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**NCERT mathematics class 12 chapter Relations and Functions Exercise 1.1 Solutions**. You can find the questions/answers/solutions for the**chapter 1/Exercise 1.1**of**NCERT class 12 mathematics**in this page. So is the case if you are looking for**NCERT class 12 Maths**related topic**Relations and Functions**. This page contains Exercise 1.1 solutions. If you’re looking for summary of the chapter or other exercise solutions, they’re available at●

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Exercise 1.1

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

i.

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

ii.

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

iii.

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

iv.

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

v.

Relation R in the set A of human beings in a town at a particular time given by

a.

R = {(x, y) : x and y work at the same place}

b.

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

c.

R = {(x, y) : x is exactly 7 cm taller than y}

d.

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

e.

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

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

⇒ 3x - y = 0

⇒ 3x = y

⇒ y = 3x

When {x = 1}, {y = 3 × x = 3 × 1 = 3}. As 3 ∈ A, (1, 3) ∈ R

When {x = 2}, {y = 3 × x = 3 × 2 = 6}. As 6 ∈ A, (2, 6) ∈ R

When {x = 3}, {y = 3 × x = 3 × 3 = 9}. As 9 ∈ A, (3, 9) ∈ R

When {x = 4}, {y = 3 × x = 3 × 4 = 12}. As 12 ∈ A, (3, 12) ∈ R

When {x = 5}, {y = 3 × x = 3 × 5 = 15}. As 15 ∉ A, (3, 15) ∉ R

These calculations can also be represented in the tabular form as follows:

x

y = 3x

(x, y) ∈ R?

1

3 × 1 = 3

✔

2

3 × 2 = 6

✔

3

3 × 3 = 9

✔

4

3 × 4 = 12

✔

5

3 × 5 = 15

❌

For higher values of x, the value of y will be more than 14. As all the elements in A are less than 14, we can stop here.

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

To Check whether R is Reflexive: The relation R in the set A is reflexive if (a, a) ∈ R, for every a ∈ A.

In this case, (1, 1) ∉ R, (2, 2) ∉ R, (3, 3) ∉ R,…. (13, 13) ∉ R, (14, 14) ∉ R.

∴ R is not Reflexive

Note: Ideally, we don’t have to check for all the elements. Even if one element for example (1,1) ∉ R, then we can say that R is not reflexive)

To Check whether R is Symmetric: The relation R in the set A is symmetric if (a_1, a_2) ∈ R implies that (a_2, a_1) ∈ R, for all a_1, a_2 ∈ A

Here we see that (1, 3) ∈ R, but (3, 1) ∉ R

∴ R is not Symmetric

To Check whether R is Transitive: The relation R in the set A is transitive if (a_1, a_2) ∈ R and (a_2, a_3) ∈ R implies that (a_1, a_3) ∈ R, for all a_1, a_2, a_3 ∈ A

Here we see that (1, 3) ∈ R and also (3, 9) ∈ R but (1, 9) ∉ R

∴ R is not Transitive

∴ R is neither reflexive nor symmetric nor transitive.

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

As it is given that {x \lt 4}, x takes the values 1, 2 and 3.

When {x = 1}, {y = x + 5 = 1 + 5 = 6}. So, (1, 6) ∈ R

When {x = 2}, {y = x + 5 = 2 + 5 = 7}. So, (2, 7) ∈ R

When {x = 3}, {y = x + 5 = 3 + 5 = 8}. So, (3, 8) ∈ R

This can also be put in the tabular form as follows:

x

y = x + 5

(x, y) ∈ R?

1

1 + 5 = 6

✔

2

2 + 5 = 7

✔

3

3 + 5 = 8

✔

∴ R = {(1, 6), (2, 7), (3, 8)}

To Check whether R is Reflexive: The relation R in the set A is reflexive if (a, a) ∈ R, for every a ∈ A

In this case, (1, 1) ∉ R, (2, 2) ∉ R and (3, 3) ∉ R.

∴ R is not Reflexive

To Check whether R is Symmetric: The relation R in the set A is symmetric if (a_1, a_2) ∈ R implies that (a_2, a_1) ∈ R, for all a_1, a_2 ∈ A

In this case, we see that (1, 6) ∈ R, but (6, 1) ∉ R

∴ R is not Symmetric

To Check whether R is Transitive: The relation R in the set A is transitive if (a_1, a_2) ∈ R and (a_2, a_3) ∈ R implies that (a_1, a_3) ∈ R, for all a_1, a_2, a_3 ∈ A

In this case, we see that (1, 3) ∈ R and also (3, 8) ∈ R but (1, 8) ∉ R

∴ R is not Transitive

∴ R is neither reflexive nor symmetric nor transitive.

