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## Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.2

**11th Maths Exercise 1.2 Solutions Question 1.**

Discuss the following relations for reflexivity, symmetricity and transitivity:

(i) The relation R defined on the set of all positive integers by “mRn if m divides n”.

Solution:

S = {set of all positive integers}

(a) mRm ⇒ ‘m’ divides’m’ ⇒ reflexive

(b) mRn ⇒ m divides n but

nRm ⇒ n does not divide m

(i.e.,) mRn ≠ nRm

It is not symmetric

(c) mRn ⇒ nRr as n divides r

It is transitive

(ii) Let P denote the set of all straight lines in a plane. The relation R defined by “lRm if l is perpendicular to m”.

Solution:

P = {set of all straight lines in a plane}

lRm ⇒ l is perpendicular to m

(a) lRl ⇒ l is not perpendicular to l

⇒ It is not reflexive

(b) lRm ⇒ l is perpendicular to m

mRl ⇒ m is perpendicular to l

It is symmetric

(c) l perpendicular to m ⇒ m perpendicular to n ⇒ l is parallel to n It is not transitive

(iii) Let A be the set consisting of all the members of a family. The relation R defined by “aRb if a is not a sister of b”.

Solution:

A = {set of all members of the family}

aRb is a is not a sister of b

(a) aRa ⇒ a is not a sister of a It is reflexive

(b) aRb ⇒ a is not a sister of b.

bRa ⇒ b is not a sister of a.

It is symmetric

(c) aRb ⇒ a is not a sister of b.

bRc ⇒ b is not a sister of c.

⇒ aRc ⇒ a can be a sister of c

It is not transitive.

(iv) Let A be the set consisting of all the female members of a family. The relation R

defined by “aRb if a is not a sister of b”.

Solution:

A = {set of all female members of a family}

(a) aRa ⇒ a is a sister of a

It is reflexive

(b) aRb ⇒ a is a sister of b

bRa ⇒ b is a sister of a

⇒ It is symmetric

(c) aRb ⇒ a is a sister of b bRc ⇒ b is a sister of c aRc ⇒ a can be sister of c It is not transitive.

(v) On the set of natural numbers the relation R defined by “xRy if x + 2y = 1”.

Solution:

N= {1, 2, 3, 4, 5,….}

xRy if x + 2y = 1 R is an empty set

(a) xRx ⇒ x + 2x = 1 ⇒ x = \(\frac{1}{3}\) ∉ N. It is not reflexive

xRy = yRx ⇒ x + 2y = 1 It does not imply that y + 2x = 1 as y = \(\frac{1-x}{2}\) It is not symmetric.

(b) -x = y ⇒ (-1, 1) ∉ N

It is not transitive.

**11th Maths Exercise 1.2 Answers Question 2.**

Let X = {a, b, c, d} and R = {(a, a), {b, b), (a, c)}. Write down the minimum number

of ordered pairs to be included to R to make it

(i) reflexive

(ii) symmetric

(iii) transitive

(iv) equivalence

Solution:

X = {a, b, c, d}

R = {(a, a), (b, b), (a, c)}

(i) To make R reflexive we need to include (c, c) and (d, d)

(ii) To make R symmetric we need to include (c, a)

(iii) R is transitive

(iv) To make R reflexive we need to include (c, c)

To make R symmetric we need to include (c, c) and (c, a) for transitive

∴ The relation now becomes

R = {(a, a), (b, b), (a, c), (c, c), (c, a)}

∴ R is equivalence relation.

**Exercise 1.2 Class 11 Maths State Board Question 3.**

Let A = {a, b, c} and R = {(a, a), (b, b), (a, c)}. Write down the minimum number of ordered pairs to be included to R to make it

(i) reflexive

(ii) symmetric

(iii) transitive

(iv) equivalence

Solution:

(i) (c, c)

(ii) (c, a)

(iii) nothing

(iv) (c, c) and (c, a)

**11th Maths Exercise 1.2 Question 4.**

Let P be the set of all triangles in a plane and R be the relation defined on P as aRb if a is

similar to b. Prove that R is an equivalence relation.

