CBSE Class 9 Syllabus 2027 Mathematics at Advanced Level

FOR CBSE CLASS 9 SYLLABUS EXAM PREPARATION

CBSE Class 9 Syllabus
2027
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Mathematics
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CBSE Class 9 Syllabus 2027 Question Paper Mathematics
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at Advanced Level

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MATHEMATICS AT m
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ADVANCED LEVEL
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GRADE 9
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MATHEMATICS AT
2026-27
ADVANCED LEVEL o m
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Academic Unit,
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of Secondary Education
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a Sector-23, Phase - I, Dwarka, New Delhi - 110077

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INDEX
S. No. Chapters Pages

1. Sets 1-21

2. Logarithms 22-37

3. Relations and Functions 38-54

4. Coordinate Geometry 55-70

5. Combinatorics 71-83

6. Exploring some more Progressions 84-94

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ACKNOWLEDGMENT

GUIDANCE AND SUPPORT:
Mr. Rahul Singh, IAS, Chairman, Central Board of Secondary Education
Dr. Praggya M. Singh, Professor and Director (Academics), Central Board of Secondary
Education

CONTENT DEVELOPMENT TEAM:
Dr. Jyoti Sharma, Professor, CIC, Delhi University
Mr. Rahul Sofat, HoD Mathematics, Air force Golden Jubilee Institute, New Delhi
Mrs. Roman Dhawan, Retired Lecturer (Sr. PGT), KHMS, Ashok Vihar, Delhi
Mr. Amit Bajaj, PGT (Mathematics), CRPF Public School, Rohini, Delhi
Mr. Shashank Vohra, Lecturer, Mathematics, CM Shri School, Janakpuri, Delhi

REVIEW TEAM:
Ms. Anjana Ghai, PGT (HOD) Mathematics, Meerut Public School for Girls, Meerut
Ms. Rajni Bhatia, PGT Mathematics, SD Public School, Pitampura, Delhi
Ms. Vaishali Taneja, TGT Mathematics, AFGJI, New Delhi

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1.1 Introduction
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Let’s begin by taking
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formally studied it. You must have heard about different blood types. The
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system used
this system
 A  B
 AB  O
Blood types have very important applications. If a person of blood type ‘A’ needs blood
transfusion, then which set of people can donate blood?
OR
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A teacher recommends a set of books in algebra to the student.
1.2 Set
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So, we may talk about a set of people a
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is often used in mathematics, we define
Definition of a Set
A set is a well-defined collection of objects. The objects in a set are the elements or
members of the set. A collection is said to be well-defined if there is no confusion in
deciding whether an object belongs to the collection or not.
“The three best students of a class” is not a set, because the term best may have
different meanings for different people.

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Some examples of set are:

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(i) A set of students participating in a quiz contest.
c. o(ii) A set of girls participating in Kho-Kho match. m .co
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set as different books

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a 1.3 Representation of a Set
 Roster Form: One method of writing a set is to list all the elements of the set
within braces separated by commas. For example, the set of vowels in English
alphabet is written as V = {a, e, i, o, u}.
Sets are denoted by capital letters.

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This method is also known as tabular form.
The fact that “e is an element of the set V” is written as e  V where the symbol
‘’ means “belongs to”.
Similarly, we use the symbol c  V to denote c is not an element of the set V.
Example 1: Write the following sets in Roster form
(i) Set of whole numbers less than or equal to 5.
(ii) Set of first 4 terms of the A.P., whose first term is –3 and common
difference is 4.
Solution: (i) A = {0, 1, 2, 3, 4, 5} (ii) B = {–3, 1, 5, 9}
Can all sets be written in roster form? Think about it!
 Set Builder Form
Another method of writing a set is set builder form. In this method we specify the
elements of the given set by its description. All the elements of the set possess a
single common property which no element outside the set possess.
For example, the set V = {a, e, i, o, u} can be written using this notation as
{x | x is a vowel in English alphabet}
which is read as, “the set of all elements x such that x is a vowel of the English
alphabet.
Example 2: Write the following sets in the set-builder form:

(ⅰ) {1, –1} (ii) 2
3
(iii) 1 1

, , 1, 2
2 4
Solution: (i) {x | x is an integer and x2 = 1}
(ii) { x | x is fraction equivalent to 0.6 }
a 
(iii)  a  1 or 4 and b  2 or 4 
b 
The theory of sets was developed by German mathematician Georg
Cantor (1845-1918). He is considered the founder of set theory.
Cantor developed interest in mathematics in his teens. He began his
university studies in Zurich but shifted to Berlin the very next year in
1863, where he studied under the eminent mathematicians Karl
Weierstrass, Ernst Kummer and Leopold Kronecker. He received his
doctorate degree in 1867 in Number theory.
Cantor was also interested in philosophy and wrote papers relating his
theory of sets to metaphysics. Cantor’s contribution includes the discovery that the
set of real numbers is uncountable. He is also noted for his contributions in analysis.

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1.4 Finite and Infinite Sets
Can you make a list of students studying in your class? Of course it can be done easily,
as the number of students are finite. Even the school’s strength though, a large
number, is finite. What about the set of natural numbers or set of real numbers
between 2 and 3. Are they finite?
No these constitute infinite sets.
Some infinite sets are given below:
A = {x |x is a multiple of 5 and greater than 20}
B = {x | x is a prime number}
Example 3: State which of the following sets are finite or infinite:
(i) {x : x is an integer lying between 5 and 91}
(ii) Set of coordinates of the points lying on a unit circle.
(iii) {x : x  N and (x, y) lies on the line 2x + y = 8}
(iv) {x : x is the number of animals on earth}

(v) {x : x is a digit in the decimal expansion of 2}
Solution: (i) Since the set is {6, 7, 8, …, 90} so it is a finite set.
(ii) There are infinite points lying on a circle so its an infinite set.
(iii) Infinite points (x, y) where x  N lie on the line 2x + y = 8. So set of
values of abscissa forms an infinite set.
(iv) Set of animals on the earth is finite.

(v) 2 is an irrational number, so its decimal expansion is non-
terminating, but the digits occurring in decimal expansion of 2 is
finite.
Example 4: Write the following sets in roster form:
(i) {x | x is an odd natural number between 3 (excluding) and 11
(including)}
(ii) { x | x is a whole number less than or equal to 3}
Solution: (ⅰ) {5, 7, 9, 11} (ii) {0, 1, 4, 9}.
Example 5: Write the following sets in roster as well as set builder form.
(i) Real numbers between 2 and 5.
(ii) Fractions whose numerator and denominator are natural numbers
and denominator exceeds the numerator by 1.
Is it possible to write in both the forms?

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Solution: (i) Real numbers between 2 and 5.
Roster Form: Not possible to write the set in roster form, as we
neither know its first element nor its last element. In fact, it’s not
possible to write two consecutive elements of this set.
Set Builder Form: {x : x  R and 2 < x < 5}
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(ii) Fractions whose numerator and denominator are natural numbers

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and denominator exceeds the numerator by 1.
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Roster Form: 1 2 3 4
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Example 6: Use  or  to indicate whether the given object is an element of the given
set or not.
(i) y … {a, x, y, z}
(ii) 4 … {1, 2, 5, 7, 11}
(iii) 2 … 

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  (iv) 5 … {x : x is a natural number  5}
Solution: (i)  (ii)  c. (iii) (iv) 
m in the alternative form (roster or set
Example 7: Write each of the sets givenebelow
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(i) {2, 3, 5, 7, 11} ag
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(ii) {1, 3, 5, 7, 9, 11}
Solution: (i) {x : x is a prime number less than 12}
(ii) {x : x is an odd natural number less than 13}
 Is the order of elements in a set important?
The elements of a set distinguish the set – not the order in which the elements are
written.
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Thus the three sets {1, 2, 3}, {1, 3, 2} and {3, 2, 1} are the same set.
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a Is it {B, A, N, A, N, A}? If it is correct, how many elements does the set contains,
6 or 3.
In fact, repetition of elements is not allowed in a set. A set contains only distinct
elements. So the set containing the letters of the word “BANANA” is {B, A, N}.
Let us define some more type of sets:

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1.5 Empty or Null Set
Consider the set {x : x is a prime number which is composite}
Is there a number which is both prime as well as composite?
Since no prime is composite, so the above set can be written in roster form as { }.
We say a set which does not contain any element is called an empty, null or void set.
An empty set is denoted by the symbol  or { }.
Example 8: Which of the following are empty sets:
(i) Set of even prime numbers.
(ii) Set of composite numbers having atmost 2 factors.
(iii) Set of numbers which are both rational and irrational.
(iv) Set of irrational numbers whose decimal expansion terminates.
Solution: (i) {2}; It is not an empty set.
(ii) Since a composite number has more than two factors, so it is an
empty set.
(iii) Since no rational number is irrational, so set of numbers which are
both rational and irrational is an empty set.
(iv) The decimal expansion of irrational number is neither terminating
nor recurring. So it is an empty set.
1.6 Equality of Sets
Let A and B be two sets. We say A = B, if A and B have the same elements.
If two sets A and B are not equal, we write A  B.
Some example of equal sets are:
{2, 5, –1, 0} and {0, –1, 5, 2} ;
{1, 2, 2, 2, 7, 7, 7, 7, 7} and {1, 2, 7} ;
{x : x is a prime divisor of 6} and {x : x is a pair of consecutive numbers that are prime}
EXERCISE 1.1
1. List the elements of the following sets :
(a) {x : x is an integer and x2 = 9}
(b) {x : x is a positive integer less than 5}
(c) {x : x is even natural number divisible by 5}
(d) {x : x  N and x < –1}

