1. RELATIONS AND FUNCTIONS

Class 12 Mathematics Chapter 1. RELATIONS AND FUNCTIONS

Chapter 1: Relations and Functions

Class 12 Mathematics Chapter 1 Relations and Functions Exercise 1.1 Questions and Answers

1. Determine whether each of the following relations are reflexive , symmetric and transitive :

(i) Relation R in the set defined as

(ii) Relation R in the set N of natural numbers defined as

(iii) Relation R in the set as

(iv) Relation R in the set Z of all integers defined as

(v) Relation R in the set A of human being in a town at a particular time given by

(a)

(b)

(c)

(d)

(e)

Solution: (i) Relation R in the set A = {1, 2, 3, ..., 13, 14} defined as R = {(x, y) : 3x – y = 0}

Reflexive: The relation is reflexive if (x, x) is in R for all x in A.

Here, 3x - x = 2x ≠ 0 for any x in A. So, it's not reflexive.

Symmetric: The relation is symmetric if whenever (x, y) is in R, then (y, x) must also be in R.

If (x, y) is in R, it means 3x - y = 0, but 3y - x ≠ 0. So, it's not symmetric.

Transitive: The relation is transitive if (x, y) and (y, z) are in R, then (x, z) must also be in R.

So, 3x - y = 0 and 3y - z = 0

But 3x - z 0. So, it's not transitive.

(ii) Relation R in the set N of natural numbers defined as R = {(x, y) : y = x + 5 and x < 4}

Reflexive: The relation is reflexive if (x, x) is in R for all x in N.

However, for any x in N, y = x + 5 would imply y > x .

So (x, x) can't be in R. It's not reflexive.

Symmetric: Since x < 4, there are no pairs (x, y) and (y, x) in R .

Therefore, it is not symmetric.

Transitive: Since there are no pairs (x, y) and (y, z) in R .

Therefore, it is not transitivity .

(iii) Relation R in the set A = {1, 2, 3, 4, 5, 6} as R = {(x, y) : y is divisible by x}

Reflexive: For any x in A, y = x is divisible by x.

Therefore, the relation is reflexive.

Symmetric: If (x, y) is in R (y is divisible by x), then (y, x) is also in R (x is divisible by y), except when x and y are not equal.

So, it is not symmetric .

Transitive: If (x, y) and (y, z) are in R, it means y is divisible by x and z is divisible by y.

This implies z is divisible by x, so (x, z) is in R. The relation is transitive.

(iv) Relation R in the set Z of all integers defined as R = {(x, y) : x – y is an integer}

Reflexive: For any integer x, x - x = 0, which is an integer(Z).

So, the relation is reflexive.

Symmetric: If (x, y) is in R, it means x - y is an integer(Z).

Then, y - x = - (x - y) is also an integer(Z).

So, the relation is symmetric.

Transitive: If (x, y) and (y, z) are in R, it means x - y and y - z are integers(Z).

Adding these integers gives x - z, which is an integer(Z).

So, the relation is transitive.

(v) Relation R in the set A of human beings in a town at a particular time: (a) R = {(x, y) : x and y work at the same place}

Reflexive: Every person works at the same place as themselves, so it's reflexive.

Symmetric: If x and y work at the same place, then y and x also work at the same place. It's symmetric.

Transitive: If x and y work at the same place, and y and z work at the same place, then x and z also work at the same place. It's transitive.

(b) R = {(x, y) : x and y live in the same locality}

Reflexive: Every person lives in the same locality as themselves, so it's reflexive.

Symmetric: If x and y live in the same locality, then y and x also live in the same locality. It's symmetric.

Transitive: If x and y live in the same locality, and y and z live in the same locality, then x and z also live in the same locality. It's transitive.

Reflexive: No one is 7 cm taller than themselves, so it's not reflexive.

Symmetric: If x is 7 cm taller than y, then y is not 7 cm taller than x. It's not symmetric.

Transitive: If x is 7 cm taller than y, and y is 7 cm taller than z, it doesn't necessarily mean x is 7 cm taller than z. It's not transitive.

(d) R = {(x, y) : x is wife of y}

Reflexive: The relation is not reflexive because a person cannot be their own wife.

Symmetric: If x is the wife of y, then y is the husband of x. It's symmetric.

Transitive: If x is the wife of y, and y is the wife of z, it doesn't imply any relationship between x and z. It's not transitive.

(e) R = {(x, y) : x is father of y}

Reflexive: The relation is not reflexive because a person cannot be their own father.

Symmetric: If x is the father of y, then y is not the father of x. It's not symmetric.

Transitive: If x is the father of y, and y is the father of z, it doesn't imply any relationship between x and z. It's not transitive.