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

To Check whether R is Reflexive: The relation R in the set A is reflexive if (a, a) ∈ R, for every a ∈ A.

As we know, every element x is divisible by itself i.e. x. In otherwords, (x, x) ∈ R

∴ R is Reflexive

To Check whether R is Symmetric: The relation R in the set A is symmetric if (a_1, a_2) ∈ R implies that (a_2, a_1) ∈ R, for all a_1, a_2 ∈ A

In this case, we can see that 6 is divisible by 3. However, 3 is not divisible by 6. In otherwords, (3, 6) ∈ R but (6, 3) ∉ R.

∴ R is not Symmetric

To Check whether R is Transitive: The relation R in the set A is transitive if (a_1, a_2) ∈ R and (a_2, a_3) ∈ R implies that (a_1, a_3) ∈ R, for all a_1, a_2, a_3 ∈ A

Here we see that (1, 3) ∈ R as 3 is divisible by 1. Also, (3, 6) ∈ R as 6 is divisible by 3. Also we see that 6 is divisible by 1 i.e. (1, 6) ∈ R

∴ R is Transitive

∴ R is both reflexive and transitive but not symmetric.

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

To Check whether R is Reflexive: The relation R in the set A is reflexive if (a, a) ∈ R, for every a ∈ A.

In this case, for any element x, {x - x = 0}. As 0 is also an integer, this property is true for every element x ∈ Z. In otherwords, (x, x) ∈ R

∴ R is Reflexive

If {x - y = k}, where k is an integer.

Now, {y - x = -(x - y) = -k}. If k is an integer, then -k is also an integer. So, we see that if (x, y) ∈ R then (y, x) ∈ R.

∴ R is Symmetric

Here we see that if {(x - y)} is an integer and {(y - z)} is another integer, then {(x - z) = (x - y) + (y - z)} is also an integer, as sum of two integers is integer. So, we can say that if (x, y) ∈ R and (y, z) ∈ R then (x, z) ∈ R.

∴ R is Transitive

∴ R is reflexive, symmetric as well as transitive.

v (a). R = {(x, y) : x and y work at the same place}

In this case, if x is working at a given place, then the same person x is also working at the same place. So, clearly (x, x) ∈ R.

∴ R is Reflexive

If x and y are working at a given place, then it implies that y and x are also working at the same place. So, we can say that if (x, y) ∈ R then (y, x) ∈ R

∴ R is Symmetric

If x and y are working at the given place and also y and z are working at that place, then it implies that both x and z are also working at the same place. In otherwords, if (x, y) ∈ R, (y, z) ∈ R then (x, y) ∈ R.

∴ R is Transitive

∴ R is reflexive, symmetric as well as transitive.

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

In this case, if x is living at a given locality, then the same person x is also living at the same locality. So, clearly (x, x) ∈ R.

∴ R is Reflexive

If x and y are living at a given locality, then it implies that y and x are also living at the same locality. This implies that if (x, y) ∈ R then (y, x) ∈ R

∴ R is Symmetric

If x and y are living at a given locality and also y and z are living at that locality, then it implies that both x and z are also living at the same locality.

In otherwords, if (x, y) ∈ R, (y, z) ∈ R then (x, z) ∈ R.

∴ R is Transitive

∴ R is reflexive, symmetric as well as transitive.

v (c). R = (x, y) : x is exactly 7 cm taller than y

As a person x can never be taller than oneself, (x, x) ∉ R.

∴ R is not Reflexive

If x is 7 cm taller than y, then we can not say that y is also taller than x. In fact, y will be shorter than x. So, in this case, if (x, y) ∈ R then (y, x) ∉ R

∴ R is not Symmetric

If x is exactly 7 cm taller than x and y is exactly 7 cm taller than z, then x is (7 cm + 7 cm) = 14 cm taller than z. So, x is not exactly 7 cm taller than z. (Though x is taller than z but not exactly by 7 cm). In otherwords, if (x, y) ∈ R, (y, z) ∈ R then (x, z) ∉ R.

∴ R is not Transitive

∴ R is neither reflexive nor symmetric nor transitive.

