Relations and Functions — Class 12 Maths Notes

Topics covered in Relations and Functions

1. 1.Introduction of Relations & Functions

   Imagine you're organizing a party and decide to assign a specific task to each of your friends to help out. This assignment of tasks creates a relationship between your friends and the tasks. In mathematics, we can think of this scenario as a relation. Now, if you ensure that each friend is assigned exactly one task, this special type of relationship becomes what we call a function.

   Relations

   A relation in mathematics is a way to connect elements from two sets. For example, if we have a set of friends ={,,ℎ}A={Alice,Bob,Charlie} and a set of tasks ={,,}B={Decorating,Cooking,Invitations}, a relation between A and B could link each friend to a task they're responsible for. Mathematically, this can be represented as a set of ordered pairs, showing which friend is linked to which task.

   For instance, a relation R might be {(,),(,),(ℎ,)}{(Alice,Decorating),(Bob,Cooking),(Charlie,Invitations)}.

   Functions

   A function is a special type of relation where each element from the first set (let's say, our set of friends) is associated with exactly one element in the second set (the tasks). The key point here is "exactly one"; no friend can be without a task, and no friend can have more than one task.

   Mathematically, we can express a function f from a set X to a set Y as:→f:X→Y, where for every element x in X, there is exactly one element y in Y such that =f(x)=y.

   For example, if ={1,2,3}X={1,2,3} and ={4,5,6}Y={4,5,6}, a function f could be represented as {(1,4),(2,5),(3,6)}{(1,4),(2,5),(3,6)}. This tells us (1)=4f(1)=4, (2)=5f(2)=5, and (3)=6f(3)=6.

   Real-world Applications and Careers

   • Programming and Computer Science: Functions in programming are essential, representing tasks or calculations that take inputs and produce outputs.
   • Economics: Functions are used to model relationships between variables, like supply and demand or cost and production levels.
   • Engineering: Engineers use functions to describe how changes in certain parameters affect systems' outputs.

   Hands-on Activity

   Try creating your own relation and function using things around you. For instance, make a list of your family members and assign each person a household chore. Ensure each person has exactly one chore for it to be a function. This activity will help you understand the concept of unique pairings in functions.

   Understanding relations and functions is not just an academic exercise; it's a way of interpreting patterns and relationships in the world, crucial for fields like computer science, economics, and engineering.

2. 2.Definations - Types of Relations

   In mathematics, relations are ways to describe how elements from two different sets are connected. There are several types of relations, each with its unique properties. Let's explore these types with examples from daily life and their mathematical expressions.

   1. Reflexive Relation

   A relation R on a set A is reflexive if every element is related to itself. In simpler terms, for every a in A, the pair (,)(a,a) is in R.

   Example: Consider the set of all students in a class, A. A reflexive relation could be "is classmates with." Every student is a classmate to themselves.

   Mathematical Expression: For all ∈a∈A, (,)∈(a,a)∈R.

   2. Symmetric Relation

   A relation R on a set A is symmetric if, whenever an element a is related to an element b, then b is also related to a.

   Example: "Friendship" among a group of people is a good example. If Alice is friends with Bob, then Bob is friends with Alice.

   Mathematical Expression: For all,∈a,b∈A, if (,)∈(a,b)∈R, then (,)∈(b,a)∈R.

   3. Transitive Relation

   A relation R on a set A is transitive if, whenever an element a is related to b, and b is related to c, then a is also related to c.

   Example: Consider the relation "is an ancestor of" in a family tree. If Alice is an ancestor of Bob, and Bob is an ancestor of Charlie, then Alice is an ancestor of Charlie.

   Mathematical Expression: For all,,∈a,b,c∈A, if (,)∈(a,b)∈R and (,)∈(b,c)∈R, then (,)∈(a,c)∈R.

   4. Antisymmetric Relation

   A relation R on a set A is antisymmetric if, whenever a is related to b and b is related to a, then a must be equal to b.

   Example: The relation "is less than or equal to" among numbers is antisymmetric. If 5≤55≤5, then 5=55=5.

   Mathematical Expression: For all,∈a,b∈A, if (,)∈(a,b)∈R and (,)∈(b,a)∈R, then =a=b.

   5. Equivalence Relation

   A relation is an equivalence relation if it is reflexive, symmetric, and transitive. This relation partitions a set into equivalence classes, where each element is equivalent to each other within the same class.

   Example: "Having the same remainder when divided by 5" among numbers. For any numbers,,a,b,c, if a and b leave the same remainder when divided by 5, and so do b and c, then a and c will also leave the same remainder.

   Mathematical Expression: A relation R is an equivalence relation if it satisfies reflexive, symmetric, and transitive properties.

   Applications and Careers

   Understanding these types of relations is crucial in fields like computer science, where data structures and databases often rely on these mathematical concepts to organize and process information efficiently. In software development, understanding relationships between objects is key to building effective algorithms and applications.

