, y): the difference between x and y is odd; x ∈ A, y ∈ B}. Write R in roster form.

Solution :

A = {1, 2, 3, 5} and B = {4, 6, 9}

R = {(x, y): the difference between x and y is odd; x ∈ A, y ∈ B}

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

Question4. Figure 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 :

According to the given figure, P = {5, 6, 7}, Q = {3, 4, 5}

(i) R = {(x, y): y = x – 2; x ∈ P} or R = {(x, y): y = x – 2 for x = 5, 6, 7}

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

Domain of R = {5, 6, 7}

Range of R = {3, 4, 5}

Question5. Let A = {1, 2, 3, 4, 6}. Let R be the relation on A defined by {(a, b): a , b ∈A, b is exactly divisible by a}.

(i) Write R in roster form

(ii) Find the domain of R

(iii) Find the range of R.

Solution :

A = {1, 2, 3, 4, 6}, R = {(a, b): a, b ∈ A, b is exactly divisible by a}

(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}

Question6. Determine the domain and range of the relation R defined by R = {(x, x + 5) : x ∈ {0, 1, 2, 3, 4, 5}}.

Solution :

R = {(x, x + 5): x ∈ {0, 1, 2, 3, 4, 5}}

∴ R = {(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)}

∴Domain of R = {0, 1, 2, 3, 4, 5}

Range of R = {5, 6, 7, 8, 9, 10}

Question7: Write the relation R = {(x, x3) : x is a prime number less than 10} in roster form.

Solution :
R = {(x, x3) : x is a prime number less than 10}

The prime numbers less than 10 are 2, 3, 5, and 7.

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

Question8. Let A = {x, y, z} and B = {1, 2}. Find the number of relations from A to B.

Solution :
It is given that A = {x, y, z} and B = {1, 2}.

∴ A × B = {(x, 1), (x, 2), (y, 1), (y, 2), (z, 1), (z, 2)}

Since n(A × B) = 6, the number of subsets of A × B is 26.

Therefore, the number of relations from A to B is 26.

Question9. Let R be the relation on Z defined by R = {(a,b): a, b ∈ Z, a – b is an integer}. Find the domain and range of R.

Solution :
R = {(a, b): a, b ∈ Z, a – b is an integer}

It is known that the difference between any two integers is always an integer.

∴Domain of R = Z

Range of R = Z

### Exercise 2.3

Question1. Which of the following 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 :We have the given relation as, {(2,1),(5,1),(8,1),(11,1),(14,1),(17,1)}

.

Thus, we can see, the domain of the relation consists of {2,5,8,11,14,17}
and range is {1}

.

And we also have, every element of the domain is having their unique images, then it is a function.

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

Ans: We have our given relation, {(2,1),(4,2),(6,3),(8,4),(10,5),(12,6),(14,7)}

.

Thus, we have our domain as, {2,4,6,8,10,12,14}
and range as, {1,2,3,4,5,6,7}

.

Every element of the domain is having their unique images, so this is a function.

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

Ans: Our given relation is, {(1,3),(1,5),(2,5)}

.

From the domain of the relation the element 1
is having two different images 3,5

.

So,every element of the domain is not having their unique images. So, this is not a function.

Question2. Find the domain and range of the following real functions:

(i) f(x) = -|x|

(ii) f(x) =  √(9 – x²)
Solution : Question3. A function f is defined by f(x) = 2x –5. Write down the values of

(i) f (0),   (ii) f (7), (iii) f (–3).

Solution :

Given:  f(x) 2x – 5

(i) Putting We have the given function as, f(x)=2x−5

So, the value of,

f(0)=2×0−5=−5

(ii) Putting

We have the given function as, f(x)=2x−5

So, the value of,

f(7)=2×7−5=14−5=9

(iii) Putting

We have the given function as, f(x)=2x−5

So, the value of,

f(−3)=2×(−3)−5=−6−5=−11

Question4. The function t which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by Find:

(i) t(0)   (ii) t(28) (iii) t(–10)   (iv) The value of C, when t(C) = 212.

Solution : Question5. Find the range of each of the following functions.

(i) f (x) = 2 – 3x, x ∈  R , x > 0.

(ii) f (x) = x 2 + 2, x is a real number.

(iii) f (x) = x, x is a real number.

Solution :

(i) f(x) = 2 – 3x, x ∈ R, x > 0

The values of f(x) for various values of real numbers x > 0 can be written in the tabular form as

Thus, it can be clearly observed that the range of f is the set of all real numbers less than 2.

i.e., range of f = (–∞, 2)

Alter:

Let x > 0

⇒ 3x > 0

⇒ 2 –3x < 2

⇒ f(x) < 2

∴Range of f = (–∞, 2)

(ii) f(x) = x2 + 2, x, is a real number

The values of f(x) for various values of real numbers x can be written in the tabular form as

Thus, it can be clearly observed that the range of f is the set of all real numbers greater than 2.

i.e., range of f = [2, ∞)

Alter:

Let x be any real number.

Accordingly,

x2 ≥ 0

⇒ x2 + 2 ≥ 0 + 2

⇒ x2 + 2 ≥ 2

⇒ f(x) ≥ 2

∴ Range of f = [2,∞)

(iii) f(x) = x, x is a real number

It is clear that the range of f is the set of all real numbers.