2. Show that the relation R in the set R of real numbers , defined as in neither reflexive nor symmetric nor transitive .

Solution : For Reflexive : A relation R is reflexive if for every element '' in the set, belongs to R.

The relation , we need to have . This isn't true for all real numbers.

Let for example. We have , which simplifies to 2 ≤ 4, but this is not true.

Therefore, the relation R is not reflexive.

For Symmetric : A relation R is symmetric if for every in R, also belongs to R.

Let, the elements in R. This is because , which is true.

If , the condition is false, so does not belong to R.

So, the relation R is not symmetric.

For Transitive : A relation R is transitive if for every and in R, also belongs to R.

Let, the elements and in R.

Both conditions are : and .

If we check , the condition is not satisfied, so does not belong to R.

Hence, the relation R is not transitive.

Therefore , the relation defined on the set of real numbers is neither reflexive, nor symmetric, nor transitive.

3. Check whether the relation R defined in the set as is reflexive , symmetric or transitive .

Solution: Given the relation defined on the set {1, 2, 3, 4, 5, 6}.

For Reflexive : A relation R is reflexive if for every element '' in the set, () belongs to R.

If then . So, (3, 4) is in R

But then . So, (6,7) is not in R

Therefore, the relation R is not reflexive.

For Symmetric :A relation R is symmetric if for every (a, b) in R, (b, a) also belongs to R.

If is in R, then must also be in R.

But, (4, 3) is not in R because 3 ≠ 4 + 1.

Similarly, (2, 3) is in R but (3, 2) is not.

Therefore, the relation R is not symmetric.

For Transitive : A relation R is transitive if for every and in R, also belongs to R.

If and Both (1, 2) and (2, 3) are in R because 2 = 1 + 1 and 3 = 2 + 1.

But is not in R because 3 ≠ 1 + 1 . Therefore, the relation R is not transitive.

Hence, R is neither reflexive nor symmetric nor transitive .

4. Show that the relation R in R defined as , is refllexive and transitive but not symmetric .

Solution: Given, the relation

For Reflexive : The relation R is reflexive because for every real number '' in the set of real numbers,() belongs to R since a is always less than or equal to itself .

For Symmetric : The relation R is not symmetric because if () is in R (),

But () is not in R () .

For Example : let is in R () .

But is not in R () .

For Transitive The relation R is transitive because if () and () are both in R ( and , then it is always true that is in R as well () .

Therefore, R is reflexive and transitive but not symmetric .

5. Check whether the relation R in R defined by , is refllexive , symmetric or transitive .

Solution: Given, the relation defined on the set of real numbers.

For Reflexive : A relation R is reflexive if for every element '' in the set, () belongs to R.

We have This is not true for all real numbers. So, R is not reflexive .

For example : If then [ i.e., (2,2) ] .

For Symmetric : A relation R is symmetric if for every () in R, () also belongs to R.

Suppose, in R, as is true.

But does not satisfy , making the relation R not symmetric.

For Transitive : A relation R is transitive if for every and in R, also belongs to R.

Let, and in R.

Both conditions hold: and .

But , the condition is not satisfied.

Therefore, R is not transitive.

Hence, R is neither reflexive nor symmetric nor transitive .

6. Show that the relation R in the set given by is symmetric but neither reflexive nor transitive .

Solution: For Symmetric: The relation R = {(1, 2), (2, 1)} is symmetric because for every pair (a, b) in R .

If (a, b) is in R , then (b, a) is in also R .

Here, (1, 2) is in R , and (2, 1) is also in R . So, R is symmetric .

For Reflexive: The relation is not reflexive because there is no pair (a, a) in R for any element 'a' in the set {1, 2, 3}.

i.e., (1,1) is not in R .

For Transitive : If (a, b) and (b, c) are in the relation, then (a, c) should also be in R .

The relation is not transitive because while (1, 2) and (2, 1) are both in R . But (1, 1) is not in R .

Hence, R is symmetric but neither reflexive nor transitive .

7. Show that the relation R in the set A of all the books in a library of a college , given by is an equivalence relation .

Solution: Given, the relation R = {(x, y) : x and y have the same number of pages}

For Reflexive: For every book 'x' in the set of all books, it's clear that 'x' has the same number of pages as itself. Therefore, the relation R is reflexive .

For Symmetric: If 'x' has the same number of pages as 'y', then 'y' also has the same number of pages as 'x'. This follows directly from the definition of the relation. Therefore, the relation R is Symmetric .

For Transitive: If 'x' has the same number of pages as 'y', and 'y' has the same number of pages as 'z', then 'x' must have the same number of pages as 'z'. This also follows from the definition of the relation. Therefore, the relation R is transitive .