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2. Determine which elements of the set

 1 2 1 19
A  5,  3,  , 0, , , 13.4, ,
2 5 3 2
, are 
(a) Natural numbers (b) Whole numbers
(c) Integers (d) Rational numbers
(e) Real numbers
3. Write the following sets in roster form:
(a) {x : x is a two digit number and the sum of the digits is 5}
(b) {x : x is an integer and |x| ≤ 9 }
(c) {x : x is letter of the word “SWEET”}
n 1
(d) { x : x  , where n is a natural number and n < 6}
n
(e) {x : x is a composite number}
4. Write the following sets in set-builder form.
(i) {2, 4, 6, 8, …} (ii) {3, 6, 9, 12, 15}
(iii) {1, 4, 9, 16, …} (iv) {8, 9, 10, 11, ...}
(v) {1, 2, 3, 6}
Can two different sets have the same roster form?
5. Which of the following pairs of sets are equal.
(i) {D, E, C, E, N, T} and {C, E, N, T, D}
(ii) a, b, , 2 and a, , 2, b
(iii) {x : x is zero of the polynomial x2} and {x : x is the root of the equation,
x2 = 0}
(iv) {x : x has numerical value less than or equal to 1} and
{x : x is the root of the equation, x2 – 1 = 0}
(v) {5, 10, 15, 20} and {5, 10, 15, 20, …}
(vi)  and {}
6. State which of the following sets are finite or infinite.
(i) {x : x  Z and (x – 1)(x + 2)(x – 3) = 0}
(ii) {x : x and 2 are coprime}
(iii) {x : x is a rational number between 3 and 4}
(iv) {x : x is an integer and |x|  5}

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1.7 Subset
Consider the set X of students who live in the vicinity of 5 km radius around your
school.
These students obviously along with others belongs to the set Y of all students in your
class.
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Since every student of set X belongs to or is contained in set Y, we say X is a subset
of Y.
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Definition: The set A is said to be a subset of B if and only if every element of A is
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also an element of B.
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Symbolically, we write it as A  B.
a a subset of set B, we write it as A  B.
If A is not
We say A  B if a  A  a  B.
Note that if A is a subset of B and A  B, then we say A is a proper subset of B and
denote it by A  B
 Is an empty set , a subset of a set P containing at least one element?
Yes, the empty set ∅ is a subset of set P. In fact, the empty set is a subset of every set.
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following problems.
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Example 9: Use the notation  to denote which set is the subset of the other in the

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(i) A = {2, 3, p}
(ii) A = {2, 3, 5, 7} g l a B = {1, 2, 3, p, q}

(iii) A = {3, 8, 9, 0}
a B = 
B = {0, 9, 8, 3}
(iv) A = {x | x = 2n, where n  N] B = {x | x = 4n, where n  N}
Solution: (i) A  B
(ii) Since  is subset of every set, So B  A
(iii) Since A = B, therefore each is a subset of the other i.e., A  B as
well as B  A.
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(iv) A = {2, 4, 6, 8, 10, 12, …}
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B = {4, 8, 12, 16, 20, ...}
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So B A.

las 1.8 Cardinality of a Set ag
ag Let A be any set. If there are exactly m distinct elements in A, we say, cardinality of
set A is m. Symbolically we write it as
n(A) = m, where m is a non-negative integer,

For example, if A = {1, 2, 3, …, 9} then n(A) = 9 and if B  ,  1

3, 7 , 0 then
2
n(B) = 4.

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1.9 Power Set
Consider the set D = {a, b}. Can you write all its subsets?
Are {a}, {b} its only subsets?
A little thinking would suggest some are still left.
It has two more subsets i.e. {a, b} and .
So the set has 4 subsets in all i.e., {a}, {b}, {a, b}, .
If these subsets are written in a set form, we call it a power set of D.
i.e., P(D) = {{a}, {b}, {a, b}, }
Note:  {a} and {a, b} are elements of the set P(D).
 Neither a, nor a, b taken together are the elements of the set P(D).
Example 10: What is the power set of A = {0, 1, 2}?
Solution: Since the power set of A is again a set containing all its subsets, so
P(A) = {, {0}, {1}, {2}, {0, 1}, {0, 2}, {1, 2}, {0, 1, 2}}
Symbolically, we denote it by P(A).
Example 11: What is the power set of an empty set? What is the power set of { }?
Solution: Since every set is a subset of itself. So  is a subset of . Hence
P() = { }.
Also the set { } has exactly two subsets  and {} itself hence
P({ }) = {, { }}.
A set containing n elements has 2n subsets. If n(A) = p, where p is a whole number
then n[P(A)] = 2p.
Consider the set A = {1, 2} then it has 2 2 i.e., 4 subsets and 22 – 1 = 3 proper subsets
(i.e. , {1} and {2}).

1.10 Universal Set
In sets, the elements that we consider are usually limited to a specific all-
encompassing set. For example, when we take sets of students belonging to a class or
different sections of the same class or students of an editorial team, they all study in
the same school. So the universal set denoted by ‘U’ is the set of all students of the
school.
Similarly, if we discuss set of natural numbers, rational numbers and irrational
numbers, then real numbers is the appropriate universal set. Universal set is denoted
by U.

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EXERCISE 1.2
1. Fill in the blanks with symbol  or .
(i) {2, 3, 4} ………… {1, 2, 3, 4, 5}
(ii) {x | x are triangles in a plane} ………… {x | x are polygons in a plane}
(iii) {x is an integer} ………… {x : x is a multiple of 4}
(iv)  ………… {}
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(v) { x |x  , where m is a non-zero integer} ………… {x | x is a rational
m
number}
(vi) {x | x = n2} ………… {x | x = n3}, (where n is a natural number)
(vii) {x | x  R} ………… {x | x = 2n}, (where R is a real number)
2. Determine whether the following statements are true or false.
(i) 1  {1} (ii) {2}  {2}
(iii) {2}  {{2}} (iv)  {1, 2, 3}
(v)  {1, 2, 3}
3. Write the power set of the following sets:
(ⅰ) {1} (ii) {p, q}
(iii) {1, 2, 5} (iv) {, {}}
4. What is the cardinality of the following sets:
(i) {a} (ii) {a, {a}}
(iii) {, 1, 2, {1, 2}} (iv) {1, {1}, {1, {1}}}
(v) {, {}, {, {}}}
5. Let A be a set and n(A) = 10, then find the value of n[P(A)]? What if A has 100
elements?
1.11 Venn Diagram
A Venn diagram is a visual or pictorial representation of sets. This representation is
known as Venn diagram after the English mathematician and philosopher John Venn
(1834-1923). He studied in Cambridge, London. These diagrams are used to illustrate
relationships between sets and draw logical results.
In the following figure two sets are represented by two circles enclosed in a rectangle.
The rectangle represents the universal set U. The circles may be overlapping or
distinct depending upon they have common elements or not.

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1.12 Set Operations
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Two sets can be combined in many different ways. For instance, if we have set of
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students who play cricket and a set of students who play soccer, then we can form a
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set of students who play either cricket or soccer. We may also consider students who
play both cricket and soccer. This is possible by performing certain operations on two
sets to give another set.

Let us study these operations in detail.

1.12.1 Union of Two Sets
Let A and B be any two sets. Then union of A and B is the set containing elements of
A or B or both A and B.
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Symbolically, A  B = {x : x  Α or x  B}

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Note: ‘A  B’ is also denoted by A or B.
Example 12: If A = {2, 3, 5, 7} and B = {1, 3, 5, 9, 11} find A B.
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Solution: Since A B consists of all the elements of A as well as B.
Hence, A B = {1, 2, 3, 5, 7, 9, 11}

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Example 13: Find the union of the following pair of sets.
(i) A = {2, 5, 9}, B = {1, 4, 7}
(ii) C = {3, 5, 6}, D = {3, 4, 5, 6, 9}
(iii) P = {a, b, d, e}, Q = {b, c, e, f, g}
Solution: (i) A  B = {1, 2, 4, 5, 7, 9}
(ii) C D = {3, 4, 5, 6, 9}
(iii) P Q = {a, b, c, d, e, f, g}
1.12.2 Intersection of Two Sets
Let A and B be two sets. Then the intersection of A and B is the set that contains
elements present in both A and B.
Symbolically, A  B = {x : x  A and x B}

Note: ‘A  B’ is also denoted by ‘A and B’.
Example 14: Find the intersection of the following two sets
A= {2, 3, 5, 7, 9} B= {3, 5, 9, 11, 13}

Solution: Since A  B consists of elements that are common to A and B.
Hence, A B = {3, 5, 9}.