   Activity

   Try creating your own examples of each type of relation using items or concepts from your daily life. For instance, for a symmetric relation, think about the "is a sibling of" relationship within a family. This hands-on approach will help solidify your understanding of these concepts.

   Relations are fundamental in mathematics, helping us understand and formalize the ways in which different objects or entities can be connected. They have wide-ranging applications in solving real-world problems across various disciplines.

3. 3.Determine whether each of the following relations are reflexive, symmetric and transitive:

   (i) Relation R in the set ={1,2,3,...,13,14}A={1,2,3,...,13,14} defined as ={(,)∣3−≥0}R={(x,y)∣3x−y≥0}

   (ii) Relation R in the set N of natural numbers defined as ={(,)∣=2+5 and <4}R={(x,y)∣y=x2+5 and x<4}

   (iii) Relation R in the set ={1,2,3,4,5,6}A={1,2,3,4,5,6} as ={(,)∣ is divisible by }R={(x,y)∣y is divisible by x}

   (iv) Relation R in the set Z of all integers defined as ={(,)∣− is an integer}R={(x,y)∣x−y is an integer}

   (v) Relation R in the set A of human beings in a town at a particular time given by

   (a) ={(,)∣ and work at the same place}R={(x,y)∣x and y work at the same place}

   (b) ={(,)∣ and live in the same locality}R={(x,y)∣x and y live in the same locality}

   (c) ={(,)∣ is exactly 7 cm taller than }R={(x,y)∣x is exactly 7 cm taller than y}

   (d) ={(,)∣ is wife of }R={(x,y)∣x is wife of y}

   (e) ={(,)∣ is father of }R={(x,y)∣x is father of y} Let's start by analyzing each relation for reflexivity, symmetry, and transitivity:

   Solutions: (i) ={(,)∣3−≥0}R={(x,y)∣3x−y≥0}

   • Reflexivity: For all ∈x∈A, 3−=2≥03x−x=2x≥0, which is true. So, R is reflexive.
   • Symmetry: If (,)∈(x,y)∈R, then 3−≥03x−y≥0. However, this does not mean that (,)∈(y,x)∈R because 3−≥03y−x≥0 does not necessarily follow. So, R is not symmetric.
   • Transitivity: If (,)∈(x,y)∈R and (,)∈(y,z)∈R, then 3−≥03x−y≥0 and 3−≥03y−z≥0. Adding these inequalities gives 3−≥03x−z≥0, which means (,)∈(x,z)∈R. So, R is transitive.

   (ii) ={(,)∣=2+5 and <4}R={(x,y)∣y=x2+5 and x<4}

   • Reflexivity: Since y depends on x and is not necessarily equal to x, not all elements are related to themselves. So, R is not reflexive.
   • Symmetry: Given (,)∈(x,y)∈R, there is no requirement that (,)∈(y,x)∈R. For example, if =2x=2, then =22+5=9y=22+5=9. But (9,2)(9,2) is not in R since 99 is not less than 4. So, R is not symmetric.
   • Transitivity: If (,)∈(x,y)∈R and (,)∈(y,z)∈R, there is no guarantee that 2+5=x2+5=y and 2+5=y2+5=z will result in (,)∈(x,z)∈R. So, R is not transitive.

   (iii) ={(,)∣ is divisible by }R={(x,y)∣y is divisible by x}

   • Reflexivity: Since any number is divisible by itself, R is reflexive.
   • Symmetry: If (,)∈(x,y)∈R, then y is divisible by x. However, this does not imply x is divisible by y. So, R is not symmetric.
   • Transitivity: If (,)∈(x,y)∈R and (,)∈(y,z)∈R, then z is divisible by y and y is divisible by x, implying z is divisible by x. Hence, R is transitive.

   (iv) ={(,)∣− is an integer}R={(x,y)∣x−y is an integer}

   Since Z is the set of all integers, subtraction of any two integers is an integer. Thus:

   • Reflexivity: −=0x−x=0, which is an integer, so R is reflexive.
   • Symmetry: If −x−y is an integer, so is −y−x (since it's just the negative of −x−y), making R symmetric.
   • Transitivity: If −x−y and −y−z are integers, so is (−)+(−)=−(x−y)+(y−z)=x−z, making R transitive.

   (v) Human Relations

   • (a) Working at the same place: Reflexive, Symmetric, Not Transitive.
   • (b) Living in the same locality: Reflexive, Symmetric, Not Transitive.
   • (c) Being exactly 7 cm taller: Not Reflexive, Not Symmetric, Not Transitive.
   • (d) Being someone's wife: Not Reflexive, Not Symmetric (generally), Not Transitive.
   • (e) Being someone's father: Not Reflexive, Not Symmetric, Not Transitive.

   Each of these human relations has its particular reasons for the classifications, based on the properties of reflexivity, symmetry, and transitivity inherent in the relationships themselves.