Hence, the relation R = {(x, y) : x and y have the same number of pages} satisfies all three properties (reflexivity, symmetry, and transitivity), it is an equivalence relation .

8. Show that the relation R in the set given by , is an equivalence relation . Show that all the elements of are related to each other and all the elements of are related to each other . But no element of is related to any element of .

Solution: For Reflexive : For a relation to be reflexive, every element should be related to itself.

Given, A = {1, 2, 3, 4, 5}. For each element 'a' in A, |a – a| = 0, which is an even number.

Therefore, every element is related to itself, and the relation is reflexive.

For Symmetric : For a relation to be symmetric, if (a, b) is in R , then (b, a) also be in R .

If (a, b) then |a - b| is even.

Again, |b - a| is also even,

Therefore, if (a, b) is in R ,then (b, a) also be in R and the relation is symmetric.

For Transitive :For a relation to be transitive, if (a, b) and (b, c) are in the relation, then (a, c) also be in R .

If |a– b| is even and |b – c| is even, then |a – c| is even.

This is because the sum or difference of two even numbers is always even.

Therefore, if (a, b) and (b, c) are in R , then (a, c) also be in R and the relation is transitive.

Therefore, R (reflexivity, symmetry, and transitivity) is an equivalence relation.

All elements in {1, 3, 5} are related to each other because the modulus of the difference between any pair of these elements is always even.

All elements in {2, 4} are related to each other because the modulus of the difference between any pair of these elements is always even.

No element from {1, 3, 5} is related to any element from {2, 4} because the modulus of the difference between any pair of elements from these two sets will always be not even .

9. Show that each of that relation R in the set , given by

(i)

(ii) is an equivalence relation . Find the set of all elements related to 1 in each case .

Solution: (i) For the relation R = {(a, b) : |a – b| is a multiple of 4}

Reflexive: Every element 'a' in the set A = {0, 1, 2, ..., 12} is related to itself since |a - a| = 0, which is a multiple of 4. So, the relation R is reflexive.

Symmetric: If (a, b) is in the relation R (|a – b| is a multiple of 4), then |b – a| is also a multiple of 4, and therefore (b, a) is also in the relation R. So, the relation R is symmetric.

Transitive: If (a, b) and (b, c) are in the relation R (|a – b| is a multiple of 4 and |b – c| is a multiple of 4), then |a – c| is also a multiple of 4. This is because the sum of two multiples of 4 is also a multiple of 4.

So, the relation R is transitive.

(ii) For the relation R = {(a, b) : a = b}

Reflexive: Every element 'a' in the set A = {0, 1, 2, ..., 12} is related to itself since a = a.

Therefore, the relation R is reflexive.

Symmetric: If (a, b) is in the relation R (a = b), then (b, a) is also in the relation R (b = a) .

Therefore, the relation R is symmetric.

Transitive: If (a, b) and (b, c) are in the relation R (a = b and b = c), then (a, c) is also in the relation R (a = c).

Therefore, the relation R is transitive.

For both cases, since the relations are reflexive, symmetric, and transitive, they are equivalence relations.

Now, the set of all elements related to 1 in each case:

(i) For the relation R = {(a, b) : |a – b| is a multiple of 4}:

Elements related to 1 are {1, 5, 9}

(ii) For the relation R = {(a, b) : a = b}:

Elements related to 1 is just {1}

10. Give an example of a relation . Which is

(i) Symmetric but neither reflexive nor transitive .

(ii) Transitive but neither reflexive nor symmetric .

(iii) Reflexive and symmetric but not transitive .

(iv) Reflexive and transitive but not symmetric .

(v) Symmetric and transitive but not reflexive .

Solution: (i) Symmetric but neither reflexive nor transitive:

Let, R = {(1, 2), (2, 1), (2, 3)}.

Symmetric: It is symmetric because for every pair (a, b) in R, if (a, b) is present, then (b, a) is also present.

Not Reflexive: It's not reflexive because, for instance, the pair (3, 3) is not in R.

Not Transitive: It's not transitive because (1, 2) and (2, 3) are in R, but (1, 3) is not in R.

(ii) Transitive but neither reflexive nor symmetric:

Let, R = {(1, 2), (2, 3), (1, 3)}.

Transitive: It's transitive because whenever (a, b) and (b, c) are present in R, (a, c) is also present.

Not Reflexive: It's not reflexive because, for example, (4, 4) is not in R.

Not Symmetric: It's not symmetric because (1, 3) is in R, but (3, 1) is not.