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

As a person x can not be wife of oneself. So, (x, x) ∉ R.

∴ R is not Reflexive

If x is wife of y then y can not be wife of x (y will be husband of x). So, if (x, y) ∈ R then (y, x) ∉ R.

∴ R is not symmetric

If x is wife of y, then y can not be wife of any other person z (as y is husband). So, when (x, y) ∈ R, (y, z) ∉ R.

∴ R is not Transitive

∴ R is neither reflexive nor symmetric nor transitive.

v (d). R = {(x, y) : x is father of y}

As a person x can not be father of oneself. So, (x, x) ∉ R.

∴ R is not Reflexive

If x is father of y then y can not be father of x (y will be child of x). So, if (x, y) ∈ R then (y, x) ∉ R.

∴ R is not symmetric

If x is father of y and y is father of z, then x is not father of z (x will be grandfather of z). So, if (x, y) ∈ R and (y, z) ∈ R then (x, z) ∉ R.

∴ R is not Transitive

∴ R is neither reflexive nor symmetric nor transitive.

2. Show that the relation R in the set R of real numbers, defined as R = {(a, b) : a \le a^2} is neither reflexive nor symmetric nor transitive.

To Check whether R is Reflexive: The relation R in the set A is reflexive if (a, a) ∈ R, for every a ∈ A

In this case, for every a ∈ R, it should satisfy the relation, a \le a^2

When {a = 2}, {a^2 = 4}. So, 2 ≤ 2²

When {a = \dfrac{1}{5}}, {a^2 = \dfrac{1}{25}}. So, {\dfrac{1}{5}\nleq\left(\dfrac{1}{25}\right)^2}

Thus (a, a) ∈ R is not valid for all real numbers a ∈ R

∴ R is not Reflexive

We have 3 ≤ 4² and also 4 ≤ 3². In this case, (a, b) ∈ R and also (b, a) ∈ R

However, 1 ≤ 2² but {2 \nleq 1^2}. In this case, (a, b) ∈ R but (b, a) ∉ R

∴ R is not symmetric

This means that if {a \le b^2} and {b \le c^2} then {a \le c^2}

We have 3 ≤ (-5)² and -5 ≤ 1² but {3 \nleq 1^2}

∴ R is not Transitive

∴ R is neither reflexive nor symmetric nor transitive.

3. Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as R = {(a, b) : b = a + 1} is reflexive, symmetric or transitive.

In this case, for every a ∈ R, it should satisfy the relation, {a = a + 1}.

But as we know, a ≠ {a + 1}.

⇒ (a, a) ∉ R

∴ R is not Reflexive

When {b = a + 1}, we know that a ≠ {b + 1} (∵ a = b - 1).

⇒ when (a, b) ∈ R, we’ve (b, a) ∉ R

∴ R is not symmetric

When {b = a + 1} and {c = b + 1},

we’ve c {= b + 1} {= (a + 1) + 1} {= a + 2} ≠ {a + 1}.

⇒ when (a, b) ∈ R and (b, c) ∈ R then (a, c) ∉ R.

∴ R is not transitive

∴ R is neither reflexive nor symmetric nor transitive.

4. Show that the relation R in R defined as R = (a, b) : a \le b, is reflexive and transitive but not symmetric.

As every element a is equal to itself, it satisfies the condition {a \le a}.

⇒ (a, a) ∈ R

∴ R is Reflexive

We have 2 ≤ 4. But 4 \nleq 2 (4 > 2). Thus for many elements when (a, b) ∈ R, we have (b, a) ≠ R.

∴ R is not symmetric.

When {a \le b} and {b \le c} then we always have {a \le c}

∴ R is Transitive

∴ R is both reflexive and transitive but not symmetric

5. Check whether the relation R in R defined by R = {(a, b) : a \le b^3} is reflexive, symmetric or transitive.

When {a = -2}, we’ve {a^3 = -8}. But -2 \nleq (-2)^3

∴ R is not Reflexive

Consider {a = 1} and {b = 2}

we’ve 1 ≤ 2³ but 2 \nleq 1^3

⇒ When (a, b) ∈ R, we have (b, a) ∉ R

∴ R is not symmetric.

Consider the case, {a = 2}, {b = 3} and {c = 4}

We’ve {b^3 = 3^3 = 27} and 4³ = 64

So, 2 ≤ 3³, 3 ≤ 4³ and 2 ≤ 4³.