4. 4.Here's are some related questions of Relations

   1. Show that the relation R in the set R of real numbers, defined as ={(,):≤2}R={(a,b):a≤b2}, is neither reflexive nor symmetric nor transitive. Solution:

   • Reflexivity: A relation is reflexive if every element is related to itself. For a to be related to itself in this relation, we would need ≤2a≤a2 to be true for all ∈a∈R. However, this is not true for all real numbers; for instance, when =−1a=−1, the inequality does not hold since −1≰(−1)2−1≤(−1)2. Hence, the relation is not reflexive.

   • Symmetry: A relation is symmetric if for every pair (,)(a,b) in the relation, the pair (,)(b,a) is also in the relation. In this case, let's take =1a=1 and =2b=2. Then (1,2)∈(1,2)∈R because 1≤221≤22. However, (2,1)∉(2,1)∈/R because 2≰122≤12. Therefore, the relation is not symmetric.

   • Transitivity: A relation is transitive if whenever (,)(a,b) and (,)(b,c) are in R, then (,)(a,c) must also be in R. Consider =1a=1, =2b=2, and =1c=1. We have (1,2)∈(1,2)∈R and (2,1)∈(2,1)∈R, but (1,1)∉(1,1)∈/R because 1≰121≤12. Hence, the relation is not transitive.

   2. Check whether the relation R defined in the set {1,2,3,4,5,6}{1,2,3,4,5,6} as ={(,):=+1}R={(a,b):b=a+1} is reflexive, symmetric or transitive. Solution:

   • Reflexivity: For reflexivity, we need (,)∈(a,a)∈R for all a. This cannot be true here because there is no a such that =+1a=a+1. Therefore, R is not reflexive.

   • Symmetry: For symmetry, we need (,)∈(b,a)∈R whenever (,)∈(a,b)∈R. However, if (,)∈(a,b)∈R, then =+1b=a+1, but it does not imply that =+1a=b+1. Therefore, R is not symmetric.

   • Transitivity: For transitivity, if (,)(a,b) and (,)(b,c) are in R, then (,)(a,c) must also be in R. Since =+1b=a+1 and =+1c=b+1, it follows that =+2c=a+2, which does not fit the condition =+1c=a+1. Therefore, R is not transitive.

   3. Show that the relation R in R defined as ={(,):≤}R={(a,b):a≤b}, is reflexive and transitive but not symmetric. Solution:

   • Reflexivity: For every ∈a∈R, ≤a≤a is true. Therefore, the relation is reflexive.

   • Symmetry: The relation is not symmetric because ≤a≤b does not imply ≤b≤a. For example, if =2a=2 and =3b=3, then (2,3)∈(2,3)∈R but (3,2)∉(3,2)∈/R.

   • Transitivity: The relation is transitive. If (,)(a,b) and (,)(b,c) are in R, then ≤a≤b and ≤b≤c, which implies ≤a≤c. Thus (,)(a,c) is also in R.

   4. Check whether the relation R in R defined by ={(,):<3}R={(a,b):a<b3} is reflexive, symmetric or transitive. Solution

   • Reflexivity: The relation is not reflexive because for any a, <3a<a3 is not always true. For example, when =−1a=−1, −1<(−1)3−1<(−1)3 is false.

   • Symmetry: The relation is not symmetric because <3a<b3 does not imply that <3b<a3. For instance, if =1a=1 and =2b=2, then 1<231<23 is true, but 2<132<13 is false.

   • Transitivity: The relation is transitive. If (,)(a,b) and (,)(b,c) are in R, then <3a<b3 and <3b<c3. Since b and c are real numbers, 3<3b3<c3 if <b<c, and it follows that <3a<c3. Therefore, (,)(a,c) is also in R.

5. 5.Show that the relation R

   1. Show that the relation R in the set {1,2,3}{1,2,3} given by ={(1,2),(2,1)}R={(1,2),(2,1)} is symmetric but neither reflexive nor transitive. Solution

   • Symmetry: The relation is symmetric because for every (,)(a,b) in R, the pair (,)(b,a) is also in R. Since (1,2)(1,2) is in R and (2,1)(2,1) is also in R, the relation is symmetric.

   • Reflexivity: A relation is reflexive if every element is related to itself. Since (1,1)(1,1), (2,2)(2,2), and (3,3)(3,3) are not in R, the relation is not reflexive.

   • Transitivity: A relation is transitive if whenever (,)(a,b) and (,)(b,c) are in R, then (,)(a,c) is also in R. We have (1,2)(1,2) and (2,1)(2,1) in R, but (1,1)(1,1) is not in R, so R is not transitive.

   2. Show that the relation R in the set A of all the books in a library of a college, given by ={(,): and have the same number of pagesR={(x,y):x and y have the same number of pages is an equivalence relation.

