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.

Suppose 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 } Reflexive Property : A relation R in a set A is said to be reflexive if x , x ∈ R , for every x ∈ A Now, 1 , 1 , 2 , 2 , 3 , 3 , 4 , 4 , 5 , 5 , 6 , 6 ∉ R ∴ R is not reflexive. Symmetric Property : A relation R in a set A is said to be symmetric if x , y ∈ R implies that y , x ∈ R for all x , y ∈ A . Now it can be observed that 1 , 2 ∈ R , but 2 , 1 ∉ R . Hence , R is not symmetric. Transitive Property : A relation R in a set A is said to be transitive if x , y ∈ R and y , z ∈ R implies that x , z ∈ R for all x , y , z ∈ A . Now, 1 , 2 , 2 , 3 ∈ R but, ( 1 , 3 ) ∉ R ∴ R is not transitive. Hence, R is neither reflexive, nor symmetric, nor transitive.

Let f x = x 1 - x 2 , x ≠ ± 1 , then find ( f o f o f ) ( x )

We have, f x = x 1 - x 2 Now, f o f x = f x 1 - f x 2 ⇒ f o f x = x 1 - x 2 1 - x 2 1 - x 2 ⇒ f o f x = x 1 - 2 x 2 , x ≠ ± 1 2 Now, f o f o f x = f f o f x ⇒ f o f o f x = x 1 - 2 x 2 1 - x 2 1 - 2 x 2 ⇒ f o f o f x = x 1 - 3 x 2 , x ≠ ± 1 3

Let f , g , h : R → R , given by f x = 2 x + 1 , g x = 1 - 4 x and h x = x 2 + 1 . Find f o g o h and g o f o h .

We have, f x = 2 x + 1 g x = 1 - 4 x h x = x 2 + 1 Now, f o g o h x = f g h x ⇒ f o g o h x = f g x 2 + 1 ⇒ f o g o h x = f 1 - 4 x 2 + 1 ⇒ f o g o h x = f - 3 - 4 x 2 ⇒ f o g o h x = 2 - 3 - 4 x 2 + 1 ⇒ f o g o h x = - 8 x 2 - 5 And, g o f o h x = g f h x ⇒ g o f o h x = g f x 2 + 1 ⇒ g o f o h x = g 2 x 2 + 1 + 1 ⇒ g o f o h x = g 2 x 2 + 3 ⇒ g o f o h x = 1 - 4 2 x 2 + 3 ⇒ g o f o h x = - 11 - 8 x 2

Let N m be the set of all ordered m - tuples of natural numbers. If x = x 1 , x 2 , x 3 , … … , x m , y = y 1 , y 2 , y 3 , … . , y m where x i , y i ∈ N , i = 1 , 2 , 3 , … . , m and an operation ' + ' is defined by x + y = x 1 + y 1 , x 2 + y 2 , … … x m + y m , then prove that given operation is commutative as well as associative.

We have, x = x 1 , x 2 , x 3 , … … , x m , y = y 1 , y 2 , y 3 , … . , y m Now, x + y = x 1 + y 1 , x 2 + y 2 , … … , x m + y m y + x = y 1 + x 1 , y 2 + x 2 , … … , y m + x m But addition is commutative, hence x 1 + y 1 = y 1 + x 1 , x 2 + y 2 = y 2 + x 2 , . . . . , x m + y m = y m + x m Therefore, x + y = y + x , so " + " is commutative. Also, let z = z 1 , z 2 , z 3 , … . , z m , then x + y + z = x 1 + y 1 + z 1 , x 2 + y 2 + z 2 , x 3 + y 3 + z 3 , . . . . , x m + y m + z m x + y + z = x 1 + y 1 + z 1 , x 2 + y 2 + z 2 , x 3 + y 3 + z 3 , . . . . , x m + y m + z m But addition is associative, hence x + y + z = x + y + z Therefore, " + " is associative.

Let a binary operation' * 'on Q (set of all rational numbers) be defined by a * b = a + 2 b for all a , b ∈ Q . Prove that : Q is closed under the given operation.

We have, a * b = a + 2 b Let x , y ∈ Q , then x * y = x + 2 y Since, x , y ∈ Q ⇔ x + 2 y ∈ Q ∀ x , y ∈ Q , hence x * y ∈ Q , hence Q is closed under given operation.

Let a binary operation' * 'on Q (set of all rational numbers) be defined by a * b = a + 2 b for all a , b ∈ Q . Prove that the given operation is not commutative.