(iii) Reflexive and symmetric but not transitive:

Let, R = {(1, 1), (1, 2), (2, 1)}.

Reflexive: It's reflexive because every element is related to itself.

Symmetric: It's symmetric because if (a, b) is in R, then (b, a) is also in R.

Not Transitive: It's not transitive because (1, 2) and (2, 1) are in R, but (1, 1) is the only related pair, and (1, 2) and (2, 1) don't lead to (1, 1).

(iv) Reflexive and transitive but not symmetric:

Let, R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)}.

Reflexive: It's reflexive because every element is related to itself.

Transitive: It's transitive because whenever (a, b) and (b, c) are present in R, (a, c) is also present.

Not Symmetric: It's not symmetric because, for example, (1, 2) is in R but (2, 1) is not.

(v) Symmetric and transitive but not reflexive:

Let, R = {(1, 1), (2, 2), (1, 2), (2, 1)}.

Symmetric: It's symmetric because for every pair (a, b) in R, if (a, b) is present, then (b, a) is also present.

Transitive: It's transitive because whenever (a, b) and (b, c) are present in R, (a, c) is also present.

Not Reflexive: It's not reflexive because, for instance, (3, 3) is not in R.

11. Show that the relation R in the set A of points in a plane given by distance of the point P from the origin is same as the distance of the point Q from the origin } , is an equivalence relation . Further , show that the set of all points related to a point is the circle passing through P with origin as centre .

Solution: Reflexive: Any point P in the plane has the same distance from itself, which is the distance of P from the origin. Therefore, the relation R is reflexive.

Symmetric: If (P, Q) is in the relation R (distance of P from the origin is the same as the distance of Q from the origin), then (Q, P) is also in the relation (distance of Q from the origin is the same as the distance of P from the origin). Therefore, the relation R is symmetric.

Transitive: If (P, Q) and (Q, R) are in the relation R (distance of P from the origin is the same as the distance of Q from the origin and distance of Q from the origin is the same as the distance of R from the origin), then (P, R) is also in the relation (distance of P from the origin is the same as the distance of R from the origin). Therefore, the relation R is transitive.

Since the relation R (reflexive, symmetric, and transitive) is an equivalence relation.

We consider a point P ≠ (0, 0) in the plane. The set of all points related to P in the relation R are those points Q for which the distance from the origin is the same as the distance from point P.

This set of points is a circle with P as its center and the origin (0, 0) as its center. The radius of this circle is the distance of point P from the origin.

12. Show that the relation R defined in the set A all triangles as , is equivalence relation . Consider three right angles triangles with sides 3 , 4 , 5 , with sides 5 , 12 , 13 and with sides 6 , 8 , 10 . Which triangles among , and are related ?

Solution: For Reflexive: Every triangle is similar to itself (according to mathematical similarity of triangles).

Therefore, the relation R is reflexive.

For Symmetric: If triangle is similar to triangle , then triangle is also similar to triangle .

Therefore, the relation R is symmetric.

For Transitive: If triangle is similar to triangle , and triangle is similar to triangle , then triangle is also similar to triangle . Therefore, the relation R is transitive.

Since, the relation R (reflexive, symmetric, and transitive) is an equivalence relation.

Similar Triangles: Two triangles are similar if their corresponding angles are equal, and the ratios of the lengths of their corresponding sides are the same.

Again, Triangles and are not similar since their corresponding sides are not proportional (i.e., ) .

Triangles and are similar since their corresponding sides are proportional ( i.e., are equal ratios).

Triangles and are not similar since their corresponding sides are not proportional.

13. Show that the relation R defined in the set A of all polygons as , is an equivalence relation . What is the set of all elements in A related to the right angle triangle T with sides 3 , 4 and 5 ?

Solution: Reflexive: Any polygon has the same number of sides as itself.

Therefore, the relation R is reflexive.

Symmetric: If polygon has the same number of sides as polygon , then polygon also has the same number of sides as polygon .

Therefore, the relation R is symmetric.

Transitive: If polygon has the same number of sides as polygon , and polygon has the same number of sides as polygon , then polygon also has the same number of sides as polygon .

Therefore, the relation R is transitive.

Since the relation R is an (reflexive, symmetric, and transitive) equivalence relation.

Set of Elements Related to the Right Angle Triangle T :

The right angle triangle T with sides 3, 4, and 5 has three sides, so we want to find the set of all polygons that also have three sides (triangles with any angles and sides).

Therefore, the set of all elements in A related to the right angle triangle T with sides 3, 4, and 5 consists of all other triangles with any angles and sides, as long as they also have three sides.

14. Let L be the set of all lines in XY plane and R be the relation in L defined as . Show that R is an equivalence relation . Find the set of all lines related to the line .