   Solution: An equivalence relation must be reflexive, symmetric, and transitive.

   • Reflexivity: Every book has the same number of pages as itself, so for every book x, (,)(x,x) is in R. Hence, R is reflexive.

   • Symmetry: If book x has the same number of pages as book y, then book y has the same number of pages as book x. Therefore, if (,)(x,y) is in R, then (,)(y,x) is also in R, so R is symmetric.

   • Transitivity: If book x has the same number of pages as book y, and book y has the same number of pages as book z, then book x must have the same number of pages as book z. Hence, R is transitive.

   Since R is reflexive, symmetric, and transitive, it is an equivalence relation.

   3. Show that the relation R in the set ={1,2,3,4,5}A={1,2,3,4,5} given by ={(,):∣−∣ is evenR={(a,b):∣a−b∣ is even is an equivalence relation. Show that all the elements of {1,3,5}{1,3,5} are related to each other and all the elements of {2,4}{2,4} are related to each other. But no element of {1,3,5}{1,3,5} is related to any element of {2,4}{2,4}.

   Solution: This relation is indeed an equivalence relation because:

   • Reflexivity: For any ∈a∈A, −=0a−a=0, which is even. So, (,)(a,a) is in R.

   • Symmetry: If ∣−∣∣a−b∣ is even, then ∣−∣∣b−a∣ is also even because absolute value disregards sign. Thus, (,)(a,b) in R implies (,)(b,a) is in R.

   • Transitivity: If ∣−∣∣a−b∣ and ∣−∣∣b−c∣ are even, then (−)+(−)=−(a−b)+(b−c)=a−c is also even (sum of two even numbers is even). Therefore, (,)(a,c) is in R.

   Regarding the subsets {1,3,5}{1,3,5} and {2,4}{2,4}:

   • The difference between any two elements in {1,3,5}{1,3,5} is even (for example, ∣1−3∣=2∣1−3∣=2), so they are all related to each other within the set.

   • The same applies to {2,4}{2,4} (for example, ∣2−4∣=2∣2−4∣=2).

   • However, the difference between any element of {1,3,5}{1,3,5} and {2,4}{2,4} is odd (for example, ∣1−2∣=1∣1−2∣=1), so no element of {1,3,5}{1,3,5} is related to any element of {2,4}{2,4} within the relation R.

6. 6.Find the set of all elements to 1 in each case.

   Show that each of the relation R in the set ={∈:0≤≤12}A={x∈Z:0≤x≤12}, given by (i) ={(,):∣−∣ is a multiple of 4}R={(a,b):∣a−b∣ is a multiple of 4} (ii) ={(,):=}R={(a,b):a=b}

   is an equivalence relation. Find the set of all elements related to 1 in each case. Solution:

   Let's examine each relation:

   For (i) ={(,):∣−∣ is a multiple of 4}R={(a,b):∣a−b∣ is a multiple of 4}:

   An equivalence relation must be reflexive, symmetric, and transitive:

   • Reflexivity: For any ∈a∈A, ∣−∣=0∣a−a∣=0, and 0 is a multiple of 4, so every element is related to itself.

   • Symmetry: If ∣−∣∣a−b∣ is a multiple of 4 (i.e., (,)∈(a,b)∈R), then ∣−∣∣b−a∣ is also a multiple of 4 because absolute values are symmetric; hence (,)∈(b,a)∈R.

   • Transitivity: If ∣−∣∣a−b∣ and ∣−∣∣b−c∣ are multiples of 4, then ∣−∣=∣(−)+(−)∣∣a−c∣=∣(a−b)+(b−c)∣ is also a multiple of 4 because the sum of multiples of any number is also a multiple of that number.

   Elements Related to 1: To find elements related to 1, we need numbers b such that ∣1−∣∣1−b∣ is a multiple of 4 within the set A. This gives us the elements 1, 5, and 9.

   For (ii) ={(,):=}R={(a,b):a=b}:

   • Reflexivity: Since =a=a for all ∈a∈A, the relation is reflexive.

   • Symmetry: If =a=b, then =b=a, so the relation is symmetric.

   • Transitivity: If =a=b and =b=c, then =a=c, hence the relation is transitive.

   Elements Related to 1: Since this relation is the equality relation, the only element related to 1 is 1 itself.

7. 7.Exercise Questions

   Question 1:

   Show that the relation R in the set A of points in a plane given by ={(,):distance of the point from the origin is same as the distance of the point from the origin}R={(P,Q):distance of the point P from the origin is same as the distance of the point Q from the origin}, is an equivalence relation. Further, show that the set of all points related to a point ≠(0,0)P=(0,0) is the circle passing through P with origin as center.