However, when {a = 4}, {b = \dfrac{7}{4}} and {c = \dfrac{5}{4}}

We’ve {b^3 = \left(\dfrac{7}{4}\right)^{3} = 5.359} and {c^3 = \left(\dfrac{5}{4}\right)^{3} = 1.953}

In this case, 4 ≤ \left(\dfrac{7}{4}\right)^3 and \dfrac{7}{4} ≤ \left(\dfrac{5}{4}\right)^3, but {4 \nleq \left(\dfrac{5}{4}\right)^3}

∴ R is not Transitive

Hence, in this case, R is neither reflexive nor symmetric nor transitive.

6. Show that the relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} is symmetric but neither reflexive nor transitive.

We have, 1 ∈ A but (1, 1) ∉ R

∴ R is not Reflexive

We see that both (1, 2) ∈ R and (2, 1) ∈ R

⇒ R is symmetric.

We see that (1, 2) ∈ R, (2, 1) ∈ R. But (1, 1) ∉ R

∴ R is not Transitive

∴ 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 R = {(x, y) : x and y have same number of pages} is an equivalence relation.

If we consider a book x, it will have the same number of pages as itself. i.e. x.

⇒ (x, x) ∈ R

∴ R is Reflexive

If book x has the same number of pages as book y, then book y will also have the same number of pages as book x.

⇒ If (x, y) ∈ R then (y, x) ∈ R

∴ R is symmetric

If book x has the same number of pages as book y and book y has the same number of pages as book z, then it implies that book x has the same number of pages as book z.

⇒ if (x, y) ∈ R and (y, z) ∈ R then (x, z) ∈ R

∴ R is Transitive

∴ R is reflexive, symmetric and transitive, it is an equivalence relation.

8. Show that the relation R in the set A = {1, 2, 3, 4, 5} given by R = {(a, b) : |a – b| is even}, is an equivalence relation. Show that all the elements of {1, 3, 5} are related to each other and all the elements of {2, 4} are related to each other. But no element of {1, 3, 5} is related to any element of {2, 4}.

For any element a, we’ve {|a - a| = 0}, which is even.

∴ R is Reflexive

As we know, for any two elements x, y, we’ve {|x - y| = |y - x|}. So, when {|x - y|} is even then {|y - x|} is also even.

⇒ If (x, y) ∈ R then (y, x) ∈ R

∴ R is Symmetric.

When {|a - b|} is even, then {a - b} is also even.

Similarly, when {|b - c|} is even, then {b - c} is also even.

⇒ {a - c = (a - b) + (b - c)} is also even (∵ The sum of two even numbers is even)

⇒ {|a - c|} is also even.

⇒ when (a, b) ∈ R and (b, c) ∈ R then (a, c) ∈ R

∴ R is reflexive, symmetric and transitive. Hence, it is an equivalence relation.

To Show that all the elements of {1, 3, 5} are related to each other

All the elements in {1, 3, 5} are odd

We know that the difference of any two odd numbers is always even.

⇒ The modulus of their difference is also even.

⇒ All the elements of {1, 3, 5} can be related to each other.

To Show that all the elements of {2, 4} are related to each other

All the elements in {2, 4} are even

We know that the difference of any two even numbers is always even

⇒ The modulus of their difference is also even.

⇒ All the elements of {2, 4} can be related to each other.

To show that no element of {1, 3, 5} is related to any element of {2, 4}.

The elements from {1, 3, 5} can not be related to {2, 4} because their combination forms an odd and an even number.

As we know the difference of an odd and even number is always odd

⇒ They can not be related to each other.

9. Show that each of the relation R in the set A = x ∈ Z : 0 ≤ x ≤ 12, given by

i.

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

ii.

R = (a, b) : a = b

is an equivalence relation. Find the set of all elements related to R in each case.

To Show that R = {(a, b) : |a – b| is a multiple of 4} is an equivalence relation.

For any element a, we’ve {|a - a| = 0}, which is a multiple of 4.

∴ R is Reflexive

As we know, for any two elements x, y, we’ve {|x - y| = |y - x|}.

⇒ When {|x - y|} is multiple of 4 then |y - x| is also multiple 4 (as both of them are equal).

⇒ If (x, y) ∈ R then (y, x) ∈ R

∴ R is Symmetric.

When {|a - b|} is multiple of 4, then {a - b} is also a multiple of 4.

