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# NCERT Solution Class 12 Maths

## Chapter 1 Relations and Functions

Exercise 1.1

### Question 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 < 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 as 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}

Answer 1:

(i) A = {1,2,3,…,12,13,14}, R = {(x, y): 3x − y = 0} R = {(1, 3), (2, 6), (3, 9), (4, 12)}

R is not reflexive since (1, 1), (2, 2) … (14, 14) ∉ R. Also, R is not symmetric as (1, 3) ∈ R, but (3, 1) ∉ R. [3(3) − 1 ≠ 0] Also R is not transitive as (1, 3) (3, 9) ∈ R but (1, 9) ∉ R.

Hence R is neither Reflexive, nor Symmetric, nor transitive.

(ii) R = {(x, y): y = x + 5 and x < 4} = {(1, 6), (2, 7), (3, 8)} It is clear that (1, 1) ∉ R. ∴ R is not reflexive.
(1, 6) ∈ R But, (1, 6) ∉ R. ∴ R is not symmetric.
Now, since there is no pair in R such that (x, y) and (y, z) ∈ R, then (x, z) cannot
belong to R. ∴ R is not transitive.
Hence, R is neither reflexive, nor symmetric, nor transitive.

(iii) A = {1, 2, 3, 4, 5, 6} R = {(x, y): y is divisible by x} We know that any number (x) is divisible by itself.
So, (x, x) ∈ R ∴ R is reflexive.
Now, (2, 4) ∈ R [as 4 is divisible by 2] But, (4, 2) ∉ R. [as 2 is not divisible by 4] ∴ R is not symmetric.
Let (x, y), (y, z) ∈ R. Then, y is divisible by x and z is divisible by y. ∴ z is divisible by x. ⇒ (x, z) ∈ R. ∴ R is transitive.

Hence, R is a reflexive, transitive but not symmetric.

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