   Answer 1:

   For R to be an equivalence relation, it must be reflexive, symmetric, and transitive:

   • Reflexivity: The distance from any point P to the origin is the same as itself, so (,)∈(P,P)∈R for all ∈P∈A.
   • Symmetry: If the distance of point P from the origin is the same as that of point Q, then the distance of Q from the origin is the same as that of P, so (,)∈(P,Q)∈R implies (,)∈(Q,P)∈R.
   • Transitivity: If (,)∈(P,Q)∈R and (,)∈(Q,R)∈R, then the distance of P from the origin is the same as that of Q, and Q has the same distance from the origin as R, so P and R have the same distance from the origin, thus (,)∈(P,R)∈R.

   The set of all points related to a point ≠(0,0)P=(0,0) that are equidistant from the origin form a circle with the origin as its center.

   Question 2:

   Show that the relation R defined in the set A of all triangles as ={(1,2):1R={(T1​,T2​):T1​ is similar to 2}T2​}, is an equivalence relation. Consider three right angle triangles 1T1​ with sides 3, 4, 5, 2T2​ with sides 5, 12, 13 and 3T3​ with sides 6, 8, 10. Which triangles among 1,2,T1​,T2​, and 3T3​ are related?

   Answer 2:

   The relation R is an equivalence relation for similarity in triangles:

   • Reflexivity: Every triangle is similar to itself.
   • Symmetry: If triangle 1T1​ is similar to 2T2​, then 2T2​ is similar to 1T1​.
   • Transitivity: If triangle 1T1​ is similar to 2T2​ and 2T2​ is similar to 3T3​, then 1T1​ is similar to 3T3​.

   Triangles 1T1​ and 3T3​ are related as they both have sides that are in proportion (3:4:5 and 6:8:10 which can be simplified to 3:4:5), meaning they are similar by the Side-Side-Side (SSS) similarity theorem. Triangle 2T2​ has different side proportions and thus is not related to 1T1​ or 3T3​.

   Question 3:

   Show that the relation R defined in the set A of all polygons as ={(1,2):1R={(P1​,P2​):P1​ and 2P2​ have the same number of sides}, is an equivalence relation. What is the set of all elements in A related to the right angle triangle T with sides 3, 4, and 5?

   Answer 3:

   The relation R is an equivalence relation:

   • Reflexivity: Every polygon has the same number of sides as itself.
   • Symmetry: If polygon 1P1​ has the same number of sides as 2P2​, then 2P2​ has the same number of sides as 1P1​.
   • Transitivity: If 1P1​ and 2P2​ have the same number of sides, and 2P2​ and 3P3​ have the same number of sides, then 1P1​ and 3P3​ have the same number of sides.

   The set of all elements related to the right angle triangle T would be the set of all triangles in A, since all triangles have three sides.

   Question 4:

   Let L be the set of all lines in XY plane and R be the relation in L defined as ={(1,2):1R={(L1​,L2​):L1​ is parallel to 2}L2​}. Show that R is an equivalence relation. Find the set of all lines related to the line =2+4y=2x+4.

   Answer 4:

   The relation R is an equivalence relation for lines being parallel:

   • Reflexivity: Every line is parallel to itself.
   • Symmetry: If line 1L1​ is parallel to line 2L2​, then 2L2​ is parallel to 1L1​.
   • Transitivity: If 1L1​ is parallel to 2L2​ and 2L2​ is parallel to 3L3​, then 1L1​ is parallel to 3L3​.

   The set of all lines related to =2+4y=2x+4 would be all lines with the slope 2, regardless of their y-intercept, since parallel lines have the same slope. So, any line of the form =2+y=2x+c where c is any real number.

8. 8.Let R be the relation in the set N given by R = {(a, b) : a = b – 2, b > 6}. Choose the correct answer.

   We have the relation R in the set N (natural numbers) defined by:

   ={(,):=−2,>6}R={(a,b):a=b−2,b>6}

   Given this definition, we need to determine which of the following ordered pairs belongs to the relation R:

   A) (2, 4) B) (3, 8) C) (6, 8) D) (8, 7)

   To belong to R, the second element must be greater than 6 and the first element must be exactly 2 less than the second element. Let's check each pair:

   A) (2, 4): Does not satisfy >6b>6, so it is not in R. B) (3, 8): Does not satisfy =−2a=b−2 (since 3 is not equal to 8 - 2), so it is not in R. C) (6, 8): This satisfies both =−2a=b−2 (since 6 is equal to 8 - 2) and >6b>6, so it is in R. D) (8, 7): Does not satisfy >6b>6 in the way required for the relation (the second number should be the larger one and the first should be two less), so it is not in R.

   Therefore, the correct answer is C) (6, 8) ∈∈R.

9. 9.Definations- Types of Functions

   Types of Functions:- 1. One-to-One Function (Injective Function)

   Mathematical Definition: A function:→f:A→B is called injective or one-to-one if (1)=(2)f(x1​)=f(x2​) implies 1=2x1​=x2​ for any 1,2∈x1​,x2​∈A. In simpler terms, every element in the range is mapped by exactly one element from the domain.