Similarly, when {|b - c|} is multiple of 4, then {b - c} is also multiple of 4.

So, we have {a - c = (a - b) + (b - c)} is also multiple of 4 as it is sum of two multiple of 4.

⇒ {|a - c|} is also multiple of 4.

⇒ when (a, b) ∈ R and (b, c) ∈ R then (a, c) ∈ R

∴ R is reflexive, symmetric and transitive. Hence, it is an equivalence relation.

To find the set of elements related to 1 for case (i)

Set A can be written in the Roster form as A = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

We have to find the elements whose difference with 1 is a multiple of 4. In otherwords, {|1 - x|} should be multiple of 4 i.e. 0 or 4 or 8

We have |1 – 1| = 0 which is divisible by 4

We have |1 – 5| = 4 which is divisible by 4

We have |1 – 9| = 8 which is divisible by 4

∴ The set of elements related to 1 are : {1, 5, 9}

To Show that R = {(a, b) : a = b} is an equivalence relation.

For any element a, we’ve {a = a}

⇒ (a, a) ∈ R

∴ R is Reflexive

As we know, for any two elements a, b when {a = b} then we’ve {b = a}

⇒ If (a, b) ∈ R then {(b, a)} ∈ R

∴ R is Symmetric.

When {a = b} and {b = c}, then we’ve {a = c}

⇒ When (a, b) ∈ R and (b, c) ∈ R then (a, c) ∈ R

∴ R is reflexive, symmetric and transitive. Hence, it is an equivalence relation.

To find the set of elements related to 1 for case (ii)

We’ve to find an element such that {1 = x} or {x = 1}

Clearly, 1 is the only element that satisfies this condition.

∴ the set of elements related to 1 is : {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.

10.i. Give an example of a relation which is Symmetric but neither reflexive nor transitive.

Consider a set A = {a, b, c} and the relation R defined as R = {(a, b), (b, a)}

The element (a, a) ∉ R

⇒ This relation is not reflexive.

In this relation, both (a, b) ∈ R and (b, a) ∈ R

⇒ This relation is symmetric.

In this example (a, b) ∈ R and (b, a) ∈ R. But (a, a) ∉ R.

⇒ This relation is not transitive.

∴ R is Symmetric but neither reflexive nor transitive.

10.ii. Give an example of a relation which is Transitive but neither reflexive nor symmetric.

Consider the relation defined by R = {(x, y): x > y}

As we know, No element greater than itself

⇒ (a, a) ∉ R.

∴ R is not reflexive.

Consider two elements a, b such that {a \gt b}. Obviously {b \ngtr a} (to be specific b \le a).

⇒ (a, b) ∈ R but (b, a) ∉ R.

∴ R is not symmetric.

If there’re 3 elements a, b and c such that {a \gt b} and {b \gt c} then obviously {a \gt c}

⇒ If (a, b) ∈ R and (b, c) ∈ R then (a, c) ∈ R

⇒ R is transitive.

∴ R is Symmetric but neither reflexive nor transitive.

10.iii. Give an example of a relation which is reflexive and symmetric but not transitive.

Consider the set A = {p, q, r}. Consider the relation on the set A = {p, q, r} defined as R = {(p, p), (q, q), (r, r), (p, q), (q, p), (q, r), (r, q)}

In the set A, we see that For every element a ∈ A, there is a corresponding (a, a) ∈ R.

⇒ (p, p) ∈ R, (q, q) ∈ R and (r, r) ∈ R

∴ R is reflexive.

In the relation R, we see that (p, q) ∈ R and (q, p) ∈ R.

Also (q, r) ∈ R and (r, q) ∈ R

∴ R is symmetric

In the relation R, we see that both (p, q) ∈ R and (q, r) ∈ R, but (p, r) ∉ R

∴ R is not transitive.

∴ R is reflexive and symmetric but not transitive.

10.iv. Give an example of a relation which is Reflexive and transitive but not symmetric.

Consider the relation R = {(x, y) : x \geq y} defined on the set of natural numbers N

For any natural number a ∈ N, we know that {a = a}.

⇒ a \geq a

∴ R is reflexive.

Consider the element (5, 4) ∈ R because 5 ≥ 4. But we see that {4 \nleq 5}.

⇒ (4, 5) ∉ R

⇒ There are few elements (a, b) ∈ R such that (b, a) ∉ R

∴ R is not symmetric.