Example 15: Find the intersection of the following pairs of sets
(i) {a, b, f } and {d, e, f, g}
(ii) {1, 2, 3, 6, 9} and {1, 4, 5, 9, 13}
(iii)  and {c, d, e}
(iv) {x : x is a natural number greater than 4 and less than 10} and
{x : x is a factor of 12}
Solution: (i) { f } (ii) {1, 9} (iii)  (iv) {6}

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1.12.3 Disjoint Sets
Two sets A and B are disjoint if they have no element in common.
Therefore, the intersection of two disjoint sets is an empty set
AB=

Example 16: Identify pair(s) of disjoint sets from the following:
A = {1, 3, 4}
B = {5, 6, 7, 8, …}
C = {x : x is a prime factor of 36}
D = {5, 7, 11, 13}
Solution: Since, A B = ; A, B are disjoint
A D = ; A, D are disjoint
C D = ; C, D are disjoint.
1.12.4 Difference of Sets
Let A and B be two sets. The difference of A and B, denoted by A – B is the set
containing elements which are in A but not in B.
Symbolically, A – B = {x : x  A and x  B}

The difference of {2, 5, 7} and {1, 2, 4} is {5, 7}.
Example 17: Find the difference of A and B from the following pair of sets.
(i) A = {1, 3, 5, 7, 9}, B = {2, 6, 8}
(ii) A = {a, b, p, q}, B = {b, q}
(iii) A = {1, 5, 9}, B = {1, 2, 4, 5, 7, 9}

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Solution: (i) A – B = {1, 3, 5, 7, 9}
(ii) A – B = {a, p}
(iii) A – B = 
1.13 Complement of a Set
Let us learn complement of a set by taking a simple example.
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Consider the students of your class as universal set. Let A be the set of boys in the . co
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class and B be the set of girls in the class. Then complement of boys is the set of
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students other than the boys i.e. the girls. Let us give the formal definition of the
complement. a
lof Complement ag
a g
Definition
The complement of set A denoted by A is defined by A= {x : x U and x  A}

m
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e
s  N : x  9}
U =a{x
Example 18: Given,
l
aAg= {1, 2, 7}
B = {1, 6, 8, 9}
find the following:
(a) A b) B (c) (A  B) d) AB
Solution: U = {1, 2, 3, 4, 5, 6, 7, 8, 9}

m
m .co
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s e g la
g la a
a (a) A = {3, 4, 5, 6, 8, 9}
(b) B = {2, 3, 4, 5, 7}
(c) (A B) = {1, 2, 6, 7, 8, 9} = {3, 4, 5}
(d) AB = {3, 4, 5, 6, 8, 9} {2, 3, 4, 5, 7} = {3, 4, 5}
Note that (AB) = Α Β. This law is known as De-Morgan’s Law.

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Example 19: Given A = {1, 3, 5, 8, 9}, B = {2, 3, 4, 7, 9, 13}
and U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 13, 15}
Use Venn diagram to verify the following De-Morgan’s Laws.
(i) (A B) (ii) (A B)
Solution: Representing given sets by Venn diagram we get

(i) (A B) = U – (A  B) = {1, 2, 4, 5, 6, 7, 8, 11, 13, 15}
AB = {2, 4, 6, 7, 11, 13, 15}  {1, 5, 6, 8, 11, 15}
= {1, 2, 4, 5, 6, 7, 8, 11, 13, 15}
Hence, (A B) = AB
(ii) (A B) = U – (A  B) = {6, 11, 15}
AB = {2, 4, 6, 7, 11, 13, 15}  {1, 5, 6, 8, 11, 15}
= {6, 11, 15}
Hence, (A B) = AB

1.14 Application of Sets
The concept of cardinal number finds many practical applications in real life.
The theory of sets and the operations on them, provides some very useful formulae.
Let us now discuss and enlist a few observations which can be very easily verified
using Venn diagrams.

1.14.1 If A and B are two finite sets, then their cardinal numbers are related
as below:
1. n(Either in A or in B) = n(A  B) = n(A) + n(B) – n(A B)
2. n(Only in A, not in B) = n(A – B) = n(A) – n(A B)
3. n(Neither in A nor in B) = n(A B) = n(A B) = n(U) – n(A B)
4. n(Only in one of them) = n[(A – B)  (B – A)] = n(A) + n(B) – 2 n(A  B)

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Example 20: In a class with 40 students, 22 play badminton, 11 play both
badminton and table tennis and 16 play neither badminton nor table
tennis. How many play table tennis but not badminton?
Solution: Let B represents the set of students who play badminton and T
represents the set of students who play table tennis.

We are given, n(U) = 40, n(B) = 22, n(B  T) = 11, n(B  T) = 16
n(B  T) = n(U) – n(B  T) = 40 – 16 = 24
Now, n(B  T) = n(B) + n(T) – n(B  T)
i.e., 24 = 22 + n(T) – 11
or 24 = 11 + n(T)
n(T) = 13 i.e. 13 students play table tennis
n(Play table tennis but not badminton)
= n(T) – n(B  T) =13 – 11 = 2 students
2 students play table tennis but not badminton.
Example 21: Each student from a group of 120 university students participated in
teaching either a language or mathematics to the needy students.
It is found that 92 of them can teach a language to the needy students
and 46 can teach mathematics.
(a) Find the number of students who can teach both the subjects.
(b) Find the number of students who can teach only one of the two
subjects.
Solution: If L denotes the set of students who can teach a language, M denotes
the set of students who can teach mathematics
n(L) = 92, n(M) = 46 and n(L  M) = 120
Applying the relation
(a) n(Either in L or in M) = n(L  M)
= n(L) + n(M) – n(L  M) we get
n(L  M) = n(L) + n(M) – n(L  M)
= 92 + 46 – 120 = 18
Therefore 18 students can teach both the subjects.

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(b) n(Exactly in one of them) = n[(L – M)  (M – L)]
= n(L) + n(M) – 2n(L  M)
= 92 + 46 – 2 × 18
= 138 – 36
= 102
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om
Therefore 102 students can teach only one of the two subjects.
. co
. cNote: You can also use the Venn diagram to solve this example. se m
Example 22: se
m l a
la subjects—Physics (P) and Chemistry (C), it is observed thata 70
In a survey of 100 students regarding their preference for two
g
g
a preferred Physics and 60 preferred Chemistry.
(i) Find the maximum possible number of students who neither
like Physics nor Chemistry.
(ii) Find the minimum possible number of students who like both
Physics and Chemistry.
Solution: We know that, n(P  C) = n(P) + n(C) – n(P C)

m
(i) For maximum number of students who like neither of the
subjects, we need (P C)
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e
n(P C) = n(U) – n(P C) will be maximum if n(P C) is
s
l a
minimum. This happens when one of the two sets is a subset of
ag
the other, so that P C becomes the maximum.
The maximum possible value of n(P C) is the cardinal number
of the smaller set, which is 60. (Means if everyone who likes
Chemistry also likes Physics).
Thus, Minimum n(P C) = 70 + 60 – 60 = 70.
Therefore, maximum number of students who neither like
Physics nor Chemistry = 100 – 70 = 30
m
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(ii) We know that n(P C) n(U)
m
m .co This means 70 + 60 – n(P C)  100
s e m
s e Therefore, n(P C) 30
g la
g la a students who like both
Thus the minimum possible number of
a Physics and Chemistry is 30.

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Alternate solution:
Let x be the number of students who like neither Physics nor
Chemistry. These students sit outside the two circles but inside the
rectangular Universal set (U).
So on the basis of information the following Venn diagram is drawn.

(i) Thus, to make x as large as possible, we must make the “space”
occupied by the circles (P  C) as small as possible. To make the
union of two circles small, we must overlap them as much as
possible. The most they can possibly overlap is the size of the
smaller set. So, we put all 60 Chemistry students inside the
Physics circle.
Therefore, Maximum value of neither Physics nor Chemistry
= 30
(ii) Since n(P  C) = n(P) + n(C) – n(P  C)
n(P  C) = n(P) + n(C) – n(P  C)
Now n(P  C) will be minimum if n(P  C) is maximum.

The maximum possible value of n(P  C) = 100 (as there are a
total of 100 students).
Hence, minimum value of n(P  C)
= n(P) + n(C) – n(P  C)
= 70 + 60 – 100 = 30

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1.14.2 If A, B and C are three finite sets, then the relation between the
cardinal numbers is given below:
n(A  B  C) = n(A) + n(B) + n(C) – [n(A  B) + n(B  C) + n(C  A)] + n(A B C)

Example 23: A market research group conducted an online survey of 1000
consumers and found that 35% of the consumers had rated a shampoo
of type A by 5-stars. While 30% of the consumers rated the shampoo
of type B by 5-stars and 250 consumers rated the shampoo of type C
by 5-stars. It was observed that 200 consumers gave 5-star rating to
the shampoos of both the types A and B, 150 consumers gave it to the
shampoos of both the types B and C and 15% gave a 5-star to both
A and C and 100 consumers rated all of them with 5-star.
(i) Find the percentage of consumers who gave 5-star rating to only
one type of shampoo.
(ii) Find the number of consumers who did not give 5-star rating to
any of the shampoos.
Solution: Let A, B and C represent the sets of consumers who rated 5-stars to
the shampoos A, B and C respectively. We observe
n(U) = 1000; n(A) = 35% of 1000 = 350;
n(B) = 30% of 1000 = 300 and n(C) = 250
n(A B) = 200; n(B C) = 150;
n(A C) = 15% of 1000 = 150 and n(A B C) = 100
Using Venn diagram we find:
n(only in A and B) = n(A  B) – n(A B C) = 200 – 100 = 100
n(only in B and C) = n(B  C) – n(A  B  C) = 150 – 100 = 50
n(only in A and C) = n(A  C) – n(A  B  C) = 150 – 100 = 50

(i)Thus, from Venn diagram, Number of consumers who gave
5-star rating to only one type of shampoo = 100 + 50 + 50 = 200
So, percentage of consumers who gave 5-star rating to only one
200
type of shampoo  100  20%
1000

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(ii) Now, number of consumers who did not give 5-star rating to any
of the shampoos n(U) – n(A  B C) = 1000 – 500 = 500
EXERCISE 1.3
1. Find the union of sets A and B i.e. A B, in each of the following pairs.