   Real-Life Example: A classroom seating arrangement where each student sits in a unique seat.

   Numeric Example: =2+1f(x)=2x+1 is injective. If (1)=(2)f(x1​)=f(x2​), then 21+1=22+12x1​+1=2x2​+1, which implies 1=2x1​=x2​.

   2. Many-to-One Function

   Mathematical Definition: A function:→f:A→B is many-to-one if different elements in the domain map to the same element in the codomain, but each element in the domain has only one image in the codomain.

   Real-Life Example: Several students having the same favorite food in the cafeteria.

   Numeric Example: =2f(x)=x2 is many-to-one, as (−2)=(2)=4f(−2)=f(2)=4.

   3. Onto Function (Surjective Function)

   Mathematical Definition: A function:→f:A→B is surjective or onto if for every ∈y∈B, there is at least one ∈x∈A such that =f(x)=y. The range of f is equal to the codomain B.

   Real-Life Example: Every slot in a parking lot being occupied by a car.

   Numeric Example: =−2f(x)=x−2 from →R→R is surjective, as for every ∈y∈R, the element =+2x=y+2 satisfies =f(x)=y.

   4. Bijective Function

   Mathematical Definition: A function:→f:A→B is bijective if it is both injective and surjective. Every element in B is an image of exactly one element in A, and all elements in A map to different elements in B.

   Real-Life Example: Perfectly pairing socks from a pile of individual socks, where each sock finds exactly one matching pair.

   Numeric Example: =2f(x)=2x from →R→R is bijective. It is injective as each input has a unique output and surjective as every real number is covered.

   5. Constant Function

   Mathematical Definition: A function:→f:A→B is constant if =f(x)=c for some fixed ∈c∈B and for all ∈x∈A. The function's graph is a horizontal line.

   Real-Life Example: The temperature of a perfectly insulated room remaining constant regardless of the outside temperature.

   Numeric Example: =7f(x)=7 for all x in the domain is a constant function.

   6. Linear Function

   Mathematical Definition: A function:→f:A→B is linear if it can be written in the form =+f(x)=mx+c where m is the slope and c is the y-intercept. The graph is a straight line.

   Real-Life Example: The total cost of goods being a linear function of the quantity if each item is the same price.

   Numeric Example: =5+2f(x)=5x+2. If you buy x items at $5 each, the total cost is linearly proportional to the quantity.

   7. Quadratic Function

   Mathematical Definition: A quadratic function is given by =2++f(x)=ax2+bx+c where a, b, and c are constants and ≠0a=0. Its graph is a parabola that opens upwards if >0a>0 and downwards if <0a<0.

   Real-Life Example: The arc of a ball thrown in the air can be modeled by a quadratic function, representing its height as a function of time.

   Numeric Example: =−162+100f(x)=−16x2+100x models the height (in feet) of a projectile x seconds after it is launched upward at 100 feet per second from the ground.

   8. Polynomial Function

   Mathematical Definition: A polynomial function has the form =+−1−1+…+1+0f(x)=an​xn+an−1​xn−1+…+a1​x+a0​ where the ai​'s are constants, n is a non-negative integer, and ≠0an​=0.

   Real-Life Example: Calculating the total cost of a car loan over time, including principal, interest, taxes, and fees.

   Numeric Example: =33−22+−5f(x)=3x3−2x2+x−5 is a cubic polynomial function.

   9. Rational Function

   Mathematical Definition: A rational function is a ratio of two polynomial functions =f(x)=q(x)p(x)​ where p(x) and q(x) are polynomial functions and ≠0q(x)=0.

   Real-Life Example: Calculating the speed of a car over time where the distance (numerator) and time (denominator) can both be represented as polynomials.

   Numeric Example: =1−1f(x)=x−11​ has a vertical asymptote at =1x=1 because the denominator equals zero.

   10. Exponential Function

   Mathematical Definition: An exponential function is defined as =f(x)=ax where a is a positive constant other than 1, and x is any real number.

   Real-Life Example: Population growth, where a population doubles every fixed amount of time, can be modeled by an exponential function.

   Numeric Example: =2f(x)=2x doubles for each increase of x by 1.

   11. Logarithmic Function

   Mathematical Definition: A logarithmic function is the inverse of an exponential function and is written as =log⁡f(x)=loga​(x) where a is the base of the logarithm.

   Real-Life Example: The decibel scale for measuring sound intensity is logarithmic.

   Numeric Example: If =log⁡2f(x)=log2​(x), then for every doubling of x, f(x) increases by 1.

   Each type of function models a different kind of real-world behavior, and understanding the properties of these functions allows mathematicians, scientists, engineers, and economists to choose the appropriate model for analyzing and solving problems in their respective fields.