∴ Range of f = R

### Miscellaneous Exercise

Question1. 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: f and

Question2. If f(x) = x2 find

Solution :

Given: f(x) = x2

At

Question3. Find the domain of the function

Solution : Question4. Find the domain and range of the real function f defined by  f(x) = √|x – 1|

Solution :

The given real function is f(x)=√|x−1

It can be seen that√| x−1
​ is defined for (x−1)≥0
i.e., f(x)= √|(x−1)
​ is defined for x≥1
Therefore the domain of f is the set of all real numbers greater than or equal to 1 i.e.,
the domain of f =[1,∞)

As x≥1
⇒(x−1)≥0
√|x−1≥0
f(x)≥0
Therefore the range of f is the set of all real numbers greater than or equal to 0
i.e., the range of f =[0,∞)

Question5. Find the domain and the range of the real function f defined by f (x) = |x – 1|.

Question6. Letbe a function from R into R. Determine the range of  f

Solution :

Here

Put

substituting values and Determing the image  , we have

Question7. Let  R → R be defined respectively by Find f(x) = x + 1, g(x) = 2x – 3. Find f + g, f – g and  f/g.

Solution :

Given:  We have the functions defined as, f,g:R→Ris defined as, f(x)=x+1,g(x)=2x−3

Thus, the function

(f+g)(x)=f(x)+g(x)

=(x+1)+(2x−3)

=3x−2

So, the function (f+g)(x)=3x−2

.

Again, the function,

(f−g)(x)=f(x)−g(x)

=(x+1)−(2x−3)

=−x+4

So, the function (f−g)(x)=−x+4

.

Similarly,

(fg)(x)=f(x)g(x)
where g(x)≠0 and also x∈R

.

Now, putting the values,

(fg)(x)=x+12x−3

where,

2x−3≠0

⇒x≠32

Question 8. Let f = {(1, 1), (2, 3), (0, –1), (–1, –3)} be a function from Z to Z defined by f(x) = ax + b, for some integers a, b. Determine a, b.

Solution :

f = {(1, 1), (2, 3), (0, –1), (–1, –3)}

f(x) = ax + b

(1, 1) ∈ f

⇒ f(1) = 1

⇒ a × 1 + b = 1

⇒ a + b = 1

(0, –1) ∈ f

⇒ f(0) = –1

⇒ a × 0 + b = –1

⇒ b = –1

On substituting b = –1 in a + b = 1, we obtain a + (–1) = 1 ⇒ a = 1 + 1 = 2.

Thus, the respective values of a and b are 2 and –1.

Question9. Let R be a relation from N to N defined by R = {(a, b): a, b ∈ N and a = b2}. Are the following true?

(i) (a, a) ∈ R, for all a ∈ N

(ii) (a, b) ∈ R, implies (b, a) ∈ R

(iii) (a, b) ∈ R, (b, c) ∈ R implies (a, c) ∈ R.

Solution :

R = {(a, b): a, b ∈ N and a = b2}

(i) It can be seen that 2 ∈ N;however, 2 ≠ 22 = 4.

Therefore, the statement “(a, a) ∈ R, for all a ∈ N” is not true.

(ii) It can be seen that (9, 3) ∈ N because 9, 3 ∈ N and 9 = 32.

Now, 3 ≠ 92 = 81; therefore, (3, 9) ∉ N

Therefore, the statement “(a, b) ∈ R, implies (b, a) ∈ R” is not true.

(iii) It can be seen that (16, 4) ∈ R, (4, 2) ∈ R because 16, 4, 2 ∈ N and 16 = 42 and 4 = 22.

Now, 16 ≠ 22 = 4; therefore, (16, 2) ∉ N

Therefore, the statement “(a, b) ∈ R, (b, c) ∈ R implies (a, c) ∈ R” is not true.

Question10. Let A = {1, 2, 3, 4}, B = {1, 5, 9, 11, 15, 16} and f = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}. Are the following true?

(i) f is a relation from A to B (ii) f is a function from A to B. Justify your answer in each case.

Solution :

A = {1, 2, 3, 4} and B = {1, 5, 9, 11, 15, 16}

∴A × B = {(1, 1), (1, 5), (1, 9), (1, 11), (1, 15), (1, 16), (2, 1), (2, 5), (2, 9), (2, 11), (2, 15), (2, 16), (3, 1), (3, 5), (3, 9), (3, 11), (3, 15), (3, 16), (4, 1), (4, 5), (4, 9), (4, 11), (4, 15), (4, 16)}

It is given that f = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}

(i) A relation from a non-empty set A to a non-empty set B is a subset of the Cartesian product A × B.

It is observed that f is a subset of A × B.

Thus, f is a relation from A to B.

(ii) Since the same first element i.e., 2 corresponds to two different images i.e., 9 and 11, relation f is not a function.

Question11. Let f be the subset of Z × Z defined by f = {(ab, a + b): a, b ∈ Z}. Is f a function from Z to Z: justify your answer.

Solution :

The relation f is defined as f = {(ab, a + b): a, b ∈ Z}

We know that a relation f from a set A to a set B is said to be a function if every element of set A has unique images in set B.

Since 2, 6, –2, –6 ∈ Z, (2 × 6, 2 + 6), (–2 × –6, –2 + (–6)) ∈ f

i.e., (12, 8), (12, –8) ∈ f

It can be seen that the same first element i.e., 12 corresponds to two different images i.e., 8 and –8. Thus, relation f is not a function.

Question12. Let A = {9, 10, 11, 12, 13} and let f: A → N be defined by f(n) = the highest prime factor of n. Find the range of f.

Solution :f(n)= the highest prime factor of n

.

The prime factor of 9 is 3

.
The prime factors of 10 is 2,5

.

The prime factor of 11 is 11

.

The prime factor of 12 is 2,3

.

The prime factor of 13 is 13

.

Thus, it can be said,

f(9) = the highest prime factor of 9=3

f(10) = the highest prime factor of 10=5

f(11) = the highest prime factor of 11=11

f(12) = the highest prime factor of 12=3

f(13)= the highest prime factor of 13=13

.Now, the range of the function will be, {3,5,11,13}

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