m
co
(i) A = {1, 2, 3, 7}, B = {2, 7, 9}

. c om
(ii) A = {a, b, d, e}, B = {a, e, i, o, u}
e m .
m
(iii) A = {x : x is natural number > 5}, B = {x : x is natural number < 5}
e l as
as
(iv) A  , B  2,
l 2,  1, 0 ag
ag each of the following.
2. Evaluate
(ⅰ) {1, 2}  {1, 2, 5}
(ii) {1, 3, 5, 7, 9}  {2, 4, 6, 8}
(iii) {g, o, a, t}  {c, a, t}
(iv) {x : x is an integer}  {x : x is a negative integer}
3. Which of the following sets are disjoint?
m
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(i) {x : x is a multiple of 2} and {x : x is a multiple of 3}
m
(ii) e, , 2, 0 and e2 ,  , as3,e1
 gl  2
a
(iii) {x : x is a real number} and {x : x is an irrational number}
4. Find A – B in each of the following.
(ⅰ) A = {1, 3, 5, 8}, B = {3, 7, 8, 9}
(ii) A = {3, 0, 8}, B = {1, 3, 0, 8, 9}
(iii) A = {2, 6}, B = {1, 3, 5, 9}
5. Use the Venn diagram given below to answer the questions that follow.
m
m .co
.co
Hint: You can find sets A, B, C and universal set U from the given Venn
diagram.
e m
e m las
las ag
ag

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(i) A (ii) B (iii) (A  B)
(iv) A B (v) A B C (vi) A (B C)
6. Verify A – B = A B using the Venn diagram given below:

7. For a competitive exam, 85% of students opted for a Mock Test in Mathematics
and 75% opted for a Mock Test in Science.
(a) What is the minimum possible percentage of students who opted for both
tests?
(b) If 10% opted for neither, how does the minimum percentage for both
change?
8. Let S be a set of 50 people. 35 people speak English and 25 speak Hindi. If k be
the number of people who speak only English, then find the possible value of k,
assuming every person speaks at least one of the two languages.
9. An organization awarded certificates to its 56 students for at least one of the
three activities of Origami, Instrumental music and Fine arts. If 17 students
received the certificates for Origami, 28 for Instrumental music, 25 for Fine arts
and only 4 students got the certificates for all the three activities. Find the
number of students who received the certificates for exactly two activities.
10. A survey of a group of 100 students in an international school revealed that
60 students could speak English, 50 students could speak German and
35 students could speak Spanish. Further 40 students could speak both English
and German, 30 could speak both German and Spanish, 25 could speak both
English and Spanish and 25 could speak all the three languages. Let E represent
the set of students who speak English, G represents the set of students who
speak German and S represent the set of students who speak Spanish. Answer
the following using Venn diagram.
(a) How many students could speak at least two languages?
(b) How many students could speak at most one language?
(c) How many students could not speak any of the three languages?

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Summary
(1) A set is a well-defined collection of objects.
(2) A set is represented in two forms
 Roster Form
 Set-Builder Form
(3) A set is finite if its number of elements is a natural number. Else it is infinite.
(4) A is subset of B if every element of A is contained in B. Symbolically it is
denoted by A  B.
(5) Cardinality of a set is the number of distinct elements in it. It is denoted by
n(A).
(6) The set of all subsets of a set A is called power set of A and is denoted by P(A).
(7) Operation of sets.
 Union of sets A and B is a set containing elements of A or elements of B or
both. It is denoted by A  B.
 Intersection of sets A and B is a set containing elements of A and B. It is
denoted by A  Β.
(8) Disjoint sets
Two sets A and B are said to be disjoint if their intersection is an empty set
i.e., A  B = 
(9) Difference of two sets A and B is a set containing elements of A which are not
in B
i.e., A – B = {x : x  A and x  B}
(10) Complement of a set A with respect to universal set U is a set containing the
elements of U which are outside A
i.e., A = U – A.
(11) Applications of Sets using cardinal relations and Venn diagrams
11.1 If A and B are two finite sets, then their cardinal numbers are related as
below:
 n(A  B) = n(A) + n(B) – n(A B)
 n(A – B) = n(A) – n(A B)
 n(A B) = n(A B) = n(U) – n(A B)
 n((A – B) (B – A)) = n(A) + n(B) – 2n(A B)
11.2 If A, B and C are three finite sets, then their cardinal numbers are
related as follows:
n(A  B  C) = n(A) + n(B) + n(C) – [n(A  B) + n(B  C) + n(C  A)]
+ n(A B C)

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o m
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2.1 Introductioncoto Logarithms
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. e m
e mbefore calculators and computers when mathematicians had to dola s
l as
Imagine a world
complex calculations
a g
involving multiplication and division of large numbers. It took
ag time and effort often involving lengthy calculations.
tremendous
Opening Puzzle: The Sound of Numbers
In a music studio, the sound engineer says:
“This speaker produces a sound that is 1000 times
more intense than the softest sound we can hear.”
Instead of saying “1000 times,” scientists say:
Sound Level = log10 1000 = 3
m
Why 3?
m .co
s e
a
Because: 103 = 1000
l
g a revolutionary idea of effectively turning
Then came John Napier with a
“multiplication into addition," and “division into subtraction” using logarithms.
Logarithms is a tool that helps to do large calculations easily.
This approach saved scientists and fellow mathematicians a lot of tedious
calculations.

Logarithms were introduced by John Napier (1550–1617), a Scottish mathematician.
His method was different from the modern approach and was based on the
m
.co
relationship between arithmetic and geometric sequences. Later, Henry Briggs
m
.co
refined this idea and developed common logarithms (base 10), which made
e m
s
calculations easier.
e m Logarithm tables were widely used in science and la
las ag
g
engineering to simplify multiplication and division until
a electronic calculators became common. Even today,
logarithms remain important in mathematics,
especially natural logarithms (base e), which are widely
used in calculus.

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2.2 Understanding Logarithms as the Inverse of Exponents
We have earlier learnt about squares and cubes. For example:
 102 = 100 (10 squared is 100)
 103 = 1000 (10 cubed is 1000)
In these cases, we have taken the base 10, their exponents as 2 and 3. We find the
value of base raised to the given exponent. But what if we know the resultant value
and the base, and we want to find the exponent?
This power or exponent to which the base is raised is called a Logarithm.
 Exponential Form: 10x  100

 Logarithmic Form: log10 100  x

We already know about exponents: 23  8
This means:
 Base = 2
 Exponent = 3
 Resultant = 8
Let us ask a different question: 2 raised to what power gives 8?
That power is 3. So we say that logarithm of 8 to the base 2 is 3.

2.2.1 Understanding Logarithms through powers of 10
Let us look at powers of 10:

Powers of 10 Expressed in logarithmic form:

100  1 log10 1  0
101  10 log10 10  1
102  100 log10 100  2
log10 1000  3
103  1000
log10 10000  4
104  10000

104  0.0001 log10 0.0001   4
103  0.001 log10 0.001   3
log10 0.01   2
102  0.01
log10 0.1   1
101  0.1 log10 1  0
100  1

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Definition: For any positive number b (where b > 0 and b  1) and a positive number
a if bx  a then logb a  x. We read this as “ logarithm of a to the base b is x” .

For example, 32 = 25 we can write 5 = log2 32. These two statements are equivalent
and we indicate this by using the symbol  . We write: 32  25  log2 32  5.

In general, bx  a where a, b > 0 and b  1 and logb a  x are equivalent statements
and we write
bx  a  logb a  x
Example 1: Write an equivalent form:
1
(a) Logarithmic form for 9 2 3
(b) Exponential form of log5 625  4

Solution:
1
1
(a) 92  3  log9 3 
2
(b) log5 625  4  54  625

Example 2: Rewrite 20  1 in logarithmic form:

Solution: 20  1  log2 1  0

EXERCISE 2.1
1. Write an equivalent logarithmic statement for:

(b) 2  32
5
(a) 53  125

1 1 1
1
(c) 7  (d) 3 2 
7 3
2. Write an equivalent exponential statement for:
(a) log2 16  4 (b) log9 81  2

1 1 
(c) log5 5  (d) log2     1
2 2
3. Find the value of
(a) log10 1000 (b) log 6 36 (c) log2 64

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2.3 Logarithms Properties
The definition and the assumptions of logarithm are as follows:

Definition of Logarithm:
For a > 0, a ≠ 1 and x > 0,
m
If y  log a x , then a y  x.
om . co
. c e m
Equivalently, loga x  y
e m
 a y  x.
l as
l a
Assumptions:s ag
>g0, a ≠ 1 (The Base)
 aa
 M > 0, N > 0 (Positive real number)
These conditions ensure all logarithms in the following proofs, are defined.
Properties of Logarithms

Product Rule
Statement: loga ( MN )  loga M  loga N
m
Proof: Let x  loga ( MN ), y  log a M , z  log a N
m .co
s e
Then, ax  MN , a y  M and az  N
l a
Therefore, a yz ag y  z  log ( MN )
 a a  MN 
y z
a

Thus, loga M  loga N  loga ( MN )
Quotient Rule
M
Statement: loga    loga M  loga N
N
M
Proof: Let x  loga   , y  loga M , z  loga N m
N
om a   M  , a  M and a  N
c.Then, m.co
m
x
 
y z
s e
s e N
g la
g la ay M M a
a Therefore, a yz 
a z
  
N
y  z  loga  
N
M
Thus, loga M  loga N  loga  
N

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Power Rule
Statement: loga  M k   k loga M