When {a \geq b} and {b \geq c}, we have {a \geq c}

⇒ For every (a, b) ∈ R and (b, c) ∈ R, there is a corresponding (a, c) ∈ R

∴ R is transitive.

∴ R is reflexive and transitive but not symmetric.

10.v. Give an example of a relation which is Symmetric and transitive but not reflexive.

Consider the relation R = {(p, q), (q, p), (p, p)} defined on the set A = {p, q}

We see that (q, q) ∉ R

∴ R is not reflexive

We see that (p, q) ∈ R and (q, p) ∈ R

∴ R is symmetric

Both (p, q) ∈ R (q, p) ∈ R and also (p, p) ∈ R.

∴ R is not transitive.

∴ R is symmetric and transitive but not reflexive

11. Show that the relation R in the set A of points in a plane given by R = {(P, Q) : 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 P ≠ (0, 0) is the circle passing through P with origin as centre.

Consider a point P.

The point P and the same point P will have same distance from the origin.

⇒ Distance of point A from Origin = Distance of point A from Origin.

⇒ (A, A) ∈ R

∴ R is reflexive.

Consider two points A and B such that

Distance of Point A from Origin = Distance of point B from Origin

⇒ Distance of point B from Origin = Distance of point A from Origin.

⇒ (A, B) ∈ R and also (B, A) ∈ R

∴ R is symmetric.

Consider 3 points A, B and C such that

Distance of Point A from Origin = Distance of Point B from Origin and

Distance of Point B from Origin = Distance of Point C from Origin

⇒ Distance of Point A from Origin = Distance of Point C from Origin

⇒ If (A, B) ∈ R, (B, C) ∈ R then (A, C) ∈ R

∴ R is transitive.

∴ R is reflexive, symmetric and transitive.

Hence, R is an equivalence relation.

Show that the set of all points related to a point P ≠ (0, 0) is the circle passing through P with origin as centre.

This means that R contains all the points whose distance from the origin O(0, 0) is constant.

This forms the set of all the points which are at a constant distance from the origin. The curve joining all these points will be a circle of radius equal to the distance of the point from the origin.

12. Show that the relation R defined in the set A of all triangles as R = {(T1, T2) : T1 is similar to T1}, is equivalence relation. Consider three right angle triangles T1 with sides 3, 4, 5, T2 with sides 5, 12, 13 and T1 with sides 6, 8, 10. Which triangles among T1, T2 and T3 are related?

We know that every triangle T is similar to itself.

⇒ (T, T) ∈ R

∴ R is reflexive.

If a triangle T1 is similar to another triangle T2

⇒ Triangle T2 is also similar to the triangle T1

⇒ If (T1, T2) ∈ R then (T2, T1) ∈ R

∴ R is symmetric.

If there are 3 triangles such that triangle T1 is similar to triangle T2 and triangle T2 is similar to triangle T3 then triangle T1 is similar to triangle T3

⇒ If (T1, T2) ∈ R and (T2, T3) ∈ R then (T1, T3) ∈ R

∴ R is transitive.

∴ R is reflexive, symmetric and transive. Hence, R is an equivalence relation.

Consider three right angle triangles T1 with sides 3, 4, 5, T2 with sides 5, 12, 13 and T1 with sides 6, 8, 10. Which triangles among T1, T2 and T3 are related?

The relation R is defined as R = {(T1, T2): T1 is similar to T2}

Triangle T1 has the sides 3, 4, 5

Triangle T2 has the sides 5, 12, 13

Triangle T3 has the sides 6, 8, 10

As per the definition, any two triangles are similar if the ratio of their corresponding sides is same.

Obviously, if we consider the triangles T1 and T3, the ratio of their sides is

{\dfrac{3}{6} = \dfrac{4}{8} = \dfrac{5}{10} = \dfrac{1}{2}}

So, the ratio of the corresponding sides is same.

⇒ The triangles T1 and T3 are similar triangles.

⇒ Among the given 3 triangles, the triangles T1 and T3 are related.

⇒ (T1, T3) ∈ R

13. Show that the relation R defined in the set A of all polygons as R = {(P1, P1) : P1 and P2 have same number of sides}, 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?

The given relation is R = {(P1, P1) : P1 and P2 have same number of sides}

Any polygon L has the same number of sides as itself.

⇒ (L, L) ∈ R

∴ R is reflexive.