Solution:

P = {set of all triangles in a plane}

aRb ⇒ a similar to b

(a) aRa ⇒ every triangle is similar to itself

∴ aRa is reflexive

(b) aRb ⇒ if a is similar to b ⇒ b is also similar to a.

⇒ It is symmetric

(c) aRb ⇒ bRc ⇒ aRc

a is similar to b and b is similar to c

⇒ a is similar to a

⇒ It is transitive

∴ R is an equivalence relation

**11 Maths Exercise 1.2 Question 5.**

On the set of natural numbers let R be the relation defined by aRb if 2a + 3b = 30. Write down the relation by listing all the pairs. Check whether it is

(i) reflexive

(ii) symmetric

(iii) transitive

(iv) equivalence

Solution:

N = {set of natural numbers}

R ={(3, 8), (6, 6), (9, 4), (12, 2)}

(a) (3, 3) ∉ R ⇒ R is not reflexive

2a + 3b = 30

3b = 30 – 2a

b = \(\frac{30-2 a}{3}\)

(b) (3, 8) ∈ R(8, 3) ∉ R

⇒ R is not symmetric

(c) (a, b) (b, c) ∉ R ⇒ R is transitive

∴ It is not equivalence relation.

**11th Maths Chapter 1 Exercise 1.2 Question 6.**

Prove that the relation “friendship” is not an equivalence relation on the set of all people in Chennai.

Solution:

(a) S = aRa (i.e. ) a person can be a friend to himself or herself.

So it is reflextive.

(b) aRb ⇒ bRa so it is symmetric

(c) aRb, bRc does not ⇒ aRc so it is not transitive

⇒ It is not an equivalence relation

**11th Maths Exercise 1.2 Answers In Tamil Question 7.**

On the set of natural numbers let R be the relation defined by aRb if a + b ≤ 6. Write down the relation by listing all the pairs. Check whether it is

(i) reflexive

(ii) symmetric

(iii) transitive

(iv) equivalence

Solution:

Set of all natural numbers aRb if a + b ≤ 6

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

(i) (5, 1) ∈ R but(5, 5) ∉ R

It is not reflexive

(ii) aRb ⇒ bRa ⇒ It is symmetric

(iii) (4, 2), (2, 3) ∈ R ⇒ (4, 3) ∉ R

∴ It is not transitive

(iv) ∴ It is not an equivalence relation

**11th Maths Exercise 1.2 Solutions In Tamil Question 8.**

Let A = {a, b, c}. What is the equivalence relation of smallest cardinality on A? What is the equivalence relation of largest cardinality on A?

Solution:

R = {{a, a), (b, b), (c, c)} is this smallest cardinality of A to make it equivalence relation n(R) = 3

(i) R = {(a, a), {a, b), (a, c), (b, c), (b, b), {b, c), (c, a), (c, b), (c, c)}

n(R) = 9 is the largest cardinality of R to make it equivalence.

**Samacheer Kalvi 11th Maths Guide Question 9.**

In the set Z of integers, define mRn if m – n is divisible by 7. Prove that R is an equivalence relation.

Solution:

mRn if m – n is divisible by 7

(a) mRm = m – m = 0

0 is divisible by 7

∴ It is reflexive

(b) mRn = {m – n) is divisible by 7

nRm = (n – m) = – {m – n) is also divisible by 7

It is symmetric

It is transitive

mRn if m – n is divisible by 7

∴ R is an equivalence relation.

### Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.2 Additional Questions

**11th Maths Samacheer Kalvi Guru Question 1.**

Find the range of the function.

f = {(1, x), (1, y), (2, x), (2, y), (3, z)}

Solution:

The range of the function is {x, y, z}.

**Samacheer 11th Maths Solution Question 2.**

For n, m ∈ N, nln means that tt is a factor of n&m. Then find whether the given relation is an equivalence relation.

Solution:

Since n is a factor of n. So the relation is reflexive.

When n is a factor of m (where m ≠ n) then m cannot be a factor of n.

So the relation is not symmetric when n is a factor of m and m is a factor of p then n will be a factor of p. So the given relation is transitive. So it is not an equivalence relation.

**Samacheer Kalvi Guru 11th Maths Question 3.**

Verify whether the relation “is greater than” is an equivalence relation.

Solution:

You can do it yourself.