Solution: Reflexive: Any line is parallel to itself, so the relation R is reflexive.

Symmetric: If line is parallel to line , then line is also parallel to line .

Therefore, the relation R is symmetric.

Transitive: If line is parallel to line , and line is parallel to line , then line is parallel to line . Therefore, the relation R is transitive.

Since the relation R is an (reflexive, symmetric, and transitive) equivalence relation.

Given, the line y = 2x + 4 and the slope of the line is 2 .

So, the lines that are parallel to it will also have the same slope of 2.

In the slope-intercept form, the equation of a line with a slope of 2 is of the form , where '' is any real number. So, the set of all lines related to can be represented as ,

15. Let R be the relation in the set given by R = { (1 , 2), (2 , 2) , (1 , 1) , (4 , 4) , (1 , 3) , (3 , 3) ,(3,2) . Choose the correct answer .

(A) R is reflexive and symmetric but not transitive .

(B) R is reflexive and transitive but not symmetric .

(D) R is an equivalence relation .

Solution: The correct Answer (B) .

[ (B) R is reflexive and transitive but not symmetric.

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

R contains (1,1), (2,2), and (4,4), which satisfy reflexivity.

R contains (1,2) but not (2,1), so it's not symmetric.

R contains (1,2), (2,2), and (1,2) implies (2,2), so it's transitive. ]

16. Let R be the relation in the set N given by . Choose the correct answer .

(A) (B) (C) (D)

Solution: The correct answer is: (C) (6, 8) ∈ R

[ Given,

According to the relation, a = b – 2. If a = 6 and b = 8, then 6 = 8 – 2, which is true.

The condition b > 6 is satisfied, as 8 is greater than 6. ]

Class 12 Mathematics Chapter 1 Relations and Functions Exercise 1.2 Questions and Answers

1. Show that the function defined by is one-one and onto, where is the set of all non-zero real numbers. Is the result true, if the domain is replaced by with co-domain being same as ?

Solution: Given, The function defined by

One-to-one (injective) : Assume for some

;

Since, and are the same, the function is injective .

Onto (Surjective) : For any , we can find such that

Hence, the function is surjective .

Again, the function defined by

In this case, the domain is the set of natural numbers N, and the codomain is .

One-to-One (Injective) : Consider two distinct natural numbers and , such that .

We know that .

and

Since, and are distinct, and also distinct .

Therefore, , and the function is injective .

Onto (Surjective) :

When the domain is replaced by N(the set of natural numbers) while the co-domain remains :

In this case, the function is not onto (surjective) because the natural numbers only cover positive integers, and the function includes real numbers between 0 and 1 (exclusive), which are not covered by natural numbers.

2. Check the injectivity and surjectivity of the following functions:
(i) given by
(ii) given by
(iii) given by
(iv) given by
(v) given by

Solution:
3. Prove that the Greatest Integer Function , given by is neither one-one nor onto, where denotes the greatest integer less than or equal to .

4. Show that the Modulus Function , given by is neither one-one nor onto, where is , if is positive or 0 and is – , if is negative.
5. Show that the Signum Function f : R → R, given is neither one-one nor onto.
6. Let and let be a function from to . Show that f is one-one.
7. In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.
(i) defined by
(ii) defined by
8. Let and be sets. Show that such that is bijective function.
9. Let be defined by for all . State whether the function f is bijective. Justify your answer.
10. Let and Consider the function defined by . Is one-one and onto? Justify your answer.
11. Let be defined as . Choose the correct answer.
(A) is one-one onto (B) is many-one onto
(C) is one-one but not onto (D) is neither one-one nor onto.
12. Let be defined as . Choose the correct answer.
(A) is one-one onto (B) is many-one onto
(C) is one-one but not onto (D) is neither one-one nor onto.

Class 12 Mathematics Chapter 1 Relations and Functions Miscellaneous Exercise on Chapter 1 Questions and Answers

1. Show that the function defined by , is one one and onto function.
2. Show that the function given by is injective.
3. Given a non-empty set X, consider the binary operation given by in where is the power set of . Show that is the identity element for this operation and X is the only invertible element in with respect to the operation ∗.
4. Find the number of all onto functions from the set to itself.
5. Let and be functions defined by and . Are and equal?
Justify your answer. (Hint: One may note that two functions and such that , are called equal functions).
6. Let . Then number of relations containing (1, 2) and (1, 3) which are reflexive and symmetric but not transitive is
(A) 1 (B) 2 (C) 3 (D) 4
7. Let . Then number of equivalence relations containing (1, 2) is
(A) 1 (B) 2 (C) 3 (D) 4