Proof: Let x  loga  M k  and y  loga M

Then, ax  M k , a y  M

Therefore, ax  aky  x  ky

Thus, loga  M k   k loga M

Change of Base Formula
logb M
Statement: loga ( M )  , For b > 0, b ≠ 1
logb (a)

Proof: Let x  loga ( M )  ax  M
Now take log with base b on both the sides,
logb (ax )  logb M  x logb a  logb M
logb ( M )
Thus, x  log a ( M ) 
logb (a)

Log of 1
Statement: log a (1)  0

Proof: Let x  loga 1  ax  1

Since, a0  1, for any a > 0, a ≠ 1

a0  1  ax  x 0
Thus, x  loga (1)  0

Log of a number to the same base
Statement: log a a  1

Proof: Let x  loga a  ax  a

Since, a1  a, for any a > 0, a ≠ 1

a1  ax  x 1
Thus, x  log a a  1

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Let us summarize the properties of logarithms we proved above in the table below:
For any base a, a > 0, a  1 and M, N > 0

Property Name Logarithmic Notation What it does

Product Rule loga  M  N  loga M  loga N Turns multiplication into
addition

Quotient Rule  M Turns division into subtraction
loga    loga M  loga N
N

Power Rule loga  Mk   k  loga M The exponent comes down

log of number to loga (a)  1 Log of any number to the same
the same base base is 1

log of 1 loga (1)  0 Logarithm of 1 to any base is
zero

Base changing logb n Base a is changed to any other
loga n 
property logb a base b (b > 0 and b  1)

Remember: For positive values of m, n, x, y and a > 0, a  1,
 loga m  n  loga m  loga n
 loga m  n  loga m  loga n
 If x  y then log a x  log a y

Example 3: Write it as a single logarithm
(a) log7 3  log7 5 (b) log2 9  log2 3

(c) log4 3  log4 6  3log4 2 (d) 1  log3 5

Solution:
(a) log7 3  log7 5 = log7 3  5
= log7 15

log2 9  log2 3 = log2  
9
(b)
3
= log 2 3

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(c) log4 3  log4 6  3log4 2 = log4 3  6  log4 23

= log4 18  log4 8

 18 
= log4  
8
m
om
9
= log4   . co
. c 4
e m
em 1  log3 5 = log3 3  log3 5 l as
(d)
l as ag
ag = log3 3  5

= log3 15

Example 4: Find the value of:

(a) log7 343 (b) log3 27 3

Solution:

(a) log7 343 = log7 73 
o m
= 3log (7)  3m
. c
s e 7

a
 3 3 
log 27 3 = log g
l 1
(b) 3 a 3
3 2

 1
 3 
= log3 3 2

7
  7
= log3 3 2  
2
Example 5: Simplify: log3 81  log3 9
m
m
Solution: log3 81  log3 9 = log3 34  log3 32
.co
m .co = 4log3 3  2log3 3
s e m
s e = 4–2=2
g la
g la 2.4 Logarithm to base 10 a
a Logarithms in base 10 are called common logarithm as they are used in many
common scales, such as the Richter scale for measuring the magnitude of
earthquakes, the pH scale for measuring acidity or alkalinity and the decibel scale
for measuring sound.
log10 x is often written as just log x, and we assume the logarithm has base 10.

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Rules of common logarithms:
The rules of common logarithm given below are similar to the one derived earlier:
 log(xy) = log x + log y
x
 log    log x  log y
 y
 log  x n   n  log x
 log 1 = 0
 log 10 = 1
Example 6: Express the following as a single logarithm
(a) log 2 + log 7 (b) log 6 – log 3
Solution:
(a) log 2 + log 7 = log (2 × 7)
= log 14
6
(b) log 6 – log 3 = log
3
= log 2
Example 7: If log3 7  a and log3 4  b. Write the following in terms of a and b.

4 7
(a) log3   (b) log3 28 (c) log3  
7 3
Solution:
4
(a) log3    log3 4  log3 7  b  a.
7
(b) log3 28  log3 4  7  log3 4  log3 7  b  a.

7
(c) log3    log3 7  log3 3  a  1.
3
Example 8: Express the following as a single logarithm:
(a) 3  log2 5 (b) 1 + log 2
Solution:
(a) 3  log2 5 = 3 1  log2 5
Substituting log2 2  1 we get
3  log2 5 = 3  log2 2  log2 5
= log2 23  log2 5

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= log2 8  log2 5
8
= log2 .
5
(b) 1 + log 2 = log 10 + log 2 (as log10 10  1 )
= log (10 × 2)
= log 20
Note:
 Common logarithms have an interesting property of scaling positive numbers that
are very small or very large. For example, if a certain quantity can take values
from 0.0000000001 to 10,000,000,000 then the common logarithms of these
numbers would lie in range –10 to 10.
 Natural logarithms are logarithm to the base e (2 < e < 3), where e like  is an
irrational number. The natural logarithm are denoted by ln x i.e., loge x  ln x.
EXERCISE 2.2
1. Express the following as a single logarithm:
(a) log 2 + 2 log 7 (b) log3 8  log3 5  log3 4
(c) log 5 + 2 log 3 – log 15 (d) 2  2log5 3
1
(e) 3  log3 9 (f) 1  2log4 3  3log4 4
2
2. Find the exact value of
(a) log11 121 (b) log7 1 (c) log5 625
(d) log8 8 (e) log 1000
3. If log2 3  p and log2 5  q. Write the following in terms of p and q
(a) log2 15 (b) log 2 45

5
(c) log2   (d) log2 10
3
4. Which of the following are true?
(a) If 2x 1  3x 2 then x + 1 = x + 2 (b) log (x + 1) = log x
(c) logb b3  3 (d) Logarithm to base 1 is not defined.
5. If log2026 x  log2026 y  a, log2026 y  log2026 z  b and log2026 z  log2026 x  c,
bc c a a b
x  y z
then find the value of       .
 y z x

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2.5 Logarithms Across Subjects
Logarithms are the “mathematical tools” used in Science, Music, and even Social
science to manage scales that grow too fast such as population or earthquakes.
1. Chemistry (The pH Scale): Scientists measure
how acidic a liquid is using the concentration of
m
co
Hydrogen ions. Because these numbers are very

. c
pH   log10 H 
om
small (like 0.00001), they use a negative log scale:
e m .
m
e Octave): When a musician goes up one octave, the frequency ofgthe
l as
a s
ldoubles (2 , 2 , 2 ) but we perceive it as equal steps (1, 2, 3). aIt is
2. Music (The

a
soundg
interesting
1 2 3

to learn that our ears hear sound logarithmically.

m
m .co
s e
l a
ag
The above graph shows all the A-note octaves from A1 to A6.

m
m .co
m .co s e m
s e g la
g la a
a
Note that frequency of each C note is written as powers of 2.
3. Social Science: Population growth often follows exponential patterns:
P  P0 1  r  . Logarithms help calculate time required for population to
t

double.

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4. Geography (Earthquakes): The Richter scale used to measure earthquakes is
logarithmic. A magnitude 7 earthquake is not “two units” stronger than a
magnitude 5 earthquake; it is 102 (100 times) more intense in amplitude!
Example 9: An earthquake measures 2 on the Richter scale, while another
earthquake measures 5 on the same scale. How many times stronger is the second
earthquake than the first?
Solution: Richter scale difference:
5–2 = 3
On the Richter scale, each 1-unit increase means 10 times stronger.
So a 3-unit increase means:
103 = 1000
The magnitude 5 earthquake is 1000 times stronger than the magnitude
2 earthquake.
EXERCISE 2.3
1. Express the following in logarithmic form:
(a) 54  625 (b) 102  0.01

(c) 70  1 (d) 81  8

2. Using the properties of logs, simplify: log2 16  log2 4
3. Evaluate:
(a) log2 256 (b) log 4 16

(c) log5 125 (d) log10 0.001

4. If log2 7  p and log2 3  q. Write in terms of p and q.

(a) log 2 21 (b) log 2 49

7
(c) log2   (d) log 2 63
3
5. Real-world Application:
(a) If a star is 100 times brighter than another, and the difference in their
magnitudes is given by 2.5  log10 (brightness ratio), find the magnitude
difference.

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(b) A solution has pH 3 and another has pH 6. How many times more acidic is
the first solution? (pH   log10[H ])

(c) A magnitude 9 earthquake occurs on the Richter Scale. How many times
stronger is it than a magnitude 4 earthquake?
6. True or False: (Explain your reasoning)
1
(a) logb x  3 x ; x  0
3
1
(b) log8 e 
ln8
(c) Logarithm of a negative number is defined.
(d) logb (M  N)  logb M  logb N

(e) The base of the logarithm can be any real number.
7. Which is the greatest integer that is less than the number log4 9  log9 28?
(Do not use calculator)
log2 5
 43  1 
8. Evaluate the value of (x + 5y), where, x  log1.43   and y    .
 40  2

2.6 Solving Logarithmic Equations: The Search for ‘x’
In our journey through logarithms, we have seen how they help us rethink the
relationship between numbers and exponents. However, the true power of a
logarithm is revealed when it becomes an active tool for solving equations where the
unknown value, x, is trapped within a logarithm or a base.
Solving a logarithmic equation is much like being a mathematical detective. You must
use the Product, Quotient, and Power rules to combine multiple logarithms into a
single expression, and then convert it into its exponential form to find the value of
the variable.
The Golden Rule of Logarithmic Equations: You must always verify your
solutions! Because the domain of a logarithmic function is strictly positive, the value
of logb a is defined if a is positive (a > 0), and base (b) must be positive and not equal
to 1 (b > 0, b  1). Sometimes, standard algebraic steps will produce an extraneous
root—a false solution that mathematically breaks these rules. If substituting your
answer back into the original equation results in the logarithm of a negative number
or zero, that solution must be rejected.
We will understand this through the following solved examples.

