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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 Exercise 1.2 Answers In Tamil 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 Chapter 1 Exercise 1.2 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 Std Maths Exercise 1.2 Answers 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.
11 Maths Samacheer Kalvi 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
Samacheer Kalvi 11th Maths Guide 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 Kalvi 11th Maths Example Sums 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.
11th Maths Solution Samacheer Question 3.
Verify whether the relation “is greater than” is an equivalence relation.
Solution:
You can do it yourself.