We have, a * b = a + 2 b b * a = b + 2 a Clearly, a * b ≠ b * a , hence given operation is not commutative.

Let a binary operation' * 'on Q (set of all rational numbers) be defined by a * b = a + 2 b for all a , b ∈ Q . Prove that the given operation is not associative.

We have, a * b = a + 2 b Let x , y , z ∈ Q , then x * y * z = x * y + 2 z = x + 2 y + 2 z = x + 2 y + 4 z x * y * z = x + 2 y * z = x + 2 y + 2 z Clearly, x * y * z ≠ x * y * z , hence given operation is not associative.

E M B I B E

Gamma Question Bank for Engineering Mathematics>Class XII>Chapter 18 - Relation and Functions>Assignment

Embibe Experts Solutions for Exercise 5: Assignment

Author:Embibe Experts

Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Exercise 5: Assignment

Attempt the practice questions from Exercise 5: Assignment with hints and solutions to strengthen your understanding. Gamma Question Bank for Engineering Mathematics solutions are prepared by Experienced Embibe Experts.

Questions from Embibe Experts Solutions for Exercise 5: Assignment with Hints & Solutions

EASY

JEE Main/Advance

IMPORTANT

Gamma Question Bank for Engineering Mathematics>Chapter 18 - Relation and Functions>Assignment>Q 20

Check whether the relation $R$ defined in the set $\{1, 2, 3, 4, 5, 6\}$ as $:$ is reflexive, symmetric or transitive.

MEDIUM

JEE Main/Advance

IMPORTANT

Gamma Question Bank for Engineering Mathematics>Chapter 18 - Relation and Functions>Assignment>Q 34

Let $f (x) = \frac{x}{\sqrt{1 - x^{2}}}, x \neq \pm 1$ , then find $R$

MEDIUM

JEE Main/Advance

IMPORTANT

Gamma Question Bank for Engineering Mathematics>Chapter 18 - Relation and Functions>Assignment>Q 35

Let $f, g, h : R \to R$ , given by $f (x) = 2 x + 1, g (x) = 1 - 4 x$ and $h (x) = x^{2} + 1$ . Find $f o g o h$ and $g o f o h$ .

HARD

JEE Main/Advance

IMPORTANT

Gamma Question Bank for Engineering Mathematics>Chapter 18 - Relation and Functions>Assignment>Q 36

Find the domain and the range of the function $f (x) = 2 x^{2} - 1$ . Also find $f (1)$ and the numbers which are associated with the number $31$ in its range. Is this function one-one.

MEDIUM

JEE Main/Advance

IMPORTANT

Gamma Question Bank for Engineering Mathematics>Chapter 18 - Relation and Functions>Assignment>Q 37

Let $N^{m}$ be the set of all ordered $m -$ tuples of natural numbers. If $x = (x_{1}, x_{2}, x_{3}, \dots \dots, x_{m}), y = (y_{1}, y_{2}, y_{3}, \dots ., y_{m})$ where $x_{i}, y_{i} \in N, i = 1, 2, 3, \dots ., m$ and an operation ' $+$ ' is defined by $x + y = (x_{1} + y_{1}, x_{2} + y_{2}, \dots \dots x_{m} + y_{m})$ , then prove that given operation is commutative as well as associative.

EASY

JEE Main/Advance

IMPORTANT

Gamma Question Bank for Engineering Mathematics>Chapter 18 - Relation and Functions>Assignment>Q 39

Let a binary operation' $*$ 'on $Q$ (set of all rational numbers) be defined by $a * b = a + 2 b$ for all $a, b \in Q$ . Prove that :
$Q$ is closed under the given operation.

EASY

JEE Main/Advance

IMPORTANT

Gamma Question Bank for Engineering Mathematics>Chapter 18 - Relation and Functions>Assignment>Q 39

Let a binary operation' $*$ 'on $Q$ (set of all rational numbers) be defined by $a * b = a + 2 b$ for all $a, b \in Q$ . Prove that the given operation is not commutative.

EASY

JEE Main/Advance

IMPORTANT

Gamma Question Bank for Engineering Mathematics>Chapter 18 - Relation and Functions>Assignment>Q 39

Let a binary operation' $*$ 'on $Q$ (set of all rational numbers) be defined by $a * b = a + 2 b$ for all $a, b \in Q$ . Prove that the given operation is not associative.

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