Show that the function f : N → N defined by f n = n + 1 is not onto, by drawing arrow diagrams.

Given, the function f : N → N defined by f n = n + 1 . N = 1 , 2 , 3 , 4 , 5 , 6 , . . . . . . . f 1 = 1 + 1 = 2 f 2 = 2 + 1 = 3 f 3 = 3 + 1 = 4 f 4 = 4 + 1 = 5 f 5 = 5 + 1 = 6 Thus, f = 1 , 2 , 2 , 3 , 3 , 4 , 4 , 5 , 5 , 6 , . . . . . . . . . . . . Different elements have different images. Here, the range is not equal to the codomain. Therefore, the given function is not an onto function.

If A = 1 , 2 , 3 , B = 4 , 5 and C = { 5 , 6 } , then find A × ( B - C ) .

Given sets, A = { 1 , 2 , 3 } , B = { 4 , 5 } and C = { 5 , 6 } . We need to find, A × ( B - C ) Now, B - C = 4 ∴ A × ( B - C ) = { 1 , 2 , 3 } × 4 ∴ A × ( B - C ) = { ( 1 , 4 ) , ( 2 , 4 ) , ( 3 , 4 ) }

Let R = { ( x , y ) ; x + 2 y < 6 , x , y ∈ N } . Find the domain of R .

Given relation, R = { ( x , y ) ; x + 2 y < 6 , x , y ∈ N } Given, x + 2 y < 6 , x , y ∈ N When x = 1 , y = 1 , then 1 + 2 < 6 , which is true, so 1 , 1 ∈ R When x = 2 , y = 1 , then 2 + 2 < 6 , which is true, so 2 , 1 ∈ R When x = 1 , y = 2 , then 1 + 4 < 6 , which is true, so 1 , 2 ∈ R When x = 2 , y = 2 , then 2 + 4 < 6 , which is not true, so 2 , 2 ∉ R When x = 3 , y = 1 , then 3 + 2 < 6 , which is true, so 3 , 1 ∈ R When x = 3 , y = 2 , then 3 + 4 < 6 , which is not true, so 3 , 2 ∉ R There is no other natural number which satisfies the given property except at x = 1 , 2 , 3 and y = 1 , 2 . ∴ R = { ( 1 , 1 ) , ( 2 , 1 ) , ( 1 , 2 ) , ( 3 , 1 ) } Hence, the domain of R is 1 , 2 , 3 .

Find the domain of the function: f x = 1 x 2 + 3 x + 2 .

Given, f x = 1 x 2 + 3 x + 2 For the domain of a function D f , f x must be a real number. ⇒ f x = 1 x 2 + 3 x + 2 must be a real number ⇒ x 2 + 3 x + 2 ≠ 0 (Denominator should not be equal to zero) ⇒ x 2 + 2 x + x + 2 ≠ 0 ⇒ x + 2 x + 1 ≠ 0 ⇒ x ≠ - 1 , - 2 ⇒ D f = set of all real numbers except at x = - 1 , - 2 Hence, D f = R - - 1 , - 2 .

If f x = 2 x + 1 and g x = x + 1 , then find f ∘ g x .

Given that, we have: f x = 2 x + 1 and g x = x + 1 Now, f ∘ g x = f g x = f x + 1 = 2 x + 1 + 1 = 2 x + 2 + 1 = 2 x + 3 Hence, f ∘ g x = 2 x + 3 .

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Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 1 - Relations and Functions>Exercise

Embibe Experts Solutions for Chapter: Relations and Functions, Exercise 1: Exercise

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

Attempt the free practice questions on Chapter 1: Relations and Functions, Exercise 1: Exercise with hints and solutions to strengthen your understanding. Mathematics Crash Course (Based on Revised Syllabus-2023) solutions are prepared by Experienced Embibe Experts.

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

EASY

10th Tamil Nadu Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 1 - Relations and Functions>Exercise>Q 1

An example for a function which is not a relation, (Domain-R, Codomain-R) is:

MEDIUM

10th Tamil Nadu Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 1 - Relations and Functions>Exercise>Q 8

If $A = \{1, 2, 3, 4\}$ and $B = \{a, b, c\}$ , $f = \{(1, a), (2, a), (3, b), (4, c)\}$ , where $f$ is the function from $A \to B$ , then find out which type of function is $f$ .