10. 10.Exercise Questions - 1, 2, 3

    Question 1:

    Show that the function:∗→f:R∗​→R defined by =1f(x)=x1​ is one-one and onto, where ∗R∗​ is the set of all non-zero real numbers. Is the result true if the domain ∗R∗​, is replaced by N with co-domain being same as R?

    Solution to Question 1:

    For a function =1f(x)=x1​, to be one-one, each element of the domain must map to a unique element of the codomain.

    • Injectivity (One-One): Suppose (1)=(2)f(x1​)=f(x2​). Then, 11=12x1​1​=x2​1​, which implies 1=2x1​=x2​. So the function is injective.

    • Surjectivity (Onto): For the function to be onto, every element ∈y∈R must have a preimage in ∗R∗​. For any ∈y∈R, =1x=y1​ will be such that =f(x)=y, provided ≠0y=0. Since ∗R∗​ excludes 0, this condition is satisfied, and the function is surjective.

    If the domain is replaced by N, the set of natural numbers, the function:→f:N→R, =1f(x)=x1​ will not be onto because negative numbers and non-reciprocal fractions of natural numbers in R will not have a preimage in N.

    Question 2:

    Check the injectivity and surjectivity of the following functions:

    i):→f:N→N given by =2f(x)=x2

    • Injectivity: Not injective because if 1=−2x1​=−x2​, then (1)=(2)f(x1​)=f(x2​), but since domain is N, negative values are not considered, hence in N it is injective.

    • Surjectivity: Not surjective because not all natural numbers are perfect squares. For instance, there is no ∈x∈N such that =2f(x)=2.

    ii):→f:Z→Z given by =2f(x)=x2

    • Injectivity: Not injective for the same reason above, but in this case, negative values are considered, and 2x2 is the same for x and −−x.

    • Surjectivity: Not surjective because there is no x that maps to negative integers in Z, since squares are always non-negative.

    iii):→f:R→R given by =2f(x)=x2

    • Injectivity: Not injective for the same reason as in ii).

    • Surjectivity: Not surjective onto R because negative numbers do not have preimages in R.

    iv):→f:N→N given by =3f(x)=x3

    • Injectivity: Injective since the cube of different natural numbers is always different.

    • Surjectivity: Surjective since every natural number is the cube of some natural number.

    v):→f:Z→Z given by =3f(x)=x3

    • Injectivity: Injective since the cube of different integers is always different.

    • Surjectivity: Surjective because for every integer y, there is an integer x such that 3=x3=y.

    Question 3:

    Prove that the Greatest Integer Function:→f:R→R, given by =[]f(x)=[x], is neither one-one nor onto, where [][x] denotes the greatest integer less than or equal to x.

    Solution to Question 3:

    • Injectivity: The function is not injective because if 1≠2x1​=x2​ but both are within the same integer interval, then [1]=[2][x1​]=[x2​]. For example, [0.5]=[0.7]=0[0.5]=[0.7]=0.

    • Surjectivity: The function is not surjective onto R because there are real numbers in R that are not integers and thus are not in the image of the function. For example, there is no x such that []=12[x]=21​.

11. 11.Practice Questions - 4, 5, 6

    Question 4:

    Show that the Modulus Function:→f:R→R, given by =∣∣f(x)=∣x∣, is neither one-one nor onto, where ∣∣∣x∣ is x, if x is positive or 00 and ∣∣∣x∣ is −−x, if x is negative.

    Solution to Question 4:

    • Injectivity (One-One): The modulus function is not injective because it maps both x and −−x to the same value when >0x>0. For example, (−2)=(2)=2f(−2)=f(2)=2, hence not one-one.

    • Surjectivity (Onto): The modulus function is not surjective onto R because negative numbers are not in the range of f. No x exists such that =−1f(x)=−1, as the modulus is always non-negative.

    Question 5:

    Show that the Signum Function:→f:R→R, given by

    ={1if >00if =0−1if <0f(x)=⎩⎨⎧​10−1​if x>0if x=0if x<0​

    is neither one-one nor onto.

    Solution to Question 5:

    • Injectivity (One-One): The signum function is not injective because more than one x can result in the same output. For example, (2)=(3)=1f(2)=f(3)=1, hence not one-one.

    • Surjectivity (Onto): The signum function is not surjective onto R because the only possible outputs are 11, 00, and −1−1. Thus, there is no x such that f(x) could be, for instance, 22 or any real number other than 11, 00, and −1−1.

    Question 6:

    Let ={1,2,3}A={1,2,3}, ={4,5,6,7}B={4,5,6,7}, and let ={(1,4),(2,5),(3,6)}f={(1,4),(2,5),(3,6)} be a function from A to B. Show that f is one-one.

    Solution to Question 6:

    A function is one-one if each element of the domain maps to a unique element in the codomain. In this function f, each element of set A is paired with a distinct element of set B:

    • 11 is paired with 44
    • 22 is paired with 55
    • 33 is paired with 66

    Since no two elements of A map to the same element of B, the function f is one-one (injective).