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Example 10: Solve for x : log2 (3x  1)  3.

Solution: log2 (3x  1) = 3
Converting logarithmic equation into its exponential form,
3x – 1 = 23
m

om
3x = 9
. co
 . c x = 3
em
em l as
l as
Check: Substitute x = 3 in log2 (3x  1) gives log2 8. Since 8 > 0, the logarithm is
ag
defined.
g
a the required solution.
x = 3 is
Example 11: Solve for x : log5 x  log5 ( x  4)  1

Solution: Using the product rule, we get
log5[ x( x  4)] = 1

 x(x – 4) = 51
m
 x 2  4x  5 = 0
m .co
 (x – 5)(x + 1) = 0
s e
 x = 5,la–1
Note that on substituting x = –1, a
g
in the equation gives log ( 1) and log ( 5). Since
5 5
log of a negative number is not defined so x = –1, is an extraneous root which is
rejected.
The only valid solution is x = 5.
Example 12: Solve for x : log3 ( x 2  8x )  2

log3 ( x 2  8x ) = 2
m
Solution:

m .co
.co
Converting logarithmic equation into its exponential form,
e m
e m x 2  8x = 32
las
las  x 2  8x  9 = 0
ag
ag  (x – 9)(x + 1) = 0
 x = 9, –1
Note that for both x = 9, –1, the expression (x 2  8x ) evaluates to 9, which is strictly
greater than zero. Therefore, both x = 9, –1 are valid solutions.

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Example 13: Solve for x : log2 (3x  4)  log2 5

Solution: log2 (3x  4) = log 2 5
 3x – 4 = 5 (as loga x  loga y  x  y)
 3x = 9
 x = 3
Example 14: Solve for x : (log3 x )2  5(log3 x )  6  0

Solution: (log3 x )2  5(log3 x )  6 = 0

Let y  log3 x , the given equation reduces to

y2  5y  6 = 0
 (y – 3)(y – 2) = 0
 y = 3 or y = 2
 log3 x = 3 or log3 x  2

 x = 33 or x  32
i.e. x = 27, 9
For both x = 27, and x = 9, the value of log3 x is defined as x > 0, so required solutions
are x = 9, 27.
Example 15: Solve for x : logb (logb Ax )  1; A  0

Solution: logb (logb Ax ) = 1
 logb Ax = b

 Ax = bb
1 b
 x = (b )
A
EXERCISE 2.4
1. Solve for x.
(a) log3 (2x  5)  2 (b) log7 (3x )  log7 2  log7 24

(c) log5 ( x  3)  log5 ( x  1)  1 (d) log2 ( x 2  7)  3
2. Solve for x.
(a) log2 ( x  3)  log2 ( x  1)  5 (b) 2log4 x  log4 (5x  4)

(c) log5 ( x  2)  log5 ( x  2)  1 (d) log10 ( x  2)  log10 ( x  1)  1

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3. Solve for x.
(a) log x (3x  10)  2, where x > 0 and x 1

(b) (log3 x )2  4 log3 x  3  0

(c) (log2 x )2  log2 x 3  10

(d) x log10 x  1000x 2
4. Solve for x.
(a) log3 ( x 2  1)  log3 (2x  1)

(b) logx 5  logx 2  logx x

1
(c) log2 x  4
log x 2
(d) log3 (3  x )  log3 (8  x )  log3 (9x  8)  2  log3 9

(e) log10 log2  log3 9   5x

1 2 3 99
5. If x  log  log  log  log , where all logs are to the base 10, then
2 3 4 100
evaluate (x + 1) (x + 2) (x + 3) … (x + 99).
Enrichment – Graph of logarithmic and Exponential functions
The graph of logarithmic function and exponential function are inverse of each other.
Note that the graph of these functions are mirror images along the line y = x.

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Summary
 Logarithm is the inverse operation of exponentiation.
 logb a  x  bx  a, where a > 0 and b > 0, b  1
 Logarithms are defined only for positive numbers.
m

om
If the base of a logarithm is not given then we assume it to be 10 and such
. co
c
logarithms are called common logarithms.
. e m

em
Important Properties:
l as
l as
For b > 0, b  1 and M > 0, N > 0 we have:
ag
 ag
Product Rule: log ( M  N )  log M  log N
b b b

M
 Quotient Rule: logb    logb M  logb N
N
 Power Rule: logb ( M k )  k  logb M

 Log of a number to the same base: logb (b)  1

m
.co
 Log of 1: logb (1)  0

 The base changing property:
e m
For any logarithmic bases b a
l s a and any positive number n,
and
ag log n  log n b
a
log a b
 Applications: Logarithms are widely used in:
 Earthquake measurement
 Acidity (pH scale)
 Sound intensity
m
m  Sports science
 Population growth
.co
m .co s e m
s e g la
g la a
a

m . c
c. o s e m
s em
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3.1 Introduction
In everyday life, we often observe associations between two quantities. The marks
obtained by a student in a MCQ test depends on the number of correct answers;
the area of a circle depends on its radius; a person’s salary may depend on the years
of experience; the distance travelled at uniform speed may depend on the time taken
etc.
All such dependencies or associations can be described mathematically using the
concepts of relations and functions.
In a school each student is allotted one house. This rule of connection — where every
student is paired with precisely one house is a function. If instead a student is
associated with one or more sports teams, we simply have a relation. The concept of
function is fundamental to all mathematics.
In this chapter, we will define ordered pairs, cartesian products, relations, some
functions and their graphs, domain, range, vertical line test and some simple
transformations.
3.2 Ordered Pairs
An ordered pair is a pair of objects written in a specific order. It is denoted by (a, b),
where a is called the first element and b is the second element. Since the order
matters, hence in general (a, b) is different from (b, a).

Definition of an Ordered Pair
An ordered pair (a, b) consists of two elements separated by comma and written in
parenthesis i.e., ( ). In (a, b) the first element is a and the second is b. Two ordered
pairs are equal if and only if both the corresponding components are equal, i.e.,
(a, b) = (c, d)  a = c and b = d

Note that (a, b)  (b, a) unless a = b. This distinguishes an ordered pair from a set
{a, b}, where order is irrelevant. You may recall that an ordered pair (x, y) is used to
represent the position of the point in the cartesian plane, where x is known as the
abscissa and y is known as the ordinate.

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Example 1: If (2x + 1, y − 3) = (9, 4), find x and y.
Solution: Since the ordered pairs are equal, equating their abscissas and
ordinates, we get
2x + 1 = 9 y–3 = 4
i.e, x = 4 i.e, y = 7

3.3 Cartesian Product of Sets
Given two non-empty sets A and B, their cartesian product A × B (read as ‘A cross B’)
is the set of all ordered pairs (a, b) where a  A and b  B.

Definition of Cartesian Product
For any two sets A and B we have, A × B = {(a, b) : a  A and b  B}
If A =  or B = , then A × B = .

Properties of Cartesian product:
 n(A × B) = n(A) × n(B).
 A × (B  C) = (A × B)  (A × C) and A × (B  C) = (A × B)  (A × C).
 If A  B then A × C  B × C for any set C.
Note: The cartesian product of two sets is not commutative i.e., A × B  B × A.

Example 2: Let A = {1, 2, 3} and B = {4, 5}. Find A × B and B × A. Are they equal?
Solution: A × B = {(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)}
B × A = {(4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (5, 3)}
Since (1, 4)  A × B but (1, 4)  B × A, So, A × B  B × A.
Example 3: If n(A) = 4 and n(B) = 3, find how many subsets will A × B have?
Solution: We know that n(A × B) = n(A) × n(B)
= 4 × 3 = 12
Number of subset of A × B = 212 = 4096.
Example 4: If A = {a, b} and B = {1, 2, 3}, write A × B and verify
n(A × B) = n(A) × n(B).
Solution: A × B = {(a, 1), (a, 2), (a, 3), (b, 1), (b, 2), (b, 3)}
n(A × B) = 6
n(A × B) = 2 × 3
n(A × B) = n(A) × n(B).
Hence verified.

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Example 5: If A = {1, 3, 5} and B = {4, 9}. Write the ordered pairs of A × B whose first
element is less than the second element. What about the ordered pairs
of B × A satisfying the same order relation? Are they equal?
Solution: The ordered pairs of A × B having first element less than the second
element are (1, 4), (1, 9), (3, 4), (3, 9) and (5, 9). The ordered pair of B ×
A satisfying the same order relation is (4, 5) only. The elements of A × B
m
om
and B × A satisfying the same rule are different.
. co
. c e m
Example 6:
e
Let A = {1, 3}, B = {2, 3, 4} and C = {4, 5}. Write the set A × (B  C) and
m l as
Solution: l as verify A × (B  C) = (A × B)  (A × C).
ag
ag
B  C = {2, 3, 4}  {4, 5}
B  C = {4}
A × (B  C) = {1, 3} × {4} = {(1, 4), (3, 4)}
(A × B)  (A × C) = {(1, 2), (1, 3), (1, 4), (3, 2), (3, 3), (3, 4)}
     {(1, 4), (1, 5), (3, 4), (3, 5)}
= {(1, 4), (3, 4)}
A × (B  C) = (A × B)  (A × C)
m
.co
So,

EXERCISE 3.1
e m
1. If (x − 5, y + 1) = (4, 6), find x anday.s
g l
2. Let A = {1, 2} and B = {2, 3, 5}. a List all elements of A × B and B × A.
3. If n(A × B) = 20 and n(A) = 4, find n(B).
4. If A = {1, 2, 3} and B = {x, y}, find A × B, B × A, A × A and B × B.
5. If A = {1, 2, 3} and B = {2, 3, 7}, find (A × B)  (B × A).
6. Verify, A × (B  C) = (A × B)  (A × C) for A = {1, 2}, B = {2, 3}, C = {4, 5}.

