Let Z be the set of integers. Show that the relation R = { ( a , b ) : a , b ∈ Z and a + b is even } is an equivalence relation.

Given that, ∀ a , b ∈ Z , R = { ( a , b ) : ( a + b ) is even } R is reflexive if ( a , a ) ∈ R ∀ a ∈ Z For any a ∈ A , we have a + a = 2 a , which is even. ⇒ ( a , a ) ∈ R Thus, R is reflexive. R is symmetric if ( a , b ) ∈ R ⇒ ( b , a ) ∈ R ∀ a , b ∈ Z ( a , b ) ∈ R ⇒ a + b is even. ⇒ b + a is even. ⇒ ( b , a ) ∈ R Thus, R is symmetric . R is transitive if ( a , b ) ∈ R and ( b , c ) ∈ R ⇒ ( a , c ) ∈ R ∀ a , b , c ∈ Z Let ( a , b ) ∈ R and ( b , c ) ∈ R ∀ a , b , c ∈ Z ⇒ a + b = 2 P and b + c = 2 Q where P , Q are any two integers Adding both, we get a + c + 2 b = 2 ( P + Q ) ⇒ a + c = 2 ( P + Q ) - 2 b ⇒ a + c is an even number ⇒ ( a , c ) ∈ R Thus, R is transitive on Z . Since R is reflexive, symmetric and transitive it is an equivalence relation on Z . Hence proved

Show that the operation * on R + (set of all positive real numbers) defined by a * b = a b 3 , ∀ a , b ∈ R + is a binary operation on R + .

Given * on R + , defined by a * b = a b 3 , ∀ a , b ∈ R + . Let a , b ∈ R + then a b ∈ R + and a b 3 ∈ R + . ⇒ a * b ∈ R + Thus, * is a binary operation on R +

Show that the relation R on N × N defined by ( a , b ) R ( c , d ) ⇔ a + d = b + c , ∀ ( a , b ) , ( c , d ) ∈ N × N is an equivalence relation.

Given that, R be the relation in N × N defined by ( a , b ) R ( c , d ) ⇔ a + d = b + c for a , b , c , d in N × N R is reflexive if ( a , b ) R ( a , b ) for a , b in N × N Let ( a , b ) R ( a , b ) ⇒ a + b = b + a which is true since addition is commutative on N . ⇒ ( a , b ) R ( a , b ) ⇒ R is reflexive. R is symmetric if ( a , b ) R ( c , d ) ⇒ ( c , d ) R ( a , b ) for a , b , c , d in N × N Let ( a , b ) R ( c , d ) ⇒ a + d = b + c ⇒ b + c = a + d ⇒ c + b = d + a [since addition is commutative on N ] ⇒ c , d R ( a , b ) ⇒ ( c , d ) R ( a , b ) ⇒ R is symmetric. R is transitive if a , b R c , d and c , d R e , f ⇔ ( a , b ) R e , f for ( a , b ) , ( c , d ) , e , f in N × N Let a , b R c , d and c , d R e , f ⇒ a + d = b + c and c + f = d + e ⇒ ( a + d ) + ( c + f ) = ( b + c ) + ( d + e ) ⇒ a + f = b + e ⇒ ( a , b ) R e , f ⇒ R is transitive. Since, R is reflexive, symmetric and transitive and hence it is an equivalence relation. Hence, proved

Let R be a relation in N defined by a R b if 5 divides a − b , ( a , b ∈ N ) . Prove that a R b ⇒ a 2 R b 2 .

It is given that a R b if 5 divides a - b , a , b ∈ N . Now, a R b ⇒ a - b = 5 k , for some k . Multiplying both sides by a + b , we get: a - b a + b = 5 k a + b ⇒ a - b a + b = 5 k a + b ⇒ a 2 - b 2 = 5 k a + b ⇒ a 2 - b 2 is divisible by 5 . Hence, a 2 R b 2 . Thus, a R b ⇒ a 2 R b 2 Hence proved.

Find the inverse of the function f : R → R defined by f ( x ) = 3 x + 1 2 ∀ x ∈ R .

In order to prove that f is invertible we will show that f is a bijection. f is an injection: Let x , y ∈ R . Then, f x = f y ⇒ 3 x + 1 2 = 3 y + 1 2 ⇒ 3 x + 1 = 3 y + 1 ⇒ 3 x = 3 y ⇒ x = y Thus, f x = f y ⇒ x = y ∀ x , y ∈ R . So, f is an injection. f is a surjection: Let y be an arbitrary element of R . Then, f x = y ⇒ 3 x + 1 2 = y ⇒ x = 2 y - 1 3 Clearly, 2 y - 1 3 ∈ R ∀ y ∈ R . Thus, for all y ∈ R , ∃ x = 2 y - 1 3 ∈ R such that f x = f 2 y - 1 3 = 3 2 y - 1 3 + 1 2 = y Thus, f is a surjection. Hence, f : R → R is a bijection and so it is invertible. Now, f ( x ) = 3 x + 1 2 Let f x = y . Then, ⇒ y = 3 x + 1 2 ⇒ 2 y = 3 x + 1 ⇒ 2 y - 1 = 3 x ⇒ x = 2 y - 1 3 ⇒ f - 1 y = 2 y - 1 3 Thus, f - 1 : R → R is given by f - 1 x = 2 x - 1 3

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 1 - Relations and Functions>Manipur Board-2020

Embibe Experts Solutions for Chapter: Relations and Functions, Exercise 1: Manipur Board-2018

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Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Relations and Functions, Exercise 1: Manipur Board-2018

Attempt the free practice questions on Chapter 1: Relations and Functions, Exercise 1: Manipur Board-2018 with hints and solutions to strengthen your understanding. EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS solutions are prepared by Experienced Embibe Experts.

Questions from Embibe Experts Solutions for Chapter: Relations and Functions, Exercise 1: Manipur Board-2018 with Hints & Solutions

MEDIUM

12th Manipur Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 1 - Relations and Functions>Manipur Board-2018>Q 3

Let $Z$ be the set of integers. Show that the relation $R = {(a, b) : a, b \in Z$ and $a + b$ is even $}$ is an equivalence relation.

MEDIUM

12th Manipur Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 1 - Relations and Functions>Manipur Board-2019>Q 4

If $f : R ⟶ R$ be given by $f (x) = {(3 - x^{3})}^{\frac{1}{3}}$ , then $f^{- 1} (x)$ equals:

EASY

12th Manipur Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 1 - Relations and Functions>Manipur Board-2019>Q 5

Show that the operation $*$ on $R_{+}$ (set of all positive real numbers) defined by $a * b = \frac{a b}{3}, \forall a, b \in R_{+}$ is a binary operation on $R_{+}$ .