12. 12.Exercise Questions - 7,8,9

    Question 7:

    In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.

    i):→f:R→R defined by =3−4f(x)=3−4x

    • One-One: Yes, the function is one-to-one (injective) because for every distinct 1x1​ and 2x2​, (1)f(x1​) is not equal to (2)f(x2​). The derivative of the function is ′=−4f′(x)=−4 which is negative, indicating the function is strictly decreasing and hence injective.

    • Onto: Yes, the function is onto (surjective) because for every ∈y∈R, there is an x such that =f(x)=y. You can solve for x from the equation 3−4=3−4x=y which gives =3−4x=43−y​.

    • Bijective: Since it is both injective and surjective, it is bijective.

    ii):→f:R→R defined by =1+2f(x)=1+x2

    • One-One: No, the function is not one-to-one (injective) because =(−)f(x)=f(−x) for any ∈x∈R. For example, (1)=(−1)=2f(1)=f(−1)=2.

    • Onto: No, the function is not onto (surjective) because there is no ∈x∈R such that f(x) is negative. The range of f is [1,∞)[1,∞), which does not cover all of R.

    • Bijective: Not bijective since it is neither injective nor surjective.

    Question 8:

    Let A and B be sets. Show that:×→×f:A×B→B×A such that (,)=(,)f(a,b)=(b,a) is bijective function.

    Solution to Question 8:

    • One-One (Injective): The function f is one-to-one because if (1,1)=(2,2)f(a1​,b1​)=f(a2​,b2​), then (1,1)=(2,2)(b1​,a1​)=(b2​,a2​), which implies 1=2a1​=a2​ and 1=2b1​=b2​.

    • Onto (Surjective): The function f is onto because for every pair (,)∈×(b,a)∈B×A, there exists a pair (,)∈×(a,b)∈A×B such that (,)=(,)f(a,b)=(b,a).

    • Bijective: The function f is bijective because it is both injective and surjective.

    Question 9:

    Let:→f:N→N be defined by ={+12if is odd2if is evenf(n)={2n+1​2n​​if n is oddif n is even​ for all ∈n∈N.

    State whether the function f is bijective. Justify your answer.

    Solution to Question 9:

    • One-One (Injective): To determine if f is injective, we observe that each natural number n maps to a unique value in N, as odd and even numbers are mapped to different outputs. No two different natural numbers will have the same output, so the function is injective.

    • Onto (Surjective): To check if f is surjective, note that every even natural number m is the image of 22m and every odd natural number m is the image of 2−12m−1. Thus, each ∈m∈N has a preimage in N, making f surjective.

    • Bijective: Since f is both injective and surjective, it is bijective.

13. 13.Composition of Functions and Invertible Function

    Composition of Functions

    Definition: The composition of two functions f and g is a third function that applies g to an input and then f to the output of g. It is denoted by (∘)(f∘g)(x), which is read as "f composed with g of x". Mathematically, this is expressed as (∘)=()(f∘g)(x)=f(g(x)).

    Numerical Example: Let's say we have two functions:

    • =2+3f(x)=2x+3
    • =2g(x)=x2

    The composition (∘)(f∘g)(x) would be:

    • (∘)=()=(2)=22+3(f∘g)(x)=f(g(x))=f(x2)=2x2+3

    So if you input =2x=2 into g, and then the output into f, you would get:

    • (2)=22=4g(2)=22=4
    • ((2))=(4)=2⋅4+3=11f(g(2))=f(4)=2⋅4+3=11

    Invertible Function

    Definition: A function:→f:A→B is invertible if there exists a function:→g:B→A such that (∘)=(f∘g)(x)=x for every x in B, and (∘)=(g∘f)(x)=x for every x in A. The function g is called the inverse of f and is denoted by −1f−1.

    For f to be invertible, it must be both one-to-one (injective) and onto (surjective), making it a bijection.

    Numerical Example: Consider the function =2+3f(x)=2x+3. Its inverse, −1f−1(x), must satisfy (−1)=f(f−1(x))=x.

    To find −1f−1, we solve the equation =f(y)=x for y:

    • =2+3x=2y+3
    • −3=2x−3=2y
    • =−32y=2x−3​

    Thus, the inverse function is:

    • −1=−32f−1(x)=2x−3​

    We can check that f and −1f−1 are inverses:

    • (∘−1)=(−32)=2(−32)+3=(f∘f−1)(x)=f(2x−3​)=2(2x−3​)+3=x
    • (−1∘)=−1(2+3)=2+3−32=(f−1∘f)(x)=f−1(2x+3)=22x+3−3​=x

    Both compositions return the input x, so f is invertible with inverse −1f−1.

    In summary, the composition of functions allows you to apply one function to the result of another, and an invertible function is one that can be "undone" by its inverse, returning each element of its codomain back to the corresponding element in its domain.