3.4 Relations
m
m .co
.co m
A relation constitutes only those elements of cartesian product which satisfies a rule

m
that governs the relation between the elements of the two sets. Any subset of the
s e
s e cartesian product A × B is a relation from A to B. Let us give a formal definition of
g la
g la the relation.
a
a Definition of a Relation
A relation R from set A to set B is a subset of A × B, i.e., R  A × B. If (a, b)  R we
write a R b and read as ‘a is related to b’. The set of all first elements of the ordered
pairs in the relation constitutes the domain of R and the set of all second elements
constitutes the range of R.

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 Domain of R: The set of all first elements of the ordered pairs is the domain,
i.e., Domain = {a  A : (a, b)  R for some b  B}. The domain of R is also called
the set of pre-images
 Range of R: The set of all second elements of the ordered pairs is the range,
i.e., Range = {b  B : (a, b)  R for some a  A}. The range of R is also called
the set of images.
 Codomain of R: The entire set B is the codomain of the relation.
Note:
 range  codomain.
 A relation R on a set A means a relation from set A to A.

Let us represent the relation, R : A  B by an arrow diagram as shown below:
Note that, R = {(2, 8), (3, 3)}
therefore, Domain = {0, 2, 3, 5}
Range = {3, 8}
Codomain = B = {1, 3, 8}
Example 7: Let A = {1, 2, 3, 4}, B = {1, 4, 9, 20, 25}. A relation R from A to B defined
by R = {(a, b) : a2 = b; a  A, b  B}. Write the relation R in roster form,
find its domain, range, and codomain.
Solution: Since, 12 = 1, 22 = 4, 32 = 9, 42 = 16, therefore R = {(1, 1), (2, 4), (3, 9)}
Domain = {1, 2, 3}, Range = {1, 4, 9}, Codomain = {1, 4, 9, 20, 25}.
Example 8: What is the domain and
range of the relation given
below:
R = {(0, 0), (1, 1), (1, –1),
(4, 2), (4, –2), (9, 3), (9, –3)}
Represent R on the graph.
Also identify the rule that
defines this relation.
Solution: Taking x coordinates we get,
Domain = {0, 1, 4, 9}
Taking y coordinates we get,
Range = {–3, –2, –1, 0, 1, 2, 3}
The relation R is given by the
equation y   x
for x = 0, 1, 4, 9.

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Example 9: Given A = {1, 3, 8}, B = {2, 3, 9, 11}
A relation, R : A  B defined by R = {(a, b) : b|a; a  A, b  B}
Write the relation R in roster form.
Also determine its domain, range and codomain.
(Note: b | a means b is a factor of a or b divides a exactly)
Solution: In roster form, R = {(3, 3), (8, 2)}
Domain = {3, 8}
Range = {2, 3}
Codomain = B = {2, 3, 9, 11}

Example 10: How many relations can be defined from
set A = {1,2,3} to B = {4, 5}?
Solution: Since, n(A × B) = 3 × 2 = 6.
Therefore, number of relations = 26 = 64.

Example 11: Gaurav studies in class XI in which there are 30 students. His sister
studies in class IX having 20 students.
How many relations are possible from set A to set B if set A represents
the set of 30 students of class XI and set B represents a set of 20
students of class IX?
Solution: We first find number of element in A × B
i.e. n(A × B) = 30 × 20 = 600
Now since relation is a subset of cartesian product, so number of
relations is equal to the number of subsets of a set containing 600
elements i.e. 2600
2600 is enormous, far beyond anything we encounter in everyday
counting.

Example 12: Let us defined a relation S from Z to Z where
S = {(x, y) : difference between x and y is odd}
Write the domain and range of the relation.
Solution: The difference between an odd integer and an even integer is odd and
the difference between an even integer and an odd integer is also odd
i.e., (odd, even) and (even, odd)  S.
So, both domain and range of the relation S are the set of integers Z.

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EXERCISE 3.2
1. Let A = {1, 2, 3}, B = {4, 5, 6, 7}. Define a relation R from A to B by
R = {(a, b) : a + b = 7; a  A, b  B}. Write the relation R in roster form and hence
find its domain and range.
2. Given A = {2, 3, 4, 5}, B = {3, 6, 7, 10}. Define R = {(a, b) : a divides b; a  A,
m
co
b  B}. Write R in roster form hence find its domain and range.
3. Let R = {(a, b) : ao+m2b = 12, a, b   Write R in roster form and hence find its .
. c e m
em
domain and range.
l as
4. Write R =s{(x, x ) : x is a prime number less than 10} in roster form. Also findgthe
lofaR. a
2

ag
range
5. Let A = {p, q, r, s} and B = {1, 2}. How many relations can be defined from set A
to set B? List any four of them.
6. Let A = {1, 2, 3, 4, 5}. Define a relation R on A by R = {(a, b) : |a − b| = 2}.
Write R in roster form and hence find its domain and range.
3.5 Functions
A relation in which each element of the domain corresponds to exactly one element of
m
.co
the range is called a function.

e m
l as
ag

The relation from set A to B, shown by the arrow diagram is also a function f : A  B.
Note: Consider a function f : A  B, where f (a) = b then b is called the image of a
m
m .co
under f and a is called the preimage of b under f.

.co
Let us give formal definition of function.
s e m
s em Definition: A relation R from set A to set B is a function fromgAlato B if every element
g la ofIn Aother
has exactly one image in set B. a
a words a relation R : A  B is a function if the following two conditions are
satisfied.
(i) every element of A has an image i.e. domain of R is A.
(ii) no element in A has more than one image i.e., the image is unique.
Note: A relation from set A to set B is a function if no two distinct ordered pairs have
the same first element.

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Example 13: Which of the following relations are functions? Justify your answer.

Solution:
(a) The relation from A to B is a function as every element of set A has a unique
image.

(b) The relation from P to Q is not a function as an element of P i.e., 1 has two
images. (Same can be said for element 4)

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(c) The relation from X to Y is a function as every element of set X has a unique
image.

(d) The relation from M to N is not a function as an element of M i.e., 1 has no image.

Example 14: Determine which of the following rules describes a function. Give
reason for each.

Domain Rule Range

(a) The set of students Teacher of each The set of
in a school student. teachers in a
school

(b) The set of Each sportsperson The set of
sportspersons country. countries.
participating in
Olympics.

(c) The set of real Square of a The set of
numbers. number. positive real
numbers.

(d) The set of real Cube of a number. The set of real
numbers. numbers.

(e) The set of integers. Square root of a The set of non-
number. negative real
numbers.

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Solution: (a) No — A student is taught by different teachers. Since the
association is not unique so it is not a function.
(b) Yes — Since each sportsperson represents his/her country, so the
association is unique. Hence the rule describes a function.
(c) No — The square of 0 is 0, which is not there in set of positive real
m
co
numbers. Since every element of domain does not have an image so

om .
the rule does not describe a function.
. c
(d) Yes — The cube of every real number is again a unique real e m
e m l as
as number, so the rule describes a function.
g
l (e) No — The square root of a negative integer is not defined. aSince
ag negative integers does not have an image so the rule does not
describe a function.
Example 15: Let A = {2, 3, 5}, B = {1, 3, 4, 10}
A relation R : A  B defined by R  {(a, b) : iff a|b ; a  A, b  B}.
Write the relation in roster form and also draw its arrow diagram.
Is R a function?
Also write its domain, co-domain and range.
m
Solution: Let us draw an arrow diagram of the
m .co
given relation.
s e
Since the element 2 of lAahas two images,
ag
so R is not a function.
We have, Domain = {2, 3, 5}
also, Range = {3, 4, 10}
and Co-domain = B = {1, 3, 4, 10}
Example 16: Which of the following relations from A = {1, 2, 3} to B = {a, b, c, d} are
functions from A to B?

m
.co
(i) R1 = {(1, a), (2, b), (3, c)} (ii) R2 = {(1, a), (1, b), (2, c), (3, d)}

om
c.Solution: em
(iii) R3 = {(1, a), (2, a), (3, a)} (iv) R4 = {(1, a), (2, c)}

e m s
(i) R1 is a function — every element of A has exactly one image.
la a and b.
las ag
(ii) R is not a function — element 1 has two images
g
2

a (iii) R3 is a function — every element of A has exactly one image.
(iv) R4 is not a function — element 3 in A has no image.
Example 17: Let f : {1, 2, 3, 4}  N be defined by f(x) = x2 + 1. Find the range of f.
Solution: f(1) = 2, f(2) = 5, f(3) = 10, f(4) = 17.
Range of f = {2, 5, 10, 17}.