If a polygon L has the same number of sides as the polygon M, then the polygon M also has the same number of sides as the polygon L

⇒ If (L, M) ∈ R then (M, L) ∈ R

∴ Relation R is symmetric

If a polygon L has the same number of sides as the polygon M and polygon M has the same number of sides as the polygon N, then the poloygon L has the same number of sides as the polygon N.

⇒ If (L, M) ∈ R and (M, N) ∈ R then (L, N) ∈ R

∴ R is transitive

∴ As the relation R is reflexive, symmetric and transitive, the relation R 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?

A triangle is also a polygon with 3 sides, the elements in A related to this triangle are all the polygons that have 3 sides i.e. all the triangles.

So, the set of all elements in A related to the right-angled triangle T with sides 3, 4 and 5 is the set of all the triangles (Note here that the triangle being right-angled and the length of the sides is redundant or not useful information)

14. Let L be the set of all lines in XY plane and R be the relation in L defined as R = {(L1, L2) : L1 is parallel to L2}. Show that R is an equivalence relation. Find the set of all lines related to the line {y = 2x + 4}.

The given relation is R = (L1, L2) : L1 is parallel to L2

As we know, any line L is parallel to itself.

⇒ (L, L) ∈ R

∴ R is reflexive.

If any line L is parallel to line M, then the line M is also parallel to line L

⇒ If (L, M) ∈ R then (M, L) ∈ R

∴ R is symmetric.

If any line L is parallel to line M and line M is parallel to line N, then line L is parallel to the line N.

⇒ If (L, M) ∈ R and (M, N) ∈ R then (L, N) ∈ R

∴ R is transitive.

∴ R is reflexive, symmetric as well as transitive.

∴ R is an equivalence relation.

Find the set of all lines related to the line {y = 2x + 4}.

Any line parallel to the line {y = 2x + 4} will be related to it.

As we know, for two lines to be parallel in a given plane, they should have the same slope m.

We know that the slope of the line {y = mx + c} is m

⇒ the slope of the line {y = 2x + 4} is 2.

⇒ the set of all the lines that are parallel to the line {y = 2x + 4} will be all the lines that have the slope 2.

⇒ The set of all lines related to the line {y = 2x + 4} will be all the lines that can be represented by the linear equation {y = 2x + k} where k is a constant

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

I.

R is reflexive and symmetric but not transitive.

II.

R is reflexive and transitive but not symmetric.

III.

R is symmetric and transitive but not reflexive.

IV.

R is an equivalence relation.

Given that R is the relation in the set {1, 2, 3, 4} given by R = {(1, 2), (2, 2), (1, 1), (4,4), (1, 3), (3, 3), (3, 2)}

To be reflexive, for every element a ∈ A we should have (a, a) ∈ R

We see that for every a ∈ {1, 2, 3, 4}, we have (1, 1) ∈ R, (2, 2) ∈ R, (3, 3) ∈ R and (4, 4) ∈ R

∴ R is reflexive.

For a relation to be symmetric, if (a, b) ∈ R then (b, a) ∈ R

We see that (1, 2) ∈ R but (2, 1) ∉ R

∴ R is not symmetric.

In this case, (1, 3) ∈ R, (3, 2) ∈ R and also (1, 2) ∈ R

∴ R is transitive.

∴ R is reflexive and transitive by not symmetric. So, option B is the correct choice.

Note:As we know that the relation R is not symmetric, we can choose that the choice B is the correct one and need not check whether R is transitive or not. This is because, only answer B considers that R is not symmetric

16. Let R be the relation in the set N given by R = {(a, b) : a = b – 2, b \gt 6}. Choose the correct answer.

(A) (2, 4) ∈ R (B) (3, 8) ∈ R (C) (6, 8) ∈ R (D) (8, 7) ∈ R

(A) (2, 4) ∈ R (B) (3, 8) ∈ R (C) (6, 8) ∈ R (D) (8, 7) ∈ R

Given that R is the relation in the set N given by R = {(a, b) : a = b~–~2, b \gt 6}

Check for option A:

2 = 4 – 2 but {4 \ngtr 6}. So, A is not correct option.

Check for option B:

3 ≠ 8 – 2. So, B is not correct option.

Check for option C:

6 = 8 – 2 and 8 > 6. So, C is correct option.

Check for option D:

8 ≠ 7 – 2. So, D is not correct option (as we already know that C is correct option, we don’t have to check for D)