NCERT Solutions for 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} Sol. A = {1, 2, 3 … 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. [3(1) − 9 ≠ 0] Hence, 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<4} Sol. 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. 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) Relation R in the set A = {1, 2, 3, 4, 5, 6) as R = {(x, y): y is divisible by x} Sol. 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 [as 4 is divisible by 2] [as 2 is not divisible by 4] 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 reflexive and transitive but not symmetric. |
(iv) Relation R in the set Z of all integers defined as R = {(x, y): x – y is as integer) Sol. R = {(x, y): x − y is an integer} Now, for every x ∈ Z, (x, x) ∈ R as x − x = 0 is an integer. ∴ R is reflexive. Now, for every x, y ∈ Z, if (x, y) ∈ R, then x − y is an integer. ⇒ −(x − y) is also an integer. ⇒ (y − x) is an integer. ∴ (y, x) ∈ R ∴ R is symmetric. Now, Let (x, y) and (y, z) ∈ R, where x, y, z ∈ Z. ⇒ (x − y) and (y − z) are integers. ⇒ x − z = (x − y) + (y − z) is an integer. ∴ (x, z) ∈ R ∴ R is transitive. Hence, R is reflexive, symmetric, and transitive. |
(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} Sol. R = {(x, y): x and y work at the same place} ⇒ (x, x) ∈ R [as x and x work at the same place] ∴ R is reflexive. If (x, y) ∈ R, then x and y work at the same place. ⇒ y and x work at the same place. ⇒ (y, x) ∈ R. ∴ R is symmetric. Now, let (x, y), (y, z) ∈ R ⇒ x and y work at the same place and y and z work at the same place. ⇒ x and z work at the same place. ⇒ (x, z) ∈ R ∴ R is transitive. Hence, R is reflexive, symmetric and transitive. (b) R = {(x, y) : x and y live in the same locality} Sol. R = {(x, y): x and y live in the same locality} Clearly, (x, x) ∈ R as x and x is the same human being. ∴ R is reflexive. If (x, y) ∈ R, then x and y live in the same locality. ⇒ y and x live in the same locality. ⇒ (y, x) ∈ R ∴ R is symmetric. Now, let (x, y) ∈ R and (y, z) ∈ R. ⇒ x and y live in the same locality and y and z live in the same locality. ⇒ x and z live in the same locality. ⇒ (x, z) ∈ R ∴ R is transitive. Hence, R is reflexive, symmetric and transitive. (c) R = {(x, y) : x is exactly 7 cm taller than y} Sol. R = {(x, y): x is exactly 7 cm taller than y} Now, (x, x) ∉ R Since human being x cannot be taller than himself. ∴ R is not reflexive. Now, let (x, y) ∈ R. ⇒ x is exactly 7 cm taller than y. Then, y is not taller than x. [Since, y is 7 cm smaller than x] ∴ (y, x) ∉ R Indeed if x is exactly 7 cm taller than y, then y is exactly 7 cm shorter than x. ∴ R is not symmetric. Now, Let (x, y), (y, z) ∈ R. ⇒ x is exactly 7 cm taller than y and y is exactly 7 cm taller than z. ⇒ x is exactly 14 cm taller than z . ∴ (x, z) ∉ R ∴ R is not transitive. Hence, R is neither reflexive, nor symmetric, nor transitive. (d) R = {(x, y) : x is wife of y} Sol. R = {(x, y): x is the wife of y} Now, (x, x) ∉ R Since x cannot be the wife of herself. ∴ R is not reflexive. Now, let (x, y) ∈ R ⇒ x is the wife of y. Clearly y is not the wife of x. ∴ (y, x) ∉ R Indeed, if x is the wife of y, then y is the husband of x. ∴ R is not transitive. Let (x, y), (y, z) ∈ R ⇒ x is the wife of y and y is the wife of z. This case is not possible. Also, this does not imply that x is the wife of z. ∴ (x, z) ∉ R ∴ R is not transitive. Hence, R is neither reflexive, nor symmetric, nor transitive. (e) R = {(x, y) : x is father of y} Sol. R = {(x, y): x is the father of y} (x, x) ∉ R As x cannot be the father of himself. ∴ R is not reflexive. Now, let (x, y) ∉ R. ⇒ x is the father of y. ⇒ y cannot be the father of y. Indeed, y is the son or the daughter of y. ∴ (y, x) ∉ R ∴ R is not symmetric. Now, let (x, y) ∈ R and (y, z) ∉ R. ⇒ x is the father of y and y is the father of z. ⇒ x is not the father of z. Indeed, x is the grandfather of z. ∴ (x, z) ∉ R ∴ R is not transitive. Hence, R is neither reflexive, nor symmetric, nor transitive. |
Question 2 : Show that the relation R in the set R of real numbers, defined as R = {(a, b): a ≤ b2} is neither reflexive nor symmetric nor transitive.Sol. (i) R = {(a, b): a ≤ b2} It can be observed that ( 1/2 , 1/2 ) ∉ R, since, 1/2 > (1/2 )² ∴ R is not reflexive. Now, (1, 4) ∈ R as 1 < 4² But, 4 is not less than 1². ∴ (4, 1) ∉ R ∴ R is not symmetric. Now, (3, 2), (2, 1.5) ∈ R [as 3 < 2² = 4 and 2 < (1.5)² = 2.25] But, 3 > (1.5)² = 2.25 ∴ (3, 1.5) ∉ R ∴ R is not transitive. Hence, R is neither reflexive, nor symmetric, nor transitive. |
Question 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.Solution 3 : Let A = {1, 2, 3, 4, 5, 6}. A relation R is defined on set A as: R = {(a, b): b = a + 1} ∴ R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)} we can find (a, a) ∉ R, where a ∈ A. For instance, (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) ∉ R ∴ R is not reflexive. It can be observed that (1, 2) ∈ R, but (2, 1) ∉ R. ∴ R is not symmetric. Now, (1, 2), (2, 3) ∈ R But, (1, 3) ∉ R ∴ R is not transitive Hence, R is neither reflexive, nor symmetric, nor transitive. |
Question 4 : Show that the relation R in R defined as R = {(a, b): a ≤ b}, is reflexive and transitive but not symmetric. Solution 4 : |
Question 5 : Check whether the relation R in R defined as R = {(a, b): a ≤ b³} is reflexive, symmetric or transitive. Solution 5 : |
Question 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.
Solution 6 : |
Question 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.
Solution 7 : |
Question 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}.
Solution 8 : |