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Example 18: Let A = {2, 4, 9, 16, 25, 36}. Define a relation R on A i.e. from set A to
A by R = {(x, y) : x = y2}. Write its domain, co-domain and range.
Solution: R = {(4, 2), (16, 4)}
Domain : x  {4, 16}
Range : y  {2, 4}
Co-domain = A = {2, 4, 9, 16, 25, 36}
Example 19: Let A = {1, 2, 3}. Define the following relations on A (i.e., from A to A).
(a) {(x, y) : y = x where x, y  A}
(b) {(x, y) : x + y  6 where x, y  A}
(c) {(x, y) : y = 4x3 where x, y  A}
Solution: (a) Since, A = {1, 2, 3}
Therefore, R = {(1, 1), (2, 2), (3, 3)}
This is an identity relation.
(b) R = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3) (3, 1), (3, 2), (3, 3)}
The relation so obtained is a universal relation as it is equal to
A × A.
(c) R = { }
Since there is no ordered pair (a, b) which satisfies b = 4a3.
So we get an empty relation.
EXERCISE 3.3
1. Which of the following relations are functions? Justify your answer.

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2. Which of the following relations from A = {3, 5, 7, 9} to B = {1, 2, 3, 4, 5} are
functions from A to B?
(a) R1 = {(3, 2), (5, 4), (7, 5), (9, 5)}
(b) R2 = {(1, 3), (3, 5), (5, 7)}
(c) R3 = {(2, 3), (2, 5), (2, 7), (3, 5), (3, 7), (5, 7)}
(d) R4 = {(3, 3), (5, 5)}
3.6 Some Functions and their Graphs
Let us draw graphs of some real valued functions. We also learn how to find the
domain and range of a function by the projection of the graph on the x-axis and y-axis
respectively. A projection of a curve on an axis is the set of points i.e. the foot of the
perpendiculars drawn from every point on the curve onto the axis. For real functions
projection is either R or its subset.
(a) Identity Function: A function f : R  R, defined by f(x) = x.
The projection of a curve along the x-axis is called the domain and the projection
of the curve along the y-axis is called the range.

(b) Constant Function: A function f : R  R, defined by f(x) = c.

Note: The foot of the perpendicular of every point on the graph of f(x) = c on the
y-axis is (0, c). So the range of the function is the set {c}.

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(c) Quadratic Function: A function f : R  R, defined by f(x) = x2.

m
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l as ag
ag

(d) Cubic Function: A function f : R  R, defined by f(x) = x3.
m
m .co
s e
l a
ag

m
m .co
m .co s e m
s e g la
g la a
a
Note:
The coloured region is not part of the graph. It is drawn to show continuum of
the projection. In fact infinitely many perpendicular lines can be drawn from the
graph to the axes.

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(e) Modulus Function: A function f : R  R, defined by f(x) = |x|.
, ≥0
We have, ∣x∣ = {
− , <0

1
(f) Reciprocal Function: A function f : R  {0}  R, defined by f ( x )  .
x
The graph of f is shown below:

(g) Square Root Function: The function f :[0, )  R, defined by f ( x )  x.
The graph of f is given below:

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(h) Greatest Integer Function: The function f : R  R, defined by
f ( x )  [ x ], x  R
is the value of the greatest integer less than or equal to x.
x 0 0.01 0.5 0.9 1 1.2 1.8 2 –1.2 –2
y 0 0 0 0 1 1 1 2 –2 –2
Note: From the definition of greatest integer function it follows:
[x] = 0 for 0  x < 1
[x] = –1 for –1  x < 0
[x] = 1 for 1  x < 2 and so on.

Note: Students are advised to take different values of x and find the
corresponding values of y in tabular form and plot the point obtained to obtain
the graph.

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Example 20: Evaluate:
(i) |−7| + |3| (ii) [4.9] – [−2.3]

(iii) 25  [4.01] (iv) [0.0001] – [1.0001]
Solution: (i) 7 + 3 = 10 (ii) 4 − (−3) = 7
m
(iii) |5 ÷ (–5)| = 1
om
(iv) 0 – 1 = –1
. co
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Example 21: Draw the graphs of the functions f, g and h on the same coordinate
e m l as
l as
axes. You may fill the tables given below to draw the graph:
ag
ag

Describe the relationship among the graph of f, g and h. Can you
m
generalise the relationship among the graphs of f (x), f (x) – c and
f (x) + c?
m .co
s e
a
Solution: Let us fill the tables
l
ag

m
m .co
m .co s e m
s e g la
g la a
a

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The graph of g(x) = x2 – 1 is obtained by a vertical shift (downwards) of
1 unit below the x-axis and graph of h(x) = x2 + 1 is obtained by a
vertical shift (upwards) of 1 unit above the x-axis.
In general y = f (x) – c where c > 0 represents a vertical translation
(shift) of c units downwards and y = f (x) + c represents a vertical
translation (shift) of c units upwards. The shape of the curve remains
identical.

EXERCISE 3.4

1. What is the domain and range of each of the relations given below? Which of
these relations are functions:
(a) R = {(5, 1), (4, 1), (3, 1), (2, 0)}
(b) R = {(1, –1), (2, –2), (3, –3), (4, –4), (5, –5)}
(c) R = {(3, –1), (3, 0), (3, 1), (3, 2)}
2. Draw a rough sketch of each of the following relations. Also write their
domain and range.

(a) R = {(x, y) : xy = 8 where x, y  Z}
(b) R = {(x, y) : x = |y| where x  Z and 0  x  5}
(c) R = {(x, y) : y   x where x  (0, )}

3. Draw the graph of the functions f, g and h on the same coordinate axes. You may
fill the tables given below to draw the graphs.

Describe the relationship among the graphs of f, g and h.

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4. Draw the graphs of the functions f, g and h on the same coordinate axes.
You may fill the tables given below to draw the graph.

Describe the relationship among the graphs of f, g and h. Are their domain and
range equal?
5. Determine the domain and range of the following functions:

(b) y  2–|x |
1
(a) y 
x2
(c) y  (x 1)3 (d) y  x
Summary
1. An ordered pair is a pair of objects generally numbers or variables written in
specific order.
For example: (3, –5), (x, y) etc.
2. The Cartesian product of two non-empty sets A and B is the set of all ordered
pairs (a, b) where a  A and b  B.
i.e., A × B = {(a, b) : a  A, b  B}
3. A relation R : A  B is a subset of the cartesian product A × B. The set of
x-coordinates of all ordered pairs constitute a domain and set of y-coordinates of
all the ordered pairs constitute the range.

4. A relation f : A  B is a function if every element of set A has a unique image.
The set A is called the domain and set of all images b where f (a) = b is called the
range.
5. The graph of y = f (x) + c can be obtained from the graph of y = f (x) by shifting it
‘c’ units above the x-axis if c > 0 and shifting it ‘c’ units below the x-axis if c < 0.
The graph of y = f (x + k) can be obtained from the graph of y = f (x) by a horizontal
shift of k units. If k > 0 the graph shifts to the left and if k < 0 the graph shifts
to the right.

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o m
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4.1 Introduction:
. coThe Language of Graphs s e m
Imagine trying etomdescribe the exact location of a single star in the sky or a specifica
as stadium. To do this accurately, we need a reference system.gIn
l
l this system is Coordinate Geometry (also known as Analytical
seat in a massive
g a
a
mathematics,
Geometry), where we use numbers to represent positions on a plane. Coordinate
geometry is the study of geometry using a coordinate system. It bridges algebra and
geometry.

4.2 The Cartesian System
The foundation of this chapter is the Cartesian Plane, named after the
mathematician René Descartes.
o m
The “Legend” of the Fly
The most famous story associated with m
. c
s e geometry came
René Descartes
a
l a fly crawl on the
(1596–1650) is that the idea for coordinate
g
ceiling. He realized that he coulda describe the fly's exact
to him while he was lying in bed watching

position at any moment by measuring its distance from two
perpendicular walls.

While this story might be apocryphal, it perfectly
illustrates the shift in thinking: position can be defined
Source of the image:
by numbers. https://learnodo-newtonic.com

Now that we know René Descartes’ big idea, let’s look at the “map” he created. m
omCartesian plane: Formed by the intersection of a horizontal line
c.The
To locate a point, we need a frame of reference. For that, following terms are defined:
m (x-axis) and a .co
s e
s em vertical line (y-axis) at the origin (0, 0). It divides the plane into
g l afour quadrants.
la Coordinates: An ordered pair (x, y) represents any point P,a where x is the abscissa
ag (horizontal distance) and y is the ordinate (vertical distance) *. (*Refer to the graph
on the next page)

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Cartesian: This term is derived directly from the Latin version of Descartes’
name, Renatus Cartesius.
Coordinates: From the idea of “ordering” or “arranging” points in a mutual
relationship to two axes.

4.3 Let Us Explore Four Quadrants
The Four Quadrants
These two axes (x-axis and y-
axis) divide the entire plane into
four regions called Quadrants.
We number them anti-clockwise
starting from the top right:
 Quadrant I: Both x and y
are positive (+, +).
 Quadrant II: x is
negative, y is positive
(–, +).
 Quadrant III: Both x and
y are negative
(–, –).

 Quadrant IV: x is
positive, y is negative
(+, –).

4.4 Moving Points: The Magic of Reflections
In Coordinate Geometry, points don't have to stay still! We can move them using
specific rules. One of the most interesting ways to move a point is through reflection,
which works exactly like a mirror.
How do mirrors work on a graph
1. The Y-axis work as a Mirror: When you stand in front of a vertical mirror
(the Y-axis), your height (y-coordinate) stays the same, but your left and right
sides (x-coordinate) swap.
2. The X-axis work as a Mirror: Imagine standing on a clear glass floor with
a mirror underneath (the X-axis). Your position on the floor (x-coordinate)
stays the same, but your “top” and “bottom” (y-coordinate) flip.

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